Page 1

SOLUCIONARIO 3/3


MAGNETIC FIELD AND MAGNETIC FORCES

27.1.

27

! IDENTIFY and SET UP: Apply Eq.(27.2) to calculate F . Use the cross products of unit vectors from Section 1.10. ! EXECUTE: v = ( +4.19 × 104 m/s ) iˆ + ( −3.85 × 104 m/s ) ˆj ! (a) B = (1.40 T ) iˆ ! ! ! F = qv × B = ( −1.24 ×10−8 C ) (1.40 T ) ⎡⎣( 4.19 ×104 m/s ) iˆ × iˆ − ( 3.85 ×104 m/s ) ˆj × iˆ ⎤⎦ iˆ × iˆ = 0, ˆj × iˆ = − kˆ ! F = ( −1.24 × 10−8 C ) (1.40 T ) ( −3.85 × 104 m/s ) − kˆ = ( −6.68 × 10−4 N ) kˆ ! ! EVALUATE: The directions of v and B are shown in Figure 27.1a. ! ! The right-hand rule gives that v × B is directed out of the paper (+z-direction). The charge is ! ! ! negative so F is opposite to v × B;

( )

Figure 27.1a ! F is in the − z - direction. This agrees with the direction calculated with unit vectors. ! (b) EXECUTE: B = (1.40 T ) kˆ ! ! ! F = qv × B = ( −1.24 ×10−8 C ) (1.40 T ) ⎡⎣( +4.19 ×104 m/s ) iˆ × kˆ − ( 3.85 ×104 m/s ) ˆj × kˆ ⎤⎦ iˆ × kˆ = − ˆj , ˆj × kˆ = iˆ ! F = ( −7.27 × 10−4 N ) − ˆj + ( 6.68 × 10−4 N ) iˆ = ⎡⎣( 6.68 × 10−4 N ) iˆ + ( 7.27 × 10−4 N ) ˆj ⎤⎦ ! ! EVALUATE: The directions of v and B are shown in Figure 27.1b.

( )

! ! ! The direction of F is opposite to v × B since ! q is negative. The direction of F computed from the right-hand rule agrees qualitatively with the direction calculated with unit vectors. Figure 27.1b 27.2.

IDENTIFY: The net force must be zero, so the magnetic and gravity forces must be equal in magnitude and opposite in direction. SET UP: The gravity force is downward so the force from the magnetic field must be upward. The charge’s velocity and the forces are shown in Figure 27.2. Since the charge is negative, the magnetic force is opposite to the ! right-hand rule direction. The minimum magnetic field is when the field is perpendicular to v . The force is also ! ! perpendicular to B , so B is either eastward or westward. ! ! EXECUTE: If B is eastward, the right-hand rule direction is into the page and FB is out of the page, as required. ! mg (0.195 × 10−3 kg)(9.80 m/s 2 ) Therefore, B is eastward. mg = q vB sin φ . φ = 90° and B = = = 1.91 T . v q (4.00 × 104 m/s)(2.50 × 10−8 C)

27-1


27-2

Chapter 27

EVALUATE: The magnetic field could also have a component along the north-south direction, that would not contribute to the force, but then the field wouldn’t have minimum magnitude.

27.3.

Figure 27.2 ! IDENTIFY: The force F on the particle is in the direction of the deflection of the particle. Apply the right-hand ! ! ! rule to the directions of v and B . See if your thumb is in the direction of F , or opposite to that direction. Use F = q vB sin φ with φ = 90° to calculate F. ! ! ! SET UP: The directions of v , B and F are shown in Figure 27.3. ! ! ! EXECUTE: (a) When you apply the right-hand rule to v and B , your thumb points east. F is in this direction, so the charge is positive. (b) F = q vB sin φ = (8.50 × 10 −6 C)(4.75 × 103 m/s)(1.25 T)sin 90° = 0.0505 N ! ! EVALUATE: If the particle had negative charge and v and B are unchanged, the particle would be deflected toward the west.

Figure 27.3 27.4.

27.5.

IDENTIFY: Apply Newton’s second law, with the force being the magnetic force. SET UP: ˆj × iˆ = −kˆ ! ! ! ! ! ! ! qv × B and EXECUTE: F = ma = qv × B gives a = m

! (1.22 × 10−8 C)(3.0 × 104 m/s)(1.63 T)( ˆj × iˆ) a= = −(0.330 m/s 2 )kˆ. 1.81 × 10−3 kg ! ! EVALUATE: The acceleration is in the − z -direction and is perpendicular to both v and B . IDENTIFY: Apply F = q vB sin φ and solve for v. SET UP: EXECUTE:

27.6.

An electron has q = −1.60 × 10−19 C . v=

F 4.60 × 10−15 N = = 9.49 × 106 m s −19 q B sin φ (1.6 × 10 C)(3.5 × 10−3 T)sin 60°

EVALUATE: Only the component B sin φ of the magnetic field perpendicular to the velocity contributes to the force. IDENTIFY: Apply Newton’s second law and F = q vB sin φ . ! ! SET UP: φ is the angle between the direction of v and the direction of B .


Magnetic Field and Magnetic Forces

27-3

EXECUTE: (a) The smallest possible acceleration is zero, when the motion is parallel to the magnetic field. The greatest acceleration is when the velocity and magnetic field are at right angles:

a=

qvB (1.6 × 10−19 C)(2.50 × 106 m s )(7.4 × 10−2 T) 2 = = 3.25 × 1016 m s . m (9.11 × 10−31 kg)

qvB sin φ , then sin φ = 0.25 and φ = 14.5°. m EVALUATE: The force and acceleration decrease as the angle φ approaches zero. ! ! ! IDENTIFY: Apply F = qv × B . ! SET UP: v = v y ˆj , with v y = − 3.80 × 103 m s . Fx = + 7.60 × 10−3 N, Fy = 0, and Fz = − 5.20 × 10−3 N . (b) If a = 14 (3.25 × 1016 m/s 2 ) =

27.7.

EXECUTE:

(a) Fx = q(v y Bz − vz By ) = qv y Bz .

Bz = Fx qv y = (7.60 × 10−3 N) ([7.80 × 10−6 C)( − 3.80 × 103 m s )] = − 0.256 T ! Fy = q (vz Bx − vx Bz ) = 0, which is consistent with F as given in the problem. There is no force component along

the direction of the velocity. Fz = q (vx By − v y Bx ) = − qv y Bx . Bx = − Fz qv y = − 0.175 T .

! ! (b) By is not determined. No force due to this component of B along v ; measurement of the force tells us nothing

27.8.

about By . ! ! (c) B ⋅ F = Bx Fx + By Fy + Bz Fz = ( −0.175 T)(+7.60 × 10−3 N) + (−0.256 T)(−5.20 × 10−3 N) ! ! ! ! B ⋅ F = 0 . B and F are perpendicular (angle is 90°) . ! ! ! ! EVALUATE: The force is perpendicular to both v and B , so v ⋅ F is also zero. ! ! ! IDENTIFY and SET UP: F = qv × B = qBz [vx (iˆ × kˆ ) + v y ( ˆj × kˆ ) + vz (kˆ × kˆ )] = qBz [vx (− ˆj ) + v y (iˆ)]. ! ! EXECUTE: (a) Set the expression for F equal to the given value of F to obtain: vx =

Fy

− qBz

=

(7.40 × 10−7 N) = −106 m s − ( − 5.60 × 10−9 C)( − 1.25 T)

Fx − (3.40 × 10−7 N) = = − 48.6 m s. qBz ( − 5.60 × 10−9 C)( − 1.25 T) ! (b) vz does not contribute to the force, so is not determined by a measurement of F . F ! ! F (c) v ⋅ F = vx Fx + v y Fy + vz Fz = y Fx + x Fy = 0; θ = 90°. − qBz qBz ! ! ! ! EVALUATE: The force is perpendicular to both v and B , so B ⋅ F is also zero. ! ! ! IDENTIFY: Apply F = qv × B to the force on the proton and to the force on the electron. Solve for the ! components of B . ! ! ! SET UP: F is perpendicular to both v and B . Since the force on the proton is in the +y-direction, By = 0 and ! ! B = Bx iˆ + Bz kˆ . For the proton, v = (1.50 km/s)iˆ . ! ! EXECUTE: (a) For the proton, F = q (1.50 ×103 m/s)iˆ × (Bx iˆ + Bz kˆ ) = q (1.50 ×103 m/s)Bz (− ˆj ). F = (2.25 ×10−16 N) ˆj , vy =

27.9.

2.25 × 10−16 N = −0.938 T . The force on the proton is independent of Bx . For the (1.60 × 10−19 C)(1.50 × 103 m/s) ! ! ! ! electron, v = (4.75 km/s)( − kˆ ) . F = qv × B = (−e)(4.75 × 103 m/s)(− kˆ ) × ( Bx iˆ + Bz kˆ ) = + e(4.75 × 103 m/s) Bx ˆj .

so Bz = −

The magnitude of the force is F = e(4.75 × 103 m/s) Bx . Since F = 8.50 × 10−16 N , Bx =

8.50 × 10−16 N = 1.12 T . Bx = ±1.12 T . The sign of Bx is not determined by measuring (1.60 × 10−19 C)(4.75 × 103 m/s)

the magnitude of the force on the electron. B = Bx2 + Bz2 = (±1.12 T) + (−0.938 T) 2 = 1.46 T . ! B −0.938 T tan θ = z = . θ = ±40° . B is in the xz-plane and is either at 40° from the +x-direction toward the Bx ±1.12 T

− z -direction or 40° from the − x-direction toward the − z -direction .


27-4

Chapter 27 ! ! (b) B = Bx iˆ + Bz kˆ . v = (3.2 km/s)( − ˆj ) . ! ! ! F = qv × B = ( −e)(3.2 km/s)( − ˆj ) × ( Bx iˆ + Bz kˆ ) = e(3.2 × 103 m/s)( Bx ( −kˆ ) + Bz iˆ) . ! F = e(3.2 × 103 m/s)(−[ ±1.12 T]kˆ − [0.938 T]iˆ) = −(4.80 × 10 −16 N) iˆ ± (5.73 × 10 −16 N) kˆ

Fz ±5.73 × 10−16 N . θ = ±50.0° . The force is in the xz-plane and is = Fx −4.80 × 10−16 N directed at 50.0° from the − x -axis toward either the +z or − z axis, depending on the sign of Bx . EVALUATE: If the direction of the force on the first electron were measured, then the sign of Bx would be determined. IDENTIFY: Magnetic field lines are closed loops, so the net flux through any closed surface is zero. SET UP: Let magnetic field directed out of the enclosed volume correspond to positive flux and magnetic field directed into the volume correspond to negative flux. EXECUTE: (a) The total flux must be zero, so the flux through the remaining surfaces must be −0.120 Wb. (b) The shape of the surface is unimportant, just that it is closed. (c) One possibility is sketched in Figure 27.10. EVALUATE: In Figure 27.10 all the field lines that enter the cube also exit through the surface of the cube. F = Fx2 + Fz2 = 7.47 × 10 −16 N . tan θ =

27.10.

Figure 27.10 27.11.

IDENTIFY and SET UP:

! ! Φ B = ∫ B ⋅ dA

! 2 Circular area in the xy-plane, so A = π r 2 = π ( 0.0650 m ) = 0.01327 m 2 and dA is in the z-direction. Use

Eq.(1.18) to calculate the scalar product. ! ! ! ! ! EXECUTE: (a) B = ( 0.230 T ) kˆ; B and dA are parallel (φ = 0°) so B ⋅ dA = B dA. ! ! B is constant over the circular area so Φ B = ∫ B ⋅ dA = ∫ B dA = B ∫ dA = BA = (0.230 T)(0.01327 m 2 ) = 3.05 ×10 −3 Wb ! ! (b) The directions of B and dA are shown in Figure 27.11a. ! ! B ⋅ dA = B cos φ dA with φ = 53.1° Figure 27.11a ! ! B and φ are constant over the circular area so Φ B = ∫ B ⋅ dA = ∫ B cos φ dA = B cos φ ∫ dA = B cos φ A Φ B = ( 0.230 T ) cos53.1° ( 0.01327 m 2 ) = 1.83 × 10 −3 Wb ! ! (c) The directions of B and dA are shown in Figure 27.11b. ! ! ! ! B ⋅ dA = 0 since dA and B are perpendicular (φ = 90°) ! ! Φ B = ∫ B ⋅ dA = 0.

Figure 27.11b

27.12.

flux is a measure of how many magnetic field lines pass through EVALUATE: Magnetic ! ! the surface. It is maximum when B is perpendicular to the plane of the loop (part a) and is zero when B is parallel to the plane of the loop (part c). ! ! ! IDENTIFY: When B is uniform across the surface, Φ B = B ⋅ A = BA cosφ . ! S!ET UP: A is normal to the! surface and is directed outward from the enclosed volume. For surface abcd, A = − Aiˆ . For surface befc, A = − Akˆ . For surface aefd, cosφ = 3/ 5 and the flux is positive.


Magnetic Field and Magnetic Forces

27.13.

27-5

! ! EXECUTE: (a) Φ B (abcd ) = B ⋅ A = 0. ! ! (b) Φ B (befc) = B ⋅ A = −(0.128 T)(0.300 m)(0.300 m) = −0.0115 Wb. ! ! (c) Φ B (aefd ) = B ⋅ A = BA cosφ = 53 (0.128 T)(0.500 m)(0.300 m) = +0.0115 Wb. (d) The net flux through the rest of the surfaces is zero since they are parallel to the x-axis. The total flux is the sum of all parts above, which is zero. EVALUATE: The total flux through any closed surface, that encloses a volume, is zero. IDENTIFY: The total flux through the bottle is zero because it is a closed surface. SET UP: The total flux through the bottle is the flux through the plastic plus the flux through the open cap, so the sum of these must be zero. Φ plastic + Φ cap = 0. Φ plastic = −Φ cap = − B A cos Φ = − B (π r 2 ) cos Φ EXECUTE:

27.14.

Substituting the numbers gives Φ plastic = – (1.75 T)π(0.0125 m)2 cos 25° = –7.8 × 10–4 Wb

EVALUATE: It would be impossible to calculate the flux through the plastic directly because of the complicated shape of the bottle, but with a little thought we can find this flux through a simple calculation. IDENTIFY: p = mv and L = Rp , since the velocity and linear momentum are tangent to the circular path. SET UP:

q vB = mv 2 / R .

⎛ RqB ⎞ −3 −19 −21 (a) p = mv = m ⎜ ⎟ = RqB = (4.68 × 10 m)(6.4 × 10 C)(1.65 T) = 4.94 × 10 kg m/s. ⎝ m ⎠ (b) L = Rp = R 2 qB = (4.68 × 10−3 m) 2 (6.4 × 10−19 C)(1.65 T) = 2.31 × 10−23 kg ⋅ m2 s. ! ! EVALUATE: p is tangent to the orbit and L is perpendicular to the orbit plane. ! ! ! (a) IDENTIFY: Apply Eq.(27.2) to relate the magnetic force F to the directions of v and B. The electron has ! ! ! negative charge so F is opposite to the direction of v × B. For motion in an arc of a circle the acceleration is ! toward the center of the arc so F must be in this direction. a = v 2 / R. SET UP:

EXECUTE:

27.15.

As the electron moves in the semicircle, its velocity is tangent !to the circular path. ! The direction of v0 × B at a point along the path is shown in Figure 27.15. Figure 27.15

! For circular motion the acceleration of the electron arad is directed in toward the center of the circle. ! Thus the force FB exerted by the magnetic field, since it is the only force on the electron, must be radially inward. ! ! ! ! Since q is negative, FB is opposite to the direction given by the right-hand rule for v0 × B. Thus B is directed ! ! ! into the page. Apply Newton's 2nd law to calculate the magnitude of B : ∑ F = ma gives ∑ Frad = ma

EXECUTE:

FB = m(v 2 /R ) FB = q vB sin φ = q vB, so q vB = m(v 2 /R ) B=

mv (9.109 × 10−31 kg)(1.41× 106 m/s) = = 1.60 × 10−4 T qR (1.602 × 10−19 C)(0.050 m)

! (b) IDENTIFY and SET UP: The speed of the electron as it moves along the path is constant. ( FB changes the ! direction of v but not its magnitude.) The time is given by the distance divided by v0 .

27.16.

The distance along the semicircular path is π R, so t =

πR

π (0.050 m)

= 1.11 × 10−7 s v0 1.41× 106 m/s EVALUATE: The magnetic field required increases when v increases or R decreases and also depends on the mass to charge ratio of the particle. IDENTIFY: Newton’s second law gives q vB = mv 2 / R . The speed v is constant and equals v0 . The direction of EXECUTE:

=

the magnetic force must be in the direction of the acceleration and is toward the center of the semicircular path. SET UP: A proton has q = +1.60 × 10−19 C and m = 1.67 × 10−27 kg . The direction of the magnetic force is given by the right-hand rule.


27-6

Chapter 27

mv (1.67 × 10−27 kg)(1.41 × 106 m s) = = 0.294 T qR (1.60 × 10−19 C)(0.0500 m) ! The direction of the magnetic field is out of the page (the charge is positive), in order for F to be directed to the right at point A. (b) The time to complete half a circle is t = π R / v0 = 1.11 × 10−7 s. EVALUATE: The magnetic field required to produce this path for a proton has a different magnitude (because of the different mass) and opposite direction (because of opposite sign of the charge) than the field required to produce the path for an electron. IDENTIFY and SET UP: Use conservation of energy ! to find the speed of the ball when it reaches the bottom of the shaft. The right-hand rule gives the direction of F and Eq.(27.1) gives its magnitude. The number of excess electrons determines the charge of the ball. EXECUTE: q = ( 4.00 × 108 )( −1.602 × 10−19 C ) = −6.408 × 10−11 C (a) B =

EXECUTE:

27.17.

speed at bottom of shaft: 12 mv 2 = mgy; v = 2 gy = 49.5 m/s ! ! ! ! ! v is downward and B is west, so v × B is north. Since q < 0, F is south. F = q vB sin θ = ( 6.408 × 10−11 C ) ( 49.5 m/s )( 0.250 T ) sin 90° = 7.93 × 10−10 N

27.18.

EVALUATE: Both the charge and speed of the ball are relatively small so the magnetic force is small, much less than the gravity force of 1.5 N. IDENTIFY: Since the particle moves perpendicular to the uniform magnetic field, the radius of its path is ! ! mv . The magnetic force is perpendicular to both v and B . R= qB SET UP:

The alpha particle has charge q = +2e = 3.20 × 10−19 C .

(6.64 × 10−27 kg)(35.6 × 103 m/s) = 6.73 × 10−4 m = 0.673 mm . The alpha particle moves in a (3.20 × 10−19 C)(1.10 T) circular arc of diameter 2 R = 1.35 mm . (b) For a very short time interval the displacement of the particle is in the direction of the velocity. The magnetic force is always perpendicular to this direction so it does no work. The work-energy theorem therefore says that the kinetic energy of the particle, and hence its speed, is constant. q vB sin φ (3.20 × 10−19 C)(35.6 × 103 m/s)(1.10 T)sin 90° F = = 1.88 ×1012 m/s 2 . We can (c) The acceleration is a = B = m m 6.64 × 10−27 kg (a) R =

EXECUTE:

v2 (35.6 × 103 m/s) 2 = 1.88 × 1012 m/s 2 , the same result. The and the result of part (a) to calculate a = 6.73 × 10−4 m R ! ! acceleration is perpendicular to v and B and so is horizontal, toward the center of curvature of the particle’s path. ! ! ! EVALUATE: (d) The unbalanced force ( FB ) is perpendicular to v , so it changes the direction of v but not its magnitude, which is the speed. IDENTIFY: In part (a), apply conservation of energy to the motion of the two nuclei. In part (b) apply q vB = mv 2 /R. also use a =

27.19.

SET UP: In part (a), let point 1 be when the two nuclei are far apart and let point 2 be when they are at their closest separation. EXECUTE: (a) K1 + U1 = K 2 + U 2 . U1 = K 2 = 0, so K1 = U 2 and 12 mv 2 = ke2 r .

v=e

!

mv (3.34 × 10−27 kg)(1.2 × 107 m/s) = = 0.10 T . (1.602 × 10−19 C)(2.50 m) qr EVALUATE: The speed calculated in part (a) is large, 4% of the speed of light. ! IDENTIFY: F = q vB sin φ . The direction of F is given by the right-hand rule.

(b) 27.20.

""!

2k 2k = (1.602 × 10−19 C) = 1.2 × 107 m s (3.34 × 10−27 kg)(1.0 × 10−15 m) mr

∑ F = ma

SET UP: EXECUTE:

gives qvB = mv 2 r . B =

An electron has q = −e . (a) F = q vB sin φ . B =

0.00320 × 10−9 N F = = 5.00 T. If the angle φ is q v sin φ 8(1.60 × 10−19 C)(500,000 m/s)sin 90°

less than 90°, a larger field is needed to produce the same force. The direction of the field must be toward the south ! ! so that v × B is downward.


Magnetic Field and Magnetic Forces

27-7

4.60 × 10−12 N F = = 1.37 × 107 m s . If φ is less than 90°, the q B sin φ (1.60 × 10−19 C)(2.10 T) sin 90° ! ! speed would have to be larger to have the same force. The force is upward, so v × B must be downward since the electron is negative, and the velocity must be toward the south. ! ! ! EVALUATE: The component of B along the direction of v produces no force and the component of v along the ! direction of B produces no force. (a) IDENTIFY and SET UP: Apply Newton's 2nd law, with a = v 2 / R since the path of the particle is circular. ! ! EXECUTE: ∑ F = ma says q vB = m ( v 2 / R ) (b) F = q vB sin φ . v =

27.21.

v=

q BR

(1.602 ×10

−19

C ) ( 2.50 T ) ( 6.96 × 10−3 m )

= 8.35 × 105 m/s 3.34 × 10−27 kg (b) IDENTIFY and SET UP: The speed is constant so t = distance/v. m

=

πR

π ( 6.96 × 10−3 m )

= 2.62 × 10−8 s 8.35 × 105 m/s v (c) IDENTIFY and SET UP: kinetic energy gained = electric potential energy lost EXECUTE: 12 mv 2 = q V EXECUTE:

t=

=

−27 5 mv 2 ( 3.34 × 10 kg )( 8.35 × 10 m/s ) = = 7.27 × 103 V = 7.27 kV 2q 2 (1.602 × 10−19 C ) 2

V=

27.22.

EVALUATE: The deutron has a much larger mass to charge ratio than an electron so a much larger B is required for the same v and R. The deutron has positive charge so gains kinetic energy when it goes from high potential to low potential. v2 IDENTIFY: For motion in an arc of a circle, a = and the net force is radially inward, toward the center of the R circle. SET UP: The direction of the force is shown in Figure 27.22. The mass of a proton is 1.67 × 10−27 kg . ! ! ! EXECUTE: (a) F is opposite to the right-hand rule direction, so the charge is negative. F = ma gives q BR 3(1.60 × 10−19 C)(0.250 T)(0.475 m) v2 q vB sin φ = m . φ = 90° and v = = = 2.84 × 106 m/s . R 12(1.67 × 10−27 kg) m (b) FB = q vB sin φ = 3(1.60 × 10−19 C)(2.84 × 106 m/s)(0.250 T)sin 90° = 3.41 × 10−13 N .

w = mg = 12(1.67 × 10−27 kg)(9.80 m/s 2 ) = 1.96 × 10−25 N . The magnetic force is much larger than the weight of the particle, so it is a very good approximation to neglect gravity. EVALUATE: (c) The magnetic force is always perpendicular to the path and does no work. The particles move with constant speed.

Figure 27.22 27.23.

IDENTIFY:

Example 27.3 shows that B =

m 2π f , where f is the frequency, in Hz, of the electromagnetic waves q

that are produced. SET UP: An electron has charge q = −e and mass m = 9.11 × 10−31 kg. A proton has charge q = + e and mass m = 1.67 × 10−27 kg. EXECUTE:

(a) B =

m 2πf (9.11 × 10−31 kg)2π (3.00 × 1012 Hz) = = 107 T. This is about 2.4 times the greatest (1.60 × 10−19 C) q

magnitude of magnetic field yet obtained on earth. (b) Protons have a greater mass than the electrons, so a greater magnetic field would be required to accelerate them with the same frequency and there would be no advantage in using them.


27-8

Chapter 27

EVALUATE:

Electromagnetic waves with frequency f = 3.0 THz have a wavelength in air of

v = 3.0 × 10−4 m. The shorter the wavelength the greater the frequency and the greater the magnetic field that f is required. B depends only on f and on the mass-to-charge ratio of the particle that moves in the circular path. IDENTIFY: The magnetic force on the beam bends it through a quarter circle. SET UP: The distance that particles in the beam travel is s = Rθ, and the radius of the quarter circle is R = mv/qB. EXECUTE: Solving for R gives R = s/θ = s/(π/2) = 1.18 cm/(π/2) = 0.751 cm. Solving for the magnetic field: B = mv/qR = (1.67 × 10–27 kg)(1200 m/s)/[(1.60 × 10–19 C)(0.00751 m)] = 1.67 × 10–3 T EVALUATE: This field is about 10 times stronger than the Earth’s magnetic field, but much weaker than many laboratory fields. IDENTIFY: When a particle of charge −e is accelerated through a potential difference of magnitude V, it gains v2 kinetic energy eV. When it moves in a circular path of radius R, its acceleration is . R SET UP: An electron has charge q = −e = −1.60 × 10−19 C and mass 9.11 × 10−31 kg .

λ=

27.24.

27.25.

EXECUTE:

1 2

q vB sin φ = m

27.26.

mv 2 = eV and v =

! ! 2eV 2(1.60 × 10−19 C)(2.00 × 103 V) = = 2.65 × 107 m/s . F = ma gives m 9.11 × 10−31 kg

v2 mv (9.11 × 10−31 kg)(2.65 × 107 m/s) = = 8.38 × 10−4 T . . φ = 90° and B = qR R (1.60 × 10−19 C)(0.180 m)

EVALUATE: The smaller the radius of the circular path, the larger the magnitude of the magnetic field that is required. IDENTIFY: After being accelerated through a potential difference V the ion has kinetic energy qV. The acceleration in the circular path is v 2 / R. SET UP: The ion has charge q = +e . EXECUTE:

K = qV = + eV . 12 mv 2 = eV and v =

2eV 2(1.60 × 10−19 C)(220 V) = = 7.79 ×104 m/s. FB = q vB sin φ . m 1.16 × 10−26 kg

! ! v2 mv (1.16 × 10−26 kg)(7.79 × 104 m/s) = = 7.81× 10−3 m = 7.81 mm. φ = 90° . F = ma gives q vB = m . R = qB R (1.60 × 10−19 C)(0.723 T)

27.27.

EVALUATE: The larger the accelerating voltage, the larger the speed of the particle and the larger the radius of its path in the magnetic field. (a) IDENTIFY and SET UP: Eq.(27.4) gives the total force on the proton. At t = 0, ! ! ! ! F = qv × B = q vx iˆ + vz kˆ × Bx iˆ = qvz Bx ˆj. F = (1.60 × 10−19 C )( 2.00 × 105 m/s ) ( 0.500 T ) ˆj = (1.60 × 10−14 N ) ˆj.

(

)

(b) Yes. The electric field exerts a force in the direction of the electric field, since the charge of the proton is positive and there is a component of acceleration ! in this direction. (c) EXECUTE: In the plane perpendicular to B (the yz-plane) the motion is circular. But there is a velocity ! component in the direction of B, so the motion is a helix. The electric field in the + iˆ direction exerts a force in the + iˆ direction. This force produces an acceleration in the + iˆ direction and this causes the pitch of the helix to

vary. The force does not affect the circular motion in the yz-plane, so the electric field does not affect the radius of the helix. (d) IDENTIFY and SET UP: Eq.(27.12) and T = 2π / ω to calculate the period of the motion. Calculate ax produced by the electric force and use a constant acceleration equation to calculate the displacement in the xdirection in time T/2. EXECUTE: Calculate the period T: ω = q B / m T= ax =

ω

=

2π (1.67 × 10−27 kg ) 2π m = = 1.312 × 10−7 s. Then t = T / 2 = 6.56 × 10−8 s. v0 x = 1.50 × 105 m/s q B (1.60 × 10−19 C ) ( 0.500 T )

−19 4 Fx (1.60 × 10 C )( 2.00 × 10 V/m ) = = +1.916 × 1012 m/s 2 m 1.67 × 10−27 kg

x − x0 = v0 xt + 12 axt 2 x − x0 = (1.50 × 105 m/s )( 6.56 × 10−8 s ) + 12 (1.916 × 1012 m/s 2 )( 6.56 × 10−8 s ) = 1.40 cm 2

EVALUATE: The electric and magnetic fields are in the same direction but produce forces that are in perpendicular directions to each other.


Magnetic Field and Magnetic Forces

27.28.

27-9

IDENTIFY: For no deflection the magnetic and electric forces must be equal in magnitude and opposite in direction. SET UP: v = E / B for no deflection. With only the magnetic force, q vB = mv 2 / R (a) v = E B = (1.56 × 104 V m ) (4.62 × 10−3 T) = 3.38 × 106 m s. ! ! ! (b) The directions of the three vectors v , E and B are sketched in Figure 27.28. mv (9.11 × 10−31 kg)(3.38 × 106 m/s) (c) R = = = 4.17 × 10−3 m. (1.60 × 10−19 C)(4.62 × 10−3 T) qB EXECUTE:

T=

2πm 2πR 2π (4.17 × 10−3 m) = = = 7.74 × 10−9 s. (3.38 × 106 m s ) qB v

EVALUATE: For the field directions shown in Figure 27.28, the electric force is toward the top of the page and the magnetic force is toward the bottom of the page.

Figure 27.28 27.29.

IDENTIFY: For the alpha particles to emerge from the plates undeflected, the magnetic force on them must exactly cancel the electric force. The battery produces an electric field between the plates, which acts on the alpha particles. SET UP: First use energy conservation to find the speed of the alpha particles as they enter the plates: qV = 1/2 mv2. The electric field between the plates due to the battery is E =Vbd. For the alpha particles not to be deflected, the magnetic force must cancel the electric force, so qvB = qE, giving B = E/v. EXECUTE: Solve for the speed of the alpha particles just as they enter the region between the plates. Their charge is 2e.

vα =

27.30.

27.31.

4 (1.60 × 10−19 C ) (1750 V) 2(2e)V = = 4.11 × 105 m/s m 6.64 × 10−27 kg

The electric field between the plates, produced by the battery, is E = Vb /d = (150 V)/(0.00820 m) = 18,300 V The magnetic force must cancel the electric force: B = E/vα = (18,300 V)/(4.11 × 105 m/s) = 0.0445 T The magnetic field is perpendicular to the electric field. If the charges are moving to the right and the electric field points upward, the magnetic field is out of the page. EVALUATE: The sign of the charge of the alpha particle does not enter the problem, so negative charges of the same magnitude would also not be deflected. IDENTIFY: For no deflection the magnetic and electric forces must be equal in magnitude and opposite in direction. SET UP: v = E / B for no deflection. EXECUTE: To pass undeflected in both cases, E = vB = (5.85 × 103 m s)(1.35 T) = 7898 N C. (a) If q = 0.640 × 10−9 C, the electric field direction is given by − ( ˆj × (− kˆ )) = iˆ, since it must point in the opposite direction to the magnetic force. (b) If q = − 0.320 × 10−9 C, the electric field direction is given by ((− ˆj ) × ( − kˆ )) = iˆ, since the electric force must point in the opposite direction as the magnetic force. Since the particle has negative charge, the electric force is opposite to the direction of the electric field and the magnetic force is opposite to the direction it has in part (a). EVALUATE: The same configuration of electric and magnetic fields works as a velocity selector for both positively and negatively charged particles. IDENTIFY and SET UP: Use the fields in the velocity selector to find the speed v of the particles that pass through. Apply Newton's 2nd law with a = v 2 / R to the circular motion in the second region of the spectrometer. Solve for the mass m of the ion. EXECUTE: In the velocity selector q E = q vB. v=

E 1.12 × 105 V/m = = 2.074 × 105 m/s 0.540 T B


27-10

Chapter 27

In the region of the circular path

!

!

∑ F = ma gives

q vB = m(v 2 /R) so m = q RB / v

Singly charged ion, so q = +e = 1.602 × 10−19 C (1.602 × 10−19 C)(0.310 m)(0.540 T) = 1.29 × 10−25 kg 2.074 × 105 m/s 1.29 × 10−25 kg = 78. Mass number = mass in atomic mass units, so is 1.66 × 10−27 kg EVALUATE: Appendix D gives the average atomic mass of selenium to be 78.96. One of its isotopes has atomic mass 78. IDENTIFY and SET UP: For a velocity selector, E = vB. For parallel plates with opposite charge, V = Ed . EXECUTE: (a) E = vB = (1.82 × 106 m s)(0.650 T) = 1.18 × 106 V m. m=

27.32.

(b) V = Ed = (1.18 × 106 V m)(5.20 × 10−3 m) = 6.14 kV.

27.33.

EVALUATE: Any charged particle with v = 1.82 × 106 m s will pass through undeflected, regardless of the sign and magnitude of its charge. IDENTIFY: The magnetic force is F = IlB sin φ . For the wire to be completely supported by the field requires that ! ! F = mg and that F and w are in opposite directions. SET UP:

The magnetic force is maximum when φ = 90°. The gravity force is downward.

mg (0.150 kg)(9.80 m/s 2 ) = = 1.34 × 104 A. This is a very large current and ohmic lB (2.00 m)(0.55 × 10−4 T) heating due to the resistance of the wire would be severe; such a current isn’t feasible. ! ! (b) The magnetic force must be upward. The directions of I, B and F are shown in Figure 27.33, where we have ! assumed that B is south to north. To produce an upward magnetic force, the current must be to the east. The wire must be horizontal and perpendicular to the earth’s magnetic field. ! EVALUATE: The magnetic force is perpendicular to both the direction of I and the direction of B. EXECUTE:

(a) IlB = mg . I =

Figure 27.33 27.34.

27.35.

IDENTIFY: Apply F = IlB sin φ . ! SET UP: l = 0.0500 m is the length of wire in the magnetic field. Since the wire is perpendicular to B, φ = 90°. EXECUTE: F = IlB = (10.8 A)(0.0500 m)(0.550 T) = 0.297 N. EVALUATE: The force per unit length of wire is proportional to both B and I. IDENTIFY: Apply F = IlB sin φ . SET UP: Label the three segments in the field as a, b, and c. Let x be the length of segment a. Segment b has length 0.300 m and segment c has length 0.600 cm − x. Figure 27.35a shows the direction of the force on each segment. For each segment, φ = 90°. The total force on the wire is the vector sum of the forces on each segment. ! ! EXECUTE: Fa = IlB = (4.50 A) x (0.240 T). Fc = (4.50 A)(0.600 m − x)(0.240 T). Since Fa and Fc are in the

same direction their vector sum has magnitude Fac = Fa + Fc = (4.50 A)(0.600 m)(0.240 T) = 0.648 N and is directed toward the bottom of the page in Figure 27.35a. Fb = (4.50 A)(0.300 m)(0.240 T) = 0.324 N and is ! ! directed to the right. The vector addition diagram for Fac and Fb is given in Figure 27.35b.

F = Fac2 + Fb2 = (0.648 N)2 + (0.324 N)2 = 0.724 N. tan θ =

Fac 0.648 N = and θ = 63.4°. The net force has Fb 0.324 N

magnitude 0.724 N and its direction is specified by θ = 63.4° in Figure 27.35b.


Magnetic Field and Magnetic Forces

27-11

EVALUATE: All three current segments are perpendicular to the magnetic field, so φ = 90° for each in the force equation. The direction of the force on a segment depends on the direction of the current for that segment.

27.36.

Figure 27.35 ! F = IlB sin φ . The direction of F is given by applying the right-hand rule to the

IDENTIFY and SET UP: ! directions of I and B . ! EXECUTE: (a) The current and field directions are shown in Figure 27.36a. The right-hand rule gives that F is directed to the south, as shown. φ = 90° and F = (1.20A)(1.00 × 10−2 m)(0.588 T) = 7.06 × 10−3 N . ! (b) The right-hand rule gives that F is directed to the west, as shown in Figure 27.36b. φ = 90° and F = 7.06 × 10 −3 N , the same as in part (a). ! (c) The current and field directions are shown in Figure 27.36c. The right-hand rule gives that F is 60.0° north of west. φ = 90° so F = 7.06 × 10−3 N , the same as in part (a). EVALUATE: In each case the current direction is perpendicular to the magnetic field. The magnitude of the magnetic force is the same in each case but its direction depends on the direction of the magnetic field.

Figure 27.36 27.37.

27.38.

27.39.

IDENTIFY: F = IlB sin φ . SET UP: Since the field is perpendicular to the rod it is perpendicular to the current and φ = 90° . 0.13 N F EXECUTE: I = = = 9.7 A lB (0.200 m)(0.067 T) EVALUATE: The force and current are proportional. We have assumed that the entire 0.200 m length of the rod is in the magnetic field. ! → ! IDENTIFY: Apply F = Il × B . SET UP: The magnetic field of a bar magnet points away from the north pole and toward the south pole. EXECUTE: Between the poles of the magnet, the magnetic field points to the right. Using the fingertips of your right hand, rotate the current vector by 90° into the direction of the magnetic field vector. Your thumb points downward—which is the direction of the magnetic force. EVALUATE If the two magnets had their poles interchanged, then the force would be upward. IDENTIFY and SET UP: The magnetic force is given by Eq.(27.19). FI = mg when the bar is just ready to levitate.

When I becomes larger, FI > mg and FI − mg is the net force that accelerates the bar upward. Use Newton's 2nd law to find the acceleration.


27-12

Chapter 27

mg ( 0.750 kg ) ( 9.80 m/s ) = = 32.67 A lB ( 0.500 m )( 0.450 T ) 2

(a) EXECUTE:

IlB = mg , I =

E = IR = ( 32.67 A )( 25.0 Ω ) = 817 V (b) R = 2.0 Ω, I = E / R = ( 816.7 V ) / ( 2.0 Ω ) = 408 A

FI = IlB = 92 N a = ( FI − mg ) / m = 113 m/s 2

27.40.

EVALUATE: I increases by over an order of magnitude when R changes to FI >> mg and a is an order of magnitude larger than g. ! ! IDENTIFY: The magnetic force FB must be upward and equal to mg. The direction of FB is determined by the direction of I in the circuit. V SET UP: FB = IlB sin φ , with φ = 90° . I = , where V is the battery voltage. R EXECUTE: (a) The forces are shown in Figure 27.40. The current I in the bar must be to the right to produce ! FB upward. To produce current in this direction, point a must be the positive terminal of the battery.

IlB VlB (175 V)(0.600 m)(1.50 T) = = = 3.21 kg . g Rg (5.00 Ω)(9.80 m/s 2 ) EVALUATE: If the battery had opposite polarity, with point a as the negative terminal, then the current would be clockwise and the magnetic force would be downward. (b) FB = mg . IlB = mg . m =

27.41.

Figure 27.40 ! ! ! IDENTIFY: Apply F = Il × B to each segment of the conductor: the straight section parallel to the x axis, the semicircular section and the straight section that is perpendicular to the plane of the figure in Example 27.8. ! ! SET UP: B = B iˆ . The force is zero when the current is along the direction of B . x

27.42.

EXECUTE: (a) The force on the straight section along the –x-axis is zero. For the half of the semicircle at negative x the force is out of the page. For the half of the semicircle at positive x the force is into the page. The net force on the semicircular section is zero. The force on the straight section that is perpendicular to the plane of the figure is in the –y-direction and has magnitude F = ILB. The total magnetic force on the conductor is ILB, in the –y-direction. EVALUATE: (b) If the semicircular section is replaced by a straight section along the x -axis, then the magnetic force on that straight section would be zero, the same as it is for the semicircle. IDENTIFY: τ = IAB sin φ . The magnetic moment of the loop is μ = IA . SET UP: Since the plane of the loop is parallel to the field, the field is perpendicular to the normal to the loop and φ = 90° . EXECUTE:

(a) τ = IAB = (6.2 A)(0.050 m)(0.080 m)(0.19 T) = 4.7 × 10−3 N ⋅ m

(b) μ = IA = (6.2 A)(0.050 m)(0.080 m) = 0.025 A ⋅ m 2 27.43.

EVALUATE: The torque is a maximum when the field is in the plane of the loop and φ = 90° . IDENTIFY: The period is T = 2π r / v , the current is Q / t and the magnetic moment is μ = IA SET UP: The electron has charge −e . The area enclosed by the orbit is π r 2 . EXECUTE: (a) T = 2πr v = 1.5 × 10−16 s (b) Charge −e passes a point on the orbit once during each period, so I = Q t = e t = 1.1 mA . (c) μ = IA = I πr 2 = 9.3 × 10−24 A ⋅ m 2 EVALUATE: Since the electron has negative charge, the direction of the current is opposite to the direction of motion of the electron.


Magnetic Field and Magnetic Forces

27.44.

27-13

! IDENTIFY: τ = IAB sin φ , where φ is the angle between B and the normal to the loop. SET UP: The coil as viewed along the axis of rotation is shown in Figure 27.44a for its original position and in Figure 27.44b after it has rotated 30.0° . ! ! ! ! EXECUTE: (a) The forces on each side of the coil are shown in Figure 27.44a. F1 + F2 = 0 and F3 + F4 = 0 . The net force on the coil is zero. φ = 0° and sin φ = 0 , so τ = 0 . The forces on the coil produce no torque. (b) The net force is still zero. φ = 30.0° and the net torque is τ = (1)(1.40 A)(0.220 m)(0.350 m)(1.50 T)sin30.0° = 0.0808 N ⋅ m . The net torque is clockwise in Figure 27.44b and is directed so as to increase the angle φ . EVALUATE: For any current loop in a uniform magnetic field the net force on the loop is zero. The torque on the loop depends on the orientation of the plane of the loop relative to the magnetic field direction.

Figure 27.44 27.45.

IDENTIFY: The magnetic field exerts a torque on the current-carrying coil, which causes it to turn. We can use the rotational form of Newton’s second law to find the angular acceleration of the coil. ! ! ! SET UP: The magnetic torque is given by τ = μ × B , and the rotational form of Newton’s second law is

∑τ = Iα . The magnetic field is parallel to the plane of the loop. EXECUTE: (a) The coil rotates about axis A2 because the only torque is along top and bottom sides of the coil. (b) To find the moment of inertia of the coil, treat the two 1.00-m segments as point-masses (since all the points in them are 0.250 m from the rotation axis) and the two 0.500-m segments as thin uniform bars rotated about their centers. Since the coil is uniform, the mass of each segment is proportional to its fraction of the total perimeter of the coil. Each 1.00-m segment is 1/3 of the total perimeter, so its mass is (1/3)(210 g) = 70 g = 0.070 kg. The mass of each 0.500-m segment is half this amount, or 0.035 kg. The result is

I = 2(0.070 kg)(0.250 m) 2 + 2 121 (0.035 kg)(0.500 m) 2 = 0.0102 kg ⋅ m 2 The torque is

!

!

!

τ = μ × B = IAB sin 90° = (2.00 A)(0.500 m)(1.00 m)(3.00 T) = 3.00 N ⋅ m

27.46.

27.47.

Using the above values, the rotational form of Newton’s second law gives τ α = = 290 rad/s 2 I EVALUATE: This angular acceleration will not continue because the torque changes as the coil turns. ! ! ! IDENTIFY: τ = μ × B and U = − μ B cosφ , where μ = NIB . τ = μ B sin φ . ! SET UP: φ is the angle between B and the normal to the plane of the loop. EXECUTE: (a) φ = 90°. τ = NIABsin(90°) = NIAB, direction kˆ × ˆj = − iˆ. U = − μ B cos φ = 0. (b) φ = 0. τ = NIABsin(0) = 0, no direction. U = − μ Bcosφ = − NIAB. (c) φ = 90°. τ = NIAB sin(90°) = NIAB, direction − kˆ × ˆj = iˆ. U = − μ B cos φ = 0. (d) φ = 180° : τ = NIAB sin(180°) = 0, no direction, U = − μ B cos(180°) = NIAB. EVALUATE: When τ is maximum, U = 0 . When U is maximum, τ = 0 . ! ! IDENTIFY and SET UP: The potential energy is given by Eq.(27.27): U = μ ⋅ B. The scalar product depends on ! ! the angle between μ and B. ! ! ! ! ! ! EXECUTE: For μ and B parallel, φ = 0° and μ ⋅ B = μ B cosφ = μ B. For μ and B antiparallel, ! ! φ = 180° and μ ⋅ B = μ B cos φ = − μ B. U1 = + μ B, U 2 = − μ B ΔU = U 2 − U1 = −2μ B = −2(1.45 A ⋅ m 2 )(0.835 T) = −2.42 J ! ! EVALUATE: U is maximum when μ and B are antiparallel and minimum when they are parallel. When the coil is rotated as specified its magnetic potential energy decreases.


27-14

Chapter 27

27.48.

IDENTIFY:

Apply Eq.(27.29) in order to calculate I. The power drawn from the line is Psupplied = IVab . The

mechanical power is the power supplied minus the I 2 r electrical power loss in the internal resistance of the motor. SET UP: Vab = 120V , E = 105 V , and r = 3.2 Ω . V − E 120 V − 105 V = = 4.7 A. EXECUTE: (a) Vab = E + Ir ⇒ I = ab r 3.2 Ω (b) Psupplied = IVab = (4.7 A)(120 V) = 564 W.

27.49.

(c) Pmech = IVab − I 2r = 564 W − (4.7 A)2 (3.2 Ω) = 493 W. EVALUATE: If the rotor isn’t turning, when the motor is first turned on or if the rotor bearings fail, then E = 0 120V = 37.5 A . This large current causes large I 2 r heating and can trip the circuit breaker. and I = 3.2 Ω IDENTIFY: The circuit consists of two parallel branches with the potential difference of 120 V applied across each. One branch is the rotor, represented by a resistance Rr and an induced emf that opposes the applied potential. Apply the loop rule to each parallel branch and use the junction rule to relate the currents through the field coil and through the rotor to the 4.82 A supplied to the motor. SET UP: The circuit is sketched in Figure 27.49.

E is the induced emf developed by the motor. It is directed so as to oppose the current through the rotor.

Figure 27.49 EXECUTE: (a) The field coils and the rotor are in parallel with the applied potential difference V , so V = I f Rf . V 120 V If = = = 1.13 A. Rf 106 Ω (b) Applying the junction rule to point a in the circuit diagram gives I − I f − I r = 0. I r = I − I f = 4.82 A − 1.13 A = 3.69 A. (c) The potential drop across the rotor, I r Rr + E , must equal the applied potential difference V : V = I r Rr + E E = V − I r Rr = 120 V − ( 3.69 A )( 5.9 Ω ) = 98.2 V (d) The mechanical power output is the electrical power input minus the rate of dissipation of electrical energy in the resistance of the motor: electrical power input to the motor Pin = IV = ( 4.82 A )(120 V ) = 578 W

electrical power loss in the two resistances 2 2 Ploss = I f2 Rf + I f2 R = (1.13 A ) (106 Ω ) + ( 3.69 A ) ( 5.9 Ω ) = 216 W mechanical power output Pout = Pin − Ploss = 578 W − 216 W = 362 W The mechanical power output is the power associated with the induced emf E Pout = PE = EI r = ( 98.2 V )( 3.69 A ) = 362 W, which agrees with the above calculation.

27.50.

EVALUATE: The induced emf reduces the amount of current that flows through the rotor. This motor differs from the one described in Example 27.12. In that example the rotor and field coils are connected in series and in this problem they are in parallel. IDENTIFY: The field and rotor coils are in parallel, so Vab = I f Rf = E + I r Rr and I = I f + I r , where I is the current

drawn from the line. The power input to the motor is P = Vab I . The power output of the motor is the power input minus the electrical power losses in the resistances and friction losses. SET UP: Vab = 120 V. I = 4.82 A. EXECUTE:

(a) Field current I f =

120 V = 0.550 A. 218 Ω

(b) Rotor current I r = I total − I f = 4.82 A − 0.550 A = 4.27 A.


Magnetic Field and Magnetic Forces

27-15

(c) V = E + I r Rr and E = V − I r Rr = 120 V − (4.27 A)(5.9 Ω) = 94.8 V. (d) Pf = I f2 Rf = (0.550 A)2 (218 Ω) = 65.9 W. (e) Pr = I r2 Rr = (4.27 A) 2 (5.9 Ω) = 108 W. (f ) Power input = (120 V) (4.82 A) = 578 W. Poutput (578 W − 65.9 W − 108 W − 45 W) 359 W = = = 0.621. (g) Efficiency = Pinput 578 W 578 W

27.51.

EVALUATE: I 2 R losses in the resistance of the rotor and field coils are larger than the friction losses for this motor. IDENTIFY: The drift velocity is related to the current density by Eq.(25.4). The electric field is determined by the requirement that the electric and magnetic forces on the current-carrying charges are equal in magnitude and opposite in direction. (a) SET UP: The section of the silver ribbon is sketched in Figure 27.51a.

J x = n q vd so vd =

Jx nq

Figure 27.51a I I 120 A = = = 4.42 × 107 A/m 2 A y1 z1 (0.23 × 10−3 m)(0.0118 m) J 4.42 ×107 A/m 2 vd = x = = 4.7 ×10−3 m/s = 4.7 mm/s n q ( 5.85 ×1028 / m3 )(1.602 ×10−19 C ) ! (b) magnitude of E

EXECUTE:

Jx =

q Ez = q vd By Ez = vd By = (4.7 ×10−3 m/s)(0.95 T) = 4.5 ×10−3 V/m ! direction of E The drift velocity of the electrons is in the opposite direction to the current, as shown in Figure 27.51b. ! ! v×B ↑ ! ! ! ! ! FB = qv × B = −ev × B ↓ Figure 27.51b

The directions of the electric and magnetic forces on an electron in the ribbon are shown in Figure 27.51c.

! ! ! FE must oppose FB so FE is in the − z -direction Figure 27.51c ! ! ! ! ! ! FE = qE = −eE so E is opposite to the direction of FE and thus E is in the + z -direction. (c) The Hall emf is the potential difference between the two edges of the strip (at z = 0 and z = z1 ) that results from

27.52.

the electric field calculated in part (b). EHall = Ez1 = (4.5 × 10−3 V/m)(0.0118 m) = 53 μ V EVALUATE: Even though the current is quite large the Hall emf is very small. Our calculated Hall emf is more than an order of magnitude larger than in Example 27.13. In this problem the magnetic field and current density are larger than in the example, and this leads to a larger Hall emf. IDENTIFY: Apply Eq.(27.30). SET UP: A = y1 z1. E = E/z1. q = e. EXECUTE:

n=

J x By

=

IBy

=

IBy z1

=

IBy

q Ez A q Ez A q E y1 q E (78.0 A)(2.29 T) n= = 3.7 × 1028 electrons / m3 (2.3 × 10−4 m)(1.6 × 10−19 C)(1.31 × 10−4 V) EVALUATE: The value of n for this metal is about one-third the value of n calculated in Example 27.12 for copper.


27-16

Chapter 27

27.53.

! ! ! Use Eq.(27.2) to relate v , B, and F . ! ! SET UP: The directions of v1 and F1 are shown in Figure 27.53a. (a) IDENTIFY:

! ! ! ! F = qv × B says that F is perpendicular ! ! to v and B. The information given here ! means that B can have no z-component. Figure 27.53a ! ! The directions of v2 and F2 are shown in Figure 27.53b.

! ! ! ! F is perpendicular to v and B, so B can have no x-component. Figure 27.53b

! ! Both pieces of information taken together say that B is in the y-direction; B = B y ˆj. ! ! ! ! EXECUTE: Use the information given about F2 to calculate Fy : F2 = F2 iˆ, v 2 = v2 kˆ , B = By ˆj. ! ! ! F2 = qv2 × B says F2 iˆ = qv2 By kˆ × ˆj = qv2 By ( − iˆ) and F2 = − qv2 By ! By = − F2 /(qv2 ) = − F2 /(qv1 ). B has the maginitude F2 /(qv1 ) and is in the − y -direction. (b) F1 = qvB sin φ = qv1 By / 2 = F2 / 2 ! ! ! ! EVALUATE: v1 = v2 . v2 is perpendicular to B whereas only the component of v1 perpendicular to B contributes 27.54.

to the force, so it is expected that F2 > F1 , as we found. ! ! ! IDENTIFY: Apply F = qv × B. SET UP:

Bx = 0.450 T, By = 0 and Bz = 0.

EXECUTE:

Fx = q(v y Bz − vz By ) = 0.

Fy = q(vz Bx − vx Bz ) = (9.45 × 10−8 C)(5.85 × 104 m/s)(0.450 T) = 2.49 × 10−3 N.

27.55.

Fz = q(vx By − v y Bx ) = − (9.45 × 10−8 C)(−3.11 × 104 m/s)(0.450 T) = 1.32 × 10−3 N. ! ! ! ! ! ! EVALUATE: F is perpendicular to both v and B. We can verify that F ⋅ v = 0. Since B is along the x-axis, vx does not affect the force components. IDENTIFY: The sum of the magnetic, electrical, and gravitational forces must be zero to aim at and hit the target. SET UP: The magnetic field must point to the left when viewed in the direction of the target for no net force. The net force is zero, so ∑ F = FB − FE − mg = 0 and qvB – qE – mg = 0. EXECUTE:

Solving for B gives

B=

27.56.

qE + mg (2500 ×10−6 C)(27.5 N/C) + (0.0050 kg)(9.80 m/s 2 ) = = 3.7 T qv (2500 ×10−6 C)(12.8 m/s)

The direction should be perpendicular to the initial velocity of the coin. EVALUATE: This is a very strong magnetic field, but achievable in some labs. IDENTIFY: Apply R = mv / q B . ω = v / R SET UP: 1 eV = 1.60 × 10−19 J EXECUTE: (a) K = 2.7 MeV = (2.7 × 106eV) (1.6 × 10−19 J/eV) = 4.32 × 10−13 J. v=

R=

2K = m

2(4.32 × 10 −13 J) = 2.27 × 107 m/s . 1.67 × 10 −27 kg

mv (1.67 × 10 −27 kg) (2.27 × 107 m/s) v 2.27 × 107 m/s = = 0.068 m. Also, ω = = = 3.34 × 108 rad/s. −19 qB R (1.6 × 10 C) (3.5 T) 0.068 m

(b) If the energy reaches the final value of 5.4 MeV, the velocity increases by The angular frequency is unchanged from part (a) so is 3.34 × 108 rad / s. EVALUATE:

2 , as does the radius, to 0.096 m.

ω = q B / m , so ω is independent of the energy of the protons. The orbit radius increases when the

energy of the proton increases.


Magnetic Field and Magnetic Forces

27.57.

27-17

(a) IDENTIFY and SET UP: The maximum radius of the orbit determines the maximum speed v of the protons. Use Newton's 2nd law and ac = v 2 /R for circular motion to relate the variables. The energy of the particle is the

kinetic energy K = 12 mv 2 . ! ! EXECUTE: ∑ F = ma gives q vB = m(v 2 /R )

v=

q BR m

=

(1.60 × 10−19 C)(0.85 T)(0.40 m) = 3.257 × 107 m/s. The kinetic energy of a proton moving with this 1.67 × 10−27 kg

speed is K = 12 mv 2 = 12 (1.67 × 10−27 kg)(3.257 × 107 m/s) 2 = 8.9 × 10−13 J = 5.6 MeV (b) The time for one revolution is the period T = 2

⎛ q BR ⎞ (c) K = 12 mv 2 = 12 m ⎜ ⎟ = ⎝ m ⎠

2π R 2π (0.40 m) = = 7.7 × 10 −8 s 3.257 × 107 m/s v

2

1 2

q B2R2 . Or, B = m

factor of 2 then B must be increased by a factor of (d) v =

2 Km . B is proportional to qR

K , so if K is increased by a

2. B = 2(0.85 T) = 1.2 T.

q BR (3.20 × 10−19 C)(0.85 T)(0.40 m) = = 1.636 × 107 m/s m 6.65 × 10−27 kg

K = 12 mv 2 = 12 (6.65 × 10 −27 kg)(1.636 × 10 7 m/s) 2 = 8.9 × 10 −13 J = 5.5 MeV, the same as the maximum energy for protons. EVALUATE: We can see that the maximum energy must be approximately the same as follows: From part (c), 2

⎛ q BR ⎞ K = 12 m ⎜ ⎟ . For alpha particles q is larger by a factor of 2 and m is larger by a factor of 4 (approximately). ⎝ m ⎠ 2

27.58.

27.59.

Thus q / m is unchanged and K is the same. ! ! ! IDENTIFY: Apply F = qv × B. ! SET UP: v = −vˆj ! EXECUTE: (a) F = −qv[ Bx ( ˆj × iˆ) + By ( ˆj × ˆj ) + Bz ( ˆj × kˆ )] = qvBx kˆ − qvBz iˆ (b) Bx > 0, Bz < 0, sign of By doesn't matter. ! ! (c) F = q vBx iˆ − q vBx kˆ and F = 2 q vBx . ! ! ! EVALUATE: F is perpendicular to v , so F has no y-component. IDENTIFY: The contact at a will break if the bar rotates about b. The magnetic field is directed out of the page, so the magnetic torque is counterclockwise, whereas the gravity torque is clockwise in the figure in the problem. The maximum current corresponds to zero net torque, in which case the torque due to gravity is just equal to the torque due to the magnetic field. SET UP: The magnetic force is perpendicular to the bar and has moment arm l / 2 , where l = 0.750 m is the ⎛l ⎞ length of the bar. The gravity torque is mg ⎜ cos60.0° ⎟ 2 ⎝ ⎠ l l EXECUTE: τ gravity = τ B and mg cos 60.0° = IlB sin 90° . This gives 2 2

I=

2 mg cos60.0° (0.458 kg) ( 9.80 m/s ) (cos 60.0°) = = 1.93 A lB sin 90° (0.750 m)(1.55 T)(1)

Once contact is broken, the magnetic torque ceases. The 90.0° angle in the expression for τ B is the ! angle between the direction of I and the direction of B. EVALUATE:

27.60.

IDENTIFY: SET UP:

Apply R =

mv . qB

Assume D << R


27-18

Chapter 27

EXECUTE: (a) The path is sketched in Figure 27.60. (b) Motion is circular: x 2 + y 2 = R 2 ⇒ x = D ⇒ y1 = R 2 − D 2 (path of deflected particle) y2 = R (equation for tangent to the circle, path of undeflected particle).

d = y2 − y1 = R − R 2 − D 2 = R − R 1 −

⎡ ⎡ ⎛ D2 D2 ⎤ 1 D2 ⎞⎤ D2 R = − − 1 1 . For a ⎢ ⎥ . If R >> D , d ≈ R ⎢1 − ⎜ 1 − ⎟⎥ = 2 2 R R ⎦ 2 R2 ⎠⎦ 2R ⎣ ⎣ ⎝

particle moving in a magnetic field, R =

mv . But qB

d≈

1 2

mv 2 = qV , so R =

1 2mV . Thus, the deflection B q

D2B q D2B e . = 2 2mV 2 2mV

(c) d =

(0.50 m) 2 (5.0 × 10 −5 T) (1.6 × 10 −19 C) = 0.067 m = 6.7 cm. d ≈ 13% of D, which is fairly 2 2(9.11 × 10 −31 kg)(750 V)

significant. 2

EVALUATE:

In part (c), R =

1 2mV D 2 ⎛ D ⎞ ⎛R⎞ = =⎜ ⎟ D = 3.7 D and ⎜ ⎟ = 14 , so the approximation made in 2d ⎝ 2d ⎠ B e ⎝D⎠

part (b) is valid.

27.61.

Figure 27.60 ! ! ! ! ! IDENTIFY and SET UP: Use Eq.(27.2) to relate q, v , B and F . The force F and a are related by Newton's 2nd law. ! ! B = −(0.120 T)kˆ , v = (1.05 ×106 m/s)( − 3iˆ + 4 ˆj + 12kˆ ), F = 1.25 N ! ! ! (a) EXECUTE: F = qv × B ! F = q( −0.120 T)(1.05 ×106 m/s)( − 3iˆ × kˆ + 4 ˆj × kˆ + 12kˆ × kˆ )

iˆ × kˆ = − ˆj , ˆj × kˆ = iˆ, kˆ × kˆ = 0 ! F = −q (1.26 × 105 N/C)(+3 ˆj + 4iˆ) = −q(1.26 × 105 N/C)( +4iˆ + 3 ˆj ) The magnitude of the vector +4iˆ + 3 ˆj is 32 + 42 = 5. Thus F = − q(1.26 × 105 N/C)(5). F 1.25 N =− = −1.98 × 10−6 C 5(1.26 × 105 N/C) 5(1.26 × 105 N/C) ! ! ! ! (b) ∑ F = ma so a = F / m ! F = −q (1.26 × 105 N/C)(+4iˆ + 3 ˆj ) = −(−1.98 × 10−6 C)(1.26 × 105 N/C)( +4iˆ + 3 ˆj ) = +0.250 N(+4iˆ + 3 ˆj ) q=−

⎛ 0.250 N ⎞ ! ! 13 2 ˆ ˆ ˆ ˆ Then a = F / m = ⎜ ⎟ (+4i + 3 j ) = (9.69 × 10 m/s )(+4i + 3 j ) −15 × 2.58 10 kg ⎝ ⎠ ! (c) IDENTIFY and SET UP: F is in the xy-plane, so in the z-direction the particle moves with constant speed ! ! 12.6 × 106 m/s. In the xy-plane the force F causes the particle to move in a circle, with F directed in towards the center of the circle. ! ! EXECUTE: ∑ F = ma gives F = m(v 2 / R ) and R = mv 2 / F v 2 = vx2 + v y2 = (−3.15 × 106 m/s) 2 + (+4.20 × 106 m/s) 2 = 2.756 × 1013 m 2 /s 2 F = Fx2 + Fy2 = (0.250 N) 42 + 32 = 1.25 N R=

mv 2 (2.58 × 10−15 kg)(2.756 × 1013 m 2 /s 2 ) = = 0.0569 m = 5.69 cm F 1.25 N


Magnetic Field and Magnetic Forces

By Eq.(27.12) the cyclotron frequency is f = ω / 2π = v / 2π R.

(d) IDENTIFY and SET UP:

The circular motion is in the xy-plane, so v = vx2 + v y2 = 5.25 × 106 m/s.

EXECUTE:

5.25 × 106 m/s = 1.47 × 107 Hz, and ω = 2π f = 9.23 × 107 rad/s 2π R 2π (0.0569 m) (e) IDENTIFY and SET UP Compare t to the period T of the circular motion in the xy-plane to find the x and y coordinates at this t. In the z-direction the particle moves with constant speed, so z = z0 + vz t. 1 1 EXECUTE: The period of the motion in the xy-plane is given by T = = = 6.80 × 10 −8 s f 1.47 × 107 Hz In t = 2T the particle has returned to the same x and y coordinates. The z-component of the motion is motion with a constant velocity of vz = +12.6 × 106 m/s. Thus z = z0 + vz t = 0 + (12.6 ×106 m/s)(2)(6.80 ×10−8 s) = +1.71 m. The coordinates at t = 2T are x = R, y = 0, z = +1.71 m. ! EVALUATE: The circular motion is in the plane perpendicular to B. The radius of this motion ! gets smaller when B increases and it gets larger when v increases. There is no magnetic force in the direction of B so the particle moves with constant velocity in that direction. The superposition of circular motion in the xy-plane and constant speed motion in the z-direction is a helical path. IDENTIFY: The net magnetic force on the wire is the vector sum of the force on the straight segment plus the force on the curved section. We must integrate to get the force on the curved section. f =

27.62.

v

=

SET UP:

27.63.

∑F = F

straight, top

π

+ Fcurved + Fstraight, bottom and Fstraight, top = Fstraight, bottom = iLstraight B. Fcurved, x = ∫ iRB sin θ dθ = 2iRB 0

(the same as if it were a straight segment 2R long) and Fy = 0 due to symmetry. Therefore, F = 2iLstraightB + 2iRB EXECUTE: Using Lstraight = 0.55 m, R = 0.95 m, i = 3.40 A, and B = 2.20 T gives F = 22 N, to right. EVALUATE: Notice that the curve has no effect on the force. In other words, the force is the same as if the wire were simply a straight wire 3.00 m long. IDENTIFY: τ = NIAB sin φ . SET UP: The area A is related to the diameter D by A = 14 π D 2 . EXECUTE:

27.64.

τ = NI ( 14 π D2 ) B sin φ . τ is proportional to D 2 . Increasing D by a factor of 3 increases τ by a factor of

32 = 9 . EVALUATE: The larger diameter means larger length of wire in the loop and also larger moment arms because parts of the loop are farther from the axis. ! ! ! IDENTIFY: Apply F = qv × B ! SET UP: v = vkˆ ! ! EXECUTE: (a) F = − qvB iˆ + qvB ˆj. But F = 3F iˆ + 4 F ˆj , so 3F = −qvB and 4 F = qvB y

x

0

0

0

y

0

x

3F 4F Therefore, By = − 0 , Bx = 0 and Bz is undetermined. qv qv 2

2

6 F0 F F 11F0 ⎛ qv ⎞ ⎛ qv ⎞ = Bx 2 + By 2 + Bz 2 = 0 9 + 16 + ⎜ ⎟ Bz 2 = 0 25 + ⎜ ⎟ Bz 2 , so Bz = ± ⋅ qv qv F qv F qv 0 0 ⎝ ⎠ ⎝ ⎠ ! EVALUATE: The force doesn’t depend on Bz , since v is along the z-direction. mv IDENTIFY: For the velocity selector, E = vB . For the circular motion in the field B′ , R = . q B′ SET UP: B = B′ = 0.701 T. E 1.88 × 104 N/C mv EXECUTE: v = = , so = 2.68 × 104 m/s. R = B 0.701 T qB′ (b) B =

27.65.

27-19

82(1.66 × 10−27 kg)(2.68 × 104 m/s) = 0.0325 m. (1.60 × 10−19 C)(0.701 T) 84(1.66 × 10−27 kg)(2.68 × 104 m/s) R84 = = 0.0333 m. (1.60 × 10−19 C)(0.701 T) 86(1.66 × 10−27 kg)(2.68 × 104 m/s) R86 = = 0.0341 m. (1.60 × 10−19 C)(0.701 T) The distance between two adjacent lines is ΔR = 1.6 mm . R82 =


27-20

27.66.

Chapter 27

EVALUATE: The distance between the 82 Kr line and the 84 Kr line is 1.6 mm and the distance between the 84 Kr line and the 86 Kr line is 1.6 mm. Adjacent lines are equally spaced since the 82 Kr versus 84 Kr and 84 Kr versus 86 Kr mass differences are the same. IDENTIFY: Apply conservation of energy to the acceleration of the ions and Newton’s second law to their motion in the magnetic field. SET UP: The singly ionized ions have q = +e . A 12 C ion has mass 12 u and a 14 C ion has mass 14 u, where

1 u = 1.66 × 10−27 kg EXECUTE:

2qV . In the magnetic field, m

(a) During acceleration of the ions, qV = 12 mv 2 and v =

qB 2 R 2 mv m 2qV / m and m = . = 2V qB qB qB 2 R 2 (1.60 × 10−19 C)(0.150 T) 2 (0.500 m) 2 (b) V = = = 2.26 × 104 V 2m 2(12)(1.66 × 10−27 kg) R=

(c) The ions are separated by the differences in the diameters of their paths. D = 2 R = 2 ΔD = D14 − D12 = 2

ΔD = 2

2Vm qB 2

−2 14

=2 12

2(2.26 × 104 V)(1.66 × 10−27 kg) (1.6 × 10−19 C)(0.150 T) 2

EVALUATE: 27.67.

2Vm qB 2

The speed of the

12

(

2V (1 u) qB 2

(

2Vm , so qB 2

)

14 − 12 .

)

14 − 12 = 8.01×10−2 m. This is about 8 cm and is easily distinguishable.

C ion is v =

2(1.60 × 10 −19 C)(2.26 × 10 4 V) = 6.0 × 105 m/s . This is very fast, but 12(1.66 × 10−27 kg)

well below the speed of light, so relativistic mechanics is not needed. IDENTIFY: The force exerted by the magnetic field is given by Eq.(27.19). The net force on the wire must be zero. SET UP: For the wire to remain at rest the force exerted on it by the magnetic field must have a component directed up the incline. To produce a force in this direction, the current in the wire must be directed from right to left in Figure 27.61 in the textbook. Or, viewing the wire from its left-hand end the directions are shown in Figure 27.67a.

Figure 27.67a

The free-body diagram for the wire is given in Figure 27.67b.

EXECUTE:

∑F

y

=0

FI cosθ − Mg sinθ = 0 FI = ILB sin φ

!

φ = 90° since B is perpendicular to the current direction. Figure 27.67b

Mg tan θ LB EVALUATE: The magnetic and gravitational forces are in perpendicular directions so their components parallel to the incline involve different trig functions. As the tilt angle θ increases there is a larger component of Mg down the incline and the component of FI up the incline is smaller; I must increase with θ to compensate. As θ → 0, I → 0 and as θ → 90°, I → ∞. IDENTIFY: The current in the bar is downward, so the magnetic force on it is vertically upwards. The net force on the bar is equal to the magnetic force minus the gravitational force, so Newton’s second law gives the acceleration. The bar is in parallel with the 10.0-Ω resistor, so we must use circuit analysis to find the initial current through it. Thus (ILB) cosθ − Mg sin θ = 0 and I =

27.68.


Magnetic Field and Magnetic Forces

27-21

SET UP: First find the current. The equivalent resistance across the battery is 30.0 Ω, so the total current is 4.00 A, half of which goes through the bar. Applying Newton’s second law to the bar gives ∑ F = ma = FB − mg = iLB − mg. EXECUTE:

27.69.

Solving for the acceleration gives iLB − mg (2.0 A)(1.50 m)(1.60 T) − 3.00 N a= = = 5.88 m/s 2 . m (3.00 N/9.80 m/s 2 )

The direction is upward. EVALUATE: Once the bar is free of the conducting wires, its acceleration will become 9.8 m/s2 downward since only gravity will be acting on it. IDENTIFY: Calculate the acceleration of the ions when they first enter the field and assume this acceleration is constant. Apply conservation of energy to the acceleration of the ions by the potential difference. ! ! SET UP: Assume v = v iˆ and neglect the y-component of v that is produced by the magnetic force. x

EXECUTE:

(a)

1 2

mv x 2 = qV , so vx = 2

2qV qv B x . Also, a y = x and t = . m m vx 2

1/ 2

1/ 2

⎛ x ⎞ 1 ⎛ qv B ⎞ ⎛ x ⎞ 1 ⎛ qBx 2 ⎞⎛ m ⎞ q ⎞ 2⎛ y = a yt = a y ⎜ ⎟ = ⎜ x ⎟ ⎜ ⎟ = ⎜ ⎟⎜ ⎟ = Bx ⎜ ⎟ . ⎝ 8mV ⎠ ⎝ vx ⎠ 2 ⎝ m ⎠ ⎝ vx ⎠ 2 ⎝ m ⎠ ⎝ 2qV ⎠ (b) This can be used for isotope separation since the mass in the denominator leads to different locations for different isotopes. EVALUATE: For B = 0.1 T, v = 1 × 104 m/s, q = +e and m = 12 u = 2.0 × 10−26 kg, y = (1.0 m −2 ) x 2 . The approximation y << x is valid as long as x is on the order of 10 cm or less. 1 2

27.70.

27.71.

2

1 2

IDENTIFY: Turning the charged loop creates a current, and the external magnetic field exerts a torque on that current. SET UP: The current is I = q/T = q/(1/f ) = qf = q(ω/2π) = qω/2π. The torque is τ = μ B sin φ . EXECUTE: In this case, φ = 90° and µ = AB, giving τ = IAB. Combining the results for the torque and current

⎛ qω ⎞ 2 2 1 and using A = πr2 gives τ = ⎜ ⎟ π r B = 2 qω r B 2 π ⎝ ⎠ EVALUATE: Any moving charge is a current, so turning the loop creates a current causing a magnetic force. mv IDENTIFY: R = . qB SET UP: After completing one semicircle the separation between the ions is the difference in the diameters of their paths, or 2( R13 − R12 ) . A singly ionized ion has charge +e . EXECUTE:

(a) B =

mv (1.99 × 10−26 kg)(8.50 × 103 m/s) = = 8.46 × 10−3 T . qR (1.60 × 10−19 C)(0.125 m)

(b) The only difference between the two isotopes is their masses.

27.72.

⎛m ⎞ ⎛ 2.16 × 10 −26 kg ⎞ R13 = R12 ⎜ 13 ⎟ = (12.5 cm) ⎜ ⎟ = 13.6 cm. The diameter is 27.2 cm. −26 ⎝ 1.99 × 10 kg ⎠ ⎝ m12 ⎠ (c) The separation is 2( R13 − R12 ) = 2(13.6 cm − 12.5 cm) = 2.2 cm. This distance can be easily observed. EVALUATE: Decreasing the magnetic field increases the separation between the two isotopes at the detector. IDENTIFY: The force exerted by the magnetic field is F = ILB sin φ . a = F / m and is constant. Apply a constant acceleration equation to relate v and d. ! SET UP: φ = 90°. The direction of F is given by the right-hand rule. EXECUTE: (a) F = ILB, to the right. v2 v 2m = (b) vx2 = v02x + 2a x ( x − x0 ) gives v 2 = 2ad and d = . 2a 2 ILB (1.12 × 104 m/s)2 (25 kg) = 3.14 × 106 m = 3140 km (c) d = 2(2000 A)(0.50 m)(0.50 T)

EVALUATE: 27.73.

R v R R = = constant and 12 = 13 . m qB m12 m13

a=

ILB (20 × 103 A)(0.50 m)(0.50 T) = = 20 m/s 2 . The acceleration due to gravity is not negligible. m 25 kg

IDENTIFY: Apply F = IlB sin φ to calculate the force on each segment of the wire that is in the magnetic field. The net force is the vector sum of the forces on each segment.


27-22

Chapter 27

SET UP: The direction ! of the ! magnetic force on each current segment in the field is shown in Figure 27.73. By symmetry, Fa = Fb . Fa and Fb are in opposite directions so their vector sum is zero. The net force equals Fc . For Fc , φ = 90° and l = 0.450 m. EXECUTE: Fc = IlB = (6.00 A)(0.450 m)(0.666 T) = 1.80 N . The net force is 1.80 N, directed to the left. EVALUATE: The shape of the region of uniform field doesn’t matter, as long as all of segment c is in the field and as long as the lengths of the portions of segments a and b that are in the field are the same.

27.74.

Figure 27.73 ! ! ! IDENTIFY: Apply F = Il × B. ! SET UP: l = lkˆ ! ! EXECUTE: (a) F = I (lkˆ ) × B = Il ⎡⎣(− By ) iˆ + ( Bx ) ˆj ⎤⎦ . This gives

Fx = − IlBy = −(9.00 A) (0.250 m)( − 0.985 T) = 2.22 N and Fy = IlBx = (9.00 A)(0.250 m)( − 0.242 T) = −0.545 N. . Fz = 0 , since the wire is in the z-direction.

27.75.

(b) F = Fx 2 + Fy 2 = (2.22 N) 2 + (0.545 N) 2 = 2.29 N. ! ! EVALUATE: F must be perpendicular to the current direction, so F has no z component. IDENTIFY: For the loop to be in equilibrium the net torque on it must be zero. Use Eq.(27.26) to calculate the torque due to the magnetic field and use Eq.(10.3) for the torque due to the gravity force. SET UP: See Figure 27.75a.

Use

∑τ

A

= 0, where

point A is at the origin.

Figure 27.75a EXECUTE:

See Figure 27.75b.

τ mg = mgr sin φ = mg (0.400 m)sin 30.0°

! The torque is clockwise; τ mg is directed into the paper. Figure 27.75b

! ! For the loop to be in equilibrium the torque due to B must be counterclockwise (opposite to τ mg ) and it must be

that τ B = τ mg . See Figure 27.75c.

!

!

!

τ B = μ × B. For this torque to be ! counterclockwise ( τ B directed out of the ! paper), B must be in the + y -direction.

Figure 27.75c


Magnetic Field and Magnetic Forces

27-23

τ B = μ B sin φ = IAB sin 60.0°

τ B = τ mg gives IAB sin 60.0° = mg ( 0.0400 m ) sin 30.0° m = ( 0.15 g/cm ) 2 ( 8.00 cm + 6.00 cm ) = 4.2 g = 4.2 × 10−3 kg A = ( 0.800 m )( 0.0600 m ) = 4.80 × 10−3 m 2 B= B=

27.76.

mg ( 0.0400 m )( sin 30.0° ) IA sin 60.0°

( 4.2 ×10

−3

kg )( 9.80 m/s 2 ) ( 0.0400 m ) sin 30.0°

(8.2 A)(4.80 × 10−3 m 2 )sin 60.0°

= 0.024 T

! EVALUATE: As the loop swings up the torque due to B decreases to zero and the torque due to mg increases from zero, so there must be an orientation of the loop where the net torque is zero. ! ! ! ! IDENTIFY: The torque exerted by the magnetic field is τ = μ × B. The torque required to hold the loop in place is −τ . ! SET UP: μ = IA. μ is normal to the plane of the loop, with a direction given by the right-hand rule that is illustrated in Figure 27.32 in the textbook. τ = IAB sin φ , where φ is the angle between the normal to the loop and ! the direction of B. EXECUTE: (a) τ = IAB sin 60° = (15.0 A)(0.060 m)(0.080 m)(0.48 T)sin 60° = 0.030 N ⋅ m , in the − ˆj direction. To keep the loop in place, you must provide a torque in the + ˆj direction. (b) τ = IAB sin 30° = (15.0 A)(0.60 m)(0.080 m)(0.48 T)sin 30° = 0.017 N ⋅ m, in the + ˆj direction. You must

provide a torque in the − ˆj direction to keep the loop in place.

27.77.

EVALUATE: (c) If the loop was pivoted through its center, then there would be a torque on both sides of the loop parallel to the rotation axis. However, the lever arm is only half as large, so the total torque in each case is identical to the values found in parts (a) and (b). IDENTIFY: Use Eq.(27.20) to calculate the force and then the torque on each small section of the rod and integrate to find the total magnetic torque. At equilibrium the torques from the spring force and from the magnetic force cancel. The spring force depends on the amount x the spring is stretched and then U = 12 kx 2 gives the energy stored in the spring. (a) SET UP:

Divide the rod into infinitesimal sections of length dr, as shown in Figure 27.77.

Figure 27.77 EXECUTE:

The magnetic force on this section is dFI = IBdr and is perpendicular to the rod. The torque dτ due to l

the force on this section is dτ = rdFI = IBr dr. The total torque is ∫ dτ = IB ∫ r dr = 12 Il 2 B = 0.0442 N ⋅ m, clockwise. 0

(b) SET UP: FI produces a clockwise torque so the spring force must produce a counterclockwise torque. The spring force must be to the left; the spring is stretched. EXECUTE: Find x, the amount the spring is stretched: ∑τ = 0, axis at hinge, counterclockwise torques positive

(kx)l sin 53° − 12 Il 2 B = 0 x=

IlB ( 6.50 A )( 0.200 m )( 0.340 T ) = 0.05765 m = 2k sin 53.0° 2 ( 4.80 N/m ) sin 53.0°

U = 12 kx 2 = 7.98 × 10−3 J EVALUATE: The magnetic torque calculated in part (a) is the same torque calculated from a force diagram in which the total magnetic force FI = IlB acts at the center of the rod. We didn't include a gravity torque since the problem said the rod had negligible mass.


27-24

27.78.

Chapter 27

! → ! IDENTIFY: Apply F = Il × B to calculate the force on each side of the loop. SET UP: The net force is the vector sum of the forces on each side of the loop. EXECUTE: (a) FPQ = (5.00 A)(0.600 m)(3.00 T)sin(0°) = 0 N . FRP = (5.00 A) (0.800 m) (3.00 T) sin(90°) = 12.0 N , into the page.

FQR = (5.00 A)(1.00 m)(3.00 T) ( 0.800 /1.00 ) = 12.0 N , out of the page. (b) The net force on the triangular loop of wire is zero. (c) For calculating torque on a straight wire we can assume that the force on a wire is applied at the wire’s center. Also, note that we are finding the torque with respect to the PR-axis (not about a point), and consequently the lever arm will be the distance from the wire’s center to the x-axis. τ = rF sin φ gives τ PQ = r (0 N) = 0 ,

τ RP = (0 m) F sin φ = 0 and τ QR = (0.300 m)(12.0 N)sin(90°) = 3.60 N ⋅ m . The net torque is 3.60 N ⋅ m . (d) According to Eq.(27.28), τ = NIAB sin φ = (1)(5.00 A) ( 12 ) (0.600 m)(0.800 m)(3.00 T)sin(90°) = 3.60 N ⋅ m ,

which agrees with part (c). (e) Since FQR is out of the page and since this is the force that produces the net torque, the point Q will be rotated

27.79.

out of the plane of the figure. EVALUATE: In the expression τ = NIAB sin φ , φ is the angle between the plane of the loop and the direction of ! B . In this problem, φ = 90° . IDENTIFY: Use Eq.(27.20) to calculate the force on a short segment of the coil and integrate over the entire coil to find the total force. SET UP: See Figure 27.79a. ! Consider the force dF on a short segment dl at the left-hand side of the coil, as viewed in Figure 27.69 in the textbook. The current at this point is ! directed out of the page. dF is perpendicular both ! to B and to the direction of I.

Figure 27.79a

See Figure 27.79b. ! Consider also the force dF ′ on a short segment on the opposite side of the coil, at the right-hand side of the coil in Figure 27.69 in the textbook. The current at this point is directed into the page.

Figure 27.79b

The two sketches show that the x-components cancel and that the y-components add. This is true for all pairs of short segments on opposite sides of the coil. The net magnetic force on the coil is in the y-direction and its magnitude is given by F = ∫ dFy . ! EXECUTE: dF = Idl B sin φ . But B is perpendicular to the current direction so φ = 90°. dFy = dF cos30.0 = IB cos30.0°dl F = ∫ dFy = IB cos30.0°∫ dl But ∫ dl = N ( 2π r ) , the total length of wire in the coil.

! F = IB cos30.0° N ( 2π r ) = ( 0.950 A )( 0.200 T )( cos30.0° )( 50 ) 2π ( 0.0078 m ) = 0.444 N and F = − ( 0.444 N ) ˆj

EVALUATE: The magnetic field makes a constant angle with the plane of the coil but has a different direction at different points around the circumference of the coil so is not uniform. The net force is proportional to the magnitude of the current and reverses direction when the current reverses direction.


Magnetic Field and Magnetic Forces

27.80.

27-25

IDENTIFY: Conservation of energy relates the accelerating potential difference V to the final speed of the ions. In mv . the magnetic field region the ions travel in an arc of a circle that has radius R = qB SET UP: The quarter-circle paths of the two ions are shown in Figure 27.80. The separation at the detector is Δr = R18 − R16 . Each ion has charge q = +e . (a) Conservation of energy gives q V = 12 mv 2 and v =

EXECUTE:

2qV m

. R=

2 q mV m 2qV . = qB m qB

2eV ( m18 − m16 ) . eB 2 2 e(Δr ) B (1.60 × 10−19 C)(4.00 × 10−2 m) 2 (0.050 T) 2 = 2 2 m18 − m16 2 2.99 × 10−26 kg − 2.66 × 10−26 kg

q = e for each ion. Δr = R18 − R16 = (b) V = 2e

(

(ΔreB) 2 m18 − m16

)

2

= 2

(

)

)

(

V = 3.32 × 10 V . EVALUATE: The speed of the 16 O ion after it has been accelerated through a potential difference of V = 3.32 × 103 V is 2.00 × 105 m/s . Increasing the accelerating voltage increases the separation of the two isotopes at the detector. But it does this by increasing the radius of the path for each ion, and this increases the required size of the magnetic field region. 3

27.81.

Figure 27.80 ! ! ! IDENTIFY: Apply dF = Idl × B to each side of the loop. ! SET UP: For each side of the loop, dl is parallel to that side of the loop and is in the direction of I. Since the loop is in the xy-plane, z = 0 at the loop and By = 0 at the loop. (a) The magnetic field lines in the yz-plane are sketched in Figure 27.81. L ! L ! ! B y dy ˆ 1 (b) Side 1, that runs from (0,0) to (0,L): F = ∫ Idl × B = I ∫ 0 i = 2 B0 LIiˆ. L 0 0 L L ! ! ! B y dx ˆ Side 2, that runs from (0,L) to (L,L): F = ∫ Idl × B = I ∫ 0 j = − IB0 Lˆj . L 0, y = L 0, y = L EXECUTE:

! Side 3, that runs from (L,L) to (L,0): F =

0

! ! Idl × B = I

L,x=L

! Side 4, that runs from (L,0) to (0,0): F =

0

L, y =0

! ! Idl × B = I

0

B0 y dy ˆ ( − i ) = − 12 IB0 Liˆ . L L, x= L

∫ 0

B0 y dx ˆ j = 0. L L, y =0

! (c) The sum of all forces is Ftotal = − IB0 Lˆj. EVALUATE: The net force on sides 1 and 3 is zero. The force on side 4 is zero, since y = 0 and z = 0 at that side and therefore B = 0 there. The net force on the loop equals the force on side 2.

Figure 27.81


27-26

27.82.

Chapter 27

! ! ! ! ! ! IDENTIFY: Apply dF = Idl × B to each side of the loop. τ = r × F . ! SET UP: For each side of the loop, dl is parallel to that side of the loop and is in the direction of I. EXECUTE: (a) The magnetic field lines in the xy-plane are sketched in Figure 27.82. L ! L ! ! B y dy ˆ (− k ) = − 12 B0 LIkˆ . (b) Side 1, that runs from (0,0) to (0,L): F = ∫ Idl × B = I ∫ 0 L 0 0 L ! L ! ! B x dx ˆ 1 Side 2, that runs from (0,L) to (L,L): F = ∫ Idl × B = I ∫ 0 k = 2 IB0 Lkˆ. L 0 0 L ! L ! ! B ydy ˆ k = + 12 IB0 Lkˆ Side 3, that runs from (L,L) to (L,0): F = ∫ Idl × B = I ∫ 0 L 0 0 L ! L ! ! B xdx ˆ (− k ) = − 12 IB0 Lkˆ. Side 4, that runs from (L,0) to (0,0): F = ∫ Idl × B = I ∫ 0 L 0 0

(c) If free to rotate about the x-axis, the torques due to the forces on sides 1 and 3 cancel and the torque due to the ! ! ! ! IB L2 forces on side 4 is zero. For side 2, r = Lˆj . Therefore, τ = r × F = 0 iˆ = 12 IAB0 iˆ . 2 (d) If free to rotate about the y-axis, the torques due to the forces on sides 2 and 4 cancel and the torque due to the ! ! ! IB L2 ! forces on side 1 is zero. For side 3, r = Liˆ . Therefore, τ = r × F = 0 ˆj = − 12 IAB0 ˆj . 2 ! ! ! EVALUATE: (e) The equation for the torque τ = μ × B is not appropriate, since the magnetic field is not constant.

27.83.

Figure 27.82 IDENTIFY: While the ends of the wire are in contact with the mercury and current flows in the wire, the magnetic field exerts an upward force and the wire has an upward acceleration. After the ends leave the mercury the electrical connection is broken and the wire is in free-fall. (a) SET UP: After the wire leaves the mercury its acceleration is g, downward. The wire travels upward a total distance of 0.350 m from its initial position. Its ends lose contact with the mercury after the wire has traveled 0.025 m, so the wire travels upward 0.325 m after it leaves the mercury. Consider the motion of the wire after it leaves the mercury. Take +y to be upward and take the origin at the position of the wire as it leaves the mercury. a y = −9.80 m/s 2 , y − y0 = +0.325 m, v y = 0 (at maximum height), v0 y = ?

v y2 = v02y + 2a y ( y − y0 ) EXECUTE:

v0 y = −2a y ( y − y0 ) = −2(−9.80 m/s 2 )(0.325 m) = 2.52 m/s

(b) SET UP: Now consider the motion of the wire while it is in contact with the mercury. Take +y to be upward and the origin at the initial position of the wire. Calculate the acceleration: y − y0 = +0.025 m, v0 y = 0 (starts from

rest), v y = +2.52 m/s (from part (a)), a y = ? v y2 = v02y + 2a y ( y − y0 ) EXECUTE:

ay =

v y2 2( y − y0 )

=

(2.52 m/s) 2 = 127 m/s 2 2(0.025 m)


Magnetic Field and Magnetic Forces

27-27

The free-body diagram for the wire is given in Figure 27.83. EXECUTE: ∑ Fy = ma y

SET UP:

FB − mg = ma y

IlB = m ( g + a y ) I=

m ( g + ay ) lB

Figure 27.83

27.84.

l is the length of the horizontal section of the wire; l = 0.150 m (5.40 × 10−5 kg)(9.80 m/s 2 + 127 m/s 2 ) I= = 7.58 A (0.150 m)(0.00650 T) (c) IDENTIFY and SET UP: Use Ohm's law. V 1.50 V EXECUTE: V = IR so R = = = 0.198 Ω I 7.58 A EVALUATE: The current is large and the magnetic force provides a large upward acceleration. During this upward acceleration the wire moves a much shorter distance as it gains speed than the distance is moves while in free-fall with a much smaller acceleration, as it loses the speed it gained. The large current means the resistance of the wire must be small. IDENTIFY and SET UP: Follow the procedures specified in the problem. ! ! EXECUTE: (a) dl = dltˆ , where tˆ is a unit vector in the tangential direction. dl = Rdθ ⎡⎣ − sin θiˆ + cos θˆj ⎤⎦ . Note that this implies that when θ = 0, the line element points in the +y-direction, and when the angle is 90°, the line element points in the –x-direction. This is in agreement with the diagram. ! ! ! dF = Idl × B = IRdθ ⎡⎣ − sin θiˆ + cos θˆj ⎤⎦ × ( Bx iˆ) = IBx Rdθ [ − cos θkˆ ] . 2π ""! 2 π (b) F = ∫ − cosθ IBx R dθkˆ = − IBx R ∫ cos θdθkˆ = 0. 0

0

! ! ! (c) dτ = r × dF = R (cos θiˆ + sin θˆj ) × ( IBx R dθ[ − cos θkˆ ]) = − R 2 IBx dθ (sin θ cos θiˆ − cos 2 θˆj )

27.85.

2π 2π ⎛ 2π ⎞ ! ! ⎛ θ sin 2θ ⎞ ˆ ! 2 2 ˆ ˆ ˆ ˆ (d) τ = ∫ dτ = − R 2 IBx ⎜ ∫ sin θ cos θdθiˆ − ∫ cos 2 θdθˆj ⎟ = IR 2 Bx ⎜ + ⎟ j . τ = IR Bx πj = IπR Bx j = IAk × Bx i 4 ⎠0 ⎝2 0 ⎝0 ⎠ ! ! ! and τ = μ × B. ! ! ! EVALUATE: Section 27.7 of the textbook derived τ = μ × B for the case of a rectangular coil. This problem shows that the same result also applies to a circular coil. ! ! ! ! (a) IDENTIFY: Use Eq.(27.27) to relate U , μ and B and use Eq.(27.26) to relate τ , μ and B. We also know that ! B02 = Bx2 + By2 + Bz2 . This gives three equations for the three components of B.

SET UP:

The loop and current are shown in Figure 27.85.

!

μ is into the plane of the paper, in the –z-direction Figure 27.85

!

μ = −μ kˆ = − IAkˆ

! (b) EXECUTE: τ = D ( +4 iˆ − 3 ˆj ), where D > 0. ! ! μ = − IAkˆ , B = Bx iˆ + By ˆj + By kˆ ! ! ! τ = μ × B = ( − IA)( Bx kˆ × iˆ + B y kˆ × ˆj + Bz kˆ × kˆ ) = IAB y iˆ − IABx ˆj ! Compare this to the expression given for τ : IABy = 4D so By = 4D / IA and − IABx = −3D so Bx = 3D / IA ! Bz doesn't contribute to the torque since μ is along the z-direction. But B = B0 and Bx2 + By2 + Bz2 = B02 ; with

B0 = 13D / IA. Thus Bz = ± B02 − Bx2 − By2 = ± ( D / IA ) 169 − 9 − 16 = ±12 ( D / IA ) ! ! ! ! That U = − μ ⋅ B is negative determines the sign of Bz : U = − μ ⋅ B = −( − IAkˆ ) ⋅ ( Bx iˆ + By ˆj + Bz kˆ ) = + IABz So U negative says that Bz is negative, and thus Bz = −12 D / IA.


27-28

Chapter 27

EVALUATE:

!

!

μ is along the z-axis so only Bx and By contribute to the torque. Bx produces a y-component of τ

! ! ! and By produces an x-component of τ . Only Bz affects U, and U is negative when μ and Bz are parallel. 27.86.

27.87.

Δq and μ = IA. Δt ! SET UP: The direction of μ is given by the right-hand rule that is illustrated in Figure 27.32 in the textbook. I is in the direction of flow of positive charge and opposite to the direction of flow of negative charge. dq Δq qu v ev EXECUTE: (a) I u = . = = = dt Δt 2π r 3π r ev evr (b) μu = I u A = π r2 = . 3π r 3 evr 2evr (c) Since there are two down quarks, each of half the charge of the up quark, μd = μu = . Therefore, μtotal = . 3 3 3μ 3(9.66 × 10−27 A ⋅ m 2 ) = = 7.55 × 107 m s. (d) v = 2er 2(1.60 × 10−19 C)(1.20 × 10−15 m) EVALUATE: The speed calculated in part (d) is 25% of the speed of light. IDENTIFY: Eq.(27.8) says that the magnetic field through any closed surface is zero. SET UP: The cylindrical Gaussian surface has its top at z = L and its bottom at z = 0 . The rest of the surface is the curved portion of the cylinder and has radius r and length L. B = 0 at the bottom of the surface, since z = 0 there. ! ! EXECUTE: (a) úB ⋅ dA = ∫ Bz dA + ∫ Br dA = ∫ ( βL) dA + ∫ Br dA = 0. This gives 0 = β Lπ r 2 + Br 2π rL , and

IDENTIFY:

I=

top

curved

top

curved

βr . 2 (b) The two diagrams in Figure 27.87 show views of the field lines from the top and side of the Gaussian surface. EVALUATE: Only a portion of each field line is shown; the field lines are closed loops. Br ( r ) = −

27.88.

IDENTIFY:

Figure 27.87 ! ! U = − μ ⋅ B . In part (b) apply conservation of energy.

SET UP: The kinetic energy of the rotating ring is K = 12 I ω 2 . ! ! ! ! ! ! "! EXECUTE: (a) ΔU = −( μf ⋅ B − μi ⋅ B ) = −( μf − μi ) ⋅ B = ⎡⎣ − μ (− kˆ − (−0.8iˆ + 0.6 ˆj )) ⎤⎦ ⋅ ⎡⎣ B0 (12iˆ + 3 ˆj − 4kˆ ) ⎤⎦ .

ΔU = IAB0 [(−0.8)( +12) + (0.6)(+3) + (+1)( −4)] = (12.5 A)(4.45 × 10−4 m 2 )(0.0115 T)( − 11.8). ΔU = −7.55 × 10−4 J .

(b) ΔK =

1 2 2ΔK 2(7.55 × 10−4 J) Iω . ω = = = 42.1 rad/s. 2 8.50 × 10−7 kg ⋅ m 2 I

EVALUATE:

27.89.

The potential energy of the ring decreases and its kinetic energy increases. mv IDENTIFY and SET UP: In the magnetic field, R = . Once the particle exits the field it travels in a straight line. qB Throughout the motion the speed of the particle is constant. mv (3.20 × 10−11 kg)(1.45 × 105 m/s) EXECUTE: (a) R = = = 5.14 m. qB (2.15 × 10−6 C)(0.420 T)


Magnetic Field and Magnetic Forces

27-29

0.35 m , so 5.14 m d 0.25 m θ = 2.78° = 0.0486 rad. d = Rθ = (5.14 m)(0.0486 rad) = 0.25 m. And t = = = 1.72 × 10−6 s. v 1.45 × 105 m/s (c) Δx1 = d tan(θ / 2) = (0.25 m)tan (2.79° / 2) = 6.08 × 10−3 m.

(b) See Figure 27.89. The distance along the curve, d , is given by d = Rθ . sin θ =

(d) Δx = Δx1 + Δx2 , where Δx2 is the horizontal displacement of the particle from where it exits the field region to where it hits the wall. Δx2 = (0.50 m) tan 2.79° = 0.0244 m. Therefore, Δx = 6.08 × 10−3 m + 0.0244 m = 0.0305 m. EVALUATE: d is much less than R, so the horizontal deflection of the particle is much smaller than the distance it travels in the y-direction.

27.90.

Figure 27.89 ! IDENTIFY: The current direction is perpendicular to B , so F = IlB . If the liquid doesn’t flow, a force (Δp ) A from the pressure difference must oppose F. SET UP: J = I/A, where A = hw. EXECUTE: (a) Δp = F / A = IlB / A = JlB. Δ p (1.00 atm)(1.013 × 105 Pa/atm) = = 1.32 × 106 A/m 2. lB (0.0350 m)(2.20 T) EVALUATE: A current of 1 A in a wire with diameter 1 mm corresponds to a current density of J = 1.36 × 106 A/m 2 , so the current density calculated in part (c) is a typical value for circuits. IDENTIFY: The electric and magnetic fields exert forces on the moving charge. The work done by the electric v2 and this acceleration must correspond to the net field equals the change in kinetic energy. At the top point, a y = R force. SET UP: The electric field is uniform so the work it does for a displacement y in the y-direction is W = Fy = qEy. ! ! At the top point, FB is in the − y -direction and FE is in the +y-direction. EXECUTE: (a) The maximum speed occurs at the top of the cycloidal path, and hence the radius of curvature is greatest there. Once the motion is beyond the top, the particle is being slowed by the electric field. As it returns to y = 0, the speed decreases, leading to a smaller magnetic force, until the particle stops completely. Then the electric field again provides the acceleration in the y-direction of the particle, leading to the repeated motion. 2qEy 1 (b) W = qEy = mv 2 and v = . 2 m

(b) J =

27.91.

mv 2 m 2qEy 2E . =− = −qE. 2qE = qvB and v = R 2y m B EVALUATE: The speed at the top depends on B because B determines the y-displacement and the work done by the electric force depends on the y-displacement.

(c) At the top, Fy = qE − qvB = −


28

SOURCES OF MAGNETIC FIELD

28.1.

! IDENTIFY and SET UP: Use Eq.(28.2) to calculate B at each point. ! ! ! ! ! μ qv × rˆ μ0 qv × r r B= 0 = , since rˆ = . 4π r 2 4π r 3 r ! ! 6 ˆ v = ( 8.00 × 10 m/s ) j and r is the vector from the charge to the point where the field is calculated.

! EXECUTE: (a) r = ( 0.500 m ) iˆ, r = 0.500 m ! ! v × r = vrˆj × iˆ = −vrkˆ

! ( 6.00 ×10−6 C )(8.00 ×106 m/s ) kˆ μ qv B = − 0 2 kˆ = − (1 × 10−7 T ⋅ m/A ) 2 4π r ( 0.500 m ) ! B = − (1.92 × 10−5 T ) kˆ ! (b) r = − ( 0.500 m ) ˆj , r = 0.500 m ! ! ! v × r = −vrˆj × ˆj = 0 and B = 0. ! (c) r = ( 0.500 m ) kˆ , r = 0.500 m ! ! v × r = vrˆj × kˆ = vriˆ

! ( 6.00 ×10−6 C )(8.00 × 106 m/s ) iˆ = + 1.92 ×10−5 T iˆ B = (1 × 10−7 T ⋅ m/A ) ( ) 2 ( 0.500 m ) ! 2 2 (d) r = − ( 0.500 m ) ˆj + ( 0.500 m ) kˆ , r = ( 0.500 m ) + ( 0.500 m ) = 0.7071 m ! ! v × r = v ( 0.500 m ) − ˆj × ˆj + ˆj × kˆ = ( 4.00 × 106 m 2 /s ) iˆ

(

28.2.

! ( 6.00 ×10−6 C )( 4.00 ×106 m / s ) iˆ = + 6.79 ×10−6 T iˆ B = (1×10−7 T ⋅ m/A ) ( ) 3 ( 0.7071 m ) ! ! ! ! EVALUATE: At each point B is perpendicular to both v and r . B = 0 along the direction of v . IDENTIFY: A moving charge creates a magnetic field as well as an electric field. 1 e μ qv sin φ SET UP: The magnetic field caused by a moving charge is B = 0 , and its electric field is E = 4π P0 r 2 4π r 2 since q = e. EXECUTE: Substitute the appropriate numbers into the above equations.

B=

μ0 qv sin φ 4π × 10−7 T ⋅ m/A (1.60 × 10−19 C)(2.2 × 106 m/s)sin 90° = = 13 T, out of the page. 4π r 2 4π (5.3 × 10−11 m) 2

E=

28.3.

)

1 e (9.00 × 109 N ⋅ m 2 /C2 )(1.60 × 10−19 C) = = 5.1 × 1011 N/C, toward the electron. 4π P0 r 2 (5.3 × 10−11 m) 2

EVALUATE: There are enormous fields within the atom! IDENTIFY: A moving charge creates a magnetic field. SET UP: The magnetic field due to a moving charge is B =

μ0 qv sin φ

r2

.

28-1


28-2

28.4.

Chapter 28

EXECUTE: Substituting numbers into the above equation gives μ qv sin φ 4π ×10−7 T ⋅ m/A (1.6 ×10−19 C)(3.0 ×107 m/s)sin 30° (a) B = 0 = . 4π r 2 4π (2.00 ×10−6 m) 2 B = 6.00 × 10–8 T, out of the paper, and it is the same at point B. (b) B = (1.00 × 10–7 T ⋅ m/A)(1.60 × 10–19 C)(3.00 × 107 m/s)/(2.00 × 10–6 m)2 B = 1.20 × 10–7 T, out of the page. (c) B = 0 T since sin(180°) = 0. EVALUATE: Even at high speeds, these charges produce magnetic fields much less than the Earth’s magnetic field. IDENTIFY: Both moving charges produce magnetic fields, and the net field is the vector sum of the two fields. SET UP: Both fields point out of the paper, so their magnitudes add, giving B = Balpha + Bel =

μ0v ( e sin 40° + 2e sin140° ) 4π r 2

EXECUTE: Factoring out an e and putting in the numbers gives B=

4π × 10−7 T ⋅ m/A (1.60 × 10−19 C)(2.50 × 105 m/s) ( sin 40° + 2sin140° ) 4π (1.75 × 10−9 m) 2

B = 2.52 × 10−3 T = 2.52 mT, out of the page. At distances very close to the charges, the magnetic field is strong enough to be important. ! μ qv! × r! IDENTIFY: Apply B = 0 . 4π r 3 ! SET UP: Since the charge is at the origin, r = xiˆ + yˆj + zkˆ. ! ! ! ! ! EXECUTE: (a) v = vi , r = riˆ; v × r = 0, B = 0 . ! ! ! ! (b) v = viˆ, r = rˆj; v × r = vrkˆ , r = 0.500 m. EVALUATE:

28.5.

−7 2 2 −6 5 ⎛ μ ⎞ q v (1.0 × 10 N ⋅ s /C )(4.80 × 10 C)(6.80 × 10 m/s) = 1.31 × 10−6 T. B=⎜ 0 ⎟ 2 = (0.500 m) 2 ⎝ 4π ⎠ r ! q is negative, so B = −(1.31 × 10−6 T)kˆ. ! ! ! ! (c) v = viˆ, r = (0.500 m)(iˆ + ˆj ); v × r = (0.500 m)vkˆ , r = 0.7071 m.

! ! (1.0 × 10−7 N ⋅ s 2 /C 2 )(4.80 × 10−6 C)(0.500 m)(6.80 × 105 m/s) ⎛μ ⎞ B = ⎜ 0 ⎟( q v × r r3 ) = . (0.7071 m)3 ⎝ 4π ⎠ ! ˆ B = 4.62 × 10 −7 T. B = −(4.62 × 10−7 T) k. ! ! ! ! (d) v = viˆ, r = rkˆ; v × r = − vrˆj , r = 0.500 m

28.6.

−7 2 2 −6 5 ⎛ μ ⎞ q v (1.0 × 10 N ⋅ s /C )(4.80 × 10 C)(6.80 × 10 m/s) = 1.31 × 10−6 T. B=⎜ 0 ⎟ 2 = 2 (0.500 m) ⎝ 4π ⎠ r ! B = (1.31 × 10−6 T ) ˆj. ! ! ! EVALUATE: In each case, B is perpendicular to both r and v . ! ! ! μ qv × r IDENTIFY: Apply B = 0 . For the magnetic force, apply the results of Example 28.1, except here the two 4π r 3 charges and velocities are different. ! ! v ×r ! ! v SET UP: In part (a), r = d and r is perpendicular to v in each case, so = 2 . For calculating the force 3 r d between the charges, r = 2d . μ ⎛ qv q′v′ ⎞ EXECUTE: (a) Btotal = B + B′ = 0 ⎜ 2 + 2 ⎟ . 4π ⎝ d d ⎠

μ0 ⎛ (8.0 × 10−6 C)(4.5 × 106 m/s) (3.0 × 10−6 C)(9.0 × 106 m/s) ⎞ −4 + ⎜ ⎟ = 4.38 × 10 T. 4π ⎝ (0.120 m) 2 (0.120 m) 2 ⎠ ! The direction of B is into the page. B=


Sources of Magnetic Field

28-3

(b) Following Example 28.1 we can find the magnetic force between the charges: μ qq′vv′ (8.00 × 10−6 C)(3.00 × 10−6 C)(4.50 × 106 m s )(9.00 × 106 m s ) −7 FB = 0 (10 T m A) = ⋅ 4π r 2 (0.240 m) 2 −3 FB = 1.69 × 10 N. The force on the upper charge points up and the force on the lower charge points down. The

Coulomb force between the charges is FC = k

28.7.

28.8.

q1q2 (8.0 × 10 −6 C)(3.0 × 10 −6 C) = (8.99 × 10 9 N ⋅ m 2 C 2 ) = 3.75 N . 2 r (0.240 m) 2

The force on the upper charge points up and the force on the lower charge points down. The ratio of the Coulomb F c2 3.75 N force to the magnetic force is C = = = 2.22 × 103 ; the Coulomb force is much larger. FB v1v2 1.69 × 10 −3 N (b) The magnetic forces are reversed in direction when the direction of only one velocity is reversed but the magnitude of the force is unchanged. EVALUATE: When two charges have the same sign and move in opposite directions, the force between them is repulsive. When two charges of the same sign move in the same direction, the force between them is attractive. ! ! μ qv! × r! ! ! IDENTIFY: Apply B = 0 . For the magnetic force on q′ , use FB = q′v × Bq and for the magnetic force on 3 4π r ! ! ! q use FB = qv × Bq ′ . ! ! v ×r v SET UP: In part (a), r = d and = 2. r3 d μ qv μ qv′ EXECUTE: (a) q′ = − q; Bq = 0 2 , into the page; Bq ′ = 0 2 , out of the page. 4πd 4πd v μ qv μ0 qv (i) v′ = gives B = 0 2 (1 − 12 ) = , into the page. (ii) v′ = v gives B = 0. 2 4πd 4π (2d 2 ) μ qv (iii) v′ = 2v gives B = 0 2 , out of the page. 4π d ! ! ! μ q 2v′v ! (b) The force that q exerts on q′ is given by F = q′v′ × Bq , so F = 0 . Bq is into the page, so the force on 4π (2d ) 2 q′ is toward q. The force that q′ exerts on q is toward q′. The force between the two charges is attractive. F μ q 2vv′ q2 (c) FB = 0 so B = μ0P0vv′ = μ0P0 (3.00 × 105 m/s)2 = 1.00 × 10−6 . , FC = 2 FC 4π (2d ) 4πP0 (2d ) 2 EVALUATE: When charges of opposite sign move in opposite directions, the force between them is attractive. For the values specified in part (c), the magnetic force between the two charges is much smaller in magnitude than the Coulomb force between them. IDENTIFY: Both moving charges create magnetic fields, and the net field is the vector sum of the two. The magnetic force on a moving charge is Fmag = qvB sin φ and the electrical force obeys Coulomb’s law.

μ0 qv sin φ . 4π r 2 EXECUTE: (a) Both fields are into the page, so their magnitudes add, giving SET UP: The magnetic field due to a moving charge is B =

B = Be + Bp = B=

μ0 ⎛ ev ev ⎞ ⎜ + ⎟ sin 90° 4π ⎜⎝ re2 rp2 ⎟⎠

⎡ ⎤ μ0 1 1 1.60 ×10−19 C ) ( 845,000 m/s ) ⎢ + ( −9 −9 2 2⎥ 4π ⎣ (5.00 ×10 m) (4.00 ×10 m) ⎦

B = 1.39 × 10–3 T = 1.39 mT, into the page.

μ0 qv sin φ , where r = 41 nm and φ = 180° − arctan(5/4) = 128.7°, we get 4π r 2 4π ×10−7 T ⋅ m/A (1.6 ×10−19 C)(845,000 m/s)sin128.7° B= = 2.58 ×10−4 T, into the page. 4π ( 41 ×10−9 m) 2

(b) Using B =

(c) Fmag = qvB sin90° = (1.60 × 10−19 C)(845,000 m/s)(2.58 × 10−4 T) = 3.48 × 10−17 N, in the +x direction.

Felec = (1/ 4π P0 )e2 /r 2 = clockwise. EVALUATE:

(9.00 ×109 N ⋅ m 2 /C2 )(1.60 ×10−19 C) 2 = 5.62 ×10−12 N, at 51.3° below the +x-axis measured ( 41 ×10−9 m) 2

The electric force is much stronger than the magnetic force.


28-4

28.9.

28.10.

Chapter 28

A current segment creates a magnetic field. μ Idl sin φ SET UP: The law of Biot and Savart gives dB = 0 . r2 4π EXECUTE: Applying the law of Biot and Savart gives 4π × 10−7 T ⋅ m/A (10.0 A)(0.00110 m) sin90° (a) dB = = 4.40 × 10–7 T, out of the paper. 4π (0.0500 m) 2

IDENTIFY:

(b) The same as above, except r = (5.00 cm) 2 + (14.0 cm) 2 and φ = arctan(5/14) = 19.65°, giving dB = 1.67 × 10–8 T, out of the page. (c) dB = 0 since φ = 0°. EVALUATE: This is a very small field, but it comes from a very small segment of current. ! ! ! μ Idl × rˆ μ0 Idl × r! IDENTIFY: Apply dB = 0 . = 4π r 2 4π r 3 μ Idl sin φ SET UP: The magnitude of the field due to the current element is dB = 0 , where φ is the angle between 4π r 2 ! r and the current direction. EXECUTE: The magnetic field at the given points is: μ Idl sin φ μ0 (200 A)(0.000100 m) dBa = 0 = = 2.00 × 10−6 T. 4π r 2 4π (0.100 m) 2

dBb =

μ0 Idl sin φ μ0 (200 A)(0.000100 m)sin 45° = = 0.705 × 10−6 T. 4π r 2 4π 2(0.100 m) 2

dBc =

μ0 Idl sin φ μ0 (200 A)(0.000100 m) = = 2.00 × 10−6 T. 4π r 2 4π (0.100 m) 2

dBd =

μ0 Idl sin φ μ0 Idl sin(0°) = = 0. 4π r 2 4π r2

μ0 Idl sin φ μ0 (200 A)(0.00100 m) 2 = = 0.545 × 10−6 T 4π r 2 4π 3(0.100 m) 2 3 The field vectors at each point are shown in Figure 28.10. ! EVALUATE: In each case dB is perpendicular to the current direction. dBe =

Figure 28.10 28.11.

IDENTIFY and SET UP: The magnetic field produced by an infinitesimal current element is given by Eq.(28.6). ! ! μ0 Il × rˆ dB = As in Example 28.2 use this equation for the finite 0.500-mm segment of wire since the 4π r 2 Δl = 0.500 mm length is much smaller than the distances to the field points. ! ! ! μ I Δl × rˆ μ0 I Δl × r! B= 0 = 4π r 2 4π r 3 ! I is in the + z -direction, so Δl = ( 0.500 × 10−3 m ) kˆ ! EXECUTE: (a) Field point is at x = 2.00 m, y = 0, z = 0 so the vector r from the source point (at the origin) to the ! field point is r = ( 2.00 m ) iˆ. ! ! Δl × r = ( 0.500 × 10−3 m ) ( 2.00 m ) kˆ × iˆ = + (1.00 × 10−3 m 2 ) ˆj ! (1× 10−7 T ⋅ m/A ) ( 4.00 A ) (1.00 ×10−3 m 2 ) ˆj = ( 5.00 ×10−11 T ) ˆj B= 3 ( 2.00 m )


Sources of Magnetic Field

28-5

! (b) r = ( 2.00 m ) ˆj , r = 2.00 m. ! ! Δl × r = ( 0.500 × 10−3 m ) ( 2.00 m ) kˆ × ˆj = − (1.00 × 10−3 m 2 ) iˆ ! (1×10−7 T ⋅ m/A ) ( 4.00 A ) (1.00 ×10−3 m2 ) iˆ = − 5.00 ×10−11 T iˆ B=− ( ) 3 ( 2.00 m ) ! (c) r = ( 2.00 m ) iˆ + ˆj , r = 2 ( 2.00 m ) . ! ! Δl × r = ( 0.500 × 10−3 m ) ( 2.00 m ) kˆ × iˆ + ˆj = (1.00 × 10−3 m 2 ) ˆj – iˆ

(

)

(

! (1× 10 B=

28.12.

−7

)

T ⋅ m/A ) ( 4.00 A ) (1.00 × 10 m −3

⎡ 2 ( 2.00 m ) ⎤ ⎣ ⎦

3

)

( ˆj – iˆ ) = ( −1.77 ×10

)

−11

(

T ) iˆ – ˆj

)

! (d) r = (2.00 m)kˆ , r = 2.00 m. ! ! ! Δl × r = (0.500 × 10−3 m)(2.00 m)kˆ × kˆ = 0; B = 0. ! ! ! EVALUATE: At each point B is perpendicular to both r and Δl . B = 0 along the length of the wire. IDENTIFY: A current segment creates a magnetic field. μ Idl sin φ SET UP: The law of Biot and Savart gives dB = 0 . r2 4π Both fields are into the page, so their magnitudes add. EXECUTE: Applying the law of Biot and Savart for the 12.0-A current gives

4π × 10−7 T ⋅ m/A dB = 4π

28.13.

(

2

⎛ 2.50 cm ⎞ (12.0 A)(0.00150 m) ⎜ ⎟ ⎝ 8.00 cm ⎠ = 8.79 × 10–8 T 2 (0.0800 m)

The field from the 24.0-A segment is twice this value, so the total field is 2.64 × 10–7 T, into the page. EVALUATE: The rest of each wire also produces field at P. We have calculated just the field from the two segments that are indicated in the problem. IDENTIFY: A current segment creates a magnetic field. μ Idl sin φ SET UP: The law of Biot and Savart gives dB = 0 . Both fields are into the page, so their magnitudes add. 4π r2

EXECUTE: Applying the Biot and Savart law, where r = 12 (3.00 cm) 2 + (3.00 cm) 2 = 2.121 cm, we have dB = 2

28.14.

28.15.

4π × 10−7 T ⋅ m/A (28.0 A)(0.00200 m)sin 45.0° = 1.76 × 10–5 T, into the paper. 4π (0.02121 m) 2

EVALUATE: Even though the two wire segments are at right angles, the magnetic fields they create are in the same direction. IDENTIFY: A current segment creates a magnetic field. μ Idl sin φ SET UP: The law of Biot and Savart gives dB = 0 . All four fields are of equal magnitude and into the 4π r2 page, so their magnitudes add. 4π × 10−7 T ⋅ m/A (15.0 A)(0.00120 m) sin90° EXECUTE: dB = 4 = 2.88 × 10–6 T, into the page. 4π (0.0500 m) 2 EVALUATE: A small current element causes a small magnetic field. IDENTIFY: We can model the lightning bolt and the household current as very long current-carrying wires. μI SET UP: The magnetic field produced by a long wire is B = 0 . 2π r EXECUTE: Substituting the numerical values gives (4π × 10−7 T ⋅ m/A)(20,000 A) = 8 × 10–4 T (a) B = 2π(5.0 m) (4π × 10−7 T ⋅ m/A)(10 A) = 4.0 × 10–5 T. 2π(0.050 m) EVALUATE: The field from the lightning bolt is about 20 times as strong as the field from the household current.

(b) B =


28-6

Chapter 28

28.16.

IDENTIFY: The long current-carrying wire produces a magnetic field. μI SET UP: The magnetic field due to a long wire is B = 0 . 2π r EXECUTE: First find the current: I = (3.50 × 1018 el/s)(1.60 × 10–19 C/el) = 0.560 A (4π × 10−7 T ⋅ m/A)(0.560 A) = 2.80 × 10–6 T Now find the magnetic field: 2π(0.0400 m) Since electrons are negative, the conventional current runs from east to west, so the magnetic field above the wire points toward the north. EVALUATE: This magnetic field is much less than that of the Earth, so any experiments involving such a current would have to be shielded from the Earth’s magnetic field, or at least would have to take it into consideration. IDENTIFY: The long current-carrying wire produces a magnetic field. μI SET UP: The magnetic field due to a long wire is B = 0 . 2π r EXECUTE: First solve for the current, then substitute the numbers using the above equation. (a) Solving for the current gives

28.17.

I = 2π rB/μ0 = 2π (0.0200 m)(1.00 × 10−4 T)/(4π × 10−7 T ⋅ m/A) = 10.0 A

28.18.

(b) The earth’s horizontal field points northward, so at all points directly above the wire the field of the wire would point northward. (c) At all points directly east of the wire, its field would point northward. EVALUATE: Even though the Earth’s magnetic field is rather weak, it requires a fairly large current to cancel this field. ! μI IDENTIFY: For each wire B = 0 (Eq.28.9), and the direction of B is given by the right-hand rule (Fig. 28.6 in 2π r the textbook). Add the field vectors for each wire to calculate the total field. (a) SET UP: The two fields at this point have the directions shown in Figure 28.18a. EXECUTE: At point P midway between ! ! the two wires the fields B1 and B2 due to the two currents are in opposite directions, so B = B2 − B1. Figure 28.18a But B1 = B2 =

(b) SET UP:

μ0 I , so B = 0. 2π a The two fields at this point have the directions shown in Figure 28.18b.

EXECUTE: At point Q above the upper ! ! wire B1 and B2 are both directed out of the page (+ z -direction), so B = B1 + B2 .

Figure 28.18b B1 =

μ0 I μ0 I , B2 = 2π a 2π (3a )

B=

μ0 I 2μ I ! 2μ I 1 + 13 ) = 0 ; B = 0 kˆ ( 2π a 3π a 3π a


Sources of Magnetic Field

28-7

(c) SET UP: The two fields at this point have the directions shown in Figure 28.18c.

EXECUTE: At point R below the lower ! ! wire B1 and B2 are both directed into the page (− z -direction), so B = B1 + B2 .

Figure 28.18c

μ0 I μI , B2 = 0 2π (3a ) 2π a μ0 I 2 μI ! 2μ I B1 = 1 + 13 ) = 0 ; B = − 0 kˆ ( 2π a 3π a 3π a B1 =

28.19.

! EVALUATE: In the figures we have drawn, B due to each wire is out of the page at points above the wire and into the page at points below the wire. If the two field vectors are in opposite directions the magnitudes subtract. IDENTIFY: The total magnetic field is the vector sum of the constant magnetic field and the wire’s magnetic field. μI SET UP: For the wire, Bwire = 0 and the direction of Bwire is given by the right-hand rule that is illustrated in 2π r ! Figure 28.6 in the textbook. B = (1.50 × 10−6 T) iˆ. 0

! ! μI μ (8.00 A) ˆ EXECUTE: (a) At (0, 0, 1 m), B = B0 − 0 iˆ = (1.50 × 10−6 T)iˆ − 0 i = −(1.0 × 10−7 T)iˆ. 2π r 2π (1.00 m) ! ! μI μ (8.00 A) ˆ (b) At (1 m, 0, 0), B = B0 + 0 kˆ = (1.50 × 10−6 T) iˆ + 0 k. 2πr 2π (1.00 m) ! B = (1.50 × 10−6 T)iˆ + (1.6 × 10−6 T)kˆ = 2.19 × 10−6 T, at θ = 46.8° from x to z.

28.20.

! ! μI μ (8.00 A) ˆ (c) At (0, 0, –0.25 m), B = B0 + 0 iˆ = (1.50 × 10−6 T)iˆ + 0 i = (7.9 × 10−6 T)iˆ. 2πr 2π (0.25 m) EVALUATE: At point c the two fields are in the same direction and their magnitudes add. At point a they are in opposite directions and their magnitudes subtract. At point b the two fields are perpendicular. ! IDENTIFY and SET UP: The magnitude of B is given by Eq.(28.9) and the direction is given by the right-hand rule. (a) EXECUTE: Viewed from above, the current is in the direction shown in Figure 28.20. Directly below the wire the direction of the magnetic field due to the current in the wire is east.

Figure 28.20 B=

μ0 I ⎛ 800 A ⎞ −5 = (2 ×10−7 T ⋅ m/A) ⎜ ⎟ = 2.91×10 T 2π r ⎝ 5.50 m ⎠

(b) EVALUATE: problem. 28.21.

B from the current is nearly equal in magnitude to the earth's field, so, yes, the current really is a

! μ0 I . The direction of B is given by the right-hand rule in Section 20.7. 2π r SET UP: Call the wires a and b, as indicated in Figure 28.21. The magnetic fields of each wire at points P1 and P2 are shown in Figure 28.21a. The fields at point 3 are shown in Figure 28.21b. EXECUTE: (a) At P1 , Ba = Bb and the two fields are in opposite directions, so the net field is zero. ! μI ! μI (b) Ba = 0 . Bb = 0 . Ba and Bb are in the same direction so 2π rb 2π ra IDENTIFY:

B=

μ0 I ⎛ 1 1 ⎞ (4π × 10−7 T ⋅ m/A)(4.00 A) ⎡ 1 1 ⎤ −6 ⎜ + ⎟= ⎢ 0.300 m + 0.200 m ⎥ = 6.67 × 10 T 2π ⎝ ra rb ⎠ 2π ⎣ ⎦ ! B has magnitude 6.67 μ T and is directed toward the top of the page. B = Ba + Bb =


28-8

Chapter 28

! ! ! ! 5 cm (c) In Figure 28.21b, Ba is perpendicular to ra and Bb is perpendicular to rb . tan θ = and θ = 14.04° . 20 cm ra = rb = (0.200 m) 2 + (0.050 m) 2 = 0.206 m and Ba = Bb . ⎛ μI ⎞ 2(4π ×10−7 T ⋅ m/A)(4.0 A)cos14.04° = 7.54 μ T B = Ba cosθ + Bb cosθ = 2 Ba cosθ = 2 ⎜ 0 ⎟ cosθ = 2π (0.206 m) ⎝ 2π ra ⎠ B has magnitude 7.53 μ T and is directed to the left. EVALUATE: At points directly to the left of both wires the net field is directed toward the bottom of the page.

Figure 28.21 28.22.

IDENTIFY: Use Eq.(28.9) and the right-hand rule to determine points where the fields of the two wires cancel. (a) SET UP: The only place where the magnetic fields of the two wires are in opposite directions is between the wires, in the plane of the wires. Consider a point a distance x from the wire carrying I 2 = 75.0 A. Btot will be zero

where B1 = B2 . EXECUTE:

μ0 I1

2π (0.400 m − x)

=

μ0 I 2 2π x

I 2 (0.400 m − x) = I1 x; I1 = 25.0 A, I 2 = 75.0 A x = 0.300 m; Btot = 0 along a line 0.300 m from the wire carrying 75.0 A and 0.100 m from the wire carrying current 25.0 A. (b) SET UP: Let the wire with I1 = 25.0 A be 0.400 m above the wire with I 2 = 75.0 A. The magnetic fields of the two wires are in opposite directions in the plane of the wires and at points above both wires or below both wires. But to have B1 = B2 must be closer to wire #1 since I1 < I 2 , so can have Btot = 0 only at points above both wires. Consider a point a distance x from the wire carrying I1 = 25.0 A. Btot will be zero where B1 = B2 . EXECUTE:

μ0 I1 μ0 I 2 = 2π x 2π (0.400 m + x)

I 2 x = I1 (0.400 m + x); x = 0.200 m Btot = 0 along a line 0.200 m from the wire carrying current 25.0 A and 0.600 m from the wire carrying current

28.23.

I 2 = 75.0 A. EVALUATE: For parts (a) and (b) the locations of zero field are in different regions. In each case the points of zero field are closer to the wire that has the smaller current. IDENTIFY: The net magnetic field at the center of the square is the vector sum of the fields due to each wire. ! μI SET UP: For each wire, B = 0 and the direction of B is given by the right-hand rule that is illustrated in 2π r Figure 28.6 in the textbook. EXECUTE: (a) and (b) B = 0 since the magnetic fields due to currents at opposite corners of the square cancel. (c) The fields due to each wire are sketched in Figure 28.23. ⎛μI⎞ B = Ba cos 45° + Bb cos 45° + Bc cos 45° + Bd cos 45° = 4 Ba cos 45° = 4 ⎜ 0 ⎟ cos 45° . ⎝ 2πr ⎠ r = (10 cm) 2 + (10 cm) 2 = 10 2 cm = 0.10 2 m , so B=4

(4π × 10−7 T ⋅ m A ) (100 A) cos 45° = 4.0 × 10−4 T, to the left. 2π (0.10 2 m)


Sources of Magnetic Field

28-9

EVALUATE: In part (c), if all four currents are reversed in direction, the net field at the center of the square would be to the right.

Figure 28.23 28.24.

IDENTIFY: Use Eq.(28.9) and the right-hand rule to determine the field due to each wire. Set the sum of the four fields equal to zero and use that equation to solve for the field and the current of the fourth wire. SET UP: The three known currents are shown in Figure 28.24.

! ! ! B1 ⊗, B2 ⊗, B3 " B=

μ0 I ; r = 0.200 m for each wire 2π r

Figure 28.24 EXECUTE: Let " be the positive z-direction. I1 = 10.0 A, I 2 = 8.0 A, I 3 = 20.0 A. Then B1 = 1.00 ×10−5 T,

B2 = 0.80 × 10−5 T, and B3 = 2.00 × 10−5 T. B1z = −1.00 × 10−5 T, B2z = −0.80 × 10−5 T, B3z = +2.00 × 10−5 T B1z + B2 z + B3 z + B4 z = 0

B4 z = −( B1z + B2 z + B3 z ) = −2.0 × 10−6 T ! To give B4 in the ⊗ direction the current in wire 4 must be toward the bottom of the page. B4 =

28.25.

28.26.

rB4 μ0 I (0.200 m)(2.0 × 10−6 T) so I 4 = = = 2.0 A 2π r ( μ0 / 2π ) (2 × 10−7 T ⋅ m/A)

EVALUATE: The fields of wires #2 and #3 are in opposite directions and their net field is the same as due to a current 20.0 A – 8.0 A = 12.0 A in one wire. The field of wire #4 must be in the same direction as that of wire #1, and 10.0 A + I 4 = 12.0 A. IDENTIFY: Apply Eq.(28.11). SET UP: Two parallel conductors carrying current in the same direction attract each other. Parallel conductors carrying currents in opposite directions repel each other. μ I I L μ (5.00 A)(2.00 A)(1.20 m) EXECUTE: (a) F = 0 1 2 = 0 = 6.00 × 10−6 N, and the force is repulsive since the 2π r 2π (0.400 m) currents are in opposite directions. (b) Doubling the currents makes the force increase by a factor of four to F = 2.40 × 10−5 N. EVALUATE: Doubling the current in a wire doubles the magnetic field of that wire. For fixed magnetic field, doubling the current in a wire doubles the force that the magnetic field exerts on the wire. IDENTIFY: Apply Eq.(28.11). SET UP: Two parallel conductors carrying current in the same direction attract each other. Parallel conductors carrying currents in opposite directions repel each other. F 2π r 2π (0.0250 m) F μ0 I1I 2 EXECUTE: (a) = gives I 2 = = (4.0 × 10−5 N m) = 8.33 A. L 2π r L μ0 I1 μ0 (0.60 A) (b) The two wires repel so the currents are in opposite directions. EVALUATE: The force between the two wires is proportional to the product of the currents in the wires.


28-10

Chapter 28

28.27.

IDENTIFY: The lamp cord wires are two parallel current-carrying wires, so they must exert a magnetic force on each other. SET UP: First find the current in the cord. Since it is connected to a light bulb, the power consumed by the bulb is μ I ′I P = IV. Then find the force per unit length using F/L = 0 . 2π r EXECUTE: For the light bulb, 100 W = I(120 V) gives I = 0.833 A. The force per unit length is

F/L =

28.28.

4π × 10−7 T ⋅ m/A (0.833 A) 2 = 4.6 × 10−5 N/m 2π 0.003 m

Since the currents are in opposite directions, the force is repulsive. EVALUATE: This force is too small to have an appreciable effect for an ordinary cord. IDENTIFY: Apply Eq.(28.11) for the force from each wire. SET UP: Two parallel conductors carrying current in the same direction attract each other. Parallel conductors carrying currents in opposite directions repel each other. F μ0 I 2 ⎛ 1 1 ⎞ μ0 I 2 EXECUTE: On the top wire = − , upward. On the middle wire, the magnetic forces cancel ⎜ ⎟= L 2π ⎝ d 2d ⎠ 4π d F μ0 I 2 ⎛ 1 1 ⎞ μ0 I 2 = − + , downward. ⎜ ⎟= L 2π ⎝ d 2d ⎠ 4π d EVALUATE: The net force on the middle wire is zero because at the location of the middle wire the net magnetic field due to the other two wires is zero. IDENTIFY: The wire CD rises until the upward force FI due to the currents balances the downward force of gravity. SET UP: The forces on wire CD are shown in Figure 28.29.

so the net force is zero. On the bottom wire

28.29.

Currents in opposite directions so the force is repulsive and FI is upward, as shown.

Figure 28.29

Eq.(28.11) says FI = EXECUTE:

μ0 I 2 L where L is the length of wire CD and h is the distance between the wires. 2π h

mg = λ Lg

μ0 I 2 L μ I2 = λ Lg and h = 0 . 2π h 2π g λ EVALUATE: The larger I is or the smaller λ is, the larger h will be. Thus FI − mg = 0 says

28.30.

IDENTIFY: The magnetic field at the center of a circular loop is B =

μ0 I 2R

. By symmetry each segment of the loop

that has length Δl contributes equally to the field, so the field at the center of a semicircle is 12 that of a full loop. μI SET UP: Since the straight sections produce no field at P, the field at P is B = 0 . EXECUTE:

B=

μ0 I 4R

4R

! ! . The direction of B is given by the right-hand rule: B is directed into the page.

EVALUATE: For a quarter-circle section of wire the magnetic field at its center of curvature is B = 28.31.

μ0 I

. 8R IDENTIFY: Calculate the magnetic field vector produced by each wire and add these fields to get the total field. SET UP: First consider the field at P produced by the current I1 in the upper semicircle of wire. See Figure 28.31a. Consider the three parts of this wire a: long straight section, b: semicircle c: long, straight section Figure 28.31a

! ! ! μ0 Idl × rˆ μ0 Idl × r! = Apply the Biot-Savart law dB = to each piece. 4π r 2 4π r 3


Sources of Magnetic Field

EXECUTE: part a See Figure 28.31b.

28-11

! ! dl × r = 0, so dB = 0

Figure 28.31b

The same is true for all the infinitesimal segments that make up this piece of the wire, so B = 0 for this piece. part c See Figure 28.31c.

! ! dl × r = 0, so dB = 0 and B = 0 for this piece. Figure 28.31c

part b See Figure 28.31d.

! ! dl × r is directed into the paper for all infinitesimal! segments that make up this semicircular piece, so B is directed into the paper and B = ∫ dB (the vector sum ! of the dB is obtained by adding their magnitudes since they are in the same direction).

Figure 28.31d ! ! ! ! ! ! dl × r = r dl sinθ . The angle θ between dl and r is 90° and r = R, the radius of the semicircle. Thus dl × r = Rdl ! ! μ0 I dl × r μ0 I1 R ⎛ μI ⎞ = dB = dl = ⎜ 0 12 ⎟ dl 3 3 4π 4π R r ⎝ 4π R ⎠ I I μ μ μI ⎛ ⎞ ⎛ ⎞ B = ∫ dB = ⎜ 0 12 ⎟ ∫ dl = ⎜ 0 12 ⎟ (π R ) = 0 1 4R ⎝ 4π R ⎠ ⎝ 4π R ⎠

∫ dl

is equal to π R , the length of wire in the semicircle.) We have shown that the two straight ! sections make zero contribution to B, so B1 = μ0 I1 / 4 R and is directed into the page.

(We used that

For current in the direction shown in Figure 28.31e, a similar analysis gives B2 = μ0 I 2 / 4 R, out of the paper Figure 28.31e

28.32.

! ! μ I −I B1 and B2 are in opposite directions, so the magnitude of the net field at P is B = B1 − B2 = 0 1 2 . 4R EVALUATE: When I1 = I 2 , B = 0. IDENTIFY: Apply Eq.(28.16). SET UP: At the center of the coil, x = 0. a is the radius of the coil, 0.0240 m. 2aBx 2(0.024 m) (0.0580 T) = = 2.77 A EXECUTE: (a) Bx = μ0 NI 2a, so I = μ0 N (4π × 10−7 T ⋅ m A) (800) (b) At the center, Bc = μ0 NI 2a . At a distance x from the center,

⎞ ⎛ ⎞ a3 a3 a3 ⎛ μ NI ⎞ ⎛ = ⎜ 0 ⎟⎜ 2 = Bc ⎜ 2 . Bx = 12 Bc says 2 = 12 , and ( x 2 + a 2 )3 = 4a 6 . 2 3/ 2 ⎟ 2 3/ 2 ⎟ 2 32 x a + 2( x + a ) ( ) ( x + a ) ⎝ 2a ⎠ ⎝ ( x + a ) ⎠ ⎝ ⎠ Since a = 0.024 m, x = 0.0184 m . EVALUATE: As shown in Figure 28.41 in the textbook, the field has its largest magnitude at the center of the coil and decreases with distance along the axis from the center. IDENTIFY: Apply Eq.(28.16). SET UP: At the center of the coil, x = 0 . a is the radius of the coil, 0.020 m. μ NI μ (600) (0.500 A) EXECUTE: (a) Bcenter = 0 = 0 = 9.42 × 10−3 T. 2a 2(0.020 m) Bx =

28.33.

μ0 NIa 2 2

2 32

μ0 NIa 2

μ0 (600)(0.500 A)(0.020 m)2

= 1.34 × 10−4 T. 2( x + a ) 2((0.080 m) 2 + (0.020 m) 2 )3/ 2 EVALUATE: As shown in Figure 28.41 in the textbook, the field has its largest magnitude at the center of the coil and decreases with distance along the axis from the center. (b) B ( x) =

2

2 3/ 2

. B (0.08 m) =


28-12

Chapter 28

28.34.

IDENTIFY and SET UP: The magnetic field at a point on the axis of N circular loops is given by μ0 NIa 2 . Solve for N and set x = 0.0600 m. Bx = 2( x 2 + a 2 )3 / 2 EXECUTE: EVALUATE:

28.35.

N=

2 Bx ( x 2 + a 2 )3 / 2 2(6.39 × 10−4 T)[(0.0600 m) 2 + (0.0600 m)2 ]3 / 2 = = 69. (4π × 10−7 T ⋅ m/A)(2.50 A)(0.0600 m)2 μ0 Ia 2

At the center of the coil the field is Bx =

μ0 NI 2a

= 1.8 × 10−3 T. The field 6.00 cm from the center is a

factor of 1/ 23 / 2 times smaller. IDENTIFY: Apply Ampere’s law. SET UP: μ0 = 4π × 10−7 T ⋅ m/A ! ! EXECUTE: (a) úB ⋅ dl = μ0 I encl = 3.83 × 10−4 T ⋅ m and I encl = 305 A.

! (b) −3.83 × 10−4 T ⋅ m since at each point on the curve the direction of dl is reversed. ! ! EVALUATE: The line integral úB ⋅ dl around a closed path is proportional to the net current that is enclosed by

28.36.

28.37.

28.38.

the path. IDENTIFY: Apply Ampere’s law. SET UP: From the right-hand rule, when going around the path in a counterclockwise direction currents out of the page are positive and currents into the page are negative. ! ! EXECUTE: Path a: I encl = 0 ⇒ úB ⋅ dl = 0. ! ! Path b: I encl = − I1 = −4.0 A ⇒ úB ⋅ dl = − μ0 (4.0 A) = −5.03 × 10−6 T ⋅ m. ! ! Path c: I encl = − I1 + I 2 = −4.0 A + 6.0 A = 2.0 A ⇒ úB ⋅ dl = μ0 (2.0 A) = 2.51 × 10−6 T ⋅ m ! ! Path d: I encl = − I1 + I 2 + I 3 = 4.0 A ⇒ úB ⋅ dl = + μ0 (4.0 A) = 5.03 × 10−6 T ⋅ m. EVALUATE: If we instead went around each path in the clockwise direction, the sign of the line integral would be reversed. IDENTIFY: Apply Ampere’s law. SET UP: To calculate the magnetic field at a distance r from the center of the cable, apply Ampere’s law to a ! ! circular path of radius r. By symmetry, úB ⋅ dl = B (2π r ) for such a path. ! ! μI EXECUTE: (a) For a < r < b, I encl = I ⇒ úB ⋅ dl = μ0 I ⇒ B 2πr = μ0 I ⇒ B = 0 . 2πr (b) For r > c, the enclosed current is zero, so the magnetic field is also zero. EVALUATE: A useful property of coaxial cables for many applications is that the current carried by the cable doesn’t produce a magnetic field outside the cable. ! IDENTIFY: Apply Ampere's law to calculate B. (a) SET UP: For a < r < b the end view is shown in Figure 28.38a. Apply Ampere's law to a circle of radius r, where a < r < b. Take currents I1 and I 2 to be directed into the page. Take this direction to be positive, so go around the integration path in the clockwise direction.

Figure 28.38a ! ! úB ⋅ dl = μ0 I encl

EXECUTE: ! ! úB ⋅ dl = B(2π r ), I encl = I1

Thus B (2π r ) = μ0 I1 and B =

μ0 I1 2π r


Sources of Magnetic Field

28-13

(b) SET UP: r > c: See Figure 28.38b. Apply Ampere's law to a circle of radius r, where r > c. Both currents are in the positive direction.

Figure 28.38b ! ! úB ⋅ dl = μ0 I encl

EXECUTE: ! ! úB ⋅ dl = B(2π r ), I encl = I1 + I 2

Thus B (2π r ) = μ0 ( I1 + I 2 ) and B =

28.39.

μ0 ( I1 + I 2 ) 2π r

EVALUATE: For a < r < b the field is due only to the current in the central conductor. For r > c both currents contribute to the total field. IDENTIFY: The largest value of the field occurs at the surface of the cylinder. Inside the cylinder, the field increases linearly from zero at the center, and outside the field decreases inversely with distance from the central axis of the cylinder. µI µI r SET UP: At the surface of the cylinder, B = 0 , inside the cylinder, Eq. 28.21 gives B = 0 2 , and outside 2π R 2π R µ0 I the field is B = . 2π r EXECUTE: For points inside the cylinder, the field is half its maximum value when

µ0 I r 1⎛ µ I ⎞ = ⎜ 0 ⎟ , which 2π R 2 2 ⎝ 2π R ⎠

µ0 I 1 ⎛ µ0 I ⎞ = ⎜ ⎟ , which gives r = 2R. 2π r 2 ⎝ 2π R ⎠ EVALUATE: The field has half its maximum value at all points on cylinders coaxial with the wire but of radius R/2 and of radius 2R. μ NI IDENTIFY: B = μ0 nI = 0 L SET UP: L = 0.150 m μ (600) (8.00 A) = 0.0402 T EXECUTE: B = 0 (0.150 m) EVALUATE: The field near the center of the solenoid is independent of the radius of the solenoid, as long as the radius is much less than the length. (a) IDENTIFY and SET UP: The magnetic field near the center of a long solenoid is given by Eq.(28.23), B = μ0 nI . gives r = R/2. Outside the cylinder, we have

28.40.

28.41.

EXECUTE: Turns per unit length n =

B

μ0 I

=

0.0270 T = 1790 turns/m (4π × 10−7 T ⋅ m/A)(12.0 A)

(b) N = nL = (1790 turns/m)(0.400 m) = 716 turns Each turn of radius R has a length 2π R of wire. The total length of wire required is N (2π R ) = (716)(2π )(1.40 × 10−2 m) = 63.0 m.

EVALUATE: resistance. 28.42.

A large length of wire is required. Due to the length of wire the solenoid will have appreciable

IDENTIFY and SET UP: At the center of a long solenoid B = μ0 nI = μ0

N I. L

BL (0.150 T)(1.40 m) = = 41.8 A μ0 N (4π × 10−7 T ⋅ m/A)(4000) EVALUATE: The magnetic field inside the solenoid is independent of the radius of the solenoid, if the radius is much less than the length, as is the case here.

EXECUTE:

I=


28-14

Chapter 28

28.43.

IDENTIFY and SET UP: configuration. EXECUTE: (a) B = (b) B =

Use the appropriate expression for the magnetic field produced by each current

2π B 2π (2.00 × 10−2 m)(37.2 T) μ0 I = = 3.72 × 106 A . so I = 4π × 10−7 T ⋅ m/A μ0 2π r

2 RB 2(0.210 m)(37.2 T) N μ0 I = = 1.24 × 105 A . so I = N μ0 (100)(4π × 10−7 T ⋅ m/A) 2R

(37.2 T)(0.320 m) BL N I so I = = = 237 A . L μ0 N (4π × 10−7 T ⋅ m/A)(40,000) EVALUATE: Much less current is needed for the solenoid, because of its large number of turns per unit length. IDENTIFY: Example 28.10 shows that outside a toroidal solenoid there is no magnetic field and inside it the μ NI magnetic field is given by B = 0 . 2πr SET UP: The torus extends from r1 = 15.0 cm to r2 = 18.0 cm. EXECUTE: (a) r = 0.12 m, which is outside the torus, so B = 0. μ NI μ (250)(8.50 A) (b) r = 0.16 m, so B = 0 = 0 = 2.66 × 10−3 T. 2π r 2π (0.160 m) (c) r = 0.20 m, which is outside the torus, so B = 0. EVALUATE: The magnetic field inside the torus is proportional to 1/ r , so it varies somewhat over the crosssection of the torus. μ NI IDENTIFY: Example 28.10 shows that inside a toroidal solenoid, B = 0 . 2π r SET UP: r = 0.070 m μ NI μ (600)(0.650 A) EXECUTE: B = 0 = 0 = 1.11× 10−3 T. 2π r 2π (0.070 m) EVALUATE: If the radial thickness of the torus is small compared to its mean diameter, B is approximately uniform inside its windings. IDENTIFY: Use Eq.(28.24), with μ0 replaced by μ = K m μ0 , with K m = 80. SET UP: The contribution from atomic currents is the difference between B calculated with μ and B calculated with μ0 .

(c) B = μ0

28.44.

28.45.

28.46.

μ NI K m μ0 NI μ0 (80)(400)(0.25 A) = = = 0.0267 T. 2π r 2π r 2π (0.060 m) (b) The amount due to atomic currents is B′ = 79 B = 79 (0.0267 T) = 0.0263 T. 80 80 EVALUATE: The presence of the core greatly enhances the magnetic field produced by the solenoid. K μ NI IDENTIFY and SET UP: B = m 0 (Eq.28.24, with μ0 replaced by K m μ0 ) 2π r EXECUTE: (a) K m = 1400 EXECUTE: (a) B =

28.47.

I=

2π rB (2.90 × 10−2 m)(0.350 T) = = 0.0725 A μ0 K m N (2 × 10−7 T ⋅ m/A)(1400)(500)

(b) K m = 5200

2π rB (2.90 × 10−2 m)(0.350 T) = = 0.0195 A μ0 K m N (2 × 10−7 T ⋅ m/A)(5200)(500) EVALUATE: If the solenoid were air-filled instead, a much larger current would be required to produce the same magnetic field. K μ NI IDENTIFY: Apply B = m 0 . 2πr SET UP: K m is the relative permeability and χ m = K m − 1 is the magnetic susceptibility. I=

28.48.

EXECUTE: (a) K m =

2πrB 2π (0.2500 m)(1.940 T) = = 2021. μ0 NI μ0 (500)(2.400 A)

(b) χ m = K m − 1 = 2020. EVALUATE: Without the magnetic material the magnetic field inside the windings would be B/2021 = 9.6 × 10−4 T. The presence of the magnetic material greatly enhances the magnetic field inside the windings.


Sources of Magnetic Field

28.49.

28-15

IDENTIFY: The magnetic field from the solenoid alone is B0 = μ0 nI . The total magnetic field is B = K m B0 . M is given by Eq.(28.29). SET UP: n = 6000 turns/m EXECUTE: (a) (i) B0 = μ0 nI = μ0 (6000 m −1 ) (0.15 A) = 1.13 × 10−3 T.

(ii) M =

Km − 1

μ0

B0 =

5199

μ0

(1.13 × 10−3 T) = 4.68 × 106 A m.

(iii) B = K m B0 = (5200)(1.13 × 10−3 T) = 5.88 T. ! ! ! ! ! (b) The directions of B , B0 and M are shown in Figure 28.49. Silicon steel is paramagnetic and B0 and M are in the same direction. EVALUATE: The total magnetic field is much larger than the field due to the solenoid current alone.

Figure 28.49 28.50.

IDENTIFY:

Curie’s law (Eq.28.32) says that 1/ M is proportional to T, so 1/χ m is proportional to T.

SET UP: The graph of 1/χ m versus the Kelvin temperature is given in Figure 28.50. EXECUTE: The material does obey Curie’s law because the graph in Figure 28.50 is a straight line. M = C

M=

B − B0

μ0

says that χ m =

B and T

T C μ0 . 1/ χ m = and the slope of 1/ χ m versus T is 1/(C μ0 ) . Therefore, from the T C μ0

1 1 = = 1.55 × 105 K ⋅ A T ⋅ m. μ0 (slope) μ0 (5.13 K −1 ) EVALUATE: For this material Curie’s law is valid over a wide temperature range.

graph the Curie constant is C =

Figure 28.50 28.51.

IDENTIFY: Moving charges create magnetic fields. The net field is the vector sum of the two fields. A charge moving in an external magnetic field feels a force. µ qv sin φ (a) SET UP: The magnetic field due to a moving charge is B = 0 . Both fields are into the paper, so 4π r 2 µ ⎛ qv sin φ q′v′ sin φ ′ ⎞ their magnitudes add, giving Bnet = B + B′ = 0 ⎜ + ⎟. r ′2 ⎠ 4π ⎝ r 2 EXECUTE: Substituting numbers gives µ ⎡ (8.00 µC)(9.00 × 104 m/s)sin 90° (5.00 µC)(6.50 × 104 m/s)sin 90° ⎤ Bnet = 0 ⎢ + ⎥ 4π ⎣ (0.300 m) 2 (0.400 m) 2 ⎦

Bnet = 1.00 × 10−6 T = 1.00μ T, into the paper.

! ! ! (b) SET UP: The magnetic force on a moving charge is F = qv × B , and the magnetic field of charge q′ at the location of charge q is into the page. The force on q is ! ! µ qv′ × rˆ ! ! ⎛ µ qv′ sin φ ⎞ ˆ ⎛ µ0 qq′vv′ sin φ ⎞ ˆ = (qv) iˆ × ⎜ 0 F = qv × B′ = (qv)iˆ × 0 ⎟ (−k ) = ⎜ ⎟j 2 4π r r2 ⎠ r2 ⎝ 4π ⎝ 4π ⎠ ! where φ is the angle between v′ and rˆ′.


28-16

Chapter 28

EXECUTE: Substituting numbers gives ! µ ⎡ ( 8.00 × 10−6 C )( 5.00 × 10−6 C )( 9.00 × 10−6 m/s )( 6.50 × 10−6 m/s ) ⎛ 0.400 ⎞ ⎤ ˆ F= 0⎢ ⎜ ⎟⎥ j 2 4π ⎢ ⎝ 0.500 ⎠ ⎦⎥ ( 0.500 m ) ⎣ ! F = ( 7.49 × 10−8 N ) ˆj .

28.52.

EVALUATE: These are small fields and small forces, but if the charge has small mass, the force can affect its motion. IDENTIFY: The wire creates a magnetic field near it, and the moving electron feels a force due to this field. µI SET UP: The magnetic field due to the wire is B = 0 , and the force on a moving charge is F = qvB sin φ . 2π r EXECUTE: F = qvB sin φ = (evμ0 I sin φ )/2π r. Substituting numbers gives

F = (1.60 × 10−19 C)(6.00 × 104 m/s)(4π × 10−7 T ⋅ m/A)(2.50 A)(sin 90°) /[2π (0.0450 m)]

28.53.

F = 1.07 × 10–19 N ! ! From the right hand rule for the cross product, the direction of v × B is opposite to the current, but since the electron is negative, the force is in the same direction as the current. EVALUATE: This force is small at an everyday level, but it would give the electron an acceleration of about 1011 m/s2. IDENTIFY: Find the force that the magnetic field of the wire exerts on the electron. SET UP: The force on a moving charge has magnitude F = q vB sin φ and direction given by the right-hand rule. ! μI For a long straight wire, B = 0 and the direction of B is given by the right-hand rule. 2π r F q vB sin φ ev ⎛ μ0 I ⎞ EXECUTE: (a) a = = = ⎜ ⎟ m m m ⎝ 2πr ⎠ a=

(1.6 × 10−17 C)(2.50 × 105 m/s)(4π × 10−7 T ⋅ m/A)(25.0 A) = 1.1 × 1013 m/s 2 , (9.11 × 10−31 kg)(2π )(0.0200 m)

away from the wire. (b) The electric force must balance the magnetic force. eE = evB , and μ I (250, 000 m/s)(4π × 10−7 T ⋅ m/A)(25.0 A) E = vB = v 0 = = 62.5 N/C . The magnetic force is directed away from 2π r 2π (0.0200 m) the wire so the force from the electric field must be toward the wire. Since the charge of the electron is negative, the electric field must be directed away from the wire to produce a force in the desired direction. EVALUATE: (c) mg = (9.11 × 10−31 kg)(9.8 m/s 2 ) ≈ 10−29 N . Fel = eE = (1.6 × 10−19 C)(62.5 N/C) ≈ 10−17 N . Fel ≈ 1012 Fgrav , so we can neglect gravity. 28.54.

IDENTIFY: Use Eq.(28.9) and the right-hand rule to calculate the magnetic field due to each wire. Add these field vectors to calculate the net field and then use Eq.(27.2) to calculate the force. SET UP: Let the wire connected to the 25.0 Ω resistor be #2 and the wire connected to the 10.0 Ω resistor be #1. Both I1 and I 2 are directed toward the right in the figure, so at the location of the proton B2 is ⊗ and B1 = ".

B1 =

μ0 I1 μI and B2 = 0 2 , with r = 0.0250 m. I1 = (100.0 V)/(10.0 Ω) = 10.0 A and I 2 = (100.0 V)/(25.0 Ω) = 4.00 A 2π r 2π r

EXECUTE:

B1 = 8.00 × 10−5 T, B2 = 3.20 × 10−5 T and B = B1 − B2 = 4.80 × 10−5 T and in the direction ". Force is to the right.

Figure 28.54

28.55.

F = qvB = (1.602 × 10−19 C)(650 × 103 m/s)(4.80 × 10−5 T) = 5.00 × 10−18 N ! ! EVALUATE: The force is perpendicular to both v and B. The magnetic force is much larger than the gravity force on the proton. IDENTIFY: Find the net magnetic field due to the two loops at the location of the proton and then find the force these fields exert on the proton. μ0 IR 2 SET UP: For a circular loop, the field on the axis, a distance x from the center of the loop is B = . 2( R 2 + x 2 )3 / 2 R = 0.200 m and x = 0.125 m .


Sources of Magnetic Field

28-17

⎡ ⎤ μ0 IR 2 EXECUTE: The fields add, so B = B1 + B2 = 2 B1 = 2 ⎢ . 2 2 3/ 2 ⎥ + 2( R x ) ⎣ ⎦ (4π × 10−7 T ⋅ m/A)(1.50 A)(0.200 m)2 B= = 5.75 × 10−6 T. [(0.200 m) 2 + (0.125 m)2 ]3/ 2 F = q vB sin φ = (1.6 × 10−19 C)(2400 m/s)(5.75 × 10−6 T) sin 90° = 2.21 × 10−21 N , perpendicular to the line ab and

28.56.

28.57.

to the velocity. EVALUATE: The weight of a proton is w = mg = 1.6 × 10−24 N , so the force from the loops is much greater than the gravity force on the proton. IDENTIFY: The net magnetic field is the vector sum of the fields due to each wire. ! μI SET UP: B = 0 . The direction of B is given by the right-hand rule. 2π r EXECUTE: (a) The currents are the same so points where the two fields are equal in magnitude are equidistant from the two wires. The net field is zero along the dashed line shown in Figure 28.56a. (b) For the magnitudes of the two fields to be the same at a point, the point must be 3 times closer to the wire with the smaller current. The net field is zero along the dashed line shown in Figure 28.56b. (c) As in (a), the points are equidistant from both wires. The net field is zero along the dashed line shown in Figure 28.56c. EVALUATE: The lines of zero net field consist of points at which the fields of the two wires have opposite directions and equal magnitudes.

Figure 28.56 ! μ0 qv!0 × rˆ IDENTIFY: B = 4π r 2 ! SET UP: rˆ = iˆ and r = 0.250 m , so v0 × rˆ = v0 z ˆj − v0 y kˆ. ! μ q μ q EXECUTE: B = 0 2 v0 z ˆj − v0 y kˆ = 6.00 × 10−6 T ˆj. v0 y = 0. 0 2 v0 z = 6.00 × 10−6 T and 4π r 4π r 4π (6.00 × 10−6 T)(0.25 m) 2 v0z = = −521 m/s. v0x = ± v02 − v0 y 2 − v0 z 2 = ± (800 m/s) 2 − (−521 m/s) 2 = ±607 m/s. μ0 ( −7.20 × 10−3 C) The sign of v0x isn’t determined. ! μ0 qv!0 × rˆ μ0 q ! ˆ v0 x kˆ − v0 z iˆ . = (b) Now r = j and r = 0.250 m . B = 4π r 2 4π r 2 μ |q| μ |q| μ (7.20 × 10−3 C) B = 0 2 v02x + v02z = 0 2 v0 = 0 800 m/s = 9.20 × 10 −6 T . 4π r 4π r 4π (0.250 m) 2 EVALUATE: The magnetic field in part (b) doesn’t depend on the sign of v0 x . ! IDENTIFY and SET UP: B = B ( x / a )iˆ

(

) (

)

(

28.58.

)

0

Apply Gauss's law for magnetic fields to a cube with side length L, one corner at the origin, and sides parallel to the x, y and z axes, as shown in Figure 28.58.

Figure 28.58


28-18

Chapter 28

! EXECUTE: Since B is parallel to the x-axis the only sides that have nonzero flux are the front side (parallel to the yz-plane at x =!L) and the back side (parallel to the yz-plane at x = 0.) ! front Φ B = ∫ B ⋅ dA = B0 ( x/a ) ∫ dA(iˆ ⋅ iˆ) = B0 ( x/a ) ∫ dA ! ! x = L on this face so B ⋅ dA = B0 ( L/a ) dA

Φ B = B0 ( L/a) ∫ dA = B0 ( L/a) L2 = B0 ( L3 /a )

28.59.

back On the back face x = 0 so B = 0 and Φ B = 0. The total flux through the cubical Gaussian surface is Φ B = B0 ( L3 /a ). ! EVALUATE: This violates Eq.(27.8), which says that Φ B = 0 for any closed surface. The claimed B is impossible because it has been shown to violate Gauss's law for magnetism. IDENTIFY: Use Eq.(28.9) and the right-hand rule to calculate the magnitude and direction of the magnetic field at P produced by each wire. Add these two field vectors to find the net field. (a) SET UP: The directions of the fields at point P due to the two wires are sketched in Figure 28.59a. ! ! EXECUTE: B1 and B2 must be equal and opposite for the resultant field at P to be zero. ! B2 is to the right so I 2 is out of the page.

Figure 28.59a

μ0 I1 μ0 ⎛ 6.00 A ⎞ μI μ ⎛ I2 ⎞ = B2 = 0 2 = 0 ⎜ ⎜ ⎟ ⎟ 2π r1 2π ⎝ 1.50 m ⎠ 2π r2 2π ⎝ 0.50 m ⎠ μ ⎛ 6.00 A ⎞ μ0 ⎛ I 2 ⎞ B1 = B2 says 0 ⎜ ⎟= ⎜ ⎟ 2π ⎝ 1.50 m ⎠ 2π ⎝ 0.50 m ⎠ ⎛ 0.50 m ⎞ I2 = ⎜ ⎟ ( 6.00 A ) = 2.00 A ⎝ 1.50 m ⎠ (b) SET UP: The directions of the fields at point Q are sketched in Figure 28.59b. μI EXECUTE: B1 = 0 1 2π r1 ⎛ 6.00 A ⎞ −6 B1 = (2 × 10−7 T ⋅ m/A) ⎜ ⎟ = 2.40 × 10 T ⎝ 0.50 m ⎠ μI B2 = 0 2 2π r2 ⎛ 2.00 A ⎞ −7 B2 = (2 × 10−7 T ⋅ m/A) ⎜ ⎟ = 2.67 × 10 T ⎝ 1.50 m ⎠ Figure 28.59b ! ! B1 and B2 are in opposite directions and B1 > B2 so ! B = B1 − B2 = 2.40 × 10−6 T − 2.67 × 10−7 T = 2.13 × 10−6 T, and B is to the right. (c) SET UP: The directions of the fields at point S are sketched in Figure 28.59c. μI EXECUTE: B1 = 0 1 2π r1 ⎛ 6.00 A ⎞ −6 B1 = (2 × 10−7 T ⋅ m/A) ⎜ ⎟ = 2.00 × 10 T ⎝ 0.60 m ⎠ μI B2 = 0 2 2π r2 ⎛ 2.00 A ⎞ −7 B2 = (2 × 10−7 T ⋅ m/A) ⎜ ⎟ = 5.00 × 10 T ⎝ 0.80 m ⎠ Figure 28.59c ! ! B1 and B2 are right angles to each other, so the magnitude of their resultant is given by B1 =

B = B12 + B22 = (2.00 × 10−6 T) 2 + (5.00 × 10−7 T) 2 = 2.06 × 10−6 T


Sources of Magnetic Field

28.60.

28-19

EVALUATE: The magnetic field lines for a long, straight wire!are concentric circles with the wire at the center. The magnetic field at each point is tangent to the field line, so B is perpendicular to the line from the wire to the point where the field is calculated. IDENTIFY: Find the vector sum of the magnetic fields due to each wire. ! μI SET UP: For a long straight wire B = 0 . The direction of B is given by the right-hand rule and is perpendicular 2π r to the line from the wire to the point where then field is calculated. EXECUTE: (a) The magnetic field vectors are shown in Figure 28.60a. μI μ0 I a μ0 Ia (b) At a position on the x-axis Bnet = 2 0 sinθ = = , in the positive x-direction. 2 2 2 2 2 2πr π ( x + a2 ) π x +a x +a (c) The graph of B versus x/a is given in Figure 28.60b. EVALUATE: (d) The magnetic field is a maximum at the origin, x = 0. μ Ia (e) When x >> a, B ≈ 0 2 . πx

Figure 28.60 28.61.

Apply F = IlB sin φ , with the magnetic field at point P that is calculated in problem 28.60. μ0 Ia SET UP: The net field of the first two wires at the location of the third wire is B = , in the +x-direction. π ( x2 + a2 ) EXECUTE: (a) Wire is carrying current into the page, so it feels a force in the − y -direction . IDENTIFY:

⎛ μ0 Ia ⎞ F μ0 (6.00 A) 2 (0.400 m) = IB = I ⎜ = = 1.11 × 10−5 N/m. ⎟ 2 2 2 2 L ⎝ π ( x + a ) ⎠ π ( (0.600 m) + (0.400 m) )

28.62.

28.63.

(b) If the wire carries current out of the page then the force felt will be in the opposite direction as in part (a). Thus the force will be 1.11× 10−5 N m, in the +y-direction. EVALUATE: We could also calculate the force exerted by each of the first two wires and find the vector sum of the two forces. IDENTIFY: The wires repel each other since they carry currents in opposite directions, so the wires will move away from each other until the magnetic force is just balanced by the force due to the spring. µ I 2L SET UP: The force of the spring is kx and the magnetic force on each wire is Fmag = 0 . 2π x EXECUTE: Call x the distance the springs each stretch. On each wire, Fspr = Fmag, and there are two spring forces µ I 2L µ0 I 2 L , which gives x = . on each wire. Therefore 2kx = 0 2π x 2π k EVALUATE: Since μ0 /2π is small, x will likely be much less than the length of the wires. ! ! IDENTIFY: Apply ∑ F = 0 to one of the wires. The force one wire exerts on the other depends on I so ∑ F = 0

gives two equations for the two unknowns T and I. SET UP: The force diagram for one of the wires is given in Figure 28.63. ⎛ μ I2 ⎞ The force one wire exerts on the other is F = ⎜ 0 ⎟ L, ⎝ 2π r ⎠ where r = 2(0.040 m)sin θ = 8.362 × 10−3 m is the distance between the two wires.

Figure 28.63


28-20

Chapter 28

EXECUTE:

∑F

x

∑F

y

= 0 gives T cosθ = mg and T = mg / cosθ

= 0 gives F = T sin θ = (mg / cosθ )sin θ = mg tan θ

And m = λ L, so F = λ Lg tan θ ⎛ μ0 I 2 ⎞ ⎜ ⎟ L = λ Lg tan θ ⎝ 2π r ⎠ λ gr tan θ I= ( μ0 / 2π ) (0.0125 kg/m)(9.80 m/s) 2 (tan 6.00°)(8.362 ×10−3 m) = 23.2 A 2 × 10−7 T ⋅ m/A EVALUATE: Since the currents are in opposite directions the wires repel. When I is increased, the angle θ from the vertical increases; a large current is required even for the small displacement specified in this problem. IDENTIFY: Consider the forces on each side of the loop. SET UP: The forces on the left and right sides cancel. The forces on the top and bottom segments of the loop are in opposite directions, so the magnitudes subtract. ⎛ μ I ⎞ ⎛ Il Il ⎞ μ IlI ⎛ 1 1 ⎞ EXECUTE: F = Ft − Fb = ⎜ 0 wire ⎟ ⎜ − ⎟ = 0 wire ⎜ − ⎟ . 2π ⎝ rr rl ⎠ ⎝ 2π ⎠ ⎝ rt rb ⎠ I=

28.64.

F=

28.65.

28.66.

28.67.

⎞ μ0 (5.00 A)(0.200 m)(14.0 A) ⎛ 1 1 −5 − ⎜ ⎟ = 7.97 × 10 N . The force on the top segment is away 2π ⎝ 0.100 m 0.026 m ⎠

from the wire, so the net force is away from the wire. EVALUATE: The net force on a current loop in a uniform magnetic field is zero, but the magnetic field of the wire is not uniform, it is stronger closer to the wire. ! ! IDENTIFY: Find the magnetic field of the first loop at the location of the second loop and apply τ = μ × B and ! ! U = − μ ⋅ B to find μ and U. SET UP: Since x is much larger than a′ , assume B is uniform over the second loop and equal to its value on the axis of the first loop. Nμ0 Ia 2 Nμ0 Ia 2 EXECUTE: (a) x >> a ⇒ B = . ≈ 2 2 3/ 2 2( x + a ) 2 x3 ⎛ Nμ Ia 2 ⎞ NN ′μ0πI I ′a 2 a′2 sin θ ! ! τ = μ × B = μB sin θ = ( N ′I ′A′) ⎜ 0 3 ⎟ sin θ = 2 x3 ⎝ 2x ⎠ ⎛ Nμ Ia 2 ⎞ NN ′μ0π II ′a 2 a′2 cosθ ! ! . (b) U = − μ ⋅ B = − μB cosθ = −( N ′I ′π a′2 ) ⎜ 0 3 ⎟ cosθ = − 2 x3 ⎝ 2x ⎠ EVALUATE: (c) Having x >> a allows us to simplify the form of the magnetic field, whereas assuming x >> a′ means we can assume that the magnetic field from the first loop is constant over the second loop. ! ! μ Idl × rˆ IDENTIFY: Apply dB = 0 . 4π r 2 SET UP: The two straight segments produce zero field at P. The field at the center of a circular loop of radius R is μI μI B = 0 , so the field at the center of curvature of a semicircular loop is B = 0 . 2R 4R EXECUTE: The semicircular loop of radius a produces field out of the page at P and the semicircular loop of 1⎛ μ I ⎞⎛ 1 1⎞ μ I ⎛ a ⎞ radius b produces field into the page. Therefore, B = Ba − Bb = ⎜ 0 ⎟ ⎜ − ⎟ = 0 ⎜ 1 − ⎟ , out of page. 2 ⎝ 2 ⎠ ⎝ a b ⎠ 4a ⎝ b ⎠ EVALUATE: If a = b , B = 0 . IDENTIFY: Find the vector sum of the fields due to each loop. μ0 Ia 2 SET UP: For a single loop B = . Here we have two loops, each of N turns, and measuring the field 2( x 2 + a 2 )3/ 2 along the x-axis from between them means that the “x” in the formula is different for each case: EXECUTE: Left coil: x → x +

a μ0 NIa 2 . ⇒ Bl = 2 2(( x + a 2) 2 + a 2 )3 2

Right coil: x → x −

a μ0 NIa 2 . ⇒ Br = 2 2(( x − a 2) 2 + a 2 )3 2


Sources of Magnetic Field

28-21

So, the total field at a point a distance x from the point between them is B=

⎞ μ0 NIa 2 ⎛ 1 1 . + ⎜ 2 2 32 2 2 32 ⎟ 2 ⎝ (( x + a 2) + a ) (( x − a 2) + a ) ⎠

(b) B versus x is graphed in Figure 28.67. Figure 28.67a is the total field and Figure 27.67b is the field from the right-hand coil. 32 ⎞ μ NIa 2 ⎛ 1 1 μ0 NIa 2 ⎛ 4 ⎞ μ0 NI (c) At point P, x = 0 and B = 0 + = = ⎜ ⎟ ⎜ ⎟ 2 ⎝ ((a 2) 2 + a 2 )3 2 ((− a 2) 2 + a 2 )3 2 ⎠ (5a 2 4)3 2 ⎝ 5 ⎠ a ⎛ 4⎞ (d) B = ⎜ ⎟ ⎝5⎠

(e) dB dx

32

μ0 NI ⎛ 4 ⎞ =⎜ ⎟ a ⎝5⎠

32

μ0 (300)(6.00 A) = 0.0202 T. (0.080 m)

⎞ dB μ0 NIa 2 ⎛ −3( x + a 2) −3( x − a 2) . At x = 0 , = + ⎜ 2 2 52 2 2 52 ⎟ 2 ⎝ (( x + a 2) + a ) (( x − a 2) + a ) ⎠ dx = x =0

⎞ μ0 NIa 2 ⎛ −3(a 2) −3( − a 2) + =0. ⎜ 2 2 52 2 2 52 ⎟ 2 ⎝ (( a 2) + a ) ((− a 2) + a ) ⎠

d B μ0 NIa 2 ⎛ −3 6( x + a 2) 2 (5 2) −3 6( x − a 2) 2 (5 2) ⎞ = + + + ⎜ ⎟ 2 2 2 52 2 2 72 2 2 5/ 2 2 ⎝ (( x + a 2) + a ) (( x + a 2) + a ) (( x − a 2) + a ) (( x − a 2) 2 + a 2 )7 / 2 ⎠ dx 2

At x = 0 ,

d 2B dx 2

= x =0

−3 −3 μ0 NIa 2 ⎛ 6(a 2) 2 (5 2) 6(− a 2) 2 (5 2) ⎞ + + + ⎜⎜ ⎟ = 0. 2 2 7/2 2 2 5/ 2 2 2 5/ 2 2 ⎝ ((a 2) + a ) ((a 2) + a ) (( a 2) + a ) ((a 2) 2 + a 2 )7 / 2 ⎠⎟

EVALUATE: Since both first and second derivatives are zero, the field can only be changing very slowly.

Figure 28.67 28.68.

IDENTIFY: A current-carrying wire produces a magnetic field, but the strength of the field depends on the shape of the wire. SET UP: The magnetic field at the center of a circular wire of radius a is B = μ0 I/2a, and the field a distance x µ0 I 2a . 4π x x 2 + a 2 EXECUTE: (a) Since the diameter D = 2a, we have B = μ0 I/2a = μ0 I/D. (b) In this case, the length of the wire is equal to the diameter of the circle, so 2a = π D, giving a = π D/2, and

from the center of a straight wire of length 2a is B =

x = D/2. Therefore B =

28.69.

2 (π D / 2 ) µ0 I µ0 I = . 4π ( D / 2) D 2 / 4 + π 2 D 2 / 4 D 1 + π 2

EVALUATE: The field in part (a) is greater by a factor of 1 + π 2 . It is reasonable that the field due to the circular wire is greater than the field due to the straight wire because more of the current is close to point A for the circular wire than it is for the straight wire. ! ! μ0 Idl × rˆ IDENTIFY: Apply dB = . 4π r 2 ! ! SET UP: The contribution from the straight segments is zero since dl × r = 0. The magnetic field from the curved wire is just one quarter of a full loop. ⎛μI⎞ μI EXECUTE: B = 14 ⎜ 0 ⎟ = 0 and is directed out of the page. ⎝ 2 R ⎠ 8R


28-22

28.70.

28.71.

Chapter 28

EVALUATE: It is very simple to calculate B at point P but it would be much more difficult to calculate B at other points. ! ! μ0 Idl × rˆ IDENTIFY: Apply dB = . 4π r 2 ! ! SET UP: The horizontal wire yields zero magnetic field since dl × r = 0. The vertical current provides the magnetic field of half of an infinite wire. (The contributions from all infinitesimal pieces of the wire point in the same direction, so there is no vector addition or components to worry about.) ⎛ μI ⎞ μI EXECUTE: B = 12 ⎜ 0 ⎟ = 0 and is directed out of the page. ⎝ 2πR ⎠ 4πR EVALUATE: In the equation preceding Eq.(28.8) the limits on the integration are 0 to a rather than −a to a and this introduces a factor of 12 into the expression for B. (a) IDENTIFY: Consider current density J for a small concentric ring and integrate to find the total current in terms of α and R. SET UP: We can't say I = JA = J π R 2 , since J varies across the cross section.

To integrate J over the cross section of the wire divide the wire cross section up into thin concentric rings of radius r and width dr, as shown in Figure 28.71. Figure 28.71 EXECUTE: The area of such a ring is dA, and the current through it is dI = J dA; dA = 2π r dr and dI = J dA = α r (2π r dr ) = 2πα r 2 dr R

I = ∫ dI = 2πα ∫ r 2 dr = 2πα ( R 3 / 3) so α = 0

3I 2π R 3

(b) IDENTIFY and SET UP: (i) r ≤ R Apply Ampere's law to a circle of radius r < R. Use the method of part (a) to find the current enclosed by the Ampere's law path. ! ! ! EXECUTE: úB ⋅ dl = úB dl = Búdl = B (2π r ), by the symmetry and direction of B. The current passing through r

the path is I encl = ∫ dl , where the integration is from 0 to r. I encl = 2πα ∫ r 2 dr = 0

!

2πα r 3 2π ⎛ 3I ⎞ 3 Ir 3 − ⎜ ⎟ r = 3 . Thus 3 3 ⎝ 2π R 3 ⎠ R

!

⎛ Ir 3 ⎞ μ Ir 2 gives B (2π r ) = μ0 ⎜ 3 ⎟ and B = 0 3 2π R ⎝R ⎠ (ii) IDENTIFY and SET UP: r ≥ R Apply Ampere’s law to a circle of radius r > R. ! ! EXECUTE: úB ⋅ dl = úB dl = Búdl = B (2π r )

úB ⋅ dl = μ I

0 encl

28.72.

! ! μI I encl = I ; all the current in the wire passes through this path. Thus úB ⋅ dl = μ0 I encl gives B (2π r ) = μ0 I and B = 0 2π r μI EVALUATE: Note that at r = R the expression in (i) (for r ≤ R ) gives B = 0 . At r = R the expression in (ii) 2π R μ0 I (for r ≥ R ) gives B = , which is the same. 2π R IDENTIFY: Apply Ampere’s law to a circle of radius r in each case. SET UP: Assume that the currents are uniform over the cross sections of the conductors. ! ! ⎛A ⎞ ⎛ r2 ⎞ ⎛ r2 ⎞ μ Ir EXECUTE: (a) r < a ⇒ I encl = I ⎜ r ⎟ = I ⎜ 2 ⎟ . úB ⋅ dl = B 2πr = μ0 I encl = μ0 I ⎜ 2 ⎟ and B = 0 2 . When 2πa ⎝a ⎠ ⎝a ⎠ ⎝ Aa ⎠

r = a, B =

μ0 I , which is just what was found in part (a) of Exercise 28.37. 2πa


Sources of Magnetic Field

⎛A ⎞ ⎛ r 2 − b2 ⎞ (b) b < r < c ⇒ I encl = I − I ⎜ b → r ⎟ = I ⎜ 1 − 2 . 2 ⎟ ⎝ c −b ⎠ ⎝ Ab → c ⎠

!

!

28-23

⎛ c2 − r 2 ⎞ r 2 − b2 ⎞ and = μ0 I ⎜ 2 2 2 ⎟ 2 ⎟ −b ⎠ ⎝c −b ⎠

úB ⋅ dl = B 2πr = μ I ⎜⎝1 − c 0

μ0 I ⎛ c 2 − r 2 ⎞ μ0 I , just as in part (a) of Exercise 28.37 and when r = c, B = 0 , just as in ⎟ . When r = b, B = ⎜ 2πb 2πr ⎝ c 2 − b 2 ⎠ part (b) of Exercise 28.37. EVALUATE: Unlike E, B is not zero within the conductors. B varies across the cross section of each conductor. ! ! IDENTIFY: Apply úB ⋅ dA = 0. B=

28.73.

SET UP: Take the closed gaussian surface to be a cylinder whose axis coincides with the wire. EXECUTE: If there is a magnetic field component in the z-direction, it must be constant because of the symmetry of the wire. Therefore the contribution to a surface integral over a closed cylinder, encompassing a long straight wire will be zero: no flux through the barrel of the cylinder, and equal but opposite flux through the ends. The radial field will have no contribution through the ends, but through the barrel: ! ! ! ! ! ! 0 = úB ⋅ dA = úBr ⋅ dA = ∫ Br ⋅ dA = ∫ Br dA = Br Abarrel = 0 . Therefore, Br = 0. barrel

28.74.

28.75.

barrel

EVALUATE: The magnetic field of a long straight wire is everywhere tangent to a circular area whose plane is perpendicular to the wire, with the wire passing through the center of the circular area. This field produces zero flux through the cylindrical gaussian surface. IDENTIFY: Apply Ampere’s law to a circular path of radius r. SET UP: Assume the current is uniform over the cross section of the conductor. EXECUTE: (a) r < a ⇒ I encl = 0 ⇒ B = 0. ! ! ⎛ A ⎞ ⎛ π (r 2 − a 2 ) ⎞ (r 2 − a 2 ) μ I (r 2 − a 2 ) (r 2 − a 2 ) . (b) a < r < b ⇒ I encl = I ⎜ a → r ⎟ = I ⎜ and B = 0 . úB ⋅ dl = B 2πr = μ0 I 2 =I 2 2 2 2 ⎟ 2 (b − a ) 2πr (b 2 − a 2 ) (b − a ) ⎝ Aa →b ⎠ ⎝ π (b − a ) ⎠ ! ! μI (c) r > b ⇒ I encl = I . úB ⋅ dl = B 2πr = μ0 I and B = 0 . 2πr EVALUATE: The expression in part (b) gives B = 0 at r = a and this agrees with the result of part (a). The μI expression in part (b) gives B = 0 at r = b and this agrees with the result of part (c). 2π b IDENTIFY: Use Ampere's law to find the magnetic field at r = 2a from the axis. The analysis of Example 28.9 shows that the field outside the cylinder is the same as for a long, straight wire along the axis of the cylinder. SET UP: EXECUTE: Apply Ampere's law to a circular path of radius 2a, as shown in Figure 28.75. B (2π ) = μ0 I encl ⎛ (2a ) 2 − a 2 ⎞ I encl = I ⎜ = 3I /8 2 2 ⎟ ⎝ (3a ) − a ⎠

Figure 28.75 3 μ0 I ; this is the magnetic field inside the metal at a distance of 2a from the cylinder axis. Outside the 16 2π a μI cylinder, B = 0 . The value of r where these two fields are equal is given by 1/ r = 3/(16a ) and r = 16a / 3. 2π r EVALUATE: For r < 3a, as r increases the magnetic field increases from zero at r = 0 to μ0 I /(2π (3a )) at r = 3a. For r > 3a the field decreases as r increases so it is reasonable for there to be a r > 3a where the field is the same as at r = 2a. IDENTIFY: The net field is the vector sum of the fields due to the circular loop and to the long straight wire. μI μI SET UP: For the long wire, B = 0 1 , and for the loop, B = 0 2 . 2π D 2R EXECUTE: At the center of the circular loop the current I 2 generates a magnetic field that is into the page, so the

B=

28.76.

current I1 must point to the right. For complete cancellation the two fields must have the same magnitude: μ0 I1 μ0 I 2 = . Thus, I1 = πD I 2 . R 2πD 2 R EVALUATE: If I1 is to the left the two fields add.


28-24

Chapter 28

28.77.

IDENTIFY: Use the current density J to find dI through a concentric ring and integrate over the appropriate cross ! section to find the current through that cross section. Then use Ampere's law to find B at the specified distance from the center of the wire. (a) SET UP: Divide the cross section of the cylinder into thin concentric rings of radius r and width dr, as shown in Figure 28.77a. The current through each ring is dI = J dA = J 2π r dr.

Figure 28.77a dI =

EXECUTE:

2I0 ⎡ 4I 2 2 1 − ( r / a ) ⎤ 2π r dr = 20 ⎡1 − ( r / a ) ⎤ r dr. The total current I is obtained by integrating dI ⎦ ⎦ a ⎣

π a2 ⎣

a

a 1 ⎛ 4I ⎞ a ⎛ 4I ⎞ ⎡ 1 ⎤ over the cross section I = ∫ dI = ⎜ 20 ⎟ ∫ (1 − r 2 / a 2 ) r dr = ⎜ 20 ⎟ ⎢ r 2 − r 4 / a 2 ⎥ = I 0 , as was to be shown. 0 4 ⎝ a ⎠ 0 ⎝ a ⎠⎣2 ⎦0 (b) SET UP: Apply Ampere's law to a path that is a circle of radius r > a, as shown in Figure 28.77b.

!

!

úB ⋅ dl = B(2π r ) I encl = I 0 (the path encloses the entire cylinder)

Figure 28.77b !

!

úB ⋅ dl = μ I

EXECUTE:

0 encl

says B (2π r ) = μ0 I 0 and B =

μ0 I 0 . 2π r

(c) SET UP: Divide the cross section of the cylinder into concentric rings of radius r ′ and width dr ′, as was done in part (a). See Figure 28.77c. The current dI through each ring 2 4 I ⎡ ⎛ r′ ⎞ ⎤ is dI = 20 ⎢1 − ⎜ ⎟ ⎥ r′ dr ′ a ⎢⎣ ⎝ a ⎠ ⎥⎦

Figure 28.77c EXECUTE: The current I is obtained by integrating dI from r ′ = 0 to r ′ = r: 2 r 4I r ⎡ ⎛ r′ ⎞ ⎤ 4I 2 4 I = ∫ dI = 20 ∫ ⎢1 − ⎜ ⎟ ⎥ r ′ dr′ = 20 ⎡ 12 ( r ′ ) − 14 ( r ′ ) /a 2 ⎤ ⎦0 a 0 ⎣⎢ ⎝ a ⎠ ⎦⎥ a ⎣ 4I I r2 ⎛ r2 ⎞ I = 20 ( r 2 / 2 − r 4 / 4a 2 ) = 0 2 ⎜ 2 − 2 ⎟ a a ⎝ a ⎠ (d) SET UP:

Apply Ampere's law to a path that is a circle of radius r < a, as shown in Figure 28.77d. ! ! úB ⋅ dl = B(2π r ) I encl =

I0r 2 ⎛ r2 ⎞ 2 − ⎜ ⎟ (from part (c)) a2 ⎝ a2 ⎠

Figure 28.77d

28.78.

!

!

I0r 2 μI r (2 − r 2 /a 2 ) and B = 0 0 2 (2 − r 2 / a 2 ) 2 2π a a μI EVALUATE: Result in part (b) evaluated at r = a: B = 0 0 . Result in part (d) evaluated at 2π a I μ0 I 0 a μ r = a: B = (2 − a 2 / a 2 ) = 0 0 . The two results, one for r > a and the other for r < a, agree at r = a. 2π a 2 2π a IDENTIFY: Apply Ampere’s law to a circle of radius r. ! ! SET UP: The current within a radius r is I = ∫ J ⋅ dA , where the integration is over a disk of radius r.

EXECUTE:

úB ⋅ dl = μ I

0 encl

says B (2π r ) = μ0


Sources of Magnetic Field

! ! a ⎛b ⎞ EXECUTE: (a) I 0 = ∫ J ⋅ dA = ∫ ⎜ e(r − a )/δ ⎟ rdrdθ = 2π b ∫ e(r − a )/δ dr = 2πbδ e(r − a )/δ 0 ⎝r ⎠

a 0

28-25

= 2π bδ (1 − e − a/δ ) .

I 0 = 2π (600 A/m) (0.025 m) (1 − e(0.050 / 0.025) ) = 81.5 A. ! ! μI (b) For r ≥ a, úB ⋅ d l = B 2πr = μ0 I encl = μ0 I 0 and B = 0 0 . 2π r r ! ! r b ⎛ ⎞ (c) For r ≤ a, I ( r ) = ∫ J ⋅ dA = ∫ ⎜ e(r ′ − a )/δ ⎟ r ′dr ′dθ = 2πb ∫ e(r − a )/δ dr = 2πbδe(r ′ − a )/δ . 0 0 ⎝ r′ ⎠

I ( r ) = 2πbδ (e(r ′− a )/δ − e − a/δ ) = 2πbδe− a/δ (er/δ − 1) and I ( r ) = I 0 (d) For r ≤ a ,

!

!

úB ⋅ dl = B(r )2πr = μ I

0 encl

(e) At r = δ = 0.025 m, B =

= μ0 I 0

(e r/δ − 1) . (e a/δ − 1)

(e r/δ − 1) μ I (e r/δ − 1) . and B = 0 0 a/δ a/ δ (e − 1) 2π r (e − 1)

μ0 I 0 (e − 1) μ (81.5 A) (e − 1) = 0 = 1.75 × 10−4 T . 2πδ (e a / δ − 1) 2π (0.025 m) (e0.050 / 0.025 − 1)

μ0 I 0 (ea / δ − 1) μ0 (81.5 A) = = 3.26 × 10−4 T. 2π a (e a / δ − 1) 2π (0.050 m) μI μ (81.5 A) At r = 2a = 0.100 m, B = 0 0 = 0 = 1.63 × 10−4 T. 2π r 2π (0.100 m)

At r = a = 0.050 m, B =

28.79.

EVALUATE: At points outside the cylinder, the magnetic field is the same as that due to a long wire running along the axis of the cylinder. IDENTIFY: Evaluate the integral as specified in the problem. μ0 Ia 2 SET UP: Eq.(28.15) says Bx = . 2 2( x + a 2 )3 / 2 EXECUTE:

∞ −∞

Bx dx = ∫ B=

μ0 Ia 2 μI dx = 0 −∞ 2( x + a 2 ) 3 / 2 2 ∞

μ0 I 2

2

∞ −∞

1 d ( x/a ). (( x/a ) 2 + 1)3/ 2

∞ dz μI ⇒ ∫ Bx dx = 0 −∞ ( z + 1)3/ 2 −∞ 2 ∞

2

π/2

∫π

− /2

π /2

cos θdθ =

μ0 I (sinθ ) = μ0 I , 2 − π /2

where we used the substitution z = tan θ to go from the first to second line. EVALUATE: This is just what Ampere’s Law tells us to expect if we imagine the loop runs along the x-axis ! ! closing on itself at infinity: úB ⋅ d l = μ0 I .

28.80.

IDENTIFY: Follow the procedure specified in the problem. SET UP: The field and integration path are sketched in Figure 28.80. ! ! ! EXECUTE: úB ⋅ d l = 0 (no currents in the region). Using the figure, let B = B0 iˆ for y < 0 and B = 0 for y > 0. ! ! Then ∫ B ⋅ d l = Bab L − Bcd L = 0. Bcd = 0 , so Bab L = 0. But we have assumed that Bab ≠ 0. This is a contradiction abcde

and violates Ampere’s Law. EVALUATE: It is often convenient to approximate B as confined to a particular region of space, but this result tells us that the boundary of such a region isn’t sharp.

Figure 28.80 28.81.

IDENTIFY: Use what we know about the magnetic field of a long, straight conductor to deduce the symmetry of the magnetic field. Then apply Ampere's law to calculate the magnetic field at a distance a above and below the current sheet.


28-26

Chapter 28

SET UP: Do parts (a) and (b) together.

Consider the individual currents in pairs, where the currents in each pair are equidistant ! on either side of the point where B is being calculated. Figure 28.81a shows that for each pair the z-components cancel, and that above the sheet the field is in the –x-direction and that below the sheet it is in the +x-direction.

Figure 28.81a

! ! Also, by symmetry the magnitude of B a distance a above the sheet must equal the magnitude of B a distance a ! below the sheet. Now that we have deduced the symmetry of B, apply Ampere's law. Use a path that is a rectangle, as shown in Figure 28.81b. !

!

ú B ⋅ dl = μ I

0 encl

Figure 28.81b I is directed out of the page, so for I to be positive the integral around the path is taken in the counterclockwise direction. ! ! ! EXECUTE: Since B is parallel to the sheet, on the sides of the rectangle that have length 2a, úB ⋅ dl = 0. On the ! long sides of length L, B is parallel to the side, in the direction we are integrating around the path, and has the ! ! same magnitude, B, on each side. Thus úB ⋅ dl = 2 BL. n conductors per unit length and current I out of the page in

28.82.

each conductor gives I encl = InL. Ampere's law then gives 2 BL = μ0 InL and B = 12 μ0 In. EVALUATE: Note that B is independent of the distance a from the sheet. Compare this result to the electric field due to an infinite sheet of charge (Example 22.7). IDENTIFY: Find the vector sum of the fields due to each sheet. ! SET UP: Problem 28.81 shows that for an infinite sheet B = 12 μ0 In . If I is out of the page, B is to the left above ! the sheet and to the right below the sheet. If I is into the page, B is to the right above the sheet and to the left below the sheet. B is independent of the distance from the sheet. The directions of the two fields at points P, R and S are shown in Figure 28.82. EXECUTE: (a) Above the two sheets, the fields cancel (since there is no dependence upon the distance from the sheets). (b) In between the sheets the two fields add up to yield B = μ0 nI , to the right. (c) Below the two sheets, their fields again cancel (since there is no dependence upon the distance from the sheets). EVALUATE: The two sheets with currents in opposite directions produce a uniform field between the sheets and zero field outside the two sheets. This is analogous to the electric field produced by large parallel sheets of charge of opposite sign.

Figure 28.82 28.83.

IDENTIFY and SET UP: Use Eq.(28.28) to calculate the total magnetic moment of a volume V of the iron. Use the density and atomic mass of iron to find the number of atoms in this volume and use that to find the magnetic dipole moment per atom.


Sources of Magnetic Field

28-27

μ total

, so μ total = MV The average magnetic moment per atom is μatom = μ total / N = MV / N , V where N is the number of atoms in volume V. The mass of volume V is m = ρV , where ρ is the density.

EXECUTE:

M=

( ρiron = 7.8 × 103 kg/m3 ). The number of moles of iron in volume V is n=

ρV m = , where 55.847 × 10−3 kg/mol is the atomic mass −3 55.847 × 10 kg/mol 55.847 × 10−3 kg/mol

of iron from appendix D. N = nN A , where N A = 6.022 × 1023 atoms/mol is Avogadro's number. Thus N = nN A =

ρVN A 55.847 × 10−3 kg/mol

.

μatom =

⎛ 55.847 × 10−3 kg/mol ⎞ M (55.847 × 10−3 kg/mol) MV . = MV ⎜ ⎟= N ρVN A ρ NA ⎝ ⎠

μatom =

(6.50 × 104 A/m)(55.847 × 10−3 kg/mol) (7.8 × 103 kg/m 3 )(6.022 × 1023 atoms/mol)

μatom = 7.73 × 10−25 A ⋅ m 2 = 7.73 × 10−25 J/T μ B = 9.274 × 10−24 A ⋅ m 2 , so μatom = 0.0834μB . 28.84.

EVALUATE: The magnetic moment per atom is much less than one Bohr magneton. The magnetic moments of each electron in the iron must be in different directions and mostly cancel each other. IDENTIFY: The force on the cube of iron must equal the weight of the iron cube. The weight is proportional to the density and the magnetic force is proportional to μ , which is in turn proportional to K m . SET UP: The densities if iron, aluminum and silver are ρ Fe = 7.8 × 103 kg/m3 , ρ Al = 2.7 × 103 kg/m3 and

ρ Ag = 10.5 × 103 kg/m3. The relative permeabilities of iron, aluminum and silver are K Fe = 1400 , K Al = 1.00022 and K Ag = 1.00 − 2.6 × 10−5 .

EXECUTE: (a) The microscopic magnetic moments of an initially unmagnetized ferromagnetic material experience torques from a magnet that aligns the magnetic domains with the external field, so they are attracted to the magnet. For a paramagnetic material, the same attraction occurs because the magnetic moments align themselves parallel to the external field. For a diamagnetic material, the magnetic moments align antiparallel to the external field so it is like two magnets repelling each other. (b) The magnet can just pick up the iron cube so the force it exerts is FFe = mFe g = ρ Fe a 3 g = (7.8 × 103 kg/m3 )(0.020 m)3 (9.8 m/s 2 ) = 0.612 N. If the magnet tries to lift the aluminum cube of the same dimensions as the iron block, then the upward force felt by the cube is K 1.000022 FAl = Al (0.612 N) = (0.612 N) = 4.37 × 10−4 N. The weight of the aluminum cube is K Fe 1400

WAl = mAl g = ρAl a3 g = (2.7 × 103 kg m3 )(0.020 m)3 (9.8 m s 2 ) = 0.212 N. Therefore, the ratio of the magnetic force 4.37 × 10−4 N = 2.1 × 10−3 and the magnet cannot lift it. 0.212 N (c) If the magnet tries to lift a silver cube of the same dimensions as the iron block, then the downward force felt K (1.00 − 2.6 × 10−5 ) by the cube is FAl = Ag (0.612 N) = (0.612 N) = 4.37 × 10−4 N. But the weight of the silver cube K Fe 1400

on the aluminum cube to the weight of the cube is

is WAg = mAg g = ρ Ag a 3 g = (10.5 × 103 kg/m 3 )(0.020 m)3 (9.8 m/s 2 ) = 0.823 N. So the ratio of the magnetic force on the silver cube to the weight of the cube is

28.85.

4.37 × 10−4 N = 5.3 × 10−4 and the magnet’s effect would not be 0.823 N

noticeable. EVALUATE: Silver is diamagnetic and is repelled by the magnet. Aluminum is paramagnetic and is attracted by the magnet. But for both these materials the force is much less that the force on a similar cube of ferromagnetic iron. IDENTIFY: The current-carrying wires repel each other magnetically, causing them to accelerate horizontally. Since gravity is vertical, it plays no initial role. F µ0 I 2 SET UP: The magnetic force per unit length is = , and the acceleration obeys the equation F/L = m/L a. L 2π d The rms current over a short discharge time is I 0 / 2 .


28-28

Chapter 28

EXECUTE: (a) First get the force per unit length: 2

2

F µ0 I 2 µ ⎛ I ⎞ µ ⎛V ⎞ µ ⎛Q ⎞ = = 0 ⎜ 0 ⎟ = 0 ⎜ ⎟ = 0 ⎜ 0 ⎟ L 2π d 2π d ⎝ 2 ⎠ 4π d ⎝ R ⎠ 4π d ⎝ RC ⎠

2

2

F m µ ⎛Q ⎞ = a = λ a = 0 ⎜ 0 ⎟ . Solving for a gives L L 4π d ⎝ RC ⎠ μ0Q02 μ0Q02 a= . From the kinematics equation vx = v0 x + axt , we have v0 = at = aRC = 2 2 4πλ R C d 4πλ RCd 2 2 ⎛ μ0Q0 ⎞ 2 ⎜ ⎟ 2 4πλ RCd ⎠ 1 ⎛ μ0Q02 ⎞ v (b) Conservation of energy gives 12 mv02 = mgh and h = 0 = ⎝ = ⎜ ⎟. 2g 2g 2 g ⎝ 4πλ RCd ⎠

Now apply Newton’s second law using the result above:

28.86.

EVALUATE: Once the wires have swung apart, we would have to consider gravity in applying Newton’s second law. IDENTIFY: Approximate the moving belt as an infinite current sheet. SET UP: Problem 28.81 shows that B = 12 μ0 In for an infinite current sheet. Let L be the width of the sheet, so n = I/L. ΔQ Δx = L σ = Lvσ . EXECUTE: The amount of charge on a length Δx of the belt is ΔQ = LΔxσ , so I = Δt Δt μ0 I μ0vσ ! Approximating the belt as an infinite sheet B = . B is directed out of the page, as shown in Figure 28.86. = 2L 2 EVALUATE: The field is uniform above the sheet, for points close enough to the sheet for it to be considered infinite.

Figure 28.86 28.87.

28.88.

IDENTIFY: The rotating disk produces concentric rings of current. Calculate the field due to each ring and integrate over the surface of the disk to find the total field. μI SET UP: At the center of a circular ring carrying current I, B = 0 . 2r 2Qrdr . If the disk rotates at n turns per EXECUTE: The charge on a ring of radius r is q = σ A = σ 2π rdr = a2 dq 2Qnrdr μ I μ 2Qnrdr μ0 nQdr = ndq = . Therefore, dB = 0 = 0 second, then the current from that ring is dI = = . dt a2 2r 2r a 2 a2 a a μ nQdr μ nQ We integrate out from the center to the edge of the disk and find B = ∫ dB = ∫ 0 2 = 0 . 0 0 a a EVALUATE: The magnetic field is proportional to the total charge on the disk and to its rotation rate. IDENTIFY: There are two parts to the magnetic field: that from the half loop and that from the straight wire segment running from − a to a. SET UP: Apply Eq.(28.14). Let the φ be the angle that locates dl around the ring. μ0 Ia 2 EXECUTE: Bx (ring ) = 12 Bloop = − . 4( x 2 + a 2 )3/ 2 μI dl x μ Iax sin φ dφ dBy ( ring ) = dB sin θ sin φ = 0 sin φ = 0 2 and 4π ( x 2 + a 2 ) ( x 2 + a 2 )1 2 4π ( x + a 2 )3 2 π

π

0

0

By ( ring ) = ∫ dBy ( ring ) = ∫

By ( rod ) = Bx = −

π

μ0 Iax sinφ dφ μ0 Iax μ0 Iax = cosφ = − . 2 + a 2 )3/ 2 4π ( x 2 + a 2 )3/ 2 4π ( x 2 + a 2 )3/ 2 2 π ( x 0

μ0 Ia , using Eq. (28.8). The total field components are: 2πx(x 2 + a 2 )1/ 2

μ0 Ia 2 4( x + a ) 2

2 3/ 2

and By =

μ0 Ia

2π x( x + a ) 2

2 1/ 2

⎛ x2 ⎞ μ0 Ia 3 = . ⎜1 − 2 2 ⎟ 2 2 3/ 2 ⎝ x + a ⎠ 2π x( x + a )

2a μI μI Bx . By decreases faster than Bx as x increases. For very small x, Bx = − 0 and By = 0 . 4a 2π a π x In this limit Bx is the field at the center of curvature of a semicircle and By is the field of a long straight wire.

EVALUATE:

By = −


ELECTROMAGNETIC INDUCTION

29.1.

29.2.

29

IDENTIFY: Altering the orientation of a coil relative to a magnetic field changes the magnetic flux through the coil. This change then induces an emf in the coil. SET UP: The flux through a coil of N turns is Φ = NBA cos φ, and by Faraday’s law the magnitude of the induced emf is E = dΦ/dt. EXECUTE: (a) ΔΦ = NBA = (50)(1.20 T)(0.250 m)(0.300 m) = 4.50 Wb (b) E = dΦ/dt = (4.50 Wb)/(0.222 s) = 20.3 V EVALUATE: This induced potential is certainly large enough to be easily detectable. ΔΦ B . Φ B = BA cos φ . Φ B is the flux through each turn of the coil. IDENTIFY: E = Δt SET UP: EXECUTE:

φi = 0°. φf = 90°. (a) Φ B ,i = BA cos0° = (6.0 × 10−5 T)(12 × 10−4 m 2 )(1) = 7.2 × 10−8 Wb. The total flux through the coil is

N Φ B ,i = (200)(7.2 × 10−8 Wb) = 1.44 × 10−5 Wb . Φ B ,f = BA cos90° = 0 . N Φ i − N Φ f 1.44 × 10−5 Wb = = 3.6 × 10−4 V = 0.36 mV . Δt 0.040 s EVALUATE: The average induced emf depends on how rapidly the flux changes. IDENTIFY and SET UP: Use Faraday’s law to calculate the average induced emf and apply Ohm’s law to the coil to calculate the average induced current and charge that flows. ΔΦI B (a) EXECUTE: The magnitude of the average emf induced in the coil is Eav = N . Initially, Δt

(b) E = 29.3.

Φ Bf − Φ Bi NBA = . The average induced current is Δt Δt E NBA NBA ⎛ NBA ⎞ I = av = . The total charge that flows through the coil is Q = I Δt = ⎜ . ⎟ Δt = R R Δt R ⎝ R Δt ⎠ EVALUATE: The charge that flows is proportional to the magnetic field but does not depend on the time Δt . (b) The magnetic stripe consists of a pattern of magnetic fields. The pattern of charges that flow in the reader coil tell the card reader the magnetic field pattern and hence the digital information coded onto the card. (c) According to the result in part (a) the charge that flows depends only on the change in the magnetic flux and it does not depend on the rate at which this flux changes. IDENTIFY and SET UP: Apply the result derived in Exercise 29.3: Q = NBA / R. In the present exercise the flux Φ Bi = BA cos φ = BA. The final flux is zero, so Eav = N

29.4.

29.5.

changes from its maximum value of Φ B = BA to zero, so this equation applies. R is the total resistance so here R = 60.0 Ω + 45.0 Ω = 105.0 Ω. NBA QR (3.56 × 10−5 C)(105.0 Ω) EXECUTE: Q = says B = = = 0.0973 T. R NA 120(3.20 × 10−4 m 2 ) EVALUATE: A field of this magnitude is easily produced. IDENTIFY: Apply Faraday’s law. ! SET UP: Let +z be the positive direction for A . Therefore, the initial flux is positive and the final flux is zero. ! ΔΦ B 0 − (1.5 T)π (0.120 m) 2 =− = +34 V. Since E is positive and A is toward us, EXECUTE: (a) and (b) E = − −3 Δt 2.0 × 10 s the induced current is counterclockwise. EVALUATE: The shorter the removal time, the larger the average induced emf.

29-1


29-2

29.6.

Chapter 29

IDENTIFY: Apply Eq.(29.4). I = E/R. SET UP: d Φ B /dt = AdB/dt. EXECUTE:

(a) E =

Nd Φ B d d = NA ( B) = NA ( (0.012 T/s)t + (3.00 × 10−5 T/s 4 )t 4 ) dt dt dt

E = NA ( (0.012 T/s) + (1.2 × 10−4 T/s 4 )t 3 ) = 0.0302 V + (3.02 × 10−4 V/s3 )t 3.

E 0.0680 V = = 1.13 × 10−4 A. R 600 Ω EVALUATE: The rate of change of the flux is increasing in time, so the induced current is not constant but rather increases in time. IDENTIFY: Calculate the flux through the loop and apply Faraday’s law. SET UP: To find the total flux integrate dΦ B over the width of the loop. The magnetic field of a long straight ! μI wire, at distance r from the wire, is B = 0 . The direction of B is given by the right-hand rule. 2π r μ 0i EXECUTE: (a) When B = , into the page. 2π r μi (b) d Φ B = BdA = 0 Ldr. 2π r b μ iL b dr μ0iL (c) Φ B = ∫ d Φ B = 0 ∫ = ln(b/a ). a 2π a r 2π d Φ B μ0 L di (d) E = = ln( b a ) . dt dt 2π μ (0.240 m) ln(0.360/0.120)(9.60 A/s) = 5.06 × 10−7 V. (e) E = 0 2π EVALUATE: The induced emf is proportional to the rate at which the current in the long straight wire is changing IDENTIFY: Apply Faraday’s law. ! SET UP: Let A be upward in Figure 29.28 in the textbook. dΦB EXECUTE: (a) Eind = = d ( B⊥ A) dt dt (b) At t = 5.00 s, E = 0.0302 V + (3.02 × 10−4 V/s3 )(5.00 s)3 = 0.0680 V. I =

29.7.

29.8.

Eind = A sin 60°

(

)

−1 −1 dB d = A sin 60° (1.4 T)e − (0.057s )t = (π r 2 )(sin 60°)(1.4 T)(0.057 s −1 )e − (0.057s )t dt dt

Eind = π (0.75 m) 2 (sin 60°)(1.4 T)(0.057 s −1 )e − (0.057s

= (0.12 V) e− (0.057 s

−1 ) t

.

− (0.057 s−1 ) t

(b) E = E0 = (0.12 V). (0.12 V) = (0.12 V) e . ln(1/10) = −(0.057 s −1 )t and t = 40.4 s. ! ! (c) B is in the direction of A so Φ B is positive. B is getting weaker, so the magnitude of the flux is decreasing and d Φ B /dt < 0. Faraday’s law therefore says E > 0. Since E > 0, the induced current must flow counterclockwise as viewed from above. EVALUATE: The flux changes because the magnitude of the magnetic field is changing. IDENTIFY and SET UP: Use Faraday’s law to calculate the emf (magnitude and direction). The direction of the induced current is the same as the direction of the emf. The flux changes because the area of the loop is changing; relate dA/dt to dc/dt, where c is the circumference of the loop. (a) EXECUTE: c = 2π r and A = π r 2 so A = c 2 /4π Φ B = BA = ( B/4π )c 2 1 10

29.9.

−1 ) t

1 10

1 10

d Φ B ⎛ B ⎞ dc =⎜ ⎟c dt ⎝ 2π ⎠ dt At t = 9.0 s, c = 1.650 m − (9.0 s)(0.120 m/s) = 0.570 m E = (0.500 T)(1/2π )(0.570 m)(0.120 m/s) = 5.44 mV E =

(b) SET UP:

The loop and magnetic field are sketched in Figure 29.9. Take into the page to be the ! positive direction for A. Then the magnetic flux is positive. Figure 29.9


Electromagnetic Induction

29-3

The positive flux is decreasing in magnitude; d Φ B / dt is negative and E is positive. By the right! hand rule, for A into the page, positive E is clockwise. EVALUATE: Even though the circumference is changing at a constant rate, dA/dt is not constant and E is not EXECUTE:

29.10.

constant. Flux ⊗ is decreasing so the flux of the induced current is ⊗ and this means that I is clockwise, which checks. IDENTIFY: A change in magnetic flux through a coil induces an emf in the coil. SET UP: The flux through a coil is Φ = NBA cos φ and the induced emf is E = d Φ/dt. EXECUTE: (a) and (c) The magnetic flux is constant, so the induced emf is zero. (b) The area inside the field is changing. If we let x be the length (along the 30.0-cm side) in the field, then A = (0.400 m)x. ΦB = BA = (0.400 m)x E = dΦ/dt = B d[(0.400 m) x]/dt = B(0.400 m)dx/dt = B(0.400 m)v E = (1.25 T)(0.400 m)(0.0200 m/s) = 0.0100 V

29.11.

29.12.

29.13.

EVALUATE: It is not a large flux that induces an emf, but rather a large rate of change of the flux. The induced emf in part (b) is small enough to be ignored in many instances. IDENTIFY: A change in magnetic flux through a coil induces an emf in the coil. SET UP: The flux through a coil is Φ = NBA cos φ and the induced emf is E = dΦ/dt. EXECUTE: (a) E = dΦ/dt = d[A(B0 + bx)]/dt = bA dx/dt = bAv (b) clockwise (c) Same answers except the current is counterclockwise. EVALUATE: Even though the coil remains within the magnetic field, the flux through it increases because the strength of the field is increasing. IDENTIFY: Use the results of Example 29.5. 2 ⎛ 2π rad/rev ⎞ SET UP: Emax = NBAω. Eav = Emax . ω = (440 rev/min) ⎜ ⎟ = 46.1 rad/s. π ⎝ 60 s/min ⎠ EXECUTE: (a) Emax = NBAω = (150)(0.060 T)π (0.025 m) 2 (46.1 rad/s) = 0.814 V 2 2 (b) Eav = Emax = (0.815 V) = 0.519 V π π EVALUATE: In Emax = NBAω , ω must be in rad/s. IDENTIFY: Apply the results of Example 29.5. SET UP: Emax = NBAω

Emax 2.40 × 10−2 V = = 10.4 rad/s NBA (120)(0.0750 T)(0.016 m) 2 EVALUATE: We may also express ω as 99.3 rev/min or 1.66 rev/s . IDENTIFY: A change in magnetic flux through a coil induces an emf in the coil. SET UP: The flux through a coil is Φ = NBA cos φ and the induced emf is E = d Φ/dt. EXECUTE: The flux is constant in each case, so the induced emf is zero in all cases. EVALUATE: Even though the coil is moving within the magnetic field and has flux through it, this flux is not changing, so no emf is induced in the coil. IDENTIFY and SET UP: The field of the induced current is directed to oppose the change in flux. EXECUTE: (a) The field is into the page and is increasing so the flux is increasing. The field of the induced current is out of the page. To produce field out of the page the induced current is counterclockwise. (b) The field is into the page and is decreasing so the flux is decreasing. The field of the induced current is into the page. To produce field into the page the induced current is clockwise. (c) The field is constant so the flux is constant and there is no induced emf and no induced current. EVALUATE: The direction of the induced current depends on the direction of the external magnetic field and whether the flux due to this field is increasing or decreasing. IDENTIFY: By Lenz’s law, the induced current flows to oppose the flux change that caused it. SET UP and EXECUTE: The magnetic field is outward through the round coil and is decreasing, so the magnetic field due to the induced current must also point outward to oppose this decrease. Therefore the induced current is counterclockwise. EVALUATE: Careful! Lenz’s law does not say that the induced current flows to oppose the magnetic flux. Instead it says that the current flows to oppose the change in flux. IDENTIFY and SET UP: Apply Lenz's law, in the form that states that the flux of the induced current tends to oppose the change in flux. EXECUTE: (a) With the switch closed the magnetic field of coil A is to the right at the location of coil B. When the switch is opened the magnetic field of coil A goes away. Hence by Lenz's law the field of the current induced in coil B is to the right, to oppose the decrease in the flux in this direction. To produce magnetic field that is to the right the current in the circuit with coil B must flow through the resistor in the direction a to b. EXECUTE: ω =

29.14.

29.15.

29.16.

29.17.


29-4

29.18.

29.19.

29.20.

Chapter 29

(b) With the switch closed the magnetic field of coil A is to the right at the location of coil B. This field is stronger at points closer to coil A so when coil B is brought closer the flux through coil B increases. By Lenz's law the field of the induced current in coil B is to the left, to oppose the increase in flux to the right. To produce magnetic field that is to the left the current in the circuit with coil B must flow through the resistor in the direction b to a. (c) With the switch closed the magnetic field of coil A is to the right at the location of coil B. The current in the circuit that includes coil A increases when R is decreased and the magnetic field of coil A increases when the current through the coil increases. By Lenz's law the field of the induced current in coil B is to the left, to oppose the increase in flux to the right. To produce magnetic field that is to the left the current in the circuit with coil B must flow through the resistor in the direction b to a. EVALUATE: In parts (b) and (c) the change in the circuit causes the flux through circuit B to increase and in part (a) it causes the flux to decrease. Therefore, the direction of the induced current is the same in parts (b) and (c) and opposite in part (a). IDENTIFY: Apply Lenz’s law. SET UP: The field of the induced current is directed to oppose the change in flux in the primary circuit. EXECUTE: (a) The magnetic field in A is to the left and is increasing. The flux is increasing so the field due to the induced current in B is to the right. To produce magnetic field to the right, the induced current flows through R from right to left. (b) The magnetic field in A is to the right and is decreasing. The flux is decreasing so the field due to the induced current in B is to the right. To produce magnetic field to the right the induced current flows through R from right to left. (c) The magnetic field in A is to the right and is increasing. The flux is increasing so the field due to the induced current in B is to the left. To produce magnetic field to the left the induced current flows through R from left to right. EVALUATE: The direction of the induced current depends on the direction of the external magnetic field and whether the flux due to this field is increasing or decreasing. IDENTIFY and SET UP: Lenz's law requires that the flux of the induced current opposes the change in flux. EXECUTE: (a) Φ B is " and increasing so the flux Φ ind of the induced current is ⊗ and the induced current is clockwise. (b) The current reaches a constant value so Φ B is constant. d Φ B / dt = 0 and there is no induced current. (c) Φ B is " and decreasing, so Φ ind is " and current is counterclockwise. EVALUATE: Only a change in flux produces an induced current. The induced current is in one direction when the current in the outer ring is increasing and is in the opposite direction when that current is decreasing. IDENTIFY: Use the results of Example 29.6. Use the three approaches specified in the problem for determining the direction of the induced current. I = E/R . ! SET UP: Let A be directed into the figure, so a clockwise emf is positive. EXECUTE: (a) E = vBl = (5.0 m/s)(0.750 T)(1.50 m) = 5.6 V (b) (i) Let q be a positive charge in the moving bar, as shown in Figure 29.20a. The magnetic force on this charge is ! ! ! F = qv × B , which points upward. This force pushes the current in a counterclockwise direction through the circuit.

(ii) Φ B is positive and is increasing in magnitude, so d Φ B / dt > 0. Then by Faraday’s law E < 0 and the emf and induced current are counterclockwise. (iii) The flux through the circuit is increasing, so the induced current must cause a magnetic field out of the paper to oppose this increase. Hence this current must flow in a counterclockwise sense, as shown in Figure 29.20b. E 5.6 V (c) E = RI . I = = = 0.22 A. R 25 Ω EVALUATE: All three methods agree on the direction of the induced current.

Figure 29.20 29.21.

IDENTIFY: A conductor moving in a magnetic field may have a potential difference induced across it, depending on how it is moving. SET UP: The induced emf is E = vBL sin φ, where φ is the angle between the velocity and the magnetic field.


Electromagnetic Induction

29.22.

29.23.

29.24.

29-5

EXECUTE: (a) E = vBL sin φ = (5.00 m/s)(0.450 T)(0.300 m)(sin 90°) = 0.675 V (b) The positive charges are moved to end b, so b is at the higher potential. ! (c) E = V/L = (0.675 V)/(0.300 m) = 2.25 V/m. The direction of E is from, b to a. (d) The positive charge are pushed to b, so b has an excess of positive charge. (e) (i) If the rod has no appreciable thickness, L = 0, so the emf is zero. (ii) The emf is zero because no magnetic force acts on the charges in the rod since it moves parallel to the magnetic field. EVALUATE: The motional emf is large enough to have noticeable effects in some cases. IDENTIFY: The moving bar has a motional emf induced across its ends, so it causes a current to flow. SET UP: The induced potential is E = vBL and Ohm’s law is E = IR. EXECUTE: (a) E = vBL = (5.0 m/s)(0.750 T)(1.50 m) = 5.6 V (b) I = E /R = (5.6 V)/(25 Ω) = 0.23 A EVALUATE: Both the induced potential and the current are large enough to have noticeable effects. IDENTIFY: E = vBL SET UP: L = 5.00 × 10−2 m. 1 mph = 0.4470 m/s. E 1.50 V = = 46.2 m/s = 103 mph. EXECUTE: v = BL (0.650 T)(5.00 × 10−2 m) EVALUATE: This is a large speed and not practical. It is also difficult to produce a 5.00 cm wide region of 0.650 T magnetic field. IDENTIFY: E = vBL. SET UP: 1 mph = 0.4470 m/s . 1 G = 10−4 T . EXECUTE:

⎛ 0.4470 m/s ⎞ −4 (a) E = (180 mph) ⎜ ⎟ (0.50 × 10 T)(1.5 m) = 6.0 mV. This is much too small to be 1 mph ⎝ ⎠

noticeable.

29.25.

29.26.

⎛ 0.4470 m/s ⎞ −4 (b) E = (565 mph) ⎜ ⎟ (0.50 × 10 T)(64.4 m) = 0.813 mV. This is too small to be noticeable. 1 mph ⎝ ⎠ EVALUATE: Even though the speeds and values of L are large, the earth’s field is small and motional emfs due to the earth’s field are not important in these situations. IDENTIFY and SET UP: E = vBL. Use Lenz's law to determine the direction of the induced current. The force Fext required to maintain constant speed is equal and opposite to the force FI that the magnetic field exerts on the rod because of the current in the rod. EXECUTE: (a) E = vBL = (7.50 m/s)(0.800 T)(0.500 m) = 3.00 V ! (b) B is into the page. The flux increases as the bar moves to the right, so the magnetic field of the induced current is out of the page inside the circuit. To produce magnetic field in this direction the induced current must be counterclockwise, so from b to a in the rod. ! E 3.00 V (c) I = = = 2.00 A. FI = ILB sin φ = (2.00 A)(0.500 m)(0.800 T)sin 90° = 0.800 N . FI is to the left. To R 1.50 Ω keep the bar moving to the right at constant speed an external force with magnitude Fext = 0.800 N and directed to the right must be applied to the bar. (d) The rate at which work is done by the force Fext is Fext v = (0.800 N)(7.50 m/s) = 6.00 W. The rate at which thermal energy is developed in the circuit is I 2 R = (2.00 A)(1.50 Ω) = 6.00 W. These two rates are equal, as is required by conservation of energy. EVALUATE: The force on the rod due to the induced current is directed to oppose the motion of the rod. This agrees with Lenz’s law. IDENTIFY: Use Faraday’s law to calculate the induced emf. Ohm’s law applied to the loop gives I. Use Eq.(27.19) to calculate the force exerted on each side of the loop. SET UP: The loop before it starts to enter the magnetic field region is sketched in Figure 29.26a. EXECUTE: For x < −3L/2 or x > 3L/2 the loop is completely outside the field dΦB region. Φ B = 0, and = 0. dt Figure 29.26a

Thus E = 0 and I = 0, so there is no force from the magnetic field and the external force F necessary to maintain constant velocity is zero.


29-6

Chapter 29

SET UP:

The loop when it is completely inside the field region is sketched in Figure 29.26b. EXECUTE: For − L/2 < x < L/2 the loop is completely inside the field region and Φ B = BL2 . Figure 29.26b

! ! ! dΦB = 0 so E = 0 and I = 0. There is no force F = Il × B from the magnetic field and the external force F dt necessary to maintain constant velocity is zero. SET UP: The loop as it enters the magnetic field region is sketched in Figure 29.26c.

But

EXECUTE: For −3L/2 < x < − L/2 the loop is entering the field region. Let x′ be the length of the loop that is within the field. Figure 29.26c dΦB dΦB = Blv. The magnitude of the induced emf is E = = BLv and the induced dt dt ! E BLv = current is I = . Direction of I: Let A be directed into the plane of the figure. Then Φ B is positive. The R R dΦB flux is positive and increasing in magnitude, so is positive. Then by Faraday’s law E is negative, and with dt ! our choice for direction of A a negative E is counterclockwise. The current induced in the loop is counterclockwise. SET UP: The induced current and magnetic force on the loop are shown in Figure 29.26d, for the situation where the loop is entering the field. ! ! ! EXECUTE: FI = Il × B gives that the ! force FI exerted on the loop by the magnetic field is to the left and has B 2 L2v ⎛ BLv ⎞ magnitude FI = ILB = ⎜ = . LB ⎟ R ⎝ R ⎠ Figure 29.26d ! The external force F needed to move the loop at constant speed is equal in magnitude and opposite in direction to ! FI so is to the right and has this same magnitude. SET UP: The loop as it leaves the magnetic field region is sketched in Figure 29.26e.

Then Φ B = BLx′ and

EXECUTE: For L/2 < x < 3L/2 the loop is leaving the field region. Let x′ be the length of the loop that is outside the field. Figure 29.26e

dΦB dΦB = BLv. The magnitude of the induced emf is E = = BLv and the induced dt dt ! E BLv current is I = = . Direction of I: Again let A be directed into the plane of the figure. Then Φ B is positive R R dΦB and decreasing in magnitude, so is negative. Then by Faraday’s law E is positive, and with our choice for dt ! direction of A a positive E is clockwise. The current induced in the loop is clockwise. Then Φ B = BL( L − x′) and


Electromagnetic Induction

29-7

SET UP: The induced current and magnetic force on the loop are shown in Figure 29.26f, for the situation where the loop is leaving the field. ! ! ! EXECUTE: FI = Il × B gives that the ! force FI exerted on the loop by the magnetic field is to the left and has B 2 L2v ⎛ BLv ⎞ magnitude FI = ILB = ⎜ . ⎟ LB = R ⎝ R ⎠ Figure 29.26f ! The ! external force F needed to move the loop at constant speed is equal in magnitude and opposite in direction to FI so is to the right and has this same magnitude. (a) The graph of F versus x is given in Figure 29.26g.

Figure 29.26g (b) The graph of the induced current I versus x is given in Figure 29.26h.

Figure 29.26h

29.27.

29.28.

EVALUATE: When the loop is either totally outside or totally inside the magnetic field region the flux isn’t changing, there is no induced current, and no external force is needed for the loop to maintain constant speed. When the loop is entering the field the external force required is directed so as to pull the loop in and when the loop is leaving the field the external force required is directed so as to pull the loop out of the field. These directions agree with Lenz’s law: the force on the induced current (opposite in direction to the required external force) is directed so as to oppose the loop entering or leaving the field. IDENTIFY: A bar moving in a magnetic field has an emf induced across its ends. SET UP: The induced potential is E = vBL sin φ. EXECUTE: Note that φ = 90° in all these cases because the bar moved perpendicular to the magnetic field. But the effective length of the bar, L sin θ, is different in each case. (a) E = vBL sin θ = (2.50 m/s)(1.20 T)(1.41 m) sin (37.0°) = 2.55 V, with a at the higher potential because positive charges are pushed toward that end. (b) Same as (a) except θ = 53.0°, giving 3.38 V, with a at the higher potential. (c) Zero, since the velocity is parallel to the magnetic field. (d) The bar must move perpendicular to its length, for which the emf is 4.23 V. For Vb > Va, it must move upward and to the left (toward the second quadrant) perpendicular to its length. EVALUATE: The orientation of the bar affects the potential induced across its ends. IDENTIFY: Use Eq.(29.10) to calculate the induced electric field E at a distance r from the center of the solenoid. Away from the ends of the solenoid, B = μ0nI inside and B = 0 outside. (a) SET UP: The end view of the solenoid is sketched in Figure 29.28.

Let R be the radius of the solenoid.

Figure 29.28 ! ! dΦB Apply úE ⋅ dl = − to an integration path that is a circle of radius r, where r < R. We need to calculate just the dt magnitude of E so we can take absolute values.


29-8

Chapter 29

EXECUTE:

!

Φ B = Bπ r 2 , − !

!

úE ⋅ dl E = 12 r

= −

!

úE ⋅ dl

= E (2π r )

dΦB dB = π r2 dt dt

dΦB dB implies E (2π r ) = π r 2 dt dt

dB dt

dB dI = μ0n dt dt dI Thus E = 12 r μ 0 n = 12 (0.00500 m)(4π × 10 −7 T ⋅ m/A)(900 m −1 )(60.0 A/s) = 1.70 × 10 −4 V/m dt (b) r = 0.0100 cm is still inside the solenoid so the expression in part (a) applies. dI E = 12 r μ 0 n = 12 (0.0100 m)(4π × 10−7 T ⋅ m/A)(900 m −1 )(60.0 A/s) = 3.39 × 10 −4 V/m dt EVALUATE: Inside the solenoid E is proportional to r, so E doubles when r doubles. IDENTIFY: Apply Eqs.(29.9) and (29.10). SET UP: Evaluate the integral if Eq.(29.10) for a path which is a circle of radius r and concentric with the solenoid. The magnetic field of the solenoid is confined to the region inside the solenoid, so B(r ) = 0 for r > R B = μ 0 nI , so

29.29.

EXECUTE: (b) E =

(a)

dΦB dB dB . =A = π r12 dt dt dt

! 1 d Φ B π r12 dB r1 dB = = . The direction of E is shown in Figure 29.29a. 2π r1 dt 2π r1 dt 2 dt

(c) All the flux is within r < R, so outside the solenoid E =

1 d Φ B π R 2 dB R 2 dB = = . 2π r2 dt 2π r2 dt 2r2 dt

(d) The graph is sketched in Figure 29.29b. dΦB dB π R 2 dB (e) At r = R 2, E = = π ( R/2) 2 = . dt dt 4 dt dΦB dB (f) At r = R , E = . = π R2 dt dt dΦB dB (g) At r = 2 R , E = . = π R2 dt dt EVALUATE: The emf is independent of the distance from the center of the cylinder at all points outside it. Even though the magnetic field is zero for r > R , the induced electric field is nonzero outside the solenoid and a nonzero emf is induced in a circular turn that has r > R.

Figure 29.29 29.30.

IDENTIFY: Use Eq.(29.10) to calculate the induced electric field E and use this E in Eq.(29.9) to calculate E between two points. (a) SET UP: Because of the axial symmetry and the absence of any electric charge, the field lines are concentric circles.


Electromagnetic Induction

29-9

(b) See Figure 29.30.

! E is tangent to the ring. The direction ! of E (clockwise or counterclockwise) is the direction in which current will be induced in the ring. Figure 29.30

! Use the sign convention for Faraday’s law to deduce this direction. Let A be into the paper. Then dΦB dΦB Φ B is positive. B decreasing then means is negative, so by E = − , E is positive and therefore dt dt ! ! ! dΦB clockwise. Thus E is clockwise around the ring. To calculate E apply úE ⋅ dl = − to a circular path that dt coincides with the ring. ! ! úE ⋅ dl = E (2π r ) EXECUTE:

Φ B = Bπ r 2 ;

dΦB dB = π r2 dt dt

dB dB 1 and E = 12 r = 2 (0.100 m)(0.0350 T/s) = 1.75 × 10−3 V/m dt dt ! ! (c) The induced emf has magnitude E = úE ⋅ dl = E (2π r ) = (1.75 × 10−3 V/m)(2π )(0.100 m) = 1.100 × 10−3 V. Then E (2π r ) = π r 2

E 1.100 × 10−3 V = = 2.75 × 10−4 A. R 4.00 Ω (d) Points a and b are separated by a distance around the ring of π r so E = E (π r ) = (1.75 × 10−3 V/m)(π )(0.100 m) = 5.50 × 10−4 V I=

29.31.

(e) The ends are separated by a distance around the ring of 2π r so E = 1.10 × 10−3 V as calculated in part (c). EVALUATE: The induced emf, calculated from Faraday’s law and used to calculate the induced current, is associated with the induced electric field integrated around the total circumference of the ring. IDENTIFY: Apply Eq.(29.1) with Φ B = μ0 niA .

A = π r 2 , where r = 0.0110 m . In Eq.(29.11), r = 0.0350 m . dΦB d d di di E 2π r EXECUTE: E = and E = E (2π r ). Therefore, = . = ( BA) = ( μ0 niA) = μ0 nA dt μ0 nA dt dt dt dt SET UP:

29.32.

di (8.00 × 10−6 V/m)2π (0.0350 m) = = 9.21 A/s. μ0 (400 m −1 )π (0.0110 m) 2 dt EVALUATE: Outside the solenoid the induced electric field decreases with increasing distance from the axis of the solenoid. IDENTIFY: A changing magnetic flux through a coil induces an emf in that coil, which means that an electric field is induced in the material of the coil. ! ! dΦB SET UP: According to Faraday’s law, the induced electric field obeys the equation úE ⋅ dl = − . dt EXECUTE: (a) For the magnitude of the induced electric field, Faraday’s law gives E2πr = d(Bπr2)/dt = πr2 dB/dt

r dB 0.0225 m = (0.250 T/s) = 2.81 × 10−3 V/m 2 dt 2 (b) The field points toward the south pole of the magnet and is decreasing, so the induced current is counterclockwise. EVALUATE: This is a very small electric field compared to most others found in laboratory equipment. ΔΦ B IDENTIFY: Apply Faraday’s law in the form Eav = N . Δt E=

29.33.

SET UP:

The magnetic field of a large straight solenoid is B = μ0 nI inside the solenoid and zero outside.

Φ B = BA , where A is 8.00 cm 2 , the cross-sectional area of the long straight solenoid.


29-10

Chapter 29

Eav = N

EXECUTE:

ΔΦ B NA( Bf − Bi ) NAμo nI = = . Δt Δt Δt

μ0 (12)(8.00 × 10−4 m 2 )(9000 m −1 )(0.350 A)

29.34.

= 9.50 × 10−4 V. 0.0400 s EVALUATE: An emf is induced in the second winding even though the magnetic field of the solenoid is zero at the location of the second winding. The changing magnetic field induces an electric field outside the solenoid and that induced electric field produces the emf. IDENTIFY: Apply Eq.(29.14). SET UP: P = 3.5 × 10−11 F/m dΦE EXECUTE: iD = P = (3.5 × 10−11 F/m)(24.0 × 103 V ⋅ m/s3 )t 2 . iD = 21 × 10−6 A gives t = 5.0 s. dt EVALUATE: iD depends on the rate at which Φ E is changing.

29.35.

IDENTIFY:

Eav =

SET UP:

Apply Eq.(29.14), where P = K P0 .

d Φ E / dt = 4(8.76 × 103 V ⋅ m/s 4 )t 3 . P0 = 8.854 × 10−12 F/m.

EXECUTE:

P=

iD 12.9 × 10−12 A = = 2.07 × 10−11 F/m. The dielectric constant is 3 ( d Φ E / dt ) 4(8.76 × 10 V ⋅ m/s 4 )(26.1× 10−3 s)3

K = P = 2.34. P0 EVALUATE: 29.36.

The larger the dielectric constant, the larger is the displacement current for a given d Φ E /dt.

IDENTIFY and SET UP: Eqs.(29.13) and (29.14) show that iC = iD and also relate iD to the rate of change of the electric field flux between the plates. Use this to calculate dE / dt and apply the generalized form of Ampere’s law (Eq.29.15) to calculate B. i i 0.280 A 0.280 A (a) EXECUTE: iC = iD , so jD = D = C = = = 55.7 A/m 2 A A π r2 π (0.0400 m)2

dE dE jD 55.7 A/m 2 = = = 6.29 × 1012 V/m ⋅ s so dt dt P0 8.854 × 10−12 C2 / N ⋅ m 2 ! ! (c) SET UP: Apply Ampere’s law úB ⋅ dl = μ0 (iC + iD )encl (Eq.(28.20)) to a circular path with radius r = 0.0200 m. (b) jD = P0

An end view of the solenoid is given in Figure 29.36. By symmetry the magnetic field is tangent to the path and constant around it.

EXECUTE:

Figure 29.36 ! ! Thus úB ⋅ dl = úBdl = B ∫ dl = B (2π r ).

iC = 0 (no conduction current flows through the air space between the plates) The displacement current enclosed by the path is jDπ r 2 . Thus B (2π r ) = μ0 ( jDπ r 2 ) and B = 12 μ0 jD r = 12 (4π × 10−7 T ⋅ m/A)(55.7 A/m 2 )(0.0200 m) = 7.00 × 10−7 T (d) B = 12 μ0 jD r. Now r is

29.37.

1 2

the value in (c), so B is

1 2

also: B = 12 (7.00 × 10−7 T) = 3.50 × 10−7 T

EVALUATE: The definition of displacement current allows the current to be continuous at the capacitor. The magnetic field between the plates is zero on the axis (r = 0) and increases as r increases. PA E IDENTIFY: q = CV . For a parallel-plate capacitor, C = , where P = K P0 . iC = dq/dt. jD = P . dt d SET UP: E = q/PA so dE/dt = iC /PA. −4

EXECUTE:

(4.70)P0 (3.00 × 10 m )(120 V) ⎛ PA ⎞ (a) q = CV = ⎜ ⎟V = = 5.99 × 10−10 C. 2.50 × 10−3 m ⎝ d ⎠ 2


Electromagnetic Induction

29-11

dq = iC = 6.00 × 10−3 A. dt dE i i (c) jD = P = K P0 C = C = jC , so iD = iC = 6.00 × 10−3 A. dt K P0 A A

(b)

iD = iC , so Kirchhoff’s junction rule is satisfied where the wire connects to each capacitor plate.

EVALUATE: 29.38.

IDENTIFY and SET UP: Use iC = q / t to calculate the charge q that the current has carried to the plates in time t. The two equations preceeding Eq.(24.2) relate q to the electric field E and the potential difference between the plates. The displacement current density is defined by Eq.(29.16). EXECUTE: (a) iC = 1.80 × 10−3 A q = 0 at t = 0 The amount of charge brought to the plates by the charging current in time t is q = iCt = (1.80 × 10−3 A)(0.500 × 10−6 s) = 9.00 × 10 −10 C

E=

σ P0

=

q 9.00 × 10−10 C = = 2.03 × 105 V/m −12 P0 A (8.854 × 10 C 2 / N ⋅ m 2 )(5.00 × 10−4 m 2 )

V = Ed = (2.03 × 105 V/m)(2.00 × 10−3 m) = 406 V (b) E = q / P0 A

dE dq / dt iC 1.80 × 10−3 A = = = = 4.07 × 1011 V/m ⋅ s −12 dt P0 A P0 A (8.854 × 10 C 2 / N ⋅ m 2 )(5.00 × 10−4 m 2 ) Since iC is constant dE/dt does not vary in time. dE (Eq.(29.16)), with P replaced by P0 since there is vacuum between the plates.) dt jD = (8.854 × 10−12 C2 / N ⋅ m 2 )(4.07 × 1011 V/m ⋅ s) = 3.60 A/m 2

(c) jD = P0

iD = jD A = (3.60 A/m 2 )(5.00 × 10−4 m 2 ) = 1.80 × 10−3 A; iD = iC EVALUATE:

iC = iD . The constant conduction current means the charge q on the plates and the electric field

between them both increase linearly with time and iD is constant. 29.39.

IDENTIFY:

Ohm’s law relates the current in the wire to the electric field in the wire. jD = P

dE . Use Eq.(29.15) to dt

calculate the magnetic fields. SET UP: Ohm’s law says E = ρ J . Apply Ohm’s law to a circular path of radius r.

ρI

(2.0 × 10−8 Ω ⋅ m)(16 A) = 0.15 V/m. 2.1 × 10−6 m 2 A dE d ⎛ ρI ⎞ ρ dI 2.0 × 10−8 Ω ⋅ m (4000 A/s) = 38 V/m ⋅ s. = ⎜ ⎟= = (b) 2.1 × 10−6 m 2 dt dt ⎝ A ⎠ A dt dE (c) jD = P0 = P0 (38 V/m ⋅ s) = 3.4 × 10−10 A/m 2 . dt (d) iD = jD A = (3.4 × 10−10 A/m 2 )(2.1 × 10−6 m 2 ) = 7.14 × 10−16 A. Eq.(29.15) applied to a circular path of radius r EXECUTE:

(a) E = ρ J =

=

μ0 I D μ0 (7.14 × 10−16 A) = = 2.38 × 10−21 T, and this is a negligible contribution. 2π r 2π (0.060 m) μI μ (16 A) BC = 0 C = 0 = 5.33 × 10−5 T. 2π r 2π (0.060 m)

gives BD =

29.40.

EVALUATE: In this situation the displacement current is much less than the conduction current. IDENTIFY: Apply Ampere's law to a circular path of radius r < R, where R is the radius of the wire. SET UP: The path is shown in Figure 29.40. !

!

ú B ⋅ dl = μ Figure 29.40

0

dΦE ⎞ ⎛ ⎜ I C + P0 ⎟ dt ⎠ ⎝


29-12

Chapter 29

EXECUTE:

There is no displacement current, so

!

!

úB ⋅ dl = μ I

0 C

The magnetic field inside the superconducting material is zero, so

29.41.

29.42.

!

!

úB ⋅ dl = 0. But then Ampere’s law says that

I C = 0; there can be no conduction current through the path. This same argument applies to any circular path with r < R, so all the current must be at the surface of the wire. EVALUATE: If the current were uniformly spread over the wire’s cross section, the magnetic field would be like that calculated in Example 28.9. IDENTIFY: A superconducting region has zero resistance. SET UP: If the superconducting and normal regions each lie along the length of the cylinder, they provide parallel conducting paths. EXECUTE: Unless some of the regions with resistance completely fill a cross-sectional area of a long type-II superconducting wire, there will still be no total resistance. The regions of no resistance provide the path for the current. EVALUATE: The situation here is like two resistors in parallel, where one has zero resistance and the other is nonzero. The equivalent resistance is!zero. ! ! IDENTIFY: Apply Eq.(28.29): B = B0 + μ0 M . SET UP: For magnetic fields less than the critical field, there is no internal magnetic field. For fields greater than ! ! the critical field, B is very nearly equal to B0 . ! EXECUTE: (a) The external field is less than the critical field, so inside the superconductor B = 0 and ! ! ! ! ! B (0.130 T ) iˆ M=− 0 =− = −(1.03 × 105 A/m) iˆ. Outside the superconductor, B = B0 = (0.130 T )iˆ and M = 0. μ0 μ0 ! ! (b) The field is greater than the critical field and B = B = (0.260 T)iˆ, both inside and outside the superconductor. 0

29.43.

29.44.

EVALUATE: Below!the !critical field the external field is expelled from the superconducting material. ! IDENTIFY: Apply B = B0 + μ0 M . ! SET UP: When the magnetic flux is expelled from the material the magnetic field B in the material is zero. When the material is completely normal,!the magnetization is close to zero. ! EXECUTE: (a) When B0 is just under Bc1 (threshold of superconducting phase), the magnetic field in the ! ! (55 × 10 −3 T)iˆ Bc1 material must be zero, and M = − =− = −(4.38 × 10 4 A/m) iˆ. μ0 μ0 ! ! ! ! (b) When B0 is just over Bc2 (threshold of normal phase), there is zero magnetization, and B = Bc2 = (15.0 T)iˆ. EVALUATE: Between Bc1 and Bc2 there are filaments of normal phase material and there is magnetic field along these filaments. IDENTIFY and SET UP: Use Faraday’s law to calculate the magnitude of the induced emf and Lenz’s law to determine its direction. Apply Ohm’s law to calculate I. Use Eq.(25.10) to calculate the resistance of the coil. ! (a) EXECUTE: The angle φ between the normal to the coil and the direction of B is 30.0°. dΦB E = = ( N π r 2 )( dB / dt ) and I = E / R. dt For t < 0 and t > 1.00 s, dB/dt = 0 E = 0 and I = 0.

For 0 ≤ t ≤ 1.00 s, dB/dt = (0.120 T) π sin π t E = ( Nπ r 2 )π (0.120 T)sin π t = (0.9475 V)sin π t

ρL = ; ρ = 1.72 × 10 −8 Ω ⋅ m, r = 0.0150 × 10 −3 m A π r2 L = Nc = N 2π r = (500)(2π )(0.0400 m) = 125.7 m Rw = 3058 Ω and the total resistance of the circuit is R = 3058 Ω + 600 Ω = 3658 Ω I = E / R = (0.259 mA)sin π t. The graph of I versus t is sketched in Figure 29.44a.

R for wire: Rw =

ρL

Figure 29.44a


Electromagnetic Induction

29-13

(b) The coil and the magnetic field are shown in Figure 29.44b.

B increasing so Φ B is " and increasing. Φ B is ⊗ so I is clockwise Figure 29.44b

29.45.

EVALUATE: The long length of small diameter wire used to make the coil has a rather large resistance, larger than the resistance of the 600-Ω resistor connected to it in the circuit. The flux has a cosine time dependence so the rate of change of flux and the current have a sine time dependence. There is no induced current for t < 0 or t > 1.00 s. IDENTIFY: Apply Faraday’s law and Lenz’s law. V SET UP: For a discharging RC circuit, i (t ) = 0 e − t / RC , where V0 is the initial voltage across the capacitor. The R resistance of the small loop is (25)(0.600 m)(1.0 Ω/m) = 15.0 Ω . EXECUTE:

(a) The large circuit is an RC circuit with a time constant of τ = RC = (10 Ω)(20 ×10−6 F) = 200 μs. Thus,

the current as a function of time is i = ( (100 V) /(10 Ω) ) e− t / 200 μ s . At t = 200 μs, we obtain i = (10 A)(e−1 ) = 3.7 A. (b) Assuming that only the long wire nearest the small loop produces an appreciable magnetic flux through the c + a μ ib μ ib ⎛ a ⎞ 0 dr = 0 ln ⎜1 + ⎟. Therefore, small loop and referring to the solution of Exercise 29.7 we obtain Φ B = ∫ c 2π r 2π ⎝ c ⎠

the emf induced in the small loop at t = 200 μs is E = −

dΦ μ b ⎛ a ⎞ di = − 0 ln ⎜1 + ⎟ . 2π ⎝ c ⎠ dt dt

(4π × 10−7 Wb/A ⋅ m 2 )(0.200 m) 3.7 A ⎞ ⎛ ln(3.0) ⎜ − = +0.81 mV. Thus, the induced current in the small −6 ⎟ 2π ⎝ 200 × 10 s ⎠ loop is i′ = E = 0.81 mV = 54μ A. R 15.0 Ω (c) The magnetic field from the large loop is directed out of the page within the small loop.The induced current will act to oppose the decrease in flux from the large loop. Thus, the induced current flows counterclockwise. EVALUATE: (d) Three of the wires in the large loop are too far away to make a significant contribution to the flux in the small loop–as can be seen by comparing the distance c to the dimensions of the large loop. IDENTIFY: A changing magnetic field causes a changing flux through a coil and therefore induces an emf in the coil. dΦB SET UP: Faraday’s law says that the induced emf is E = − and the magnetic flux through a coil is defined dt as Φ B = BA cos φ . E=−

29.46.

EXECUTE: In this case, Φ B = BA, where A is constant. So the emf is proportional to the negative slope of the magnetic field. The result is shown in Figure 29.46. EVALUATE: It is the rate at which the magnetic field is changing, not the field’s magnitude, that determines the induced emf. When the field is constant, even though it may have a large value, the induced emf is zero.

Figure 29.46 29.47.

IDENTIFY: Follow the steps specified in the problem. SET UP: Let the flux through the loop due to the current be positive. μi μ iπ a EXECUTE: (a) Φ B = BA = 0 π a 2 = 0 . 2a 2


29-14

Chapter 29

(b) E = −

dΦB d ⎛ μ iπ a ⎞ μ0π a di di 2R = iR ⇒ − ⎜ 0 = iR ⇒ = −i ⎟=− dt dt ⎝ 2 ⎠ dt 2 dt μ0π a

(c) Solving

di 2R = − dt for i (t ) yields i (t ) = i0e −t (2 R/μ0π a ) . i μ0π a

(d) We want i (t ) = i0 (0.010) = i0e− t (2 R/μ0π a ) , so ln(0.010) = −t (2 R/μ0π a ) and

29.48.

μ0π a

μ0π (0.50 m)

ln(0.010) = 4.55 × 10−5 s. 2(0.10 Ω) EVALUATE: (e) We can ignore the self-induced currents because it takes only a very short time for them to die out. IDENTIFY: A changing magnetic field causes a changing flux through a coil and therefore induces an emf in the coil. dΦB SET UP: Faraday’s law says that the induced emf is E = − and the magnetic flux through a coil is defined dt as Φ B = BA cos φ . t=−

2R

ln(0.010) = −

EXECUTE: In this case, Φ B = BA, where A is constant. So the emf is proportional to the negative slope of the magnetic field. The result is shown in Figure 29.48. EVALUATE: It is the rate at which the magnetic field is changing, not the field’s magnitude, that determines the induced emf. When the field is constant, even though it may have a large value, the induced emf is zero.

Figure 29.48 29.49.

dΦB (a) IDENTIFY: (i) E = . The flux is changing because the magnitude of the magnetic field of the wire decreases dt

with distance from the wire. Find the flux through a narrow strip of area and integrate over the loop to find the total flux. SET UP:

Consider a narrow strip of width dx and a distance x from the long wire, as shown in Figure 29.49a. The magnetic field of the wire at the strip is B = μ0 I/2π x. The flux through the strip is d Φ B = Bb dx = ( μ0 Ib/2π )(dx/x)

Figure 29.49a

⎛ μ Ib ⎞ r + a dx The total flux through the loop is Φ B = ∫ dΦ B = ⎜ 0 ⎟ ∫ ⎝ 2π ⎠ r x ⎛ μ Ib ⎞ ⎛ r + a ⎞ Φ B = ⎜ 0 ⎟ ln ⎜ ⎟ ⎝ 2π ⎠ ⎝ r ⎠ ⎞ d Φ B d Φ B dr μ0 Ib ⎛ a = = ⎜− ⎟v 2π ⎜⎝ r ( r + a ) ⎟⎠ dt dt dt μ0 Iabv E = 2π r ( r + a )

EXECUTE:

(ii) IDENTIFY: E = Bvl for a bar of length l moving at speed v perpendicular to a magnetic field B. Calculate the induced emf in each side of the loop, and combine the emfs according to their polarity.


Electromagnetic Induction

SET UP:

29-15

The four segments of the loop are shown in Figure 29.49b.

The emf in each side ⎛μI⎞ of the loop is E1 = ⎜ 0 ⎟ vb, ⎝ 2π r ⎠ EXECUTE:

⎛ ⎞ μ0 I E2 = ⎜ ⎟ vb, E2 = E4 = 0 + 2 r ( r a ) π ⎝ ⎠

Figure 29.49b

Both emfs E1 and E2 are directed toward the top of the loop so oppose each other. The net emf is E = E1 − E2 =

μ0 Ivb ⎛ 1 1 ⎞ μ0 Iabv ⎜ − ⎟= 2π ⎝ r r + a ⎠ 2π r (r + a )

This expression agrees with what was obtained in (i) using Faraday’s law. (b) (i) IDENTIFY and SET UP: The flux of the induced current opposes the change in flux. ! EXECUTE: B is ⊗ . Φ B is ⊗ and decreasing, so the flux Φ ind of the induced current is ⊗ and the current is clockwise. (ii) IDENTIFY and SET UP: Use the right-hand rule to find the force on the positive charges in each side of the loop. The forces on positive charges in segments 1 and 2 of the loop are shown in Figure 29.49c.

Figure 29.49c

29.50.

29.51.

EXECUTE: B is larger at segment 1 since it is closer to the long wire, so FB is larger in segment 1 and the induced current in the loop is clockwise. This agrees with the direction deduced in (i) using Lenz’s law. (c) EVALUATE: When v = 0 the induced emf should be zero; the expression in part (a) gives this. When a → 0 the flux goes to zero and the emf should approach zero; the expression in part (a) gives this. When r → ∞ the magnetic field through the loop goes to zero and the emf should go to zero; the expression in part (a) gives this. IDENTIFY: Apply Faraday’s law. SET UP: For rotation about the y-axis the situation is the same as in Examples 29.4 and 29.5 and we can apply the results from those examples. EXECUTE: (a) Rotating about the y-axis: the flux is given by Φ B = BA cosφ and dΦB Emax = = ω BA = (35.0 rad/s)(0.450 T)(6.00 × 10−2 m) = 0.945 V. dt dΦB (b) Rotating about the x-axis: = 0 and E = 0. dt (c) Rotating about the z-axis: the flux is given by Φ B = BA cosφ and dΦB Emax = = ω BA = (35.0 rad/s)(0.450 T)(6.00 × 10−2 m) = 0.945 V. dt EVALUATE: The maximum emf is the same if the loop is rotated about an edge parallel to the z-axis as it is when it is rotated about the z-axis. IDENTIFY: Apply the results of Example 29.4, so Emax = Nω BA for N loops. SET UP: For the minimum ω , let the rotating loop have an area equal to the area of the uniform magnetic field, so A = (0.100 m) 2 . EXECUTE: N = 400 , B = 1.5 T , A = (0.100 m) 2 and Emax = 120 V gives ω = Emax /NBA = (20 rad/s)(1 rev/2π rad)(60 s/1 min) = 190 rpm. EVALUATE:

In Emax = ω BA, ω is in rad/s.


29-16

Chapter 29

29.52.

IDENTIFY: SET UP:

Apply the results of Example 29.4, generalized to N loops: Emax = N ω BA. v = rω . In the expression for Emax , ω must be in rad/s. 30 rpm = 3.14 rad/s

E0 9.0 V = = 18 m 2 . ω NB (3.14 rad/s)(2000 turns)(8.0 × 10−5 T) (b) Assuming a point on the coil at maximum distance from the axis of rotation we have EXECUTE:

(a) Solving for A we obtain A =

A 18 m 2 ω= (3.14 rad/s) = 7.5 m s. π π EVALUATE: The device is not very feasible. The coil would need a rigid frame and the effects of air resistance would be appreciable. ΔΦ B IDENTIFY: Apply Faraday’s law in the form Eav = − N to calculate the average emf. Apply Lenz’s law to Δt calculate the direction of the induced current. SET UP: Φ B = BA . The flux changes because the area of the loop changes. v = rω =

29.53.

EXECUTE:

29.54.

(a) Eav =

ΔΦ B ΔA π r2 π (0.0650/2 m) 2 =B =B = (0.950 T) = 0.0126 V. Δt Δt Δt 0.250 s

(b) Since the magnetic field is directed into the page and the magnitude of the flux through the loop is decreasing, the induced current must produce a field that goes into the page. Therefore the current flows from point a through the resistor to point b . ! EVALUATE: Faraday’s law can be used to find the direction of the induced current. Let A be into the page. Then Φ B is positive and decreasing in magnitude, so d Φ B /dt < 0. Therefore E > 0 and the induced current is clockwise around the loop. IDENTIFY: By Lenz’s law, the induced current flows to oppose the flux change that caused it. SET UP: When the switch is suddenly closed with an uncharged capacitor, the current in the outer circuit immediately increases from zero to its maximum value. As the capacitor gets charged, the current in the outer circuit gradually decreases to zero. EXECUTE: (a) (i) The current in the outer circuit is suddenly increasing and is in a counterclockwise direction. The magnetic field through the inner circuit is out of the paper and increasing. The magnetic flux through the inner circuit is increasing, so the induced current in the inner circuit is clockwise (a to b) to oppose the flux increase. (ii) The current in the outer circuit is still counterclockwise but is now decreasing, so the magnetic field through the inner circuit is out of the page but decreasing. The flux through the inner circuit is now decreasing, so the induced current is counterclockwise (b to a) to oppose the flux decrease. (b) The graph is sketched in Figure 29.54. EVALUATE: Even though the current in the outer circuit does not change direction, the current in the inner circuit does as the flux through it changes from increasing to decreasing.

Figure 29.54 29.55.

IDENTIFY: Use Faraday’s law to calculate the induced emf and Ohm’s law to find the induced current. Use Eq.(27.19) to calculate the magnetic force FI on the induced current. Use the net force F − FI in Newton’s 2nd law to calculate the acceleration of the rod and use that to describe its motion.


Electromagnetic Induction

(a) SET UP:

29-17

The forces in the rod are shown in Figure 29.55a. EXECUTE:

I=

E =

dΦB = BLv dt

BLv R

Figure 29.55a

! dΦB to find the direction of I: Let A be into the page. Then Φ B > 0. The area of the circuit is dt ! dΦB > 0. Then E < 0 and with our direction for A this means that E and I are counterclockwise, increasing, so dt ! ! ! ! as shown in the sketch. The force FI on the rod due to the induced current is given by FI = Il × B. This gives FI ! to the left with magnitude FI = ILB = ( BLv / R) LB = B 2 L2v / R. Note that FI is directed to oppose the motion of the rod, as required by Lenz’s law. EVALUATE: The net force on the rod is F − FI , so its acceleration is a = ( F − FI ) / m = ( F − B 2 L2v / R) / m. The rod starts with v = 0 and a = F/m. As the speed v increases the acceleration a decreases. When a = 0 the rod has reached its terminal speed vt . The graph of v versus t is sketched in Figure 29.55b.

Use E = −

(Recall that a is the slope of the tangent to the v versus t curve.)

Figure 29.55b (b) EXECUTE: EVALUATE: 29.56.

F − B 2 L2vt / R RF = 0 and vt = 2 2 . m B L A large F produces a large vt . If B is larger, or R is smaller, the induced current is larger at a given v v = vt when a = 0 so

so FI is larger and the terminal speed is less. IDENTIFY: Apply Newton’s 2nd law to the bar. The bar will experience a magnetic force due to the induced current in the loop. Use a = dv / dt to solve for v. At the terminal speed, a = 0 . SET UP: The induced emf in the loop has a magnitude BLv . The induced emf is counterclockwise, so it opposes the voltage of the battery, E . EXECUTE: (a) The net current in the loop is I = E − BLv . The acceleration of the bar is R ILB sin(90 ° ) ( E − BLv ) LB (E − BLv) LB = a= F = . To find v(t ) , set dv = a = and solve for v using the method m m mR dt mR of separation of variables:

v

0

t LB 2 2 dv E = (1 − e− B L t /mR) = (10 m/s)(1 − e −t /3.1 s ) dt → v = (E − BLv) ∫ 0 mR BL

The graph of v versus t is sketched in Figure 29.56. Note that the graph of this function is similar in appearance to that of a charging capacitor. (b) Just after the switch is closed, v = 0 and I = E/R = 2.4 A, F = ILB = 2.88 N and a = F/m = 3.2 m/s 2 . [12 V − (1.5 T)(0.8 m)(2.0 m/s)](0.8 m)(1.5 T) = 2.6 m/s 2 . (0.90 kg)(5.0 Ω) (d) Note that as the speed increases, the acceleration decreases. The speed will asymptotically approach the 12 V = 10 m/s, which makes the acceleration zero. terminal speed E = BL (1.5 T)(0.8 m) (c) When v = 2.0 m/s, a =


29-18

Chapter 29

EVALUATE: The current in the circuit is counterclockwise and the magnetic force on the bar is to the right. The energy that appears as kinetic energy of the moving bar is supplied by the battery.

29.57.

29.58.

29.59.

Figure 29.56 ! ! IDENTIFY: Apply E = BvL. Use ∑ F = ma applied to the satellite motion to find the speed v of the satellite. mm SET UP: The gravitational force on the satellite is Fg = G 2 E , where m is the mass of the satellite and r is the r radius of its orbit. mm v2 GmE EXECUTE: B = 8.0 ×10−5 T, L = 2.0 m. G 2 E = m and r = 400 ×103 m + RE gives v = = 7.665 ×103 m/s. r r r Using this v in E = vBL gives E = (8.0 × 10−5 T)(7.665 × 103 m/s)(2.0 m) = 1.2 V. EVALUATE: The induced emf is large enough to be measured easily. IDENTIFY: The induced emf is E = BvL, where L is measured in a direction that is perpendicular to both the magnetic field and the velocity of the bar. SET UP: The magnetic force pushed positive charge toward the high potential end of the bullet. EXECUTE: (a) E = BLv = (8 × 10−5 T)(0.004 m)(300 m/s) = 96 μ V. Since a positive charge moving to the east would be deflected upward, the top of the!bullet will be at a higher potential. ! (b) For a bullet that travels south, v and B are along the same line, there is no magnetic force and the induced emf is zero. ! (c) If v is horizontal, the magnetic force on positive charges in the bullet is either upward or downward, perpendicular to the line between the front and back of the bullet. There is no emf induced between the front and back of the bullet. EVALUATE: Since the velocity of a bullet is always in the direction from the back to the front of the bullet, and since the magnetic force is perpendicular to the velocity, there is never an induced emf between the front and back of the bullet, no matter what the direction of the magnetic field is. IDENTIFY: Find the magnetic field at a distance r from the center of the wire. Divide the rectangle into narrow strips of width dr, find the flux through each strip and integrate to find the total flux. SET UP: Example 28.8 uses Ampere’s law to show that the magnetic field inside the wire, a distance r from the axis, is B (r ) = μ0 Ir 2π R 2 . EXECUTE: Consider a small strip of length W and width dr that is a distance r from the axis of the wire, as shown μ IW in Figure 29.59. The flux through the strip is d Φ B = B ( r )W dr = 0 2 r dr . The total flux through the rectangle is 2π R μ0 IW ⎛ μ0 IW ⎞ R ΦB = ∫ dΦB = ⎜ r dr = . 2 ⎟∫ 4π ⎝ 2π R ⎠ 0 EVALUATE: Note that the result is independent of the radius R of the wire.

Figure 29.59


Electromagnetic Induction

29.60.

29-19

IDENTIFY: Apply ! Faraday’s law to calculate the magnitude and direction of the induced emf. SET UP: Let A be directed out of the page in Figure 29.50 in the textbook. This means that counterclockwise emf is positive. EXECUTE: (a) Φ B = BA = B0πr0 2 (1 − 3(t t0 ) 2 + 2(t t0 )3 ). 2 dΦB d B πr 2 6 B0πr0 2 ⎛ ⎛ t ⎞ ⎛ t ⎞ ⎞ = ⎜ ⎜ ⎟ − ⎜ ⎟ ⎟ . At = − B0πr0 2 (1 − 3(t/t0 ) 2 + 2(t/t0 )3 ) = − 0 0 ( −6(t/t0 ) + 6(t/t0 ) 2 ). E = − ⎜ ⎝ t0 ⎠ ⎝ t0 ⎠ ⎟ t0 dt dt t0 ⎝ ⎠ 2 −3 −3 2 ⎛ ⎞ 6 B π (0.0420 m) ⎛ 5.0 × 10 s ⎞ ⎛ 5.0 × 10 s ⎞ ⎜ ⎟ = 0.0665 V. E is positive so it is − t = 5.0 × 10−3 s, E = − 0 ⎜ ⎜⎝ 0.010 s ⎟⎠ ⎜⎝ 0.010 s ⎟⎠ ⎟ 0.010 s ⎝ ⎠ counterclockwise. 0.0665 V E E ⇒ Rtotal = r + R = ⇒ r = − 12 Ω = 10.2 Ω. (c) I = 3.0 × 10−3 A Rtotal I (d) Evaluating the emf at t = 1.21 × 10−2 s and using the equations of part (b), E = −0.0676 V, and the current flows clockwise, from b to a through the resistor. ⎛ ⎛ t ⎞2 ⎛ t ⎞ ⎞ t (e) E = 0 when 0 = ⎜ ⎜ ⎟ − ⎜ ⎟ ⎟ . 1 = and t = t0 = 0.010 s. ⎜ ⎝ t 0 ⎠ ⎝ t0 ⎠ ⎟ t0 ⎝ ⎠ ! EVALUATE: At t = t0 , B = 0 . At t = 5.00 × 10 −3 s , B is in the + kˆ direction and is decreasing in magnitude. Lenz’s ! law therefore says E is counterclockwise. At t = 0.0121 s , B is in the + kˆ direction and is increasing in magnitude.

(b) E = −

29.61.

Lenz’s law therefore says E is clockwise. These results for the direction of E agree with the results we obtained from Faraday’s law. (a) and (b) IDENTIFY and Set Up: The magnetic field of the wire is given by μI B = 0 and varies along the length of the 2π r ! bar. At every point along the bar B has direction into the page. Divide the bar up into thin slices, as shown in Figure 29.61a. Figure 29.61a ! ! ! ! ! The emf dE induced in each slice is given by d E = v × B ⋅ dl . v × B is directed toward the wire, so ⎛μI⎞ d E = −vB dr = −v ⎜ 0 ⎟ dr. The total emf induced in the bar is ⎝ 2π r ⎠ b d + L ⎛ μ Iv ⎞ μ Iv d + L dr μ Iv d +L Vba = ∫ d E = − ∫ ⎜ 0 ⎟ dr = − 0 ∫ = − 0 [ ln( r ) ]d a d d r 2π 2π ⎝ 2π r ⎠ μ0 Iv μ0 Iv Vba = − (ln( d + L ) − ln(d )) = − ln(1 + L/d ) 2π 2π EVALUATE: The minus sign means that Vba is negative, point a is at higher potential than point b. (The force ! ! ! F = qv × B on positive charge carriers in the bar is towards a, so a is at higher potential.) The potential difference increases when I or v increase, or d decreases. (c) IDENTIFY: Use Faraday’s law to calculate the induced emf. SET UP: The wire and loop are sketched in Figure 29.61b. EXECUTE:

EXECUTE: As the loop moves to the right the magnetic flux through it doesn’t change. Thus dΦB E =− = 0 and I = 0. dt Figure 29.61b EVALUATE: This result can also be understood as follows. The induced emf in section ab puts point a at higher potential; the induced emf in section dc puts point d at higher potential. If you travel around the loop then these two induced emf’s sum to zero. There is no emf in the loop and hence no current.


29-20

Chapter 29

29.62.

IDENTIFY: E = vBL, where v is the component of velocity perpendicular to the field direction and perpendicular to the bar. SET UP: Wires A and C have a length of 0.500 m and wire D has a length of 2(0.500 m) 2 = 0.707 m. ! ! EXECUTE: Wire A: v is parallel to B , so the induced emf is zero. ! ! ! Wire C: v is perpendicular to B. The component of v perpendicular to the bar is v cos 45° . E = (0.350 m/s)(cos 45°)(0.120 T)(0.500 m) = 0.0148 V. ! ! ! Wire D: v is perpendicular to B. The component of v perpendicular to the bar is v cos 45° . E = (0.350 m/s)(cos 45°)(0.120 T)(0.707 m) = 0.0210 V. ! ! ! EVALUATE: The induced emf depends on the angle between v and B and also on the angle between v and the bar. (a) IDENTIFY: Use the expression for motional emf to calculate the emf induced in the rod. SET UP: The rotating rod is shown in Figure 29.63a.

29.63.

The emf induced !in a!thin ! slice is d E = v × B ⋅ dl .

Figure 29.63a ! ! ! EXECUTE: Assume that B is directed out of the page. Then v × B is directed radially outward and ! ! ! dl = dr , so v × B ⋅ dl = vB dr v = rω so d E = ω Br dr. The d E for all the thin slices that make up the rod are in series so they add: L

E = ∫ d E = ∫ ω Br dr = 12 ω BL2 = 12 (8.80 rad/s)(0.650 T)(0.240 m) 2 = 0.165 V 0

EVALUATE: E increases with ω , B or L2 . (b) No current flows so there is no IR drop in potential. Thus the potential difference between the ends equals the emf of 0.165 V calculated in part (a). (c) SET UP: The rotating rod is shown in Figure 29.63b.

Figure 29.63b EXECUTE: The emf between the center of the rod and each end is E = 12 ω B( L / 2) 2 = 14 (0.165 V) = 0.0412 V, with the direction of the emf from the center of the rod toward each end. The emfs in each half of the rod thus oppose each other and there is no net emf between the ends of the rod. EVALUATE: ω and B are the same as in part (a) but L of each half is 12 L for the whole rod. E is proportional to 29.64.

L2 , so is smaller by a factor of 14 . IDENTIFY: The power applied by the person in moving the bar equals the rate at which the electrical energy is dissipated in the resistance. ( BLv) 2 SET UP: From Example 29.7, the power required to keep the bar moving at a constant velocity is P = . R EXECUTE:

(a) R =

( BLv) 2 [(0.25 T)(3.0 m)(2.0 m s)]2 = = 0.090 Ω. P 25 W

(b) For a 50 W power dissipation we would require that the resistance be decreased to half the previous value. (c) Using the resistance from part (a) and a bar length of 0.20 m, ( BLv) 2 [(0.25 T)(0.20 m)(2.0 m s)]2 = = 0.11 W . P= R 0.090 Ω EVALUATE: When the bar is moving to the right the magnetic force on the bar is to the left and an applied force directed to the right is required to maintain constant speed. When the bar is moving to the left the magnetic force on the bar is to the right and an applied force directed to the left is required to maintain constant speed.


Electromagnetic Induction

29.65.

29-21

(a) IDENTIFY: Use Faraday’s law to calculate the induced emf, Ohm’s law to calculate I, and Eq.(27.19) to calculate the force on the rod due to the induced current. SET UP: The force on the wire is shown in Figure 29.65. EXECUTE: When the wire has speed v the induced emf is E = Bva and the Bva induced current is I = E / R = R Figure 29.65

! ! ! The induced current flows upward in the wire as shown, so the force F = Il × B exerted by the magnetic field on ! the induced current is to the left. F opposes the motion of the wire, as it must by Lenz’s law. The magnitude of the force is F = IaB = B 2 a 2v / R. ! ! (b) Apply ∑ F = ma to the wire. Take +x to be toward the right and let the origin be at the location of the wire at

t = 0, so x0 = 0.

∑F

x

= max says − F = max

F B 2 a 2v =− m mR Use this expression to solve for v(t): dv B 2 a 2v dv B2a 2 =− and =− ax = dt dt mR v mR v dv′ B2a 2 t = − dt ′ ∫v0 v′ mR ∫ 0 B 2 a 2t ln(v) − ln(v0 ) = − mR ⎛v⎞ 2 2 B 2 a 2t ln ⎜ ⎟ = − and v = v0e − B a t / mR mR ⎝ v0 ⎠ ax = −

Note: At t = 0, v = v0 and v → 0 when t → ∞ Now solve for x(t): 2 2 2 2 dx = v0e − B a t / mR so dx = v0e − B a t / mR dt v= dt

x 0

t

dx′ = ∫ v0e − B 0

2 a 2 t /mR

dt ′

(

)

t 2 2 2 2 mRv ⎛ mR ⎞ x = v0 ⎜ − 2 2 ⎟ ⎡e − B a t ′/mR ⎤ = 2 20 1 − e − B a t / mR ⎣ ⎦ 0 Ba ⎝ Ba ⎠ Comes to rest implies v = 0. This happens when t → ∞. mRv t → ∞ gives x = 2 20 . Thus this is the distance the wire travels before coming to rest. Ba EVALUATE: The motion of the slide wire causes an induced emf and current. The magnetic force on the induced current opposes the motion of the wire and eventually brings it to rest. The force and acceleration depend on v and are constant. If the acceleration were constant, not changing from its initial value of ax = − B 2 a 2v0 / mR, then the

29.66.

stopping distance would be x = −v02 / 2ax = mRv0 / 2 B 2 a 2 . The actual stopping distance is twice this. ! ! ! IDENTIFY: Since the bar is straight and the magnetic field is uniform, integrating dε = v × B ⋅ dl along the length ! ! ! of the bar gives E = (v × B) ⋅ L ! ! SET UP: v = (4.20 m/s)iˆ . L = (0.250 m)(cos36.9°iˆ + sin 36.9° ˆj ). ! ! ! ! EXECUTE: (a) E = (v × B ) ⋅ L = (4.20 m/s)iˆ × ((0.120 T)iˆ − ( 0.220 T ) ˆj − (0.0900 T)kˆ ) ⋅ L.

(

)(

)

E = ( 0.378 V/m ) ˆj – ( 0.924 V/m ) kˆ ⋅ (0.250 m)(cos 36.9°iˆ + sin 36.9° ˆj ) . E = (0.378 V/m)(0.250 m)sin 36.9° = 0.0567 V. (b) The higher potential end is the end to which positive charges in the rod are pushed by the magnetic force. ! ! v × B has a positive y-component, so the end of the rod marked + in Figure 29.66 is at higher potential.


29-22

Chapter 29

! ! ! EVALUATE: Since v × B has nonzero ˆj and kˆ components, and L has nonzero iˆ and ˆj components, only the ! kˆ component of B contributes to E . In fact, E = vx Bz Ly = (4.20 m/s)(0.0900 T)(0.250 m)sin 36.9° = 0.0567 V.

Figure 29.66 29.67.

IDENTIFY: SET UP:

! ! Use Eq.(29.10) to calculate the induced electric field at each point and then use F = qE . !

!

dΦB to a concentric circle of dt ! radius r, as shown in Figure 29.67a. Take A to ! be into the page, in the direction of B.

Apply

ú E ⋅ dl = −

Figure 29.67a ! ! dΦB > 0, so úE ⋅ dl is negative. This means that E is tangent to the circle in dt the counterclockwise direction, as shown in Figure 29.67b.

EXECUTE: B increasing then gives

!

!

úE ⋅ dl = − E (2π r ) dΦB dB = π r2 dt dt

Figure 29.67b dB dB so E = 12 r dt dt The induced electric field and the force on q are shown in Figure 29.67c. dB F = qE = 12 qr dt ! F is to the left ! ! ( F is in the same direction as E since q is positive.) Figure 29.67c

− E (2π r ) = −π r 2

point a

point b The induced electric field and the force on q are shown in Figure 29.67d. dB F = qE = 12 qr dt ! F is toward the top of the page.

Figure 29.67d

29.68.

point c r = 0 here, so E = 0 and F = 0. EVALUATE: If there were a concentric conducting ring of radius r in the magnetic field region, Lenz’s law tells us that the increasing magnetic field would induce a counterclockwise current in the ring. This agrees with the direction of the force we calculated for the individual positive point charges. IDENTIFY: A bar moving in a magnetic field has an emf induced across its ends. The propeller acts as such a bar. SET UP: Different parts of the propeller are moving at different speeds, so we must integrate to get the total induced emf. The potential induced across an element of length dx is d E = vBdx, where B is uniform.


Electromagnetic Induction

29-23

EXECUTE: (a) Call x the distance from the center to an element of length dx, and L the length of the propeller. The speed of dx is xω, giving d E = vBdx = xω Bdx. E = ∫

L/ 2 0

xω Bdx = ω BL2 /8.

(b) The potential difference is zero since the potential is the same at both ends of the propeller. (2.0 m)3 ⎛ 220 rev ⎞ −4 (c) E = 2π ⎜ =5.8 × 10−4 V = 0.58 mV ⎟ ( 0.50 × 10 T ) 60 s 8 ⎝ ⎠ 29.69.

EVALUATE: A potential difference of about 12 mV is not large enough to be concerned about in a propeller. IDENTIFY: Follow the steps specified in the problem. SET UP: The electric field region is sketched in Figure 29.69. ! ! ! ! ! ! ! dΦB dΦB EXECUTE: úE ⋅ dl = − . If B is constant then = 0, so úE ⋅ dl = 0. ∫ E ⋅ dl = Eab L − Ecd L = 0. But abcda dt dt Ecd = 0 , so Eab L = 0. But since we assumed Eab ≠ 0, this contradicts Faraday’s law. Thus, we can’t have a uniform electric field abruptly drop to zero in a region in which the magnetic field is constant. ! ! EVALUATE: If the magnetic field in the region is constant, then the integral úE ⋅ dl must be zero.

Figure 26.69 29.70.

IDENTIFY and SET UP: At the terminal speed vt , the upward force FI exerted on the loop due to the induced current equals the downward force of gravity: FI = mg . Use Eq.(29.6) to find the induced emf in the side of the loop that is totally within the magnetic field. There is no induced emf in the other sides of the loop. EXECUTE: E = Bvs, I = Bvs / R and FI = IsB − B 2 s 2v / R B 2 s 2vt mgR = mg and vt = 2 2 R Bs m = ρ mV = ρ m (4s )π ( d / 2) 2 = ρ mπ d 2 R=

ρL A

=

ρ R 4 s 16 ρ R s = πd2 πd2

1 4

Using these expressions for m and R gives vt = 16 ρ m ρ R g / B 2

EVALUATE: 29.71.

We know ρ m = 8900 kg/m3 (Table 14.1) and ρ R = 1.72 × 10−8 Ω ⋅ m

(Table 25.1). Taking B = 0.5 T gives vt = 9.6 cm/s. IDENTIFY: Follow the steps specified in the problem. SET UP: (a) The magnetic field region is sketched in Figure 29.71. ! ! ! EXECUTE: (b) úB ⋅ dl = 0 (no currents in the region). Using the figure, let B = B0iˆ for y < 0 and B = 0 for y > 0. ! ! ∫ B ⋅ dl = Bab L − Bcd L = 0 but Bcd = 0. Bab L = 0, but Bab ≠ 0. This is a contradiction and violates Ampere’s Law. abcde

EVALUATE: We often describe a magnetic field as being confined to a region, but this result shows that the edges of such a region can't be sharp.

Figure 29.71


29-24

Chapter 29

29.72.

IDENTIFY and SET UP: Apply Ohm’s law to the dielectric to relate the current in the dielectric to the charge on the plates. Use Eq.(25.1) for the current and obtain a differential equation for q(t). Integrate this equation to obtain q(t) and i(t). Use E = q / PA and Eq.(29.16) to calculate jD . EXECUTE: (a) Apply Ohm’s law to the dielectric: The capacitor is sketched in Figure 29.72. v(t ) R q (t ) PA v(t ) = and C = K 0 C d

i (t ) =

Figure 29.72 ⎛ d ⎞ v(t ) = ⎜ ⎟ q (t ) ⎝ K P0 A ⎠ v(t ) ⎛ q (t ) d ⎞ ⎛ A ⎞ q (t ) . But the current i(t) =⎜ ⎟⎜ ⎟= R ⎝ K P0 A ⎠ ⎝ ρ d ⎠ K P0 ρ in the dielectric is related to the rate of change dq/dt of the charge q(t) on the plates by i(t) = –dq/dt (a positive i in the direction from the + to the – plate of the capacitor corresponds to a decrease in the charge). Using this in the above dq ⎛ 1 ⎞ dq dt gives − =− . Integrate both sides of this equation from t = 0, where q = Q0 , to a later =⎜ ⎟ q (t ). q K ρ P0 dt ⎝ K ρ P0 ⎠

The resistance R of the dielectric slab is R = ρ d / A. Thus i (t ) =

⎛ 1 ⎞ t ⎛ q ⎞ dq t and q (t ) = Q0e−t / K ρ P0 . Then = −⎜ ⎟ ∫ 0 dt. ln ⎜ ⎟ = − ρ P ρ P q K Q K 0 0 0 ⎝ ⎠ ⎝ ⎠ dq ⎛ Q0 ⎞ −t / K ρ P0 i (t ) ⎛ Q0 ⎞ −t / K ρ P0 and jC = i (t ) = − . The conduction current flows from the positive to =⎜ =⎜ ⎟e ⎟e dt ⎝ K ρ P0 ⎠ A ⎝ AK ρ P0 ⎠ the negative plate of the capacitor. q (t ) q (t ) = (b) E (t ) = PA K P0 A dE dE dq (t ) / dt i (t ) jD (t ) = P = K P0 = K P0 = − C = − jC (t ) dt dt K P0 A A ! The minus sign means that jD (t ) is directed from the negative to the positive plate. E is from + to – but dE/dt is negative (E decreases) so jD (t ) is from – to +. EVALUATE: There is no conduction current to and from the plates so the concept of displacement current, with ! ! jD = − jC in the dielectric, allows the current to be continuous at the capacitor. IDENTIFY: The conduction current density is related to the electric field by Ohm's law. The displacement current density is related to the rate of change of the electric field by Eq.(29.16). SET UP: dE/dt = ω E0 cos ωt E 0.450 V/m = 1.96 × 10−4 A/m 2 EXECUTE: (a) jC (max) = 0 = ρ 2300 Ω ⋅ m ⎛ dE ⎞ −9 2 (b) jD (max) = P0 ⎜ ⎟ = P0ω E0 = 2π P0 fE0 = 2π P0 (120 Hz)(0.450 V/m) = 3.00 × 10 A/m dt ⎝ ⎠ max E0 1 = ωP0 E0 and ω = = 4.91 × 107 rad/s (c) If jC = jD then ρP0 ρ time t when the charge is q(t).

29.73.

q

Q0

f =

29.74.

ω 4.91× 107 rad s = = 7.82 × 106 Hz. 2π 2π

EVALUATE: (d) The two current densities are out of phase by 90° because one has a sine function and the other has a cosine, so the displacement current leads the conduction current by 90°. IDENTIFY: A current is induced in the loop because of its motion and because of this current the magnetic field exerts a torque on the loop. SET UP: Each side of the loop has mass m / 4 and the center of mass of each side is at the center of each side. The flux through the loop is Φ B = BA cos φ .


Electromagnetic Induction

29-25

! ! EXECUTE: (a) τ g = ∑ rcm × mg summed over each leg. !

⎛ L ⎞⎛ m ⎞

⎛ L ⎞⎛ m ⎞

⎛m⎞

τ g = ⎜ ⎟⎜ ⎟g sin(90° − φ ) + ⎜ ⎟⎜ ⎟ g sin(90° − φ ) + ( L) ⎜ ⎟ g sin(90° − φ ) ⎝ 2 ⎠⎝ 4 ⎠ ⎝ 2 ⎠⎝ 4 ⎠ ⎝4⎠ mgL τg = cos φ (clockwise). 2 ! ! τ B = τ × B = IAB sin φ (counterclockwise). E BA d BA dφ BAω = cos φ = − sin φ = sin φ . The current is going counterclockwise looking to the − kˆ direction. R R dt R dt R mgL B 2 L4ω 2 B 2 A2ω 2 B 2 L4ω 2 Therefore, τ B = sin φ = sin φ . The net torque is τ = cos φ − sin φ , opposite to the 2 R R R direction of the rotation. 5 (b) τ = Iα (I being the moment of inertia). About this axis I = mL2 . Therefore, 12 12 1 ⎡ mgL B 2 L4ω 2 ⎤ 6 g 12 B 2 L2ω 2 α= cos φ − sin φ ⎥ = cos φ − sin φ . ⎢ 5 mL2 ⎣ 2 R 5mR ⎦ 5L I=

29.75.

EVALUATE: (c) The magnetic torque slows down the fall (since it opposes the gravitational torque). (d) Some energy is lost through heat from the resistance of the loop. IDENTIFY: Apply Eq.(29.10). SET UP: Use an integration path that is a circle of radius r. By symmetry the induced electric field is tangent to this path and constant in magnitude at all points on the path. EXECUTE: (a) The induced electric field at these points is shown in Figure 29.75a. (b) To work out the amount of the electric field that is in the direction of the loop at a general position, we will use E E E cosθ = = . Therefore, the geometry shown in Figure 29.75b. Eloop = E cosθ but E = 2π r 2π ( a /cosθ ) 2π a E cos 2 θ π a 2 dB a dB dΦB dB dB π a 2 dB , so Eloop = = . This is exactly the value . But E = =A = π r2 = 2 dt dt dt cos θ dt 2π a dt 2 dt 2π a for a ring, obtained in Exercise 29.30, and has no dependence on the part of the loop we pick. E A dB L2 dB (0.20 m)2 (0.0350 T/s) (c) I = = = = = 7.37 × 10−4 A. R R dt R dt 1.90 Ω 1 1 dB (0.20 m)2 (0.0350 T/s) (d) Eab = E = L2 = = 1.75 × 10−4 V. But there is potential drop V = IR = −1.75 × 10−4 V, 8 8 dt 8 so the potential difference is zero. EVALUATE: The magnitude of the induced emf between any two points equals the magnitudes of the potential drop due to the current through the resistance of that portion of the loop.

Eloop =

Figure 29.75 29.76.

IDENTIFY: Apply Eq.(29.10). SET UP: Use an integration path that is a circle of radius r. By symmetry the induced electric field is tangent to this path and constant in magnitude at all points on the path. EXECUTE: (a) The induced emf at these points is shown in Figure 29.76. (b) The induced emf on the side ac is zero, because the electric field is always perpendicular to the line ac. dΦB dB dB (c) To calculate the total emf in the loop, E = . E = (0.20 m)2 (0.035 T/s) = 1.40 × 10−3 V. =A = L2 dt dt dt


29-26

Chapter 29

E 1.40 ×10−3 V = = 7.37 ×10−4 A R 1.90 Ω (e) Since the loop is uniform, the resistance in length ac is one quarter of the total resistance. Therefore the potential difference between a and c is Vac = IRac = (7.37 ×10−4 A)(1.90 Ω 4) = 3.50 ×10−4 V and the point a is at a higher potential since the current is flowing from a to c. EVALUATE: This loop has the same resistance as the loop in Challenge Problem 29.75 and the induced current is the same.

(d) I =

Figure 29.76 29.77.

IDENTIFY: The motion of the bar produces an induced current and that results in a magnetic force on the bar. ! ! SET UP: FB is perpendicular to B , so is horizontal. The vertical component of the normal force equals mg cosφ , so the horizontal component of the normal force equals mg tan φ . EXECUTE: (a) As the bar starts to slide, the flux is decreasing, so the current flows to increase the flux, which LB LB d Φ B LB dA LB 2 vL2 B 2 E= (vL cos φ ) = cos φ . At the terminal means it flows from a to b. FB = iLB = = B = R R dt R dt R R v L2 B 2 Rmg tan φ speed the horizontal forces balance, so mg tan φ = t cos φ and vt = 2 2 . R L B cos φ E 1 d Φ B 1 dA B vLB cos φ mg tan φ = = B = (vL cos φ ) = = . R R dt R dt R R LB Rm 2 g 2 tan 2 φ (d) P = i 2 R = . L2 B 2 ⎛ Rmg tan φ ⎞ Rm 2 g 2 tan 2 φ (e) Pg = Fv cos(90° − φ ) = mg ⎜ 2 2 . ⎟ sin φ and Pg = L2 B 2 ⎝ L B cos φ ⎠

(c) i =

29.78.

EVALUATE: The power in part (e) equals that in part (d), as is required by conservation of energy. IDENTIFY: Follow the steps indicated in the problem. SET UP: The primary assumption throughout the problem is that the square patch is small enough so that the velocity is constant over its whole area, that is, v = ω r ≈ ω d . E EA ω dBA ! ! EXECUTE: (a) ω → clockwise, B → into page. E = vBL = ω d BL . I = = = . Since v × B points R ρL ρ ω dBLt ! ! flowing radially outward since v × B points outward, A is just the cross-sectional area tL. Therefore, I =

ρ

outward. ! ! ! ω d 2 B 2 L2t ! ! ! (b) τ = d × F and FB = IL × B = ILB pointing counterclockwise. So τ = pointing out of the page (a

ρ

counterclockwise torque opposing the clockwise rotation). ! ! (c) If ω → counterclockwise and B → into page, then I → inward radially since v × B points inward. τ → clockwise (again opposing the motion). If ω → counterclockwise and B → out of the page, then I → radially outward. τ → clockwise (opposing the motion) The magnitudes of I and τ are the same as in part (a). EVALUATE: In each case the magnetic torque due to the induced current opposes the rotation of the disk, as is required by conservation of energy.


30

INDUCTANCE

30.1.

IDENTIFY and SET UP:

Apply Eq.(30.4). di (a) E2 = M 1 = (3.25 ×10−4 H)(830 A/s) = 0.270 V; yes, it is constant. dt

EXECUTE:

di2 ; M is a property of the pair of coils so is the same as in part (a). Thus E1 = 0.270 V. dt EVALUATE: The induced emf is the same in either case. A constant di/dt produces a constant emf. (b) E1 = M

30.2.

E1 = M

IDENTIFY:

Δi2 Δi NΦ and E2 = M 1 . M = 2 B 2 , where Φ B 2 is the flux through one turn of the second Δt Δt i1

coil. SET UP:

M is the same whether we consider an emf induced in coil 1 or in coil 2. 1.65 ×10−3 V E2 EXECUTE: (a) M = = = 6.82 ×10−3 H = 6.82 mH Δi1 / Δt 0.242 A/s (b) Φ B 2 =

Mi1 (6.82 ×10−3 H)(1.20 A) = = 3.27 × 10−4 Wb N2 25

Δi2 = (6.82 ×10−3 H)(0.360 A/s) = 2.46 ×10−3 V = 2.46 mV Δt EVALUATE: We can express M either in terms of the total flux through one coil produced by a current in the other coil, or in terms of the emf induced in one coil by a changing current in the other coil. IDENTIFY: Replace units of Wb, A and Ω by their equivalents. SET UP: 1 Wb = 1 T ⋅ m 2 . 1 T = 1 N/(A ⋅ m). 1 N ⋅ m = 1 J. 1 A = 1 C/s. 1 V = 1 J/C. 1 V/A = 1 Ω. (c) E1 = M

30.3.

30.4.

EXECUTE: EVALUATE: IDENTIFY: (a) SET UP: EXECUTE:

1 H = 1 Wb/A = 1 T ⋅ m 2 /A = 1 N ⋅ m/A 2 = 1 J/A 2 = 1 (J/[A ⋅ C])s = 1 (V/A)s = 1 Ω ⋅ s. We may use whichever equivalent unit is the most convenient in a particular problem. Changing flux from one object induces an emf in another object. The magnetic field due to a solenoid is B = μ 0 nI . The above formula gives B1 =

( 4π ×10

−7

T ⋅ m/A ) (300)(0.120 A)

0.250 m The average flux through each turn of the inner solenoid is therefore

=1.81×10−4 T

Φ B = B1 A = (1.81 × 10−4 T ) π (0.0100 m) 2 = 5.68 × 10−8 Wb (b) SET UP:

The flux is the same through each turn of both solenoids due to the geometry, so M=

(25) ( 5.68 × 10−8 Wb )

N 2Φ B ,2 N 2Φ B ,1 = i1 i1

= 1.18 ×10−5 H

EXECUTE:

M=

(c) SET UP:

The induced emf is E2 = − M

EXECUTE:

E2 = − (1.18 × 10−5 H ) (1750 A/s) = −0.0207 V

EVALUATE:

0.120 A

di1 . dt

A mutual inductance around 10−5 H is not unreasonable. 30-1


30-2

30.5.

Chapter 30

IDENTIFY and SET UP: Apply Eq.(30.5). 400 ( 0.0320 Wb ) NΦ EXECUTE: (a) M = 2 B 2 = = 1.96 H 6.52 A i1

N1Φ B1 Mi (1.96 H)(2.54 A) so Φ B1 = 2 = = 7.11×10−3 Wb i2 N1 700 EVALUATE: M relates the current in one coil to the flux through the other coil. Eq.(30.5) shows that M is the same for a pair of coils, no matter which one has the current and which one has the flux. IDENTIFY: A changing current in an inductor induces an emf in it. μ N2A (a) SET UP: The self-inductance of a toroidal solenoid is L = 0 . 2π r (4π × 10−7 T ⋅ m/A)(500) 2 (6.25 × 10−4 m 2 ) EXECUTE: L = = 7.81 × 10−4 H 2π (0.0400 m) (b) M =

30.6.

(b) SET UP:

The magnitude of the induced emf is E = L

di . dt

⎛ 5.00 A − 2.00 A ⎞ E = ( 7.81 × 10−4 H ) ⎜ ⎟ = 0.781 V −3 ⎝ 3.00 × 10 s ⎠ (c) The current is decreasing, so the induced emf will be in the same direction as the current, which is from a to b, making b at a higher potential than a. EVALUATE: This is a reasonable value for self-inductance, in the range of a mH. Δi NΦB IDENTIFY: E = L and L = . i Δt EXECUTE:

30.7.

Δi = 0.0640 A/s Δt E 0.0160 V EXECUTE: (a) L = = = 0.250 H Δi / Δt 0.0640 A/s

SET UP:

Li (0.250 H)(0.720 A) = = 4.50 × 10−4 Wb. N 400 EVALUATE: The self-induced emf depends on the rate of change of flux and therefore on the rate of change of the current, not on the value of the current. IDENTIFY: Combine the two expressions for L: L = N Φ B / i and L = E/(di/dt ).

(b) The average flux through each turn is Φ B =

30.8.

SET UP:

30.9.

30.10.

30.11.

Φ B is the average flux through one turn of the solenoid.

(12.6 × 10−3 V)(1.40 A) = 238 turns. (0.00285 Wb)(0.0260 A/s) EVALUATE: The induced emf depends on the time rate of change of the total flux through the solenoid. IDENTIFY and SET UP: Apply E = L di/dt . Apply Lenz’s law to determine the direction of the induced emf in EXECUTE:

Solving for N we have N = Ei/Φ B (di/dt ) =

the coil. EXECUTE:

(a) E = L( di/dt ) = (0.260 H)(0.0180 A/s) = 4.68 × 10−3 V

(b) Terminal a is at a higher potential since the coil pushes current through from b to a and if replaced by a battery it would have the + terminal at a. EVALUATE: The induced emf is directed so as to oppose the decrease in the current. di IDENTIFY: Apply E = − L . dt SET UP: The induced emf points from low potential to high potential across the inductor. EXECUTE: (a) The induced emf points from b to a, in the direction of the current. Therefore, the current is decreasing and the induced emf is directed to oppose this decrease. (b) E = L Δi / Δt , so Δi/Δt = Vab /L = (1.04 V) /(0.260 H) = 4.00 A/s. In 2.00 s the decrease in i is 8.00 A and the

current at 2.00 s is 12.0 A − 8.0 A = 4.0 A. EVALUATE: When the current is decreasing the end of the inductor where the current enters is at the lower potential. This agrees with our result and with Figure 30.6d in the textbook. IDENTIFY and SET UP: Use Eq.(30.6) to relate L to the flux through each turn of the solenoid. Use Eq.(28.23) for the magnetic field through the solenoid.


Inductance

NΦB . If the magnetic field is uniform inside the solenoid Φ B = BA. From Eq.(28.23), i μ NiA N ⎛ μ NiA ⎞ μ 0 N 2 A ⎛N⎞ . Then L = ⎜ 0 . B = μ 0 ni = μ 0 ⎜ ⎟ i so Φ B = 0 ⎟= l i ⎝ l ⎠ l ⎝ l ⎠ EVALUATE: Our result is the same as L for a torodial solenoid calculated in Example 30.3, except that the average circumference 2π r of the toroid is replaced by the length l of the straight solenoid. IDENTIFY and SET UP: The stored energy is U = 12 LI 2 . The rate at which thermal energy is developed is P = I 2 R. EXECUTE: (a) U = 12 LI 2 = 12 (12.0 H)(0.300 A)2 = 0.540 J (b) P = I 2 R = (0.300 A)2 (180 Ω) = 16.2 W = 16.2 J/s EVALUATE: (c) No. If I is constant then the stored energy U is constant. The energy being consumed by the resistance of the inductor comes from the emf source that maintains the current; it does not come from the energy stored in the inductor. IDENTIFY and SET UP: Use Eq.(30.9) to relate the energy stored to the inductance. Example 30.3 gives the μ N 2A inductance of a toroidal solenoid to be L = 0 , so once we know L we can solve for N. 2π r 2U 2(0.390 J) EXECUTE: U = 12 LI 2 so L = 2 = = 5.417 × 10−3 H (12.0 A) 2 I EXECUTE:

30.12.

30.13.

N= 30.14.

L=

2π rL 2π (0.150 m)(5.417 ×10−3 H) = = 2850. (4π ×10−7 T ⋅ m/A)(5.00 × 10−4 m 2 ) μ0 A

EVALUATE: L and hence U increase according to the square of N. IDENTIFY: A current-carrying inductor has a magnetic field inside of itself and hence stores magnetic energy. μ NI (a) SET UP: The magnetic field inside a toroidal solenoid is B = 0 . 2π r μ0 (300)(5.00 A) −3 EXECUTE: B = = 2.50 × 10 T = 2.50 mT 2π (0.120 m) μ N2A (b) SET UP: The self-inductance of a toroidal solenoid is L = 0 . 2π r (4π × 10−7 T ⋅ m/A)(300) 2 (4.00 × 10−4 m 2 ) EXECUTE: L = = 6.00 × 10−5 H 2π (0.0120 m) (c) SET UP: The energy stored in an inductor is U L = 12 LI 2 . EXECUTE:

U L = 12 (6.00 × 10−5 H)(5.00 A) 2 = 7.50 ×10−4 J

(d) SET UP:

The energy density in a magnetic field is u =

B2 . 2μ 0

(2.50 × 10−3 T) 2 = 2.49 J/m3 2(4π × 10−7 T ⋅ m/A) energy energy 7.50 × 10−4 J (e) u = = = = 2.49 J/m3 volume 2π rA 2π (0.120 m)(4.00 × 10−4 m 2 ) EVALUATE: An inductor stores its energy in the magnetic field inside of it. IDENTIFY: A current-carrying inductor has a magnetic field inside of itself and hence stores magnetic energy. (a) SET UP: The magnetic field inside a solenoid is B = μ0 nI . (4π × 10−7 T ⋅ m/A)(400)(80.0 A) = 0.161 T EXECUTE: B = 0.250 m B2 (b) SET UP: The energy density in a magnetic field is u = . 2μ0 (0.161 T) 2 EXECUTE: u = = 1.03 ×104 J/m3 2(4π ×10−7 T ⋅ m/A) (c) SET UP: The total stored energy is U = uV. EXECUTE: U = uV = u (lA) = (1.03 ×104 J/m3 )(0.250 m)(0.500 ×10−4 m2 ) = 0.129 J EXECUTE:

30.15.

30-3

u=

(d) SET UP: The energy stored in an inductor is U = 12 LI 2 . EXECUTE: Solving for L and putting in the numbers gives 2U 2(0.129 J) L= 2 = = 4.02 ×10−5 H I (80.0 A) 2 EVALUATE: An inductor stores its energy in the magnetic field inside of it.


30-4

Chapter 30

30.16.

IDENTIFY: Energy = Pt . U = 12 LI 2 . SET UP: P = 200 W = 200 J/s EXECUTE: (a) Energy = (200 W)(24 h)(3600 s/h) = 1.73 ×107 J 2U 2(1.73 × 107 J) = = 5.41×103 H I2 (80.0 A) 2 EVALUATE: A large value of L and a large current would be required, just for one light bulb. Also, the resistance of the inductor would have to be very small, to avoid a large P = I 2 R rate of electrical energy loss. IDENTIFY and SET UP: Starting with Eq. (30.9), follow exactly the same steps as in the text except that the magnetic permeability μ is used in place of μ 0 .

(b) L =

30.17.

EXECUTE:

30.18.

B2 μN 2A μ NI and B = gives u = . 2π r 2π r 2μ

EVALUATE: For a given value of B, the energy density is less when μ is larger than μ 0 . IDENTIFY and SET UP: The energy density (energy per unit volume) in a magnetic field (in vacuum) is given by U B2 (Eq.30.10). u= = V 2μ0 EXECUTE: (b) u =

B=

30.19.

Using L =

(a) V =

2μ 0U 2(4π ×10−7 T ⋅ m/A)(3.60 ×106 J) = = 25.1 m3 . B2 (0.600 T) 2

U B2 = V 2μ0

2μ 0U 2(4π ×10−7 T ⋅ m/A)(3.60 ×106 J) = = 11.9 T 3 V ( 0.400 m )

EVALUATE: Large-scale energy storage in a magnetic field is not practical. The volume in part (a) is quite large and the field in part (b) would be very difficult to achieve. IDENTIFY: Apply Kirchhoff’s loop rule to the circuit. i(t) is given by Eq.(30.14). SET UP: The circuit is sketched in Figure 30.19. di is positive as the current dt increases from its initial value of zero.

Figure 30.19 EXECUTE:

E − vR − v L = 0

di E E − iR − L = 0 so i = (1 − e− ( R / L )t ) dt R di (a) Initially (t = 0), i = 0 so E − L = 0 dt di E 6.00 V = = = 2.40 A/s dt L 2.50 H di (b) E − iR − L = 0 (Use this equation rather than Eq.(30.15) since i rather than t is given.) dt di E − iR 6.00 V − (0.500 A)(8.00 Ω) Thus = = = 0.800 A/s dt L 2.50 H E ⎛ 6.00 V ⎞ − (8.00 Ω/ 2.50 H)(0.250 s) (c) i = (1 − e − ( R/L )t ) = ⎜ ) = 0.750 A(1 − e−0.800 ) = 0.413 A ⎟ (1 − e R ⎝ 8.00 Ω ⎠ (d) Final steady state means t → ∞ and

di → 0, so E − iR = 0. dt

E 6.00 V = = 0.750 A R 8.00 Ω EVALUATE: Our results agree with Fig.30.12 in the textbook. The current is initially zero and increases to its final value of E/R. The slope of the current in the figure, which is di/dt, decreases with t.

i=


Inductance

30.20.

IDENTIFY: The current decays exponentially. SET UP: After opening the switch, the current is i = I 0e−tR/L , and the time constant is τ = L /R. EXECUTE:

(a) The initial current is I 0 = (6.30 V)/(15.0 Ω) = 0.420 A. Now solve for L and put in the numbers.

L=

30.21.

30-5

−tR −(2.00 ms)(15.0 Ω) = = 43.3 mH ln(i / I 0 ) ⎛ 0.210 A ⎞ ln ⎜ ⎟ ⎝ 0.420 A ⎠

(b) τ = L/R = (43.3 mH)/(15.0 Ω) = 2.89 ms (c) Solve i = I 0e − t/τ for t, giving t = −τ ln(i/I 0 ) = −(2.89 ms)ln(0.0100) = 13.3 ms. EVALUATE: In less than 5 time constants, the current is only 1% of its initial value. IDENTIFY: i = E/R (1 − e − t/τ ), with τ = L/R. The energy stored in the inductor is U = 12 Li 2 . SET UP: The maximum current occurs after a long time and is equal to E/R. EXECUTE: (a) imax = E/R so i = imax /2 when (1 − e −t/τ ) = 12 and e− t/τ = 12 . −t/τ = ln ( 12 ) .

t=

L ln 2 (ln 2)(1.25 × 10−3 H) = = 17.3 μs R 50.0 Ω

(b) U = 12 U max when i = imax

2 . 1 − e − t/τ = 1

2 , so e− t/τ = 1 − 1

2 = 0.2929. t = − L ln (0.2929)/R = 30.7 μs.

EVALUATE: τ = L/R = 2.50 × 10 s = 25.0 μs. The time in part (a) is 0.692τ and the time in part (b) is 1.23τ . IDENTIFY: With S1 closed and S 2 open, i (t ) is given by Eq.(30.14). With S1 open and S 2 closed, i (t ) is given by Eq.(30.18). SET UP: U = 12 Li 2 . After S1 has been closed a long time, i has reached its final value of I = E/R. −5

30.22.

EXECUTE:

(a) U = 12 LI 2 and I =

2U 2(0.260 J) = = 2.13 A. E = IR = (2.13 A)(120 Ω) = 256 V. L 0.115 H

(b) i = Ie − ( R/L )t and U = 12 Li 2 = 12 LI 2e−2( R/L ) t = 12 U 0 =

t=−

30.23.

30.24.

30.25.

1 2

(

1 2

LI 2 ) . e−2( R/L )t = 12 , so

L 0.115 H ln ( 12 ) = − ln ( 12 ) = 3.32 × 10−4 s. 2R 2(120 Ω)

EVALUATE: τ = L/R = 9.58 × 10−4 s. The time in part (b) is τ ln(2) / 2 = 0.347τ . IDENTIFY: L has units of H and R has units of Ω . SET UP: 1 H = 1 Ω ⋅ s EXECUTE: Units of L/R = H / Ω = (Ω ⋅ s) / Ω = s = units of time. EVALUATE: Rt/L = t/τ is dimensionless. IDENTIFY: Apply the loop rule. SET UP: In applying the loop rule, go around the circuit in the direction of the current. The voltage across the inductor is − Ldi / dt. i di′ R t R di R EXECUTE: − Ldi/dt − iR = 0. = − ∫ dt ′ and ln(i/I 0 ) = − t. i = I 0e− t ( R/L ) . = − i gives ∫ I 0 0 i′ dt L L L EVALUATE: di / dt is negative, so there is a potential rise across the inductor; point c is at higher potential than point b. There is a potential drop across the resistor. IDENTIFY: Apply the concepts of current decay in an R-L circuit. Apply the loop rule to the circuit. i(t) is given by Eq.(30.18). The voltage across the resistor depends on i and the voltage across the inductor depends on di/dt. SET UP: The circuit with S1 closed and S 2 open is sketched in Figure 30.25a.

E − iR − L

Figure 30.25a

Constant current established means EXECUTE:

i=

di = 0. dt

E 60.0 V = = 0.250 A R 240 Ω

di =0 dt


30-6

Chapter 30

(a) SET UP:

The circuit with S 2 closed and S1 open is shown in Figure 30.25b. i = I 0e − ( R / L ) t At t = 0, i = I 0 = 0.250 A Figure 30.25b

The inductor prevents an instantaneous change in the current; the current in the inductor just after S 2 is closed and S1 is opened equals the current in the inductor just before this is done. (b) EXECUTE: i = I 0e− ( R / L )t = (0.250 A)e− (240 Ω / 0.160 H)(4.00×10 (c) SET UP: See Figure 30.25c.

−4

s)

= (0.250 A)e−0.600 = 0.137 A

Figure 30.25c EXECUTE: If we trace around the loop in the direction of the current the potential falls as we travel through the resistor so it must rise as we pass through the inductor: vab > 0 and vbc < 0. So point c is at higher potential than point b. vab + vbc = 0 and vbc = −vab

Or, vcb = vab = iR = (0.137 A)(240 Ω) = 32.9 V (d) i = I 0e − ( R/L )t

i = 12 I 0 says 12 I 0 = I 0e− ( R/L ) t and

1 2

= e − ( R/L )t

Taking natural logs of both sides of this equation gives ln( 12 ) = − Rt / L ⎛ 0.160 H ⎞ −4 t =⎜ ⎟ ln 2 = 4.62 × 10 s ⎝ 240 Ω ⎠

30.26.

EVALUATE: The current decays, as shown in Fig. 30.13 in the textbook. The time constant is τ = L/R = 6.67 × 10−4 s. The values of t in the problem are less than one time constant. At any instant the potential drop across the resistor (in the direction of the current) equals the potential rise across the inductor. IDENTIFY: Apply Eq.(30.14). di SET UP: vab = iR. vbc = L . The current is increasing, so di / dt is positive. dt EXECUTE: (a) At t = 0, i = 0. vab = 0 and vbc = 60 V. (b) As t → ∞, i → E/R and di / dt → 0. vab → 60 V and vbc → 0. (c) When i = 0.150 A, vab = iR = 36.0 V and vbc = 60.0 V − 36.0 V = 24.0 V.

30.27.

EVALUATE: At all times, E = vab + vbc , as required by the loop rule. IDENTIFY: i (t ) is given by Eq.(30.14). SET UP: The power input from the battery is Ei. The rate of dissipation of energy in the resistance is i 2 R. The voltage across the inductor has magnitude Ldi/dt , so the rate at which energy is being stored in the inductor is iLdi/dt. E2 (6.00 V) 2 (1 − e− (8.00 Ω / 2.50 H) t ). EXECUTE: (a) P = Ei = EI 0 (1 − e − ( R/L )t ) = (1 − e − ( R/L )t ) = 8.00 Ω R P = (4.50 W)(1 − e − (3.20 s

−1

)t

).

−1 E (6.00 V) 2 (1 − e − ( R/L )t ) 2 = (1 − e − (8.00 Ω/ 2.50 H) t ) 2 = (4.50 W)(1 − e− (3.20 s ) t ) 2 8.00 Ω R 2 di E ⎛E ⎞ E (c) PL = iL = (1 − e − ( R/L )t ) L ⎜ e − ( R/L )t ⎟ = (e − ( R/L )t − e −2( R/L )t ) dt R ⎝L ⎠ R

(b) PR = i 2 R =

2

−1

−1

PL = (4.50 W)(e− (3.20 s ) t − e− (6.40 s ) t ). EVALUATE: (d) Note that if we expand the square in part (b), then parts (b) and (c) add to give part (a), and the total power delivered is dissipated in the resistor and inductor. Conservation of energy requires that this be so.


Inductance

30.28.

30-7

An L-C circuit oscillates, with the energy going back and forth between the inductor and capacitor. 1 1 ω (a) SET UP: The frequency is f = and ω = , giving f = . 2π 2π LC LC 1 EXECUTE: f = = 2.13 × 103 Hz = 2.13 kHz −3 2π ( 0.280 × 10 H )( 20.0 ×10−6 F ) IDENTIFY:

(b) SET UP:

The energy stored in a capacitor is U = 12 CV 2 .

EXECUTE:

U = 12 (20.0 ×10−6 F)(150.0 V) 2 = 0.225 J

(c) SET UP:

The current in the circuit is i = −ωQ sin ωt , and the energy stored in the inductor is U = 12 Li 2 .

First find ω and Q. ω = 2π f = 1.336 × 104 rad/s. Q = CV = (20.0 × 10–6 F)(150.0 V) = 3.00 × 10–3 C Now calculate the current: i = − (1.336 × 104 rad/s)(3.00 × 10–3 C) sin[(1.336 × 104 rad/s)(1.30 × 10–3 s)] Notice that the argument of the sine is in radians, so convert it to degrees if necessary. The result is i = − 39.92 A Now find the energy in the inductor: U = 12 Li 2 = 12 (0.280 ×10−3 H)( −39.92 A) 2 = 0.223 J EVALUATE: At the end of 1.30 ms, nearly all the energy is now in the inductor, leaving very little in the capacitor. IDENTIFY: The energy moves back and forth between the inductor and capacitor. 1 1 2π (a) SET UP: The period is T = = = = 2π LC . f ω / 2π ω EXECUTE: Solving for L gives T2 (8.60 × 10−5 s) 2 L= 2 = 2 = 2.50 × 10−2 H = 25.0 mH 4π C 4π (7.50 × 10−9 C) EXECUTE:

30.29.

(b) SET UP: The charge on a capacitor is Q = CV. EXECUTE: Q = CV = (7.50 × 10–9 F)(12.0 V) = 9.00 × 10–8 C (c) SET UP: The stored energy is U = Q2/2C.

( 9.00 ×10 C ) = 5.40 ×10 U= 2 ( 7.50 ×10 F ) −8

EXECUTE:

−9

2

−7

J

(d) SET UP: The maximum current occurs when the capacitor is discharged, so the inductor has all the initial energy. U L + U C = U Total . 12 LI 2 + 0 = U Total . EXECUTE: Solve for the current: I=

30.30.

2 ( 5.40 × 10−7 J ) 2U Total = = 6.58 × 10 −3 A = 6.58 mA 2.50 × 10 −2 H L

EVALUATE: The energy oscillates back and forth forever. However if there is any resistance in the circuit, no matter how small, all this energy will eventually be dissipated as heat in the resistor. IDENTIFY: The circuit is described in Figure 30.14 of the textbook. SET UP: The energy stored in the inductor is U L = 12 Li 2 and the energy stored in the capacitor is U C = q 2 /2C. Initially,

U C = 12 CV 2, with V = 12.0 V. The period of oscillation is T = 2π LC = 2π (12.0 ×10−3 H)(18.0 ×10−6 F) = 2.92 ms . EXECUTE:

(a) Energy conservation says U L (max) = U C (max), and

imax = V C/L = (22.5 V)

1 2

2 Limax = 12 CV 2 .

18 × 10−6 F = 0.871 A. The charge on the capacitor is zero because all the energy is in 12 × 10−3 H

the inductor. (b) From Figure 30.14 in the textbook, q = 0 at t = T/4 = 0.730 ms and at t = 3T/4 = 2.19 ms. (c) q0 = CV = (18 μ F) (22.5 V) = 405 μ C is the maximum charge on the plates. The graphs are sketched in Figure 30.30. q refers to the charge on one plate and the sign of i indicates the direction of the current. EVALUATE: If the capacitor is fully charged at t = 0 it is fully charged again at t = T/2, but with the opposite polarity.

Figure 30.30


30-8

Chapter 30

30.31.

IDENTIFY and SET UP: The angular frequency is given by Eq.(30.22). q(t) and i(t) are given by Eqs.(30.21) and (30.23). The energy stored in the capacitor is U C = 12 CV 2 = q 2 /2C. The energy stored in the inductor is U L = 12 Li 2 . EXECUTE:

(a) ω =

1 1 = = 105.4 rad/s, which rounds to 105 rad/s. The period is LC (1.50 H)(6.00 ×10−5 F)

2π = = 0.0596 s ω 105.4 rad/s (b) The circuit containing the battery and capacitor is sketched in Figure 30.31.

given by T =

Q =0 C Q = EC = (12.0 V)(6.00 ×10−5 F) = 7.20 ×10−4 C

E−

Figure 30.31 (c) U = CV = 12 (6.00 × 10−5 F)(12.0 V) 2 = 4.32 × 10 −3 J (d) q = Q cos(ω t + φ ) (Eq.30.21) q = Q at t = 0 so φ = 0 1 2

2

q = Q cos ω t = (7.20 ×10−4 C)cos([105.4 rad/s][0.0230 s]) = −5.42 ×10−4 C The minus sign means that the capacitor has discharged fully and then partially charged again by the current maintained by the inductor; the plate that initially had positive charge now has negative charge and the plate that initially had negative charge now has positive charge. (e) i = −ω Q sin(ω t + φ ) (Eq.30.23) i = −(105 rad/s)(7.20 ×10 −4 C)sin([105.4 rad/s][0.0230 s]) = −0.050 A The negative sign means the current is counterclockwise in Figure 30.15 in the textbook. or q2 Q2 1 2 1 gives i = ± = Q 2 − q 2 (Eq.30.26) 2 Li + 2C 2C LC

i = ±(105 rad/s) (7.20 ×10−4 C) 2 − (−5.42 ×10−4 C) 2 = ±0.050 A, which checks. q 2 (−5.42 ×10−4 C) 2 = = 2.45 ×10−3 J 2C 2(6.00 ×10−5 F) U L = 12 Li 2 = 12 (1.50 H)(0.050 A) 2 = 1.87 × 10−3 J

(f ) U C =

EVALUATE:

Note that U C + U L = 2.45 ×10−3 J + 1.87 ×10−3 J = 4.32 ×10−3 J.

Q 2 (7.20 ×10−4 C) 2 = = 4.32 ×10−3 J. 2C 2(6.00 ×10−5 F) Energy is conserved. At some times there is energy stored in both the capacitor and the inductor. When i = 0 all the energy is stored in the capacitor and when q = 0 all the energy is stored in the inductor. But at all times the total energy stored is the same. 1 IDENTIFY: ω = = 2πf LC SET UP: ω is the angular frequency in rad/s and f is the corresponding frequency in Hz. 1 1 EXECUTE: (a) L = 2 2 = 2 = 2.37 × 10−3 H. 4π f C 4π (1.6 ×106 Hz) 2 (4.18 ×10 −12 F) (b) The maximum capacitance corresponds to the minimum frequency. 1 1 Cmax = = = 3.67 × 10−11 F = 36.7 pF 2 2 2 5 4π f min L 4π (5.40 × 10 Hz) 2 (2.37 ×10 −3 H) EVALUATE: To vary f by a factor of three (approximately the range in this problem), C must be varied by a factor of nine. IDENTIFY: Apply energy conservation and Eqs. (30.22) and (30.23). Q2 SET UP: If I is the maximum current, 12 LI 2 = . For the inductor, U L = 12 Li 2 . 2C Q2 EXECUTE: (a) 12 LI 2 = gives Q = i LC = (0.750 A) (0.0800 H)(1.25 ×10−9 F) = 7.50 ×10−6 C . 2C 1 1 ω (b) ω = = = 1.00 ×105 rad/s . f = = 1.59 ×104 Hz . −9 2 π LC (0.0800 H)(1.25 ×10 F) This agrees with the total energy initially stored in the capacitor, U =

30.32.

30.33.


Inductance

(c) q = Q at t = 0 means φ = 0 . i = −ω Q sin(ω t ) , so

i = −(1.00 ×105 rad/s)(7.50 ×10−6 C)sin([1.00 ×105 rad/s][2.50 ×10−3 s]) = −0.7279 A . U L = 12 Li 2 = 12 (0.0800 H)( − 0.7279 A)2 = 0.0212 J .

30.34.

EVALUATE: The total energy of the system is 12 LI 2 = 0.0225 J . At t = 2.50 ms , the current is close to its maximum value and most of the system’s energy is stored in the inductor. IDENTIFY: Apply Eq.(30.25). SET UP: q = Q when i = 0 . i = imax when q = 0 . 1/ LC = 1917 s −1 . EXECUTE:

(a)

1 2

2 Limax =

Q2 . Q = imax LC = (0.850 ×10−3 A) (0.0850 H)(3.20 × 10−6 F) = 4.43 ×10−7 C 2C 2

30.35.

⎛ 5.00 ×10−4 A ⎞ −7 (b) q = Q 2 − LCi 2 = (4.43 ×10−7 C) 2 − ⎜ ⎟ = 3.58 ×10 C . −1 1917s ⎝ ⎠ EVALUATE: The value of q calculated in part (b) is less than the maximum value Q calculated in part (a). IDENTIFY: q = Q cos(ω t + φ ) and i = −ω Q sin(ω + φ ) q2 . U L = 12 Li 2 . 2C q 2 1 Q 2 cos 2 (ωt + φ ) EXECUTE: (a) U C = 12 = . C 2 C 1 Q 2 sin 2 (ωt + φ ) . , since ω2 = U L = 12 Li 2 = 12 Lω2Q 2 sin 2 (ωt + φ ) = 12 C LC 1 Q2 1 (b) U Total = U C + U L = cos 2 (ωt + φ ) + Lω2Q 2 sin 2 (ωt + φ ) 2 C 2 2 Q 1 Q2 Q2 ⎛ ⎞ 2 2 1 U total = 12 cos 2 (ωt + φ ) + 12 L ⎜ (cos 2 (ωt + φ ) + sin 2 (ωt + φ )) = 12 ⎟ Q sin (ωt + φ ) = 2 C C C ⎝ LC ⎠ U Total is a constant. EVALUATE: Eqs.(30.21) and (30.23) are consistent with conservation of energy in the L-C circuit. d 2q and insert into Eq.(20.20). IDENTIFY: Evaluate dt 2 d 2q 1 SET UP: Equation (30.20) is + q = 0. dt 2 LC dq d 2q EXECUTE: q = Q cos(ωt + φ ) ⇒ = −ωQ sin(ωt + φ ) ⇒ 2 = −ω2Q cos(ωt + φ ). dt dt 2 d q 1 Q 1 1 q = −ω2Q cos(ωt + φ ) + cos(ωt + φ ) = 0 ⇒ ω 2 = + ⇒ω= . dt 2 LC LC LC LC EVALUATE: The value of φ depends on the initial conditions, the value of q at t = 0 . IDENTIFY: The unit of L is H and the unit of C is F. SET UP: C = q / Vab says 1 F = 1 C/V . 1 H = 1 V ⋅ s/A = 1 V ⋅ s 2 /C . SET UP:

30.36.

30.37.

EXECUTE: 30.38.

UC =

1 H ⋅ F = (1 V ⋅ s 2 /C)(1 C/V) = 1 s 2 . Therefore, LC has units of s 2 and

LC has units of s.

EVALUATE: Our result shows that ω t is dimensionless, since ω = 1/ LC . IDENTIFY: The presence of resistance in an L-R-C circuit affects the frequency of oscillation and causes the amplitude of the oscillations to decrease over time. 1 R2 − 2 . (a) SET UP: The frequency of damped oscillations is ω ′ = LC 4 L EXECUTE:

ω′ =

1 (75.0 Ω) 2 − = 5.5 ×104 rad/s ( 22 ×10−3 H )(15.0 ×10−9 F) 4 ( 22 ×10−3 H )2

The frequency f is f =

ω 5.50 ×104 rad/s = = 8.76 ×103 Hz = 8.76 kHz . 2π 2π

(b) SET UP: The amplitude decreases as A(t) = A0 e–(R/2L)t. EXECUTE: Solving for t and putting in the numbers gives:

−2 L ln( A / A0 ) −2 ( 22.0 ×10 H ) ln(0.100) = = 1.35 ×10−3 s = 1.35 ms R 75.0 Ω −3

t=

30-9


30-10

30.39.

Chapter 30

(c) SET UP:

At critical damping, R = 4 L / C .

EXECUTE:

R=

4 ( 22.0 ×10−3 H ) 15.0 ×10−9 F

= 2420 Ω

EVALUATE: The frequency with damping is almost the same as the resonance frequency of this circuit (1/ LC ), which is plausible because the 75-Ω resistance is considerably less than the 2420 Ω required for critical damping. IDENTIFY: Follow the procedure specified in the problem. 1 SET UP: Make the substitutions x → q, m → L, b → R, k → . C d 2 x b dx kx d 2 q R dq q EXECUTE: (a) Eq. (13.41): 2 + + = 0 . This becomes + + = 0 , which is Eq.(30.27). dt m dt m dt 2 L dt LC

1 R2 k b2 − . This becomes ω′ = − 2 , which is Eq.(30.29). 2 m 4m LC 4 L − (b / 2 m )t cos(ω′t + φ ) . This becomes q = Ae− ( R / 2 L )t cos(ω′t + φ ) , which is Eq.(30.28). (c) Eq. (13.42): x = Ae EVALUATE: Equations for the L-R-C circuit and for a damped harmonic oscillator have the same form. IDENTIFY: For part (a), evaluate the derivatives as specified in the problem. For part (b) set q = Q in Eq.(30.28) and set dq / dt = 0 in the expression for dq / dt . (b) Eq. (13.43): ω′ =

30.40.

SET UP:

In terms of ω ′ , Eq.(30.28) is q (t ) = Ae − ( R / 2 L )t cos(ω ′t + φ ) .

EXECUTE:

(a) q = Ae− ( R / 2 L ) t cos(ω′t + φ ) .

dq R = − A e − ( R / 2 L ) t cos(ω′t + φ ) − ω′Ae − ( R / 2 L )t sin(ω′t + φ ). 2L dt

2

d 2q R ⎛ R ⎞ = A ⎜ ⎟ e − ( R / 2 L ) t cos(ω′t + φ ) + 2ω′A e − ( R / 2 L ) t sin(ω′t + φ ) − ω′2 Ae − ( R / 2 L )t cos(ω′t + φ ) 2 2L dt ⎝ 2L ⎠ ⎛ ⎛ R ⎞2 d 2 q R dq q R2 1 ⎞ 1 R2 2 2 + + = q − 2. ⎜⎜ ⎜ ⎟ − ω ′ − 2 + ⎟⎟ = 0 , so ω′ = 2 dt L dt LC 2 L LC ⎠ LC 4 L ⎝ ⎝ 2L ⎠ dq dq R Q = 0 , so q = Acos φ = Q and =− Acos φ − ω′Asin φ = 0 . This gives A = and (b) At t = 0, q = Q, i = dt dt 2L cos φ R R . =− 2 Lω′ 2 L 1/ LC − R 2 / 4 L2 EVALUATE: If R = 0 , then A = Q and φ = 0 . IDENTIFY: Evaluate Eq.(30.29). SET UP: The angular frequency of the circuit is ω ′ . 1 1 EXECUTE: (a) When R = 0, ω0 = = = 298 rad s. LC (0.450 H) (2.50 ×10−5 F)

tan φ = −

30.41.

(b) We want

R=

30.42.

ω (1 LC − R 2 4 L2 ) R 2C = 0.95 , so = 1− = (0.95) 2 . This gives ω0 1 LC 4L

4L 4(0.450 H)(0.0975) = 83.8 Ω. (1 − (0.95) 2 ) = C (2.50 ×10−5 F)

EVALUATE: When R increases, the angular frequency decreases and approaches zero as R → 2 L / C . IDENTIFY: L has units of H and C has units of F. SET UP: 1 H = 1 Ω ⋅ s . C = q / V says 1 F = 1 C/V. V = IR says 1 V/A = 1 Ω . EXECUTE:

The units of L / C are

H Ω⋅s Ω⋅ V = = = Ω 2 . Therefore, the unit of F CV A

L / C is Ω.

1 R2 and must have the same units, so R and LC 4 L2 same units, and we have shown that this is indeed the case. EVALUATE:

30.43.

IDENTIFY:

For Eq.(30.28) to be valid,

L / C must have the

The emf E2 in solenoid 2 produced by changing current i1 in solenoid 1 is given by E2 = M

Δi1 . The Δt

mutual inductance of two solenoids is derived in Example 30.1. For the two solenoids in this problem μ AN N M = 0 1 2 , where A is the cross-sectional area of the inner solenoid and l is the length of the outer solenoid. l


Inductance

SET UP:

30-11

μ 0 = 4π ×10−7 T ⋅ m/A . Let the outer solenoid be solenoid 1.

EXECUTE:

(a) M =

(4π ×10−7 T ⋅ m/A)π (6.00 ×10−4 m) 2 (6750)(15) = 2.88 ×10−7 H = 0.288 μ H 0.500 m

Δi1 = (2.88 ×10−7 H)(37.5 A/s) = 1.08 ×10−5 V Δt EVALUATE: If current in the inner solenoid changed at 37.5 A/s, the emf induced in the outer solenoid would be 1.08 ×10−5 V. di IDENTIFY: Apply E = − L and Li = N Φ B . dt SET UP: Φ B is the flux through one turn. (b) E2 =

30.44.

EXECUTE:

(a) E = − L

ε = (3.50 ×10

−3

di d = −(3.50 ×10−3 H) ((0.680 A)cos(π t /[0.0250 s])). dt dt

H)(0.680 A)

π 0.0250 s

Emax = (3.50 ×10−3 H)(0.680 A)

sin(π t /[0.0250 s]) . Therefore,

π 0.0250 s

= 0.299 V.

Limax (3.50 ×10−3 H)(0.680 A) = = 5.95 ×10−6 Wb. N 400 di (c) E (t ) = − L = − (3.50 ×10−3 H)(0.680 A)(π / 0.0250 s)sin(π t / 0.0250 s). dt E (t ) = − (0.299 V)sin((125.6 s −1 )t ) .Therefore, at t = 0.0180 s , (b) Φ B max =

E (0.0180 s) = − (0.299 V)sin((125.6 s −1 )(0.0180 s)) = 0.230 V . The magnitude of the induced emf is 0.230 V. EVALUATE: 30.45.

The maximum emf is when i = 0 and at this instant Φ B = 0.

di . dt SET UP: During an interval in which the graph of i versus t is a straight line, di / dt is constant and equal to the slope of that line. EXECUTE: (a) The pattern on the oscilloscope is sketched in Figure 30.45. EVALUATE: (b) Since the voltage is determined by the derivative of the current, the V versus t graph is indeed proportional to the derivative of the current graph. IDENTIFY:

E = −L

Figure 30.45 30.46.

IDENTIFY:

Apply E = − L

di . dt

d cos(ω t ) = −ω sin(ω t ) dt di d EXECUTE: (a) E = − L = − L ((0.124 A)cos[(240 π s)t ]. dt dt SET UP:

E = + (0.250 H) (0.124 A) (240 π / s)sin((240π s)t ) = +(23.4 V) sin ((240 π s)t ).

The graphs are given in Figure 30.46. (b) Emax = 23.4 V; i = 0, since the emf and current are 90° out of phase. (c) imax = 0.124 A; E = 0, since the emf and current are 90° out of phase.


30-12

Chapter 30

EVALUATE:

The induced emf depends on the rate at which the current is changing.

Figure 30.46 30.47.

di to the series and parallel combinations. dt SET UP: In series, i1 = i2 and the voltages add. In parallel the voltages are the same and the currents add. di di di di di di EXECUTE: (a) Series: L1 1 + L2 2 = Leq , but i1 = i2 = i for series components so 1 = 2 = and dt dt dt dt dt dt L1 + L2 = Leq . IDENTIFY:

Apply E = − L

(b) Parallel: Now L1

di di1 di2 di1 di di di L di = + = L2 2 = Leq , where i = i1 + i2 . Therefore, . But 1 = eq and dt dt dt dt dt dt dt L1 dt −1

30.48.

⎛1 1⎞ di2 Leq di di Leq di Leq di = . and Leq = ⎜ + ⎟ . = + dt L2 dt dt L1 dt L2 dt ⎝ L1 L2 ⎠ EVALUATE: Inductors in series and parallel combine in the same way as resistors. IDENTIFY: Follow the steps outlined in the problem. SET UP: The energy stored is U = 12 Li 2 . ! ! μi EXECUTE: (a) AB ⋅ dl = μ0 I encl ⇒ B 2πr = μ0i ⇒ B = 0 . 2πr μ0i ldr. (b) d Φ B = BdA = 2πr b b μ il dr μ0il = ln ( b a ). (c) Φ B = ∫ d Φ B = 0 ∫ 2π a r 2π a

NΦB μ = l 0 ln( b a). i 2π 1 2 1 μ0 μ li 2 (e) U = Li = l ln( b a )i 2 = 0 ln( b a ). 2 2 2π 4π EVALUATE: The magnetic field between the conductors is due only to the current in the inner conductor. (a) IDENTIFY and SET UP: An end view is shown in Figure 30.49. (d) L =

30.49.

Apply Ampere’s law to a circular path of radius r. ! ! AB ⋅ dl = μ 0 I encl Figure 30.49 ! ! EXECUTE: AB ⋅ dl = B (2π r ) I encl = i, the current in the inner conductor μi Thus B(2π r ) = μ 0i and B = 0 . 2π r


Inductance

30-13

(b) IDENTIFY and SET UP: Follow the procedure specified in the problem. B2 EXECUTE: u = 2μ 0 dU = u dV , where dV = 2π rl dr 2

dU =

1 ⎛ μ 0i ⎞ μ 0i 2 l π rl dr dr (2 ) = ⎜ ⎟ 2μ0 ⎝ 2π r ⎠ 4π r

μ 0i 2l b dr μ 0i 2l = [ln r ]ba 4π ∫ a r 4π μ i 2l μ i 2l ⎛ b ⎞ U = 0 (ln b − ln a ) = 0 ln ⎜ ⎟ 4π 4π ⎝a⎠

(c) U = ∫ dU =

(d) Eq.(30.9): U = 12 Li 2

Part (c): U = 1 2

Li 2 =

L=

30.50.

μ 0i 2 l ⎛ b ⎞ ln ⎜ ⎟ 4π ⎝a⎠

μ 0i 2l ⎛ b ⎞ ln ⎜ ⎟ 4π ⎝a⎠

μ 0l ⎛ b ⎞ ln ⎜ ⎟ . 2π ⎝ a ⎠

EVALUATE: The value of L we obtain from these energy considerations agrees with L calculated in part (d) of Problem 30.48 by considering flux and Eq.(30.6) NΦ NΦB IDENTIFY: Apply L = to each solenoid, as in Example 30.3. Use M = 2 B 2 to calculate the mutual i1 i inductance M. SET UP: The magnetic field produced by solenoid 1 is confined to the space within its windings and is equal to μ Ni B1 = 0 1 1 . 2π r N1Φ B1 N1 A ⎛ μ0 N1i1 ⎞ μ0 N12 A N 2Φ B2 N 2 A ⎛ μ0 N 2i2 ⎞ μ0 N 2 2 A . EXECUTE: (a) L1 = = , L2 = = ⎜ ⎟= ⎜ ⎟= 2π r i1 i1 ⎝ 2π r ⎠ i2 i2 ⎝ 2π r ⎠ 2π r 2

⎛ μ N N A⎞ μ N 2 A μ0 N 2 2 A N 2 AB1 μ 0 N1 N 2 A = . M2 =⎜ 0 1 2 ⎟ = 0 1 = L1L2 . 2π r 2π r 2π r i1 ⎝ 2π r ⎠ EVALUATE: If the two solenoids are identical, so that N1 = N 2 , then M = L . (b) M =

30.51.

IDENTIFY:

U = 12 LI 2 . The self-inductance of a solenoid is found in Exercise 30.11 to be L =

μ 0 AN 2 l

.

The length l of the solenoid is the number of turns divided by the turns per unit length. 2U 2(10.0 J) EXECUTE: (a) L = 2 = = 8.89 H I (1.50 A) 2 μ AN 2 (b) L = 0 . If α is the number of turns per unit length, then N = α l and L = μ 0 Aα 2l . For this coil l L 8.89 H = = 56.3 m . α = 10 coils/mm = 10 × 103 coils/m . l = 2 −7 (4π ×10 T ⋅ m/A)π (0.0200 m) 2 (10 ×103 coils/m) 2 μ 0 Aα This is not a practical length for laboratory use. EVALUATE: The number of turns is N = (56.3 m)(10 ×103 coils/m) = 5.63 ×105 turns . The length of wire in the SET UP:

solenoid is the circumference C of one turn times the number of turns. C = π d = π (4.00 ×10−2 m) = 0.126 m . The length of wire is (0.126 m)(5.63 × 105 ) = 7.1× 104 m = 71 km . This length of wire will have a large resistance and 30.52.

I 2 R electrical energy loses will be very large. IDENTIFY: This is an R-L circuit and i (t ) is given by Eq.(30.14). SET UP: When t → ∞ , i → if = V / R . V 12.0 V EXECUTE: (a) R = = = 1860 Ω. if 6.45 ×10−3 A Rt − Rt −(1860 Ω)(7.25 ×10−4 s) (b) i = if (1 − e − ( R / L )t ) so = −ln(1 − i / if ) and L = = = 0.963 H . L ln(1 − i / if ) ln(1 − (4.86 / 6.45))


30-14

Chapter 30

30.53.

EVALUATE: The current after a long time depends only on R and is independent of L. The value of R / Ldetermines how rapidly the final value of i is reached. IDENTIFY and SET UP: Follow the procedure specified in the problem. L = 2.50 H, R = 8.00 Ω,

E = 6.00 V. i = (E / R)(1 − e −t / τ ), τ = L / R EXECUTE: (a) Eq.(30.9): U L = 12 Li 2 t = τ so i = (E / R )(1 − e −1 ) = (6.00 V/8.00 Ω)(1 − e −1 ) = 0.474 A

Then U L = 12 Li 2 = 12 (2.50 H)(0.474 A) 2 = 0.281 J dU L di Exercise 30.27 (c): PL = = Li dt dt di ⎛ E ⎞ − ( R / L ) t E −t /τ ⎛E⎞ = ⎜ ⎟e = e i = ⎜ ⎟ (1 − e −t /τ ); R dt L ⎝ ⎠ ⎝ L⎠ 2 ⎛E ⎞⎛ E ⎞ E PL = L ⎜ (1 − e− t / τ ) ⎟⎜ e− t / τ ⎟ = (e− t / τ − e −2 ± / τ ) R L ⎝ ⎠⎝ ⎠ R τ

E 2 τ −t /τ E2 ⎡ τ ⎤ −τ e −t /τ + e −2t /τ ⎥ (e − e −2t /τ ) dt = ∫ 0 R 0 R ⎢⎣ 2 ⎦0 2 τ E2 E τ 2 / 1 2 −t /τ − t − − ⎤⎦ = τ ⎡⎣1 − 12 − e + 12 e ⎤⎦ U L = − τ ⎡⎣e − 12 e 0 R R 2 2 ⎛ E ⎞⎛ L ⎞ E⎞ −1 −2 −1 −2 1⎛ − + = (1 2 ) UL = ⎜ e e ⎟⎜ ⎟ ⎟ L(1 − 2e + e ) 2⎜ 2 R R R ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ τ

U L = ∫ PL dt =

2

⎛ 6.00 V ⎞ U L = 12 ⎜ ⎟ (2.50 H)(0.3996) = 0.281 J, which checks. ⎝ 8.00 Ω ⎠ 2

⎛E ⎞ E (b) Exercise 30.27(a): The rate at which the battery supplies energy is PE = Ei = E ⎜ (1 − e − t /τ ) ⎟ = (1 − e − t /τ ) ⎝R ⎠ R τ

U E = ∫ PE dt = 0

τ ⎛ E2 ⎞ E2 τ E2 ⎡t + τ e −t /τ ⎦⎤ = ⎜ ⎟ (τ + τ e−1 − τ ) (1 − e −t /τ )dt = ∫ ⎣ 0 R 0 R ⎝ R⎠ 2

⎛ E2 ⎞ ⎛ E 2 ⎞⎛ L ⎞ ⎛E⎞ U E = ⎜ ⎟τ e −1 = ⎜ ⎟⎜ ⎟ e −1 = ⎜ ⎟ Le −1 ⎝R⎠ ⎝ R⎠ ⎝ R ⎠⎝ R ⎠ 2

⎛ 6.00 V ⎞ UE = ⎜ ⎟ (2.50 H)(0.3679) = 0.517 J ⎝ 8.00 Ω ⎠ ⎛ E2 ⎞ E2 (c) PR = i 2 R = ⎜ ⎟ (1 − e−t /τ ) 2 = (1 − 2e−t /τ + e−2t /τ ) R ⎝ R⎠ τ

U R = ∫ PR dt = 0

UR =

τ

E2 τ E2 ⎡ τ ⎤ − t /τ −2 t / τ − + = (1 2 e e ) dt t + 2τ e− t /τ − e−2t /τ ⎥ R ∫0 R ⎢⎣ 2 ⎦0

E2 ⎡ τ τ ⎤ E2 ⎡ τ τ ⎤ τ + 2τ e−1 − e−2 − 2τ + ⎥ = ⎢ − + 2τ e −1 − e −2 ⎥ ⎢ R⎣ 2 2⎦ R ⎣ 2 2 ⎦

⎛ E 2 ⎞⎛ L ⎞ −1 −2 UR = ⎜ ⎟⎜ ⎟ ⎡⎣ −1 + 4e − e ⎤⎦ ⎝ 2R ⎠ ⎝ R ⎠ 2

30.54.

2

⎛E⎞ ⎛ 6.00 V ⎞ 1 U R = ⎜ ⎟ ( 12 L) ⎡⎣ −1 + 4e−1 − e −2 ⎤⎦ = ⎜ ⎟ 2 (2.50 H)(0.3362) = 0.236 J R ⎝ ⎠ ⎝ 8.00 Ω ⎠ (d) EVALUATE: U E = U R + U L . (0.517 J = 0.236 J + 0.281 J) The energy supplied by the battery equals the sum of the energy stored in the magnetic field of the inductor and the energy dissipated in the resistance of the inductor. IDENTIFY: This is a decaying R-L circuit with I 0 = E / R . i (t ) = I 0e − ( R / L )t . SET UP: E = 60.0 V , R = 240 Ω and L = 0.160 H . The rate at which energy stored in the inductor is decreasing is iLdi / dt . 2

EXECUTE:

(a) U =

2

⎛ 60 V ⎞ 1 1 ⎛E⎞ 1 −3 LI 0 2 = L ⎜ ⎟ = (0.160 H) ⎜ ⎟ = 5.00 ×10 J. Ω 2 2 ⎝ R⎠ 2 240 ⎝ ⎠


Inductance

(b) i =

30-15

E − ( R / L )t E2 di R dU L di dU L (60 V) 2 −2(240 / 0.160)(4.00×10−4 ) ⇒ =− i⇒ = iL = − Ri 2 = e−2( R / L ) t . =− = −4.52 W. e e R dt L dt dt R dt 240 Ω

(c) In the resistor, PR =

dU R 2 (60 V) 2 −2(240 / 0.160)(4.00×10−4 ) E2 = i R = e −2( R / L ) t = = 4.52 W . e dt R 240 Ω ∞

(d) PR (t ) = i 2 R =

30.55.

E 2 −2( R / L )t E 2 −2( R / L ) t E 2 L (60 V) 2 (0.160 H) e . UR = = = = 5.00 ×10−3 J, which is the same as e R R ∫0 R 2R 2(240 Ω) 2

part (a). EVALUATE: During the decay of the current all the electrical energy originally stored in the inductor is dissipated in the resistor. IDENTIFY and SET UP: Follow the procedure specified in the problem. 12 Li 2 is the energy stored in the inductor di q − = 0. dt C d d d di di qi ⎛ di ⎞ EXECUTE: Multiplying by –i gives i 2 R + Li + = 0. U L = ( 12 Li 2 ) = 12 L ( i 2 ) = 12 L ⎜ 2i ⎟ = Li , the dt dt dt dt dt C ⎝ dt ⎠ and q 2 / 2C is the energy stored in the capacitor. The equation is −iR − L

d d ⎛ q2 ⎞ 1 d 2 1 dq qi UC = ⎜ (q ) = (2q ) = , the third term. i 2 R = PR , the rate at which ⎟= dt dt ⎝ 2C ⎠ 2C dt 2C dt C d electrical energy is dissipated in the resistance. U L = PL , the rate at which the amount of energy stored in the dt d inductor is changing. U C = PC , the rate at which the amount of energy stored in the capacitor is changing. dt EVALUATE: The equation says that PR + PL + PC = 0; the net rate of change of energy in the circuit is zero. Note second term.

that at any given time one of PC or PL is negative. If the current and U L are increasing the charge on the capacitor and U C are decreasing, and vice versa. 30.56.

IDENTIFY: The energy stored in a capacitor is U C = 12 Cv 2 . The energy stored in an inductor is U L = 12 Li 2 . Energy conservation requires that the total stored energy be constant. SET UP: The current is a maximum when the charge on the capacitor is zero and the energy stored in the capacitor is zero. EXECUTE: (a) Initially v = 16.0 V and i = 0 . U L = 0 and U C = 12 Cv 2 = 12 (5.00 ×10−6 F)(16.0 V)2 = 6.40 ×10−4 J . The total energy stored is 0.640 mJ . (b) The current is maximum when q = 0 and U C = 0 . U C + U L = 6.40 ×10−4 J so U L = 6.40 ×10−4 J .

2(6.40 ×10−4 J) = 0.584 A . 3.75 ×10−3 H EVALUATE: The maximum charge on the capacitor is Q = CV = 80.0 μ C . 1 2

30.57.

2 Limax = 6.40 ×10−4 J and imax =

IDENTIFY and SET UP: Use U C = 12 CVC2 (energy stored in a capacitor) to solve for C. Then use Eq.(30.22) and ω = 2π f to solve for the L that gives the desired current oscillation frequency. EXECUTE:

1 (2 π f )2 C 2π LC f = 3500 Hz gives L = 9.31 μ H EVALUATE: f is in Hz and ω is in rad/s; we must be careful not to confuse the two. IDENTIFY: Apply energy conservation to the circuit. SET UP: For a capacitor V = q / C and U = q 2 / 2C . For an inductor U = 12 Li 2 f =

30.58.

1

VC = 12.0 V; U C = 12 CVC2 so C = 2U C / VC2 = 2(0.0160 J)/(12.0 V) 2 = 222 μ F

EXECUTE: (b)

1 2 Limax 2

(c) U max =

so L =

Q 6.00 ×10−6 C = = 0.0240 V. C 2.50 ×10−4 F Q2 Q 6.00 ×10−6 C = , so imax = = = 1.55 ×10−3 A 2C LC (0.0600 H)(2.50 ×10−4 F) (a) Vmax =

1 2 1 Limax = (0.0600 H)(1.55 ×10−3 A) 2 = 7.21×10−8 J. 2 2


30-16

Chapter 30

30.59.

3 1 1 (d) If i = imax then U L = U max = 1.80 × 10 −8 J and U C = U max = 4 2 4 3 q= Q = 5.20 ×10−6 C. 4 1 1 q2 EVALUATE: U max = Li 2 + for all times. 2 2C IDENTIFY: Set U B = K , where K = 12 mv 2 .

(

(3/ 4)Q 2C

)

2

=

q2 . This gives 2C

SET UP: The energy density in the magnetic field is u B = B 2 / 2μ 0 . Consider volume V = 1 m 3 of sunspot material. EXECUTE: The energy density in the sunspot is uB = B 2 /2 μ0 = 6.366 ×104 J /m3. The total energy stored in

volume V of the sunspot is U B = u BV . The mass of the material in volume V of the sunspot is m = ρV . K = U B so 30.60.

1 2

mv 2 = U B .

1 2

ρVv 2 = uBV . The volume divides out, and v = 2uB / ρ = 2 ×104 m/s .

EVALUATE: The speed we calculated is about 30 times smaller than the escape speed. IDENTIFY: i (t ) is given by Eq.(30.14). SET UP: The graph shows V = 0 at t = 0 and V approaches the constant value of 25 V at large times. EXECUTE: (a) The voltage behaves the same as the current. Since VR is proportional to i, the scope must be across the 150 Ω resistor. (b) From the graph, as t → ∞, VR → 25 V, so there is no voltage drop across the inductor, so its internal resistance ⎛ 1⎞ must be zero. VR = Vmax (1 − e −t / r ) . When t = τ , VR = Vmax ⎜1 − ⎟ ≈ 0.63Vmax . From the graph, V = 0.63Vmax = 16 V at ⎝ e⎠ t ≈ 0.5 ms . Therefore τ = 0.5 ms . L / R = 0.5 ms gives L = (0.5 ms) (150 Ω) = 0.075 H . (c) The graph if the scope is across the inductor is sketched in Figure 30.60. EVALUATE: At all times VR + VL = 25.0 V . At t = 0 all the battery voltage appears across the inductor since i = 0 . At t → ∞ all the battery voltage is across the resistance, since di / dt = 0 .

30.61.

Figure 30.60 IDENTIFY and SET UP: The current grows in the circuit as given by Eq.(30.14). In an R-L circuit the full emf initially is across the inductance and after a long time is totally across the resistance. A solenoid in a circuit is represented as a resistance in series with an inductance. Apply the loop rule to the circuit; the voltage across a resistance is given by Ohm’s law. EXECUTE: (a) In the R-L circuit the voltage across the resistor starts at zero and increases to the battery voltage. The voltage across the solenoid (inductor) starts at the battery voltage and decreases to zero. In the graph, the voltage drops, so the oscilloscope is across the solenoid. (b) At t → ∞ the current in the circuit approaches its final, constant value. The voltage doesn’t go to zero because the solenoid has some resistance RL . The final voltage across the solenoid is IRL , where I is the final current in the circuit. (c) The emf of the battery is the initial voltage across the inductor, 50 V. Just after the switch is closed, the current is zero and there is no voltage drop across any of the resistance in the circuit. (d) As t → ∞, E − IR − IRL = 0

E = 50 V and from the graph IRL = 15 V (the final voltage across the inductor), so IR = 35 V and I = (35 V)/R = 3.5 A (e) IRL = 15 V, so RL = (15 V)/(3.5 A) = 4.3 Ω E − VL − iR = 0, where VL includes the voltage across the resistance of the solenoid. E VL = E − iR, i = (1 − e−t /τ ) , so VL = E ⎡⎢1 − RR (1 − e−t /τ ) ⎤⎥ Rtot tot ⎣ ⎦ E = 50 V, R = 10 Ω, Rtot = 14.3 Ω, so when t = τ , VL = 27.9 V. From the graph, VL has this value when t = 3.0 ms (read approximately from the graph), so τ = L / Rtot = 3.0 ms. Then L = (3.0 ms)(14.3 Ω ) = 43 mH. EVALUATE: At t = 0 there is no current and the 50 V measured by the oscilloscope is the induced emf due to the inductance of the solenoid. As the current grows, there are voltage drops across the two resistances in the circuit.


Inductance

30-17

We derived an equation for VL , the voltage across the solenoid. At t = 0 it gives VL = E and at t → ∞ it gives 30.62.

30.63.

VL = ER / Rtot = iR. IDENTIFY: At t = 0 , i = 0 through each inductor. At t → ∞ , the voltage is zero across each inductor. SET UP: In each case redraw the circuit. At t = 0 replace each inductor by a break in the circuit and at t → ∞ replace each inductor by a wire. EXECUTE: (a) Initially the inductor blocks current through it, so the simplified equivalent circuit is shown in E 50 V = 0.333 A . V1 = (100 Ω)(0.333 A) = 33.3 V . V4 = (50 Ω)(0.333 A) = 16.7 V . V3 = 0 Figure 30.62a. i = = R 150 Ω since no current flows through it. V2 = V4 = 16.7 V , since the inductor is in parallel with the 50 Ω resistor. A1 = A3 = 0.333 A, A2 = 0 . (b) Long after S is closed, steady state is reached, so the inductor has no potential drop across it. The simplified 50 V = 0.385 A . V1 = (100 Ω)(0.385 A) = 38.5 V ; V2 = 0 ; circuit is sketched in Figure 30.62b. i = E/R = 130 Ω 11.5 V 11.5 V = 0.153 A; i3 = = 0.230 A . V3 = V4 = 50 V − 38.5 V = 11.5 V . i1 = 0.385 A; i2 = 75 Ω 50 Ω EVALUATE: Just after the switch is closed the current through the battery is 0.333 A. After a long time the current through the battery is 0.385 A. After a long time there is an additional current path, the equivalent resistance of the circuit is decreased and the current has increased.

Figure 30.62 IDENTIFY and SET UP: Just after the switch is closed, the current in each branch containing an inductor is zero and the voltage across any capacitor is zero. The inductors can be treated as breaks in the circuit and the capacitors can be replaced by wires. After a long time there is no voltage across each inductor and no current in any branch containing a capacitor. The inductors can be replaced by wires and the capacitors by breaks in the circuit. EXECUTE: (a) Just after the switch is closed the voltage V5 across the capacitor is zero and there is also no current through the inductor, so V3 = 0. V2 + V3 = V4 = V5 , and since V5 = 0 and V3 = 0, V4 and V2 are also zero.

V4 = 0 means V3 reads zero. V1 then must equal 40.0 V, and this means the current read by A1 is (40.0 V)/(50.0 Ω) = 0.800 A. A2 + A3 + A4 = A1 , but A2 = A3 = 0 so A4 = A1 = 0.800 A. A1 = A4 = 0.800 A; all other

ammeters read zero. V1 = 40.0 V and all other voltmeters read zero. (b) After a long time the capacitor is fully charged so A4 = 0. The current through the inductor isn’t changing, so V2 = 0. The currents can be calculated from the equivalent circuit that replaces the inductor by a short circuit, as shown in Figure 30.63a.

Figure 30.63a I = (40.0 V)/(83.33 Ω) = 0.480 A; A1 reads 0.480 A V1 = I (50.0 Ω) = 24.0 V The voltage across each parallel branch is 40.0 V – 24.0 V = 16.0 V V2 = 0, V3 = V4 = V5 = 16.0 V

V3 = 16.0 V means A2 reads 0.160 A. V4 = 16.0 V means A3 reads 0.320 A. A4 reads zero. Note that A2 + A3 = A1. (c) V5 = 16.0 V so Q = CV = (12.0 μ F)(16.0 V) = 192 μ C


30-18

Chapter 30

(d) At t = 0 and t → ∞, V2 = 0. As the current in this branch increases from zero to 0.160 A the voltage V2 reflects the rate of change of the current. The graph is sketched in Figure 30.63b.

Figure 30.63b

30.64.

30.65.

EVALUATE: This reduction of the circuit to resistor networks only apply at t = 0 and t → ∞. At intermediate times the analysis is complicated. IDENTIFY: At all times v1 + v2 = 25.0 V . The voltage across the resistor depends on the current through it and the voltage across the inductor depends on the rate at which the current through it is changing. SET UP: Immediately after closing the switch the current thorough the inductor is zero. After a long time the current is no longer changing. EXECUTE: (a) i = 0 so v1 = 0 and v2 = 25.0 V . The ammeter reading is A = 0 . v 25.0 V (b) After a long time, v2 = 0 and v1 = 25.0 V. v1 = iR and i = 1 = = 1.67 A. The ammeter reading is R 15.0 Ω A = 1.67A. (c) None of the answers in (a) and (b) depend on L so none of them would change. EVALUATE: The inductance L of the circuit affects the rate at which current reaches its final value. But after a long time the inductor doesn’t affect the circuit and the final current does not depend on L. IDENTIFY: At t = 0 , i = 0 through each inductor. At t → ∞ , the voltage is zero across each inductor. SET UP: In each case redraw the circuit. At t = 0 replace each inductor by a break in the circuit and at t → ∞ replace each inductor by a wire. EXECUTE: (a) Just after the switch is closed there is no current through either inductor and they act like breaks in the circuit. The current is the same through the 40.0 Ω and 15.0 Ω resistors and is equal to

(25.0 V) (40.0 Ω + 15.0 Ω) = 0.455 A. A1 = A4 = 0.455 A; A2 = A3 = 0. (b) After a long time the currents are constant, there is no voltage across either inductor, and each inductor can be treated as a short-circuit . The circuit is equivalent to the circuit sketched in Figure 30.65. I = (25.0 V) (42.73 Ω) = 0.585 A . A1 reads 0.585 A. The voltage across each parallel branch is 25.0 V − (0.585 A)(40.0 Ω) = 1.60 V . A2 reads (1.60 V) /(5.0 Ω) = 0.320 A . A3 reads (1.60 V) (10.0 Ω) = 0.160 A . A4 reads (1.60 V) (15.0 Ω) = 0.107 A. EVALUATE: Just after the switch is closed the current through the battery is 0.455 A. After a long time the current through the battery is 0.585 A. After a long time there are additional current paths, the equivalent resistance of the circuit is decreased and the current has increased.

Figure 30.65 30.66.

IDENTIFY: Closing S 2 and simultaneously opening S1 produces an L-C circuit with initial current through the inductor of 3.50 A. When the current is a maximum the charge q on the capacitor is zero and when the charge q is 2 a maximum the current is zero. Conservation of energy says that the maximum energy 12 Limax stored in the inductor 2 qmax stored in the capacitor. C = 3.50 A , the current in the inductor just after the switch is closed.

equals the maximum energy SET UP:

imax

1 2


Inductance

EXECUTE:

30.67.

(a)

1 2

2 Limax = 12

30-19

2 qmax . C

qmax = ( LC )imax = (2.0 ×10−3 H)(5.0 ×10−6 F)(3.50 A) = 3.50 ×10−4 C = 0.350 mC . (b) When q is maximum, i = 0 . EVALUATE: In the final circuit the current will oscillate. IDENTIFY: Apply the loop rule to each parallel branch. The voltage across a resistor is given by iR and the voltage across an inductor is given by L di / dt . The rate of change of current through the inductor is limited. SET UP:

With S closed the circuit is sketched in Figure 30.67a. The rate of change of the current through the inductor is limited by the induced emf. Just after the switch is closed the current in the inductor has not had time to increase from zero, so i2 = 0. Figure 30.67a

EXECUTE: (a) E − vab = 0, so vab = 60.0 V (b) The voltage drops across R, as we travel through the resistor in the direction of the current, so point a is at higher potential. (c) i2 = 0 so vR2 = i2 R2 = 0

E − vR2 − vL = 0 so vL = E = 60.0 V (d) The voltage rises when we go from b to a through the emf, so it must drop when we go from a to b through the inductor. Point c must be at higher potential than point d. di (e) After the switch has been closed a long time, 2 → 0 so vL = 0. Then E − vR2 = 0 and i2 R2 = E dt E 60.0 V so i2 = = = 2.40 A. R2 25.0 Ω SET UP: The rate of change of the current through the inductor is limited by the induced emf. Just after the switch is opened again the current through the inductor hasn’t had time to change and is still i2 = 2.40 A. The circuit is sketched in Figure 30.67b. EXECUTE: The current through R1 is i2 = 2.40 A, in the direction b to a. Thus vab = −i2 R1 = −(2.40 A)(40.0 Ω)

vab = −96.0 V Figure 30.67b (f ) Point where current enters resistor is at higher potential; point b is at higher potential. (g) vL − vR1 − vR2 = 0

vL = vR1 + vR2 vR1 = −vab = 96.0 V; vR2 = i2 R2 = (2.40 A)(25.0 Ω) = 60.0 V

Then vL = vR1 + vR2 = 96.0 V + 60.0 V = 156 V. As you travel counterclockwise around the circuit in the direction of the current, the voltage drops across each resistor, so it must rise across the inductor and point d is at higher potential than point c. The current is decreasing, so the induced emf in the inductor is directed in the direction of the current. Thus, vcd = −156 V. (h) Point d is at higher potential. EVALUATE: The voltage across R1 is constant once the switch is closed. In the branch containing R2 , just after S is closed the voltage drop is all across L and after a long time it is all across R2 . Just after S is opened the same current flows in the single loop as had been flowing through the inductor and the sum of the voltage across the resistors equals the voltage across the inductor. This voltage dies away, as the energy stored in the inductor is dissipated in the resistors.


30-20

Chapter 30

30.68.

Apply the loop rule to the two loops. The current through the inductor doesn't change abruptly. di SET UP: For the inductor E = L and E is directed to oppose the change in current. dt EXECUTE: (a) Switch is closed, then at some later time di di = 50.0 A/s ⇒ vcd = L = (0.300 H) (50.0 A/s) = 15.0 V. dt dt 60.0 V = 1.50 A. The top circuit loop: 60.0 V = i1R1 ⇒ i1 = 40.0 Ω 45.0 V = 1.80 A. The bottom loop: 60 V − i2 R2 − 15.0 V = 0 ⇒ i2 = 25.0 Ω 60.0 V = 2.40 A, and immediately when the switch is opened, the inductor maintains (b) After a long time: i2 = 25.0 Ω this current, so i1 = i2 = 2.40 A. IDENTIFY:

EVALUATE: 30.69.

The current through R1 changes abruptly when the switch is closed.

IDENTIFY and SET UP: The circuit is sketched in Figure 30.69a. Apply the loop rule. Just after S1 is closed, i = 0. After a long time i has reached its final value and di/dt = 0. The voltage across a resistor depends on i and the voltage across an inductor depends on di/dt.

Figure 30.69a EXECUTE: (a) At time t = 0, i0 = 0 so vac = i0 R0 = 0. By the loop rule E − vac − vcb = 0, so vcb = E − vac = E = 36.0 V.

( i0 R = 0 so this potential difference of 36.0 V is across the inductor and is an induced emf produced by the changing current.) di di (b) After a long time 0 → 0 so the potential − L 0 across the inductor becomes zero. The loop rule gives dt dt E − i0 ( R0 + R) = 0. E 36.0 V i0 = = = 0.180 A R0 + R 50.0 Ω + 150 Ω vac = i0 R0 = (0.180 A)(50.0 Ω) = 9.0 V di Thus vcb = i0 R + L 0 = (0.180 A)(150 Ω) + 0 = 27.0 V (Note that vac + vcb = E. ) dt (c) E − vac − vcb = 0 di E − iR0 − iR − L = 0 dt ⎛ L ⎞ di di E L = E − i( R0 + R) and ⎜ ⎟ = −i + dt R R dt R R0 + + 0 ⎠ ⎝

di ⎛ R + R0 ⎞ =⎜ ⎟ dt −i + E /( R + R0 ) ⎝ L ⎠ Integrate from t = 0, when i = 0, to t, when i = i0 :

i0

0

i0

⎡ di R + R0 t E ⎤ ⎛ R + R0 ⎞ = dt = − ln ⎢ −i + ⎥ =⎜ ⎟ t, ∫ 0 −i + E /( R + R0 ) L R + R0 ⎦ 0 ⎝ L ⎠ ⎣

⎛ ⎛ E ⎞ E ⎞ ⎛ R + R0 ⎞ so ln ⎜ −i0 + ⎟ − ln ⎜ ⎟ = −⎜ ⎟t R R R R + + ⎝ L ⎠ 0 ⎠ 0 ⎠ ⎝ ⎝ ⎛ −i + E /( R + R0 ) ⎞ ⎛ R + R0 ⎞ ln ⎜ 0 ⎟ = −⎜ ⎟t ⎝ L ⎠ ⎝ E /( R + R0 ) ⎠ E −i0 + E /( R + R0 ) 1 − e− ( R + R0 ) t / L = e− ( R + R0 ) t / L and i0 = R + R0 E /( R + R0 ) 36.0 V Substituting in the numerical values gives i0 = (1 − e−(200 Ω / 4.00 H)t ) = (0.180 A) (1 − e−t / 0.020 s ) 50 Ω + 150 Ω

Taking exponentials of both sides gives

(

)


Inductance

30-21

At t → 0, i0 = (0.180 A)(1 − 1) = 0 (agrees with part (a)). At t → ∞, i0 = (0.180 A)(1 − 0) = 0.180 A (agrees with part (b)). ER0 vac = i0 R0 = 1 − e− ( R + R0 )t / L = 9.0 V (1 − e−t / 0.020 s ) R + R0

(

)

vcb = E − vac = 36.0 V − 9.0 V(1 − e−t / 0.020 s ) = 9.0 V(3.00 + e−t / 0.020 s ) At t → 0, vac = 0, vcb = 36.0 V (agrees with part (a)). At t → ∞, vac = 9.0 V, vcb = 27.0 V (agrees with part (b)). The graphs are given in Figure 30.69b.

30.70.

Figure 30.69b EVALUATE: The expression for i(t) we derived becomes Eq.(30.14) if the two resistors R0 and R in series are replaced by a single equivalent resistance R0 + R. IDENTIFY: Apply the loop rule. The current through the inductor doesn't change abruptly. SET UP: With S2 closed, vcb must be zero. EXECUTE: (a) Immediately after S 2 is closed, the inductor maintains the current i = 0.180 A through R. The loop rule around the outside of the circuit yields 36 V = 0.720 A . E + EL − iR − i0 R0 = 36.0 V + (0.18 A)(150 Ω ) − (0.18 A)(150 Ω ) − i0 (50 Ω ) = 0 . i0 = 50 Ω vac = (0.72 A)(50 V) = 36.0 V and vcb = 0 . E 36.0 V = = 0.720 A , iR = 0 and is 2 = 0.720 A . (b) After a long time, vac = 36.0 V, and vcb = 0. Thus i0 = R0 50 Ω −1 E − ( R L )t e (c) i0 = 0.720 A , iR (t ) = and iR (t ) = (0.180 A)e − (12.5 s ) t . Rtotal

is 2 (t ) = (0.720 A) − (0.180 A)e − (12.5 s

−1

)t

(

= (0.180 A) 4 − e − (12.5 s

−1

)t

). The graphs of the currents are given in Figure 30.70.

EVALUATE: R0 is in a loop that contains just E and R0, so the current through R0 is constant. After a long time the current through the inductor isn't changing and the voltage across the inductor is zero. Since vcb is zero, the voltage across R must be zero and iR becomes zero.

Figure 30.70


30-22

Chapter 30

30.71.

IDENTIFY: The current through an inductor doesn't change abruptly. After a long time the current isn't changing and the voltage across each inductor is zero. SET UP: Problem 30.47 shows how to find the equivalent inductance of inductors in series and parallel. EXECUTE: (a) Just after the switch is closed there is no current in the inductors. There is no current in the resistors so there is no voltage drop across either resistor. A reads zero and V reads 20.0 V. (b) After a long time the currents are no longer changing, there is no voltage across the inductors, and the inductors can be replaced by short-circuits. The circuit becomes equivalent to the circuit shown in Figure 30.71a. I = (20.0 V) (75.0 Ω) = 0.267 A . The voltage between points a and b is zero, so the voltmeter reads zero. (c) Use the results of Problem 30.49 to combine the inductor network into its equivalent, as shown in Figure 30.71b. R = 75.0 Ω is the equivalent resistance. Eq.(30.14) says i = (E R)(1 − e −t τ ) with

τ = L / R = (10.8 mH) (75.0 Ω) = 0.144 ms . E = 20.0 V , R = 75.0 Ω , t = 0.115 ms so i = 0.147 A . VR = iR = (0.147 A)(75.0 Ω) = 11.0 V . 20.0 V − VR − VL = 0 and VL = 20.0 V − VR = 9.0 V . The ammeter reads 0.147 A and the voltmeter reads 9.0 V. EVALUATE: The current through the battery increases from zero to a final value of 0.267 A. The voltage across the inductor network drops from 20.0 V to zero.

Figure 30.71 30.72.

IDENTIFY: At steady state with the switch in position 1, no current flows to the capacitors and the inductors can be replaced by wires. Apply conservation of energy to the circuit with the switch in position 2. SET UP: Replace the series combinations of inductors and capacitors by their equivalents. For the inductors use the results of Problem 30.47. E 75.0 V EXECUTE: (a) At steady state i = = = 0.600 A . R 125 Ω (b) The equivalent circuit capacitance of the two capacitors is given by

1 1 1 = + and Cs = 14.6 μ F . Cs 25 μ F 35 μ F

Ls = 15.0 mH + 5.0 mH = 20.0 mH . The equivalent circuit is sketched in Figure 30.72a. q2 1 2 = Li0 . q = i0 LC = (0.600 A) (20 ×10−3 H)(14.6 ×10−6 F) = 3.24 ×10−4 C . As shown 2C 2 in Figure 30.72b, the capacitors have their maximum charge at t = T / 4 .

Energy conservation:

t = 14 T = 14 (2π LC ) =

π

π

LC = (20 ×10−3 H)(14.6 × 10−6 F) = 8.49 ×10−4 s 2 2 EVALUATE: With the switch closed the battery stores energy in the inductors. This then is the energy in the L-C circuit when the switch is in position 2.

Figure 30.72 30.73.

IDENTIFY: SET UP:

Follow the steps specified in the problem. Find the flux through a ring of height h, radius r and thickness dr. Example 28.19 shows that B =

inside the toroid.

μ 0 Ni 2π r


Inductance

μ Nih dr μ0 Nih ⎛ μ Ni ⎞ = (a) Φ B = ∫ B (h dr ) = ∫ ⎜ 0 ⎟ ( h dr ) = 0 ln(b a ). π r 2 2π ∫a r 2π ⎠ a a ⎝ b

EXECUTE: (b) L =

30-23

b

b

N Φ B μ0 N 2h ln(b a ). = i 2π

b − a (b − a ) 2 μ0 N 2h ⎛ b − a ⎞ + + ⋅⋅⋅ ⇒ ≈ L ⎜ ⎟. a 2a 2 2π ⎝ a ⎠ EVALUATE: h(b − a ) is the cross-sectional area A of the toroid and a is approximately the radius r, so this result is approximately the same as the result derived in Example 30.3. IDENTIFY: The direction of the current induced in circuit A is given by Lenz’s law. SET UP: When the switch is closed current flows counterclockwise in the circuit on the left, from the positive plate of the capacitor. The current decreases as a function of time, as the charge and voltage of the capacitor decrease. EXECUTE: At loop A the magnetic field from the wire of the other circuit adjacent to A is into the page. The magnetic field of this current is decreasing, as the current decreases. Therefore, the magnetic field of the induced current in A is directed into the page inside A and to produce a magnetic field in this direction the induced current is clockwise. EVALUATE: The magnitude of the emf induced in circuit A decreases with time after the switch is closed, because the rate of change of the current in the other circuit decreases. (a) IDENTIFY and SET UP: With switch S closed the circuit is shown in Figure 30.75a. (c) ln(b / a ) = ln(1 − (b − a ) / a ) ≈

30.74.

30.75.

Apply the loop rule to loops 1 and 2. EXECUTE: loop 1 E − i1R1 = 0 i1 =

E (independent of t) R1

Figure 30.75a

loop (2) E − i2 R2 − L

di2 =0 dt

E (1 − e− R2t / L ) R2 E E di The expressions derived in part (a) give that as t → ∞, i1 = and i2 = . Since 2 → 0 at dt R1 R2

This is in the form of equation (30.12), so the solution is analogous to Eq.(30.14): i2 = (b) EVALUATE:

steady-state, the inductance then has no effect on the circuit. The current in R1 is constant; the current in R2 starts at zero and rises to E / R2 . (c) IDENTIFY and SET UP:

The circuit now is as shown in Figure 30.75b. Let t = 0 now be when S is opened. E At t = 0, i = . R2

Figure 30.75b Apply the loop rule to the single current loop. di di EXECUTE: −i ( R1 + R2 ) − L = 0. (Now is negative.) dt dt di di ⎛ R + R2 ⎞ = −⎜ 1 L = −i ( R1 + R2 ) gives ⎟ dt dt i ⎝ L ⎠ Integrate from t = 0, when i = I 0 = E / R2 , to t.

i I0

⎛ i ⎞ di ⎛ R1 + R2 ⎞ ⎛ R + R2 ⎞ t = −⎜ 1 ⎟t ⎟ ∫ 0 dt and ln ⎜ ⎟ = − ⎜ I i L ⎝ L ⎠ ⎝ ⎠ ⎝ 0⎠

Taking exponentials of both sides of this equation gives i = I 0e − ( R1 + R2 )t / L =

E − ( R1 + R2 )t / L e R2


30-24

Chapter 30

(d) IDENTIFY and SET UP: Use the equation derived in part (c) and solve for R2 and E . EXECUTE: L = 22.0 H V2 V 2 (120 V) 2 RR1 = = 40.0 W gives R1 = = = 360 Ω. 40.0 W R1 PR1

We are asked to find R2 and E . Use the expression derived in part (c). I 0 = 0.600 A so E / R2 = 0.600 A i = 0.150 A when t = 0.080 s, so i = 1 4

E − ( R1 + R2 )t / L e gives 0.150 A = (0.600 A)e − ( R1 + R2 )t / L R2

= e − ( R1 + R2 ) t / L so ln 4 = ( R1 + R2 )t / L

L ln 4 (22.0 H)ln 4 − R1 = − 360 Ω = 381.2 Ω − 360 Ω = 21.2 Ω t 0.080 s Then E = (0.600 A)R2 = (0.600 A)(21.2 Ω) = 12.7 V. (e) IDENTIFY and SET UP: Use the expressions derived in part (a). R2 =

E 12.7 V = = 0.0353 A R1 360 Ω EVALUATE: When the switch is opened the current through the light bulb jumps from 0.0353 A to 0.600 A. Since the electrical power dissipated in the bulb (brightness) depend on i 2 , the bulb suddenly becomes much brighter. IDENTIFY: Follow the steps specified in the problem. SET UP: The current in an inductor does not change abruptly. EXECUTE: (a) Using Kirchhoff’s loop rule on the left and right branches: di di Left: E − (i1 + i2 ) R − L 1 = 0 ⇒ R (i1 + i2 ) + L 1 = E . dt dt q2 q2 Right: E − (i1 + i2 ) R − = 0 ⇒ R (i1 + i2 ) + = E . C C E (b) Initially, with the switch just closed, i1 = 0, i2 = and q2 = 0. R (c) The substitution of the solutions into the circuit equations to show that they satisfy the equations is a somewhat tedious exercise but straightforward exercise. We will show that the initial conditions are satisfied: E −βt E At t = 0, q2 = e sin(ω t ) = sin(0) = 0. ωR ωR E E i1 (t ) = (1 − e − β t [(2ω RC ) −1 sin(ω t ) + cos(ω t )] ⇒ i1 (0) = (1 − [cos(0)]) = 0. R R 1 1 − = 625 rad/s . (d) When does i2 first equal zero? ω = LC (2 RC ) 2 E i2 (t ) = 0 = e − β t [ −(2ω RC ) −1 sin(ω t ) + cos(ω t )] ⇒ − (2ω RC ) −1 tan(ω t ) + 1 = 0 and R tan(ω t ) = +2ω RC = +2(625 rad/s)(400 Ω)(2.00 ×10−6 F) = +1.00. 0.785 ω t = arctan( + 1.00) = +0.785 ⇒ t = = 1.256 ×10−3 s. 625 rad/s EVALUATE: As t → ∞ , i1 → E / R , q2 → 0 and i2 → 0 . NΦB IDENTIFY: Apply L = to calculate L. i μ Ni μ Ni SET UP: In the air the magnetic field is BAir = 0 . In the liquid, BL = W W μ 0 Ni K μ 0 Ni EXECUTE: (a) Φ B = BA = BL AL + BAir AAir = (( D − d )W ) + ( dW ) = μ 0 Ni[( D − d ) + Kd ] . W W NΦB d d ⎛ L − L0 ⎞ = μ 0 N 2 [( D − d ) + Kd ] = L0 − L0 + Lf = L0 + ⎜ f L= ⎟d . i D D ⎝ D ⎠ ⎛ L − L0 ⎞ 2 2 d =⎜ ⎟ D, where L0 = μ 0 N D, and L f = K μ 0 N D. L L − 0 ⎠ ⎝ f EXECUTE:

30.76.

30.77.

The current through the light bulb before the switch is opened is i1 =


Inductance

30-25

d⎞ ⎛ (b) Using K = χ m + 1 we can find the inductance for any height L = L0 ⎜ 1 + χ m ⎟ . D⎠ ⎝ _______________________________________________________________________________ Height of Fluid Inductance of Liquid Oxygen Inductance of Mercury d=D 4 0.63024 H 0.63000 H d=D 2

30.78.

30.79.

0.63048 H

0.62999 H

d = 3D 4 0.63072 H 0.62999 H d=D 0.63096 H 0.62998 H ________________________________________________________________________________ The values χ m (O 2 ) = 1.52 ×10−3 and χ m (Hg) = −2.9 ×10−5 have been used. EVALUATE: (d) The volume gauge is much better for the liquid oxygen than the mercury because there is an easily detectable spread of values for the liquid oxygen, but not for the mercury. IDENTIFY: The induced emf across the two coils is due to both the self-inductance of each and the mutual inductance of the pair of coils. di SET UP: The equivalent inductance is defined by E = Leq , where E and i are the total emf and current across dt the combination. di di di di di EXECUTE: Series: L1 1 + L2 2 + M 21 1 + M 12 2 ≡ Leq . dt dt dt dt dt di di1 di2 di di + and M 12 = M 21 ≡ M , so ( L1 + L2 + 2M ) = Leq and Leq = L1 + L2 + 2M . But i = i1 + i2 ⇒ = dt dt dt dt dt di di di di di di di di di and L2 2 + M 21 1 = Leq , with 1 + 2 = and M 12 = M 21 ≡ M . Parallel: We have L1 1 + M 12 2 = Leq dt dt dt dt dt dt dt dt dt di di di To simplify the algebra let A = 1 , B = 2 , and C = . So L1 A + MB = LeqC , L2 B + MA = LeqC , A + B = C. Now dt dt dt solve for A and B in terms of C. ( L1 − M ) A + ( M − L2 ) B = 0 using A = C − B . ( L1 − M )(C − B) + ( M − L2 ) B = 0. ( M − L1 ) ( L1 − M )C − ( L1 − M ) B + ( M − L2 ) B = 0. (2M − L1 − L2 ) B = ( M − L1 )C and B = C. (2M − L1 − L2 ) M − L2 ( M − L1 )C (2M − L1 − L2 ) − M + L1 = C , or A = C . Substitute A in B back But A = C − B = C − 2M − L1 − L2 (2M − L1 − L2 ) (2M − L1 − L2 ) into original equation: L1 ( M − L2 )C M ( M − L1 ) M 2 − L1L2 LL −M2 + C = LeqC and . C = LeqC . Finally, Leq = 1 2 2M − L1 − L2 (2M − L1 − L2 ) 2 M − L1 − L2 L1 + L2 − 2 M EVALUATE: If the flux of one coil doesn't pass through the other coil, so M = 0 , then the results reduce to those of problem 30.47. IDENTIFY: Apply Kirchhoff’s loop rule to the top and bottom branches of the circuit. SET UP: Just after the switch is closed the current through the inductor is zero and the charge on the capacitor is zero. di E q di i E EXECUTE: E − i1R1 − L 1 = 0 ⇒ i1 = (1 − e− ( R1 L ) t ). E − i2 R2 − 2 = 0 ⇒ − 2 R2 − 2 = 0 ⇒ i2 = e − (1 R2C )t ) . dt R1 C dt C R2 t

q2 = ∫ i2 dt ′ = − 0

t

E R2Ce − (1/ R2C )t ′ = EC (1 − e − (1/ R2C ) t ) . R2 0

E E 48.0 V (1 − e0 ) = 0, i2 = e0 = = 9.60 ×10−3 A. R1 R2 5000 Ω E E 48.0 V E = 1.92 A, i2 = e−∞ = 0. A good definition of a “long time” is (c) As t → ∞ : i1 (∞) = (1 − e−∞ ) = = R1 R1 25.0 Ω R2 many time constants later. (b) i1 (0) =


30-26

Chapter 30

(d) i1 = i2 ⇒

E (1 − e− ( R1 R1

L)t

)=

E − (1 R2C )t e ⇒ (1 − e− ( R1 R2

L )t

)=

R1 − (1 R2C )t e . Expanding the exponentials like R2

2

⎞ R R ⎛ t t2 1⎛ R ⎞ x 2 x3 + 2 2 − "⎟ and + + ", we find: 1 t − ⎜ 1 ⎟ t 2 + " = 1 ⎜1 − L R2 ⎝ RC 2 R C 2⎝ L ⎠ 2 3! ⎠ ⎛R R ⎞ R t ⎜ 1 + 21 ⎟ + O(t 2 ) + ⋅⋅⋅ = 1 , if we have assumed that t << 1. Therefore: L R C R ⎝ ⎠ 2 2 ex = 1 + x +

t≈

⎞ ⎛ LR2C ⎞ ⎛ (8.0 H)(5000 Ω)(2.0 ×10−5 F) ⎞ 1 ⎛ 1 −3 ⎜ ⎟=⎜ ⎟=⎜ ⎟ = 1.6 ×10 s. R2 ⎝ (1 L) + (1 R2 2C ) ⎠ ⎝ L + R2 2C ⎠ ⎝ 8.0 H + (5000 Ω) 2 (2.0 ×10−5 F) ⎠

(e) At t = 1.57 × 10−3 s : i1 =

E (1 − e − ( R1 R1

L )t

)=

48 V (1 − e − (25 8)t ) = 9.4 ×10−3 A. 25 Ω

(f ) We want to know when the current is half its final value. We note that the current i2 is very small to begin with, and just gets smaller, so we ignore it and find: E L 8.0 H i1 2 = 0.960 A = i1 = (1 − e − ( R1 L )t ) = (1.92 A)(1 − e − ( R1 L )t ). e − ( R1 L )t = 0.500 ⇒ t = ln(0.5) = ln(0.5) = 0.22 s . R1 R1 25 Ω EVALUATE:

i1 is initially zero and rises to a final value of 1.92 A. i2 is initially 9.60 mA and falls to zero, q2 is

initially zero and rises to q2 = EC = 960 μ C .


31

ALTERNATING CURRENT

31.1.

i = I cos ω t and I rms = I/ 2.

IDENTIFY:

The specified value is the root-mean-square current; I rms = 0.34 A.

SET UP:

(a) I rms = 0.34 A

EXECUTE:

31.2.

(b) I = 2 I rms = 2(0.34 A) = 0.48 A. (c) Since the current is positive half of the time and negative half of the time, its average value is zero. 2 (d) Since I rms is the square root of the average of i 2, the average square of the current is I rms = (0.34 A) 2 = 0.12 A 2. EVALUATE: The current amplitude is larger than its rms value. IDENTIFY and SET UP: Apply Eqs.(31.3) and (31.4) EXECUTE: (a) I = 2 I rms = 2(2.10 A) = 2.97 A. (b) I rav =

31.3.

31.4.

2

π

I=

2

π

(2.97 A) = 1.89 A.

EVALUATE: (c) The root-mean-square voltage is always greater than the rectified average, because squaring the current before averaging, and then taking the square root to get the root-mean-square value will always give a larger value than just averaging. IDENTIFY and SET UP: Apply Eq.(31.5). 45.0 V V EXECUTE: (a) Vrms = = = 31.8 V. 2 2 (b) Since the voltage is sinusoidal, the average is zero. EVALUATE: The voltage amplitude is larger than Vrms . IDENTIFY: SET UP:

V = IX C with X C =

1

ωC

.

ω is the angular frequency, in rad/s. I

so I = V ω C = (60.0 V)(100 rad s)(2.20 ×10−6 F) = 0.0132 A. ωC (b) I = V ω C = (60.0 V)(1000 rad s)(2.20 ×10−6 F) = 0.132 A. EXECUTE:

(a) V = IX C =

(c) I = V ω C = (60.0 V)(10,000 rad s)(2.20 ×10−6 F) = 1.32 A. (d) The plot of log I versus log ω is given in Figure 31.4. EVALUATE: I = ωVC so log I = log(VC ) + log ω . A graph of log I versus log ω should be a straight line with slope +1, and that is what Figure 31.4 shows.

Figure 31.4

31-1


31-2

31.5.

Chapter 31

IDENTIFY: V = IX L with X L = ω L . SET UP: ω is the angular frequency, in rad/s. V 60.0 V EXECUTE: (a) V = IX L = Iω L and I = = = 0.120 A. ω L (100 rad s)(5.00 H) (b) I =

V 60.0 V = = 0.0120 A . ω L (1000 rad s)(5.00 H)

V 60.0 V = = 0.00120 A . ω L (10,000 rad s)(5.00 H) (d) The plot of log I versus log ω is given in Figure 31.5. (c) I =

V

so log I = log(V / L ) − log ω . A graph of log I versus log ω should be a straight line with ωL slope −1, and that is what Figure 31.5 shows. EVALUATE:

I=

Figure 31.5 31.6.

IDENTIFY: The reactance of capacitors and inductors depends on the angular frequency at which they are operated, as well as their capacitance or inductance. SET UP: The reactances are X C = 1/ ω C and X L = ω L . EXECUTE:

(a) Equating the reactances gives ω L =

(b) Using the numerical values we get ω =

31.7.

1

ωC

⇒ω=

1 LC

1 1 = 7560 rad/s = (5.00 mH)(3.50 µF) LC

XC = XL = ωL = (7560 rad/s)(5.00 mH) = 37.8 Ω EVALUATE: At other angular frequencies, the two reactances could be very different. IDENTIFY and SET UP: For a resistor vR = iR. For an inductor, vL = V cos(ω t + 90°). For a capacitor,

vC = V cos(ω t − 90°). EXECUTE: The graphs are sketched in Figures 31.7a-c. The phasor diagrams are given in Figure 31.7d. EVALUATE: For a resistor only in the circuit, the current and voltage in phase. For an inductor only, the voltage leads the current by 90°. For a capacitor only, the voltage lags the current by 90°.

Figure 31.7a and b


Alternating Current

31-3

Figure 31.7c

Figure 31.7d 31.8.

The reactance of an inductor is X L = ω L = 2π fL . The reactance of a capacitor is X C =

IDENTIFY:

1

ωC

=

1 . 2π fC

SET UP: The frequency f is in Hz. EXECUTE: (a) At 60.0 Hz, X L = 2π (60.0 Hz)(0.450 H) = 170 Ω. X L is proportional to f so at 600 Hz, X L = 1700 Ω. (b) At 60.0 Hz, X C =

1 = 1.06 ×103 Ω. X C is proportional to 1/ f , so at 600 Hz, X C = 106 Ω. 2π (60.0 Hz)(2.50 × 10−6 F)

(c) X L = X C says 2π fL =

31.9.

EVALUATE: X L increases when f increases. X C increases when f increases. IDENTIFY and SET UP: Use Eqs.(31.12) and (31.18). EXECUTE: (a) X L = ω L = 2π fL = 2π (80.0 Hz)(3.00 H) = 1510 Ω (b) X L = 2π fL gives L = (c) X C = (d) X C =

1

ωC

IDENTIFY: SET UP:

1 1 = = 497 Ω 2π fC 2π (80.0 Hz)(4.00 ×10−6 F)

X L increases when L increases; X C decreases when C increases. VL = Iω L

ω is the angular frequency, in rad/s. f =

ω is the frequency in Hz. 2π

VL (12.0 V) = = 1.63 × 106 Hz. 2ω IL 2π (2.60 × 10−3 A)(4.50 × 10−4 H) EVALUATE: When f is increased, I decreases. IDENTIFY and SET UP: Apply Eqs.(31.18) and (31.19). V 170 V EXECUTE: V = IX C so X C = = = 200 Ω I 0.850 A 1 1 1 XC = = = 1.33 ×10−5 F = 13.3 μ F gives C = 2π f X C 2π (60.0 Hz)(200 Ω) ωC EVALUATE: The reactance relates the voltage amplitude to the current amplitude and is similar to Ohm’s law.

EXECUTE: 31.11.

=

XL 120 Ω = = 0.239 H 2π f 2π (80.0 Hz)

1 1 1 = = 1.66 ×10−5 F gives C = 2π fC 2π f X C 2π (80.0 Hz)(120 Ω)

EVALUATE: 31.10.

1 1 1 and f = = = 150 Hz. 2π fC 2π LC 2π (0.450 H)(2.50 ×10−6 F)

VL = Iω L so f =


31-4

Chapter 31

31.12.

IDENTIFY:

Compare vC that is given in the problem to the general form vC =

I

ωC

sin ω t and determine ω .

1 . vR = iR and i = I cos ω . ωC 1 1 = = 1736 Ω EXECUTE: (a) X C = ω C (120 rad s)(4.80 ×10−6 F) V 7.60 V = 4.378 ×10−3 A and i = I cos ω t = (4.378 × 10−3 A)cos[(120 rad/s)t ]. Then (b) I = C = X C 1736 Ω

SET UP:

31.13.

31.14.

31.15.

XC =

vR = iR = (4.38 ×10−3 A)(250 Ω)cos((120 rad s)t ) = (1.10 V)cos((120 rad s)t ). EVALUATE: The voltage across the resistor has a different phase than the voltage across the capacitor. IDENTIFY and SET UP: The voltage and current for a resistor are related by vR = iR. Deduce the frequency of the voltage and use this in Eq.(31.12) to calculate the inductive reactance. Eq.(31.10) gives the voltage across the inductor. EXECUTE: (a) vR = (3.80 V)cos[(720 rad/s)t ] v ⎛ 3.80 V ⎞ vR = iR, so i = R = ⎜ ⎟ cos[(720 rad/s)t ] = (0.0253 A)cos[(720 rad/s)t ] R ⎝ 150 Ω ⎠ (b) X L = ω L ω = 720 rad/s, L = 0.250 H, so X L = ω L = (720 rad/s)(0.250 H) = 180 Ω (c) If i = I cos ω t then vL = VL cos(ω t + 90°) (from Eq.31.10). VL = Iω L = IX L = (0.02533 A)(180 Ω) = 4.56 V vL = (4.56 V)cos[(720 rad/s)t + 90°] But cos( a + 90°) = − sin a (Appendix B), so vL = −(4.56 V)sin[(720 rad/s)t ]. EVALUATE: The current is the same in the resistor and inductor and the voltages are 90° out of phase, with the voltage across the inductor leading. IDENTIFY: Calculate the reactance of the inductor and of the capacitor. Calculate the impedance and use that result to calculate the current amplitude. X V SET UP: With no capacitor, Z = R 2 + X L2 and tan φ = L . X L = ω L. I = . VL = IX L and VR = IR. For an Z R inductor, the voltage leads the current. EXECUTE: (a) X L = ω L = (250 rad/s)(0.400 H) = 100 Ω. Z = (200 Ω) 2 + (100 Ω) 2 = 224 Ω. V 30.0 V (b) I = = = 0.134 A Z 224 Ω (c) VR = IR = (0.134 A)(200 Ω) = 26.8 V. VL = IX L = (0.134 A)(100 Ω) = 13.4 V. X 100 Ω (d) tan φ = L = and φ = +26.6°. Since φ is positive, the source voltage leads the current. R 200 Ω (e) The phasor diagram is sketched in Figure 31.14. EVALUATE: Note that VR + VL is greater than V. The loop rule is satisfied at each instance of time but the voltages across R and L reach their maxima at different times.

Figure 31.14 IDENTIFY: vR (t ) is given by Eq.(31.8). vL (t ) is given by Eq.(31.10). SET UP: From Exercise 31.14, V = 30.0 V , VR = 26.8 V , VL = 13.4 V and φ = 26.6° . EXECUTE: (a) The graph is given in Figure 31.15. (b) The different voltages are v = (30.0 V)cos(250t + 26.6°), vR = (26.8 V)cos(250t ),

vL = (13.4 V)cos(250t + 90°). At t = 20 ms: v = 20.5 V, vR = 7.60 V, vL = 12.85 V. Note that vR + vL = v.


Alternating Current

31-5

(c) At t = 40 ms: v = −15.2 V, vR = −22.49 V, vL = 7.29 V. Note that vR + vL = v. EVALUATE: It is important to be careful with radians versus degrees in above expressions!

31.16.

Figure 31.15 IDENTIFY: Calculate the reactance of the inductor and of the capacitor. Calculate the impedance and use that result to calculate the current amplitude. X − XC 1 SET UP: With no resistor, Z = ( X L − X C ) 2 = X L − X C . tan φ = L . X L = ω L. For an . XC = zero ωC inductor, the voltage leads the current. For a capacitor, the voltage lags the current. 1 1 = = 667 Ω. EXECUTE: (a) X L = ω L = (250 rad/s)(0.400 H) = 100 Ω. X C = ω C (250 rad/s)(6.00 × 10−6 F) Z = X L − X C = 100 Ω − 667 Ω = 567 Ω. V 30.0 V = = 0.0529 A Z 567 Ω (c) VC = IX C = (0.0529 A)(667 Ω) = 35.3 V. VL = IX L = (0.0529 A)(100 Ω) = 5.29 V. X − X C 100 Ω − 667 Ω (d) tan φ = L = = −∞ and φ = −90°. The phase angle is negative and the source voltage lags zero zero the current. (e) The phasor diagram is sketched in Figure 31.16. EVALUATE: When X C > X L the phase angle is negative and the source voltage lags the current.

(b) I =

Figure 31.16 31.17.

IDENTIFY and SET UP: Calculate the impendance of the circuit and use Eq.(31.22) to find the current amplitude. The voltage amplitudes across each circuit element are given by Eqs.(31.7), (31.13), and (31.19). The phase angle is calculated using Eq.(31.24). The circuit is shown in Figure 31.17a.

No inductor means X L = 0 R = 200 Ω, C = 6.00 ×10−6 F, V = 30.0 V, ω = 250 rad/s Figure 31.17a


31-6

Chapter 31

EXECUTE:

(a) X C =

1 1 = = 666.7 Ω ω C (250 rad/s)(6.00 ×10−6 F)

Z = R 2 + ( X L − X C ) 2 = (200 Ω) 2 + (666.7 Ω) 2 = 696 Ω V 30.0 V = = 0.0431 A = 43.1 mA Z 696 Ω (c) Voltage amplitude across the resistor: VR = IR = (0.0431 A)(200 Ω) = 8.62 V Voltage amplitude across the capacitor: VC = IX C = (0.0431 A)(666.7 Ω) = 28.7 V X − X C 0 − 666.7 Ω (d) tan φ = L = = −3.333 so φ = −73.3° R 200 Ω The phase angle is negative, so the source voltage lags behind the current. (e) The phasor diagram is sketched qualitatively in Figure 31.17b.

(b) I =

31.18.

31.19.

Figure 31.17b EVALUATE: The voltage across the resistor is in phase with the current and the capacitor voltage lags the current by 90°. The presence of the capacitor causes the source voltage to lag behind the current. Note that VR + VC > V . The instantaneous voltages in the circuit obey the loop rule at all times but because of the phase differences the voltage amplitudes do not. IDENTIFY: vR (t ) is given by Eq.(31.8). vC (t )is given by Eq.(31.16). SET UP: From Exercise 31.17, V = 30.0 V, VR = 8.62 V, VC = 28.7 V and φ = −73.3°. EXECUTE: (a) The graph is given in Figure 31.18. (b) The different voltage are: v = (30.0 V)cos(250t − 73.3°), vR = (8.62 V)cos(250t ), vC = (28.7 V)cos(250t − 90°). At t = 20 ms : v = −25.1 V, vR = 2.45 V, vC = −27.5 V. Note that vR + vC = v. (c) At t = 40 ms: v = −22.9 V, vR = −7.23 V, vC = −15.6 V. Note that vR + vC = v. EVALUATE: It is important to be careful with radians vs. degrees!

Figure 31.18 IDENTIFY: Apply the equations in Section 31.3. SET UP: ω = 250 rad/s, R = 200 Ω, L = 0.400 H, C = 6.00 μ F and V = 30.0 V. EXECUTE:

(a) Z = R 2 + (ω L − 1/ ω C ) 2 .

Z = (200 Ω) 2 + ((250 rad/s)(0.0400 H) − 1/((250 rad/s)(6.00 ×10−6 F))) 2 = 601 Ω V 30 V = 0.0499 A. (b) I = = Z 601 Ω

⎛ 100 Ω − 667 Ω ⎞ ⎛ ω L − 1/ ω C ⎞ (c) φ = arctan ⎜ ⎟ = −70.6°, and the voltage lags the current. ⎟ = arctan ⎜ R 200 Ω ⎝ ⎠ ⎝ ⎠


Alternating Current

31-7

(d) VR = IR = (0.0499 A)(200 Ω) = 9.98 V;

VL = Iω L = (0.0499 A)(250 rad s)(0.400 H) = 4.99 V; VC = EVALUATE:

I (0.0499 A) = = 33.3 V. ω C (250 rad/s)(6.00 × 10−6 F)

(e) At any instant, v = vR + vC + vL . But vC and vL are 180° out of phase, so vC can be larger than v

at a value of t, if vL + vR is negative at that t. 31.20.

IDENTIFY:

vR (t ) is given by Eq.(31.8). vC (t ) is given by Eq.(31.16). vL (t ) is given by Eq.(31.10).

SET UP: From Exercise 31.19, V = 30.0 V, VL = 4.99 V, VR = 9.98 V, VC = 33.3 V and φ = −70.6°. EXECUTE: (a) The graph is sketched in Figure 31.20. The different voltages plotted in the graph are: v = (30 V)cos(250t − 70.6°), vR = (9.98 V)cos(250t ), vL = (4.99 V)cos(250t + 90°) and vC = (33.3 V)cos(250t − 90°). (b) At t = 20 ms: v = −24.3 V, vR = 2.83 V, vL = 4.79 V, vC = − 31.9 V. (c) At t = 40 ms: v = −23.8 V, vR = −8.37 V, vL = 2.71 V, vC = −18.1 V. EVALUATE: In both parts (b) and (c), note that the source voltage equals the sum of the other voltages at the given instant. Be careful with degrees versus radians!

Figure 31.20 31.21.

IDENTIFY and SET UP: The current is largest at the resonance frequency. At resonance, X L = X C and Z = R. For part (b), calculate Z and use I = V / Z . 1 EXECUTE: (a) f 0 = = 113 Hz. I = V / R = 15.0 mA. 2π LC (b) X C = 1/ ω C = 500 Ω. X L = ω L = 160 Ω. Z = R 2 + ( X L − X C ) 2 = (200 Ω) 2 + (160 Ω − 500 Ω)2 = 394.5 Ω.

I = V / Z = 7.61 mA. X C > X L so the source voltage lags the current.

31.22.

EVALUATE: ω 0 = 2π f 0 = 710 rad/s. ω = 400 rad/s and is less than ω 0 . When ω < ω 0 , X C > X L . Note that I in part (b) is less than I in part (a). IDENTIFY: The impedance and individual reactances depend on the angular frequency at which the circuit is driven. 2

1 ⎞ ⎛ The impedance is Z = R 2 + ⎜ ω L − ⎟ , the current amplitude is I = V/Z, and the instantaneous C⎠ ω ⎝ values of the potential and current are v = V cos(ωt + φ), where tan φ = (XL – XC)/R, and i = I cos ωt. 1 1 1 , which gives ω = = 3162 rad/s = EXECUTE: (a) Z is a minimum when ω L = = ωC (8.00 mH)(12.5 µF) LC 3160 rad/s and Z = R = 175 Ω. (b) I = V/Z = (25.0 V)/(175 Ω) = 0.143 A (c) i = I cos ωt = I/2, so cosωt = 12 , which gives ωt = 60° = π/3 rad. v = V cos(ωt + φ), where tan φ = (XL – XC)/R = SET UP:

0/R =0. So, v = (25.0 V) cosωt = (25.0 V)(1/2) = 12.5 V. vR = Ri = (175 Ω)(1/2)(0.143 A) = 12.5 V. 0.143 A cos(60° − 90°) = +3.13 V. vC = VC cos(ωt – 90°) = IXC cos(ωt – 90°) = (3162 rad/s)(12.5 µF)


31-8

Chapter 31

31.23.

vL = VL cos(ωt + 90°) = IXL cos(ωt + 90°) = (0.143 A)(3162 rad/s)(8.00 mH) cos(60° + 90°). vL = –3.13 V. (d) vR + vL + vC = 12.5 V + (–3.13 V) + 3.13 V = 12.5 V = vsource EVALUATE: The instantaneous potential differences across all the circuit elements always add up to the value of the source voltage at that instant. In this case (resonance), the potentials across the inductor and capacitor have the same magnitude but are 180° out of phase, so they add to zero, leaving all the potential difference across the resistor. IDENTIFY and SET UP: Use the equation that preceeds Eq.(31.20): V 2 = VR2 + (VL − VC ) 2

31.24.

EXECUTE: V = (30.0 V) 2 + (50.0 V − 90.0 V) 2 = 50.0 V EVALUATE: The equation follows directly from the phasor diagrams of Fig.31.13 (b or c). Note that the voltage amplitudes do not simply add to give 170.0 V for the source voltage. 1 IDENTIFY and SET UP: X L = ω L and X C = . ωC 1 L 1 1 , then X = ω L − EXECUTE: (a) If ω = ω 0 = and X = − = 0. ωC LC LC C LC (b) When ω > ω 0 , X > 0 (c) When ω > ω0 , X < 0 (d) The graph of X versus ω is given in Figure 31.24. EVALUATE:

Z = R 2 + X 2 and tan φ = X / R.

Figure 31.24 31.25.

2 R. IDENTIFY: For a pure resistance, Pav = Vrms I rms = I rms SET UP: 20.0 W is the average power Pav . EXECUTE: (a) The average power is one-half the maximum power, so the maximum instantaneous power is 40.0 W. P 20.0 W = 0.167 A (b) I rms = av = Vrms 120 V P 20.0 W = 720 Ω (c) R = 2av = I rms (0.167 A) 2

31.26.

VR2,rms

2 (120 V) 2 Vrms = = 20.0 W. 750 Ω R R IDENTIFY: The average power supplied by the source is P = Vrms I rms cos φ . The power consumed in the resistance

EVALUATE:

We can also calculate the average power as Pav =

=

2 R. is P = I rms

SET UP:

ω = 2π f = 2π (1.25 × 103 Hz) = 7.854 ×103 rad/s. X L = ω L = 157 Ω. X C =

1

ωC

= 909 Ω.

EXECUTE: (a) First, let us find the phase angle between the voltage and the current: X − X C 157 Ω − 909 Ω and φ = −65.04°. The impedance of the circuit is tan φ = L = R 350 Ω Z = R 2 + ( X L − X C ) 2 = (350 Ω) 2 + ( −752 Ω) 2 = 830 Ω. The average power provided by the generator is then P = Vrms I rms cos(φ ) =

2 Vrms (120 V) 2 cos(φ ) = cos( −65.04°) = 7.32 W Z 830 Ω


Alternating Current

31-9

2

31.27.

⎛ 120 V ⎞ 2 (b) The average power dissipated by the resistor is PR = I rms R=⎜ ⎟ (350 Ω) = 7.32 W. ⎝ 830 Ω ⎠ EVALUATE: Conservation of energy requires that the answers to parts (a) and (b) are equal. IDENTIFY: The power factor is cos φ , where φ is the phase angle in Fig.31.13. The average power is given by Eq.(31.31). Use the result of part (a) to rewrite this expression. (a) SET UP: The phasor diagram is sketched in Figure 31.27.

EXECUTE: From the diagram V IR R cos φ = R = = , V IZ Z as was to be shown.

31.28.

31.29.

Figure 31.27 V ⎛ R ⎞ ⎛V ⎞ 2 (b) Pav = Vrms I rms cos φ = Vrms I rms ⎜ ⎟ = ⎜ rms ⎟ I rms R. But rms = I rms , so Pav = I rms R. Z ⎝Z⎠ ⎝ Z ⎠ EVALUATE: In an L-R-C circuit, electrical energy is stored and released in the inductor and capacitor but none is dissipated in either of these circuit elements. The power delivered by the source equals the power dissipated in the resistor. V R IDENTIFY and SET UP: Pav = Vrms I rms cosφ . I rms = rms . cos φ = . Z Z 80.0 V 75.0 Ω EXECUTE: I rms = = 0.714. Pav = (80.0 V)(0.762 A)(0.714) = 43.5 W. = 0.762 A. cos φ = 105 Ω 105 Ω EVALUATE: Since the average power consumed by the inductor and by the capacitor is zero, we can also 2 R = (0.762 A) 2 (75.0 Ω) = 43.5 W. calculate the average power as Pav = I rms IDENTIFY and SET UP: Use the equations of Section 31.3 to calculate φ , Z and Vrms . The average power 2 R delivered by the source is given by Eq.(31.31) and the average power dissipated in the resistor is I rms EXECUTE: (a) X L = ω L = 2π f L = 2π (400 Hz)(0.120 H) = 301.6 Ω 1 1 1 XC = = = = 54.51 Ω ω C 2π fC 2π (400 Hz)(7.3 ×10−6 Hz) X − X C 301.6 Ω − 54.41 Ω tan φ = L = , so φ = +45.8°. The power factor is cos φ = +0.697. R 240 Ω (b) Z = R 2 + ( X L − X C ) 2 = (240 Ω) 2 + (301.6 Ω − 54.51 Ω) 2 = 344 Ω

(c) Vrms = I rms Z = (0.450 A)(344 Ω) = 155 V (d) Pav = I rmsVrms cos φ = (0.450 A)(155 V)(0.697) = 48.6 W

31.30.

2 R = (0.450 A)2 (240 Ω) = 48.6 W (e) Pav = I rms EVALUATE: The average electrical power delivered by the source equals the average electrical power consumed in the resistor. (f ) All the energy stored in the capacitor during one cycle of the current is released back to the circuit in another part of the cycle. There is no net dissipation of energy in the capacitor. (g) The answer is the same as for the capacitor. Energy is repeatedly being stored and released in the inductor, but no net energy is dissipated there. IDENTIFY: The angular frequency and the capacitance can be used to calculate the reactance X C of the capacitor. The angular frequency and the inductance can be used to calculate the reactance X L of the inductor. Calculate the phase angle φ and then the power factor is cos φ . Calculate the impedance of the circuit and then the rms current in the circuit. The average power is Pav = Vrms I rms cos φ . On the average no power is consumed in the capacitor or the inductor, it is all consumed in the resistor. V 45 V SET UP: The source has rms voltage Vrms = = = 31.8 V. 2 2


31-10

Chapter 31

EXECUTE:

X L = ω L = (360 rad/s)(15 × 10−3 H) = 5.4 Ω. X C =

1 1 = = 794 Ω. ω C (360 rad/s)(3.5 × 10−6 F)

X L − X C 5.4 Ω − 794 Ω = and φ = −72.4°. The power factor is cos φ = 0.302. R 250 Ω V 31.8 V (b) Z = R 2 + ( X L − X C ) 2 = (250 Ω) 2 + (5.4 Ω − 794 Ω) 2 = 827 Ω. I rms = rms = = 0.0385 A. Z 827 Ω Pav = Vrms I rms cos φ = (31.8 V)(0.0385 A)(0.302) = 0.370 W. tan φ =

31.31.

2 (c) The average power delivered to the resistor is Pav = I rms R = (0.0385 A) 2 (250 Ω) = 0.370 W. The average power delivered to the capacitor and to the inductor is zero. EVALUATE: On average the power delivered to the circuit equals the power consumed in the resistor. The capacitor and inductor store electrical energy during part of the current oscillation but each return the energy to the circuit during another part of the current cycle. IDENTIFY and SET UP: At the resonance frequency, Z = R. Use that V = IZ, VR = IR, VL = IX L and VC = IX C . Pav is given by Eq.(31.31). (a) EXECUTE: V = IZ = IR = (0.500 A)(300 Ω) = 150 V

(b) VR = IR = 150 V X L = ω L = L(1/ LC ) = L / C = 2582 Ω; VL = IX L = 1290 V

X C = 1/(ω C ) = L / C = 2582 Ω; VC = IX C = 1290 V (c) Pav = 12 VI cos φ = 12 I 2 R, since V = IR and cos φ = 1 at resonance.

31.32.

31.33.

31.34.

Pav = 12 (0.500 A) 2 (300 Ω) = 37.5 W EVALUATE: At resonance VL = VC . Note that VL + VC > V . However, at any instant vL + vC = 0. IDENTIFY: The current is maximum at the resonance frequency, so choose C such that ω = 50.0 rad/s is the resonance frequency. At the resonance frequency Z = R. SET UP: VL = Iω L V EXECUTE: (a) The amplitude of the current is given by I = . Thus, the current will have a 2 R + (ω L − 1 ) 2 ωC 1 maximum amplitude when ω L = 1 . Therefore, C = 12 = = 44.4 μ F. ωC ω L (50.0 rad/s) 2 (9.00 H) (b) With the capacitance calculated above we find that Z = R, and the amplitude of the current is 120 V I =V = = 0.300 A. Thus, the amplitude of the voltage across the inductor is R 400 Ω VL = I (ω L) = (0.300 A)(50.0 rad/s)(9.00 H) = 135 V. EVALUATE: Note that VL is greater than the source voltage amplitude. IDENTIFY and SET UP: At resonance X L = X C , φ = 0 and Z = R. R = 150 Ω, L = 0.750 H, C = 0.0180 μ F, V = 150 V X − XC EXECUTE: (a) At the resonance frequency X L = X C and from tan φ = L we have that φ = 0° and the R power factor is cos φ = 1.00. (b) Pav = 12 VI cos φ (Eq.31.31) V V At the resonance frequency Z = R, so I = = Z R 2 2 V V (150 V) ⎛ ⎞ =1 = 75.0 W Pav = 12 V ⎜ ⎟ cos φ = 12 R 2 150 Ω ⎝R⎠ (c) EVALUATE: When C and f are changed but the circuit is kept on resonance, nothing changes in Pav = V 2 /(2 R ), so the average power is unchanged: Pav = 75.0 W. The resonance frequency changes but since Z = R at resonance the current doesn’t change. 1 IDENTIFY: ω 0 = . VC = IX C . V = IZ . LC SET UP: At resonance, Z = R.


Alternating Current

(a) ω 0 =

EXECUTE:

31.35.

31-11

1 1 = = 1.54 ×104 rad/s LC (0.350 H)(0.0120 ×10−6 F)

⎛V ⎞ ⎛V ⎞ 1 1 (b) V = IZ = ⎜ C ⎟ Z = ⎜ C ⎟ R. X C = = = 5.41× 103 Ω. −6 4 ω C × × X X (1.54 10 rad/s)(0.0120 10 F) C C ⎝ ⎠ ⎝ ⎠ ⎛ 550 V ⎞ V =⎜ ⎟ (400 Ω) = 40.7 V. 3 ⎝ 5.41 × 10 Ω ⎠ EVALUATE: The voltage amplitude for the capacitor is more than a factor of 10 times greater than the voltage amplitude of the source. 1 1 . X L = ω L. X C = IDENTIFY and SET UP: The resonance angular frequency is ω 0 = and ωC LC Z = R 2 + ( X L − X C ) 2 . At the resonance frequency X L = X C and Z = R. (a) Z = R = 115 Ω 1

EXECUTE: (b) ω 0 =

−3

(4.50 × 10 H)(1.26 ×10

−6

F)

= 1.33 ×104 rad/s . ω = 2ω 0 = 2.66 × 104 rad/s.

X L = ω L = (2.66 × 104 rad/s)(4.50 × 10−3 H) = 120 Ω. X C =

1 1 = = 30 Ω 4 ω C (2.66 ×10 rad/s)(1.25 × 10−6 F)

Z = (115 Ω) 2 + (120 Ω − 30 Ω) 2 = 146 Ω (c) ω = ω 0 / 2 = 6.65 × 103 rad/s. X L = 30 Ω. X C =

31.36.

1

ωC

= 120 Ω. Z = (115 Ω) 2 + (30 Ω − 120 Ω) 2 = 146 Ω, the

same value as in part (b). EVALUATE: For ω = 2ω 0 , X L > X C . For ω = ω 0 / 2, X L < X C . But ( X L − X C ) 2 has the same value at these two frequencies, so Z is the same. IDENTIFY: At resonance Z = R and X L = X C . SET UP:

1 . V = IZ . VR = IR, VL = IX L and VC = VL . LC 1 1 (a) ω 0 = = = 945 rad s. LC ( 0.280 H ) ( 4.00 × 10−6 F )

ω0 =

EXECUTE: (b)

I = 1.20 A at resonance, so R = Z =

V 120 V = = 70.6 Ω I 1.70 A

(c) At resonance, VR = 120 V, VL = VC = Iω L = (1.70 A )( 945 rad s )( 0.280 H ) = 450 V. 31.37.

EVALUATE: At resonance, VR = V and VL − VC = 0. IDENTIFY and SET UP: Eq.(31.35) relates the primary and secondary voltages to the number of turns in each. I = 2 2 V/R and the power consumed in the resistive load is I rms = Vrms / R. EXECUTE:

(a)

V2 N 2 N V 120 V = so 1 = 1 = = 10 V1 N1 N 2 V2 12.0 V

V2 12.0 V = = 2.40 A R 5.00 Ω (c) Pav = I 22 R = (2.40 A) 2 (5.00 Ω) = 28.8 W (d) The power drawn from the line by the transformer is the 28.8 W that is delivered by the load. (b) I 2 =

Pav = 2

V2 V 2 (120 V) 2 so R = = = 500 Ω R Pav 28.8 W

⎛N ⎞ And ⎜ 1 ⎟ (5.00 Ω) = (10) 2 (5.00 Ω) = 500 Ω, as was to be shown. ⎝ N2 ⎠ EVALUATE: The resistance is “transformed”. A load of resistance R connected to the secondary draws the same power as a resistance ( N1 / N 2 )2 R connected directly to the supply line, without using the transformer.


31-12

Chapter 31

31.38.

IDENTIFY: SET UP: EXECUTE:

31.39.

Pav = VI and Pav,1 = Pav,2 .

N1 V1 = . N 2 V2

V1 = 120 V. V2 = 13,000 V. (a)

N 2 V2 13,000 V = = = 108 N1 V1 120 V

(b) Pav = V2 I 2 = (13,000 V)(8.50 ×10−3 A) = 110 W P 110 W = 0.917 A (c) I1 = av = V1 120 V EVALUATE: Since the power supplied to the primary must equal the power delivered by the secondary, in a stepup transformer the current in the primary is greater than the current in the secondary. V N IDENTIFY: A transformer transforms voltages according to 2 = 2 . The effective resistance of a secondary V1 N1 R V2 . Resistance R is related to Pav and V by Pav = . Conservation of energy circuit of resistance R is Reff = 2 ( N 2 / N1 ) R requires Pav,1 = Pav,2 so V1I1 = V2 I 2 . SET UP:

Let V1 = 240 V and V2 = 120 V, so P2,av = 1600 W. These voltages are rms.

(a) V1 = 240 V and we want V2 = 120 V, so use a step-down transformer with N 2 / N1 = 12 . P 1600 W (b) Pav = VI , so I = av = = 6.67 A. V 240 V V 2 (120 V) 2 (c) The resistance R of the blower is R = = = 9.00 Ω. The effective resistance of the blower is P 1600 W 9.00 Ω Reff = = 36.0 Ω. (1/ 2)2 EVALUATE: I 2V2 = (13.3 A)(120 V) = 1600 W. Energy is provided to the primary at the same rate that it is consumed in the secondary. Step-down transformers step up resistance and the current in the primary is less than the current in the secondary. 1 IDENTIFY: Z = R 2 + ( X L − X C ) 2 , with X L = ω L and X C = . ωC SET UP: The woofer has a R and L in series and the tweeter has a R and C in series. EXECUTE: (a) Z tweeter = R 2 + (1 ω C ) 2 EXECUTE:

31.40.

(b) Z woofer = R 2 + (ω L )

2

(c) If Z tweeter = Z woofer , then the current splits evenly through each branch. (d) At the crossover point, where currents are equal, R 2 + (1 ω C 2 ) = R 2 + (ω L ) . ω = 2

f =

31.41.

31.42.

ω 1 . = 2π 2π LC

1 and LC

EVALUATE: The crossover frequency corresponds to the resonance frequency of a R-C-L circuit, since the crossover frequency is where X L = X C . IDENTIFY and SET UP: Use Eq.(31.24) to relate L and R to φ . The voltage across the coil leads the current in it by 52.3°, so φ = +52.3°. X − XC X EXECUTE: tan φ = L . But there is no capacitance in the circuit so X C = 0. Thus tan φ = L and X L = R R XL 62.1 Ω = = 0.124 H. R tan φ = (48.0 Ω) tan 52.3° = 62.1 Ω. X L = ω L = 2π f L so L = 2π f 2π (80.0 Hz) EVALUATE: φ > 45° when ( X L − X C ) > R, which is the case here. IDENTIFY: SET UP:

Z = R 2 + ( X L − X C ) 2 . I rms = Vrms =

V 30.0 V = = 21.2 V. 2 2

Vrms . Vrms = I rms R. VC ,rms = I rms X C . VL ,rms = I rms X L . Z


Alternating Current

31-13

(a) ω = 200 rad/s , so X L = ω L = (200 rad/s)(0.400 H) = 80.0 Ω and

EXECUTE:

1 1 XC = = = 833 Ω. Z = (200 Ω) 2 + (80.0 Ω − 833 Ω) 2 = 779 Ω. ω C (200 rad/s)(6.00 × 10−6 F) I rms = VL ,rms

Vrms 21.2 V = = 0.0272 A. V1 reads VR ,rms = I rms R = (0.0272 A)(200 Ω) = 5.44 V. V2 reads Z 779 Ω = I rms X L = (0.0272 A)(80.0 Ω) = 2.18 V. V3 reads VC ,rms = I rms X C = (0.0272 A)(833 Ω) = 22.7 V. V4 reads

VL − VC = VL ,rms − VC ,rms = 2.18 V − 22.7 V = 20.5 V. V5 reads Vrms = 21.2 V. 2 1 833 Ω (b) ω = 1000 rad/s so X L = ω L = (5)(80.0 Ω) = 400 Ω and X C = = = 167 Ω. 5 ωC V 21.2 V Z = (200 Ω) 2 + (400 Ω − 167 Ω) 2 = 307 Ω. I rms = rms = = 0.0691 A. V1 reads VR ,rms = 13.8 V. V2 reads 307 Ω Z VL ,rms = 27.6 V. V3 reads VC ,rms = 11.5 V. V4 reads VL ,rms − VC ,rms = 27.6 V − 11.5 V = 16.1 V. V5 reads Vrms = 21.2 V. 1 = 645 rad/s. 200 rad/s is less than the LC resonance frequency and X C > X L . 1000 rad/s is greater than the resonance frequency and X L > X C . IDENTIFY and SET UP: The rectified current equals the absolute value of the current i. Evaluate the integral as specified in the problem. EXECUTE: (a) From Fig.31.3b, the rectified current is zero at the same values of t for which the sinusoidal current is zero. At these t, cos ω t = 0 and ω t = ±π / 2, ± 3π / 2,…. The two smallest positive times are t1 = π / 2ω , t2 = 3π / 2ω . EVALUATE:

31.43.

(b) A =

t2 t1

The resonance frequency for this circuit is ω 0 =

t

2 t2 I ⎡1 ⎤ idt = − ∫ I cos ω tdt = − I ⎢ sin ω t ⎥ = − (sin ω t2 − sin ω t1 ) t1 ω ⎣ω ⎦ t1

sin ω t1 = sin[ω (π / 2ω )] = sin(π / 2) = 1 sin ω t2 = sin[ω (3π / 2ω )] = sin(3π / 2) = −1 2I ⎛I⎞ A = ⎜ ⎟ (1 − ( −1)) = ω ⎝ω ⎠ (c ) I rav (t2 − t1 ) = 2 I / ω 2I 2I 2I = = , which is Eq.(31.3). ω (t2 − t1 ) ω (3π / 2ω − π / 2ω ) π EVALUATE: We have shown that Eq.(31.3) is correct. The average rectified current is less than the current amplitude I, since the rectified current varies between 0 and I. The average of the current is zero, since it has both positive and negative values. IDENTIFY: X L = ω L. Pav = Vrms I rms cos φ SET UP: f = 120 Hz; ω = 2π f . I rav =

31.44.

EXECUTE:

(a) X L = ω L ⇒ L =

(b) Z = R 2 + X L 2 =

Vrms = Z

2

ω

=

250 Ω = 0.332 Ω 2π (120 Hz )

+ ( 250 Ω ) = 472 Ω. cos φ = 2

V R V2 R and I rms = rms . Pav = rms , so Z Z Z Z

800 W Pav = ( 472 Ω ) = 668 V. 400 Ω R

EVALUATE: 31.45.

( 400 Ω )

XL

I rms =

Vrms 668 V 2 R = (1.415 A)2 (400 Ω) = 800 W, which = = 1.415 A. We can calculate Pav as I rms 472 Ω Z

checks. (a) IDENTIFY and SET UP: Source voltage lags current so it must be that X C > X L and we must add an inductor in series with the circuit. When X C = X L the power factor has its maximum value of unity, so calculate the additional L needed to raise X L to equal X C .


31-14

Chapter 31

power factor cosφ equals 1 so φ = 0 and X C = X L . Calculate the present value of X C − X L to

(b) EXECUTE:

see how much more X L is needed: R = Z cosφ = (60.0 Ω)(0.720) = 43.2 Ω X L − XC so X L − X C = R tan φ R cos φ = 0.720 gives φ = −43.95° ( φ is negative since the voltage lags the current) tan φ =

Then X L − X C = R tan φ = (43.2 Ω) tan(−43.95°) = −41.64 Ω. Therefore need to add 41.64 Ω of X L . XL 41.64 Ω = = 0.133 H, amount of inductance to add. 2π f 2π (50.0 Hz) EVALUATE: From the information given we can’t calculate the original value of L in the circuit, just how much to add. When this L is added the current in the circuit will increase. 2 IDENTIFY: Use Vrms = I rms Z to calculate Z and then find R. Pav = I rms R X L = ω L = 2π f L and L =

31.46.

SET UP:

X C = 50.0 Ω

EXECUTE: R=

Z=

Vrms 240 V 2 = = 80.0 Ω = R 2 + X C2 = R 2 + ( 50.0 Ω ) . Thus, I rms 3.00 A

( 80.0 Ω ) − ( 50.0 Ω ) 2

2

= 62.4 Ω. The average power supplied to this circuit is equal to the power dissipated

2 by the resistor, which is P = I rms R = ( 3.00 A ) ( 62.4 Ω ) = 562 W. 2

X L − X C −50.0 Ω and φ = −38.7°. = R 62.4 Ω Pav = Vrms I rms cos φ = (240 V)(3.00 A)cos(−38.7°) = 562 W, which checks. IDENTIFY: The voltage and current amplitudes are the maximum values of these quantities, not necessarily the instantaneous values. SET UP: The voltage amplitudes are VR = RI, VL = XLI, and VC = XCI, where I = V/Z and EVALUATE:

31.47.

tan φ =

2

1 ⎞ ⎛ Z = R2 + ⎜ω L − . ω C ⎟⎠ ⎝ EXECUTE: (a) ω = 2πf = 2π(1250 Hz) = 7854 rad/s. Carrying extra figures in the calculator gives XL = ωL = (7854 rad/s)(3.50 mH) = 27.5 Ω; XC = 1/ωC = 1/[(7854 rad/s)(10.0 µF)] = 12.7 Ω; Z = R 2 + ( X L − X C )2 =

(50.0 Ω)2 + (27.5 Ω − 12.7 Ω) 2 = 52.1 Ω;

I = V/Z = (60.0 V)/(52.1 Ω) = 1.15 A; VR = RI = (50.0 Ω)(1.15 A) = 57.5 V; VL = XLI = (27.5 Ω)(1.15 A) = 31.6 V; VC = XCI = (12.7 Ω)(1.15 A) = 14.7 V. The voltage amplitudes can add to more than 60.0 V because these voltages do not all occur at the same instant of time. At any instant, the instantaneous voltages all add to 60.0 V. (b) All of them will change because they all depend on ω. XL = ωL will double to 55.0 Ω, and XC = 1/ωC will

31.48.

decrease by half to 6.35 Ω. Therefore Z = (50.0 Ω) 2 + (55.0 Ω − 6.35 Ω) 2 = 69.8 Ω; I = V/Z = (60.0 V)/(69.8 Ω) = 0.860 A; VR = IR = (0.860 A)(50.0 Ω) = 43.0 V; VL = IXL = (0.860 A)(55.0 Ω) = 47.3 V; VC = IXC = (0.860 A)(6.35 Ω) = 5.47 V. EVALUATE: The new amplitudes in part (b) are not simple multiples of the values in part (a) because the impedance and reactances are not all the same simple multiple of the angular frequency. 1 IDENTIFY and SET UP: X C = . X L = ω L. ωC 1 XL ωL 1 1 EXECUTE: (a) = ω1L and LC = 2 . At angular frequency ω 2 , = 2 = ω 22 LC = (2ω1 ) 2 2 = 4. ω1C ω1 ω1 X C 1/ ω 2C X L > XC. 2

XL ⎛ω ⎞ ⎛ 1 ⎞ 1 = ω32 LC = ⎜ 1 ⎟ ⎜ 2 ⎟ = . X C > X L . XC ⎝ 3 ⎠ ⎝ ω1 ⎠ 9 When ω increases, X L increases and X C decreases. When ω decreases, X L decreases and X C

(b) At angular frequency ω3 ,

EVALUATE: increases. (c) The resonance angular frequency ω 0 is the value of ω for which X C = X L , so ω 0 = ω1.


Alternating Current

31.49.

31-15

Express Z and I in terms of ω , L, C and R. The voltages across the resistor and the

IDENTIFY and SET UP:

inductor are 90° out of phase, so Vout = VR2 + VL2 . The circuit is sketched in Figure 31.49.

EXECUTE:

X L = ω L, X C =

1

ωC 2

1 ⎞ ⎛ Z = R2 + ⎜ω L − ⎟ ωC ⎠ ⎝ V Vs I= s = 2 Z 1 ⎞ ⎛ R2 + ⎜ω L − ⎟ ωC ⎠ ⎝

Figure 31.49

Vout = I R 2 + X L2 = I R 2 + ω 2 L2 = Vs

Vout = Vs

R 2 + ω 2 L2 1 ⎞ ⎛ R2 + ⎜ω L − ⎟ ωC ⎠ ⎝

2

R 2 + ω 2 L2 1 ⎞ ⎛ R2 + ⎜ω L − ⎟ ωC ⎠ ⎝

2

ω small 2

1 ⎞ 1 ⎛ 2 2 2 2 As ω gets small, R 2 + ⎜ ω L − ⎟ → 2 2 ,R +ω L → R ωC ⎠ ωC ⎝

Therefore

Vout R2 → = ω RC as ω becomes small. Vs (1/ ω 2C 2 )

ω large 2

1 ⎞ ⎛ 2 2 2 2 2 2 2 2 2 2 As ω gets large, R 2 + ⎜ ω L − ⎟ → R +ω L → ω L , R +ω L → ω L ωC ⎠ ⎝ Therefore,

31.50.

Vout ω 2 L2 → = 1 as ω becomes large. Vs ω 2 L2

EVALUATE: Vout / Vs → 0 as ω becomes small, so there is Vout only when the frequency ω of Vs is large. If the source voltage contains a number of frequency components, only the high frequency ones are passed by this filter. IDENTIFY: V = VC = IX C . I = V / Z . SET UP:

X L = ω L, X C =

EXECUTE:

31.51.

Vout = VC =

1

ωC

.

I V 1 ⇒ out = . 2 Vs ω C R + (ω L − 1 ω C )2 ωC

If ω is large:

Vout 1 1 1 = ≈ = . 2 2 Vs ω C R 2 + (ω L − 1 ω C ) ( LC )ω 2 ω C (ω L )

If ω is small:

Vout ≈ Vs ω C

EVALUATE:

When ω is large, X C is small and X L is large so Z is large and the current is small. Both factors in

1

(1 ω C )

2

=

ωC = 1. ωC

VC = IX C are small. When ω is small, X C is large and the voltage amplitude across the capacitor is much larger than the voltage amplitudes across the resistor and the inductor. IDENTIFY: I = V / Z and Pav = 12 I 2 R. Z = R 2 + (ω L − 1/ ω C ) 2 V V EXECUTE: (a) I = = . 2 2 Z R + (ω L − 1 ω C )

SET UP:


31-16

Chapter 31 2

(b) Pav =

1 2 1 ⎛V ⎞ V 2R 2 . I R= ⎜ ⎟ R= 2 2 2 2⎝Z⎠ R + (ω L − 1 ω C )

(c) The average power and the current amplitude are both greatest when the denominator is smallest, which occurs 1 1 . , so ω 0 = for ω 0 L = ω 0C LC

25ω 2 (100 V ) ( 200 Ω ) 2 = . 2 2 2 ( 200 Ω ) + (ω ( 2.00 H ) − 1 [ω (0.500 ×10−6 F)]) 40,000ω 2 + ( 2ω 2 − 2,000,000 ) 2

(d) Pav =

The graph of Pav versus ω is sketched in Figure 31.51. EVALUATE: Note that as the angular frequency goes to zero, the power and current are zero, just as they are when the angular frequency goes to infinity. This graph exhibits the same strongly peaked nature as the light purple curve in Figure 31.19 in the textbook.

Figure 31.51 31.52.

IDENTIFY:

Problem 31.51 shows that I =

SET UP: EXECUTE: (b) VC =

I VL = Iω L and VC = . ωC

(a) VL = I ω L =

V R + (ω L − 1/[ω C ]) 2 2

.

V ωL V ωL = . 2 Z R 2 + (ω L − 1/[ωC ])

I I 1 = = . 2 ωC ω CZ ω C R + (ω L − 1 [ω C ])2

(c) The graphs are given in Figure 31.52. EVALUATE: (d) When the angular frequency is zero, the inductor has zero voltage while the capacitor has voltage of 100 V (equal to the total source voltage). At very high frequencies, the capacitor voltage goes to zero, 1 while the inductor’s voltage goes to 100 V. At resonance, ω 0 = = 1000 rad s, the two voltages are equal, and LC are a maximum, 1000 V.

Figure 31.52


Alternating Current

31.53.

U B = 12 Li 2 . U E = 12 Cv 2 .

IDENTIFY:

Let 〈 x〉 denote the average value of the quantity x. 〈i 2 〉 = 12 I 2 and 〈vC2 〉 = 12 VC2 . Problem 31.51 shows

SET UP:

that I =

31-17

V R + (ω L − 1/[ω C ]) 2

2

. Problem 31.52 shows that VC =

V

ω C R + (ω L − 1/[ω C ]) 2 2

.

2

EXECUTE:

⎛ I ⎞ 1 2 2 (a) U B = 12 Li 2 ⇒ U B = 12 L i 2 = 12 LI rms = 12 L ⎜ ⎟ = 4 LI . ⎝ 2⎠ 2

1 ⎛V ⎞ U E = 12 CvC2 ⇒ U E = C vC2 = 12 CVC2,rms = 12 C ⎜ C ⎟ = 14 CVC2 2 ⎝ 2⎠ (b) Using Problem 31.51a UB

⎛ 1 1 V2 = LI 2 = L ⎜ 4 4 ⎜ R 2 + (ω L − 1 ω C ) 2 ⎝

2

⎞ LV 2 ⎟ = . ⎟ 4 R 2 + (ω L − 1 ω C ) 2 ⎠

(

)

2

1 1 V V2 Using Problem (31.47b): U E = CVC 2 = C . = 2 4 4 ω 2C 2 R 2 + (ω L − 1 ω C ) 2 4ω 2C R 2 + (ω L − 1 ω C )

(

)

(

)

(c) The graphs of the magnetic and electric energies are given in Figure 31.53. EVALUATE: (d) When the angular frequency is zero, the magnetic energy stored in the inductor is zero, while the electric energy in the capacitor is U E = CV 2 4. As the frequency goes to infinity, the energy noted in both

inductor and capacitor go to zero. The energies equal each other at the resonant frequency where ω 0 = UB = UE =

1 and LC

LV 2 . 4R2

Figure 31.53 31.54.

IDENTIFY: At any instant of time the same rules apply to the parallel ac circuit as to parallel dc circuit: the voltages are the same and the currents add. SET UP: For a resistor the current and voltage in phase. For an inductor the voltage leads the current by 90° and for a capacitor the voltage lags the current by 90°. EXECUTE: (a) The parallel L-R-C circuit must have equal potential drops over the capacitor, inductor and resistor, so vR = vL = vC = v. Also, the sum of currents entering any junction must equal the current leaving the junction. Therefore, the sum of the currents in the branches must equal the current through the source: i = iR + iL + iC . (b) iR = v is always in phase with the voltage. iL = v lags the voltage by 90°, and iC = vωC leads the voltage R ωL by 90°. The phase diagram is sketched in Figure 31.54. (c) From the diagram, I = I R + ( I C − I L ) 2

2

2

2

2

(d) From part (c):

I =V

2

V ⎞ ⎛V ⎞ ⎛ = ⎜ ⎟ + ⎜ V ωC − ⎟ . R L⎠ ω ⎝ ⎠ ⎝

V 1 ⎛ 1 ⎞ 1 + ⎜ ωC − = ⎟ . But I = , so 2 Z Z R ωL ⎠ ⎝

2

1 ⎛ 1 ⎞ + ⎜ ωC − . 2 R ⎝ ω L ⎟⎠


31-18

Chapter 31

1 . The current in the capacitor branch is much larger than the current in the ωC other branches. For small ω , Z → ω L. The current in the inductive branch is much larger than the current in the other branches. EVALUATE:

For large ω , Z →

Figure 31.54 31.55.

Apply the expression for I from problem 31.54 when ω 0 = 1/ LC .

IDENTIFY:

2

SET UP:

From Problem 31.54, I = V

EXECUTE:

(a) At resonance, ω 0 =

1 ⎛ 1 ⎞ + ⎜ ωC − 2 R ω L ⎟⎠ ⎝ 1 1 V ⇒ ω 0C = ⇒ I C = V ω 0C = = I so I = I R and I is a minimum. ω0 L ω0 L L LC

2 Vrms V2 cos φ = at resonance where R < Z so power is a maximum. Z R (c) At ω = ω 0 , I and V are in phase, so the phase angle is zero, which is the same as a series resonance.

(b) Pav =

EVALUATE:

(d) The parallel circuit is sketched in Figure 31.55. At resonance, iC = iL and at any instant of time

these two currents are in opposite directions. Therefore, the net current between a and b is always zero. (e) If the inductor and capacitor each have some resistance, and these resistances aren’t the same, then it is no longer true that iC + iL = 0 and the statement in part (d) isn’t valid.

Figure 31.55 31.56.

IDENTIFY: Refer to the results and the phasor diagram in Problem 31.54. The source voltage is applied across each parallel branch. SET UP: V = 2Vrms = 311 V EXECUTE:

(a) I R =

V 311 V = = 0.778 A. R 400 Ω

(b) I C = V ωC = ( 311 V )( 360 rad s ) ( 6.00 × 10−6 F ) = 0.672 A.

⎛I ⎞ ⎛ 0.672 A ⎞ (c) φ = arctan ⎜ C ⎟ = arctan ⎜ ⎟ = 40.8°. I ⎝ 0.778 A ⎠ ⎝ R⎠ (d) I = I R2 + I C2 =

31.57.

( 0.778 A )

2

+ ( 0.672 A ) = 1.03 A. 2

(e) Leads since φ > 0. EVALUATE: The phasor diagram shows that the current in the capacitor always leads the source voltage. IDENTIFY and SET UP: Refer to the results and the phasor diagram in Problem 31.54. The source voltage is applied across each parallel branch. V V EXECUTE: (a) I R = ; I C = V ωC ; I L = . R ωL (b) The graph of each current versus ω is given in Figure 31.57a. (c) ω → 0 : I C → 0; I L → ∞. ω → ∞: I C → ∞; I L → 0. At low frequencies, the current is not changing much so the inductor’s back-emf doesn’t “resist.” This allows the current to pass fairly freely. However, the current in the capacitor goes to zero because it tends to “fill up” over the slow period, making it less effective at passing charge. At high frequency, the induced emf in the inductor resists the violent changes and passes little current. The capacitor never gets a chance to fill up so passes charge freely.


Alternating Current

(d) ω =

31-19

1 1 = = 1000 rad sec and f = 159 Hz. The phasor diagram is sketched in LC (2.0 H)(0.50 ×10−6 F)

Figure 31.57b. 2

2

V ⎞ ⎛V ⎞ ⎛ (e) I = ⎜ ⎟ + ⎜ V ωC − ⎟ . R L⎠ ω ⎝ ⎝ ⎠ 2

2

⎛ 100 V ⎞ ⎛ ⎞ 100 V −1 −6 I= ⎜ ⎟ + ⎜ (100 V)(1000 s )(0.50 ×10 F) − ⎟ = 0.50 A −1 (1000 s )(2.0 H) ⎠ ⎝ 200 Ω ⎠ ⎝ (f ) At resonance I L = I C = V ωC = (100 V)(1000 s −1 )(0.50 ×10−6 F) = 0.0500 A and I R = EVALUATE: resistor.

V 100 V = = 0.50 A. R 200 Ω

At resonance iC = iL = 0 at all times and the current through the source equals the current through the

Figure 31.57 31.58.

IDENTIFY:

The average power depends on the phase angle φ. 2

SET UP: EXECUTE:

1 ⎞ ⎛ The average power is Pav = VrmsIrmscos φ, and the impedance is Z = R 2 + ⎜ ω L − . ω C ⎟⎠ ⎝

(a) Pav = VrmsIrmscos φ =

1 2

(VrmsIrms), which gives cos φ = 12 , so φ = π/3 = 60°. tan φ = (XL – XC)/R,

which gives tan 60° = (ωL – 1/ωC)/R. Using R = 75.0 Ω, L = 5.00 mH, and C = 2.50 µF and solving for ω we get ω = 28760 rad/s = 28,800 rad/s. (b) Z = R 2 + ( X L − X C ) 2 , where XL = ωL = (28,760 rad/s)(5.00 mH) = 144 Ω and

XC = 1/ωC = 1/[(28,760 rad/s)(2.50 µF)] = 13.9 Ω, giving Z = (75 Ω) 2 + (144 Ω − 13.9 Ω) 2 = 150 Ω;

31.59.

I = V/Z = (15.0 V)/(150 Ω) = 0.100 A and Pav = 12 VI cos φ = 12 (15.0 V)(0.100 A)(1/2) = 0.375 W. EVALUATE: All this power is dissipated in the resistor because the average power delivered to the inductor and capacitor is zero. 2 R from Exercise 31.27 to calculate IDENTIFY: We know R, X C and φ so Eq.(31.24) tells us X L . Use Pav = I rms I rms . Then calculate Z and use Eq.(31.26) to calculate Vrms for the source. SET UP: Source voltage lags current so φ = −54.0°. X C = 350 Ω, R = 180 Ω, Pav = 140 W X − XC EXECUTE: (a) tan φ = L R X L = R tan φ + X C = (180 Ω) tan(−54.0°) + 350 Ω = −248 Ω + 350 Ω = 102 Ω 2 R (Exercise 31.27). I rms = (b) Pav = Vrms I rms cosφ = I rms

Pav 140 W = = 0.882 A R 180 Ω

(c) Z = R 2 + ( X L − X C ) 2 = (180 Ω) 2 + (102 Ω − 350 Ω) 2 = 306 Ω

Vrms = I rms Z = (0.882 A)(306 Ω) = 270 V. EVALUATE: We could also use Eq.(31.31): Pav = Vrms I rms cos φ Pav 140 W Vrms = = = 270 V, which agrees. The source voltage lags the current when I rms cos φ (0.882 A)cos( −54.0°) X C > X L , and this agrees with what we found.


31-20

Chapter 31

31.60.

IDENTIFY and SET UP: Calculate Z and I = V / Z . EXECUTE: (a) For ω = 800 rad s: Z = R 2 + (ω L − 1 ω C ) 2 = (500 Ω) 2 + ((800 rad/s)(2.0 H) − 1 ((800 rad/s)(5.0 ×10 −7 F))) 2 . Z = 1030 Ω . V 100 V 1 0.0971 A = = 243 V I= = = 0.0971 A. VR = IR = (0.0971 A)(500 Ω) = 48.6 V, VC = ω C (800 rad s)(5.0 ×10−7 F) Z 1030 Ω ⎛ ω L − 1 (ωC ) ⎞ and VL = IωL = (0.0971 A)(800 rad s)(2.00 H) = 155 V. φ = arctan ⎜ ⎟ = −60.9°. The graph of each R ⎝ ⎠ voltage versus time is given in Figure 31.60a. (b) Repeating exactly the same calculations as above for ω = 1000 rad/s: Z = R = 500 Ω; φ = 0; I = 0.200 A; VR = V = 100 V; VC = VL = 400 V. The graph of each voltage versus time is given in Figure 31.60b. (c) Repeating exactly the same calculations as part (a) for ω = 1250 rad/s : Z = R = 1030 Ω; φ = +60.9°; I = 0.0971 A; VR = 48.6 V; VC = 155 V; VL = 243 V. The graph of each voltage versus time is given in Figure 31.60c. 1 1 EVALUATE: The resonance frequency is ω 0 = = = 1000 rad/s. For ω < ω 0 the phase LC (2.00 H)(0.500 μ F)

angle is negative and for ω > ω 0 the phase angle is positive.

Figure 31.60


Alternating Current

31.61.

31-21

Eq.(31.19) allows us to calculate I and then Eq.(31.22) gives Z. Solve Eq.(31.21) for L. V 360 V = 0.750 A (a) VC = IX C so I = C = X C 480 Ω

IDENTIFY and SET UP: EXECUTE:

V 120 V = = 160 Ω I 0.750 A (c) Z 2 = R 2 + ( X L − X C ) 2

(b) V = IZ so Z =

X L − X C = ± Z 2 − R 2 , so X L = X C ± Z 2 − R 2 = 480 Ω ± (160 Ω) 2 − (80.0 Ω) 2 = 480 Ω ± 139 Ω X L = 619 Ω or 341 Ω 1 and X L = ω L. At resonance, X C = X L . As the frequency is lowered below the ωC resonance frequency X C increases and X L decreases. Therefore, for ω < ω 0 , X L < X C . So for X L = 341 Ω the XC =

(d) EVALUATE:

angular frequency is less than the resonance angular frequency. ω is greater than ω 0 when X L = 619 Ω. But at these two values of X L , the magnitude of X L − X C is the same so Z and I are the same. In one case ( X L = 691 Ω) 31.62.

the source voltage leads the current and in the other ( X L = 341 Ω) the source voltage lags the current. IDENTIFY and SET UP: The maximum possible current amplitude occurs at the resonance angular frequency because the impedance is then smallest. EXECUTE: (a) At the resonance angular frequency ω 0 = 1/ LC , the current is a maximum and Z = R, giving Imax = V/R. At the required frequency, I = Imax/3. I = V/Z = Imax/3 = (V/R)/3, which means that Z = 3R. Squaring gives R2 + (ωL – 1/ωC)2 = 9R2 . Solving for ω gives ω = 3.192 × 105 rad/s and ω = 8.35 × 104 rad/s. I V 49.5 V (b) V = 2Vrms = 2(35.0 V) = 49.5 V. I = max = = = 0.132 A. 3 3R 3(125 Ω) For ω = 8.35 × 104 rad/s: R = 125 Ω and VR = IR = 16.5 Ω; X L = ω L = 125 Ω and VL = 16.5 V; 1 = 479 Ω and VC = 63.2 V. ωC For ω = 3.192 × 105 rad/s: R = 125 Ω and VR = IR = 16.5 Ω; X L = ω L = 479 Ω and VL = 63.2 V; XC =

XC =

1

ωC

= 125 Ω and VC = 16.5 V.

For the lower frequency, X C > X L and VC > VL . For the higher frequency, X L > X C and VL > VC .

EVALUATE: 31.63.

IDENTIFY and SET UP:

this interval i =

2I0

τ

Consider the cycle of the repeating current that lies between t1 = τ / 2 and t2 = 3τ / 2. In

(t − τ ). I av =

t2 t2 1 1 2 i dt and I rms = i 2 dt t2 − t1 ∫ t1 t2 − t1 ∫t1 3τ / 2

I av =

EXECUTE:

t2 1 1 3τ / 2 2 I 0 2I ⎡ 1 ⎤ (t − τ ) dt = 20 ⎢ t 2 − τ t ⎥ i dt = ∫ ∫ t τ / 2 1 t2 − t1 τ τ τ ⎣2 ⎦τ /2

2 3τ 2 τ 2 τ 2 ⎞ I ⎛ 2 I ⎞ ⎛ 9τ − − + ⎟ = (2 I 0 ) 18 (9 − 12 − 1 + 4) = 0 (13 − 13) = 0. I av = ⎜ 20 ⎟ ⎜ τ 8 2 8 2 4 ⎝ ⎠⎝ ⎠ 2 = ( I 2 )av = I rms

2 = I rms

4 I 02

τ

3

t2 1 1 3τ / 2 4 I 02 (t − τ ) 2 dt i 2 dt = ∫ ∫ t τ τ /2 τ 2 t2 − t1 1

3τ / 2

∫τ

/2

(t − τ ) 2 dt =

4 I 02 ⎡⎛ τ ⎞ ⎛ τ ⎞ 3 3τ / 2 1 ⎡ ⎤ ( ) − = t τ 3 ⎦τ / 2 3τ 3 ⎢⎜⎝ 2 ⎟⎠ − ⎜⎝ − 2 ⎟⎠ τ3 ⎣ ⎢⎣

4 I 02

3

3

⎤ ⎥ ⎥⎦

I 02 [1 + 1] = 13 I 02 6 I 2 I rms = I rms = 0 . 3 EVALUATE: In each cycle the current has as much negative value as positive value and its average is zero. i 2 is always positive and its average is not zero. The relation between I rms and the current amplitude for this current is different from that for a sinusoidal current (Eq.31.4). 2 = I rms


31-22

Chapter 31

31.64.

IDENTIFY:

Apply Vrms = I rms Z 1 and Z = R 2 + ( X L − X C ) 2 . LC 1 1 = = 786 rad s. (a) ω 0 = LC (1.80 H)(9.00 ×10−7 F)

ω0 =

SET UP: EXECUTE:

(b) Z = R 2 + (ωL − 1 ωC ) 2 . Z = (300 Ω) 2 + ((786 rad s)(1.80 H) − 1 ((786 rad s)(9.00 ×10−7 F))) 2 = 300 Ω.

I rms-0 =

Vrms 60 V = = 0.200 A. Z 300 Ω

(c) We want I =

ω 2 L2 +

1 V I rms-0 = rms = 2 Z

Vrms R 2 + (ω L − 1 ωC ) 2

. R 2 + (ω L −1 ωC ) 2 =

2 4Vrms . 2 I rms-0

2 ⎛ ⎞ 1 2 L 4Vrms 1 2L 4V 2 − 2 ⎟ + 2 = 0. − + R 2 − 2 rms = 0 and (ω 2 ) 2 L2 + ω 2 ⎜ R 2 − 2 ωC C I rms-0 C I C rms-0 ⎠ ⎝ 2

Substituting in the values for this problem, the equation becomes (ω 2 ) 2 (3.24) + ω 2 (−4.27 × 106 ) + 1.23 × 1012 = 0. Solving this quadratic equation in ω 2 we find ω 2 = 8.90 ×105 rad 2 s 2 or 4.28 ×105 rad 2 /s 2 and

ω = 943 rad s or 654 rad s. (d) (i) R = 300 Ω, I rms-0 = 0.200 A, ω1 − ω2 = 289 rad s. (ii) R = 30 Ω, I rms-0 = 2A, ω1 − ω 2 = 28 rad/s.

(iii) R = 3 Ω, I rms-0 = 20 A, ω1 − ω 2 = 2.88 rad/s. 31.65.

EVALUATE: The width gets smaller as R gets smaller; I rms-0 gets larger as R gets smaller. IDENTIFY: The resonance frequency, the reactances, and the impedance all depend on the values of the circuit elements. SET UP:

The resonance frequency is ω0 = 1/ LC , the reactances are XL = ωL and XC = 1/ωC, and the impedance

is Z = R 2 + ( X L − X C ) 2 . 1 → 1/ 2, so ω0 decreases by 12 . 2 L 2C (b) Since XL = ωL, if L is doubled, XL increases by a factor of 2. (c) Since XC = 1/ωC, doubling C decreases XC by a factor of 12 .

EXECUTE:

(a) ω0 = 1/ LC becomes

(d) Z = R 2 + ( X L − X C ) 2 → Z = (2 R ) 2 + (2 X L − 12 X C ) 2 , so Z does not change by a simple factor of 2 or

1 2

.

The impedance does not change by a simple factor, even though the other quantities do. V N IDENTIFY: A transformer transforms voltages according to 2 = 2 . The effective resistance of a secondary V1 N1 R . circuit of resistance R is Reff = ( N 2 / N1 ) 2 SET UP: N 2 = 275 and V1 = 25.0 V. EXECUTE: (a) V2 = V1 ( N 2 / N1 ) = (25.0 V)(834 / 275) = 75.8 V

EVALUATE: 31.66.

R 125 Ω = = 13.6 Ω ( N 2 / N1 ) 2 (834 / 275) 2 EVALUATE: The voltage across the secondary is greater than the voltage across the primary since N 2 > N1. The effective load resistance of the secondary is less than the resistance R connected across the secondary. 1 1 . IDENTIFY: The resonance angular frequency is ω 0 = and the resonance frequency is f 0 = LC 2π LC SET UP: ω 0 is independent of R. (b) Reff =

31.67.

EXECUTE:

31.68.

(a) ω 0 (or f 0 ) depends only on L and C so change these quantities.

(b) To double ω 0 , decrease L and C by multiplying each of them by 12 . EVALUATE: Increasing L and C decreases the resonance frequency; decreasing L and C increases the resonance frequency. IDENTIFY: At resonance, Z = R. I = V / R. VR = IR, VC = IX C and VL = IX L . U E = 12 CVC2 and U L = 12 LI 2 . SET UP: The amplitudes of each time dependent quantity correspond to the maximum values of those quantities.


Alternating Current

EXECUTE:

(a) I =

(b) VC = IX C = (c) VL = IX L =

31.69.

V Rω 0C

=

V R + (ω L − 1/ ω C ) 2

2

. At resonance ω L =

1

ωC

and I max =

V . R

V L . R C

V V L ω0 L = . R R C

1 1 V2 L 1 V2 (d) U C = CVC2 = C 2 = L 2 . 2 2 R C 2 R 1 1 V2 (e) U L = LI 2 = L 2 . 2 2 R EVALUATE: At resonance VC = VL and the maximum energy stored in the inductor equals the maximum energy stored in the capacitor. IDENTIFY: I = V / R. VR = IR, VC = IX C and VL = IX L . U E = 12 CVC2 and U L = 12 LI 2 . SET UP: The amplitudes of each time dependent quantity correspond to the maximum values of those quantities. EXECUTE: (a) I =

V = Z

ω0

ω=

2

.

V ⎛ω L ⎞ R 2 + ⎜ 0 − 2 / ω 0C ⎟ 2 ⎝ ⎠

(b) VC = IX C =

2 ω 0C

ω0 L

V R2 +

2

=

9 L 4C

=

V 9 L R + 4C

.

2

L C

2V R2 +

9 L 4C

.

L V 2 = . C L 9 9 L 2 2 R + R + 4C 4C 1 2 LV 2 2 . (d) U C = CVC = 9 L 2 R2 + 4C 1 2 1 LV 2 . (e) U L = LI = 2 2 R2 + 9 L 4C EVALUATE: For ω < ω 0 , VC > VL and the maximum energy stored in the capacitor is greater than the maximum energy stored in the inductor. IDENTIFY: I = V / R . VR = IR , VC = IX C and VL = IX L . U E = 12 CVC2 and U L = 12 LI 2 . SET UP: The amplitudes of each time dependent quantity correspond to the maximum values of those quantities. EXECUTE: ω = 2ω 0 . V V V (a) I = = = . 2 2 Z 9 L R + (2ω 0 L − 1/ 2ω 0C ) 2 R + 4C (c) VL = IX L =

31.70.

V = Z

31-23

(b) VC = IX C =

V

2

1 2ω 0C

(c) VL = IX L = 2ω 0 L

V R2 +

9 L 4C

V

9 L R2 + 4C 1 LV 2 2 (d) U C = CVC = . 2 9 L 2 8 R + 4C

=

=

L C L C

V 2 . 9 L R2 + 4C 2V 9 L R + 4C 2

.


31-24

Chapter 31

31.71.

LV 2 . 9 L 2 R2 + 4C EVALUATE: For ω > ω 0 , VL > VC and the maximum energy stored in the inductor is greater than the maximum energy stored in the capacitor. IDENTIFY and SET UP: Assume the angular frequency ω of the source and the resistance R of the resistor are known. V Iω L ω L EXECUTE: Connect the source, capacitor, resistor, and inductor in series. Measure VR and VL . L = = VR IR R and L can be calculated. EVALUATE: There are a number of other approaches. The frequency could be varied until VC = VL , and then this

31.72.

frequency is equal to 1/ LC . If C is known, then L can be calculated. V IDENTIFY: Pav = Vrms I rms cos φ and I rms = rms . Calculate Z. R = Z cos φ . Z SET UP: f = 50.0 Hz and ω = 2π f . The power factor is cosφ .

(e) U L =

1 2 LI = 2

2 Vrms V 2 cos φ (120 V) 2 (0.560) cos φ . Z = rms = = 36.7 Ω. Z Pav (220 W) R = Z cos φ = (36.7 Ω)(0.560) = 20.6 Ω.

(a) Pav =

EXECUTE:

(b) Z = R 2 + X L 2 ⋅ X L = Z 2 − R 2 = (36.7 Ω) 2 − (20.6 Ω) 2 = 30.4 Ω. But φ = 0 is at resonance, so the inductive

and capacitive reactances equal each other. Therefore we need to add X C = 30.4 Ω. X C = C=

1

ωX C

=

31.73.

IDENTIFY: SET UP:

therefore gives

1 1 = = 1.05 ×10−4 F. 2π f X C 2π (50.0 Hz)(30.4 Ω)

(c) At resonance, Pav = EVALUATE: resonance.

1

ωC

V 2 (120 V) 2 = = 699 W. R 20.6 Ω

2 Pav = I rms R and I rms is maximum at resonance, so the power drawn from the line is maximum at

pR = i 2 R. pL = iL i = I cos ω t

di q . pC = i. C dt

1 (a) pR = i 2 R = I 2 cos 2 (ωt ) R = VR I cos 2 (ωt ) = VR I (1 + cos(2ωt )). 2 1 T VR I T VR I T 1 Pav ( R ) = ∫ pR dt = (1 + cos(2ω t )) dt = [t ]0 = 2 VR I . T 0 2T ∫ 0 2T T di (b) pL = Li = −ω LI 2 cos(ω t )sin(ω t ) = − 12 VL I sin(2ω t ). But ∫ sin(2ω t ) dt = 0 ⇒ Pav ( L) = 0. 0 dt T q (c) pC = i = vC i = VC I sin(ω t )cos(ω t ) = 12 VC I sin(2ω t ). But ∫ sin(2ω t )dt = 0 ⇒ Pav (C ) = 0. 0 C (d) p = pR + pL + pc = VR I cos 2 (ω t ) − 12 VL I sin(2ω t ) + 12 VC I sin(2ω t ) and

EXECUTE:

V −V VR and sin φ = L C , so V V p = VI cos(ω t )(cos φ cos(ω t ) − sin φ sin(ω t )), at any instant of time. EVALUATE: At an instant of time the energy stored in the capacitor and inductor can be changing, but there is no net consumption of electrical energy in these components. dVL dVC IDENTIFY: VL = IX L . = 0 at the ω where VL is a maximum. VC = IX C . = 0 at the ω where VC is a dω dω maximum. V . SET UP: Problem 31.51 shows that I = 2 R + (ω L − 1/ ω C ) 2 p = I cos(ω t )(VR cos(ω t ) − VL sin(ω t ) + VC sin(ω t )). But cos φ =

31.74.

EXECUTE:

(a) VR =maximum when VC = VL ⇒ ω = ω 0 =

1 . LC


Alternating Current

31-25

⎞ dVL d ⎛ V ωL dVL ⎜ ⎟. =0= = 0.Therefore: dω dω ⎜⎝ R 2 + (ω L − 1 ωC ) 2 ⎟⎠ dω VL V ω 2 L( L − 1 ω 2C )( L + 1 ω 2C ) − . R 2 + (ω L − 1 ω C ) 2 = ω 2 ( L2 − 1 ω 4C 2 ) . 2 2 3 2 2 2 ( R + ( ω L − 1 ω C ) ) R + (ω L − 1 ωC )

(b) VL =maximum when 0=

R2 +

1 R 2C 2 1 2L 1 LC = − and ω = . − = − 2 ω 2C 2 C ω 2C 2 ω 2

1 LC − R 2C 2 2

.

⎞ dVC d ⎛ V dVC ⎜ ⎟. =0= = 0. Therefore: dω dω ⎜ ω C R 2 + (ω L − 1 ω C ) 2 ⎟ dω ⎝ ⎠ V V ( L − 1 ω 2C )( L + 1 ω 2C ) 0=− − . R 2 + (ω L − 1 ω C ) 2 = −ω 2 ( L2 − 1 ω 4C 2 ). 2 2 3 2 ω 2C R 2 + (ω L − 1 ω C ) 2 C ( R + (ω L − 1 ω C ) )

(c) VC = maximum when

1 R2 2L − 2. = −ω 2 L2 and ω = C LC 2 L 2L 2 2 2 2 2 = −ω L . R +ω L − C EVALUATE: VL is maximum at a frequency greater than the resonance frequency and X C is a maximum at a frequency less than the resonance frequency. These frequencies depend on R , as well as on L and on C. IDENTIFY: Follow the steps specified in the problem. SET UP: In part (a) use Eq.(31.23) to calculate Z and then I = V / Z . φ is given by Eq.(31.24). In part (b) let Z = R + iX . R 2 + ω 2 L2 −

31.75.

EXECUTE:

(a) From the current phasors we know that Z = R 2 + (ω L − 1 ω C ) 2 . 2

⎛ ⎞ 1 Z = (400 Ω) + ⎜ (1000 rad s)(0.50 H) − ⎟ = 500 Ω. −6 (1000 rad s)(1.25 10 F) × ⎝ ⎠ V 200 V I= = = 0.400 A. Z 500 Ω 2

⎛ (1000 rad s)(0.500 H) − 1 (1000 rad s)(1.25 × 10−6 F) ⎞ ⎛ ω L − 1 (ω C ) ⎞ arctan (b) φ = arctan ⎜ φ = . ⎜ ⎟ = +36.9° ⎟ 400 Ω R ⎝ ⎠ ⎝ ⎠ ⎛ ⎞ 1 1 ⎞ ⎛ (c) Z cpx = R + i ⎜ ω L − ⎟ = 400 Ω − 300 Ωi. ⎟ . Z cpx = 400 Ω − i ⎜ (1000 rad s)(0.50 H) − −6 (1000 rad s)(1.25 10 F) × ω C ⎝ ⎠ ⎝ ⎠ Z = (400 Ω) 2 + ( −300 Ω) 2 = 500 Ω.

200 V V ⎛ 8 + 6i ⎞ ⎛ 8 + 6i ⎞ ⎛ 8 − 6i ⎞ = =⎜ ⎟ ⎜ ⎟ = 0.400 A. ⎟ A = (0.320 A) + (0.240 A)i. I = ⎜ Z cpx (400 − 300i ) Ω ⎝ 25 ⎠ ⎝ 25 ⎠ ⎝ 25 ⎠ Im( I cpx ) 6 25 = = 0.75 ⇒ φ = +36.9°. (e) tan φ = Re( I cpx ) 8 25 (d) I cpx =

⎛ 8 + 6i ⎞ (f ) VRcpx = I cpx R = ⎜ ⎟ (400 Ω) = (128 + 96i)V. ⎝ 25 ⎠ ⎛ 8 + 6i ⎞ VLcpx = iI cpxω L = i ⎜ ⎟ (1000 rad s)(0.500 H) = ( − 120 + 160i ) V. ⎝ 25 ⎠ VCcpx = i (g) Vcpx

1 ⎛ 8 + 6i ⎞ = i⎜ = (+192 − 256i ) V. ⎟ ω C ⎝ 25 ⎠ (1000 rad s)(1.25 × 10−6 F) = VRcpx + VLcpx + VCcpx = (128 + 96i ) V + (−120 + 160i)V + (192 − 256i ) V = 200 V. I cpx

EVALUATE:

Both approaches yield the same value for I and for φ .


ELECTROMAGNETIC WAVES

32.1.

IDENTIFY: Since the speed is constant, distance x = ct. SET UP: The speed of light is c = 3.00 ×108 m/s . 1 yr = 3.156 ×107 s.

x 3.84 ×108 m = = 1.28 s c 3.00 ×108 m/s (b) x = ct = (3.00 ×108 m/s)(8.61 yr)(3.156 ×107 s/yr) = 8.15 × 1016 m = 8.15 × 1013 km EVALUATE: The speed of light is very great. The distance between stars is very large compared to terrestrial distances. IDENTIFY: Since the speed is constant the difference in distance is cΔt. SET UP: The speed of electromagnetic waves in air is c = 3.00 × 108 m/s. EXECUTE: A total time difference of 0.60 μ s corresponds to a difference in distance of (a) t =

EXECUTE:

32.2.

32.3.

32

cΔt = (3.00 ×108 m/s)(0.60 ×10−6 s) = 180 m. EVALUATE: The time delay doesn’t depend on the distance from the transmitter to the receiver, it just depends on the difference in the length of the two paths. IDENTIFY: Apply c = f λ. c = 3.00 ×108 m/s c 3.0 ×108 m s EXECUTE: (a) f = = = 6.0 ×104 Hz. 5000 m λ

SET UP:

(b) f = (c) f =

λ c

λ

=

3.0 ×108 m s = 6.0 ×107 Hz. 5.0 m

=

3.0 ×108 m s = 6.0 ×1013 Hz. 5.0 ×10−6 m

3.0 ×108 m s = 6.0 ×1016 Hz. λ 5.0 × 10−9 m EVALUATE: f increases when λ decreases. 2π IDENTIFY: c = f λ and k = . (d) f =

32.4.

c

c

=

λ

c = 3.00 ×108 m/s . c EXECUTE: (a) f = . UVA: 7.50 ×1014 Hz to 9.38 ×1014 Hz . UVB: 9.38 ×1014 Hz to 1.07 ×1015 Hz . SET UP:

λ

(b) k = 32.5.

λ

. UVA: 1.57 ×107 rad/m to 1.96 × 107 rad/m . UVB: 1.96 × 107 rad/m to 2.24 ×107 rad/m .

EVALUATE: Larger λ corresponds to smaller f and k. IDENTIFY: c = f λ . Emax = cBmax . k = 2π / λ . ω = 2π f .

Since the wave is traveling in empty space, its wave speed is c = 3.00 ×108 m/s . c 3.00 ×108 m/s = 6.94 ×1014 Hz EXECUTE: (a) f = = λ 432 ×10−9 m (b) Emax = cBmax = (3.00 ×108 m/s)(1.25 ×10−6 T) = 375 V/m SET UP:

32-1


32-2

Chapter 32

2π rad = 1.45 ×107 rad/m . ω = (2π rad)(6.94 ×1014 Hz) = 4.36 ×1015 rad/s . 432 ×10−9 m E = Emax cos( kx − ω t ) = (375 V/m)cos([1.45 ×107 rad/m]x − [4.36 ×1015 rad/s] t )

(c) k =

32.6.

=

λ

B = Bmax cos(kx − ωt ) = (1.25 × 10−6 T)cos([1.45 × 107 rad/m]x − [4.36 × 1015 rad/s]t ) EVALUATE: The cos(kx − ω t ) factor is common to both the electric and magnetic field expressions, since these two fields are in phase. IDENTIFY: c = f λ . Emax = cBmax . Apply Eqs.(32.17) and (32.19). The speed of the wave is c = 3.00 × 108 m/s. c 3.00 ×108 m/s = 6.90 ×1014 Hz EXECUTE: (a) f = = λ 435 ×10−9 m E 2.70 ×10−3 V/m = 9.00 ×10−12 T (b) Bmax = max = 3.00 ×108 m/s c ! 2π (c) k = = 1.44 ×107 rad/m . ω = 2π f = 4.34 ×1015 rad/s . If E ( z, t ) = iˆEmax cos(kz + ωt ) , then λ ! ! ! B ( z , t ) = − ˆjBmax cos(kz + ωt ) , so that E × B will be in the − kˆ direction. ! E ( z, t ) = iˆ(2.70 × 10−3 V/m)cos([1.44 × 107 rad/s) z + [4.34 × 1015 rad/s] t ) and ! B ( z , t ) = − ˆj (9.00 × 10−12 T)cos([1.44 × 107 rad/s) z + [4.34 × 1015 rad/s] t ) . ! ! EVALUATE: The directions of E and B and of the propagation of the wave are all mutually perpendicular. The argument of the cosine is kz + ω t since the wave is traveling in the − z -direction . Waves for visible light have very high frequencies. ! IDENTIFY and SET UP: The equations are of the form of Eqs.(32.17), with x replaced by z. B is along the y-axis; ! deduce the direction of E . EXECUTE: ω = 2π f = 2π (6.10 ×1014 Hz) = 3.83 ×1015 rad/s SET UP:

32.7.

2π f ω 3.83 × 1015 rad/s = = = 1.28 ×107 rad/m λ c c 3.00 × 108 m/s Bmax = 5.80 ×10−4 T

k=

=

Emax = cBmax = (3.00 × 108 m/s)(5.80 ×10−4 T) = 1.74 ×105 V/m ! ! ! B is along the y-axis. E × B is in the direction of propagation (the +z-direction). From this we can deduce the ! direction of E , as shown in Figure 32.7.

! E is along the x-axis.

32.8.

Figure 32.7 ! E = Emax iˆ cos(kz − ω t ) = (1.74 ×105 V/m)iˆ cos[(1.28 × 107 rad/m)z − (3.83 × 1015 rad/s)t ] ! B = Bmax ˆj cos( kz − ω t ) = ( 5.80 × 10−4 T ) ˆj cos[(1.28 × 107 rad/m)z − (3.83 × 1015 rad/s)t ] ! ! EVALUATE: E and B are perpendicular and oscillate in phase. IDENTIFY: For an electromagnetic wave propagating in the negative x direction, E = Emax cos( kx + ω t ) . ω = 2π f

and k = SET UP:

λ

.T=

1 . Emax = cBmax . f

The wave specified in the problem has a different phase, so E = − Emax sin(kx + ω t ) . Emax = 375 V/m ,

k = 1.99 × 107 rad/m and ω = 5.97 ×1015 rad/s . E EXECUTE: (a) Bmax = max = 1.25 μ T . c


Electromagnetic Waves

ω 2π 1 = 9.50 ×1014 Hz . λ = = 3.16 ×10−7 m = 316 nm . T = = 1.05 × 10 −15 s . This wavelength is too short 2π k f

(b) f =

32.9.

32-3

to be visible. (c) c = f λ = (9.50 × 1014 Hz)(3.16 × 10 −7 m) = 3.00 × 108 m/s . This is what the wave speed should be for an electromagnetic wave propagating in vacuum. ⎛ ω ⎞⎛ 2π ⎞ ω EVALUATE: c = f λ = ⎜ ⎟⎜ ⎟ = is an alternative expression for the wave speed. ⎝ 2π ⎠⎝ k ⎠ k ! IDENTIFY and SET UP: Compare the E ( y , t ) given in the problem to the general form given by Eq.(32.17). Use ! ! the direction of propagation and of E to find the direction of B. (a) EXECUTE: The equation for the electric field contains the factor sin(ky − ω t ) so the wave is traveling in the ! +y-direction. The equation for E ( y , t ) is in terms of sin(ky − ω t ) rather than cos(ky − ωt ); the wave is shifted in phase by 90° relative to one with a cos(ky − ω t ) factor. ! (b) E ( y, t ) = −(3.10 ×105 V/m)kˆ sin[ ky − (2.65 ×1012 rad/s)t ] Comparing to Eq.(32.17) gives ω = 2.65 ×1012 rad/s 2π c 2π c 2π (2.998 × 108 m/s) = = 7.11 × 10−4 m ω = 2π f = so λ = λ ω (2.65 × 1012 rad/s) (c) ! ! E × B must be in the +ydirection (the direction in which the wave is traveling). ! When E is in the –z-direction ! then B must be in the –xdirection, as shown in Figure 32.9. Figure 32.9

ω

2.65 × 1012 rad/s = 8.84 ×103 rad/m λ c 2.998 ×108 m/s Emax = 3.10 × 105 V/m

k=

=

=

Emax 3.10 ×105 V/m = = 1.03 ×10−3 T c 2.998 ×108 m/s ! ! Using Eq.(32.17) and the fact that B is in the − iˆ direction when E is in the − kˆ direction, ! B = −(1.03 ×10−3 T)iˆ sin[(8.84 ×103 rad/m)y − (2.65 ×1012 rad/s)t ] ! ! EVALUATE: E and B are perpendicular and oscillate in phase. IDENTIFY: Apply Eqs.(32.17) and (32.19). f = c / λ and k = 2π / λ . Then Bmax =

32.10.

SET UP:

32.11.

The wave in this problem has a different phase, so By ( z , t ) = Bmax sin(kx + ωt ).

EXECUTE: (a) The phase of the wave is given by kx + ω t , so the wave is traveling in the − x direction. 2π 2π f kc (1.38 ×104 rad m)(3.0 ×108 m s) (b) k = = . f = = = 6.59 ×1011 Hz. c λ 2π 2π (c) Since the magnetic field is in the + y -direction, and the wave is propagating in the − x -direction, then the ! ! electric field is in the + z -direction so that E × B will be in the − x -direction. ! E ( x, t ) = + cB ( x, t )kˆ = cBmax sin(kx + ωt )kˆ. ! E ( x, t ) = (c(3.25 × 10−9 T))sin ( (1.38 × 104 rad/m) x + (4.14 × 1012 rad/s)t ) kˆ. ! E ( x, t ) = + (2.48 V m)sin ( (1.38 × 104 rad/m) x + (4.14 × 1012 rad/s)t ) kˆ. ! ! EVALUATE: E and B have the same phase and are in perpendicular directions. IDENTIFY and SET UP: c = f λ allows calculation of λ . k = 2π / λ and ω = 2π f . Eq.(32.18) relates the electric and magnetic field amplitudes. c 2.998 ×108 m/s EXECUTE: (a) c = f λ so λ = = = 361 m f 830 ×103 Hz


32-4

Chapter 32

2π rad = = 0.0174 rad/m λ 361 m (c) ω = 2π f = (2π )(830 ×103 Hz) = 5.22 ×106 rad/s (b) k =

32.12.

(d) Eq.(32.18): Emax = cBmax = (2.998 ×108 m/s)(4.82 × 10−11 T) = 0.0144 V/m EVALUATE: This wave has a very long wavelength; its frequency is in the AM radio braodcast band. The electric and magnetic fields in the wave are very weak. IDENTIFY: Emax = cBmax .

The magnetic field of the earth is about 10−4 T. E 3.85 ×10−3 V/m = 1.28 ×10−11 T. EXECUTE: B = = c 3.00 ×108 m/s EVALUATE: The field is much smaller than the earth's field. IDENTIFY and SET UP: v = f λ relates frequency and wavelength to the speed of the wave. Use Eq.(32.22) to calculate n and K. v 2.17 ×108 m/s = 3.81×10−7 m EXECUTE: (a) λ = = f 5.70 ×1014 Hz SET UP:

32.13.

c 2.998 × 108 m/s = = 5.26 ×10−7 m f 5.70 ×1014 Hz

(b) λ = (c) n =

c 2.998 × 108 m/s = = 1.38 v 2.17 × 108 m/s

(d) n = KK m ≈ K so K = n 2 = (1.38) 2 = 1.90

32.14.

EVALUATE: In the material v < c and f is the same, so λ is less in the material than in air. v < c always, so n is always greater than unity. IDENTIFY: Apply Eq.(32.21). Emax = cBmax . v = f λ . Apply Eq.(32.29) with μ = K m μ 0 in place of μ 0 .

K = 3.64 . K m = 5.18

SET UP:

(a) v =

EXECUTE: (b) λ =

(3.00 ×108 m s) c = = 6.91×107 m s. (3.64)(5.18) KK m

v 6.91×107 m s = = 1.06 ×106 m. f 65.0 Hz

(c) Bmax =

Emax 7.20 ×10−3 V m = = 1.04 ×10−10 T. 6.91× 107 m s v

Emax Bmax (7.20 ×10−3 V m)(1.04 ×10−10 T) = = 5.75 ×10−8 W m 2 . 2Km μ0 2(5.18) μ 0 EVALUATE: The wave travels slower in this material than in air. 2 . Emax = cBmax . IDENTIFY: I = P / A . I = 12 P0cEmax (d) I =

32.15.

SET UP:

32.16.

The surface area of a sphere of radius r is A = 4π r 2 . P0 = 8.85 × 10−12 C 2 /N ⋅ m 2 . P (0.05)(75 W) = = 330 W/m 2 . A 4π (3.0 ×10−2 m) 2

EXECUTE:

(a) I =

(b) Emax =

2I 2(330 W/m 2 ) E = = 500 V/m . Bmax = max = 1.7 ×10−6 T = 1.7 μ T . −12 c P0c (8.85 × 10 C 2 /N ⋅ m 2 )(3.00 × 108 m/s)

EVALUATE: At the surface of the bulb the power radiated by the filament is spread over the surface of the bulb. Our calculation approximates the filament as a point source that radiates uniformly in all directions. ! ! IDENTIFY and SET UP: The direction of propagation is given by E × B . EXECUTE: (a) Sˆ = iˆ × (− ˆj ) = − kˆ. (b) Sˆ = ˆj × iˆ = −kˆ. (c) Sˆ = ( − kˆ ) × ( − iˆ) = ˆj .

(d) Sˆ = iˆ × ( − kˆ ) = ˆj. EVALUATE:

! ! In each case the directions of E , B and the direction of propagation are all mutually perpendicular.


Electromagnetic Waves

32.17.

Emax

IDENTIFY: SET UP:

c = 3.00 ×108 m/s . Emax = 4.00 V/m.

! ! ! ! Bmax = Emax c = 1.33 × 10−8 T . For E in the +x-direction, E × B is in the +z-direction when B is in

EXECUTE:

32.18.

32.19.

the +y-direction. ! ! EVALUATE: E , B and the direction of propagation are all mutually perpendicular. 2 2 IDENTIFY: The intensity of the electromagnetic wave is given by Eq.(32.29): I = 12 P0cEmax = P0cErms . The total energy passing through a window of area A during a time t is IAt. SET UP: P0 = 8.85 × 10−12 F/m 2 EXECUTE: Energy = P0cErms At = (8.85 × 10 −12 F m)(3.00 × 108 m s)(0.0200 V m) 2 (0.500 m 2 )(30.0 s) = 15.9 μ J EVALUATE: The intensity is proportional to the square of the electric field amplitude. IDENTIFY and SET UP: Use Eq.(32.29) to calculate I, Eq.(32.18) to calculate Bmax , and use I = Pav / 4π r 2 to

calculate Pav . (a) EXECUTE:

2 I = 12 P0 Emax ; Emax = 0.090 V/m, so I = 1.1× 10−5 W/m 2

(b) Emax = cBmax so Bmax = Emax / c = 3.0 ×10−10 T

32.20.

(c) Pav = I (4π r 2 ) = (1.075 ×10−5 W/m 2 )(4π )(2.5 ×103 m) 2 = 840 W (d) EVALUATE: The calculation in part (c) assumes that the transmitter emits uniformly in all directions. 2 IDENTIFY and SET UP: I = Pav / A and I = P0cErms . (a) The average power from the beam is Pav = IA = (0.800 W m 2 )(3.0 × 10−4 m 2 ) = 2.4 × 10−4 W .

EXECUTE:

0.800 W m 2 I = =17.4 V m −12 (8.85 × 10 F m)(3.00 × 108 m s) P0c

(b) Erms = 32.21:

32.22.

EVALUATE: The laser emits radiation only in the direction of the beam. IDENTIFY: I = Pav / A SET UP: At a distance r from the star, the radiation from the star is spread over a spherical surface of area A = 4π r 2 . EXECUTE: Pav = I (4π r 2 ) = (5.0 ×103 W m 2 )(4π )(2.0 ×1010 m) 2 = 2.5 ×1025 J EVALUATE: The intensity decreases with distance from the star as 1/ r 2 . IDENTIFY and SET UP: c = f λ , Emax = cBmax and I = Emax Bmax / 2μ 0 (a) f =

EXECUTE: (b) Bmax =

EVALUATE:

3.00 ×108 m s = 8.47 ×108 Hz. 0.354 m

Emax Bmax (0.0540 V m)(1.80 × 10−10 T) = = 3.87 × 10−6 W m 2 . 2 μ0 2μ0 2 Alternatively, I = 12 P0cEmax .

The surface area of a sphere is A = 4π r 2 .

⎛ E2 ⎞ Pav = Sav A = ⎜ max ⎟ (4π r 2 ) . Emax = ⎝ 2cμ 0 ⎠ E 12.0 V m = max = = 4.00 × 10−8 T. c 3.00 × 108 m s

EXECUTE:

Bmax

λ

=

2 Pav = IA and I = 12 P0cEmax

IDENTIFY: SET UP:

c

0.0540 V m Emax = = 1.80 ×10−10 T. 3.00 ×108 m s c

(c) I = Sav =

32.23.

Pavcμ 0 (60.0 W)(3.00 ×108 m s) μ 0 = = 12.0 V m. 2 2π r 2π (5.00 m) 2

Emax and Bmax are both inversely proportional to the distance from the source. ! ! ! IDENTIFY: The Poynting vector is S = E × B. SET UP: The electric field is in the +y-direction, and the magnetic field is in the +z-direction. cos 2 φ = 12 (1 + cos 2φ ) EVALUATE:

32.24.

32-5

! ! = cBmax . E × B is in the direction of propagation.

EXECUTE: (a) Sˆ = Eˆ × Bˆ = (− ˆj ) × kˆ = − iˆ. The Poynting vector is in the –x-direction, which is the direction of propagation of the wave.


32-6

Chapter 32

(b) S ( x, t ) =

E ( x, t ) B ( x, t )

μ0

=

Emax Bmax

μ0

cos 2 (kx + ωt ) =

Emax Bmax (1 + cos(2(ωt + kx)) ). But over one period, the 2μ0

Emax Bmax . This is Eq.(32.29). 2μ0 EVALUATE: We can also show that these two results also apply to the wave represented by Eq.(32.17). IDENTIFY: Use the radiation pressure to find the intensity, and then Pav = I (4π r 2 ). cosine function averages to zero, so we have Sav =

32.25.

SET UP:

I c = 2.70 × 103 W/m 2 . Then

For a perfectly absorbing surface, prad = prad = I c so I = cprad

EXECUTE:

Pav = I (4π r ) = (2.70 ×103 W m 2 )(4π )(5.0 m) 2 = 8.5 ×105 W. EVALUATE: Even though the source is very intense the radiation pressure 5.0 m from the surface is very small. IDENTIFY: The intensity and the energy density of an electromagnetic wave depends on the amplitudes of the electric and magnetic fields. 2 SET UP: Intensity is I = Pav / A , and the average power is Pav = 2I / c, where I = 12 P0cEmax . The energy density is 2

32.26.

u = P0 E 2 . EXECUTE:

(a) I = Pav /A =

316,000 W 2(0.00201 W/m 2 ) = 1.34 × 10−11 Pa = 0.00201 W/m2. Pav = 2I / c = 2 3.00 ×108 m/s 2π (5000 m)

2 (b) I = 12 P0cEmax gives

Emax =

2(0.00201 W/m 2 ) = 1.23 N/C (8.85 ×10 −12 C 2 /N ⋅ m 2 )(3.00 ×108 m/s)

2I = P0c

Bmax = Emax /c = (1.23 N/C)/(3.00 × 108 m/s) = 4.10 × 10−9 T (c) u = P0 E 2 , so uav = P0 ( Eav ) 2 and Eav =

Emax , so 2

2 (8.85 ×10−12 C2 /N ⋅ m2 ) (1.23 N/C)2 = 6.69 × 10−12 J/m3 P0 Emax = 2 2 (d) As was shown in Section 32.4, the energy density is the same for the electric and magnetic fields, so each one has 50% of the energy density. EVALUATE: Compared to most laboratory fields, the electric and magnetic fields in ordinary radiowaves are extremely weak and carry very little energy. IDENTIFY and SET UP: Use Eqs.(32.30) and (32.31). dp Sav I = = EXECUTE: (a) By Eq.(32.30) the average momentum density is dV c 2 c 2 dp 0.78 × 103 W/m 2 = = 8.7 × 10−15 kg/m 2 ⋅ s dV (2.998 × 108 m/s) 2

uav =

32.27.

Sav I 0.78 ×103 W/m 2 = = = 2.6 × 10−6 Pa c c 2.998 × 108 m/s The radiation pressure that the sunlight would exert on an absorbing or reflecting surface is very

(b) By Eq.(32.31) the average momentum flow rate per unit area is

32.28.

EVALUATE: small. IDENTIFY: Apply Eqs.(32.32) and (32.33). The average momentum density is given by Eq.(32.30), with S replaced by Sav = I . SET UP: EXECUTE:

prad =

1 atm = 1.013 ×105 Pa (a) Absorbed light: prad =

8.33 × 10−6 Pa = 8.23 × 10−11 atm. 1.013 × 105 Pa atm

(b) Reflecting light: prad = prad =

I 2500 W m 2 = = 8.33 ×10−6 Pa. Then c 3.0 ×108 m s

2 I 2(2500 W m 2 ) = = 1.67 ×10−5 Pa. Then 3.0 ×108 m s c

1.67 × 10−5 Pa = 1.65 × 10−10 atm. 1.013 × 105 Pa atm


Electromagnetic Waves

(c) The momentum density is

32.29.

2500 W m 2 dp Sav = 2 = = 2.78 ×10−14 kg m 2 ⋅ s. dV c (3.0 ×108 m s) 2

EVALUATE: The factor of 2 in prad for the reflecting surface arises because the momentum vector totally reverses direction upon reflection. Thus the change in momentum is twice the original momentum. IDENTIFY: Apply Eq.(32.4) and (32.9). SET UP: Eq.(32.26) is S = P0cE 2 . EXECUTE:

S=

P0 E2 = P0 μ0

P0

μ0

E2 =

P0

μ0

Ec

P E 1 = c 0 EB = μ0 c P0 μ0

32.32.

32.33.

μ0

EB =

EB

μ0

=

E2 = P0cE 2 μ 0c

λ

3

2

3.00 × 108 m s c = = 0.200 m = 20.0 cm. There must be nodes at the planes, which 2 2 f 2(7.50 ×108 Hz) are 80.0 cm apart, and there are two nodes between the planes, each 20.0 cm from a plane. It is at 20 cm, 40 cm, and 60 cm from one plane that a point charge will remain at rest, since the electric fields there are zero. EVALUATE: The magnetic field amplitude at these points isn’t zero, but the magnetic field doesn’t exert a force on a stationary charge. IDENTIFY and SET UP: Apply Eqs.(32.36) and (32.37). ! EXECUTE: (a) By Eq.(32.37) we see that the nodal planes of the B field are a distance λ / 2 apart, so λ / 2 = 3.55 mm and λ = 7.10 mm. ! (b) By Eq.(32.36) we see that the nodal planes of the E field are also a distance λ / 2 = 3.55 mm apart. (c) v = f λ = (2.20 ×1010 Hz)(7.10 ×10−3 m) = 1.56 ×108 m/s. ! ! EVALUATE: The spacing between the nodes of E is the same as the spacing between the nodes of B. Note that v < c, as it must. ! ! IDENTIFY: The nodal planes of E and B are located by Eqs.(32.26) and (32.27). c 3.00 ×108 m/s SET UP: λ = = = 4.00 m f 75.0 ×106 Hz EXECUTE:

32.31.

P0

EVALUATE: We can also write S = P0c(cB) = P0c B . S can be written solely in terms of E or solely in terms of B. IDENTIFY: The electric field at the nodes is zero, so there is no force on a point charge placed at a node. SET UP: The location of the nodes is given by Eq.(32.36), where x is the distance from one of the planes. λ = c/ f . 2

32.30.

32-7

Δxnodes =

EXECUTE:

(a) Δx =

EXECUTE:

λ=

=

λ

= 2.00 m. 2 (b) The distance between the electric and magnetic nodal planes is one-quarter of a wavelength, so is λ Δx 2.00 m = = =1.00 m. 4 2 2 ! ! EVALUATE: The nodal planes of B are separated by a distance λ / 2 and are midway between the nodal planes of E . ! (a) IDENTIFY and SET UP: The distance between adjacent nodal planes of B is λ / 2. There is an antinodal plane ! of B midway between any two adjacent nodal planes, so the distance between a nodal plane and an adjacent antinodal plane is λ / 4. Use v = f λ to calculate λ .

λ

v 2.10 × 108 m/s = = 0.0175 m f 1.20 ×1010 Hz

0.0175 m = 4.38 ×10−3 m = 4.38 mm 4 4 ! ! (b) IDENTIFY and SET UP: The nodal planes of E are at x = 0, λ / 2, λ , 3λ /2, . . . , so the antinodal planes of E ! are at x = λ / 4, 3λ /4, 5λ /4, . . . . The nodal planes of B are at x = λ / 4, 3λ / 4, 5λ /4, . . . , so the antinodal planes ! of B are at λ / 2, λ , 3λ /2, . . . . ! ! EXECUTE: The distance between adjacent antinodal planes of E and antinodal planes of B is therefore λ / 4 = 4.38 mm. ! ! (c) From Eqs.(32.36) and (32.37) the distance between adjacent nodal planes of E and B is λ / 4 = 4.38 mm. ! ! ! EVALUATE: The nodes of E coincide with the antinodes of B , and conversely. The nodes of B and the nodes ! of E are equally spaced. =


32-8

Chapter 32

32.34.

IDENTIFY:

Evaluate the derivatives of the expressions for E y ( x, t ) and Bz ( x, t ) that are given in Eqs.(32.34) and

(32.35). ∂ ∂ ∂ ∂ sin kx = k cos kx , sin ω t = ω cos ω t . cos kx = −k sin kx , cos ω t = −ω sin ω t . ∂t ∂t ∂x ∂x ∂ 2 E y ( x, t ) ∂ 2 ∂ = 2 ( −2 Emax sin kx sin ωt ) = (−2kEmax cos kx sin ωt ) and EXECUTE: (a) ∂x 2 ∂x ∂x 2 ∂ 2 E y ( x, t ) ∂ 2 E y ( x, t ) ω 2 2 k E sin kx sin t 2 E sin kx sin t . P ω ω μ = = = max max 0 0 ∂x 2 ∂t 2 c2 ∂ 2 Bz ( x, t ) ∂ 2 ∂ = 2 (−2 Bmax cos kx cos ωt ) = (+2kBmax sin kx cos ωt ) and Similarly: ∂x 2 ∂x ∂x ∂ 2 Bz ( x, t ) ω2 ∂ 2 Bz ( x, t ) = 2k 2 Bmax cos kx cos ωt = 2 2 Bmax cos kx cos ωt = P0 μ0 . 2 ∂x ∂t 2 c ∂E y ( x, t ) ∂ = ( −2 Emax sin kx sin ωt ) = −2kEmax cos kx sin ωt . (b) ∂x ∂x ∂E y ( x, t ) ω E = − 2 Emax cos kx sin ωt = −ω 2 max cos kx sin ωt = −ω 2 Bmax cos kx sin ωt . ∂x c c ∂E y ( x, t ) ∂ ∂Bz ( x, t ) = + (2 Bmax cos kx cos ωt ) = − . ∂x ∂t ∂t ∂B ( x, t ) ∂ = (+2 Bmax cos kx cos ωt ) = −2kBmax sin kx cos ωt . Similarly: − z ∂x ∂x ∂Bz ( x, t ) ω ω − = − 2 Bmax sin kx cos ωt = − 2 2cBmax sin kx cos ωt . c c ∂x ∂E ( x, t ) ∂B ( x, t ) ∂ − z = −P0 μ0ω 2 Emax sin kx cos ωt = P0 μ0 (−2 Emax sin kx sin ωt ) = P0 μ0 y . ∂x ∂t ∂t EVALUATE: The standing waves are linear superpositions of two traveling waves of the same k and ω . IDENTIFY: The nodal and antinodal planes are each spaced one-half wavelength apart. SET UP: 2 12 wavelengths fit in the oven, so ( 2 12 ) λ = L, and the frequency of these waves obeys the equation fλ = c.

SET UP:

32.35.

EXECUTE:

(a) Since ( 2 12 ) λ = L, we have L = (5/2)(12.2 cm) = 30.5 cm.

(b) Solving for the frequency gives f = c/λ = (3.00 × 108 m/s)/(0.122 m) = 2.46 × 109 Hz. (c) L = 35.5 cm in this case. ( 2 12 ) λ = L, so λ = 2L/5 = 2(35.5 cm)/5 = 14.2 cm.

32.36.

f = c/λ = (3.00 × 108 m/s)/(0.142 m) = 2.11 × 109 Hz EVALUATE: Since microwaves have a reasonably large wavelength, microwave ovens can have a convenient size for household kitchens. Ovens using radiowaves would need to be far too large, while ovens using visible light would have to be microscopic. IDENTIFY: Evaluate the partial derivatives of the expressions for E y ( x, t ) and Bz ( x, t ) . SET UP:

∂ ∂ ∂ sin( kx − ω t ) = k cos(kx − ω t ) , sin(kx − ω t ) = −ω cos(kx − ω t ) . cos(kx − ω t ) = −k sin( kx − ω t ) , ∂x ∂t ∂x

∂ cos( kx − ω t ) = ω sin( kx − ω t ) ∂t ! ! EXECUTE: Assume E = Emax ˆj sin( kx − ω t ) and B = Bmax kˆ sin( kx − ω t + φ ), with − π < φ < π . Eq. (32.12) is

∂E y

∂Bz . This gives kEmax cos(kx − ω t ) = +ω Bmax cos( kx − ω t + φ ) , so φ = 0 , and kEmax = ω Bmax , so ∂x ∂t ∂E y ∂B ω 2π f gives Emax = Bmax = Bmax = fω Bmax = cBmax . Similarly for Eq.(32.14), − z = P0 μ0 k 2π / ω ∂x ∂t − kBmax cos( kx − ωt + φ ) = −P0 μ0ω Emax cos( kx − ωt ) , so φ = 0 and kBmax = P0 μ0ω Emax , so =−

P0 μ0ω 2π f 1 fω Emax = 2 Emax = 2 Emax = Emax . k c 2π / ω c c ! ! EVALUATE: The E and B fields must oscillate in phase.

Bmax =


Electromagnetic Waves

32.37.

32-9

IDENTIFY and SET UP:

Take partial derivatives of Eqs.(32.12) and (32.14), as specified in the problem. ∂E y ∂B =− z EXECUTE: Eq.(32.12): ∂x ∂t ∂2Ey ∂E ∂2B ∂B ∂ ∂ = − 2 z . Eq.(32.14) says − z = P0 μ 0 y . Taking Taking of both sides of this equation gives of ∂x∂t ∂t ∂x ∂t ∂x ∂t ∂2Ey ∂2Ey ∂ 2 Ey ∂ 2 Ey ∂2B 1 ∂ 2 Bz =− = both sides of this equation gives − 2z = P0 μ 0 . But (The order in , so P0 μ 0 ∂x 2 ∂x ∂t ∂x ∂t ∂x ∂x∂t ∂t∂x which the partial derivatives are taken doesn't change the result.) So −

∂ 2 Bz 1 ∂ 2 Bz ∂ 2 Bz ∂2B and = P0 μ 0 2 z , =− 2 2 2 ∂t P0 μ 0 ∂x ∂x ∂t

as was to be shown. EVALUATE: Both fields, electric and magnetic, satisfy the wave equation, Eq.(32.10). We have also shown that both fields propagate with the same speed v = 1/ P0 μ 0 . 32.38.

The average energy density in the electric field is u E ,av = 12 P0 ( E 2 )av and the average energy density in

IDENTIFY:

the magnetic field is u B ,av = SET UP:

( cos (kx − ωt ) )

EXECUTE:

2

av

1 ( B 2 )av . 2μ0

= 12 .

2 2 E y ( x, t ) = Emax cos(kx − ωt ) . u E = 12 P0 E y2 = 12 P0 Emax . cos 2 (kx − ωt ) and u E ,av = 14 P0 Emax

Bz ( x, t ) = Bmax cos( kx − ωt ) , so u B = 2 u E ,av = 14 P0c 2 Bmax . c=

1 1 2 , so u E ,av = Bmax , which equals u B ,av . 2μ0 P0 μ0

1 2 Bmax . 2μ 0 IDENTIFY: The intensity of an electromagnetic wave depends on the amplitude of the electric and magnetic fields. Such a wave exerts a force because it carries energy. 2 SET UP: The intensity of the wave is I = Pav / A = 12 P0cEmax , and the force is F = Pav A where Pav = I / c . 2 EXECUTE: (a) I = Pav /A = (25,000 W)/[4π(575 m) ] = 0.00602 W/m2 EVALUATE:

32.39.

1 2 1 2 1 2 Bz = Bmax cos 2 (kx − ω t ) and u B ,av = Bmax . Emax = cBmax , so 4μ0 2μ 0 2μ0

2 Our result allows us to write uav = 2u E ,av = 12 P0 Emax and uav = 2u B ,av =

2 (b) I = 12 P0cEmax , so Emax =

2I = P0c

2(0.00602 W/m 2 ) = 2.13 N/C. (8.85 ×10−12 C2 /N ⋅ m 2 )(3.00 ×108 m/s)

Bmax = Emax/c = (2.13 N/C)/(3.00 × 108 m/s) = 7.10 × 10−9 T

32.40.

(c) F =Pav A = ( I / c) A = (0.00602 W/m2)(0.150 m)(0.400 m)/(3.00 × 108 m/s) = 1.20 × 10−12 N EVALUATE: The fields are very weak compared to ordinary laboratory fields, and the force is hardly worth worrying about! I 2 IDENTIFY: c = f λ . Emax = cBmax . I = 12 P0cEmax . For a totally absorbing surface the radiation pressure is . c SET UP: The wave speed in air is c = 3.00 ×108 m/s . c 3.00 ×108 m/s EXECUTE: (a) f = = = 7.81×109 Hz λ 3.84 ×10−2 m E 1.35 V/m (b) Bmax = max = = 4.50 ×10−9 T c 3.00 ×108 m/s 2 (c) I = 12 P0cEmax = 12 (8.854 × 10−12 C2 /N ⋅ m 2 )(3.00 × 108 m/s)(1.35 V/m) 2 = 2.42 × 10−3 W/m 2

IA (2.42 × 10−3 W/m 2 )(0.240 m 2 ) = = 1.94 ×10−12 N 3.00 ×108 m/s c EVALUATE: The intensity depends only on the amplitudes of the electric and magnetic fields and is independent of the wavelength of the light. (d) F = (pressure)A =


32-10

Chapter 32

32.41.

(a) IDENTIFY and SET UP: EXECUTE: I=

Calculate I and then use Eq.(32.29) to calculate Emax and Eq.(32.18) to calculate Bmax .

The intensity is power per unit area: I =

3.20 ×10−3 W P = = 652 W/m 2 . A π (1.25 ×10−3 m) 2

2 Emax , so Emax = 2 μ 0cI 2μ 0 c

Emax = 2(4π ×10−7 T ⋅ m/A)(2.998 ×108 m/s)(652 W/m 2 ) = 701 V/m 701 V/m Emax = = 2.34 ×10−6 T 2.998 ×108 m/s c EVALUATE: The magnetic field amplitude is quite small. (b) IDENTIFY and SET UP: Eqs.(24.11) and (30.10) give the energy density in terms of the electric and magnetic field values at any time. For sinusoidal fields average over E 2 and B 2 to get the average energy densities. EXECUTE: The energy density in the electric field is uE = 12 P0 E 2 . E = Emax cos( kx − ω t ) and the average value of Bmax =

cos 2 (kx − ω t ) is 12 . The average energy density in the electric field then is 2 u E ,av = 14 P0 Emax = 14 (8.854 ×10−12 C2 / N ⋅ m 2 )(701 V/m) 2 = 1.09 ×10−6 J/m3 . The energy density in the magnetic field

B2 B2 (2.34 ×10−6 T) 2 . The average value is u B ,av = max = = 1.09 ×10−6 J/m 3 . 2μ 0 4μ 0 4(4π ×10−7 T ⋅ m/A) EVALUATE: Our result agrees with the statement in Section 32.4 that the average energy density for the electric field is the same as the average energy density for the magnetic field. (c) IDENTIFY and SET UP: The total energy in this length of beam is the total energy density uav = u E ,av + u B ,av = 2.18 × 10−6 J/m3 times the volume of this part of the beam.

is u B =

32.42.

EXECUTE: U = uav LA = (2.18 ×10−6 J/m3 )(1.00 m)π (1.25 ×10−3 m) 2 = 1.07 ×10−11 J. EVALUATE: This quantity can also be calculated as the power output times the time it takes the light to travel L = 1.00 m ⎛ L⎞ ⎛ ⎞ −11 1.00 m: U = P ⎜ ⎟ = (3.20 ×10−3 W) ⎜ ⎟ = 1.07 ×10 J, which checks. 8 × ⎝c⎠ ⎝ 2.998 10 m/s ⎠ IDENTIFY: Use the gaussian surface specified in the hint. ! ! SET UP: The wave is in free space, so in Gauss’s law for the electric field, Qencl = 0 and AE ⋅ dA = 0. Gauss’s law ! ! for the magnetic field says AB ⋅ dA = 0 EXECUTE: Use a gaussian surface such that the front surface is ahead of the wave front (no electric or magnetic ! ! Q fields) and the back face is behind the wave front, as shown in Figure 32.42. AE ⋅ dA = Ex A = encl = 0 , so Ex = 0.

ε0

! ! AB ⋅ dA = Bx A = 0 and Bx = 0. EVALUATE: The wave must be transverse, since there are no components of the electric or magnetic field in the direction of propagation.

Figure 32.42 32.43.

I c SET UP: Assume the electromagnetic waves are formed at the center of the sun, so at a distance r from the center of the sun I = Pav /(4π r 2 ).

IDENTIFY:

I = Pav / A . For an absorbing surface, the radiation pressure is prad =


Electromagnetic Waves

EXECUTE:

(a) At the sun’s surface: I =

32-11

3.9 ×1026 W Pav = = 6.4 ×107 W m 2 and 2 4π R 4π (6.96 ×108 m) 2

I 6.4 ×107 W m 2 = = 0.21 Pa. c 3.00 ×108 m s (b) Halfway out from the sun’s center, the intensity is 4 times more intense, and so is the radiation pressure: I = 2.6 × 108 W/m 2 and prad = 0.85 Pa. At the top of the earth’s atmosphere, the measured sunlight intensity is prad =

32.44.

1400 W m 2 = 5 × 10−6 Pa , which is about 100,000 times less than the values above. EVALUATE: (b) The gas pressure at the sun’s surface is 50,000 times greater than the radiation pressure, and halfway out of the sun the gas pressure is believed to be about 6 × 1013 times greater than the radiation pressure. Therefore it is reasonable to ignore radiation pressure when modeling the sun’s interior structure. P 2 IDENTIFY: I = . I = 12 P0cEmax . A SET UP: 3.00 ×108 m/s P 2.80 ×103 W EXECUTE: I = = = 77.8 W/m 2 . 36.0 m 2 A Emax =

32.45.

2I 2(77.8 W/m 2 ) = = 242 N/C . −12 (8.854 × 10 C 2 /N ⋅ m 2 )(3.00 × 108 m/s) P0c

EVALUATE: This value of Emax is similar to the electric field amplitude in ordinary light sources. IDENTIFY: The same intensity light falls on both reflectors, but the force on the reflecting surface will be twice as great as the force on the absorbing surface. Therefore there will be a net torque about the rotation axis. SET UP: For a totally absorbing surface, F = Pav A = ( I/c) A, while for a totally reflecting surface the force will be 2 twice as great. The intensity of the wave is I = 12 P0cEmax . Once we have the torque, we can use the rotational form

of Newton’s second law, τnet = Iα, to find the angular acceleration. 2 P cEmax A 1 2 = 2 P0 AEmax c 2 For a totally reflecting surface, the force will be twice as great, which is P0cEmax . The net torque is therefore

EXECUTE:

The force on the absorbing reflector is FAbs = pav A = ( I/c) A =

1 2 0

2 τ net = FRefl ( L/2) − FAbs ( L/2) = P0 AEmax L/4 2 Newton’s 2nd law for rotation gives τ net = Iα . P0 AEmax L/4 = 2m( L/2) 2 α 2 Solving for α gives α = P0 AEmax /(2mL) =

32.46.

(8.85 ×10

−12

C 2 /N ⋅ m 2 ) (0.0150 m) 2 (1.25 N/C) 2

= 3.89 × 10−13 rad/s 2 (2)(0.00400 kg)(1.00 m) EVALUATE: This is an extremely small angular acceleration. To achieve a larger value, we would have to greatly increase the intensity of the light wave or decrease the mass of the reflectors. IDENTIFY: For light of intensity I abs incident on a totally absorbing surface, the radiation pressure is I abs 2I . For light of intensity I refl incident on a totally reflecting surface, prad,refl = refl . c c SET UP: The total radiation pressure is prad = prad,abs + prad,refl . I abs = wI and I refl = (1 − w) I prad,abs =

I abs 2 I refl wI 2(1 − w) I (2 − w) I . + = + = c c c c c I 2I (b) (i) For totally absorbing w = 1 so prad = . (ii) For totally reflecting w = 0 so prad = . These are just equations c c 32.32 and 32.33. (2 − 0.9)(1.40 × 103 W m 2 ) = 5.13 × 10−6 Pa. For w = 0.1 and (c) For w = 0.9 and I = 1.40 × 102 W/m 2 , prad = 3.00 × 108 m/s (2 − 0.1)(1.40 × 102 W m 2 ) = 8.87 × 10−6 Pa. I = 1.40 × 103 W m 2 , prad = 3.00 × 108 m/s EVALUATE: The radiation pressure is greater when a larger fraction is reflected. IDENTIFY and SET UP: In the wire the electric field is related to the current density by Eq.(25.7). Use Ampere’s ! law to calculate B. The Poynting vector is given by Eq.(32.28) and the equation that follows it relates the energy ! flow through a surface to S .

EXECUTE:

32.47.

(a) prad = prad,abs + prad,refl =


32-12

Chapter 32

! EXECUTE: (a) The direction of E is parallel to the axis of the cylinder, in the direction of the current. From Eq.(25.7), E = ρ J = ρ I / π a 2 . (E is uniform across the cross section of the conductor.) (b) A cross-sectional view of the conductor is given in Figure 32.47a; take the current to be coming out of the page. Apply Ampere’s law to a circle of radius a. ! ! AB ⋅ dl = B(2π a ) I encl = I Figure 32.47a

! ! μ0I AB ⋅ dl = μ 0 I encl gives B (2π a) = μ 0 I and B = 2π a ! The direction of B is counterclockwise around the circle. ! ! (c) The directions of E and B are shown in Figure 32.47b. ! 1 ! ! The direction of S = E×B

μ0

is radially inward. 1 1 ⎛ ρ I ⎞⎛ μ I ⎞ S= EB = ⎜ 2 ⎟⎜ 0 ⎟ μ0 μ 0 ⎝ π a ⎠⎝ 2π a ⎠ ρI 2 S= 2 3 2π a Figure 32.47b

Since S is constant over the surface of the conductor, the rate of energy flow P is given by S ρI 2 ρ lI 2 ρl . But R = 2 , so the times the surface of a length l of the conductor: P = SA = S (2π al ) = 2 3 (2π al ) = 2π a π a2 πa result from the Poynting vector is P = RI 2 . This agrees with PR = I 2 R, the rate at which electrical energy is being ! dissipated by the resistance of the wire. Since S is radially inward at the surface of the wire and has magnitude equal to the rate at which electrical energy is being dissipated in the wire, this energy can be thought of as entering through the cylindrical sides of the conductor. IDENTIFY: The intensity of the wave, not the electric field strength, obeys an inverse-square distance law. SET UP: The intensity is inversely proportional to the distance from the source, and it depends on the amplitude of the electric field by I = Sav = 12 P0 cEmax2. (d) EVALUATE:

32.48.

EXECUTE:

32.49.

Since I = 12 P0 cEmax2, Emax ∝ I . A point at 20.0 cm (0.200 m) from the source is 50 times closer to

the source than a point that is 10.0 m from it. Since I ∝ 1/r2 and (0.200 m)/(10.0 m) = 1/50, we have I0.20 = 502 I10. Since Emax ∝ I , we have E0.20 = 50E10 = (50)(1.50 N/C) = 75.0 N/C. EVALUATE: While the intensity increases by a factor of 502 = 2500, the amplitude of the wave only increases by a factor of 50. Recall that the intensity of any wave is proportional to the square of its amplitude. dΦB IDENTIFY and SET UP: The magnitude of the induced emf is given by Faraday’s law: E = . To calculate dt d Φ B / dt we need dB / dt at the antenna. Use the total power output to calculate I and then combine Eq.(32.29) and (32.18) to calculate Bmax . The time dependence of B is given by Eq.(32.17). Φ B = Bπ R 2 , where R = 0.0900 m is the radius of the loop. (This assumes that the magnetic field is dB uniform across the loop, an excellent approximation.) E = π R 2 dt dB B = Bmax cos(kx − ω t ) so = Bmaxω sin( kx − ω t ) dt

EXECUTE:

The maximum value of

dB is Bmaxω , so E max = π R 2 Bmaxω . dt

R = 0.0900 m, ω = 2π f = 2π (95.0 ×106 Hz) = 5.97 ×108 rad/s


Electromagnetic Waves

32-13

Calculate the intensity I at this distance from the source, and from that the magnetic field amplitude Bmax : I=

2 P 55.0 ×103 W Emax (cBmax ) 2 c 2 −4 2 = = = = = 7.00 × 10 W/m . I Bmax 4π r 2 4π (2.50 × 103 m) 2 2μ 0c 2μ 0c 2μ0

Thus Bmax =

2 μ0 I 2(4π × 10−7 T ⋅ m/A)(7.00 × 10−4 W/m 2 ) = = 2.42 × 10−9 T. Then c 2.998 × 108 m/s

E max = π R 2 Bmaxω = π (0.0900 m) 2 (2.42 ×10−9 T)(5.97 ×108 rad/s) = 0.0368 V.

32.50.

32.51.

EVALUATE: An induced emf of this magnitude is easily detected. IDENTIFY: The nodal planes are one-half wavelength apart. SET UP: The nodal planes of B are at x = λ/4, 3λ/4, 5λ/4, …, which are λ/2 apart. EXECUTE: (a) The wavelength is λ = c/ f = (3.000 × 108 m/s)/(110.0 × 106 Hz) = 2.727 m. So the nodal planes are at (2.727 m)/2 = 1.364 m apart. (b) For the nodal planes of E, we have λn = 2L/n, so L = nλ/2 = (8)(2.727 m)/2 = 10.91 m EVALUATE: Because radiowaves have long wavelengths, the distances involved are easily measurable using ordinary metersticks. IDENTIFY and SET UP: Find the force on you due to the momentum carried off by the light. Express this force in terms of the radiated power of the flashlight. Use this force to calculate your acceleration and use a constant acceleration equation to find the time. (a) EXECUTE: prad = I / c and F = prad A gives F = IA / c = Pav / c

ax = F / m = Pav /(mc) = (200 W)/[(150 kg)(3.00 × 108 m/s)] = 4.44 ×10−9 m/s 2 Then x − x0 = v0 xt + 12 axt 2 gives t = 2( x − x0 )/ax = 2(16.0 m)/(4.44 × 10−9 m/s 2 ) = 8.49 × 104 s = 23.6 h

32.52.

EVALUATE: The radiation force is very small. In the calculation we have ignored any other forces on you. (b) You could throw the flashlight in the direction away from the ship. By conservation of linear momentum you would move toward the ship with the same magnitude of momentum as you gave the flashlight. 2 IDENTIFY: Pav = IA and I = 12 P0cEmax . Emax = cBmax SET UP: The power carried by the current i is P = Vi . EXECUTE: Bmax =

Pav 1 = P0cE 2 and Emax = A 2

2 Pav = AP0c

2Vi 2(5.00 × 105 V)(1000 A) = = 6.14 × 104 V m. AP0c (100 m 2 )P0 (3.00 × 108 m s)

Emax 6.14 × 104 V m = = 2.05 × 10−4 T. c 3.00 × 108 m s

EVALUATE: 32.53.

I=

I = Vi / A =

(5.00 ×105 V)(1000 A) = 5.00 ×106 W/m 2 . This is a very intense beam spread over a 100 m 2

large area. IDENTIFY: The orbiting satellite obeys Newton’s second law of motion. The intensity of the electromagnetic waves it transmits obeys the inverse-square distance law, and the intensity of the waves depends on the amplitude of the electric and magnetic fields. SET UP: Newton’s second law applied to the satellite gives mv2/R = GmM/r2, where M is the mass of the Earth and m is the mass of the satellite. The intensity I of the wave is I = Sav = 12 P0 cEmax2, and by definition, I = Pav /A. EXECUTE: (a) The period of the orbit is 12 hr. Applying Newton’s 2nd law to the satellite gives mv2/R = GmM/r2, m ( 2π r / T ) GmM = . Solving for r, we get r r2 2

which gives

1/ 3 ⎡ ( 6.67 × 10−11 N ⋅ m 2 /kg 2 )( 5.97 × 1024 kg ) (12 × 3600 s )2 ⎤ ⎛ GMT 2 ⎞ ⎢ ⎥ = 2.66 × 107 m r =⎜ = ⎟ 4π 2 ⎝ 4π 2 ⎠ ⎣ ⎦ The height above the surface is h = 2.66 × 107 m – 6.38 × 106 m = 2.02 × 107 m. The satellite only radiates its energy to the lower hemisphere, so the area is 1/2 that of a sphere. Thus, from the definition of intensity, the intensity at the ground is 1/ 3

I = Pav /A = Pav /(2πh2) = (25.0 W)/[2π(2.02 × 107 m)2] = 9.75 × 10−15 W/m2 (b) I = Sav = 12 P0 cEmax2, so Emax =

2I 2(9.75 × 10−15 W/m 2 ) = = 2.71 × 10−6 N/C (8.85 × 10−12 C 2 /N ⋅ m 2 )(3.00 × 108 m/s) P0c

Bmax = Emax /c = (2.71 × 10−6 N/C)/(3.00 × 108 m/s) = 9.03 × 10−15 T t = d/c = (2.02 × 107 m)/(3.00 × 108 m/s) = 0.0673 s


32-14

32.54.

Chapter 32

(c) Pav = I / c = (9.75 × 10–15 W/m2)/(3.00 × 108 m/s) = 3.25 × 10 – 2 3 Pa (d) λ = c / f = (3.00 × 108 m/s)/(1575.42 × 106 Hz) = 0.190 m EVALUATE: The fields and pressures due to these waves are very small compared to typical laboratory quantities. 2I IDENTIFY: For a totally reflective surface the radiation pressure is . Find the force due to this pressure and c mM sun express the force in terms of the power output P of the sun. The gravitational force of the sun is Fg = G . r2 SET UP: The mass of the sum is M sun = 1.99 × 1030 kg. G = 6.67 ×10−11 N ⋅ m 2 /kg 2 . EXECUTE: (a) The sail should be reflective, to produce the maximum radiation pressure. P ⎛ 2I ⎞ (b) Frad = ⎜ ⎟ A , where A is the area of the sail. I = , where r is the distance of the sail from the sun. 4π r 2 ⎝ c ⎠

PA PA mM sun ⎛ 2 A ⎞⎛ P ⎞ = Frad = ⎜ . Frad = Fg so . =G ⎟⎜ 2 ⎟ 2 2π r 2c r2 ⎝ c ⎠⎝ 4π r ⎠ 2π r c 2π cGmM sun 2π (3.00 ×108 m/s)(6.67 × 10−11 N ⋅ m 2 /kg 2 )(10,000 kg)(1.99 × 1030 kg) = . P 3.9 ×1026 W A = 6.42 × 106 m 2 = 6.42 km 2 . (c) Both the gravitational force and the radiation pressure are inversely proportional to the square of the distance from the sun, so this distance divides out when we set Frad = Fg . A=

32.55.

EVALUATE: A very large sail is needed, just to overcome the gravitational pull of the sun. IDENTIFY and SET UP: The gravitational force is given by Eq.(12.2). Express the mass of the particle in terms of its density and volume. The radiation pressure is given by Eq.(32.32); relate the power output L of the sun to the intensity at a distance r. The radiation force is the pressure times the cross sectional area of the particle. mM EXECUTE: (a) The gravitational force is Fg = G 2 . The mass of the dust particle is m = ρV = ρ 43 π R 3 . Thus r 4 ρ Gπ MR 3 Fg = . 3r 2 I (b) For a totally absorbing surface prad = . If L is the power output of the sun, the intensity of the solar radiation c L L a distance r from the sun is I = . Thus prad = . The force Frad that corresponds to prad is in the 2 4π cr 2 4π r direction of propagation of the radiation, so Frad = prad A⊥ , where A⊥ = π R 2 is the component of area of the particle

LR 2 ⎛ L ⎞ perpendicular to the radiation direction. Thus Frad = ⎜ (π R 2 ) = . 2 ⎟ 4cr 2 ⎝ 4π cr ⎠ (c) Fg = Frad 4 ρ Gπ MR 3 LR 2 = 3r 2 4cr 2 L 3L ⎛ 4 ρ Gπ M ⎞ and R = ⎜ ⎟R = 3 4 c 16 c ρ Gπ M ⎝ ⎠ R=

3(3.9 ×1026 W) 16(2.998 ×10 m/s)(3000 kg/m3 )(6.673 ×10−11 N ⋅ m 2 / kg 2 )π (1.99 × 1030 kg) 8

R = 1.9 ×10−7 m = 0.19 μ m. EVALUATE: The gravitational force and the radiation force both have a r −2 dependence on the distance from the sun, so this distance divides out in the calculation of R. ⎛ LR 2 ⎞⎛ ⎞ F 3r 2 3L . Frad is proportional to R 2 and Fg is proportional to R 3 , so this (d) rad = ⎜ ⎟= 2 ⎟⎜ Fg ⎝ 4cr ⎠⎝ 4 ρ Gπ mR 3 ⎠ 16c ρ Gπ MR

ratio is proportional to 1/R. If R < 0.20 μ m then Frad > Fg and the radiation force will drive the particles out of the solar system. 32.56.

IDENTIFY: SET UP:

v2 . R C . An electron has q = e = 1.60 ×10−19 C .

The electron has acceleration a = 1 eV = 1.60 ×10−19


Electromagnetic Waves

32.57.

32-15

EXECUTE: For the electron in the classical hydrogen atom, its acceleration is v 2 1 mv 2 2(13.6 eV)(1.60 ×10−19 J eV ) a = = 21 = = 9.03 ×1022 m/s 2 . Then using the formula for the rate of energy R 2 mR (9.11× 10−31 kg)(5.29 ×10−11 m) emission given in Problem 32.57: dE q 2a 2 (1.60 × 10−19 C) 2 (9.03 × 1022 m s 2 ) 2 dE = = = 4.64 × 10−8 J s = 2.89 × 1011 eV s. This large value of 3 dt dt 6π P0c 6π P0 (3.00 × 108 m s)3 would mean that the electron would almost immediately lose all its energy! EVALUATE: The classical physics result in Problem 32.57 must not apply to electrons in atoms. v2 IDENTIFY: The orbiting particle has acceleration a = . R SET UP: K = 12 mv 2 . An electron has mass me = 9.11×10−31 kg and a proton has mass mp = 1.67 × 10−27 kg . 2 ⎡ q 2a 2 ⎤ C2 ( m s )2 N⋅m J ⎡ dE ⎤ (a) ⎢ = = = = W = ⎢ ⎥. 3⎥ 2 2 3 ⋅ π 6 c (C N m )( m s) s s P ⎣ dt ⎦ 0 ⎣ ⎦ (b) For a proton moving in a circle, the acceleration is v 2 1 mv 2 2(6.00 ×106 eV) (1.6 × 10−19 J eV ) 2 a = = 21 = = 1.53 × 1015 m s . The rate at which it emits energy because of R 2 mR (1.67 × 10−27 kg) (0.75 m) dE q 2a 2 (1.6 × 10 −19 C) 2 (1.53 × 1015 m s 2 ) 2 its acceleration is = = = 1.33 × 10 −23 J s = 8.32 × 10−5 eV s. dt 6π P0c 3 6π P0 (3.0 × 108 m s)3 (dE dt )(1 s) 8.32 ×10−5 eV Therefore, the fraction of its energy that it radiates every second is = = 1.39 ×10−11. E 6.00 ×106 eV (c) Carry out the same calculations as in part (b), but now for an electron at the same speed and radius. That means the electron’s acceleration is the same as the proton, and thus so is the rate at which it emits energy, since they also have the same charge. However, the electron’s initial energy differs from the proton’s by the ratio of their masses: m (9.11×10−31 kg) Ee = Ep e = (6.00 ×106 eV) = 3273 eV. Therefore, the fraction of its energy that it radiates every mp (1.67 ×10−27 kg)

EXECUTE:

second is

( dE dt )(1 s) 8.32 ×10 −5 eV = = 2.54 × 10−8. E 3273 eV

EVALUATE:

32.58.

The proton has speed v =

2E 2(6.0 ×106 eV)(1.60 × 10−19 J/eV) = = 3.39 ×107 m/s . The electron mp 1.67 ×10−27 kg

has the same speed and kinetic energy 3.27 keV. The particles in the accelerator radiate at a much smaller rate than the electron in Problem 32.56 does, because in the accelerator the orbit radius is very much larger than in the atom, so the acceleration is much less. IDENTIFY and SET UP: Follow the steps specified in the problem. EXECUTE: (a) E y ( x, t ) = Emax e − kC x sin ( kC x − ω t ). ∂E y

= Emax (− kC )e − kC x sin( kC x − ω t ) + Emax ( + kC )e − kC x cos(kC x − ω t ) ∂x ∂ 2 Ey = Emax (+ kC2 )e − kC x sin(kC x − ωt ) + Emax (− kC2 )e − kC x cos(kC x − ωt ) ∂x 2 + Emax (− kC2 )e − kC x cos(kC x − ω t ) + Emax (− kC2 )e − kC x sin(kC x − ω t ) .

∂2Ey ∂x 2

= −2 Emax kC2e − kC x cos(kC x − ω t ) .

Setting 2kC2

ω

=

∂ 2 Ey ∂x

2

=

∂E y ∂t

= Emax e − kC x ω cos(kC x − ω t ).

μ∂E y gives 2 Emax kC2e − k x cos(kC x − ω t ) = Emax e − k x ω cos(kC x − ω t ) . This will only be true if ρ∂t C

C

μ ωμ , or kC = . 2ρ ρ

(b) The energy in the wave is dissipated by the i 2 R heating of the conductor. E 1 2ρ 2(1.72 ×10−8 Ω ⋅ m) (c) E y = y 0 ⇒ kC x = 1, x = = = = 6.60 ×10−5 m. e kC ωμ 2π (1.0 ×106 Hz)μ 0 EVALUATE: The lower the frequency of the waves, the greater is the distance they can penetrate into a conductor. A dielectric (insulator) has a much larger resistivity and these waves can penetrate a greater distance in these materials.


THE NATURE AND PROPAGATION OF LIGHT

33.1.

33

IDENTIFY: For reflection, θ r = θ a . SET UP: The desired path of the ray is sketched in Figure 33.1. 14.0 cm EXECUTE: tan φ = , so φ = 50.6° . θ r = 90° − φ = 39.4° and θ r = θ a = 39.4° . 11.5 cm EVALUATE: The angle of incidence is measured from the normal to the surface.

Figure 33.1 33.2

IDENTIFY: For reflection, θ r = θ a . SET UP: The angles of incidence and reflection at each reflection are shown in Figure 33.2. For the rays to be perpendicular when they cross, α = 90° . α EXECUTE: (a) θ + φ = 90° and β + φ = 90° , so β = θ . + β = 90° and α = 180° − 2θ . 2 (b) θ = 12 (180° − α ) = 12 (180° − 90°) = 45° . EVALUATE: As θ → 0° , α → 180° . This corresponds to the incident and reflected rays traveling in nearly the same direction. As θ → 90° , α → 0° . This corresponds to the incident and reflected rays traveling in nearly opposite directions.

Figure 33.2 33-1


33-2

33.3.

33.4.

33.5.

33.6.

Chapter 33

IDENTIFY and SET UP: Use Eqs.(33.1) and (33.5) to calculate v and λ. c c 2.998 × 108 m/s EXECUTE: (a) n = so v = = = 2.04 × 108 m/s v n 1.47 λ 650 nm (b) λ = 0 = = 442 nm 1.47 n EVALUATE: Light is slower in the liquid than in vacuum. By v = f λ , when v is smaller, λ is smaller. λ IDENTIFY: In air, c = f λ0 . In glass, λ = 0 . n SET UP: c = 3.00 × 108 m/s c 3.00 × 108 m/s = 517 nm EXECUTE: (a) λ0 = = f 5.80 × 1014 Hz λ 517 nm (b) λ = 0 = = 340 nm 1.52 n EVALUATE: In glass the light travels slower than in vacuum and the wavelength is smaller. c λ IDENTIFY: n = . λ = 0 , where λ0 is the wavelength in vacuum. v n SET UP: c = 3.00 × 108 m/s . n for air is only slightly larger than unity. c 3.00 × 108 m/s = 1.54. EXECUTE: (a) n = = v 1.94 × 108 m/s (b) λ0 = nλ = (1.54)(3.55 × 10−7 m) = 5.47 × 10−7 m. EVALUATE: In quartz the speed is lower and the wavelength is smaller than in air. IDENTIFY: SET UP:

λ=

λ0

. n From Table 33.1, nwater = 1.333 and nbenzene = 1.501 .

⎛ n ⎞ ⎛ 1.333 ⎞ (a) λwater nwater = λbenzene nbenzene = λ0 . λbenzene = λwater ⎜ water ⎟ = (438 nm) ⎜ ⎟ = 389 nm . n ⎝ 1.501 ⎠ ⎝ benzene ⎠ (b) λ0 = λwater nwater = (438 nm)(1.333) = 584 nm EVALUATE: λ is smallest in benzene, since n is largest for benzene. IDENTIFY: Apply Eqs.(33.2) and (33.4) to calculate θ r and θ b . The angles in these equations are measured with respect to the normal, not the surface. (a) SET UP: The incident, reflected and refracted rays are shown in Figure 33.7. EXECUTE:

33.7.

EXECUTE: θ r = θ a = 42.5° The reflected ray makes an angle of 90.0° − θ r = 47.5° with the surface of the glass. Figure 33.7

33.8.

(b) na sinθ a = nb sinθ b , where the angles are measured from the normal to the interface. n sin θ a (1.00)(sin 42.5°) sin θ b = a = = 0.4070 nb 1.66 θ b = 24.0° The refracted ray makes an angle of 90.0° − θ b = 66.0° with the surface of the glass. EVALUATE: The light is bent toward the normal when the light enters the material of larger refractive index. c IDENTIFY: Use the distance and time to find the speed of light in the plastic. n = . v SET UP: c = 3.00 × 108 m/s c 3.00 × 108 m/s d 2.50 m EXECUTE: v = = = 1.38 . = 2.17 × 108 m s . n = = −9 v 2.17 × 108 m/s t 11.5 × 10 s 2.50 m EVALUATE: In air light travels this same distance in = 8.3 ns . 3.00 × 108 m/s


The Nature and Propagation of Light

33.9.

33.10.

33-3

IDENTIFY and SET UP: Use Snell’s law to find the index of refraction of the plastic and then use Eq.(33.1) to calculate the speed v of light in the plastic. EXECUTE: na sinθ a = nb sinθ b

⎛ sin θ a ⎞ ⎛ sin 62.7° ⎞ nb = na ⎜ ⎟ = 1.00 ⎜ ⎟ = 1.194 sin θ sin 48.1° ⎠ ⎝ b ⎠ ⎝ c c n = so v = = (3.00 × 108 m/s) /1.194 = 2.51 × 108 m/s v n EVALUATE: Light is slower in plastic than in air. When the light goes from air into the plastic it is bent toward the normal. IDENTIFY: Apply Snell’s law at both interfaces. SET UP: The path of the ray is sketched in Figure 33.10. Table 33.1 gives n = 1.329 for the methanol. EXECUTE: (a) At the air-glass interface (1.00)sin 41.3° = nglass sin α . At the glass-methanol interface

nglass sin α = (1.329)sinθ . Combining these two equations gives sin 41.3° = 1.329sin θ and θ = 29.8° . (b) The same figures applies as for part (a), except θ = 20.2° . (1.00)sin 41.3° = n sin 20.2° and n = 1.91. EVALUATE: The angle α is 25.2° . The index of refraction of methanol is less than that of the glass and the ray is bent away from the normal at the glass → methanol interface. The unknown liquid has an index of refraction greater than that of the glass, so the ray is bent toward the normal at the glass → liquid interface.

33.11.

33.12.

Figure 33.10 IDENTIFY: Apply Snell’s law to each refraction. SET UP: Let the light initially be in the material with refractive index na and let the final slab have refractive index nb . In part (a) let the middle slab have refractive index n1. EXECUTE: (a) 1st interface: na sin θ a = n1 sinθ1 . 2nd interface: n1 sin θ1 = nb sin θ b . Combining the two equations gives na sinθ a = nb sinθ b . This is the equation that would apply if the middle slab were absent. (b) For N slabs, na sin θ a = n1 sinθ1 , n1 sinθ1 = n2 sin θ 2 , …, nN − 2 sinθ N − 2 = nb sin θ b . Combining all these equations gives na sinθ a = nb sinθ b . EVALUATE: The final direction of travel depends on the angle of incidence in the first slab and the refractive indices of the first and last slabs. IDENTIFY: Apply Snell's law to the refraction at each interface. SET UP: nair = 1.00. nwater = 1.333.

⎛ n ⎞ ⎛ 1.00 ⎞ (a) θ water = arcsin ⎜ air sinθ air ⎟ = arcsin ⎜ sin 35.0° ⎟ = 25.5°. n 1.333 ⎝ ⎠ ⎝ water ⎠ EVALUATE: (b) This calculation has no dependence on the glass because we can omit that step in the chain: nair sin θ air = nglass sin θ glass = nwater sin θ water . EXECUTE:

33.13.

IDENTIFY: When a wave passes from one material into another, the number of waves per second that cross the boundary is the same on both sides of the boundary, so the frequency does not change. The wavelength and speed of the wave, however, do change. λ SET UP: In a material having index of refraction n, the wavelength is λ = 0 , where λ0 is the wavelength in n vacuum, and the speed is c . n λ EXECUTE: (a) The frequency is the same, so it is still f. The wavelength becomes λ = 0 , so λ0 = nλ. The speed n is v = c , so c = nv. n (b) The frequency is still f. The wavelength becomes λ ′ = λ0 = nλ = ⎜⎛ n ⎟⎞ λ and the speed becomes n′ n′ ⎝ n′ ⎠ c nv ⎛ n ⎞ v′ = = = ⎜ ⎟v n′ n′ ⎝ n′ ⎠ EVALUATE: These results give the speed and wavelength in a new medium in terms of the original medium without referring them to the values in vacuum (or air).


33-4

Chapter 33

33.14.

IDENTIFY: Apply the law of reflection. SET UP: The mirror in its original position and after being rotated by an angle θ are shown in Figure 33.14. α is the angle through which the reflected ray rotates when the mirror rotates. The two angles labeled φ are equal and the two angles labeled φ ′are equal because of the law of reflection. The two angles labeled θ are equal because the lines forming one angle are perpendicular to the lines forming the other angle. EXECUTE: From the diagram, α = 2φ ′ − 2φ = 2(φ ′ − φ ) and θ = φ ′ − φ . α = 2θ , as was to be shown. EVALUATE: This result is independent of the initial angle of incidence.

33.15.

Figure 33.14 IDENTIFY: Apply na sin θ a = nb sin θ b . SET UP: The light refracts from the liquid into the glass, so na = 1.70 , θ a = 62.0° . nb = 1.58 . ⎛n ⎞ ⎛ 1.70 ⎞ sin θb = ⎜ a ⎟ sin θ a = ⎜ ⎟ sin 62.0° = 0.950 and θ b = 71.8° . ⎝ 1.58 ⎠ ⎝ nb ⎠ EVALUATE: The ray refracts into a material of smaller n, so it is bent away from the normal. IDENTIFY: Apply Snell's law. SET UP: θ a and θ b are measured relative to the normal to the surface of the interface. θ a = 60.0° − 15.0° = 45.0° .

EXECUTE: 33.16.

33.17.

⎛n ⎞ ⎛ 1.33 ⎞ EXECUTE: θ b = arcsin ⎜ a sinθ a ⎟ = arcsin ⎜ sin 45.0° ⎟ = 38.2° . But this is the angle from the normal to the surface, ⎝ 1.52 ⎠ ⎝ nb ⎠ so the angle from the vertical is an additional 15° because of the tilt of the surface. Therefore, the angle is 53.2°. EVALUATE: Compared to Example 33.1, θ a is shifted by 15° but the shift in θ b is only 53.2° − 49.3° = 3.9° . IDENTIFY: The critical angle for total internal reflection is θ a that gives θ b = 90° in Snell's law. SET UP: In Figure 33.17 the angle of incidence θ a is related to angle θ by θ a + θ = 90° . EXECUTE: (a) Calculate θ a that gives θ b = 90°. na = 1.60 , nb = 1.00 so na sin θ a = nb sin θ b gives 1.00 and θ a = 38.7° . θ = 90° − θ a = 51.3° . (1.60)sin θ a = (1.00)sin 90° . sin θ a = 1.60 1.333 (b) na = 1.60 , nb = 1.333 . (1.60)sin θ a = (1.333)sin 90°. sin θ a = and θ a = 56.4° . θ = 90° − θ a = 33.6° . 1.60 n EVALUATE: The critical angle increases when the ratio a increases. nb

Figure 33.17


The Nature and Propagation of Light

33.18.

33.19.

33-5

IDENTIFY: Since the refractive index of the glass is greater than that of air or water, total internal reflection will occur at the cube surface if the angle of incidence is greater than or equal to the critical angle. SET UP: At the critical angle θc, Snell’s law gives nglass sin θc = nair sin 90° and likewise for water. EXECUTE: (a) At the critical angle θc , nglass sin θc = nair sin 90°. 1.53 sin θc = (1.00)(1) andθc = 40.8°. (b) Using the same procedure as in part (a), we have 1.53 sin θc = 1.333 sin 90° and θc = 60.6°. EVALUATE: Since the refractive index of water is closer to the refractive index of glass than the refractive index of air is, the critical angle for glass-to-water is greater than for glass-to-air. IDENTIFY: Use the critical angle to find the index of refraction of the liquid. SET UP: Total internal reflection requires that the light be incident on the material with the larger n, in this case the liquid. Apply na sinθ a = nb sinθ b with a = liquid and b = air, so na = nliq and nb = 1.0. EXECUTE:

nliq =

θ a = θ crit when θ b = 90°, so nliq sin θ crit = (1.0)sin 90°

1 1 = = 1.48. sin θ crit sin 42.5°

(a) na sinθ a = nb sinθ b (a = liquid, b = air)

sin θ b =

na sinθ a (1.48)sin 35.0° = = 0.8489 and θ b = 58.1° 1.0 nb

(b) Now na sin θ a = nb sin θ b with a = air, b = liquid

na sin θ a (1.0)sin 35.0° = = 0.3876 and θ b = 22.8° 1.48 nb EVALUATE: For light traveling liquid → air the light is bent away from the normal. For light traveling air → liquid the light is bent toward the normal. IDENTIFY: The largest angle of incidence for which any light refracts into the air is the critical angle for water → air . SET UP: Figure 33.20 shows a ray incident at the critical angle and therefore at the edge of the ring of light. The radius of this circle is r and d = 10.0 m is the distance from the ring to the surface of the water. EXECUTE: From the figure, r = d tan θ crit . θ crit is calculated from na sinθ a = nb sinθ b with na = 1.333 , θ a = θ crit , (1.00)sin 90° nb = 1.00 and θ b = 90°. sinθ crit = and θ crit = 48.6°. r = (10.0 m) tan 48.6° = 11.3 m . 1.333 A = π r 2 = π (11.3 m) 2 = 401 m 2 . EVALUATE: When the incident angle in the water is larger than the critical angle, no light refracts into the air. sin θ b =

33.20.

Figure 33.20 33.21.

IDENTIFY and SET UP:

For glass → water, θ crit = 48.7°. Apply Snell’s law with θ a = θ crit to calculate the index

of refraction na of the glass. nb 1.333 = = 1.77 sin θ crit sin 48.7° EVALUATE: For total internal reflection to occur the light must be incident in the material of larger refractive index. Our results give nglass > nwater , in agreement with this.

EXECUTE:

na sin θ crit = nb sin 90°, so na =


33-6

Chapter 33

33.22.

IDENTIFY: If no light refracts out of the glass at the glass to air interface, then the incident angle at that interface is θ crit . SET UP: The ray has an angle of incidence of 0° at the first surface of the glass, so enters the glass without being bent, as shown in Figure 33.22. The figure shows that α + θ crit = 90°. EXECUTE: (a) For the glass-air interface θ a = θ crit , na = 1.52, nb = 1.00 and θb = 90°. na sinθ a = nb sinθ b gives (1.00)(sin 90°) sinθ crit = and θ crit = 41.1°. α = 90° − θ crit = 48.9°. 1.52 (1.333)(sin 90°) (b) Now the second interface is glass → water and nb = 1.333 . na sinθ a = nb sinθ b gives sinθ crit = 1.52 and θ crit = 61.3°. α = 90° − θ crit = 28.7° . EVALUATE: The critical angle increases when the air is replaced by water and rays are bent as they refract out of the glass.

Figure 33.22 33.23.

Apply na sinθ a = nb sinθ b .

IDENTIFY: SET UP:

The light is in diamond and encounters an interface with air, so na = 2.42 and nb = 1.00 . The largest

θ a is when θ b = 90° .

33.24.

1 and θ a = 24.4° . 2.42 EVALUATE: Diamond has an usually large refractive index, and this results in a small critical angle. c IDENTIFY: Snell's law is na sin θ a = nb sin θ b . v = . n SET UP: a = air , b = glass . EXECUTE:

(2.42)sin θ a = (1.00)sin 90° . sin θ a =

EXECUTE:

(a) red: nb =

na sinθ a (1.00)sin 57.0° (1.00)sin 57.0° = = 1.36 . violet: nb = = 1.40 . sin θ b sin 38.1° sin 36.7°

c 3.00 × 108 m/s c 3.00 × 108 m/s = = 2.21 × 108 m/s ; violet: v = = = 2.14 × 108 m/s . 1.36 1.40 n n EVALUATE: n is larger for the violet light and therefore this light is bent more toward the normal, and the violet light has a smaller speed in the glass than the red light. IDENTIFY: When unpolarized light passes through a polarizer the intensity is reduced by a factor of 12 and the (b) red: v =

33.25.

transmitted light is polarized along the axis of the polarizer. When polarized light of intensity I max is incident on a polarizer, the transmitted intensity is I = I max cos 2 φ , where φ is the angle between the polarization direction of the incident light and the axis of the filter. SET UP: For the second polarizer φ = 60° . For the third polarizer, φ = 90° − 60° = 30° . EXECUTE:

(a) At point A the intensity is I 0 / 2 and the light is polarized along the vertical direction. At point B

the intensity is ( I 0 / 2)(cos60°) 2 = 0.125I 0 , and the light is polarized along the axis of the second polarizer. At point C the intensity is (0.125 I 0 )(cos30°) 2 = 0.0938 I 0 .

33.26.

(b) Now for the last filter φ = 90° and I = 0 . EVALUATE: Adding the middle filter increases the transmitted intensity. IDENTIFY: Apply Snell's law. SET UP: The incident, reflected and refracted rays are shown in Figure 33.26. sin θ a sin 53° EXECUTE: From the figure, θ b = 37.0° and nb = na = 1.33 = 1.77. sin θ b sin 37°


The Nature and Propagation of Light

EVALUATE: refracts.

33-7

The refractive index of b is greater than that of a, and the ray is bent toward the normal when it

Figure 33.26 33.27.

IDENTIFY and SET UP: Reflected beam completely linearly polarized implies that the angle of incidence equals the polarizing angle, so θ p = 54.5°. Use Eq.(33.8) to calculate the refractive index of the glass. Then use Snell’s

law to calculate the angle of refraction. n EXECUTE: (a) tanθ p = b gives nglass = nair tanθ p = (1.00) tan 54.5° = 1.40. na (b) na sinθ a = nb sinθ b

na sinθ a (1.00)sin 54.5° = = 0.5815 and θ b = 35.5° 1.40 nb EVALUATE:

sin θ b =

Note: φ = 180.0° − θ r − θ b and θ r = θ a . Thus φ = 180.0° − 54.5° − 35.5° = 90.0°; the reflected ray and the refracted ray are perpendicular to each other. This agrees with Fig.33.28. Figure 33.27 33.28.

IDENTIFY: SET UP:

Set I = I 0 /10 , where I is the intensity of light passed by the second polarizer. When unpolarized light passes through a polarizer the intensity is reduced by a factor of

1 2

and the

transmitted light is polarized along the axis of the polarizer. When polarized light of intensity I max is incident on a polarizer, the transmitted intensity is I = I max cos 2 φ , where φ is the angle between the polarization direction of the incident light and the axis of the filter. I EXECUTE: (a) After the first filter I = 0 and the light is polarized along the vertical direction. After the second 2 I0 I0 ⎛ I0 ⎞ filter we want I = , so = ⎜ ⎟ (cos φ ) 2 . cos φ = 2 /10 and φ = 63.4° . 10 ⎝ 2 ⎠ 10 I (b) Now the first filter passes the full intensity I 0 of the incident light. For the second filter 0 = I 0 (cos φ ) 2 . 10

33.29.

cos φ = 1/10 and φ = 71.6° . EVALUATE: When the incident light is polarized along the axis of the first filter, φ must be larger to achieve the same overall reduction in intensity than when the incident light is unpolarized. IDENTIFY: From Malus’s law, the intensity of the emerging light is proportional to the square of the cosine of the angle between the polarizing axes of the two filters. SET UP: If the angle between the two axes is θ, the intensity of the emerging light is I = Imax cos2θ. 1 EXECUTE: At angle θ, I = Imax cos2θ, and at the new angle α, I = Imax cos2α. Taking the ratio of the intensities 2 cosθ I max cos 2 α 12 I ⎛ cosθ ⎞ = , which gives us cosα = . Solving for α yields α = arccos ⎜ gives ⎟. I max cos 2 θ I 2 ⎝ 2 ⎠ EVALUATE: Careful! This result is not cos2θ.


33-8

Chapter 33

33.30.

IDENTIFY:

The reflected light is completely polarized when the angle of incidence equals the polarizing angle n θ p , where tanθ p = b . na

nb = 1.66 .

SET UP: EXECUTE:

(a) na = 1.00 . tan θ p =

1.66 and θ p = 58.9° . 1.00

1.66 and θ p = 51.2° . 1.333 EVALUATE: The polarizing angle depends on the refractive indicies of both materials at the interface. IDENTIFY: When unpolarized light of intensity I 0 is incident on a polarizing filter, the transmitted light has (b) na = 1.333 . tan θ p =

33.31.

intensity

1 2

I 0 and is polarized along the filter axis. When polarized light of intensity I 0 is incident on a polarizing

filter the transmitted light has intensity I 0 cos 2 φ . SET UP: EXECUTE:

For the second filter, φ = 62.0° − 25.0° = 37.0° . After the first filter the intensity is

1 2

I 0 = 10.0 W m 2 and the light is polarized along the axis of the

first filter. The intensity after the second filter is I = I 0cos 2φ , where I 0 = 10.0 W m 2 and φ = 37.0° . This

33.32.

gives I = 6.38 W m 2 . EVALUATE: The transmitted intensity depends on the angle between the axes of the two filters. IDENTIFY: After passing through the first filter the light is linearly polarized along the filter axis. After the second filter, I = I max (cos φ ) 2 , where φ is the angle between the axes of the two filters. SET UP: The maximum amount of light is transmitted when φ = 0 . EXECUTE:

(a) I = I 0 (cos 22.5°) 2 = 0.854 I 0

(b) I = I 0 (cos 45.0°) 2 = 0.500 I 0 (c) I = I 0 (cos67.5°) 2 = 0.146 I 0

33.33.

EVALUATE: As φ increases toward 90° the axes of the two filters are closer to being perpendicular to each other and the transmitted intensity decreases. IDENTIFY and SET UP: Apply Eq.(33.7) to polarizers #2 and #3. The light incident on the first polarizer is unpolarized, so the transmitted light has half the intensity of the incident light, and the transmitted light is polarized. (a) EXECUTE: The axes of the three filters are shown in Figure 33.33a.

I = I max cos 2 φ

Figure 33.33a

After the first filter the intensity is I1 = 12 I 0 and the light is linearly polarized along the axis of the first polarizer. After the second filter the intensity is I 2 = I1 cos 2 φ = ( 12 I 0 )(cos 45.0°) 2 = 0.250 I 0 and the light is linearly polarized along the axis of the second polarizer. After the third filter the intensity is I 3 = I 2 cos 2 φ = 0.250 I 0 (cos 45.0°) 2 = 0.125I 0 and the light is linearly polarized along the axis of the third polarizer. (b) The axes of the remaining two filters are shown in Figure 33.33b. After the first filter the intensity is I1 = 12 I 0 and the light is linearly polarized along the axis of the first polarizer.

Figure 33.33b

After the next filter the intensity is I 3 = I1 cos 2 φ = ( 12 I 0 )(cos90.0°) 2 = 0. No light is passed. EVALUATE: Light is transmitted through all three filters, but no light is transmitted if the middle polarizer is removed.


The Nature and Propagation of Light

33.34.

33-9

IDENTIFY: Use the transmitted intensity when all three polairzers are present to solve for the incident intensity I 0 . Then repeat the calculation with only the first and third polarizers. SET UP:

For unpolarized light incident on a filter, I = 12 I 0 and the light is linearly polarized along the filter axis.

For polarized light incident on a filter, I = I max (cos φ ) 2 , where I max is the intensity of the incident light, and the emerging light is linearly polarized along the filter axis. EXECUTE: With all three polarizers, if the incident intensity is I 0 the transmitted intensity is I 75.0 W/cm 2 = = 293 W/cm 2 . With only the first 0.256 0.256 and third polarizers, I = ( 12 I 0 )(cos62.0°) 2 = 0.110 I 0 = (0.110)(293 W/cm 2 ) = 32.2 W/cm 2 . EVALUATE: The transmitted intensity is greater when all three filters are present. IDENTIFY: The shorter the wavelength of light, the more it is scattered. The intensity is inversely proportional to the fourth power of the wavelength. SET UP: The intensity of the scattered light is proportional to 1/λ4, we can write it as I = (constant)/ λ4. EXECUTE: (a) Since I is proportional to 1/λ4, we have I = (constant)/ λ4. Taking the ratio of the intensity of the 4 4 I R (constant) / λR4 ⎛ λG ⎞ ⎛ 520 nm ⎞ = 0.374, so IR = 0.374I. = = = red light to that of the green light gives ⎜ ⎟ ⎜ ⎟ I (constant) / λG4 ⎝ λR ⎠ ⎝ 665 nm ⎠

I = ( 12 I 0 )(cos 23.0°) 2 (cos[62.0° − 23.0°]) 2 = 0.256 I 0 . I 0 =

33.35.

4

33.36.

33.37.

33.38.

4

⎛ λ ⎞ ⎛ 520 nm ⎞ I (b) Following the same procedure as in part (a) gives V = ⎜ G ⎟ = ⎜ ⎟ = 2.35, so IV = 2.35I. I ⎝ λV ⎠ ⎝ 420 nm ⎠ EVALUATE: In the scattered light, the intensity of the short-wavelength violet light is about 7 times as great as that of the red light, so this scattered light will have a blue-violet color. IDENTIFY: As the wave front reaches the sharp object, every point on the front will act as a source of secondary wavelets. SET UP: Consider a wave front that is just about to go past the corner. Follow it along and draw the successive wave fronts. EXECUTE: The path of the wavefront is drawn in Figure 33.36. EVALUATE: The wave fronts clearly bend around the sharp point, just as water waves bend around a rock and light waves bend around the edge of a slit.

Figure 33.36 IDENTIFY: Reflection reverses the sign of the component of light velocity perpendicular to the reflecting surface but leaves the other components unchanged. SET UP: Consider three mirrors, M1 in the (x,y)-plane, M2 in the (y,z)-plane, and M3 in the (x,z)-plane. EXECUTE: A light ray reflecting from M1 changes the sign of the z-component of the velocity, reflecting from M2 changes the x-component, and from M3 changes the y-component. Thus the velocity, and hence also the path, of the light beam flips by 180° EVALUATE: Example 33.3 discusses some uses of corner reflectors. IDENTIFY: The light travels slower in the jelly than in the air and hence will take longer to travel the length of the tube when it is filled with jelly than when it contains just air. SET UP: The definition of the index of refraction is n = c/v, where v is the speed of light in the jelly. EXECUTE: First get the length L of the tube using air. In the air, we have L = ct = (3.00 × 108 m/s)(8.72 ns) = 2.616 m. c L The speed in the jelly is v = = (2.616 m)/(8.72 ns + 2.04 ns) = 2.431 × 108 m/s. n = = v t (3.00 × 108 m/s)/(2.431 × 108 m/s) = 1.23 EVALUATE: A high-speed timer would be needed to measure times as short as a few nanoseconds.


33-10

Chapter 33

33.39.

IDENTIFY and SET UP: Apply Snell's law at each interface. EXECUTE: (a) n1 sinθ1 = n2 sinθ 2 and n2 sinθ 2 = n3 sinθ 3 , so n1 sin θ1 = n3 sinθ 3 and sin θ 3 = ( n1 sinθ1 ) / n3 . (b) n3 sinθ 3 = n2 sinθ 2 and n2 sinθ 2 = n1 sinθ1 , so n1 sin θ1 = n3 sinθ 3 and the light makes the same angle with respect to the normal in the material that has refractive index n1 as it did in part (a). (c) For reflection, θ r = θ a . These angles are still equal if θ r becomes the incident angle; reflected rays are also reversible. EVALUATE: Both the refracted and reflected rays are reversible, in the sense that if the direction of the light is reversed then each of these rays follow the path of the incident ray. c λ IDENTIFY: Use the change in transit time to find the speed v of light in the slab, and then apply n = and λ = 0 . n v SET UP: It takes the light an additional 4.2 ns to travel 0.840 m after the glass slab is inserted into the beam. 0.840 m 0.840 m 0.840 m EXECUTE: − = ( n − 1) = 4.2 ns. We can now solve for the index of refraction: c/n c c −9 8 490 nm (4.2 × 10 s) (3.00 × 10 m s) = 196 nm . n= + 1 = 2.50. The wavelength inside of the glass is λ = 2.50 0.840 m EVALUATE: Light travels slower in the slab than in air and the wavelength is shorter. IDENTIFY: The angle of incidence at A is to be the critical angle. Apply Snell’s law at the air to glass refraction at the top of the block. SET UP: The ray is sketched in Figure 33.41. EXECUTE: For glass → air at point A, Snell's law gives (1.38)sin θ crit = (1.00)sin 90°and θ crit = 46.4° . θ b = 90° − θ crit = 43.6° . Snell's law applied to the refraction from air to glass at the top of the block gives (1.00)sinθ a = (1.38)sin(43.6°) and θ a = 72.1° . EVALUATE: If θ a is larger than 72.1° then the angle of incidence at point A is less than the initial critical angle and total internal reflection doesn’t occur.

33.40.

33.41.

33.42.

Figure 33.41 IDENTIFY: As the light crosses the glass-air interface along AB, it is refracted and obeys Snell’s law. SET UP: Snell’s law is na sin θa = nb sin θb and n = 1.000 for air. At point B the angle of the prism is 30.0° . EXECUTE: Apply Snell’s law at AB. The prism angle at A is 60.0°, so for the upper ray, the angle of incidence at AB is 60.0° + 12.0° = 72.0°. Using this value gives n1 sin 60.0° = sin 72.0° and n1 = 1.10. For the lower ray, the angle of incidence at AB is 60.0° + 12.0° + 8.50° = 80.5°, giving n2 sin 60.0° = sin 80.5° and n2 = 1.14. EVALUATE: The lower ray is deflected more than the upper ray because that wavelength has a slightly greater index of refraction than the upper ray.


The Nature and Propagation of Light

33.43.

33.44.

33.45.

33-11

IDENTIFY: Circularly polarized light consists of the superposition of light polarized in two perpendicular directions, with a quarter-cycle ( 90° ) phase difference between the two polarization components. SET UP: A quarter-wave plate shifts the relative phase of the two perpendicular polarization components by 90° . EXECUTE: In the circularly polarized light the two perpendicular polarization components are 90° out of phase. The quarter-wave plate shifts the relative phase by ±90° and then the two components are either in phase or 180° out of phase. Either corresponds to linearly polarized light. EVALUATE: Either left circularly polarized light or right circularly polarized light is converted to linearly polarized light by the quarter-wave plate. λ d IDENTIFY: Apply λ = 0 . The number of wavelengths in a distance d of a material is where λ is the n λ wavelength in the material. SET UP: The distance in glass is d glass = 0.00250 m . The distance in air is

d air = 0.0180 m − 0.00250 m = 0.0155 m EXECUTE: number of wavelengths = number in air + number in glass. d d 0.0155 m 0.00250 m + number of wavelengths = air + glass n = (1.40) = 3.52 × 104 . λ λ 5.40 × 10 −7 m 5.40 × 10−7 m EVALUATE: Without the glass plate the number of wavelengths between the source and screen is 0.0180 m = 3.33 × 104 . The wavelength is shorter in the glass so there are more wavelengths in a distance in 5.40 × 10−3 m glass than there are in the same distance in air. IDENTIFY: Find the critical angle for glass → air. Light incident at this critical angle is reflected back to the edge of the halo. SET UP: The ray incident at the critical angle is sketched in Figure 33.45.

Figure 33.45 2.67 mm = 0.8613; θ crit = 40.7°. 3.10 mm Apply Snell’s law to the total internal reflection to find the refractive index of the glass: na sinθ a = nb sinθ b

EXECUTE:

From the distances given in the sketch, tan θ crit =

nglass sin θ crit = 1.00sin 90° nglass =

1 1 = = 1.53 sinθ crit sin 40.7°

EVALUATE: 33.46.

Light incident on the back surface is also totally reflected if it is incident at angles greater than θ crit .

If it is incident at less than θ crit it refracts into the air and does not reflect back to the emulsion. IDENTIFY: Apply Snell's law to the refraction of the light as it passes from water into air. ⎛ 1.5 m ⎞ SET UP: θ a = arctan ⎜ ⎟ = 51° . na = 1.00 . nb = 1.333 . ⎝ 1.2 m ⎠ EXECUTE:

⎛ na ⎞ ⎛ 1.00 ⎞ sinθ a ⎟ = arcsin ⎜ sin 51° ⎟ = 36°. Therefore, the distance along the bottom of the n 1.333 ⎝ ⎠ b ⎝ ⎠

θ b = arcsin ⎜

pool from directly below where the light enters to where it hits the bottom is x = (4.0 m)tanθb = (4.0 m)tan36° = 2.9 m. xtotal = 1.5 m + x = 1.5 m + 2.9 m = 4.4 m. EVALUATE: The light ray from the flashlight is bent toward the normal when it refracts into the water.


33-12

Chapter 33

33.47.

IDENTIFY: Use Snell’s law to determine the effect of the liquid on the direction of travel of the light as it enters the liquid. SET UP: Use geometry to find the angles of incidence and refraction. Before the liquid is poured in the ray along your line of sight has the path shown in Figure 33.47a.

8.0 cm = 0.500 16.0 cm θ a = 26.57°

tan θ a =

Figure 33.47a After the liquid is poured in, θ a is the same and the refracted ray passes through the center of the bottom of the glass, as shown in Figure 33.47b.

4.0 cm = 0.250 16.0 cm θ b = 14.04°

tanθ b =

Figure 33.47b

33.48.

EXECUTE: Use Snell’s law to find nb , the refractive index of the liquid: na sinθ a = nb sinθ b n sin θ a (1.00)(sin 26.57°) nb = a = = 1.84 sin θ b sin14.04° EVALUATE: When the light goes from air to liquid (larger refractive index) it is bent toward the normal. IDENTIFY: Apply Snell’s law to each refraction and apply the law of reflection at the mirrored bottom. SET UP: The path of the ray is sketched in Figure 33.48. The problem asks us to calculate θ b′ . EXECUTE: Apply Snell's law to the air → liquid refraction. (1.00)sin(42.5°) = (1.63)sin θ b and θ b = 24.5° .

θ b = φ and φ = θ a′ , so θ a′ = θ b = 24.5°. Snell's law applied to the liquid → air refraction gives (1.63)sin(24.5°) = (1.00)sin θ b′and θ b′ = 42.5° .


The Nature and Propagation of Light

EVALUATE:

33.49.

33-13

The light emerges from the liquid at the same angle from the normal as it entered the liquid.

Figure 33.48 IDENTIFY: Apply Snell’s law to the water → ice and ice → air interfaces. (a) SET UP: Consider the ray shown in Figure 33.49.

We want to find the incident angle θ a at the water-ice interface that causes the incident angle at the ice-air interface to be the critical angle. Figure 33.49 EXECUTE:

ice-air interface: nice sin θ crit = 1.0sin 90°

nice sin θ crit = 1.0 so sin θ crit =

1 nice

But from the diagram we see that θ b = θ crit , so sin θ b =

1 . nice

water-ice interface: nw sin θ a = nice sin θ b 1 1 1 so nw sin θ a = 1.0. sin θ a = = = 0.7502 and θ a = 48.6°. nice nw 1.333 (b) EVALUATE: The angle calculated in part (a) is the critical angle for a water-air interface; the answer would be the same if the ice layer wasn’t there! IDENTIFY: The incident angle at the prism → water interface is to be the critical angle. SET UP: The path of the ray is sketched in Figure 33.50. The ray enters the prism at normal incidence so is not bent. For water, nwater = 1.333 .

But sin θ b =

33.50.

EXECUTE: nglass =

From the figure, θ crit = 45° . na sinθ a = nb sinθ b gives nglass sin 45° = (1.333)sin 90° .

1.333 = 1.89 . sin 45°


33-14

Chapter 33

EVALUATE: For total internal reflection the ray must be incident in the material of greater refractive index. nglass > nwater , so that is the case here.

33.51.

Figure 33.50 IDENTIFY: Apply Snell’s law to the refraction of each ray as it emerges from the glass. The angle of incidence equals the angle A = 25.0°. SET UP: The paths of the two rays are sketched in Figure 33.51.

EXECUTE: na sinθ a = nb sinθ b nglas sin 25.0° = 1.00sin θ b

Figure 33.51

sin θ b = nglass sin 25.0°

sin θ b = 1.66sin 25.0° = 0.7015 θ b = 44.55° β = 90.0° − θ b = 45.45° Then δ = 90.0° − A − β = 90.0° − 25.0° − 45.45° = 19.55°. The angle between the two rays is 2δ = 39.1°. 33.52.

EVALUATE: The light is incident normally on the front face of the prism so the light is not bent as it enters the prism. IDENTIFY: The ray shown in the figure that accompanies the problem is to be incident at the critical angle. SET UP: θ b = 90° . The incident angle for the ray in the figure is 60° . ⎛ n sin θ a ⎞ ⎛ 1.62sin 60° ⎞ na sinθ a = nb sinθ b gives nb = ⎜ a ⎟=⎜ ⎟ = 1.40. ⎝ sin θ b ⎠ ⎝ sin 90° ⎠ EVALUATE: Total internal reflection occurs only when the light is incident in the material of the greater refractive index. IDENTIFY: No light enters the gas because total internal reflection must have occurred at the water-gas interface. SET UP: At the minimum value of S, the light strikes the water-gas interface at the critical angle. We apply Snell’s law, na sinθa = nb sinθb, at that surface. S EXECUTE: (a) In the water, θ = = (1.09 m)/(1.10 m) = 0.991 rad = 56.77°. This is the critical angle. So, using R the refractive index for water from Table 33.1, we get n = (1.333) sin 56.77° = 1.12 (b) (i) The laser beam stays in the water all the time, so

EXECUTE:

33.53.

Dnwater ⎛ c ⎞ t = 2R/v = 2R/ ⎜ = = (2.20 m)(1.333)/(3.00 × 108 m/s) = 9.78 ns ⎟ c n ⎝ water ⎠


The Nature and Propagation of Light

33.54.

33.55.

33-15

(ii) The beam is in the water half the time and in the gas the other half of the time. Rngas tgas = = (1.10 m)(1.12)/(3.00 × 108 m/s) = 4.09 ns c The total time is 4.09 ns + (9.78 ns)/2 = 8.98 ns EVALUATE: The gas must be under considerable pressure to have a refractive index as high as 1.12. IDENTIFY: No light enters the water because total internal reflection must have occurred at the glass-water surface. SET UP: A little geometry tells us that θ is the angle of incidence at the glass-water face in the water. Also, θ = 59.2° must be the critical angle at that surface, so the angle of refraction is 90.0°. Snell’s law, na sin θa = nb sin θb, c applies at that glass-water surface, and the index of refraction is defined as n = . v c EXECUTE: Snell’s law at the glass-water surface gives n sin 59.2° = (1.333)(1.00), which gives n = 1.55. v = = n (3.00 × 108 m/s)/1.55 = 1.93 × 108 m/s. EVALUATE: Notice that θ is not the angle of incidence at the reflector, but it is the angle of incidence at the glasswater surface. (a) IDENTIFY: Apply Snell’s law to the refraction of the light as it enters the atmosphere. SET UP: The path of a ray from the sun is sketched in Figure 33.55.

δ = θ a − θb From the diagram sin θ b =

R R+h

⎛ R ⎞

θ b = arcsin ⎜ ⎟ ⎝ R+h⎠ Figure 33.55

33.56.

EXECUTE: Apply Snell’s law to the refraction that occurs at the top of the atmosphere: na sinθ a = nb sinθ b (a = vacuum of space, refractive, index 1.0; b = atmosphere, refractive index n) ⎛ R ⎞ ⎛ nR ⎞ sin θ a = n sin θ b = n ⎜ ⎟ so θ a = arcsin ⎜ ⎟ + R h ⎝ ⎠ ⎝R+h⎠ ⎛ nR ⎞ ⎛ R ⎞ δ = θ a − θ b = arcsin ⎜ ⎟ − arcsin ⎜ ⎟ + R h ⎝ ⎠ ⎝ R+h⎠ R 6.38 × 106 m (b) = = 0.99688 R + h 6.38 × 106 m + 20 × 103 m nR = 1.0003(0.99688) = 0.99718 R+h ⎛ R ⎞ θ b = arcsin ⎜ ⎟ = 85.47° ⎝ R+h⎠ ⎛ nR ⎞ θ b = arcsin ⎜ ⎟ 85.70° ⎝ R+h⎠ δ = θ a − θ b = 85.70° − 85.47° = 0.23° EVALUATE: The calculated δ is about the same as the angular radius of the sun. IDENTIFY and SET UP: Follow the steps specified in the problem. EXECUTE: (a) The distance traveled by the light ray is the sum of the two diagonal segments: d ( x 2 + y12 )1/ 2 + ((l − x ) 2 + y22 )1/ 2 d = ( x 2 + y12 )1/ 2 + ((l − x)2 + y22 )1/ 2 . Then the time taken to travel that distance is t = = . c c (b) Taking the derivative with respect to x of the time and setting it to zero yields 1 2 dt −1 2 dt 1 d ⎡ 2 1 ⎤ = 0 . This gives = ( x + y12 )1 2 + (l − x)2 + y22 ⎤⎥ = ⎡⎢ x( x 2 + y12 )−1 2 − (l − x) (l − x)2 + y22 ⎦ dx c ⎣ ⎦⎥ dx c dt ⎣⎢ x (l − x) = , sinθ1 = sinθ 2 and θ1 = θ 2 . x 2 + y12 (l − x)2 + y22

(

)

(

)

EVALUATE: For any other path between points 1 and 2, that includes a point on the reflective surface, the distance traveled and therefore the travel time is greater than for this path.


33-16

Chapter 33

33.57.

IDENTIFY and SET UP: Find the distance that the ray travels in each medium. The travel time in each medium is the distance divided by the speed in that medium. (a) EXECUTE:

The light travels a distance

h12 + x 2 in traveling from point A to the interface. Along this path

the speed of the light is v1 , so the time it takes to travel this distance is t1 =

h22 + (l − x) 2 in traveling from the interface to point B. Along this path the speed of the light is v2 ,

distance

so the time it takes to travel this distance is t2 =

h22 + (l − x ) 2 . The total time to go from A to B is v2

h12 + x 2 h 2 + (l + x ) 2 + 2 . v1 v2

t = t1 + t2 =

(b)

h12 + x 2 . The light travels a v1

1 ⎛1⎞ dt 1 ⎛ 1 ⎞ 2 = ⎜ ⎟ (h1 + x 2 )−1/ 2 (2 x) + ⎜ ⎟ (h22 + (l − x)2 )−1/ 2 2(l − x)(−1) = 0 dx v1 ⎝ 2 ⎠ v2 ⎝ 2 ⎠ x

v1 h + x 2 1

2

=

l−x v2 h + (l − x)2 2 2

Multiplying both sides by c gives

c x c = v1 h12 + x 2 v2

l−x h + (l − x)2 2 2

c c = n1 and = n2 (Eq.33.1) v1 v2 From Fig.33.55 in the textbook, sin θ1 =

33.58.

33.59.

x h12 + x 2

and sin θ 2 =

l−x h22 + (l − x) 2

.

So n1 sin θ1 = n2 sin θ 2 , which is Snell’s law. EVALUATE: Snell’s law is a result of a change in speed when light goes from one material to another. IDENTIFY: Apply Snell's law to each refraction. SET UP: Refer to the angles and distances defined in the figure that accompanies the problem. EXECUTE: (a) For light in air incident on a parallel-faced plate, Snell’s Law yields: n sin θ a = n′ sin θ b′ = n′ sin θ b = n sin θ a′ ⇒ sin θ a = sin θ a′ ⇒ θ a = θ a′ . (b) Adding more plates just adds extra steps in the middle of the above equation that always cancel out. The requirement of parallel faces ensures that the angle θ n′ = θ n and the chain of equations can continue. (c) The lateral displacement of the beam can be calculated using geometry: t t sin(θ a − θ b′ ) ⇒d = . d = L sin(θ a − θ b′ ) and L = cosθ b′ cosθ b′ (2.40 cm)sin(66.0° − 30.5°) ⎛ n sinθ a ⎞ ⎛ sin 66.0° ⎞ (d) θ b′ = arcsin ⎜ = 1.62 cm. ⎟ = arcsin ⎜ ⎟ = 30.5° and d = ′ cos30.5° ⎝ n ⎠ ⎝ 1.80 ⎠ EVALUATE: The lateral displacement in part (d) is large, of the same order as the thickness of the plate. IDENTIFY: Apply Snell’s law to each refraction and apply the law of reflection to each reflection. SET UP: The paths of rays A and B are sketched in Figure 33.59. Let θ be the angle of incidence for the combined ray. EXECUTE: For ray A its final direction of travel is at an angle θ with respect to the normal, by the law of reflection. Let the final direction of travel for ray B be at angle φ with respect to the normal. At the upper surface,

Snell’s law gives n1 sin θ = n2 sin α . The lower surface reflects ray B at angle α . Ray B returns to the upper surface of the film at an angle of incidence α . Snell’s law applied to the refraction as ray B leaves the film gives n2 sin α = n1 sin φ . Combining the two equations gives n1 sin θ = n1 sin φ and θ = φ ; the two rays are parallel after they emerge from the film.


The Nature and Propagation of Light

EVALUATE: the film.

33-17

Ray B is bent toward the normal as it enters the film and away from the normal as it refracts out of

Figure 33.59 33.60.

IDENTIFY: Apply Snell's law and the results of Problem 33.58. SET UP: From Figure 33.58 in the textbook, nr = 1.61 for red light and nv = 1.66 for violet. In the notation of Problem 33.58, t is the thickness of the glass plate and the lateral displacement is d. We want the difference in d for (1.00)sin 70.0° the two colors of light to be 1.0 mm. θ a = 70.0° . For red light, na sinθ a = nb sinθ b′ gives sin θ b′ = 1.61 (1.00)sin 70.0° and θ b′ = 34.48° . and θ b′ = 35.71° . For violet light, sin θ b′ = 1.66 EXECUTE: (a) n decreases with increasing λ , so n is smaller for red than for blue. So beam a is the red one. sin(θ a − θ b′ ) sin(70° − 35.71°) (b) Problem 33.58 says d = t . For red light, d r = t = 0.6938t and for violet light, cosθ b′ cos35.71° sin(70° − 35.48°) 0.10 cm = 0.7048t . d v − d r = 1.0 mm gives t = = 9.1 cm . cos35.48° 0.7048 − 0.6958 EVALUATE: Our calculation shown that the violet light has greater lateral displacement and this is ray b. IDENTIFY: Apply Snell's law to the two refractions of the ray. SET UP: Refer to the figure that accompanies the problem. A + 2α A A A ⎛A ⎞ = n sin . EXECUTE: (a) na sinθ a = nb sinθ b gives sin θ a = nb sin . But θ a = + α , so sin ⎜ + α ⎟ = sin 2 2 2 2 ⎝2 ⎠ dv = t

33.61.

A+δ A = n sin . 2 2 A⎞ 60.0° ⎞ ⎛ ⎛ (b) From part (a), δ = 2arcsin ⎜ n sin ⎟ − A . δ = 2arcsin ⎜ (1.52)sin ⎟ − 60.0° = 38.9°. 2⎠ 2 ⎠ ⎝ ⎝ (c) If two colors have different indices of refraction for the glass, then the deflection angles for them will differ: At each face of the prism the deviation is α , so 2α = δ and sin

⎛ ⎝

δ red = 2arcsin ⎜ (1.61)sin ⎛ ⎝

60.0° ⎞ ⎟ − 60.0° = 47.2° 2 ⎠

δ violet = 2arcsin ⎜ (1.66)sin

33.62.

60.0° ⎞ ⎟ − 60.0° = 52.2° ⇒ Δδ = 52.2° − 47.2° = 5.0° 2 ⎠

EVALUATE: The violet light has a greater refractive index and therefore the angle of deviation is greater for the violet light. IDENTIFY: The reflected light is totally polarized when light strikes a surface at Brewster’s angle. SET UP: At the plastic wall, Brewster’s angle obeys the equation tan θp = nb /na, and Snell’s law, na sinθa = nb sinθb, applies at the air-water surface. EXECUTE: To be totally polarized, the reflected sunlight must have struck the wall at Brewster’s angle. tan θp = nb /na = (1.61)/(1.00) and θp = 58.15° This is the angle of incidence at the wall. A little geometry tells us that the angle of incidence at the water surface is 90.00° – 58.15° = 31.85°. Applying Snell’s law at the water surface gives (1.00) sin31.85° = 1.333 sin θ and θ = 23.3° EVALUATE: We have two different principles involved here: Reflection at Brewster’s angle at the wall and Snell’s law at the water surface.


33-18

Chapter 33

33.63.

IDENTIFY and SET UP:

The polarizer passes

1 2

of the intensity of the unpolarized component, independent of φ .

Out of the intensity I p of the polarized component the polarizer passes intensity I p cos 2 (φ − θ ), where φ − θ is the angle between the plane of polarization and the axis of the polarizer. (a) Use the angle where the transmitted intensity is maximum or minimum to find θ . See Figure 33.63.

Figure 33.63

The total transmitted intensity is I = 12 I 0 + I p cos 2 (φ − θ ). This is maximum when θ = φ and from the

EXECUTE:

table of data this occurs for φ between 30° and 40°, say at 35° and θ = 35°. Alternatively, the total transmitted intensity is minimum when φ − θ = 90° and from the data this occurs for φ = 125°. Thus, θ = φ − 90° = 125° − 90° = 35°, in agreement with the above. (b) IDENTIFY and SET UP:

I = 12 I 0 + I p cos 2 (φ − θ )

Use data at two values of φ to determine the two constants I 0 and I p . Use data where the I p term is large (φ = 30°) and where it is small (φ = 130°) to have the greatest sensitivity to both I 0 and I p :

φ = 30° gives 24.8 W/m 2 = 12 I 0 + I p cos 2 (30° − 35°)

EXECUTE:

24.8 W/m 2 = 0.500 I 0 + 0.9924 I p

φ = 130° gives 5.2 W/m 2 = 12 I 0 + I p cos 2 (130° − 35°) 5.2 W/m 2 = 0.500 I 0 + 0.0076 I p Subtracting the second equation from the first gives 19.6 W/m 2 = 0.9848I p and I p = 19.9 W/m 2 . And then I 0 = 2(5.2 W/m 2 − 0.0076(19.9 W/m 2 )) = 10.1 W/m 2 .

Now that we have I 0 , I p and θ we can verify that I = 12 I 0 + I p cos 2 (φ − θ ) describes that data in the

EVALUATE: 33.64.

table. IDENTIFY: The number of wavelengths in a distance D of material is D / λ , where λ is the wavelength of the light in the material. D D 1 = + , where we have assumed n1 > n2 so λ2 > λ1 . SET UP: The condition for a quarter-wave plate is λ1 λ2 4 EXECUTE:

λ0

n1D

λ0

=

n2 D

λ0

+

1 λ0 and D = . 4 4(n1 − n2 )

5.89 × 10−7 m = 6.14 × 10−7 m. 4(n1 − n2 ) 4(1.875 − 1.635) EVALUATE: The thickness of the quarter-wave plate in part (b) is 614 nm, which is of the same order as the wavelength in vacuum of the light. IDENTIFY: Follow the steps specified in the problem. SET UP: cos(α − β ) = sin α sin β + cosα cos β . sin(α − β ) = sin α cos β − cosα sin β . EXECUTE: (a) Multiplying Eq.(1) by sin β and Eq.(2) by sin α yields: (b) D =

33.65.

(a)

=

x y sin β = sin ω t cosα sin β − cos ω t sin α sin β and (2): sinα = sin ω t cos β sin α − cos ω t sin β sin α . a a x sin β − y sin α Subtracting yields: = sin ω t (cosα sin β − cos β sin α ). a

(1):


The Nature and Propagation of Light

33-19

(b) Multiplying Eq. (1) by cos β and Eq. (2) by cos α yields:

x y (1) : cos β = sin ω t cosα cos β − cos ω t sin α cos β and (2) : cosα = sin ω t cos β cosα − cos ω t sin β cosα . a a x cos β − y cosα Subtracting yields: = − cos ω t (sin α cos β − sin β cosα ). a (c) Squaring and adding the results of parts (a) and (b) yields: ( x sin β − y sin α ) 2 + ( x cos β − y cosα ) 2 = a 2 (sin α cos β − sin β cosα ) 2 (d) Expanding the left-hand side, we have:

x 2 (sin 2 β + cos 2 β ) + y 2 (sin 2 α + cos 2 α ) − 2 xy (sin α sin β + cosα cos β )

= x 2 + y 2 − 2 xy (sin α sin β + cosα cos β ) = x 2 + y 2 − 2 xy cos(α − β ). The right-hand side can be rewritten: a 2 (sin α cos β − sin β cosα ) 2 = a 2 sin 2 (α − β ). Therefore, x 2 + y 2 − 2 xy cos(α − β ) = a 2 sin 2 (α − β ). Or, x 2 + y 2 − 2 xy cos δ = a 2 sin 2 δ , where δ = α − β .

EVALUATE:

δ= δ= 33.66.

π 4 π 2

(e) δ = 0 : x 2 + y 2 − 2 xy = ( x − y ) 2 = 0 ⇒ x = y , which is a straight diagonal line

: x 2 + y 2 − 2 xy =

a2 , which is an ellipse 2

: x 2 + y 2 = a 2 ,which is a circle. This pattern repeats for the remaining phase differences.

IDENTIFY: Apply Snell's law to each refraction. SET UP: Refer to the figure that accompanies the problem. EXECUTE: (a) By the symmetry of the triangles, θ bA = θ aB , and θ aC = θ rB = θ aB = θ bA . Therefore,

sinθ bC = n sinθ aC = n sinθ bA = sinθ aA = θ bC = θ aA . (b) The total angular deflection of the ray is Δ = θ aA − θ bA + π − 2θ aB + θ bC − θ aC = 2θ aA − 4θ bA + π .

⎛1 ⎞ (c) From Snell’s Law, sin θ aA = n sin θ bA ⇒ θ bA = arcsin ⎜ sinθ aA ⎟ . n ⎝ ⎠ ⎛1 ⎞ Δ = 2θ aA − 4θ bA + π = 2θ aA − 4arcsin ⎜ sin θ aA ⎟ + π . ⎝n ⎠

(d)

dΔ d ⎛ ⎛1 ⎞⎞ = 0 = 2 − 4 A ⎜ arcsin ⎜ sin θ aA ⎟ ⎟ ⇒ 0 = 2 − dθ aA dθ a ⎝ n ⎝ ⎠⎠

2 2 ⎛ cosθ1 ⎞ ⎛ sin θ1 ⎞ ⎛ 16cos θ1 ⎞ ⋅⎜ =⎜ . 4 ⎜1 − ⎟ ⎟. ⎟ 2 2 n ⎠ ⎝ n sin 2 θ1 ⎝ n ⎠ ⎝ ⎠ 1− n2

4

1 4cos 2 θ1 = n 2 − 1 + cos 2 θ1 . 3cos 2 θ1 = n 2 − 1 . cos 2 θ1 = (n 2 − 1). 3 ⎛ 1 2 ⎞ ⎛ 1 ⎞ (e) For violet: θ1 = arccos ⎜⎜ ( n − 1) ⎟⎟ = arccos ⎜⎜ (1.3422 − 1) ⎟⎟ = 58.89° . ⎝ 3 ⎠ ⎝ 3 ⎠ Δ violet = 139.2° ⇒ θ violet = 40.8°.

33.67.

⎛ 1 ⎞ ⎛ 1 ⎞ For red: θ1 = arccos ⎜⎜ ( n 2 − 1) ⎟⎟ = arccos ⎜⎜ (1.3302 − 1) ⎟⎟ = 59.58° . Δ red = 137.5° ⇒ θ red = 42.5°. ⎝ 3 ⎠ ⎝ 3 ⎠ EVALUATE: The angles we have calculated agree with the values given in Figure 37.20d in the textbook. θ1 is larger for red than for violet, so red in the rainbow is higher above the horizon. IDENTIFY: Follow similar steps to Challenge Problem 33.66. SET UP: Refer to Figure 33.20e in the textbook. EXECUTE: The total angular deflection of the ray is Δ = θ aA − θ bA + π − 2θ bA + π − 2θ bA + θ aA − θ bA = 2θ aA − 6θ bA + 2π , where we have used the fact from the previous problem that all the internal angles are equal and the two external equals are equal. Also using the Snell’s Law ⎛1 ⎞ ⎛1 ⎞ relationship, we have: θ bA = arcsin ⎜ sinθ aA ⎟ . Δ = 2θ aA − 6θ bA + 2π = 2θ aA − 6arcsin ⎜ sin θ aA ⎟ + 2π . ⎝n ⎠ ⎝n ⎠


33-20

Chapter 33

(b)

dΔ d ⎛ ⎛1 ⎞⎞ = 0 = 2 − 6 A ⎜ arcsin ⎜ sin θ aA ⎟ ⎟ ⇒ 0 = 2 − dθ aA dθ a ⎝ ⎝n ⎠⎠

6 sin θ 1− 2 2 n

⎛ cosθ 2 ⎞ .⎜ ⎟. ⎝ n ⎠

⎛ sin 2 θ 2 ⎞ 1 2 2 2 2 2 n 2 ⎜1 − ⎟ = (n − 1 + cos θ 2 ) = 9cos θ 2 . cos θ 2 = (n − 1) . 2 8 n ⎠ ⎝ ⎛ 1 ⎞ ⎛ 1 ⎞ (c) For violet, θ 2 = arccos ⎜⎜ ( n 2 − 1) ⎟⎟ = arccos ⎜⎜ (1.3422 − 1) ⎟⎟ = 71.55° . Δ violet = 233.2° and θ violet = 53.2°. ⎝ 8 ⎠ ⎝ 8 ⎠ ⎛ 1 ⎞ ⎛ 1 ⎞ For red, θ 2 = arccos ⎜⎜ (n 2 − 1) ⎟⎟ = arccos ⎜⎜ (1.3302 − 1) ⎟⎟ = 71.94°. Δ red = 230.1° and θ red = 50.1°. ⎝ 8 ⎠ ⎝ 8 ⎠ EVALUATE: The angles we calculated agree with those given in Figure 37.20e in the textbook. The color that appears higher above the horizon is violet. The colors appear in reverse order in a secondary rainbow compared to a primary rainbow.


34

GEOMETRIC OPTICS

34.1.

IDENTIFY and SET UP: Plane mirror: s = − s′ (Eq.34.1) and m = y′ / y = − s′ / s = +1 (Eq.34.2). We are given s and y and are asked to find s′ and y′. EXECUTE: The object and image are shown in Figure 34.1. s′ = − s = −39.2 cm y′ = m y = ( +1)(4.85 cm)

y′ = 4.85 cm Figure 34.1

34.2.

The image is 39.2 cm to the right of the mirror and is 4.85 cm tall. EVALUATE: For a plane mirror the image is always the same distance behind the mirror as the object is in front of the mirror. The image always has the same height as the object. h d IDENTIFY: Similar triangles say tree = tree . hmirror d mirror SET UP:

d tree 28.0 m + 0.350 m = 0.040 m = 3.24 m. d mirror 0.350 m EVALUATE: The image of the tree formed by the mirror is 28.0 m behind the mirror and is 3.24 m tall. IDENTIFY: Apply the law of reflection. SET UP: If up is the +y-direction and right is the +x-direction, then the object is at (− x0 , − y0 ) and P2′ is at EXECUTE:

34.3.

d mirror = 0.350 m, hmirror = 0.0400 m and d tree = 28.0 m + 0.350 m. htree = hmirror

( x0 , − y0 ) . EXECUTE: 34.4.

34.5.

Mirror 1 flips the y-values, so the image is at ( x0 , y0 ) which is P3′.

EVALUATE: Mirror 2 uses P1′ as an object and forms an image at P3′ . IDENTIFY: f = R / 2 SET UP: For a concave mirror R > 0. R 34.0 cm EXECUTE: (a) f = = = 17.0 cm 2 2 EVALUATE: (b) The image formation by the mirror is determined by the law of reflection and that is unaffected by the medium in which the light is traveling. The focal length remains 17.0 cm. IDENTIFY and SET UP: Use Eq.(34.6) to calculate s′ and use Eq.(34.7) to calculate y′. The image is real if s′ is positive and is erect if m > 0. Concave means R and f are positive, R = +22.0 cm; f = R / 2 = +11.0 cm. EXECUTE: (a)

Three principal rays, numbered as in Sect. 34.2, are shown in Figure 34.5. The principal ray diagram shows that the image is real, inverted, and enlarged. Figure 34.5

34-1


34-2

Chapter 34

1 1 1 + = s s′ f 1 1 1 s− f sf (16.5 cm)(11.0 cm) = − = so s′ = = = +33.0 cm s′ f s sf s − f 16.5 cm − 11.0 cm s′ > 0 so real image, 33.0 cm to left of mirror vertex s′ 33.0 cm = −2.00 (m < 0 means inverted image) y′ = m y = 2.00(0.600 cm) = 1.20 cm m=− =− s 16.5 cm EVALUATE: The image is 33.0 cm to the left of the mirror vertex. It is real, inverted, and is 1.20 cm tall (enlarged). The calculation agrees with the image characterization from the principal ray diagram. A concave mirror used alone always forms a real, inverted image if s > f and the image is enlarged if f < s < 2f. s′ 1 1 1 IDENTIFY: Apply + = and m = − . s s′ f s R SET UP: For a convex mirror, R < 0 . R = −22.0 cm and f = = −11.0 cm . 2 EXECUTE: (a) The principal-ray diagram is sketched in Figure 34.6. s′ −6.6 cm 1 1 1 sf (16.5 cm)(−11.0 cm) = +0.400 . = = −6.6 cm . m = − = − (b) + = . s′ = s − f 16.5 cm − (−11.0 cm) s s′ f s 16.5 cm y′ = m y = (0.400)(0.600 cm) = 0.240 cm . The image is 6.6 cm to the right of the mirror. It is 0.240 cm tall. (b)

34.6.

s′ < 0 , so the image is virtual. m > 0 , so the image is erect. EVALUATE: The calculated image properties agree with the image characterization from the principal-ray diagram.

Figure 34.6 34.7.

y′ 1 1 1 s′ . Find m and calculate y′ . + = . m=− . m = s s′ f s y SET UP: f = +1.75 m . EXECUTE: s ! f so s′ = f = 1.75 m .

IDENTIFY:

1.75 m s′ =− = −3.14 × 10−11. y′ = m y = (3.14 × 10−11 )(6.794 × 106 m) = 2.13 × 10−4 m = 0.213 mm . 5.58 × 1010 m s EVALUATE: The image is real and is 1.75 m in front of the mirror. 1 1 1 s′ IDENTIFY: Apply + = and m = − . s s′ f s SET UP: The mirror surface is convex so R = −3.00 cm . s = 24.0 cm − 3.00 cm = 21.0 cm . R 1 1 1 sf (21.0 cm)( −1.50 cm) EXECUTE: f = = −1.50 cm . + = . s′ = = = −1.40 cm . The image is 2 s − f 21.0 cm − (−1.50 cm) s s′ f 1.40 cm behind the surface so it is 3.00 cm − 1.40 cm = 1.60 cm from the center of the ornament, on the same side s′ −1.40 cm as the object. m = − = − = +0.0667 . y′ = m y = (0.0667)(3.80 mm) = 0.253 mm . s 21.0 cm EVALUATE: The image is virtual, upright and smaller than the object. IDENTIFY: The shell behaves as a spherical mirror. SET UP: The equation relating the object and image distances to the focal length of a spherical mirror is 1 1 1 s′ + = , and its magnification is given by m = − . s s′ f s m=−

34.8.

34.9.


Geometric Optics

EXECUTE:

1 1 1 1 2 1 + = ⇒ = − ⇒ s = 18.0 cm from the vertex. s s′ f s −18.0 cm −6.00 cm

s′ −6.00 cm 1 1 =− = ⇒ y′ = (1.5 cm) = 0.50 cm . The image is 0.50 cm tall, erect, and virtual. s 18.0 cm 3 3 EVALUATE: Since the magnification is less than one, the image is smaller than the object. IDENTIFY: The bottom surface of the bowl behaves as a spherical convex mirror. SET UP: The equation relating the object and image distances to the focal length of a spherical mirror is 1 1 1 s′ + = , and its magnification is given by m = − . ′ s s s f m=−

34.10.

EXECUTE:

1 1 1 1 −2 1 + = ⇒ = − ⇒ s′ = −15 cm behind bowl. ′ ′ s s f s 35 cm 90 cm

s′ 15 cm = = 0.167 ⇒ y′ = (0.167)(2.0 cm) = 0.33 cm . The image is 0.33 cm tall, erect, and virtual. s 90 cm EVALUATE: Since the magnification is less than one, the image is smaller than the object. IDENTIFY: We are dealing with a spherical mirror. SET UP: The equation relating the object and image distances to the focal length of a spherical mirror is 1 1 1 s′ + = , and its magnification is given by m = − . s s′ f s 1 1 1 1 1 1 s− f sf s′ f EXECUTE: (a) + = ⇒ = − = ⇒ s′ = . Also m = − = . s s′ f s′ f s fs s− f s f −s (b) The graph is given in Figure 34.11a. (c) s′ > 0 for s > f , s < 0. (d) s′ < 0 for 0 < s < f . (e) The image is at negative infinity, “behind” the mirror. (f ) At the focal point, s = f. (g) The image is at the mirror, s′ = 0 . (h) The graph is given in Figure 34.11b. (i) Erect and larger if 0 < s < f. (j) Inverted if s > f . (k) The image is smaller if s > 2 f or s < 0. (l) As the object is moved closer and closer to the focal point, the magnification increases to infinite values. EVALUATE: As the object crosses the focal point, both the image distance and the magnification undergo discontinuities. m=−

34.11.

Figure 34.11

34-3


34-4

34.12.

Chapter 34

sf f and m = . s− f f −s s f f SET UP: With f = − f , s′ = − and m = . s+ f s+ f EXECUTE: The graphs are given in Figure 34.12. (a) s′ > 0 for − f < s < 0. (b) s′ < 0 for s < − f and s < 0. IDENTIFY:

s′ =

(c) If the object is at infinity, the image is at the outward going focal point. (d) If the object is next to the mirror, then the image is also at the mirror (e) The image is erect (magnification greater than zero) for s > − f . (f) The image is inverted (magnification less than zero) for s < − f . (g) The image is larger than the object (magnification greater than one) for −2 f < s < 0. (h) The image is smaller than the object (magnification less than one) for s > 0 and s < − 2 f . EVALUATE: object.

For a real image ( s > 0) , the image formed by a convex mirror is always virtual and smaller than the

Figure 34.12 34.13.

1 1 1 y′ s′ + = and m = = − . ′ y s s s f SET UP: m = +2.00 and s = 1.25 cm . An erect image must be virtual. sf f EXECUTE: (a) s′ = and m = − . For a concave mirror, m can be larger than 1.00. For a convex mirror, s− f s− f f and m is always less than 1.00. The mirror must be concave ( f > 0) . f = − f so m = + s+ f 1 s′ + s ss′ s′ s (−2.00s ) = (b) . f = . m = − = +2.00 and s′ = −2.00s . f = = +2.00s = +2.50 cm . s s − 2.00s f ss′ s + s′ R = 2 f = +5.00 cm . (c) The principal ray diagram is drawn in Figure 34.13. IDENTIFY:


Geometric Optics

EVALUATE:

34-5

The principal-ray diagram agrees with the description from the equations.

Figure 34.13 34.14.

IDENTIFY: SET UP:

Apply

1 1 1 s′ + = and m = − . s s′ f s

For a concave mirror, R > 0 . R = 32.0 cm and f =

R = 16.0 cm . 2

1 1 1 sf (12.0 cm)(16.0 cm) s′ −48.0 cm = = −48.0 cm . m = − = − = +4.00 . + = . s′ = s 12.0 cm s − f 12.0 cm − 16.0 cm s s′ f (b) s′ = −48.0 cm , so the image is 48.0 cm to the right of the mirror. s′ < 0 so the image is virtual. (c) The principal-ray diagram is sketched in Figure 34.14. The rules for principal rays apply only to paraxial rays. Principal ray 2, that travels to the mirror along a line that passes through the focus, makes a large angle with the optic axis and is not described well by the paraxial approximation. Therefore, principal ray 2 is not included in the sketch. EVALUATE: A concave mirror forms a virtual image whenever s < f . EXECUTE:

(a)

Figure 34.14 34.15.

IDENTIFY:

Apply Eq.(34.11), with R → ∞. s′ is the apparent depth.

SET UP The image and object are shown in Figure 34.15.

na nb nb − na + = ; s s′ R R → ∞ (flat surface), so na nb + =0 s s′ Figure 34.15

nb s (1.00)(3.50 cm) =− = −2.67 cm na 1.309 The apparent depth is 2.67 cm. EVALUATE: When the light goes from ice to air (larger to smaller n), it is bent away from the normal and the virtual image is closer to the surface than the object is. EXECUTE:

s′ = −


34-6

34.16.

Chapter 34

na nb + =0. s s′ SET UP: The light travels from the fish to the eye, so na = 1.333 and nb = 1.00 . When the fish is viewed, s = 7.0 cm . The fish is 20.0 cm − 7.0 cm = 13.0 cm above the mirror, so the image of the fish is 13.0 cm below the mirror and 20.0 cm + 13.0 cm = 33.0 cm below the surface of the water. When the image is viewed, s = 33.0 cm . IDENTIFY:

The surface is flat so R → ∞ and

⎛n ⎞ ⎛ 1.00 ⎞ (a) s′ = − ⎜ b ⎟ s = − ⎜ ⎟ (7.0 cm) = −5.25 cm . The apparent depth is 5.25 cm. ⎝ 1.333 ⎠ ⎝ na ⎠ ⎛n ⎞ ⎛ 1.00 ⎞ (b) s′ = − ⎜ b ⎟ s = − ⎜ ⎟ (33.0 cm) = −24.8 cm . The apparent depth of the image of the fish in the mirror is 24.8 cm. n ⎝ 1.333 ⎠ ⎝ a⎠ EVALUATE: In each case the apparent depth is less than the actual depth of what is being viewed. n s′ na nb nb − na + = . m = − a . Light comes from the fish to the person’s eye. IDENTIFY: nb s s s′ R EXECUTE:

34.17.

SET UP: R = −14.0 cm . s = +14.0 cm . na = 1.333 (water). nb = 1.00 (air). Figure 34.17 shows the object and the refracting surface. 1.333 1.00 1.00 − 1.333 (1.333)( −14.0 cm) = +1.33 . + = EXECUTE: (a) . s′ = −14.0 cm . m = − 14.0 cm s′ −14.0 cm (1.00)(14.0 cm) The fish’s image is 14.0 cm to the left of the bowl surface so is at the center of the bowl and the magnification is 1.33. n n − na (b) The focal point is at the image location when s → ∞ . b = b . na = 1.00 . nb = 1.333 . R = +14.0 cm . s′ R 1.333 1.333 − 1.00 = . s′ = +56.0 cm . s′ is greater than the diameter of the bowl, so the surface facing the sunlight s′ 14.0 cm does not focus the sunlight to a point inside the bowl. The focal point is outside the bowl and there is no danger to the fish. EVALUATE: In part (b) the rays refract when they exit the bowl back into the air so the image we calculated is not the final image.

34.18.

Figure 34.17 na nb nb − na IDENTIFY: Apply + = . s s′ R SET UP: For a convex surface, R > 0 . R = +3.00 cm . na = 1.00 , nb = 1.60 . EXECUTE:

34.19.

(a) s → ∞ .

⎛ nb ⎞ nb nb − na ⎛ 1.60 ⎞ = . s′ = ⎜ ⎟R = ⎜ ⎟ ( +3.00 cm) = +8.00 cm . The image is s′ R n − n ⎝ 1.60 − 1.00 ⎠ a ⎠ ⎝ b

8.00 cm to the right of the vertex. 1.00 1.60 1.60 − 1.00 + = (b) s = 12.0 cm . . s′ = +13.7 cm . The image is 13.7 cm to the right of the vertex. 12.0 cm s′ 3.00 cm 1.00 1.60 1.60 − 1.00 (c) s = 2.00 cm . + = . s′ = −5.33 cm . The image is 5.33 cm to the left of the vertex. 2.00 cm s′ 3.00 cm EVALUATE: The image can be either real ( s′ > 0 ) or virtual ( s′ < 0 ), depending on the distance of the object from the refracting surface. IDENTIFY: The hemispherical glass surface forms an image by refraction. The location of this image depends on the curvature of the surface and the indices of refraction of the glass and oil. SET UP: The image and object distances are related to the indices of refraction and the radius of curvature by the n n n − na equation a + b = b . s s′ R


Geometric Optics

34-7

na nb nb − na 1.45 1.60 0.15 + = ⇒ + = ⇒ s = 0.395 cm s s′ R s 1.20 m 0.0300 m EVALUATE: The presence of the oil changes the location of the image. na nb nb − na n s′ + = IDENTIFY: . m=− a . ′ s s R nb s EXECUTE:

34.20.

SET UP: EXECUTE:

R = +4.00 cm . na = 1.00 . nb = 1.60 . s = 24.0 cm . (1.00)(14.8 cm) 1 1.60 1.60 − 1.00 = −0.385 . + = . s′ = +14.8 cm . m = − (1.60)(24.0 cm) 24.0 cm s′ 4.00 cm

y′ = m y = (0.385)(1.50 mm) = 0.578 mm . The image is 14.8 cm to the right of the vertex and is 0.578 mm tall.

34.21.

m < 0 , so the image is inverted. EVALUATE: The image is real. IDENTIFY: Apply Eqs.(34.11) and (34.12). Calculate s′ and y′. The image is erect if m > 0. SET UP: The object and refracting surface are shown in Figure 34.21.

Figure 34.21

na nb nb − na + = s s′ R 1.00 1.60 1.60 − 1.00 + = 24.0 cm s′ −4.00 cm

EXECUTE:

Multiplying by 24.0 cm gives 1.00 +

38.4 = −3.60 s′

38.4 cm 38.4 cm = −4.60 and s′ = − = −8.35 cm s′ 4.60 n s′ (1.00)( −8.35 cm) = +0.217 Eq.(34.12): m = − a = − nb s (1.60)( +24.0 cm) y′ = m y = (0.217)(1.50 mm) = 0.326 mm

34.22.

34.23.

EVALUATE: The image is virtual ( s′ < 0) and is 8.35 cm to the left of the vertex. The image is erect (m > 0) and is 0.326 mm tall. R is negative since the center of curvature of the surface is on the incoming side. IDENTIFY: The hemispherical glass surface forms an image by refraction. The location of this image depends on the curvature of the surface and the indices of refraction of the glass and liquid. SET UP: The image and object distances are related to the indices of refraction and the radius of curvature by the n n n − na equation a + b = b . s s′ R na nb nb − na na 1.60 1.60 − na + = ⇒ + = ⇒ na = 1.24 . EXECUTE: ′ s s R 14.0 cm 9.00 cm 4.00 cm EVALUATE: The result is a reasonable refractive index for liquids. ⎛1 1 1 1 y′ s′ 1 1 ⎞ and m = = − . = ( n − 1) ⎜ − ⎟ to calculate f. The apply + = IDENTIFY: Use y s f s s′ f ⎝ R1 R2 ⎠ SET UP:

R1 → ∞ . R2 = −13.0 cm . If the lens is reversed, R1 = +13.0 cm and R2 → ∞ .

1 1 1 s− f 1 1 0.70 ⎛1 ⎞ = − = . = (0.70) ⎜ − and f = 18.6 cm . ⎟= f s′ f s sf ⎝ ∞ −13.0 cm ⎠ 13.0 cm sf (22.5 cm)(18.6 cm) s′ 107 cm s′ = = = 107 cm . m = − = − = −4.76 . s− f 22.5 cm − 18.6 cm s 22.5 cm y′ = my = ( −4.76)(3.75 mm) = −17.8 mm . The image is 107 cm to the right of the lens and is 17.8 mm tall. The image is real and inverted. 1 1 1⎞ ⎛ = ( n − 1) ⎜ − ⎟ and f = 18.6 cm . The image is the same as in part (a). (b) f ⎝ 13.0 cm ∞ ⎠ EVALUATE: Reversing a lens does not change the focal length of the lens. EXECUTE:

(a)


34-8

34.24.

Chapter 34

1 1 1 + = . The sign of f determines whether the lens is converging or diverging. s s′ f SET UP: s = 16.0 cm . s′ = −12.0 cm . ss′ (16.0 cm)(−12.0 cm) = = −48.0 cm . f < 0 and the lens is diverging. EXECUTE: (a) f = s + s′ 16.0 cm + ( −12.0 cm) s′ −12.0 cm (b) m = − = − = +0.750 . y′ = m y = (0.750)(8.50 mm) = 6.38 mm . m > 0 and the image is erect. s 16.0 cm (c) The principal-ray diagram is sketched in Figure 34.24. EVALUATE: A diverging lens always forms an image that is virtual, erect and reduced in size. IDENTIFY:

Figure 34.24 34.25.

IDENTIFY: SET UP: EXECUTE:

The liquid behaves like a lens, so the lensmaker’s equation applies. ⎛1 1 1 1 ⎞ s′ The lensmaker’s equation is + = (n − 1) ⎜ − ⎟ , and the magnification of the lens is m = − . s s′ R R s 2 ⎠ ⎝ 1 (a)

⎛1 ⎛ ⎞ 1 1 1 ⎞ 1 1 1 1 + = ( n − 1) ⎜ − ⎟ ⇒ + = (1.52 − 1) ⎜ − ⎟ ′ s s′ R R 24.0 cm s − 7.00 cm − 4.00 cm ⎝ ⎠ 2 ⎠ ⎝ 1 ⇒ s′ = 71.2 cm , to the right of the lens.

s′ 71.2 cm =− = −2.97 s 24.0 cm EVALUATE: Since the magnification is negative, the image is inverted. y′ s′ 1 1 1 IDENTIFY: Apply m = = − to relate s′ and s and then use + = . s s′ f y s SET UP: Since the image is inverted, y′ < 0 and m < 0 . (b) m = −

34.26.

EXECUTE:

34.27.

34.28.

m=

s′ 1 1 1 y′ −4.50 cm gives = = −1.406 . m = − gives s′ = +1.406 s . + = s s s′ f 3.20 cm y

1 1 1 + = and s = 154 cm . s′ = (1.406)(154 cm) = 217 cm . The object is 154 cm to the left of the s 1.406 s 90.0 cm lens. The image is 217 cm to the right of the lens and is real. EVALUATE: For a single lens an inverted image is always real. IDENTIFY: The thin-lens equation applies in this case. 1 1 1 s′ y ′ SET UP: The thin-lens equation is + = , and the magnification is m = − = . s s′ f s y y′ 34.0 mm s′ −12.0 cm EXECUTE: m = = = 4.25 = − = − ⇒ s = 2.82 cm . The thin-lens equation gives y 8.00 mm s s 1 1 1 + = ⇒ f = 3.69 cm . s s′ f EVALUATE: Since the focal length is positive, this is a converging lens. The image distance is negative because the object is inside the focal point of the lens. s′ 1 1 1 IDENTIFY: Apply m = − to relate s and s′ . Then use + = . s s s′ f SET UP: Since the image is to the right of the lens, s′ > 0 . s′ + s = 6.00 m . EXECUTE: (a) s′ = 80.0 s and s + s′ = 6.00 m gives 81.00s = 6.00 m and s = 0.0741 m . s′ = 5.93 m .


Geometric Optics

34-9

(b) The image is inverted since both the image and object are real ( s′ > 0, s > 0).

1 1 1 1 1 = + = + ⇒ f = 0.0732 m, and the lens is converging. f s s′ 0.0741 m 5.93 m EVALUATE: The object is close to the lens and the image is much farther from the lens. This is typical for slide projectors. ⎛1 1 1 ⎞ IDENTIFY: Apply = ( n − 1) ⎜ − ⎟ . f R R 2 ⎠ ⎝ 1 (c)

34.29.

SET UP:

For a distant object the image is at the focal point of the lens. Therefore, f = 1.87 cm . For the double-

convex lens, R1 = + R and R2 = − R , where R = 2.50 cm . EXECUTE: EVALUATE: 34.30.

1 1 ⎞ 2( n − 1) R 2.50 cm ⎛1 +1 = + 1 = 1.67 . = ( n − 1) ⎜ − . n= ⎟= f R 2f 2(1.87 cm) ⎝ R −R ⎠ f > 0 and the lens is converging. A double-convex lens is always converging.

⎛1 1 1 ⎞ = ( n − 1) ⎜ − ⎟ f R R 2 ⎠ ⎝ 1 EXECUTE: We have a converging lens if the focal length is positive, which requires ⎛1 ⎛1 1 ⎞ 1 1 ⎞ = ( n − 1) ⎜ − ⎟ > 0 ⇒ ⎜ − ⎟ > 0. This can occur in one of three ways: f ⎝ R1 R2 ⎠ ⎝ R1 R2 ⎠ IDENTIFY and SET UP:

Apply

(i) R1 and R2 both positive and R1 < R2 . (ii) R1 ≥ 0, R2 ≤ 0 (double convex and planoconvex). (iii) R1 and R2 both negative and R1 > R2 (meniscus). The three lenses in Figure 35.32a in the textbook fall into these categories. We have a diverging lens if the focal length is negative, which requires ⎛1 ⎛1 1 1 ⎞ 1 ⎞ = ( n − 1) ⎜ − ⎟ < 0 ⇒ ⎜ − ⎟ < 0. This can occur in one of three ways: f R R R R 2 ⎠ 2 ⎠ ⎝ 1 ⎝ 1 (i) R1 and R2 both positive and R1 > R2 (meniscus). (ii) R1 and R2 both negative and R2 > R1 . (iii) R1 ≤ 0, R2 ≥ 0

34.31.

34.32.

(planoconcave and double concave). The three lenses in Figure 34.32b in the textbook fall into these categories. EVALUATE: The converging lenses in Figure 34.32a are all thicker at the center than at the edges. The diverging lenses in Figure 34.32b are all thinner at the center than at the edges. 1 1 1 s′ IDENTIFY and SET UP: The equations + = and m = − apply to both thin lenses and spherical mirrors. s s′ f s EXECUTE: (a) The derivation of the equations in Exercise 34.11 is identical and one gets: 1 1 1 1 1 1 s− f sf s′ f + = ⇒ = − = ⇒ s′ = , and also m = − = . s s′ f s′ f s fs s− f s f −s (b) Again, one gets exactly the same equations for a converging lens rather than a concave mirror because the equations are identical. The difference lies in the interpretation of the results. For a lens, the outgoing side is not that on which the object lies, unlike for a mirror. So for an object on the left side of the lens, a positive image distance means that the image is on the right of the lens, and a negative image distance means that the image is on the left side of the lens. (c) Again, for Exercise 34.12, the change from a convex mirror to a diverging lens changes nothing in the exercises, except for the interpretation of the location of the images, as explained in part (b) above. EVALUATE: Concave mirrors and converging lenses both have f > 0 . Convex mirrors and diverging lenses both have f < 0 . IDENTIFY: SET UP:

1 1 1 y′ s′ + = and m = = − . s s′ f y s f = +12.0 cm and s′ = −17.0 cm . Apply

1 1 1 1 1 1 + = ⇒ = − ⇒ s = 7.0 cm. s s′ f s 12.0 cm −17.0 cm (−17.0) s′ y′ 0.800 cm m=− =− = +2.4 ⇒ y = = = +0.34 cm, so the object is 0.34 cm tall, erect, same side as the 7.2 s m +2.4 image. The principal-ray diagram is sketched in Figure 34.32.

EXECUTE:


34-10

Chapter 34

EVALUATE:

When the object is inside the focal point, a converging lens forms a virtual, enlarged image.

Figure 34.32 34.33.

IDENTIFY: Use Eq.(34.16) to calculate the object distance s. m calculated from Eq.(34.17) determines the size and orientation of the image. SET UP: f = −48.0 cm. Virtual image 17.0 cm from lens so s′ = −17.0 cm. EXECUTE:

1 1 1 1 1 1 s− f + = , so = − = s s′ f s′ f s sf

(−17.0 cm)(−48.0 cm) s′f = = +26.3 cm s′ − f −17.0 cm − (−48.0 cm) s′ −17.0 cm m=− =− = +0.646 s +26.3 cm y′ 8.00 mm y′ so y = m= = = 12.4 mm 0.646 y m The principal-ray diagram is sketched in Figure 34.33. EVALUATE: Virtual image, real object (s > 0) so image and object are on same side of lens. m > 0 so image is erect with respect to the object. The height of the object is 12.4 mm. s=

34.34.

Figure 34.33 1 1 1 IDENTIFY: Apply + = . s s′ f SET UP: The sign of f determines whether the lens is converging or diverging. s = 16.0 cm . s′ = +36.0 cm . Use s′ m = − to find the size and orientation of the image. s (16.0 cm)(36.0 cm) ss′ EXECUTE: (a) f = = = 11.1 cm . f > 0 and the lens is converging. s + s′ 16.0 cm + 36.0 cm s′ 36.0 cm (b) m = − = − = −2.25 . y′ = m y = (2.25)(8.00 mm) = 18.0 mm . m < 0 so the image is inverted. s 16.0 cm (c) The principal-ray diagram is sketched in Figure 34.34. EVALUATE: The image is real so the lens must be converging.

Figure 34.34


Geometric Optics

34.35.

34.36.

34-11

1 1 1 + = . s s′ f SET UP: The image is to be formed on the film, so s′ = +20.4 cm . 1 1 1 1 1 1 EXECUTE: + = ⇒ + = ⇒ s = 1020 cm = 10.2 m. s s′ f s 20.4 cm 20.0 cm EVALUATE: The object distance is much greater than f, so the image is just outside the focal point of the lens. 1 1 1 y′ s′ IDENTIFY: Apply + = and m = = − . ′ s s f y s SET UP: s = 3.90 m . f = 0.085 m . Apply

IDENTIFY:

1 1 1 1 1 1 0.0869 s′ 1750 mm = −39.0 mm, so + = ⇒ + = ⇒ s′ = 0.0869 m. y′ = − y = − 3.90 s s′ f 3.90 m s′ 0.085 m s it will not fit on the 24-mm × 36-mm film. EVALUATE: The image is just outside the focal point and s′ ≈ f . To have y′ = 36 mm , so that the image will fit EXECUTE:

on the film, s = − 34.37.

IDENTIFY: SET UP:

s′ . s s ! f , so s′ ≈ f .

EXECUTE:

m =

(a) m =

s′ ≈ s s′ (c) m = ≈ s EVALUATE: (b) m =

34.38.

28 mm s′ f ≈ ⇒ m = = 1.4 × 10−4. 200,000 mm s s

105 mm f ⇒m = = 5.3 × 10−4. 200,000 mm s

300 mm f ⇒m = = 1.5 × 10−3. 200,000 mm s The magnitude of the magnification increases when f increases. y′ s′ IDENTIFY: m = = s y SET UP: s ! f , so s′ ≈ f . 5.00 m s′ f (70.7 m) = 0.0372 m = 37.2 mm. y≈ y= 9.50 × 103 m s s EVALUATE: A very long focal length lens is needed to photograph a distant object. IDENTIFY and SET UP: Find the lateral magnification that results in this desired image size. Use Eq.(34.17) to relate m and s′ and Eq.(34.16) to relate s and s′ to f. 24 × 10−3 m 36 × 10−3 m EXECUTE: (a) We need m = − = −1.5 × 10−4. Alternatively, m = − = −1.5 × 10−4. 160 m 240 m s ! f so s′ ≈ f EXECUTE:

34.39.

(0.085 m)(1.75 m) s′y ≈− = 4.1 m . The person would need to stand about 4.1 m from the lens. y′ −0.036 m

y′ =

s′ f = − = −1.5 × 10−4 and f = (1.5 × 10−4 )(600 m) = 0.090 m = 90 mm. s s A smaller f means a smaller s′ and a smaller m, so with f = 85 mm the object’s image nearly fills the picture area. 36 × 10−3 m f (b) We need m = − = 3.75 × 10−3 and = −3.75 × 10−3. Then, as in part (a), s 9.6 m f = (40.0 m)(3.75 × 10−3 ) = 0.15 m = 150 mm. Therefore use the 135 mm lens. EVALUATE: When s ! f and s′ ≈ f , y′ = − f ( y / s ). For the mobile home y/s is smaller so a larger f is needed. Note that m is very small; the image is much smaller than the object. 1 1 1 IDENTIFY: Apply + = to each lens. The image of the first lens serves as the object for the second lens. s s′ f SET UP: For a distant object, s → ∞ EXECUTE: (a) s1 = ∞ ⇒ s1′ = f1 = 12 cm. Then m = −

34.40.

(b) s2 = 4.0 cm − 12 cm = −8 cm.


34-12

Chapter 34

(c)

1 1 1 1 1 1 + = ⇒ + = ⇒ s2′ = 24 cm, to the right. s s′ f −8 cm s2′ −12 cm

(d) s1 = ∞ ⇒ s′1 = f1 = 12 cm. s2 = 8.0 cm − 12 cm = −4 cm. 34.41.

34.42.

1 1 1 1 1 1 + = ⇒ + = ⇒ s2′ = 6 cm. s s′ f −4 cm s2′ −12 cm

EVALUATE: In each case the image of the first lens serves as a virtual object for the second lens, and s2 < 0 . IDENTIFY: The f-number of a lens is the ratio of its focal length to its diameter. To maintain the same exposure, the amount of light passing through the lens during the exposure must remain the same. SET UP: The f-number is f/D. 180.0 mm f EXECUTE: (a) f -number = ⇒ f -number = ⇒ f -number = f /11 . (The f-number is an integer.) 16.36 mm D (b) f/11 to f/2.8 is four steps of 2 in intensity, so one needs 1/16th the exposure. The exposure should be 1/480 s = 2.1 × 10−3 s = 2.1 ms. EVALUATE: When opening the lens from f/11 to f/2.8, the area increases by a factor of 16, so 16 times as much light is allowed in. Therefore the exposure time must be decreased by a factor of 1/16 to maintain the same exposure on the film or light receptors of a digital camera. IDENTIFY and SET UP: The square of the aperture diameter is proportional to the length of the exposure time required. 2

⎛ 1 ⎞ ⎛ 8 mm ⎞ ⎛ 1 ⎞ s ⎟⎜ s⎟ ⎟ ≈⎜ ⎜ ⎝ 30 ⎠ ⎝ 23.1 mm ⎠ ⎝ 250 ⎠ EVALUATE: An increase in the aperture diameter decreases the exposure time. 1 1 1 IDENTIFY and SET UP: Apply + = to calculate s′ . s s′ f EXECUTE: (a) A real image is formed at the film, so the lens must be convex. 1 1 1 1 s− f sf (b) + = so = and s′ = , with f = +50.0.0 mm . For s = 45 cm = 450 mm, s′ = 56 mm. For s s′ f s′ sf s− f s = ∞, s′ = f = 50 mm. The range of distances between the lens and film is 50 mm to 56 mm. EVALUATE: The lens is closer to the film when photographing more distant objects. IDENTIFY: The projector lens can be modeled as a thin lens. 1 1 1 s′ SET UP: The thin-lens equation is + = , and the magnification of the lens is m = − . s s′ f s EXECUTE:

34.43.

34.44.

EXECUTE:

(a)

1 1 1 1 1 1 + = ⇒ = + ⇒ f = 147.5 mm , so use a f = 148 mm lens. s s′ f f 0.150 m 9.00 m

s′ ⇒ m = 60 ⇒ Area = 1.44 m × 2.16 m . s EVALUATE: The lens must produce a real image to be viewed on the screen. Since the magnification comes out negative, the slides to be viewed must be placed upside down in the tray. (a) IDENTIFY: The purpose of the corrective lens is to take an object 25 cm from the eye and form a virtual image at the eye’s near point. Use Eq.(34.16) to solve for the image distance when the object distance is 25 cm. 1 1 SET UP: m = +0.3636 m (converging lens) = +2.75 diopters means f = + 2.75 f f = 36.36 cm; s = 25 cm; s′ = ? (b) m = −

34.45.

1 1 1 + = so s s′ f (25 cm)(36.36 cm) sf s′ = = = −80.0 cm 25 cm − 36.36 cm s− f The eye’s near point is 80.0 cm from the eye. (b) IDENTIFY: The purpose of the corrective lens is to take an object at infinity and form a virtual image of it at the eye’s far point. Use Eq.(34.16) to solve for the image distance when the object is at infinity. 1 1 SET UP: m = −0.7692 m (diverging lens) = −1.30 diopters means f = − 1.30 f f = −76.92 cm; s = ∞; s′ = ? EXECUTE:

EXECUTE:

1 1 1 1 1 + = and s = ∞ says = and s′ = f = −76.9 cm. The eye’s far point is 76.9 cm from the eye. s s′ f s′ f


Geometric Optics

34.46.

34-13

EVALUATE: In each case a virtual image is formed by the lens. The eye views this virtual image instead of the object. The object is at a distance where the eye can’t focus on it, but the virtual image is at a distance where the eye can focus. na nb nb − na IDENTIFY: + = s s′ R SET UP: na = 1.00 , nb = 1.40 . s = 40.0 cm , s′ = 2.60 cm .

1 1.40 0.40 + = and R = 0.710 cm . 40.0 cm 2.60 cm R EVALUATE: The cornea presents a convex surface to the object, so R > 0 . IDENTIFY: In each case the lens forms a virtual image at a distance where the eye can focus. Power in diopters equals 1/ f , where f is in meters. SET UP: In part (a), s = 25 cm and in part (b), s → ∞ . 1 1 1 1 1 1 EXECUTE: (a) = + = + ⇒ power = = +2.33 diopters. ′ f s s 0.25 m −0.600 m f EXECUTE:

34.47.

1 1 1 1 1 1 = + = + ⇒ power = = −1.67 diopters. ′ f s s ∞ −0.600 m f EVALUATE: A converging lens corrects the near vision and a diverging lens corrects the far vision. 25.0 cm 1 1 1 to calculate s for IDENTIFY: When the object is at the focal point, M = . In part (b), apply + = s s′ f f s′ = −25.0 cm . SET UP: Our calculation assumes the near point is 25.0 cm from the eye. 25.0 cm 25.0 cm EXECUTE: (a) Angular magnification M = = = 4.17. f 6.00 cm (b)

34.48.

(b)

1 1 1 1 1 1 + = ⇒ + = ⇒ s = 4.84 cm. s s′ f s −25.0 cm 6.00 cm

y y 25.0 cm 25.0 cm ,θ= and M = = = 5.17 . M is greater when the image s 25.0 cm s 4.84 cm is at the near point than when the image is at infinity. IDENTIFY: Use Eqs.(34.16) and (34.17) to calculate s and y′. (a) SET UP: f = 8.00 cm; s′ = −25.0 cm; s = ?

EVALUATE: 34.49.

34.50.

34.51.

In part (b), θ ′ =

1 1 1 1 1 1 s′ − f + = , so = − = s s′ f s f s′ s′f ( −25.0 cm)(+8.00 cm) s′f EXECUTE: s = = = +6.06 cm s′ − f −25.0 cm − 8.00 cm s′ −25.0 cm (b) m = − = − = +4.125 s 6.06 cm y′ so y′ = m y = (4.125)(1.00 mm) = 4.12 mm m = y EVALUATE: The lens allows the object to be much closer to the eye than the near point. The lense allows the eye to view an image at the near point rather than the object. s′ y ′ y′ y IDENTIFY: For a thin lens, − = , so = , and the angular size of the image equals the angular size of the s y s′ s object. y SET UP: The object has angular size θ = , with θ in radians. f y y 2.00 mm EXECUTE: θ = ⇒ f = = = 80.0 mm = 8.00 cm. θ 0.025 rad f 2.00 mm EVALUATE: If the insect is at the near point of a normal eye, its angular size is = 0.0080 rad . 250 mm IDENTIFY: The thin-lens equation applies to the magnifying lens. 1 1 1 SET UP: The thin-lens equation is + = . s s′ f


34-14

Chapter 34

EXECUTE:

34.52.

34.53.

The image is behind the lens, so s′ < 0 . The thin-lens equation gives 1 1 1 1 1 1 + = ⇒ = − ⇒ s = 4.17 cm , on the same side of the lens as the ant. s s′ f s 5.00 cm −25.0 cm

EVALUATE: Since s′ < 0 , the image will be erect. IDENTIFY: Apply Eq.(34.24). SET UP: s1′ = 160 mm + 5.0 mm = 165 mm (250 mm)s1′ (250 mm)(165 mm) EXECUTE: (a) M = = = 317. (5.00 mm)(26.0 mm) f1 f 2 0.10 mm 0.10 mm (b) The minimum separation is = = 3.15 × 10−4 mm. 317 M EVALUATE: The angular size of the image viewed by the eye when looking through the microscope is 317 times larger than if the object is viewed at the near-point of the unaided eye. (a) IDENTIFY and SET UP:

Figure 34.53

Final image is at ∞ so the object for the eyepiece is at its focal point. But the object for the eyepiece is the image of the objective so the image formed by the objective is 19.7 cm – 1.80 cm = 17.9 cm to the right of the lens. Apply Eq.(34.16) to the image formation by the objective, solve for the object distance s. f = 0.800 cm; s′ = 17.9 cm; s = ? 1 1 1 1 1 1 s′ − f + = , so = − = s s′ f s f s′ s′f s′f (17.9 cm)(+0.800 cm) EXECUTE: s = = = +8.37 mm 17.9 cm − 0.800 cm s′ − f (b) SET UP: Use Eq.(34.17). s′ 17.9 cm EXECUTE: m1 = − = − = −21.4 0.837 cm s The linear magnification of the objective is 21.4. (c) SET UP: Use Eq.(34.23): M = m1M 2 25 cm 25 cm EXECUTE: M 2 = = = 13.9 1.80 cm f2 M = m1M 2 = (−21.4)(13.9) = −297 EVALUATE: M is not accurately given by (25 cm) s1′ / f1 f 2 = 311, because the object is not quite at the focal point of the objective ( s1 = 0.837 cm and f1 = 0.800 cm). 34.54.

Eq.(34.24) can be written M = m1 M 2 =

IDENTIFY: SET UP:

s1′ M2. f1

s1′ = f1 + 120 mm

EXECUTE:

f = 16 mm : s′ = 120 mm + 16 mm = 136 mm; s = 16 mm . m1 =

s′ 136 mm = = 8.5. s 16 mm

s′ 124 mm = = 31. 4 mm s s′ 122 mm = 64 . f = 1.9 mm : s′ = 120 mm + 1.9 mm = 122 mm; s = 1.9 mm ⇒ m1 = = s 1.9 mm The eyepiece magnifies by either 5 or 10, so: (a) The maximum magnification occurs for the 1.9-mm objective and 10x eyepiece: M = m1 M e = (64)(10) = 640. f = 4 mm : s′ = 120 mm + 4 mm = 124 mm; s = 4 mm ⇒ m1 =

(b) The minimum magnification occurs for the 16-mm objective and 5x eyepiece: M = m1 M e = (8.5)(5) = 43. EVALUATE:

The smaller the focal length of the objective, the greater the overall magnification.


Geometric Optics

34.55.

34.56.

IDENTIFY: f -number = f / D SET UP: D = 1.02 m f EXECUTE: = 19.0 ⇒ f = (19.0) D = (19.0)(1.02 m) = 19.4 m. D EVALUATE: Camera lenses can also have an f-number of 19.0. For a camera lens, both the focal length and lens diameter are much smaller, but the f-number is a measure of their ratio. f IDENTIFY: For a telescope, M = − 1 . f2 SET UP:

34.57.

34-15

f 2 = 9.0 cm . The distance between the two lenses equals f1 + f 2 .

EXECUTE:

f1 + f 2 = 1.80 m ⇒ f1 = 1.80 m − 0.0900 m = 1.71 m . M = −

EVALUATE:

For a telescope, f1 ! f 2 .

(a) IDENTIFY and SET UP:

f1 171 =− = −19.0. 9.00 f2

Use Eq.(34.24), with f1 = 95.0 cm (objective) and f 2 = 15.0 cm (eyepiece).

f1 95.0 cm =− = −6.33 15.0 cm f2 (b) IDENTIFY and SET UP: Use Eq.(34.17) to calculate y′. M =−

EXECUTE:

SET UP: s = 3.00 × 103 m s′ = f1 = 95.0 cm (since s is very large, s′ ≈ f )

s′ 0.950 m =− = −3.167 × 10−4 3.00 × 103 m s y′ = m y = (3.167 × 10−4 )(60.0 m) = 0.0190 m = 1.90 cm m=−

EXECUTE:

(c) IDENTIFY: Use Eq.(34.21) and the angular magnification M obtained in part (a) to calculate θ ′. The angular size θ of the image formed by the objective (object for the eyepiece) is its height divided by its distance from the objective. 0.0190 m EXECUTE: The angular size of the object for the eyepiece is θ = = 0.0200 rad. 0.950 m 60.0 m (Note that this is also the angular size of the object for the objective: θ = = 0.0200 rad. For a thin lens 3.00 × 103 m the object and image have the same angular size and the image of the objective is the object for the eyepiece.) θ′ M= (Eq.34.21) so the angular size of the image is θ ′ = M θ = −(6.33)(0.0200 rad) = −0.127 rad (The minus

θ

34.58.

34.59.

sign shows that the final image is inverted.) EVALUATE: The lateral magnification of the objective is small; the image it forms is much smaller than the object. But the total angular magnification is larger than 1.00; the angular size of the final image viewed by the eye is 6.33 times larger than the angular size of the original object, as viewed by the unaided eye. IDENTIFY: The angle subtended by Saturn with the naked eye is the same as the angle subtended by the image of Saturn formed by the objective lens (see Fig. 34.53 in the textbook). diameter of Saturn y′ SET UP: The angle subtended by Saturn is θ = = . distance to Saturn f1 y′ 1.7 mm 0.0017 m EXECUTE: Putting in the numbers gives θ = = = = 9.4 × 10−5 rad = 0.0054° 18 m 18 m f1 EVALUATE: The angle subtended by the final image, formed by the eyepiece, would be much larger than 0.0054°. f IDENTIFY: f = R / 2 and M = − 1 . f2 SET UP:

For object and image both at infinity, f1 + f 2 equals the distance d between the two mirrors.

f 2 = 1.10 cm . R1 = 1.30 m . R1 = 0.650 m ⇒ d = f1 + f 2 = 0.661 m. 2 f 0.650 m (b) M = 1 = = 59.1. f 2 0.011 m EXECUTE:

EVALUATE:

(a) f1 =

For a telescope, f1 ! f 2 .


34-16

Chapter 34

34.60.

IDENTIFY: The primary mirror forms an image which then acts as the object for the secondary mirror. SET UP: The equation relating the object and image distances to the focal length of a spherical mirror is 1 1 1 + = . s s′ f EXECUTE: For the first image (formed by the primary mirror):

1 1 1 1 1 1 + = ⇒ = − ⇒ s′ = 2.5 m . s s′ f s′ 2.5 m ∞ For the second image (formed by the secondary mirror), the distance between the two vertices is x. Assuming that the image formed by the primary mirror is to the right of the secondary mirror, the object distance is s = x – 2.5 m and the image distance is s′ = x + 0.15 m. Therefore we have 1 1 1 1 1 1 + = ⇒ + = s s′ f x − 2.5 m x + 0.15 m −1.5 m

34.61.

34.62.

The positive root of the quadratic equation gives x = 1.7 m, which is the distance between the vertices. EVALUATE: Some light is blocked by the secondary mirror, but usually not enough to make much difference. ds ds′ IDENTIFY and SET UP: For a plane mirror s′ = − s. v = and v′ = , so v′ = −v. dt dt EXECUTE: The velocities of the object and image relative to the mirror are equal in magnitude and opposite in direction. Thus both you and your image are receding from the mirror surface at 2.40 m/s, in opposite directions. Your image is therefore moving at 4.80 m/s relative to you. EVALUATE: The result derives from the fact that for a plane mirror the image is the same distance behind the mirror as the object is in front of the mirror. IDENTIFY: Apply the law of reflection. SET UP: The image of one mirror can serve as the object for the other mirror. EXECUTE: (a) There are three images formed, as shown in Figure 34.62a. (b) The paths of rays for each image are sketched in Figure 34.62b. EVALUATE: Our results agree with Figure 34.9 in the textbook.

Figure 34.62 34.63.

IDENTIFY: Apply the law of reflection for rays from the feet to the eyes and from the top of the head to the eyes. SET UP: In Figure 34.63, ray 1 travels from the feet of the woman to her eyes and ray 2 travels from the top of her head to her eyes. The total height of the woman is h. EXECUTE: The two angles labeled θ1 are equal because of the law of reflection, as are the two angles labeled

θ 2 . Since these angles are equal, the two distances labeled y1 are equal and the two distances labeled y2 are equal. The height of the woman is hw = 2 y1 + 2 y2 . As the drawing shows, the height of the mirror is hm = y1 + y2 . Comparing, we find that hm = hw / 2 . The minimum height required is half the height of the woman.


Geometric Optics

34-17

EVALUATE: The height of the image is the same as the height of the woman, so the height of the image is twice the height of the mirror.

34.64.

IDENTIFY:

1 1 2 s′ Apply + = and m = − . s s′ R s

Figure 34.63

s′ so m is negative and s m = −2.25 . The object, mirror and wall are sketched in Figure 34.64. The sketch shows that s′ − s = 400 cm . s′ EXECUTE: m = −2.25 = − and s′ = 2.25s . s′ − s = 2.25s − s = 400 cm and s = 320 cm . s 1 1 2 1 1 2 + = . s′ = 400 cm + 320 cm = 720 cm . The mirror should be 7.20 m from the wall. + = . s s′ R 320 cm 720 cm R R = 4.43 m. EVALUATE: The focal length of the mirror is f = R / 2 = 222 cm . s > f , as it must if the image is to be real. SET UP:

34.65.

Since the image is projected onto the wall it is real and s′ > 0 . m = −

Figure 34.64 IDENTIFY: We are given the image distance, the image height, and the object height. Use Eq.(34.7) to calculate the object distance s. Then use Eq.(34.4) to calculate R. (a) SET UP: Image is to be formed on screen so is real image; s′ > 0. Mirror to screen distance is 8.00 m, so s′ s′ = +800 cm. m = − < 0 since both s and s′ are positive. s


34-18

Chapter 34

y′ 36.0 m s′ s′ 800 cm = = 60.0 and m = −60.0. Then m = − gives s = − = − = +13.3 cm. s y 0.600 cm m −60.0 1 1 2 2 s + s′ (b) + = , so = s s′ R R ss′ ⎛ ss′ ⎞ ⎛ (13.3 cm)(800 cm) ⎞ R = 2⎜ ⎟ = 2⎜ ⎟ = 26.2 cm ⎝ s + s′ ⎠ ⎝ 800 cm + 13.3 cm ⎠ EVALUATE: R is calculated to be positive, which is correct for a concave mirror. Also, in part (a) s is calculated to be positive, as it should be for a real object. 1 1 1 s′ y ′ IDENTIFY: Apply + = to calculate s′ and then use m = − = to find the height of the image. s s′ f s y R SET UP: For a convex mirror, R < 0 , so R = −18.0 cm and f = = −9.00 cm . 2 1 1 1 sf (1300 cm)(−9.00 cm) s′ −8.94 cm EXECUTE: (a) + = . s′ = = = −8.94 cm . m = − = − = 6.88 × 10−3. 1300 cm s s′ f s s − f 1300 cm − (−9.00 cm) y′ = m y = (6.88 × 10−3 )(1.5 m) = 0.0103 m = 1.03 cm . EXECUTE:

34.66.

34.67.

34.68.

34.69.

m =

(b) The height of the image is much less than the height of the car, so the car appears to be farther away than its actual distance. EVALUATE: The image formed by a convex mirror is always virtual and smaller than the object. 1 1 2 s′ IDENTIFY: Apply + = and m = − . ′ s s R s SET UP: R = +19.4 cm . 1 1 2 1 1 2 EXECUTE: (a) + = ⇒ + = ⇒ s′ = −46 cm, so the image is virtual. 8.0 cm s′ 19.4 cm s s′ R s′ −46 (b) m = − = − = 5.8, so the image is erect, and its height is y′ = (5.8) y = (5.8)(5.0 mm) = 29 mm. 8.0 s EVALUATE: (c) When the filament is 8 cm from the mirror, the image is virtual and cannot be projected onto a wall. 1 1 2 s′ IDENTIFY: Combine + = and m = − . s s′ R s SET UP: m = +2.50 . R > 0 . s′ 1 1 2 0.600 2 EXECUTE: m = − = +2.50. s′ = −2.50s . + = . = and s = 0.300 R . s −2.50 s R s R s s′ = −2.50 s = (−2.50)(0.300 R ) = −0.750 R . The object is a distance of 0.300R in front of the mirror and the image is a distance of 0.750 R behind the mirror. EVALUATE: For a single mirror an erect image is always virtual. IDENTIFY and SET UP: Apply Eqs.(34.6) and (34.7). For a virtual object s < 0. The image is real if s′ > 0. EXECUTE: (a) Convex implies R < 0; R = −24.0 cm; f = R / 2 = −12.0 cm 1 1 1 1 1 1 s− f + = , so = − = s s′ f s′ f s sf sf (−12.0 cm)s s′ = = s− f s + 12.0 cm (12.0 cm) s s is negative, so write as s = − s ; s′ = + . Thus s′ > 0 (real image) for s < 12.0 cm. Since s is negative 12.0 cm − s

this means −12.0 cm < s < 0. A real image is formed if the virtual object is closer to the mirror than the focus. s′ (b) m = − ; real image implies s′ > 0; virtual object implies s < 0. Thus m > 0 and the image is erect. s


Geometric Optics

34-19

(c) The principal-ray diagram is given in Figure 34.69.

34.70.

Figure 34.69 EVALUATE: For a real object, only virtual images are formed by a convex mirror. The virtual object considered in this problem must have been produced by some other optical element, by another lens or mirror in addition to the convex one we considered. n n n − na IDENTIFY: Apply a + b = b , with R → ∞ since the surfaces are flat. s s′ R SET UP: The image formed by the first interface serves as the object for the second interface. EXECUTE: For the water-benzene interface to get the apparent water depth:

na nb 1.33 1.50 + =0⇒ + = 0 ⇒ s′ = −7.33 cm. 6.50 cm s s′ s′ For the benzene-air interface, to get the total apparent distance to the bottom:

na nb 1.50 1 + =0⇒ + = 0 ⇒ s′ = −6.62 cm. (7.33 cm + 2.60 cm) s′ s s′

34.71.

EVALUATE: At the water-benzene interface the light refracts into material of greater refractive index and the overall effect is that the apparent depth is greater than the actual depth. ⎛1 1 ⎞ 1 IDENTIFY: The focal length is given by = ( n − 1) ⎜ − ⎟ . f ⎝ R1 R2 ⎠ SET UP: R1 = ±4.0 cm or ±8.0 cm . R2 = ±8.0 cm or ±4.0 cm . The signs are determined by the location of the center of curvature for each surface. ⎛ ⎞ 1 1 1 EXECUTE: = (0.60) ⎜ − ⎟ , so f = ±4.44 cm, ± 13.3 cm. The possible lens shapes are f ⎝ ±4.00 cm ±8.00 cm ⎠

sketched in Figure 34.71. f1 = +13.3 cm; f 2 = +4.44 cm; f3 = 4.44 cm; f 4 = −13.3 cm; f5 = −13.3 cm; f 6 = +13.3 cm;

f 7 = −4.44 cm; f8 = −4.44 cm. EVALUATE: f is the same whether the light travels through the lens from right to left or left to right, so for the pairs (1,6), (4,5) and (7,8) the focal lengths are the same.

Figure 34.71 34.72.

1 1 1 + = and the concept of principal rays. s s′ f SET UP: s = 10.0 cm . If extended backwards the ray comes from a point on the optic axis 18.0 cm from the lens and the ray is parallel to the optic axis after it passes through the lens. EXECUTE: (a) The ray is bent toward the optic axis by the lens so the lens is converging. (b) The ray is parallel to the optic axis after it passes through the lens so it comes from the focal point; f = 18.0 cm . IDENTIFY:

Apply


34-20

Chapter 34

(c) The principal ray diagram is drawn in Figure 34.72. The diagram shows that the image is 22.5 cm to the left of the lens. 1 1 1 sf (10.0 cm)(18.0 cm) (d) + = gives s′ = = = −22.5 cm . The calculated image position agrees with the s s′ f s − f 10.0 cm − 18.0 cm principal ray diagram. EVALUATE: The image is virtual. A converging lens produces a virtual image when the object is inside the focal point.

Figure 34.72 34.73.

IDENTIFY: Since the truck is moving toward the mirror, its image will also be moving toward the mirror. SET UP: The equation relating the object and image distances to the focal length of a spherical mirror is 1 1 1 + = , where f = R/2. s s′ f EXECUTE: Since the mirror is convex, f = R/2 = (–1.50 m)/2 = –0.75 m. Applying the equation for a spherical 1 1 1 fs . mirror gives + = ⇒ s′ = s s′ f s− f

Using the chain rule from calculus and the fact that v = ds/dt, we have v′ = 2

34.74.

34.75.

ds′ ds′ ds f2 = =v ( s − f )2 dt ds dt

2

⎛s− f ⎞ ⎡ 2.0 m − ( −0.75 m) ⎤ Solving for v gives v = v′ ⎜ ⎟ = (1.5 m/s) ⎢ ⎥ = 20.2 m/s . −0.75 m ⎣ ⎦ ⎝ f ⎠ This is the velocity of the truck relative to the mirror, so the truck is approaching the mirror at 20.2 m/s. You are traveling at 25 m/s, so the truck must be traveling at 25 m/s + 20.2 m/s = 45 m/s relative to the highway. EVALUATE: Even though the truck and car are moving at constant speed, the image of the truck is not moving at constant speed because its location depends on the distance from the mirror to the truck. n n IDENTIFY: In this context, the microscope just looks at an image or object. Apply a + b = 0 to the image s s′ formed by refraction at the top surface of the second plate. In this calculation the object is the bottom surface of the second plate. SET UP: The thickness of the second plate is 2.50 mm + 0.78 mm , and this is s. The image is 2.50 mm below the top surface, so s′ = −2.50 mm . na nb n 1 s 2.50 mm + 0.780 mm EXECUTE: + =0⇒ + =0⇒n=− =− = 1.31. s s′ s s′ s′ −2.50 mm EVALUATE: The object and image distances are measured from the front surface of the second plate, and the image is virtual. 1 1 1 s′ IDENTIFY and SET UP: In part (a) use + = to evaluate ds′ / ds . Compare to m = − . In part (b) use s s′ f s 1 1 2 + = to find the location of the image of each face of the cube. s s′ R 1 1 1 d ⎛1 1 1 ⎞ 1 1 ds′ and taking its derivative with respect to s we have 0 = ⎜ + − ⎟ = − 2 − 2 EXECUTE: (a) + = ds ⎝ s s′ f ⎠ s s′ ds s s′ f ds′ s′ 2 ds′ and = − 2 = − m 2 . But = m′ , so m′ = − m 2 . Images are always inverted longitudinally. ds s ds 1 1 2 1 1 2 (b) (i) Front face: + = ⇒ + = ⇒ s′ = 120.00 cm. s s′ R 200.000 cm s′ 150.000 cm


Geometric Optics

34-21

1 1 2 1 1 2 + = ⇒ + = ⇒ s′ = 119.96 cm. ′ ′ s s R 200.100 cm s 150.000 cm s′ 120.000 2 (ii) m = − = − = −0.600 . m′ = − m 2 = − ( −0.600 ) = −0.360. s 200.000 (iii) The two faces perpendicular to the axis (the front and rear faces): squares with side length 0.600 mm. The four faces parallel to the axis (the side faces): rectangles with sides of length 0.360 mm parallel to the axis and 0.600 mm perpendicular to the axis. EVALUATE: Since the lateral and longitudinal magnifications have different values the image of the cube is not a cube. n s′ IDENTIFY: m′ = ds′ / ds and m = − a . nb s na nb nb − na SET UP: Use + = to evaluate ds′ / ds . s s′ R na nb nb − na EXECUTE: + = and taking its derivative with respect to s we have s s′ R ⎛ s′ 2 n 2 ⎞ n d ⎛ n n n − na ⎞ n n ds′ ds′ s′ 2 n n = − a2 − b2 and 0= ⎜ a + b − b = − 2 a = − ⎜ 2 a2 ⎟ b = − m 2 b . ⎟ ′ ′ ds s nb na ds ⎝ s s R ⎠ s s ds ⎝ s nb ⎠ na n ds′ But = m′, so m′ = − m 2 b . na ds EVALUATE: m′ is always negative. This means that images are always inverted longitudinally. s′ IDENTIFY and SET UP: Rays that pass through the hole are undeflected. All other rays are blocked. m = − . s EXECUTE: (a) The ray diagram is drawn in Figure 34.77. The ray shown is the only ray from the top of the object that reaches the film, so this ray passes through the top of the image. An inverted image is formed on the far side of the box, no matter how far this side is from the pinhole and no matter how far the object is from the pinhole. 20.0 cm s′ (b) s = 1.5 m . s′ = 20.0 cm . m = − = − = −0.133 . y′ = my = ( −0.133)(18 cm) = −2.4 cm . The image is 150 cm s 2.4 cm tall. EVALUATE: A defect of this camera is that not much light energy passes through the small hole each second, so long exposure times are required. Rear face:

34.76.

34.77.

34.78.

Figure 34.77 na s′ na nb nb − na to each refraction. The overall magnification is m = m1m2 . IDENTIFY: Apply + = and m = − s s′ R nb s SET UP: For the first refraction, R = +6.0 cm , na = 1.00 and nb = 1.60 . For the second refraction, R = −12.0 cm , na = 1.60 and nb = 1.00 . EXECUTE: (a) The image from the left end acts as the object for the right end of the rod. n n n − na 1 1.60 0.60 ⇒ + = ⇒ s′ = 28.3 cm. (b) a + b = b s s′ R 23.0 cm s′ 6.0 cm n s′ 28.3 So the second object distance is s2 = 40.0 cm − 28.3 cm = 11.7 cm. m1 = − a = − = −0.769. nb s 1.60 ( )( 23.0 ) (c) The object is real and inverted. n n n − na 1.60 1 −0.60 ⇒ + = ⇒ s′ = −11.5 cm. (d) a + b = b s2 s′2 R 11.7 cm s′2 −12.0 cm n s′ (1.60 )( −11.5) = 1.57 ⇒ m = m m = −0.769 1.57 = −1.21. m2 = − a = − ( )( ) 1 2 nb s 11.7 (e) The final image is virtual, and inverted.


34-22

Chapter 34

(f) y′ = (1.50 mm )( −1.21) = −1.82 mm.

34.79.

EVALUATE: The first image is to the left of the second surface, so it serves as a real object for the second surface, with positive object distance. IDENTIFY: Apply Eqs.(34.11) and (34.12) to the refraction as the light enters the rod and as it leaves the rod. The image formed by the first surface serves as the object for the second surface. The total magnification is mtot = m1m2 , where m1 and m2 are the magnifications for each surface. SET UP: The object and rod are shown in Figure 34.79.

Figure 34.79 (a) image formed by refraction at first surface (left end of rod): s = +23.0 cm; na = 1.00; nb = 1.60; R = +6.00 cm na nb nb − na + = s s′ R 1 1.60 1.60 − 1.00 + = 23.0 cm s′ 6.00 cm 1.60 1 1 23 − 10 13 = − = = s′ 10.0 cm 23.0 cm 230 cm 230 cm ⎛ 230 cm ⎞ s′ = 1.60 ⎜ ⎟ = +28.3 cm; image is 28.3 cm to right of first vertex. ⎝ 13 ⎠ This image serves as the object for the refraction at the second surface (right-hand end of rod). It is 28.3 cm − 25.0 cm = 3.3 cm to the right of the second vertex. For the second surface s = −3.3 cm (virtual object). (b) EVALUATE: Object is on side of outgoing light, so is a virtual object. (c) SET UP: Image formed by refraction at second surface (right end of rod): s = −3.3 cm; na = 1.60; nb = 1.00; R = −12.0 cm

EXECUTE:

na nb nb − na + = s s′ R 1.60 1.00 1.00 − 1.60 EXECUTE: + = −3.3 cm s′ −12.0 cm s′ = +1.9 cm; s′ > 0 so image is 1.9 cm to right of vertex at right-hand end of rod. (d) s′ > 0 so final image is real. Magnification for first surface: n s′ (1.60)(+28.3 cm) m=− a =− = −0.769 nb s (1.00)(+23.0 cm) Magnification for second surface: n s′ (1.60)(+1.9 cm) m=− a =− = +0.92 nb s (1.00)( −3.3 cm)

34.80.

The overall magnification is mtot = m1m2 = (−0.769)(+0.92) = −0.71 mtot < 0 so final image is inverted with respect to the original object. (e) y′ = mtot y = ( −0.71)(1.50 mm) = −1.06 mm The final image has a height of 1.06 mm. EVALUATE: The two refracting surfaces are not close together and Eq.(34.18) does not apply. 1 1 1 y′ s′ and m = = − . The type of lens determines the sign of f. The sign of IDENTIFY: Apply + = y s s s′ f s′ determines whether the image is real or virtual. SET UP: s = +8.00 cm . s′ = −3.00 cm . s′ is negative because the image is on the same side of the lens as the object. 1 s + s′ ss′ (8.00 cm)(−3.00 cm) = = −4.80 cm . f is negative so the lens is = and f = EXECUTE: (a) ′ ′ s+s 8.00 cm − 3.00 cm f ss diverging.


Geometric Optics

34-23

s′ −3.00 cm =− = +0.375 . y′ = my = (0.375)(6.50 mm) = 2.44 mm . s′ < 0 and the image is virtual. s 8.00 cm EVALUATE: A converging lens can also form a virtual image, if the object distance is less than the focal length. But in that case s′ > s and the image would be farther from the lens than the object is. (b) m = −

34.81.

34.82.

1 1 1 y′ s′ + = . The type of lens determines the sign of f. m = = − . The sign of s′ depends on y s s s′ f whether the image is real or virtual. s = 16.0 cm . SET UP: s′ = −22.0 cm ; s′ is negative because the image is on the same side of the lens as the object. 1 s + s′ ss′ (16.0 cm)(−22.0 cm) and f = EXECUTE: (a) = = = +58.7 cm . f is positive so the lens is converging. 16.0 cm − 22.0 cm s + s′ f ss′ s′ −22.0 cm = 1.38 . y′ = my = (1.38)(3.25 mm) = 4.48 mm . s′ < 0 and the image is virtual. (b) m = − = − s 16.0 cm EVALUATE: A converging lens forms a virtual image when the object is closer to the lens than the focal point. n n n − na . Use the image distance when viewed from the flat end to determine the IDENTIFY: Apply a + b = b s s′ R refractive index n of the rod. SET UP: When viewing from the flat end, na = n , nb = 1.00 and R → ∞ . When viewing from the curved end, IDENTIFY:

na = n , nb = 1.00 and R = −10.0 cm . EXECUTE:

na nb n 1 15.0 + =0⇒ + =0⇒n= = 1.58. When viewed from the curved end of the s s′ 15.0 cm −9.50 cm 9.50

na nb nb − na n 1 1− n 1.58 1 −0.58 + = ⇒ + = ⇒ + = , and s′ = −21.1 cm . The image is 21.1 cm s s′ R s s′ R 15.0 cm s′ −10.0 cm within the rod from the curved end. EVALUATE: In each case the image is virtual and on the same side of the surface as the object. (a) IDENTIFY: Apply Snell’s law to the refraction of a ray at each side of the beam to find where these rays strike the table. SET UP: The path of a ray is sketched in Figure 34.83. rod

34.83.

Figure 34.83

The width of the incident beam is exaggerated in the sketch, to make it easier to draw. Since the diameter of the beam is much less than the radius of the hemisphere, angles θ a and θb are small. The diameter of the circle of light formed on the table is 2 x. Note the two right triangles containing the angles θ a and θb . r = 0.190 cm is the radius of the incident beam. R = 12.0 cm is the radius of the glass hemisphere. r x′ x EXECUTE: θ a and θb small imply x ≈ x′; sin θ a = , sin θ b = ≈ R R R Snell’s law: na sin θ a = nb sin θb Using the above expressions for sin θ a and sin θb gives na

r x = nb R R

na r 1.00(0.190 cm) = = 0.1267 cm nb 1.50 The diameter of the circle on the table is 2 x = 2(0.1267 cm) = 0.253 cm. (b) EVALUATE: R divides out of the expression; the result for the diameter of the spot is independent of the radius R of the hemisphere. It depends only on the diameter of the incident beam and the index of refraction of the glass. na r = nb x so x =


34-24

Chapter 34

34.84.

IDENTIFY and SET UP:

Treating each of the goblet surfaces as spherical surfaces, we have to pass, from left to n n n − na to each surface. The image formed by one surface serves as right, through four interfaces. Apply a + b = b ′ s s R the object for the next surface. EXECUTE: (a) For the empty goblet: na nb nb − na 1 1.50 0.50 + = ⇒ + = ⇒ s1′ = 12 cm . ′ ′ s s R s1 4.00 cm ∞ s2 = 0.60 cm − 12 cm = −11.4 cm ⇒

1.50 1 −0.50 + = ⇒ s′2 = −64.6 cm. ′ −11.4 cm s2 3.40 cm

s3 = 64.6 cm + 6.80 cm = 71.4 cm ⇒

1 1.50 0.50 + = ⇒ s3′ = −9.31 cm. ′ 71.4 cm s3 −3.40 cm

1.50 1 −0.50 + = ⇒ s′4 = −37.9 cm. The final image is ′ 9.91 cm s4 −4.00 cm 37.9 cm − 2(4.0 cm) = 29.9 cm to the left of the goblet. (b) For the wine-filled goblet: na nb nb − na 1 1.50 0.50 + = ⇒ + = ⇒ s1′ = 12 cm . s s′ R s1′ 4.00 cm ∞ s4 = 9.31 cm + 0.60 cm = 9.91 cm ⇒

s2 = 0.60 cm − 12 cm = −11.4 cm ⇒

1.50 1.37 −0.13 + = ⇒ s′2 = 14.7 cm. s2′ 3.40 cm −11.4 cm

s3 = 6.80 cm − 14.7 cm = −7.9 cm ⇒

1.37 1.50 0.13 + = ⇒ s3′ = 11.1 cm. s3′ −7.9 cm −3.40 cm

s4 = 0.60 cm − 11.1 cm = −10.5 cm ⇒

34.85.

34.86.

1.50 1 −0.50 + = ⇒ s′4 = 3.73 cm . The final image is 3.73 cm to the −10.5 cm s′4 −4.00 cm

right of the goblet. EVALUATE: If the object for a surface is on the outgoing side of the light, then the object is virtual and the object distance is negative. n n n − na IDENTIFY: The image formed by refraction at the surface of the eye is located by a + b = b . s s′ R 1 SET UP: na = 1.00 , nb = 1.35 . R > 0 . For a distant object, s ≈ ∞ and ≈ 0 . s 1.35 1.35 − 1.00 = and R = 0.648 cm = 6.48 mm . EXECUTE: (a) s ≈ ∞ and s′ = 2.5 cm : 2.5 cm R 1.00 1.35 1.35 − 1.00 1.35 + = (b) R = 0.648 cm and s = 25 cm : . = 0.500 and s′ = 2.70 cm = 27.0 mm . The s′ s′ 25 cm 0.648 image is formed behind the retina. 1.35 1.35 − 1.00 = (c) Calculate s′ for s ≈ ∞ and R = 0.50 cm : . s′ = 1.93 cm = 19.3 mm . The image is formed in s′ 0.50 cm front of the retina. EVALUATE: The cornea alone cannot achieve focus of both close and distant objects. n s′ n n n − na IDENTIFY: Apply a + b = b and m = − a to each surface. The overall magnification is m = m1m2 . The nb s s s′ R image formed by the first surface is the object for the second surface. SET UP: For the first surface, na = 1.00, nb = 1.60 and R = +15.0 cm . For the second surface, na = 1.60, nb = 1.00 and R =→ ∞ . n n n − na 1 1.60 0.60 EXECUTE: (a) a + b = b ⇒ + = ⇒ s′ = −36.9 cm. The object distance for the far end ′ ′ s s R 12.0 cm s 15.0 cm of the rod is 50.0 cm − (−36.9 cm) = 86.9 cm. The final image is 4.3 cm to the left of the vertex of the n n n − na 1.60 1 hemispherical surface. a + b = b ⇒ + = 0 ⇒ s′ = −54.3 cm. s s′ R 86.9 cm s′ (b) The magnification is the product of the two magnifications: n s′ −36.9 m1 = − a = − = 1.92, m2 = 1.00 ⇒ m = m1m2 = 1.92. nb s (1.60)(12.0) EVALUATE: The final image is virtual, erect and larger than the object.


Geometric Optics

34.87.

34.88.

34-25

IDENTIFY: Apply Eq.(34.11) to the image formed by refraction at the front surface of the sphere. SET UP: Let ng be the index of refraction of the glass. The image formation is shown in Figure 34.87. s=∞ s′ = +2r , where r is the radius of the sphere na = 1.00, nb = ng , R = + r Figure 34.87 na nb nb − na + = s s′ R 1 ng ng − 1.00 + = EXECUTE: ∞ 2r r n g ng 1 ng 1 = − ; = and ng = 2.00 2r r r 2r r EVALUATE: The required refractive index of the glass does not depend on the radius of the sphere. n n n − na to each surface. The image of the first surface is the object for the second IDENTIFY: Apply a + b = b ′ s s R surface. The relation between s1′ and s2 involves the length d of the rod. SET UP:

For the first surface, na = 1.00 , nb = 1.55 and R = +6.00 cm . For the second surface, na = 1.55 ,

nb = 1.00 and R = −6.00 cm . EXECUTE: We have images formed from both ends. From the first surface: na nb nb − na 1 1.55 0.55 + = ⇒ + = ⇒ s′ = 30.0 cm. s s′ R 25.0 cm s′ 6.00 cm This image becomes the object for the second end:

34.89.

d − 30.0 cm = 20.3 cm ⇒ d = 50.3 cm. EVALUATE: The final image is real. The first image is 20.3 cm to the right of the second surface and serves as a real object. IDENTIFY: The first lens forms an image which then acts as the object for the second lens. 1 1 1 s′ SET UP: The thin-lens equation is + = and the magnification is m = − . s s s′ f

EXECUTE:

34.90.

na nb nb − na 1.55 1 −0.55 + = ⇒ + = . s s′ R d − 30.0 cm 65.0 cm −6.00 cm

(a) For the first lens:

1 1 1 1 1 1 + = ⇒ + = ⇒ s′ = −3.75 cm , to the left of the lens s s′ f 5.00 cm s′ −15.0 cm

(virtual image). (b) For the second lens, s = 12.0 cm + 3.75 cm = 15.75 cm. 1 1 1 1 1 1 + = ⇒ + = ⇒ s′ = 315 cm , or 332 cm from the object. s s′ f 15.75 cm s′ 15.0 cm (c) The final image is real. s′ (d) m = − , m1 = 0.750, m2 = −20.0, mtotal = −15.0 ⇒ y′ = −6.00 cm, inverted. s EVALUATE: Note that the total magnification is the product of the individual magnifications. ⎛1 1 ⎞ 1 IDENTIFY and SET UP: Use = ( n − 1) ⎜ − ⎟ to calculate the focal length of the lenses. The image formed f ⎝ R1 R2 ⎠ 1 1 1 sf by the first lens serves at the object for the second lens. mtot = m1m2 . + = gives s′ = . s s′ f s− f 1 1 1 ⎞ ⎛ EXECUTE: (a) = (0.60) ⎜ − ⎟ and f = +35.0 cm . f 12.0 cm 28.0 cm ⎝ ⎠ s f (45.0 cm)(35.0 cm) Lens 1: f1 = +35.0 cm . s1 = +45.0 cm . s1′ = 1 1 = = +158 cm . s1 − f1 45.0 cm − 35.0 cm s1′ 158 cm =− = −3.51 . y1′ = m1 y1 = (3.51)(5.00 mm) = 17.6 mm . The image of the first lens is 158 cm s1 45.0 cm to the right of lens 1 and is 17.6 mm tall. m1 = −


34-26

Chapter 34

(b) The image of lens 1 is 315 cm − 158 cm = 157 cm to the left of lens 2. f 2 = +35.0 cm . s2 = +157 cm .

s2 f 2 (157 cm)(35.0 cm) s′ 45.0 cm = −0.287 . = = +45.0 cm . m2 = − 2 = − s2 − f 2 157 cm − 35.0 cm s2 157 cm mtot = m1m2 = ( −3.51)(−0.287) = +1.00 . The final image is 45.0 cm to the right of lens 2. The final image is 5.00 s2′ =

34.91.

mm tall. mtot > 0 . So the final image is erect. EVALUATE: The final image is real. It is erect because each lens produces an inversion of the image, and two inversions return the image to the orientation of the object. IDENTIFY and SET UP: Apply Eq.(34.16) for each lens position. The lens to screen distance in each case is the image distance. There are two unknowns, the original object distance x and the focal length f of the lens. But each lens position gives an equation, so there are two equations for these two unknowns. The object, lens and screen before and after the lens is moved are shown in Figure 34.91. s = x; s′ = 30.0 cm 1 1 1 + = s s′ f 1 1 1 + = x 30.0 cm f

34.92.

34.93.

Figure 34.91 s = x + 4.00 cm; s′ = 22.0 cm 1 1 1 1 1 1 + = gives + = s s′ f x + 4.00 cm 22.0 cm f EXECUTE: Equate these two expressions for 1/f: 1 1 1 1 + = + x 30.0 cm x + 4.00 cm 22.0 cm 1 1 1 1 − = − x x + 4.00 cm 22.0 cm 30.0 cm x + 4.00 cm − x 30.0 − 22.0 4.00 cm 8 = and = x( x + 4.00 cm) 660 cm x( x + 4.00 cm) 660 cm 1 x 2 + (4.00 cm)x − 330 cm 2 = 0 and x = (−4.00 ± 16.0 + 4(330)) cm 2 1 x must be positive so x = (−4.00 + 36.55) cm = 16.28 cm 2 1 1 1 1 1 1 Then + = + = and f 16.28 cm 30.0 cm x 30.0 cm f f = +10.55 cm, which rounds to 10.6 cm. f > 0; the lens is converging. EVALUATE: We can check that s = 16.28 cm and f = 10.55 cm gives s′ = 30.0 cm and that s = (16.28 + 4.0) cm = 20.28 cm and f = 10.55 cm gives s′ = 22.0 cm. n n n − na IDENTIFY and SET UP: Apply a + b = b . s s′ R n n n − na n n n −n n n n −n ⇒ a + b = b a and a + b = b a . EXECUTE: (a) a + b = b ′ ′ s s R f ∞ R ∞ f R na nb − na nb nb − na na nb f = = . Therefore, and = and na / nb = . f R f′ R f f′ f′ na nb nb − na nb f nb nb (1 − f f ′) f f ′ f ′(1 − f f ′) f ′ − f + = ⇒ + = . Therefore, + = (b) = = 1. s s′ R sf ′ s′ R s s′ R R EVALUATE: For a thin lens the first and second focal lengths are equal. (a) IDENTIFY: Use Eq.(34.6) to locate the image formed by each mirror. The image formed by the first mirror serves as the object for the 2nd mirror.


Geometric Optics

SET UP:

34-27

The positions of the object and the two mirrors are shown in Figure 34.93a.

R = 0.360 m f = R / 2 = 0.180 m

Figure 34.93a EXECUTE: Image formed by convex mirror (mirror #1): convex means f1 = −0.180 m; s1 = L − x s1 f1 ( L − x )( −0.180 m) ⎛ 0.600 m − x ⎞ = = −(0.180 m) ⎜ ⎟<0 s1 − f1 L − x + 0.180 m ⎝ 0.780 m − x ⎠ ⎛ 0.600 m − x ⎞ The image is (0.180 m) ⎜ ⎟ to the left of mirror #1 so is ⎝ 0.780 m − x ⎠ 2 ⎛ 0.600 m − x ⎞ 0.576 m − (0.780 m)x to the left of mirror #2. 0.600 m + (0.180 m) ⎜ = ⎟ 0.780 m − x ⎝ 0.780 m − x ⎠ Image formed by concave mirror (mirror #2); concave implies f 2 = +0.180 m s1′ =

s2 =

0.576 m 2 − (0.780 m) x 0.780 m − x

Rays return to the source implies s2′ = x. Using these expressions in s2 =

s′2 f 2 gives s2′ − f 2

0.576 m 2 − (0.780 m)x (0.180 m)x = 0.780 m − x x − 0.180 m 0.600 x 2 − (0.576 m)x + 0.10368 m2 = 0 1 1 x= (0.576 ± (0.576)2 − 4(0.600)(0.10368) ) m = (0.576 ± 0.288) m 1.20 1.20 x = 0.72 m (impossible; can’t have x > L = 0.600 m) or x = 0.24 m. (b) SET UP: Which mirror is #1 and which is #2 is now reversed form part (a). This is shown in Figure 34.93b.

Figure 34.93b EXECUTE: Image formed by concave mirror (mirror #1): concave means f1 = +0.180 m; s1 = x s f (0.180 m)x s1′ = 1 1 = s1 − f1 x − 0.180 m (0.180 m)x (0.180 m) x (0.420 m)x − 0.180 m 2 The image is to the left of mirror #1, so s2 = 0.600 m − = x − 0.180 m x − 0.180 m x − 0.180 m Image formed by convex mirror (mirror #2): convex means f 2 = −0.180 m rays return to the source means s′2 = L − x = 0.600 m − x 1 1 1 + = gives s s′ f x − 0.180 m 1 1 + =− 2 (0.420 m)x − 0.180 m 0.600 m − x 0.180 m

⎛ ⎞ x − 0.180 m 0.780 m − x = −⎜ ⎟ 2 2 (0.420 m)x − 0.180 m ⎝ 0.180 m − (0.180 m)x ⎠ 0.600 x 2 − (0.576 m)x + 0.1036 m 2 = 0 This is the same quadratic equation as obtained in part (a), so again x = 0.24 m.


34-28

34.94.

Chapter 34

EVALUATE: For x = 0.24 m the image is at the location of the source, both for rays that initially travel from the source toward the left and for rays that travel from the source toward the right. 1 1 1 sf + = gives s′ = IDENTIFY: , for both the mirror and the lens. s s′ f s− f SET UP: For the second image, the image formed by the mirror serves as the object for the lens. For the mirror, f m = +10.0 cm . For the lens, f = 32.0 cm . The center of curvature of the mirror is R = 2 f m = 20.0 cm to the right of the mirror vertex. EXECUTE: (a) The principal-ray diagrams from the two images are sketched in Figures 34.94a-b. In Figure 34.94b, only the image formed by the mirror is shown. This image is at the location of the candle so the principal ray diagram that shows the image formation when the image of the mirror serves as the object for the lens is analogous to that in Figure 34.94a and is not drawn. (b) Image formed by the light that passes directly through the lens: The candle is 85.0 cm to the left of the lens. sf (85.0 cm)(32.0 cm) s′ 51.3 cm s′ = = = +51.3 cm . m = − = − = −0.604 . This image is 51.3 cm to the right of s− f 85.0 cm − 32.0 cm s 85.0 cm the lens. s′ > 0 so the image is real. m < 0 so the image is inverted. Image formed by the light that first reflects off the mirror: First consider the image formed by the mirror. The candle is 20.0 cm to the right of the mirror, so sf (20.0 cm)(10.0 cm) s′ 20.0 cm = −1.00 . The image formed by s = +20.0 cm . s′ = = = 20.0 cm . m1 = − 1 = − s− f 20.0 cm − 10.0 cm s1 20.0 cm

the mirror is at the location of the candle, so s2 = +85.0 cm and s2′ = 51.3 cm . m2 = −0.604 . mtot = m1m2 =

(−1.00)(−0.604) = 0.604 . The second image is 51.3 cm to the right of the lens. s2′ > 0, so the final image is real. mtot > 0 , so the final image is erect. EVALUATE: The two images are at the same place. They are the same size. One is erect and one is inverted.

34.95.

Figure 34.94 na nb nb − na IDENTIFY: Apply to each case. + = s s′ R SET UP: s = 20.0 cm . R > 0 . Use s′ = +9.12 cm to find R. For this calculation, na = 1.00 and nb = 1.55 . Then

repeat the calculation with na = 1.33 . EXECUTE:

na nb nb − na 1.00 1.55 1.55 − 1.00 + = gives + = . R = 2.50 cm . s s′ R 20.0 cm 9.12 cm R


Geometric Optics

34-29

1.33 1.55 1.55 − 1.33 + = gives s′ = −72.1 cm . The image is 72.1 cm to the left of the surface vertex. 20.0 cm s′ 2.50 cm EVALUATE: With the rod in air the image is real and with the rod in water the image is virtual. 1 1 1 to each lens. The image formed by the first lens serves as the object for the second IDENTIFY: Apply + = s s′ f Then

34.96.

lens. The focal length of the lens combination is defined by

⎛1 1 ⎞ 1 1 1 1 = ( n − 1) ⎜ − ⎟ to + = . In part (b) use ′ s1 s2 f f ⎝ R1 R2 ⎠

calculate f for the meniscus lens and for the CCl 4 , treated as a thin lens. SET UP: With two lenses of different focal length in contact, the image distance from the first lens becomes exactly minus the object distance for the second lens. 1 1 1 1 1 1 1 1 1 1 ⎛1 1⎞ 1 1 + = + = ⎜ − ⎟ + = . But overall for the lens EXECUTE: (a) + = ⇒ = − and ′ ′ ′ ′ s2 s2 − s1 s′2 ⎝ s1 f1 ⎠ s′2 f 2 s1 s1 f1 s1 f1 s1 1 1 1 1 1 1 + = ⇒ = + . s1 s2′ f f f 2 f1 (b) With carbon tetrachloride sitting in a meniscus lens, we have two lenses in contact. All we need in order to calculate the system’s focal length is calculate the individual focal lengths, and then use the formula from part (a). ⎛1 ⎛ ⎞ 1 1 ⎞ 1 1 −1 = (nb − na ) ⎜ − ⎟ = (0.55)⎜ − For the meniscus lens ⎟ = 0.061 cm and f m = 16.4 cm . fm ⎝ 4.50 cm 9.00 cm ⎠ ⎝ R1 R2 ⎠ system,

For the CCl 4 :

⎛1 ⎛ 1 1 ⎞ 1 1⎞ = (nb − na )⎜ − ⎟ = (0.46)⎜ − ⎟ = 0.051 cm −1 and f w = 19.6 cm . fw ⎝ 9.00 cm ∞ ⎠ ⎝ R1 R2 ⎠

1 1 1 = + = 0.112 cm −1 and f = 8.93 cm . f fw fm f1 f 2 , so f for the combination is less than either f1 or f 2 . f1 + f 2 IDENTIFY: Apply Eq.(34.11) with R → ∞ to the refraction at each surface. For refraction at the first surface the point P serves as a virtual object. The image formed by the first refraction serves as the object for the second refraction. SET UP: The glass plate and the two points are shown in Figure 37.97. plane faces means R → ∞ and na nb + =0 s s′ n s′ = − b s na EVALUATE:

34.97.

f =

Figure 34.97 EXECUTE: refraction at the first (left-hand) surface of the piece of glass: The rays converging toward point P constitute a virtual object for this surface, so s = −14.4 cm. na = 1.00, nb = 1.60.

1.60 (−14.4 cm) = +23.0 cm 1.00 This image is 23.0 cm to the right of the first surface so is a distance 23.0 cm − t to the right of the second surface. This image serves as a virtual object for the second surface. refraction at the second (right-hand) surface of the piece of glass: n The image is at P′ so s′ = 14.4 cm + 0.30 cm − t = 14.7 cm − t. s = −(23.0 cm − t ); na = 1.60; nb = 1.00 s′ = − b s na s′ = −

⎛ 1.00 ⎞ gives 14.7 cm − t = − ⎜ ⎟ (−[23.0 cm − t ]). 14.7 cm − t = +14.4 cm − 0.625t. ⎝ 1.60 ⎠ 0.375t = 0.30 cm and t = 0.80 cm EVALUATE: The overall effect of the piece of glass is to diverge the rays and move their convergence point to the right. For a real object, refraction at a plane surface always produces a virtual image, but with a virtual object the image can be real.


34-30

Chapter 34

34.98.

IDENTIFY: SET UP:

na nb nb − na n n n −n and b + c = c b . + = s1 s1′ R1 s2 s2′ R2 ′ , n = n , and s = − s . na = nliq = nc b 1 2

Apply the two equations

⎛1 1 ⎞ n n − nliq n nliq nliq − n 1 1 1 1 1 + = + = = (n nliq − 1)⎜ − ⎟ . = + = and . s1 s′2 s s′ f ′ R R s1′ R1 − s1′ s2′ R2 ⎝ 1 2 ⎠ (b) Comparing the equations for focal length in and out of air we have: EXECUTE:

(a)

nliq s1

+

⎛ n − nliq ⎞ ⎡ n (n − 1) ⎤ f ( n − 1) = f ′(n nliq − 1) = f ′ ⎜ ⇒ f ′ = ⎢ liq ⎥ f. ⎜ nliq ⎟⎟ ⎢⎣ n − nliq ⎥⎦ ⎝ ⎠ EVALUATE:

When nliq = 1 , f ′ = f , as it should.

1 1 1 + = . s s′ f SET UP: The image formed by the converging lens is 30.0 cm from the converging lens, and becomes a virtual object for the diverging lens at a position 15.0 cm to the right of the diverging lens. The final image is projected 15 cm + 19.2 cm = 34.2 cm from the diverging lens. 1 1 1 1 1 1 EXECUTE: + = ⇒ + = ⇒ f = −26.7 cm. s s′ f −15.0 cm 34.2 cm f EVALUATE: Our calculation yields a negative value of f, which should be the case for a diverging lens. 34.100. IDENTIFY: The spherical mirror forms an image of the object. It forms another image when the image of the plane mirror serves as an object. SET UP: For the convex mirror f = −24.0 cm . The image formed by the plane mirror is 10.0 cm to the right of the plane mirror, so is 20.0 cm + 10.0 cm = 30.0 cm from the vertex of the spherical mirror. EXECUTE: The first image formed by the spherical mirror is the one where the light immediately strikes its surface, without bouncing from the plane mirror. 1 1 1 1 1 1 + = ⇒ + = ⇒ s′ = −7.06 cm, and the image height s s′ f 10.0 cm s′ −24.0 cm s′ −7.06 is y′ = − y = − (0.250 cm) = 0.177 cm. s 10.0 The second image is of the plane mirror image is located 30.0 cm from the vertex of the spherical mirror. 1 1 1 1 1 1 + = ⇒ + = ⇒ s′ = −13.3 cm and the image height is s s′ f 30.0 cm s′ −24.0 cm s′ −13.3 y′ = − y = − (0.250 cm) = 0.111 cm. s 30.0 EVALUATE: Other images are formed by additional reflections from the two mirrors. 34.101. IDENTIFY: In the sketch in Figure 34.101 the light travels upward from the object. Apply Eq.(34.11) with R → ∞ to the refraction at each surface. The image formed by the first surface serves as the object for the second surface. SET UP: The locations of the object and the glass plate are shown in Figure 34.101. 34.99.

IDENTIFY:

Apply

For a plane (flat) surface n n R → ∞ so a + b = 0 s s′ n s′ = − b s na Figure 34.101 EXECUTE:

First refraction (air → glass):

na = 1.00; nb = 1.55; s = 6.00 cm nb 1.55 (6.00 cm) = −9.30 cm s=− na 1.00 The image is 9.30 cm below the lower surface of the glass, so is 9.30 cm + 3.50 cm = 12.8 cm below the upper surface. s′ = −


Geometric Optics

34-31

Second refraction (glass → air): na = 1.55; nb = 1.00; s = +12.8 cm nb 1.00 (12.8 cm) = −8.26 cm s=− na 1.55 The image of the page is 8.26 cm below the top surface of the glass plate and therefore 9.50 cm − 8.26 cm = 1.24 cm above the page. EVALUATE: The image is virtual. If you view the object by looking down from above the plate, the image of the page that you see is closer to your eye than the page is. 34.102. IDENTIFY: Light refracts at the front surface of the lens, refracts at the glass-water interface, reflects from the plane mirror and passes through the two interfaces again, now traveling in the opposite direction. SET UP: Use the focal length in air to find the radius of curvature R of the lens surfaces. ⎛1 1 1 ⎞ 1 ⎛2⎞ = ( n − 1)⎜ − ⎟ ⇒ = 0.52 ⎜ ⎟ ⇒ R = 41.6 cm. EXECUTE: (a) f R R 40 cm ⎝R⎠ ⎝ 1 2 ⎠ s′ = −

At the air–lens interface:

na nb nb − na 1 1.52 0.52 + = ⇒ + = and s1′ = −851 cm and s2 = 851 cm. 70.0 cm 41.6 cm s s′ R s1′

−0.187 1.52 1.33 and s′2 = 491 cm . + = 851 cm s2′ −41.6 cm The mirror reflects the image back (since there is just 90 cm between the lens and mirror.) So, the position of the image is 401 cm to the left of the mirror, or 311 cm to the left of the lens. 1.33 1.52 0.187 and s3′ = +173 cm . At the water–lens interface: ⇒ + = 41.6 cm −311 cm s3′

At the lens–water interface: ⇒

At the lens–air interface: ⇒

1.52 1 −0.52 and s′4 = +47.0 cm , to the left of lens. + = −173 cm s4′ −41.6 cm

⎛ n s′ ⎞⎛ n s′ ⎞⎛ n s′ ⎞⎛ n s′ ⎞ ⎛ −851 ⎞⎛ 491 ⎞⎛ +173 ⎞⎛ +47.0 ⎞ m = m1m2 m3m4 = ⎜ a1 1 ⎟⎜ a 2 2 ⎟⎜ a 3 3 ⎟⎜ a 4 4 ⎟ = ⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟ = −1.06. ⎝ nb1s1 ⎠⎝ nb 2 s2 ⎠⎝ nb 3 s3 ⎠⎝ nb 4 s4 ⎠ ⎝ 70 ⎠⎝ −851 ⎠⎝ −311 ⎠⎝ −173 ⎠ (Note all the indices of refraction cancel out.) (b) The image is real. (c) The image is inverted. (d) The final height is y′ = my = (1.06)(4.00 mm) = 4.24 mm. EVALUATE: The final image is real even though it is on the same side of the lens as the object! 34.103. IDENTIFY: The camera lens can be modeled as a thin lens that forms an image on the film. 1 1 1 s′ SET UP: The thin-lens equation is + = , and the magnification of the lens is m = − . s s′ f s s′ y′ 1 (0.0360 m) ⇒ s′ = (7.50 × 10−4 ) s , EXECUTE: (a) m = − = = s y 4 (12.0 m)

1 1 1 1 1⎛ 1 1 ⎞ 1 + = + = ⎜1 + ⇒ s = 46.7 m . ⎟= = s s′ s (7.50 × 10−4 ) s s ⎝ 7.50 × 10−4 ⎠ f 0.0350 m (b) To just fill the frame, the magnification must be 3.00 × 10−3 so:

1⎛ 1 1 ⎞ 1 ⇒ s = 11.7 m . ⎜1 + ⎟= = s ⎝ 3.00 × 10−3 ⎠ f 0.0350 m Since the boat is originally 46.7 m away, the distance you must move closer to the boat is 46.7 m – 11.7 m = 35.0 m. EVALUATE: This result seems to imply that if you are 4 times as far, the image is ¼ as large on the film. However this result is only an approximation, and would not be true for very close distances. It is a better approximation for large distances. 1 1 1 s′ 34.104. IDENTIFY: Apply + = and m = − . s s′ f s ′ SET UP: s + s = 18.0 cm 1 1 1 + = . ( s′) 2 − (18.0 cm) s′ + 54.0 cm 2 = 0 so s′ = 14.2 cm or 3.80 cm . 18.0 cm − s′ s′ 3.00 cm s = 3.80 cm or 14.2 cm , so the screen must either be 3.80 cm or 14.2 cm from the object.

EXECUTE:

(a)


34-32

Chapter 34

s′ 3.80 s′ 14.2 =− = −0.268. s = 14.2 cm : m = − = − = −3.74. s s 14.2 3.80 EVALUATE: Since the image is projected onto the screen, the image is real and s′ is positive. We assumed this when we wrote the condition s + s′ = 18.0 cm . 34.105. IDENTIFY: Apply Eq.(34.16) to calculate the image distance for each lens. The image formed by the 1st lens serves as the object for the 2nd lens, and the image formed by the 2nd lens serves as the object for the 3rd lens. SET UP: The positions of the object and lenses are shown in Figure 34.105. 1 1 1 + = s s′ f (b) s = 3.80 cm : m = −

1 1 1 s− f = − = s′ f s sf s′ =

sf s− f

Figure 34.105 EXECUTE: lens #1 s = +80.0 cm; f = +40.0 cm

sf (+80.0 cm)( +40.0 cm) = = +80.0 cm +80.0 cm − 40.0 cm s− f The image formed by the first lens is 80.0 cm to the right of the first lens, so it is 80.0 cm − 52.0 cm = 28.0 cm to the right of the second lens. lens #2 s = −28.0 cm; f = +40.0 cm s′ =

sf (−28.0 cm)( +40.0 cm) = = +16.47 cm s− f −28.0 cm − 40.0 cm The image formed by the second lens is 16.47 cm to the right of the second lens, so it is 52.0 cm − 16.47 cm = 35.53 cm to the left of the third lens. lens #3 s = +35.53 cm; f = +40.0 cm s′ =

sf (+35.53 cm)( +40.0 cm) = = −318 cm s− f +35.53 cm − 40.0 cm The final image is 318 cm to the left of the third lens, so it is 318 cm − 52 cm − 52 cm − 80 cm = 134 cm to the left of the object. EVALUATE: We used the separation between the lenses and the sign conventions for s and s′ to determine the object distances for the 2nd and 3rd lenses. The final image is virtual since the final s′ is negative. 1 1 1 34.106. IDENTIFY: Apply + = and calculate s′ for each s. s s′ f SET UP: f = 90 mm s′ =

1 1 1 1 1 1 + = ⇒ + = ⇒ s′ = 96.7 mm. s s′ f 1300 mm s′ 90 mm 1 1 1 1 1 1 + = ⇒ + = ⇒ s′ = 91.3 mm. s s′ f 6500 mm s′ 90 mm ⇒ Δ s′ = 96.7 mm − 91.3 mm = 5.4 mm toward the film sf EVALUATE: s′ = . For f > 0 and s > f , s′ decreases as s increases. s− f near point near point 34.107. IDENTIFY and SET UP: The generalization of Eq.(34.22) is M = . , so f = f M EXECUTE: (a) age 10, near point = 7 cm 7 cm f = = 3.5 cm 2.0 (b) age 30, near point = 14 cm 14 cm f = = 7.0 cm 2.0 EXECUTE:


Geometric Optics

34-33

(c) age 60, near point = 200 cm 200 cm f = = 100 cm 2.0 (d) f = 3.5 cm (from part (a)) and near point = 200 cm (for 60-year-old)

200 cm = 57 3.5 cm (e) EVALUATE: No. The reason f = 3.5 cm gives a larger M for a 60-year-old than for a 10-year-old is that the eye of the older person can’t focus on as close an object as the younger person can. The unaided eye of the 60year-old must view a much smaller angular size, and that is why the same f gives a much larger M. The angular size of the image depends only on f and is the same for the two ages. 1 1 1 θ′ 34.108. IDENTIFY: Use + = to calculate s that gives s′ = −25 cm . M = . θ s s′ f M=

SET UP: EXECUTE:

y y and θ = . s 25 cm 1 1 1 1 1 1 f (25 cm) (a) + = ⇒ + . = ⇒s= ′ s s f s −25 cm f f + 25 cm

Let the height of the object be y , so θ ′ =

⎛ y ( f + 25 cm) ⎞ y ( f + 25 cm) ⎛ y⎞ . (b) θ ′ = arctan ⎜ ⎟ = arctan ⎜ ⎟≈ f (25 cm) ⎝s⎠ ⎝ f (25 cm) ⎠ θ ′ y ( f + 25 cm) 1 f + 25 cm = . (c) M = = θ f (25 cm) y / 25 cm f (d) If f = 10 cm ⇒ M =

10 cm + 25 cm = 3.5. This is 1.4 times greater than the magnification obtained if the image 10 cm

25 cm = 2.5). f EVALUATE: (e) Having the first image form just within the focal length puts one in the situation described above, where it acts as a source that yields an enlarged virtual image. If the first image fell just outside the second focal point, then the image would be real and diminished. 1 1 1 34.109. IDENTIFY: Apply + = . The near point is at infinity, so that is where the image must be formed for any s s′ f objects that are close. 1 SET UP: The power in diopters equals , with f in meters. f if formed at infinity ( M ∞ =

1 1 1 1 1 1 = + = + = = 4.17 diopters. f s s′ 24 cm −∞ 0.24 m EVALUATE: To focus on closer objects, the power must be increased. n n n − na 34.110. IDENTIFY: Apply a + b = b . s s′ R SET UP: na = 1.00 , nb = 1.40 . EXECUTE:

1 1.40 0.40 + = ⇒ s′ = 2.77 cm. s′ 36.0 cm 0.75 cm EVALUATE: This distance is greater than the normal eye, which has a cornea vertex to retina distance of about 2.6 cm. 34.111. IDENTIFY: Use similar triangles in Figure 34.63 in the textbook and Eq.(34.16) to derive the expressions called for in the problem. (a) SET UP: The effect of the converging lens on the ray bundle is sketched in Figure 34.111. EXECUTE: From similar triangles in Figure 34.111a, r0 r′ = 0 f1 f1 − d EXECUTE:

Figure 34.111a


34-34

Chapter 34

⎛ f −d ⎞ Thus r0′ = ⎜ 1 ⎟ r0 , as was to be shown. ⎝ f1 ⎠ (b) SET UP: The image at the focal point of the first lens, a distance f1 to the right of the first lens, serves as the

object for the second lens. The image is a distance f1 − d to the right of the second lens, so s2 = −( f1 − d ) = d − f1. EXECUTE:

s′2 =

s2 f 2 (d − f1 ) f 2 = s2 − f 2 d − f1 − f 2

( f1 − d ) f 2 , as was to be shown. f 2 − f1 + d (c) SET UP: The effect of the diverging lens on the ray bundle is sketched in Figure 34.111b. f 2 < 0 so f 2 = − f 2 and s′2 =

From similar r r′ triangles in the sketch, 0 = 0 f s2′

EXECUTE:

Thus

r0 f = r0′ s2′

Figure 34.111b

From the results of part (a),

r0 f1 f1 f . Combining the two results gives = = f1 − d s2′ r0′ f1 − d

( f1 − d ) f 2 f1 f1 f 2 ⎛ f ⎞ = f = s2′ ⎜ 1 ⎟ = , as was to be shown. − − + − f d f f d f d f ( ) ) 1 1 2 − f1 + d ⎝ 1 ⎠ ( 2 (d) SET UP: Put the numerical values into the expression derived in part (c). f1 f 2 EXECUTE: f = f 2 − f1 + d

216 cm 2 6.0 cm + d d = 0 gives f = 36.0 cm; maximum f d = 4.0 cm gives f = 21.6 cm; minimum f f1 = 12.0 cm, f 2 = 18.0 cm, so f =

216 cm 2 6.0 cm + d 6.0 cm + d = 7.2 cm and d = 1.2 cm EVALUATE: Changing d produces a range of effective focal lengths. The effective focal length can be both smaller and larger than f1 + f 2 . f = 30.0 cm says 30.0 cm =

34.112. IDENTIFY: SET UP:

M =

y′ y′ y′ f θ′ . θ = 1 , and θ ′ = 2 . This gives M = 2 . 1 . θ f1 s′2 s2′ y1′

Since the image formed by the objective is used as the object for the eyepiece, y1′ = y2 .

y′2 f1 y′ f s′ f f f 48.0 cm = 2 . 1 = 2 . 1 = 1 . Therefore, s2 = 1 = . = 1.33 cm, and this is just M 36 s′2 y2 y2 s′2 s2 s′2 s2 outside the eyepiece focal point. Now the distance from the mirror vertex to the lens is f1 + s2 = 49.3 cm, and so

EXECUTE:

M =

−1

⎛ 1 1 1 1 1 ⎞ + = ⇒ s2′ = ⎜ − ⎟ = 12.3 cm. Thus we have a final image which is real and 12.3 cm from s2 s2′ f 2 1.20 cm 1.33 cm ⎠ ⎝ the eyepiece. (Take care to carry plenty of figures in the calculation because two close numbers are subtracted.) EVALUATE: Eq.(34.25) gives M = 40 , somewhat larger than M for this telescope. 34.113. IDENTIFY and SET UP: The image formed by the objective is the object for the eyepiece. The total lateral magnification is mtot = m1m2 . f1 = 8.00 mm (objective); f 2 = 7.50 cm (eyepiece)


Geometric Optics

34-35

(a) The locations of the object, lenses and screen are shown in Figure 34.113.

Figure 34.113 EXECUTE:

Find the object distance s1 for the objective:

s1′ = +18.0 cm, f1 = 0.800 cm, s1 = ?

1 1 1 1 1 1 s1′ − f1 + = , so = − = s1 s1′ f1 s1 f1 s1′ s1′ f1 ′ s f (18.0 cm)(0.800 cm) = 0.8372 cm s1 = 1 1 = s1′ − f1 18.0 cm − 0.800 cm Find the object distance s2 for the eyepiece: s2′ = +200 cm, f 2 = 7.50 cm, s2 = ?

1 1 1 + = s2 s′2 f 2 s′ f (200 cm)(7.50 cm) = 7.792 cm s2 = 2 2 = s′2 − f 2 200 cm − 7.50 cm Now we calculate the magnification for each lens: s′ 18.0 cm m1 = − 1 = − = −21.50 0.8372 cm s1 s′ 200 cm = −25.67 m2 = − 2 = − s2 7.792 cm mtot = m1m2 = (−21.50)( −25.67) = 552. (b) From the sketch we can see that the distance between the two lenses is s1′ + s2 = 18.0 cm + 7.792 cm = 25.8 cm. EVALUATE: The microscope is not being used in the conventional way; it merely serves as a two-lens system. In particular, the final image formed by the eyepiece in the problem is real, not virtual as is the case normally for a microscope. Eq.(34.23) does not apply here, and in any event gives the angular not the lateral magnification. u′ 34.114. IDENTIFY: For u and u′ as defined in Figure 34.64 in the textbook, M = . u SET UP: f 2 is negative. From Figure 34.64, the length of the telescope is f1 + f 2 . u′ f y y y EXECUTE: (a) From the figure, u = and u′ = = − . The angular magnification is M = = − 1 . f1 f2 f2 u f2 f1 f1 95.0 cm (b) M = − ⇒ f 2 = − =− = −15.0 cm. f2 M 6.33 (c) The length of the telescope is 95.0 cm − 15.0 cm = 80.0 cm, compared to the length of 110 cm for the telescope in Exercise 34.57. EVALUATE: An advantage of this construction is that the telescope is somewhat shorter. 1 1 1 34.115. IDENTIFY: Use + = to calculate s′ (the distance of each point from the lens), for points A, B and C. s s′ f SET UP: The object and lens are shown in Figure 34.115a. 1 1 1 1 1 1 + = ⇒ s′ = 36.0 cm. EXECUTE: (a) For point C : + = ⇒ ′ ′ s s f 45.0 cm s 20.0 cm s′ 36.0 y′ = − y = − (15.0 cm) = −12.0 cm , so the image of point C is 36.0 cm to the right of the lens, and s 45.0 12.0 cm below the axis. For point A: s = 45.0 cm + 8.00 cm(cos 45°) = 50.7 cm . 1 1 1 1 1 1 s′ 33.0 + = ⇒ + = ⇒ s′ = 33.0 cm. y′ = − y = − (15.0 cm − 8.00 cm(sin 45°)) = −6.10 cm, s s s′ f 45.0 50.7 cm s′ 20.0 cm so the image of point A is 33.0 cm to the right of the lens, and 6.10 cm below the axis.


34-36

Chapter 34

For point B: s = 45.0 cm − 8.00 cm(cos 45°) = 39.3 cm .

1 1 1 1 1 1 + = ⇒ + = ⇒ s′ = 40.7 cm. ′ ′ s s f 39.3 cm s 20.0 cm

s′ 40.7 y=− (15.0 cm + 8.00 cm(sin 45°)) = −21.4 cm, so the image of point B is 40.7 cm to the right of the s 39.3 lens, and 21.4 cm below the axis. The image is shown in Figure 34.115b. (b) The length of the pencil is the distance from point A to B: L = ( x A − xB ) 2 + ( y A − yB ) 2 = (33.0 cm − 40.7 cm) 2 + (6.10 cm − 21.4 cm) 2 = 17.1 cm y′ = −

EVALUATE:

The image is below the optic axis and is larger than the object.

Figure 34.115 34.116. IDENTIFY and SET UP:

Consider the ray diagram drawn in Figure 34.116. h sin θ sin α EXECUTE: (a) Using the diagram and law of sines, = but sin θ = = sin α (law of (R − f ) g R R2 R reflection), g = ( R − f ). Bisecting the triangle: cosθ = ⇒ R cosθ − f cosθ = . (R − f ) 2 R⎡ 1 ⎤ 1 ⎤ R ⎡ f = ⎢2 − = f0 ⎢ 2 − ⎥ . f 0 = 2 is the value of f for θ near zero (incident ray near the axis). When θ 2⎣ cosθ ⎥⎦ cos θ ⎣ ⎦ increases, (2 − 1/ cosθ ) decreases and f decreases. f − f0 f 1 1 = −0.02 ⇒ = 0.98 so 2 − = 0.98 . cosθ = (b) = 0.98 and θ = 11.4°. f0 f0 cosθ 2 − 0.98 EVALUATE: For θ = 45° , f = 0.586 f 0 , and f approaches zero as θ approaches 60° .

Figure 34.116


Geometric Optics

34-37

34.117. IDENTIFY: The distance between image and object can be calculated by taking the derivative of the separation distance and minimizing it. SET UP: For a real image s′ > 0 and the distance between the object and the image is D = s + s′ . For a real image must have s > f . EXECUTE:

D = s + s′ but s′ =

sf sf s2 ⇒D=s+ = . s− f s− f s− f

dD d ⎛ s 2 ⎞ 2s s2 s 2 − 2sf = ⎜ − = = 0 . s 2 − 2sf = 0 . s ( s − 2 f ) = 0 . s = 2 f is the solution for which ⎟= ds ds ⎝ s − f ⎠ s − f ( s − f ) 2 ( s − f ) 2 s > f . For s = 2 f , s′ = 2 f . Therefore, the minimum separation is 2 f + 2 f = 4 f . (b) A graph of D / f versus s / f is sketched in Figure 34.117. Note that the minimum does occur for D = 4 f . EVALUATE: If, for example, s = 3 f / 2 , then s′ = 3 f and D = s + s′ = 4.5 f , greater than the minimum value.

Figure 34.117 34.118. IDENTIFY and SET UP: For a plane mirror, s′ = − s . EXECUTE: (a) By the symmetry of image production, any image must be the same distance D as the object from the mirror intersection point. But if the images and the object are equal distances from the mirror intersection, they lie on a circle with radius equal to D. (b) The center of the circle lies at the mirror intersection as discussed above. (c) The diagram is sketched in Figure 34.118. EVALUATE: To see the image, light from the object must be able to reflect from each mirror and reach the person's eyes.

Figure 34.118 na nb nb − na + = to refraction at the cornea to find where the object for the cornea must be in 34.119. IDENTIFY: Apply s s′ R 1 1 1 order for the image to be at the retina. Then use + = to calculate f so that the lens produces an image of a s s′ f distant object at this point. SET UP: For refraction at the cornea, na = 1.33 and nb = 1.40 . The distance from the cornea to the retina in this model of the eye is 2.60 cm. From Problem 34.46, R = 0.71 cm . EXECUTE: (a) People with normal vision cannot focus on distant objects under water because the image is unable to be focused in a short enough distance to form on the retina. Equivalently, the radius of curvature of the normal eye is about five or six times too great for focusing at the retina to occur. (b) When introducing glasses, let’s first consider what happens at the eye: na nb nb − na 1.33 1.40 0.07 + = ⇒ + = ⇒ s2 = −3.02 cm. That is, the object for the cornea must be 3.02 cm s2 s′2 R s2 2.6 cm 0.71 cm behind the cornea. Now, assume the glasses are 2.00 cm in front of the eye, so s1′ = 2.00 cm + s2 = 5.02 cm .


34-38

Chapter 34

1 1 1 1 1 1 + = gives + = and f1′ = 5.02 cm. This is the focal length in water, but to get it in air, we use s1 s1′ f1′ ∞ 5.02 cm f1′ ⎡ n − nliq ⎤ ⎡ 1.52 − 1.333 ⎤ the formula from Problem 34.98: f1 = f1′⎢ ⎥ = (5.02 cm) ⎢ ⎥ = 1.35 cm . ⎢⎣ nliq (n − 1) ⎥⎦ ⎣1.333(1.52 − 1) ⎦ EVALUATE: A converging lens is needed.


35

INTERFERENCE

35.1.

35.2.

IDENTIFY: Compare the path difference to the wavelength. SET UP: The separation between sources is 5.00 m, so for points between the sources the largest possible path difference is 5.00 m. EXECUTE: (a) For constructive interference the path difference is mλ , m = 0, ± 1, ± 2, . . . Thus only the path difference of zero is possible. This occurs midway between the two sources, 2.50 m from A. (b) For destructive interference the path difference is (m + 12 )λ , m = 0, ± 1, ± 2, . . . A path difference of ± λ / 2 = 3.00 m is possible but a path difference as large as 3λ / 2 = 9.00 m is not possible. For a point a distance x from A and 5.00 − x from B the path difference is x − (5.00 m − x). x − (5.00 m − x) = +3.00 m gives x = 4.00 m. x − (5.00 m − x) = −3.00 m gives x = 1.00 m . EVALUATE: The point of constructive interference is midway between the points of destructive interference. IDENTIFY: For destructive interference the path difference is (m + 12 )λ , m = 0, ±1, ±2, … . The longest wavelength is for m = 0 . For constructive interference the path difference is mλ , m = 0, ±1, ±2, … The longest wavelength is for m = 1 . SET UP: The path difference is 120 m.

35.3.

λ

= 120 m ⇒ λ = 240 m. 2 (b) The longest wavelength for constructive interference is λ = 120 m. EVALUATE: The path difference doesn't depend on the distance of point Q from B. IDENTIFY: Use c = f λ to calculate the wavelength of the transmitted waves. Compare the difference in the distance from A to P and from B to P. For constructive interference this path difference is an integer multiple of the wavelength. SET UP: Consider Figure 35.3 The distance of point P from each coherent source is rA = x and EXECUTE:

(a) For destructive interference

rB = 9.00 m − x. Figure 35.3 EXECUTE:

The path difference is rB − rA = 9.00 m − 2 x.

rB − rA = mλ , m = 0, ± 1, ± 2, …

λ=

c 2.998 × 108 m/s = = 2.50 m f 120 × 106 Hz

9.00 m − m(2.50 m) = 4.50 m − (1.25 m)m. x must lie in the range 0 to 2 9.00 m since P is said to be between the two antennas. m = 0 gives x = 4.50 m m = +1 gives x = 4.50 m − 1.25 m = 3.25 m m = +2 gives x = 4.50 m − 2.50 m = 2.00 m m = +3 gives x = 4.50 m − 3.75 m = 0.75 m m = −1 gives x = 4.50 m + 1.25 m = 5.75 m m = −2 gives x = 4.50 m + 2.50 m = 7.00 m m = −3 gives x = 4.50 m + 3.75 m = 8.25 m

Thus 9.00 m − 2 x = m(2.50 m) and x =

35-1


35-2

35.4.

Chapter 35

All other values of m give values of x out of the allowed range. Constructive interference will occur for x = 0.75 m, 2.00 m, 3.25 m, 4.50 m, 5.75 m, 7.00 m, and 8.25 m. EVALUATE: Constructive interference occurs at the midpoint between the two sources since that point is the same distance from each source. The other points of constructive interference are symmetrically placed relative to this point. IDENTIFY: For constructive interference the path difference d is related to λ by d = mλ , m = 0,1,2, … For destructive interference d = ( m + 12 )λ , m = 0,1,2, … SET UP: d = 2040 nm EXECUTE: (a) The brightest wavelengths are when constructive interference occurs: d 2040 nm 2040 nm d = mλm ⇒ λm = ⇒ λ3 = = 680 nm, λ4 = = 510 nm and m 3 4 2040 nm λ5 = = 408 nm. 5 (b) The path-length difference is the same, so the wavelengths are the same as part (a). d 2040 nm (c) d = ( m + 12 )λm so λm = . The visible wavelengths are λ3 = 583 nm and λ4 = 453 nm . = m + 12 m + 12

35.5.

EVALUATE: The wavelengths for constructive interference are between those for destructive interference. IDENTIFY: If the path difference between the two waves is equal to a whole number of wavelengths, constructive interference occurs, but if it is an odd number of half-wavelengths, destructive interference occurs. SET UP: We calculate the distance traveled by both waves and subtract them to find the path difference. EXECUTE: Call P1 the distance from the right speaker to the observer and P2 the distance from the left speaker to the observer. (a) P1 = 8.0 m and P2 = (6.0 m) 2 + (8.0 m) 2 = 10. 0 m . The path distance is

ΔP = P2 − P1 = 10.0 m – 8.0 m = 2.0 m (b) The path distance is one wavelength, so constructive interference occurs.

35.6.

(c) P1 = 17.0 m and P2 = (6.0 m) 2 + (17.0 m) 2 = 18.0 m . The path difference is 18.0 m – 17.0 m = 1.0 m, which is one-half wavelength, so destructive interference occurs. EVALUATE: Constructive interference also occurs if the path difference 2 λ , 3 λ , 4 λ , etc., and destructive interference occurs if it is λ /2, 3 λ /2, 5 λ /2, etc. IDENTIFY: At an antinode the interference is constructive and the path difference is an integer number of wavelengths; path difference = mλ , m = 0, ±1, ±2, … at an antinode. SET UP: The maximum magnitude of the path difference is the separation d between the two sources. EXECUTE: (a) At S1 , r2 − r1 = 4λ , and this path difference stays the same all along the y -axis, so

m = +4. At S 2 , r2 − r1 = −4λ , and the path difference below this point, along the negative y-axis, stays the same, so m = −4. (b) The wave pattern is sketched in Figure 35.6. d (c) The maximum and minimum m-values are determined by the largest integer less than or equal to .

λ

1 (d) If d = 7 λ ⇒ −7 ≤ m ≤ +7, so there will be a total of 15 antinodes between the sources. 2 EVALUATE: We are considering points close to the two sources and the antinodal curves are not straight lines.

Figure 35.6


Interference

35.7.

35.8.

IDENTIFY: At an antinodal point the path difference is equal to an integer number of wavelengths. SET UP: For m = 3 , the path difference is 3λ . EXECUTE: Measuring with a ruler from both S1 and S 2 to the different points in the antinodal line labeled m = 3 , we find that the difference in path length is three times the wavelength of the wave, as measured from one crest to the next on the diagram. EVALUATE: There is a whole curve of points where the path difference is 3λ . mλ IDENTIFY: The value of y20 is much smaller than R and the approximate expression ym = R is accurate. d SET UP: y20 = 10.6 × 10−3 m .

d=

EXECUTE:

20 Rλ (20)(1.20 m)(502 × 10−9 m) = = 1.14 × 10−3 m = 1.14 mm y20 10.6 × 10−3 m

y20 so θ 20 = 0.51° and the approximation sin θ 20 ≈ tan θ 20 is very accurate. R IDENTIFY and SET UP: The dark lines correspond to destructive interference and hence are located by Eq.(35.5): 1⎞ ⎛ ⎜ m + ⎟λ 1⎞ 2⎠ ⎛ ⎝ , m = 0, ± 1, ± 2,… d sin θ = ⎜ m + ⎟ λ so sinθ = 2⎠ d ⎝ Solve for θ that locates the second and third dark lines. Use y = R tan θ to find the distance of each of the dark lines from the center of the screen. EXECUTE: 1st dark line is for m = 0 3λ 3(500 × 10−9 m) 2nd dark line is for m = 1 and sin θ1 = = = 1.667 × 10−3 and θ1 = 1.667 × 10−3 rad 2d 2(0.450 × 10−3 m) EVALUATE:

35.9.

35-3

tan θ 20 =

3rd dark line is for m = 2 and sin θ 2 =

5λ 5(500 × 10−9 m) = = 2.778 × 10−3 and θ 2 = 2.778 × 10−3 rad 2d 2(0.450 × 10−3 m)

(Note that θ1 and θ 2 are small so that the approximation θ ≈ sin θ ≈ tan θ is valid.) The distance of each dark line from the center of the central bright band is given by ym = R tan θ , where R = 0.850 m is the distance to the screen. tan θ ≈ θ so ym = Rθ m

y1 = Rθ1 = (0.750 m)(1.667 × 10−3 rad) = 1.25 × 10−3 m y2 = Rθ 2 = (0.750 m)(2.778 × 10−3 rad) = 2.08 × 10−3 m Δy = y2 − y1 = 2.08 × 10−3 m − 1.25 × 10−3 m = 0.83 mm EVALUATE:

35.10.

1⎞ ⎛ interference: ym = R ⎜ m + ⎟ λ / d . 2⎠ ⎝ IDENTIFY: Since the dark fringes are eqully spaced, R " ym , the angles are small and the dark bands are located by ym + 1 = R 2

SET UP:

(m + 12 )λ . d

The separation between adjacent dark bands is Δy =

Rλ . d

Rλ Rλ (1.80 m) (4.50 × 10−7 m) ⇒d = = = 1.93 × 10−4 m = 0.193 m. d Δy 4.20 × 10−3 m EVALUATE: When the separation between the slits decreases, the separation between dark fringes increases. IDENTIFY and SET UP: The positions of the bright fringes are given by Eq.(35.6): ym = R (mλ / d ). For each EXECUTE:

35.11.

Since θ1 and θ 2 are very small we could have used Eq.(35.6), generalized to destructive

Δy =

fringe the adjacent fringe is located at ym +1 = R(m + 1)λ / d . Solve for λ. EXECUTE:

The separation between adjacent fringes is Δy = ym +1 − ym = Rλ / d .

d Δy (0.460 × 10−3 m)(2.82 × 10−3 m) = = 5.90 × 10−7 m = 590 nm R 2.20 m EVALUATE: Eq.(35.6) requires that the angular position on the screen be small. The angular position of bright fringes is given by sin θ = mλ / d . The slit separation is much larger than the wavelength (λ / d = 1.3 × 10−3 ), so θ is small so long as m is not extremely large.

λ=


35-4

Chapter 35

35.12.

The width of a bright fringe can be defined to be the distance between its two adjacent destructive ( m + 12 )λ minima. Assuming the small angle formula for destructive interference ym = R . d SET UP: d = 0.200 × 10−3 m . R = 4.00 m . EXECUTE: The distance between any two successive minima λ (400 × 10−9 m) is ym +1 − ym = R = (4.00 m) = 8.00 mm. Thus, the answer to both part (a) and part (b) is that the (0.200 × 10−3 m) d width is 8.00 mm. EVALUATE: For small angles, when ym # R , the interference minima are equally spaced.

35.13.

IDENTIFY:

1⎞ ⎛ The dark lines are located by d sin θ = ⎜ m + ⎟ λ . The distance of each line from the 2⎠ ⎝ center of the screen is given by y = R tan θ . IDENTIFY and SET UP:

First dark line is for m = 0 and d sin θ1 = λ / 2.

EXECUTE:

sin θ1 =

λ 2d

=

550 × 10−9 m = 0.1528 and θ1 = 8.789°. Second dark line is for m = 1 and d sin θ 2 = 3λ / 2. 2(1.80 × 10−6 m)

⎛ 550 × 10−9 m ⎞ 3λ = 3⎜ ⎟ = 0.4583 and θ 2 = 27.28°. −6 2d ⎝ 2(1.80 × 10 m) ⎠ y1 = R tan θ1 = (0.350 m) tan8.789° = 0.0541 m

sin θ 2 =

y2 = R tan θ 2 = (0.350 m) tan 27.28° = 0.1805 m The distance between the lines is Δy = y2 − y1 = 0.1805 m − 0.0541 m = 0.126 m = 12.6 cm.

35.14.

EVALUATE: sin θ1 = 0.1528 and tan θ1 = 0.1546. sin θ 2 = 0.4583 and tan θ 2 = 0.5157. As the angle increases, sin θ ≈ tan θ becomes a poorer approximation. mλ IDENTIFY: Using Eq.(35.6) for small angles: ym = R . d SET UP: First-order means m = 1 . EXECUTE: The distance between corresponding bright fringes is

Δy =

35.15.

Rm (5.00 m)(1) (660 − 470) × (10−9 m) = 3.17 mm. Δλ = d (0.300 × 10−3 m)

EVALUATE: The separation between these fringes for different wavelengths increases when the slit separation decreases. IDENTIFY and SET UP: Use the information given about the bright fringe to find the distance d between the two slits. Then use Eq.(35.5) and y = R tan θ to calculate λ for which there is a first-order dark fringe at this same place on the screen. Rλ Rλ (3.00 m)(600 × 10−9 m) EXECUTE: y1 = 1 , so d = 1 = = 3.72 × 10−4 m. (R is much greater than d, so Eq.35.6 d y1 4.84 × 10−3 m

1⎞ ⎛ is valid.) The dark fringes are located by d sin θ = ⎜ m + ⎟ λ , m = 0, ± 1, ± 2, … The first order dark fringe is located 2⎠ ⎝ by sin θ = λ2 / 2d , where λ2 is the wavelength we are seeking. y = R tan θ ≈ R sin θ =

λ2 R 2d

Rλ1 Rλ2 = and λ2 = 2λ1 = 1200 nm. d 2d EVALUATE: For λ = 600 nm the path difference from the two slits to this point on the screen is 600 nm. For this same path difference (point on the screen) the path difference is λ / 2 when λ = 1200 nm. mλ IDENTIFY: Bright fringes are located at ym = R , when ym # R . Dark fringes are at d sin θ = ( m + 12 )λ and d y = R tan θ . We want λ2 such that y = y1. This gives

35.16.

c 3.00 × 108 m/s = = 4.75 × 10−7 m . For the third bright fringe (not counting the central bright f 6.32 × 1014 Hz spot), m = 3 . For the third dark fringe, m = 2 . SET UP:

λ=


Interference

EXECUTE:

(a) d =

mλ R 3(4.75 × 10−7 m)(0.850 m) = = 3.89 × 10−5 m = 0.0389 mm ym 0.0311 m

λ

⎛ 4.75 × 10−7 m ⎞ = (2.5) ⎜ ⎟ = 0.0305 and θ = 1.75° . y = R tan θ = (85.0 cm) tan1.75° = 2.60 cm . −5 d ⎝ 3.89 × 10 m ⎠ EVALUATE: The third dark fringe is closer to the center of the screen than the third bright fringe on one side of the central bright fringe. IDENTIFY: Bright fringes are located at angles θ given by d sin θ = mλ . SET UP: The largest value sin θ can have is 1.00. d 0.0116 × 10−3 m d sin θ EXECUTE: (a) m = . For sin θ = 1 , m = = = 19.8 . Therefore, the largest m for fringes λ λ 5.85 × 10−7 m on the screen is m = 19 . There are 2(19) + 1 = 39 bright fringes, the central one and 19 above and 19 below it. (b) sin θ = (2 + 12 )

35.17.

λ

⎛ 5.85 × 10−7 m ⎞ = ±19 ⎜ ⎟ = ±0.958 and θ = ±73.3° . −3 d ⎝ 0.0116 × 10 m ⎠ EVALUATE: For small θ the spacing Δy between adjacent fringes is constant but this is no longer the case for larger angles. IDENTIFY: At large distances from the antennas the equation d sin θ = mλ , m = 0, ±1, ±2, … gives the angles where (b) The most distant fringe has m = ±19 . sin θ = m

35.18.

35-5

maximum intensity is observed and d sin θ = (m + 12 )λ , m = 0, ±1, ±2, … gives the angles where minimum intensity is observed. c SET UP: d = 12.0 m . λ = . f EXECUTE:

35.19.

(a) λ =

c 3.00 × 108 m/s mλ ⎛ 2.78 m ⎞ = m⎜ = = 2.78 m . sin θ = ⎟ = m(0.232) . 6 f 107.9 × 10 Hz d ⎝ 12.0 m ⎠

θ = ±13.4°, ± 27.6°, ± 44.1°, ± 68.1° . λ (b) sin θ = (m + 12 ) = (m + 12 )(0.232) . θ = ±6.66°, ± 20.4°, ± 35.5°, ± 54.3° . d EVALUATE: The angles for zero intensity are approximately midway between those for maximum intensity. IDENTIFY: Eq.(35.10): I = I 0 cos 2 (φ 2) . Eq.(35.11): φ = (2π / λ )(r2 − r1 ) . SET UP: φ is the phase difference and (r2 − r1 ) is the path difference. EXECUTE:

(a) I = I 0 (cos 30.0°) 2 = 0.750 I 0

(b) 60.0° = (π / 3) rad . (r2 − r1 ) = (φ / 2π )λ = [ (π / 3) / 2π ] λ = λ / 6 = 80 nm . EVALUATE: 35.20.

IDENTIFY:

φ = 360° / 6 and (r2 − r1 ) = λ / 6 . Δφ path difference relates the path difference to the phase difference Δφ . = 2π λ

The sources and point P are shown in Figure 35.20. ⎛ 524 cm − 486 cm ⎞ EXECUTE: Δφ = 2π ⎜ ⎟ = 119 radians 2 cm ⎝ ⎠

SET UP:

EVALUATE:

The distances from B to P and A to P aren't important, only the difference in these distances.

Figure 35.20 35.21.

IDENTIFY and SET UP:

The phase difference φ is given by φ = (2π d / λ )sin θ (Eq.35.13.)

EXECUTE: φ = [2π (0.340 × 10−3 m)/(500 × 10 −9 m)]sin 23.0° = 1670 rad EVALUATE: The mth bright fringe occurs when φ = 2π m, so there are a large number of bright fringes within 23.0° from the centerline. Note that Eq.(35.13) gives φ in radians.


35-6

Chapter 35

35.22.

IDENTIFY: The maximum intensity occurs at all the points of constructive interference. At these points, the path difference between waves from the two transmitters is an integral number of wavelengths. SET UP: For constructive interference, sin θ = mλ/d. EXECUTE: (a) First find the wavelength of the UHF waves: λ = c/f = (3.00 × 108 m/s)/(1575.42 MHz) = 0.1904 m For maximum intensity (πd sin θ )/λ = mπ, so sin θ = mλ/d = m[(0.1904 m)/(5.18 m)] = 0.03676m The maximum possible m would be for θ = 90°, or sin θ = 1, so mmax = d/λ = (5.18 m)/(0.1904 m) = 27.2 which must be ±27 since m is an integer. The total number of maxima is 27 on either side of the central fringe, plus the central fringe, for a total of 27 + 27 + 1 = 55 bright fringes. (b) Using sin θ = mλ/d, where m = 0, ±1, ±2, and ±3, we have sin θ = mλ/d = m[(0.1904 m)/(5.18 m)] = 0.03676m

m = 0: sin θ = 0, which gives θ = 0° m = ±1: sin θ = ±(0.03676)(1), which gives θ = ±2.11° m = ±2: sin θ = ±(0.03676)(2), which gives θ = ±4.22° m = ±3: sin θ = ±(0.03676)(3), which gives θ = ±6.33°

35.23.

⎛ π d sin θ ⎞ 2 2 2 ⎡ π (5.18 m)sin(4.65°) ⎤ (c) I = I 0 cos 2 ⎜ ⎟ = ( 2.00 W/m ) cos ⎢ ⎥ = 1.28 W/m . 0.1904 m ⎣ ⎦ ⎝ λ ⎠ EVALUATE: Notice that sinθ increases in integer steps, but θ only increases in integer steps for small θ. 1⎞ ⎛ (a) IDENTIFY and SET UP: The minima are located at angles θ given by d sin θ = ⎜ m + ⎟ λ . The first minimum 2⎠ ⎝ corresponds to m = 0. Solve for θ . Then the distance on the screen is y = R tan θ . sin θ =

EXECUTE:

λ 2d

=

660 × 10−9 m = 1.27 × 10−3 and θ = 1.27 × 10−3 rad 2(0.260 × 10−3 m)

y = (0.700 m) tan(1.27 ×10−3 rad) = 0.889 mm. (b) IDENTIFY and SET UP: Eq.(35.15) given the intensity I as a function of the position y on the screen: ⎛ π dy ⎞ I = I 0 cos 2 ⎜ ⎟ . Set I = I 0 / 2 and solve for y. ⎝ λR ⎠ I=

EXECUTE:

1 ⎛ π dy ⎞ 1 I 0 says cos 2 ⎜ ⎟= 2 ⎝ λR ⎠ 2

π dy π ⎛ π dy ⎞ 1 so = rad cos ⎜ ⎟= λR 4 2 ⎝ λR ⎠ λ R (660 × 10−9 m)(0.700 m) = = 0.444 mm y= 4d 4(0.260 × 10−3 m) EVALUATE: 35.24.

IDENTIFY: SET UP: EXECUTE:

35.25.

I = I 0 / 2 at a point on the screen midway between where I = I 0 and I = 0. ⎛πd ⎞ Eq. (35.14): I = I 0 cos 2 ⎜ sin θ ⎟ . ⎝ λ ⎠

The intensity goes to zero when the cosine’s argument becomes an odd integer multiple of

πd sin θ = ( m + 1/ 2)π gives d sin θ = λ (m + 1/ 2), which is Eq. (35.5). λ

π 2

EVALUATE: Section 35.3 shows that the maximum-intensity directions from Eq.(35.14) agree with Eq.(35.4). IDENTIFY: The intensity decreases as we move away from the central maximum. ⎛ π dy ⎞ SET UP: The intensity is given by I = I 0 cos 2 ⎜ ⎟. ⎝ λR ⎠ EXECUTE: First find the wavelength: λ = c/f = (3.00 × 108 m/s)/(12.5 MHz) = 24.00 m At the farthest the receiver can be placed, I = I0/4, which gives I0 1 1 ⎛ π dy ⎞ ⎛ π dy ⎞ 2 ⎛ π dy ⎞ = I 0 cos 2 ⎜ ⎟ ⇒ cos ⎜ ⎟ = ⇒ cos ⎜ ⎟=± 4 2 ⎝ λR ⎠ ⎝ λR ⎠ 4 ⎝ λR ⎠


Interference

35.26.

35-7

The solutions are πdy/λR = π/3 and 2π/3. Using π/3, we get y = λR/3d = (24.00 m)(500 m)/[3(56.0 m)] = 71.4 m It must remain within 71.4 m of point C. EVALUATE: Using πdy/λR = 2π/3 gives y = 142.8 m. But to reach this point, the receiver would have to go beyond 71.4 m from C, where the signal would be too weak, so this second point is not possible. 2π IDENTIFY: The phase difference φ and the path difference r1 − r2 are related by φ = (r1 − r2 ) . The intensity is

λ

⎛φ ⎞ given by I = I 0 cos 2 ⎜ ⎟ . ⎝2⎠

SET UP: λ = EXECUTE:

35.27.

c 3.00 × 108 m/s = = 2.50 m . When the receiver measures zero intensity I 0 , φ = 0 . f 1.20 × 108 Hz

(a) φ =

λ

( r1 − r2 ) =

2π (1.8 m) = 4.52 rad. 2.50 m

⎛φ ⎞ ⎛ 4.52 rad ⎞ (b) I = I 0 cos 2 ⎜ ⎟ = I 0 cos 2 ⎜ ⎟ = 0.404 I 0 . 2 ⎝2⎠ ⎝ ⎠ EVALUATE: (r1 − r2 ) is greater than λ / 2 , so one minimum has been passed as the receiver is moved. IDENTIFY: Consider interference between rays reflected at the upper and lower surfaces of the film. Consider phase difference due to the path difference of 2t and any phase differences due to phase changes upon reflection. SET UP: Consider Figure 35.27. Both rays (1) and (2) undergo a 180° phase change on reflection, so these is no net phase difference introduced and the condition for destructive interference is 1⎞ ⎛ 2t = ⎜ m + ⎟ λ. 2⎠ ⎝ Figure 35.27

35.28.

35.29.

1⎞ ⎛ ⎜ m + ⎟λ λ 2⎠ ⎝ ; thinnest film says m = 0 so t = EXECUTE: t = 2 4 −9 λ0 650 × 10 m λ = = 1.14 × 10−7 m = 114 nm λ = 0 and t = 1.42 4(1.42) 4(1.42) EVALUATE: We compared the path difference to the wavelength in the film, since that is where the path difference occurs. IDENTIFY: Require destructive interference for light reflected at the front and rear surfaces of the film. SET UP: At the front surface of the film, light in air ( n = 1.00 ) reflects from the film ( n = 2.62 ) and there is a 180° phase shift due to the reflection. At the back surface of the film, light in the film ( n = 2.62 ) reflects from glass ( n = 1.62 ) and there is no phase shift due to reflection. Therefore, there is a net 180° phase difference produced by the reflections. The path difference for these two rays is 2t, where t is the thickness of the film. The 505 nm wavelength in the film is λ = . 2.62 EXECUTE: (a) Since the reflection produces a net 180° phase difference, destructive interference of the reflected ⎛ 505 nm ⎞ light occurs when 2t = mλ . t = m ⎜ ⎟ = (96.4 nm)m . The minimum thickness is 96.4 nm. ⎝ 2[2.62] ⎠ (b) The next three thicknesses are for m = 2 , 3 and 4: 192 nm, 289 nm and 386 nm. EVALUATE: The minimum thickness is for t = λ / 2n . Compare this to Problem 35.27, where the minimum thickness for destructive interference is t = λ / 4n . IDENTIFY: The fringes are produced by interference between light reflected from the top and bottom surfaces of the air wedge. The refractive index of glass is greater than that of air, so the waves reflected from the top surface of the air wedge have no reflection phase shift and the waves reflected from the bottom surface of the air wedge do


35-8

Chapter 35

have a half-cycle reflection phase shift. The condition for constructive interference (bright fringes) is therefore 2t = (m + 12 )λ . SET UP: The geometry of the air wedge is sketched in Figure 35.29. At a distance x from the point of contact of the two plates, the thickness of the air wedge is t. t λ λ λ EXECUTE: tan θ = so t = x tan θ . tm = (m + 12 ) . xm = (m + 12 ) and xm +1 = ( m + 32 ) . The 2 2 tan θ 2 tan θ x λ 1.00 distance along the plate between adjacent fringes is Δx = xm +1 − xm = . 15.0 fringes/cm = and Δx 2 tan θ 1.00 λ 546 × 10−9 m Δx = = 0.0667 cm . tan θ = = = 4.09 × 10−4 . The angle of the wedge is 2Δx 2(0.0667 × 10−2 m) 15.0 fringes/cm 4.09 × 10−4 rad = 0.0234° . EVALUATE: The fringes are equally spaced; Δx is independent of m.

Figure 35.29 35.30.

IDENTIFY: The fringes are produced by interference between light reflected from the top and from the bottom surfaces of the air wedge. The refractive index of glass is greater than that of air, so the waves reflected from the top surface of the air wedge have no reflection phase shift and the waves reflected from the bottom surface of the air wedge do have a half-cycle reflection phase shift. The condition for constructive interference (bright fringes) therefore is 2t = (m + 12 )λ . SET UP: The geometry of the air wedge is sketched in Figure 35.30. t 0.0800 mm λ EXECUTE: tan θ = = 8.89 × 10−4 . tan θ = so t = (8.89 × 10−4 ) x . tm = (m + 12 ) . 2 x 90.0 mm

xm = (m + 12 )

λ 2(8.89 × 10−4 )

is Δx = xm +1 − xm =

and xm +1 = (m + 32 )

λ 2(8.89 × 10 −4 )

=

λ 2(8.89 × 10−4 )

. The distance along the plate between adjacent fringes

656 × 10−9 m = 3.69 × 10−4 m = 0.369 mm . The number of fringes per cm is 2(8.89 × 10−4 )

1.00 1.00 = = 27.1 fringes/cm . Δx 0.0369 cm EVALUATE: As t → 0 the interference is destructive and there is a dark fringe at the line of contact between the two plates.

35.31.

Figure 35.30 IDENTIFY: The light reflected from the top of the TiO2 film interferes with the light reflected from the top of the glass surface. These waves are out of phase due to the path difference in the film and the phase differences caused by reflection. SET UP: There is a π phase change at the TiO2 surface but none at the glass surface, so for destructive interference the path difference must be mλ in the film. EXECUTE: (a) Calling T the thickness of the film gives 2T = mλ0/n, which yields T = mλ0/(2n). Substituting the numbers gives T = m (520.0 nm)/[2(2.62)] = 99.237m


Interference

35.32.

35.33.

35-9

T must be greater than 1036 nm, so m = 11, which gives T = 1091.6 nm, since we want to know the minimum thickness to add. ΔT = 1091.6 nm – 1036 nm = 55.6 nm (b) (i) Path difference = 2T = 2(1092 nm) = 2184 nm = 2180 nm. (ii) The wavelength in the film is λ = λ0/n = (520.0 nm)/2.62 = 198.5 nm. Path difference = (2180 nm)/[(198.5 nm)/wavelength] = 11.0 wavelengths EVALUATE: Because the path difference in the film is 11.0 wavelengths, the light reflected off the top of the film will be 180° out of phase with the light that traveled through the film and was reflected off the glass due to the phase change at reflection off the top of the film. IDENTIFY: Consider the phase difference produced by the path difference and by the reflections. For destructive interference the total phase difference is an integer number of half cycles. SET UP: The reflection at the top surface of the film produces a half-cycle phase shift. There is no phase shift at the reflection at the bottom surface. EXECUTE: (a) Since there is a half-cycle phase shift at just one of the interfaces, the minimum thickness for λ λ 550 nm = 74.3 nm. constructive interference is t = = 0 = 4 4n 4(1.85) (b) The next smallest thickness for constructive interference is with another half wavelength thickness added: 3λ 3λ0 3 ( 550 nm ) = = = 223 nm. t= 4 4n 4(1.85) EVALUATE: Note that we must compare the path difference to the wavelength in the film. IDENTIFY: Consider the interference between rays reflected from the two surfaces of the soap film. Strongly reflected means constructive interference. Consider phase difference due to the path difference of 2t and any phase difference due to phase changes upon reflection. (a) SET UP: Consider Figure 35.33. There is a 180° phase change when the light is reflected from the outside surface of the bubble and no phase change when the light is reflected from the inside surface. Figure 35.33 EXECUTE: The reflections produce a net 180° phase difference and for there to be constructive interference the path difference 2t must correspond to a half-integer number of wavelengths to compensate for the λ / 2 shift due to 1⎞ ⎛ the reflections. Hence the condition for constructive interference is 2t = ⎜ m + ⎟ (λ0 / n), m = 0,1,2,… Here λ0 is 2⎠ ⎝

the wavelength in air and (λ0 / n) is the wavelength in the bubble, where the path difference occurs.

2tn 2(290 nm)(1.33) 771.4 nm = = 1 1 1 m+ m+ m+ 2 2 2 for m = 0, λ = 1543 nm; for m = 1, λ = 514 nm; for m = 2, λ = 308 nm;… Only 514 nm is in the visible region; the color for this wavelength is green. 2tn 2(340 nm)(1.33) 904.4 nm (b) λ0 = = = 1 1 1 m+ m+ m+ 2 2 2 for m = 0, λ = 1809 nm; for m = 1, λ = 603 nm; for m = 2, λ = 362 nm;… Only 603 nm is in the visible region; the color for this wavelength is orange. EVALUATE: The dominant color of the reflected light depends on the thickness of the film. If the bubble has varying thickness at different points, these points will appear to be different colors when the light reflected from the bubble is viewed. IDENTIFY: The number of waves along the path is the path length divided by the wavelength. The path difference and the reflections determine the phase difference.

λ0 =

35.34.

SET UP:

The path length is 2t = 17.52 × 10−6 m . The wavelength in the film is λ =

λ0 n

.


35-10

Chapter 35

2t 17.52 × 10−6 m 648 nm = = 36.5 . = 480 nm . The number of waves is 480 × 10−9 m 1.35 λ (b) The path difference introduces a λ / 2 , or 180° , phase difference. The ray reflected at the top surface of the film undergoes a 180° phase shift upon reflection. The reflection at the lower surface introduces no phase shift. Both rays undergo a 180° phase shift, one due to reflection and one due to reflection. The two effects cancel and the two rays are in phase as they leave the film. EVALUATE: Note that we must use the wavelength in the film to determine the number of waves in the film. IDENTIFY: Require destructive interference between light reflected from the two points on the disc. SET UP: Both reflections occur for waves in the plastic substrate reflecting from the reflective coating, so they both have the same phase shift upon reflection and the condition for destructive interference (cancellation) is EXECUTE:

35.35.

(a) λ =

2t = (m + 12 )λ , where t is the depth of the pit. λ =

35.37.

35.38.

35.39.

λ

λ0

n

. The minimum pit depth is for m = 0 .

790 nm = 110 nm = 0.11 μ m . 2 4 4n 4(1.8) EVALUATE: The path difference occurs in the plastic substrate and we must compare the wavelength in the substrate to the path difference. IDENTIFY: Consider light reflected at the front and rear surfaces of the film. SET UP: At the front surface of the film, light in air ( n = 1.00 ) reflects from the film ( n = 2.62 ) and there is a 180° phase shift due to the reflection. At the back surface of the film, light in the film ( n = 2.62 ) reflects from glass ( n = 1.62 ) and there is no phase shift due to reflection. Therefore, there is a net 180° phase difference produced by the reflections. The path difference for these two rays is 2t, where t is the thickness of the film. The 505 nm wavelength in the film is λ = . 2.62 EXECUTE: (a) Since the reflection produces a net 180° phase difference, destructive interference of the reflected ⎛ 505 nm ⎞ light occurs when 2t = mλ . t = m ⎜ ⎟ = (96.4 nm)m . The minimum thickness is 96.4 nm. ⎝ 2[2.62] ⎠ EXECUTE:

35.36.

λ

λ0

2t =

. t=

=

=

(b) The next three thicknesses are for m = 2 , 3 and 4: 192 nm, 289 nm and 386 nm. EVALUATE: The minimum thickness is for t = λ / 2n . Compare this to Problem 34.27, where the minimum thickness for destructive interference is t = λ / 4n . IDENTIFY and SET UP: Apply Eq.(35.19) and calculate y for m = 1800. EXECUTE: Eq.(35.19): y = m(λ / 2) = 1800(633 × 10−9 m) / 2 = 5.70 × 10−4 m = 0.570 mm EVALUATE: A small displacement of the mirror corresponds to many wavelengths and a large number of fringes cross the line. IDENTIFY: Apply Eq.(35.19). SET UP: m = 818 . Since the fringes move in opposite directions, the two people move the mirror in opposite directions. mλ 818(6.06 × 10−7 m) EXECUTE: (a) For Jan, the total shift was y1 = 1 = = 2.48 × 10−4 m. For Linda, the total shift 2 2 mλ2 818(5.02 × 10−7 m) was y2 = = = 2.05 × 10−4 m. 2 2 (b) The net displacement of the mirror is the difference of the above values: Δy = y1 − y2 = 0.248 mm − 0.205 mm = 0.043 mm. EVALUATE: The person using the larger wavelength moves the mirror the greater distance. IDENTIFY: Consider the interference between light reflected from the top and bottom surfaces of the air film between the lens and the glass plate. 1⎞ ⎛ SET UP: For maximum intensity, with a net half-cycle phase shift due to reflections, 2t = ⎜ m + ⎟ λ . 2⎠ ⎝ t = R − R2 − r 2 . (2m + 1)λ (2m + 1)λ = R − R2 − r 2 ⇒ R2 − r 2 = R − EXECUTE: 4 4 2

(2m + 1)λ R ⎡ (2m + 1)λ ⎤ ⎡ (2m + 1)λ ⎤ (2m + 1)λ R ⇒ R2 − r 2 = R2 + ⎢ − ⇒r= −⎢ ⎥ ⎥⎦ 4 2 2 4 ⎣ ⎦ ⎣ ⇒r≈

(2m + 1)λ R , for R " λ . 2

2


Interference

35-11

The second bright ring is when m = 1: r≈

(2(1) + 1) (5.80 × 10−7 m) (0.952 m) = 9.10 × 10−4 m = 0.910 mm. 2

So the diameter of the second bright ring is 1.82 mm. EVALUATE: The diameter of the m th ring is proportional to increases. This agrees with Figure 35.17b in the textbook. 35.40.

As found in Problem 35.39, the radius of the mth bright ring is r ≈

IDENTIFY:

λair n

n

, where n

.

EXECUTE: EVALUATE:

35.43.

λ

is the refractive index of the liquid. (2m + 1)λ R r 0.850 mm = = = 0.737 mm. EXECUTE: r (n) ≈ 2n n 1.33 EVALUATE: The refractive index of the water is less than that of the glass plate, so the phase changes on reflection are the same as when air is in the space. IDENTIFY: The liquid alters the wavelength of the light and that affects the locations of the interference minima. SET UP: The interference minima are located by d sin θ = (m + 12 )λ . For a liquid with refractive index n,

λliq =

35.42.

(2m + 1)λ R , for R " λ . 2

Introducing a liquid between the lens and the plate just changes the wavelength from λ to

SET UP:

35.41.

2m + 1 , so the rings get closer together as m

sin θ

λ

=

(m + 12 ) sin θ air sin θ liq sin θ air sin θ liq sin θ air sin 35.20° = . = and n = = = 1.730 . = constant , so sin θ liq sin19.46° λair λliq λair λair / n d

In the liquid the wavelength is shorter and sin θ = (m + 12 )

λ d

gives a smaller θ than in air, for the

same m. IDENTIFY: As the brass is heated, thermal expansion will cause the two slits to move farther apart. SET UP: For destructive interference, d sin θ = λ/2. The change in separation due to thermal expansion is dw = αw0 dT, where w is the distance between the slits. EXECUTE: The first dark fringe is at d sin θ = λ/2 ⇒ sin θ = λ/2d. Call d ≡ w for these calculations to avoid confusion with the differential. sin θ = λ/2w Taking differentials gives d(sin θ) = d(λ/2w) and cosθ dθ = − λ/2 dw/w2. λ α w0 dT λα dT =− . Solving for dθ gives For thermal expansion, dw = αw0 dT, which gives cosθ dθ = − 2 2 w0 2w0 λα dT dθ = − . Get λ: w0 sin θ0 = λ/2 → λ = 2w0 sinθ0. Substituting this quantity into the equation for dθ gives 2 w0 cosθ 0 2 w sin θ 0α dT dθ = − 0 = − tan θ 0 α dT . 2w0 cosθ 0 dθ = − tan(32.5°)(2.0 × 10−5 K −1 )(115 K) = −0.001465 rad = −0.084° The minus sign tells us that the dark fringes move closer together. EVALUATE: We can also see that the dark fringes move closer together because sinθ is proportional to 1/d, so as d increases due to expansion, θ decreases. IDENTIFY: Both frequencies will interfere constructively when the path difference from both of them is an integral number of wavelengths. SET UP: Constructive interference occurs when sinθ = mλ/d. EXECUTE: First find the two wavelengths. λ1 = v/f1 = (344 m/s)/(900 Hz) = 0.3822 m

λ2 = v/f2 = (344 m/s)/(1200 Hz) = 0.2867 m To interfere constructively at the same angle, the angles must be the same, and hence the sines of the angles must be equal. Each sine is of the form sin θ = mλ/d, so we can equate the sines to get m1λ1/d = m2λ2/d

m1(0.3822 m) = m2(0.2867 m) m2 = 4/3 m1


35-12

Chapter 35

Since both m1 and m2 must be integers, the allowed pairs of values of m1 and m2 are m1 = m2 = 0

m1 = 3, m2 = 4 m1 = 6, m2 = 8 m1 = 9, m2 = 12 etc.

35.44.

For m1 = m2 = 0, we have θ = 0. For m1 = 3, m2 = 4, we have sin θ1 = (3)(0.3822 m)/(2.50 m), giving θ1 = 27.3° For m1 = 6, m2 = 8, we have sin θ1 = (6)(0.3822 m)/(2.50 m), giving θ1 = 66.5° For m1 = 9, m2 = 12, we have sin θ1 = (9)(0.3822 m)/(2.50 m) = 1.38 > 1, so no angle is possible. EVALUATE: At certain other angles, one frequency will interfere constructively, but the other will not. 1⎞ ⎛ IDENTIFY: For destructive interference, d = r2 − r1 = ⎜ m + ⎟ λ . 2⎠ ⎝ SET UP:

r2 − r1 = (200 m) 2 + x 2 − x 2

⎡⎛ 1⎞ ⎤ 1⎞ ⎛ (200 m) 2 + x 2 = x 2 + ⎢⎜ m + ⎟ λ ⎥ + 2 x ⎜ m + ⎟ λ . 2 2⎠ ⎠ ⎦ ⎝ ⎣⎝ 2 20,000 m 1 ⎛ 1⎞ c 3.00 × 108 m s − ⎜ m + ⎟ λ . The wavelength is calculated by λ = = x= = 51.7 m. 1⎞ f 2⎝ 2⎠ 5.80 × 106 Hz ⎛ ⎜ m + ⎟λ 2⎠ ⎝ m = 0 : x = 761 m; m = 1: x = 219 m; m = 2 : x = 90.1 m; m = 3; x = 20.0 m. EVALUATE: For m = 3 , d = 3.5λ = 181 m . The maximum possible path difference is the separation of 200 m between the sources. IDENTIFY: The two scratches are parallel slits, so the light that passes through them produces an interference pattern. However the light is traveling through a medium (plastic) that is different from air. SET UP: The central bright fringe is bordered by a dark fringe on each side of it. At these dark fringes, d sin θ = ½ λ/n, where n is the refractive index of the plastic. EXECUTE: First use geometry to find the angles at which the two dark fringes occur. At the first dark fringe tanθ = [(5.82 mm)/2]/(3250 mm), giving θ = ±0.0513° For destructive interference, we have d sin θ = ½ λ/n and n = λ/(2dsin θ) = (632.8 nm)/[2(0.000225 m)(sin 0.0513°)] = 1.57 EVALUATE: The wavelength of the light in the plastic is reduced compared to what it would be in air. IDENTIFY: Interference occurs due to the path difference of light in the thin film. SET UP: Originally the path difference was an odd number of half-wavelengths for cancellation to occur. If the path difference decreases by ½ wavelength, it will be a multiple of the wavelength, so constructive interference will occur. EXECUTE: Calling ΔT the thickness that must be removed, we have path difference = 2ΔT = ½ λ/n and ΔT = λ/4n = (525 nm)/[4(1.40)] = 93.75 nm, At 4.20 nm/yr, we have (4.20 nm/yr)t = 93.75 nm and t = 22.3 yr. EVALUATE: If you were giving a warranty on this film, you certainly could not give it a “lifetime guarantee”! IDENTIFY and SET UP: If the total phase difference is an integer number of cycles the interference is constructive and if it is a half-integer number of cycles it is destructive. EXECUTE: (a) If the two sources are out of phase by one half-cycle, we must add an extra half a wavelength to the path difference equations Eq.(35.1) and Eq.(35.2). This exactly changes one for the other, for m → m + 12 and m + 12 → m, since m in any integer. φ (b) If one source leads the other by a phase angle φ , the fraction of a cycle difference is . Thus the path length 2π difference for the two sources must be adjusted for both destructive and constructive interference, by this amount. So for constructive inference: r1 − r2 = ( m + φ 2π )λ , and for destructive interference, r1 − r2 = ( m + 1 2 + φ 2π )λ , where in each case m = 0, ±1, ±2, … EVALUATE: If φ = 0 these results reduce to Eqs.(35.1) and (35.2). IDENTIFY: Follow the steps specified in the problem. SET UP: Use cos(ωt + φ / 2) = cos(ωt ) cos(φ / 2) − sin(ωt )sin(φ / 2) . Then 1 + cos(φ ) 2cos(φ / 2)cos(ωt + φ / 2) = 2cos(ωt )cos 2 (φ / 2) − 2sin(ωt )sin(φ / 2)cos(φ / 2) . Then use cos 2 (φ / 2) = and 2 EXECUTE:

35.45.

35.46.

35.47.

35.48.


Interference

35-13

2sin(φ / 2)cos(φ / 2) = sin φ . This gives cos(ωt ) + (cos(ωt )cos(φ ) − sin(ωt )sin(φ )) = cos(ωt ) + cos(ωt + φ ) , using again the trig identity for the cosine of the sum of two angles. EXECUTE: (a) The electric field is the sum of the two fields and can be written as EP (t ) = E2 (t ) + E1 (t ) = E cos(ωt ) + E cos(ωt + φ ) . EP (t ) = 2 E cos(φ / 2)cos(ωt + φ / 2) . (b) E p (t ) = A cos(ωt + φ / 2), so comparing with part (a), we see that the amplitude of the wave (which is always

positive) must be A = 2 E | cos(φ / 2) | . (c) To have an interference maximum,

φ = 2π m . So, for example, using m = 1, the relative phases are 2

φ = 2π , and all waves are in phase. 2 φ 1⎞ ⎛ (d) To have an interference minimum, = π ⎜ m + ⎟ . So, for example using m = 0, relative phases are E2 : 0; E1: φ = 4π ; E p :

2 2⎠ ⎝ E2 : 0; E1: φ = π ; E p : φ /2 = π /2, and the resulting wave is out of phase by a quarter of a cycle from both of the

original waves. (e) The instantaneous magnitude of the Poynting vector is $% | S |= ε 0cE p2 (t ) = ε 0c (4 E 2 cos 2 (φ 2)cos 2 (ωt + φ 2)).

35.49.

35.50.

1 For a time average, cos 2 (ωt + φ 2) = , so Sav = 2ε 0cE 2 cos 2 (φ 2). 2 EVALUATE: The result of part (e) shows that the intensity at a point depends on the phase difference φ at that point for the waves from each source. IDENTIFY: Follow the steps specified in the problem. SET UP: The definition of hyperbola is the locus of points such that the difference between P to S2 and P to S1 is a constant. EXECUTE: (a) Δr = mλ . r1 = x 2 + ( y − d ) 2 and r2 = x 2 + ( y + d ) 2 . Δr = x 2 + ( y + d ) 2 − x 2 + ( y − d ) 2 = mλ . (b) For a given m and λ , Δr is a constant and we get a hyperbola. Or, in the case of all m for a given λ , a family of hyperbolas. (c) x 2 + ( y + d ) 2 − x 2 + ( y − d ) 2 = ( m + 12 )λ . EVALUATE: The hyperbolas approach straight lines at large distances from the source. IDENTIFY: Follow the derivation of Eq.(35.7), but with different amplitudes for the two waves. SET UP: cos(π − φ ) = − cos φ

EXECUTE:

(a) E p2 = E12 + E22 − 2 E1E2 cos(π − φ ) = E 2 + 4 E 2 + 4 E 2 cos φ = 5E 2 + 4 E 2 cos φ

1 ⎡⎛ 5 ⎞ ⎛ 4 ⎞ ⎤ 9 ⎡5 4 ⎤ I = ε 0cE p 2 = ε 0c ⎢⎜ E 2 ⎟ + ⎜ E 2 ⎟ cos φ ⎥ . φ = 0 ⇒ I 0 = ε 0cE 2 . Therefore, I = I 0 ⎢ + cos φ ⎥ . 9 9 2 2 2 2 ⎣ ⎦ ⎝ ⎠ ⎝ ⎠ ⎣ ⎦ 1 (b) The graph is shown in Figure 35.50. I min = I 0 which occurs when φ = nπ (n odd). 9 EVALUATE: The maxima and minima occur at the same points on the screen as when the two sources have the same amplitude, but when the amplitudes are different the intensity is no longer zero at the minima.

Figure 35.50


35-14

Chapter 35

35.51.

IDENTIFY and SET UP: Consider interference between rays reflected from the upper and lower surfaces of the film to relate the thickness of the film to the wavelengths for which there is destructive interference. The thermal expansion of the film changes the thickness of the film when the temperature changes. EXECUTE: For this film on this glass, there is a net λ / 2 phase change due to reflection and the condition for destructive interference is 2t = m(λ / n), where n = 1.750. Smallest nonzero thickness is given by t = λ / 2n. At 20.0°C, t0 = (582.4 nm) /[(2)(1.750)] = 166.4 nm. At 170°C, t0 = (588.5 nm) /[(2)(1.750)] = 168.1 nm. t = t0 (1 + αΔT ) so

35.52.

35.53.

34.54.

35.55.

α = (t − t0 ) /(t0 ΔT ) = (1.7 nm) /[(166.4 nm)(150°C)] = 6.8 × 10−5 (C°) −1 EVALUATE: When the film is heated its thickness increases, and it takes a larger wavelength in the film to equal 2t.The value we calculated for α is the same order of magnitude as those given in Table 17.1. IDENTIFY and SET UP: At the m = 3 bright fringe for the red light there must be destructive interference at this same θ for the other wavelength. EXECUTE: For constructive interference: d sin θ = mλ1 ⇒ d sin θ = 3(700 nm) = 2100 nm. For destructive 1⎞ d sin θ 2100 nm ⎛ interference: d sin θ = ⎜ m + ⎟ λ2 ⇒ λ2 = = . So the possible wavelengths are 2⎠ m + 12 m + 12 ⎝ λ2 = 600 nm, for m = 3, and λ2 = 467 nm, for m = 4. EVALUATE: Both d and θ drop out of the calculation since their combination is just the path difference, which is the same for both types of light. ⎛πd ⎞ IDENTIFY: Apply I = I 0 cos ⎜ sin θ ⎟ . ⎝ λ ⎠ πd π 3π sin θ is rad , rad ,…. SET UP: I = I 0 / 2 when 4 4 λ EXECUTE: First we need to find the angles at which the intensity drops by one-half from the value of the m th πd π dθ m π ⎛πd ⎞ I bright fringe. I = I 0 cos 2 ⎜ = (m + 1 2) . sin θ ⎟ = 0 ⇒ sin θ ≈ 2 λ λ ⎝ λ ⎠ 2 λ 3λ λ − + m = 0 :θ = θm = ⇒ Δθ m = . ; m = 1: θ = θ m = 4d 4d 2d EVALUATE: There is no dependence on the m-value of the fringe, so all fringes at small angles have the same half-width. IDENTIFY: Consider the phase difference produced by the path difference and by the reflections. SET UP: There is just one half-cycle phase change upon reflection, so for constructive interference 2t = (m1 + 12 )λ1 = (m2 + 12 )λ2 , where these wavelengths are in the glass. The two different wavelengths differ by just one m-value, m2 = m1 − 1. 1⎞ 1⎞ λ + λ2 λ + λ2 ⎛ ⎛ EXECUTE: ⎜ m1 + ⎟ λ1 = ⎜ m1 − ⎟ λ2 ⇒ m1 (λ2 − λ1 ) = 1 . ⇒ m1 = 1 λ2 − λ1 ) 2 2 2 2( ⎝ ⎠ ⎝ ⎠ 477.0 nm + 540.6 nm 1⎞λ 17(477.0 nm) ⎛ m1 = = 8. 2t = ⎜ 8 + ⎟ 01 ⇒ t = = 1334 nm. n 2(540.6 nm − 477.0 nm) 2 4(1.52) ⎝ ⎠ EVALUATE: Now that we have t we can calculate all the other wavelengths for which there is constructive interference. IDENTIFY: Consider the phase difference due to the path difference and due to the reflection of one ray from the glass surface. (a) SET UP: Consider Figure 35.55 path difference = 2 h2 + x2 / 4 − x =

4h 2 + x 2 − x

Figure 35.55


Interference

35-15

Since there is a 180° phase change for the reflected ray, the condition for constructive interference is path 1⎞ ⎛ difference = ⎜ m + ⎟ λ and the condition for destructive interference is path difference = mλ . 2⎠ ⎝ (b) EXECUTE:

1⎞ 4h 2 + x 2 − x ⎛ Constructive interference: ⎜ m + ⎟ λ = 4h 2 + x 2 − x and λ = . Longest λ is for 1 2⎠ ⎝ m+ 2

m = 0 and then λ = 2 35.56.

35.57.

35.58.

(

) (

4h 2 + x 2 − x = 2

EVALUATE: For λ = 0.72 m the path difference is λ / 2. IDENTIFY: Require constructive interference for the reflection from the top and bottom surfaces of each cytoplasm layer and each guanine layer. SET UP: At the water (or cytoplasm) to guanine interface, there is a half-cycle phase shift for the reflected light, but there is not one at the guanine to cytoplasm interface. Therefore there will always be one half-cycle phase difference between two neighboring reflected beams, just due to the reflections. EXECUTE: For the guanine layers: 2t n 1 λ 2(74 nm) (1.80) 266 nm = ⇒ λ = 533 nm (m = 0). 2tg = ( m + ) ⇒ λ = g g1 = 2 ng (m + 2 ) (m + 12 ) (m + 12 )

For the cytoplasm layers: 1⎞ λ 2tc nc 2(100 nm) (1.333) 267 nm ⎛ = = ⇒ λ = 533 nm (m = 0). 2tc = ⎜ m + ⎟ ⇒ λ = 1 + 2 ( ) (m + 12 ) (m + 12 ) n m ⎝ ⎠ c 2 (b) By having many layers the reflection is strengthened, because at each interface some more of the transmitted light gets reflected back, increasing the total percentage reflected. (c) At different angles, the path length in the layers changes (always to a larger value than the normal incidence case). If the path length changes, then so do the wavelengths that will interfere constructively upon reflection. EVALUATE: The thickness of the guanine and cytoplasm layers are inversely proportional to their refractive ⎛ 100 1.80 ⎞ indices ⎜ = ⎟ , so both kinds of layers produce constructive interference for the same wavelength in air. ⎝ 74 1.333 ⎠ IDENTIFY: The slits will produce an interference pattern, but in the liquid, the wavelength of the light will be less than it was in air. SET UP: The first bright fringe occurs when d sin θ = λ/n. EXECUTE: In air: dsin18.0° = λ. In the liquid: dsin12.6° = λ/n. Dividing the equations gives n = (sin 18.0°)/(sin 12.6°) = 1.42 EVALUATE: It was not necessary to know the spacing of the slits, since it was the same in both air and the liquid. IDENTIFY: Consider light reflected at the top and bottom surfaces of the film. Wavelengths that are predominant in the transmitted light are those for which there is destructive interference in the reflected light. SET UP: For the waves reflected at the top surface of the oil film there is a half-cycle reflection phase shift. For the waves reflected at the bottom surface of the oil film there is no reflection phase shift. The condition for constructive interference is 2t = (m + 12 )λ . The condition for destructive interference is 2t = mλ . The range of visible wavelengths is approximately 400 nm to 700 nm. In the oil film, λ = EXECUTE:

35.59.

)

4(0.24 m) 2 + (0.14 m) 2 − 0.14 m = 0.72 m

λ0

. n λ 2tn 2(380 nm)(1.45) 1102 nm . (a) 2t = (m + 12 )λ = (m + 12 ) 0 . λ0 = = = 1 m+ 2 m + 12 m + 12 n

m = 0 : λ0 = 2200 nm . m = 1 : λ0 = 735 nm . m = 2 : λ0 = 441 nm . m = 3 : λ0 = 315 nm . The visible wavelength for which there is constructive interference in the reflected light is 441 nm. λ 2tn 1102 nm (b) 2t = mλ = m 0 . λ0 = = . m = 1 : λ0 = 1102 nm . m = 2 : λ0 = 551 nm . m = 3 : λ0 = 367 nm . m m n The visible wavelength for which there is destructive interference in the reflected light is 551 nm. This is the visible wavelength predominant in the transmitted light. EVALUATE: At a particular wavelength the sum of the intensities of the reflected and transmitted light equals the intensity of the incident light. (a) IDENTIFY: The wavelength in the glass is decreased by a factor of 1/ n, so for light through the upper slit a shorter path is needed to produce the same phase at the screen. Therefore, the interference pattern is shifted downward on the screen. (b) SET UP: Consider the total phase difference produced by the path length difference and also by the different wavelength in the glass.


35-16

Chapter 35

At a point on the screen located by the angle θ the difference in path length is d sin θ . This

EXECUTE:

⎛ 2π ⎞ introduces a phase difference of φ = ⎜ ⎟ (d sin θ ), where λ0 is the wavelength of the light in air or vacuum. ⎝ λ0 ⎠ L nL In the thickness L of glass the number of wavelengths is = . A corresponding length L of the path of the ray

λ

λ0

through the lower slit, in air, contains L / λ0 wavelengths. The phase difference this introduces is ⎛ nL

φ = 2π ⎜

⎝ λ0

L⎞ ⎟ and φ = 2π (n − 1)( L / λ0 ). The total phase difference is the sum of these two, λ0 ⎠

⎛ 2π ⎞ ⎜ ⎟ (d sin θ ) + 2π (n − 1)( L / λ0 ) = (2π / λ0 )( d sin θ + L(n − 1)). Eq.(35.10) then gives ⎝ λ0 ⎠ ⎡⎛ π ⎞ ⎤ I = I 0 cos 2 ⎢⎜ ⎟ (d sin θ + L( n − 1)) ⎥ . ⎣⎢⎝ λ0 ⎠ ⎦⎥ (c) Maxima means cos φ / 2 = ±1 and φ / 2 = mπ , m = 0, ± 1, ± 2, … (π / λ0 )(d sin θ + L( n − 1)) = mπ d sin θ + L(n − 1) = mλ0 mλ0 − L(n − 1) d EVALUATE: When L → 0 or n → 1 the effect of the plate goes away and the maxima are located by Eq.(35.4). IDENTIFY: Dark fringes occur because the path difference is one-half of a wavelength. ⎛ π d sin θ ⎞ SET UP: At the first dark fringe, dsinθ = λ/2. The intensity at any angle θ is given by I = I 0 cos 2 ⎜ ⎟. ⎝ λ ⎠ (a) At the first dark fringe, we have d sin θ = λ/2 sin θ =

35.60.

d/λ = 2/(2 sin 15.0°) = 1.93 1 ⎛ π d sin θ ⎞ I 0 ⎛ π d sin θ ⎞ (b) I = I 0 cos 2 ⎜ ⇒ cos ⎜ ⎟= ⎟= λ λ 10 10 ⎝ ⎠ ⎝ ⎠

π d sin θ ⎛ 1 ⎞ = arccos ⎜ ⎟ = 71.57° = 1.249 rad λ ⎝ 10 ⎠

35.61.

Using the result from part (a), that d/λ = 1.93, we have π(1.93)sin θ = 1.249. sin θ = 0.2060 and θ = ±11.9° EVALUATE: Since the first dark fringes occur at ±15.0°, it is reasonable that at ≈12° the intensity is reduced to only 1/10 of its maximum central value. IDENTIFY: There are two effects to be considered: first, the expansion of the rod, and second, the change in the rod’s refractive index.

λ=

λ0

and Δn = n0 (2.50 × 10−5 (C°) −1 ) ΔT . ΔL = L0 (5.00 × 10−6 (C°) −1 ) ΔT . n EXECUTE: The extra length of rod replaces a little of the air so that the change in the number of wavelengths due 2n ΔL 2n ΔL 2(nglass − 1) L0αΔT = and to this is given by: ΔN1 = glass − air SET UP:

λ0

λ0

λ0

−6

2(1.48 − 1)(0.030 m)(5.00 × 10 C°)(5.00 C°) = 1.22. 5.89 × 10−7 m The change in the number of wavelengths due to the change in refractive index of the rod is: 2Δnglass L0 2(2.50 × 10 −5 C°)(5.00 C° min)(1.00 min)(0.0300 m) ΔN 2 = = = 12.73. λ0 5.89 × 10−7 m So, the total change in the number of wavelengths as the rod expands is ΔN = 12.73 + 1.22 = 14.0 fringes/minute. EVALUATE: Both effects increase the number of wavelengths along the length of the rod. Both ΔL and Δnglass are very small and the two effects can be considered separately. ΔN1 =

35.62.

IDENTIFY:

Apply Snell's law to the refraction at the two surfaces of the prism. S1 and S2 serve as coherent

sources so the fringe spacing is Δy =

Rλ , where d is the distance between S1 and S2 . d


Interference

35-17

SET UP: For small angles, sin θ ≈ θ , with θ expressed in radians. EXECUTE: (a) Since we can approximate the angles of incidence on the prism as being small, Snell’s Law tells us that an incident angle of θ on the flat side of the prism enters the prism at an angle of θ n , where n is the index of refraction of the prism. Similarly on leaving the prism, the in-going angle is θ / n − A from the normal, and the outgoing angle, relative to the prism, is n(λ n − A). So the beam leaving the prism is at an angle of

θ ′ = n(θ n − A) + A from the optical axis. So θ − θ ′ = (n − 1) A. At the plane of the source S0 , we can calculate the d = tan(θ − θ ′) a ≈ (θ − θ ′)a = (n − 1) Aa ⇒ d = 2aA(n − 1). 2 (b) To find the spacing of fringes on a screen, we use (2.00 m + 0.200 m) (5.00 × 10−7 m) Rλ Rλ Δy = = = = 1.57 × 10−3 m. 2aA( n − 1) 2(0.200 m) (3.50 × 10−3 rad) (1.50 − 1.00) d EVALUATE: The fringe spacing is proportional to the wavelength of the light. The biprism serves as an alternative to two closely spaced narrow slits. height of one image above the source:


36

DIFFRACTION

36.1.

IDENTIFY: Use y = x tan θ to calculate the angular position θ of the first minimum. The minima are located by mλ , m = ±1, ± 2,… First minimum means m = 1 and sin θ1 = λ / a and λ = a sin θ1. Use this Eq.(36.2): sin θ = a equation to calculate λ . SET UP: The central maximum is sketched in Figure 36.1. EXECUTE: y1 = x tan θ1 y tan θ1 = 1 = x 1.35 × 10−3 m = 0.675 × 10−3 2.00 m θ1 = 0.675 × 10−3 rad Figure 36.1

λ = a sin θ1 = (0.750 × 10−3 m)sin(0.675 × 10−3 rad) = 506 nm EVALUATE: θ1 is small so the approximation used to obtain Eq.(36.3) is valid and this equation could have been used. 36.2.

IDENTIFY: The angle is small, so ym = x SET UP:

y1 = 10.2 mm

xλ xλ (0.600 m)(5.46 × 10−7 m) ⇒a = = 3.21 × 10−5 m. a y1 10.2 × 10−3 m EVALUATE: The diffraction pattern is observed at a distance of 60.0 cm from the slit. mλ IDENTIFY: The dark fringes are located at angles θ that satisfy sin θ = , m = ±1, ± 2, .... a SET UP: The largest value of sin θ is 1.00. EXECUTE:

36.3.

mλ . a

y1 =

EXECUTE: (a) Solve for m that corresponds to sin θ = 1 : m =

a

λ

=

0.0666 × 10−3 m = 113.8. The largest value m 585 × 10−9 m

can have is 113. m = ±1 , ±2 , …, ±113 gives 226 dark fringes. ⎛ 585 × 10−9 m ⎞ (b) For m = ±113, sin θ = ±113 ⎜ ⎟ = ±0.9926 and θ = ±83.0° . −3 ⎝ 0.0666 × 10 m ⎠

36.4.

EVALUATE: When the slit width a is decreased, there are fewer dark fringes. When a < λ there are no dark fringes and the central maximum completely fills the screen. mλ IDENTIFY and SET UP: λ / a is very small, so the approximate expression ym = R is accurate. The distance a between the two dark fringes on either side of the central maximum is 2 y1 .

36.5.

λR

(633 × 10−9 m)(3.50 m) = 2.95 × 10 −3 m = 2.95 mm . 2 y1 = 5.90 mm . a 0.750 × 10 −3 m EVALUATE: When a is decreased, the width 2 y1 of the central maximum increases. mλ IDENTIFY: The minima are located by sin θ = a SET UP: a = 12.0 cm . x = 40.0 cm . EXECUTE:

y1 =

=

36-1


36-2

36.6.

Chapter 36

⎛ 9.00 cm ⎞ ⎛λ⎞ EXECUTE: The angle to the first minimum is θ = arcsin ⎜ ⎟ = arcsin ⎜ ⎟ = 48.6°. a ⎝ ⎠ ⎝ 12.00 cm ⎠ So the distance from the central maximum to the first minimum is just y1 = x tan θ = (40.0 cm) tan(48.6°) = ±45.4 cm. EVALUATE: 2λ / a is greater than 1, so only the m = 1 minimum is seen. IDENTIFY: The angle that locates the first diffraction minimum on one side of the central maximum is given by λ 1 v sin θ = . The time between crests is the period T. f = and λ = . a T f SET UP: The time between crests is the period, so T = 1.0 h . 1 1 v 800 km/h EXECUTE: (a) f = = = 1.0 h −1 . λ = = = 800 km . T 1.0 h f 1.0 h −1 800 km and θ = 10.2° . 4500 km 800 km Australia-Antarctica: sin θ = and θ = 12.5° . 3700 km EVALUATE: Diffraction effects are observed when the wavelength is about the same order of magnitude as the dimensions of the opening through which the wave passes. IDENTIFY: We can model the hole in the concrete barrier as a single slit that will produce a single-slit diffraction pattern of the water waves on the shore. SET UP: For single-slit diffraction, the angles at which destructive interference occurs are given by sinθm = mλ/a, where m = 1, 2, 3, …. EXECUTE: (a) The frequency of the water waves is f = 75.0 min −1 = 1.25 s −1 = 1.25 Hz, so their wavelength is λ = v/f = (15.0 cm/s)/(1.25 Hz) = 12.0 cm. At the first point for which destructive interference occurs, we have tan θ = (0.613 m)/(3.20 m) ⇒ θ = 10.84°. a sin θ = λ and a = λ/sin θ = (12.0 cm)/(sin 10.84°) = 63.8 cm. (b) First find the angles at which destructive interference occurs. sin θ2 = 2λ/a = 2(12.0 cm)/(63.8 cm) → θ2 = ±22.1°

(b) Africa-Antarctica: sin θ =

36.7.

sin θ3 = 3λ/a = 3(12.0 cm)/(63.8 cm) → θ3 = ±34.3° sin θ4 = 4λ/a = 4(12.0 cm)/(63.8 cm) → θ4 = ±48.8°

36.8.

sin θ5 = 5λ/a = 5(12.0 cm)/(63.8 cm) → θ5 = ±70.1° EVALUATE: These are large angles, so we cannot use the approximation that θm ≈ mλ/a. mλ IDENTIFY: The minima are located by sin θ = . For part (b) apply Eq.(36.7). a SET UP: For the first minimum, m = 1 . The intensity at θ = 0 is I 0 . mλ mλ λ = sin 90.0° = 1 = = . Thus a = λ = 580 nm = 5.80 × 10 −4 mm. a a a (b) According to Eq.(36.7),

EXECUTE: (a) sinθ =

I ⎧⎪ sin [π a (sin θ ) λ ] ⎪⎫ ⎪⎧ sin [π (sin π / 4) ] ⎪⎫ =⎨ ⎬ =⎨ ⎬ = 0.128. I 0 ⎪⎩ π a (sin θ ) λ ⎭⎪ ⎪⎩ π (sin π / 4) ⎭⎪ 2

2

I ⎧⎪ sin [ (π / 2)(sin π / 4) ] ⎫⎪ =⎨ ⎬ = 0.81. As a / λ decreases, I 0 ⎪⎩ (π / 2)(sin π / 4) ⎪⎭ 2

EVALUATE: If a = λ / 2 , for example, then at θ = 45° , 36.9.

the screen becomes more uniformly illuminated. IDENTIFY and SET UP: v = f λ gives λ . The person hears no sound at angles corresponding to diffraction minima. The diffraction minima are located by sin θ = mλ / a, m = ±1, ± 2,… Solve for θ . EXECUTE: λ = v / f = (344 m/s) /(1250 Hz) = 0.2752 m; a = 1.00 m. m = ±1, θ = ±16.0°; m = ±2, θ = ±33.4°; m = ±3, θ = ±55.6°; no solution for larger m EVALUATE: λ / a = 0.28 so for the large wavelength sound waves diffraction by the doorway is a large effect. Diffraction would not be observable for visible light because its wavelength is much smaller and λ / a V 1.


Diffraction

36.10.

IDENTIFY: Compare E y to the expression E y = Emax sin( kx − ωt ) and determine k, and from that calculate λ . mλ . a SET UP: c = 3.00 × 108 m/s . The first dark band corresponds to m = 1. 2π 2π 2π EXECUTE: (a) E = Emax sin( kx − ωt ) . k = ⇒λ = = = 5.24 × 10−7 m . λ k 1.20 × 107 m −1 c 3.0 × 108 m s fλ =c⇒ f = = = 5.73 × 1014 Hz . λ 5.24 × 10−7 m λ 5.24 × 10−7 m a= = = 1.09 × 10−6 m . (b) a sin θ = λ . sin θ sin 28.6° −7 (c) a sin θ = mλ (m = 1, 2, 3, . . .). sin θ 2 = ±2 λ = ±2 5.24 × 10−6 m and θ 2 = ±74D . a 1.09 × 10 m mλ EVALUATE: For m = 3 , is greater than 1 so only the first and second dark bands appear. a IDENTIFY and SET UP: sinθ = λ / a locates the first minimum. y = x tan θ . EXECUTE: tanθ = y x = (36.5 cm) (40.0 cm) and θ = 42.38°.

f = c / λ . The dark bands are located by sin θ =

36.11.

36.12.

a = λ sinθ = (620 × 10 −9 m) (sin 42.38°) = 0.920 μ m EVALUATE: θ = 0.74 rad and sin θ = 0.67 , so the approximation sin θ ≈ θ would not be accurate. mλ IDENTIFY: The angle is small, so ym = x applies. a SET UP: The width of the central maximum is 2y1 , so y1 = 3.00 mm .

xλ xλ (2.50 m)(5.00 × 10−7 m) ⇒a= = = 4.17 × 10−4 m. a y1 3.00 × 10−3m xλ (2.50 m)(5.00 × 10−5 m) = = 4.17 × 10−2 m = 4.2 cm. (b) a = y1 3.00 × 10 −3 m xλ (2.50 m)(5.00 × 10−10 m) = = 4.17 × 10−7 m. (c) a = y1 3.00 × 10−3 m EVALUATE: The ratio a / λ stays constant, so a is smaller when λ is smaller. IDENTIFY: Calculate the angular positions of the minima and use y = x tan θ to calculate the distance on the screen between them. (a) SET UP: The central bright fringe is shown in Figure 36.13a. EXECUTE: The first minimum is located by EXECUTE: (a) y1 =

36.13.

sin θ1 =

λ

= a 633 × 10−9 m = 1.809 × 10−3 0.350 × 10−3 m θ1 = 1.809 × 10 −3 rad Figure 36.13a y1 = x tan θ1 = (3.00 m) tan(1.809 × 10−3 rad) = 5.427 × 10 −3 m w = 2 y1 = 2(5.427 × 10−3 m) = 1.09 × 10 −2 m = 10.9 mm (b) SET UP: The first bright fringe on one side of the central maximum is shown in Figure 36.13b. EXECUTE: w = y2 − y1 y1 = 5.427 × 10−3 m (part (a)) 2λ sin θ 2 = = 3.618 × 10−3 a θ 2 = 3.618 × 10−3 rad y2 = x tan θ 2 = 1.085 × 10−2 m

Figure 36.13b w = y2 − y1 = 1.085 × 10 −2 m − 5.427 × 10 −3 m = 5.4 mm EVALUATE: The central bright fringe is twice as wide as the other bright fringes.

36-3


36-4

Chapter 36

36.14.

IDENTIFY:

2

SET UP: EXECUTE:

⎛ sin( β / 2) ⎞ 2π I = I0 ⎜ a sin θ . ⎟ . β= λ ⎝ β /2 ⎠ The angle θ is small, so sin θ ≈ tan θ ≈ y / x .

β=

2π a

λ

(a) y = 1.00 × 10−3 m :

sin θ ≈

β 2

=

2π a y 2π (4.50 × 10−4 m) = y = (1520 m −1 ) y. λ x (6.20 × 10−7 m)(3.00 m)

(1520 m −1 )(1.00 × 10−3 m) = 0.760. 2 2

⎛ sin( β 2) ⎞ ⎛ sin(0.760) ⎞ ⇒ I = I0 ⎜ ⎟ = I0 ⎜ ⎟ = 0.822 I 0 ⎝ 0.760 ⎠ ⎝ β 2 ⎠

(b) y = 3.00 × 10−3 m :

β 2

=

2

(1520 m −1 )(3.00 × 10−3 m) = 2.28. 2 2

⎛ sin( β 2) ⎞ ⎛ sin(2.28) ⎞ ⇒ I = I0 ⎜ ⎟ = I0 ⎜ ⎟ = 0.111I 0 . β 2 ⎝ 2.28 ⎠ ⎝ ⎠

(c) y = 5.00 × 10−3 m :

β 2

=

2

(1520 m −1 )(5.00 × 10−3 m) = 3.80. 2 2

⎛ sin( β 2) ⎞ ⎛ sin(3.80) ⎞ ⇒ I = I0 ⎜ ⎟ = I0 ⎜ ⎟ = 0.0259 I 0 . β 2 ⎝ 3.80 ⎠ ⎝ ⎠

(6.20 × 10−7 m)(3.00 m) = 4.1 mm . The distances in parts (a) a 4.50 × 10−4 m and (b) are within the central maximum. y = 5.00 mm is within the first secondary maximum. (a) IDENTIFY: Use Eq.(36.2) with m = 1 to locate the angular position of the first minimum and then use y = x tan θ to find its distance from the center of the screen. SET UP: The diffraction pattern is sketched in Figure 36.15. EVALUATE: The first minimum occurs at y1 =

36.15.

λx

2

=

sin θ1 =

λ

= a 540 × 10−9 m = 2.25 × 10−3 0.240 × 10−3 m θ1 = 2.25 × 10−3 rad

Figure 36.15 y1 = x tan θ1 = (3.00 m) tan(2.25 × 10 −3 rad) = 6.75 × 10−3 m = 6.75 mm (b) IDENTIFY and SET UP: Use Eqs.(36.5) and (36.6) to calculate the intensity at this point. EXECUTE: Midway between the center of the central maximum and the first minimum implies 1 y = (6.75 mm) = 3.375 × 10−3 m. 2 y 3.375 × 10−3 m tan θ = = = 1.125 × 10−3 ; θ = 1.125 × 10−3 rad x 3.00 m The phase angle β at this point on the screen is 2π ⎛ 2π ⎞ (0.240 × 10−3 m)sin(1.125 × 10 −3 rad) = π . ⎟ a sin θ = −9 540 10 m λ × ⎝ ⎠

β =⎜

2

⎛ sin β / 2 ⎞ 2 ⎛ sin π / 2 ⎞ −6 Then I = I 0 ⎜ ⎟ = (6.00 × 10 W/m ) ⎜ ⎟ ⎝ π /2 ⎠ ⎝ β /2 ⎠ ⎛ 4 ⎞ I = ⎜ 2 ⎟ (6.00 × 10−6 W/m 2 ) = 2.43 × 10 −6 W/m 2 . ⎝π ⎠ EVALUATE: The intensity at this point midway between the center of the central maximum and the first minimum is less than half the maximum intensity. Compare this result to the corresponding one for the two-slit pattern, Exercise 35.23. 2


Diffraction

36.16.

36-5

IDENTIFY: In the single-slit diffraction pattern, the intensity is a maximum at the center and zero at the dark spots. At other points, it depends on the angle at which one is observing the light. SET UP: Dark fringes occur when sin θm = mλ/a, where m = 1, 2, 3, …, and the intensity is given by 2

⎛ sin β / 2 ⎞ π a sin θ . I0 ⎜ ⎟ , where β / 2 = λ ⎝ β /2 ⎠ EXECUTE: (a) At the maximum possible angle, θ = 90°, so

mmax = (asin90°)/λ = (0.0250 mm)/(632.8 nm) = 39.5 Since m must be an integer and sin θ must be ≤ 1, mmax = 39. The total number of dark fringes is 39 on each side of the central maximum for a total of 78. (b) The farthest dark fringe is for m = 39, giving sinθ39 = (39)(632.8 nm)/(0.0250 mm) ⇒ θ39 = ±80.8° (c) The next closer dark fringe occurs at sinθ38 = (38)(632.8 nm)/(0.0250 mm) ⇒θ38 = 74.1°. The angle midway these two extreme fringes is (80.8° + 74.1°)/2 = 77.45°, and the intensity at this angle is I = 2

⎛ sin β / 2 ⎞ π a sin θ π (0.0250 mm)sin(77.45°) = = 121.15 rad, which I0 ⎜ ⎟ , where β / 2 = λ 632.8 nm β / 2 ⎝ ⎠ 2

36.17.

⎡ sin(121.15 rad) ⎤ -4 2 gives I = ( 8.50 W/m 2 ) ⎢ ⎥ = 5.55 × 10 W/m ⎣ 121.15 rad ⎦ EVALUATE: At the angle in part (c), the intensity is so low that the light would be barely perceptible. IDENTIFY and SET UP: Use Eq.(36.6) to calculate λ and use Eq.(36.5) to calculate I. θ = 3.25°, β = 56.0 rad, a = 0.105 × 10 −3 m. ⎛ 2π ⎞ ⎟ a sin θ so ⎝ λ ⎠

β =⎜

(a) EXECUTE:

λ=

2π a sin θ

β

=

2π (0.105 × 10−3 m)sin 3.25° = 668 nm 56.0 rad 2

⎛ sin β / 2 ⎞ ⎛ 4 ⎞ 4 2 (b) I = I 0 ⎜ [sin(28.0 rad)]2 = 9.36 × 10 −5 I 0 ⎟ = I 0 ⎜ 2 ⎟ (sin( β / 2)) = I 0 (56.0 rad) 2 ⎝ β /2 ⎠ ⎝β ⎠ EVALUATE: At the first minimum β = 2π rad and at the point considered in the problem β = 17.8π rad, so the point is well outside the central maximum. Since β is close to mπ with m = 18, this point is near one of the minima. The intensity here is much less than I 0 .

36.18.

IDENTIFY: Use β =

2π a

λ

sin θ to calculate β .

SET UP: The total intensity is given by drawing an arc of a circle that has length E0 and finding the length of the chord which connects the starting and ending points of the curve. E 2 2π a 2π a λ EXECUTE: (a) β = = π . From Figure 36.18a, π p = E0 ⇒ E p = E0 . sin θ = λ λ 2a 2 π 2

4I ⎛2⎞ The intensity is I = ⎜ ⎟ I 0 = 20 = 0.405 I 0 . This agrees with Eq.(36.5). π ⎝π ⎠ 2π a 2π a λ (b) β = = 2π . From Figure 36.18b, it is clear that the total amplitude is zero, as is the intensity. sin θ = λ λ a This also agrees with Eq.(36.5). E 2 2π a 2π a 3λ E0 . The intensity is (c) β = = 3π . From Figure 36.18c, 3π p = E0 ⇒ E p = sin θ = λ λ 2a 2 3π 2

4 ⎛ 2 ⎞ I = ⎜ ⎟ I 0 = 2 I 0 . This agrees with Eq.(36.5). 9π ⎝ 3π ⎠


36-6

Chapter 36

EVALUATE: In part (a) the point is midway between the center of the central maximum and the first minimum. In part (b) the point is at the first maximum and in (c) the point is approximately at the location of the first secondary maximum. The phasor diagrams help illustrate the rapid decrease in intensity at successive maxima.

Figure 36.18 36.19.

IDENTIFY: The space between the skyscrapers behaves like a single slit and diffracts the radio waves. SET UP: Cancellation of the waves occurs when a sin θ = mλ, m = 1, 2, 3, …, and the intensity of the waves is 2

⎛ sin β / 2 ⎞ π a sin θ . given by I 0 ⎜ ⎟ , where β / 2 = λ β / 2 ⎝ ⎠ EXECUTE: (a) First find the wavelength of the waves: λ = c/f = (3.00 × 108 m/s)/(88.9 MHz) = 3.375 m For no signal, a sin θ = mλ. m = 1: sin θ1 = (1)(3.375 m)/(15.0 m) ⇒ θ1 = ±13.0° m = 2: sin θ2 = (2)(3.375 m)/(15.0 m) ⇒ θ2 = ±26.7° m = 3: sin θ3 = (3)(3.375 m)/(15.0 m) ⇒ θ3 = ±42.4° m = 4: sin θ4 = (4)(3.375 m)/(15.0 m) ⇒ θ4 = ±64.1° 2

⎛ sin β / 2 ⎞ π a sin θ π (15.0 m)sin(5.00°) = = 1.217 rad (b) I 0 ⎜ ⎟ , where β / 2 = λ 3.375 m β / 2 ⎝ ⎠ 2

⎡ sin(1.217 rad) ⎤ 2 I = ( 3.50 W/m 2 ) ⎢ ⎥ = 2.08 W/m ⎣ 1.217 rad ⎦

36.20.

EVALUATE: The wavelength of the radio waves is very long compared to that of visible light, but it is still considerably shorter than the distance between the buildings. IDENTIFY: The net intensity is the product of the factor due to single-slit diffraction and the factor due to double slit interference. 2

φ⎞ ⎛ sin β / 2 ⎞ ⎛ SET UP: The double-slit factor is I DS = I 0 ⎜ cos 2 ⎟ and the single-slit factor is ISS = ⎜ ⎟ . 2⎠ ⎝ ⎝ β /2 ⎠ EXECUTE: (a) d sinθ = mλ ⇒ sinθ = mλ/d. sinθ1 = λ/d, sinθ2 = 2λ/d, sinθ3 = 3λ/d, sinθ4 = 4λ/d π a sin θ π (d / 3)sin θ (b) At the interference bright fringes, cos2φ/2 = 1 and β / 2 = = .

λ λ π ( d / 3)(λ / d ) = π / 3 . The intensity is therefore At θ1, sin θ1 = λ/d, so β / 2 = λ 2

φ ⎞ ⎛ sin β / 2 ⎞ ⎛ sin π / 3 ⎞ ⎛ I1 = I 0 ⎜ cos 2 ⎟ ⎜ ⎟ = I 0 (1) ⎜ ⎟ = 0.684 I0 β 2 / 2 ⎝ ⎠⎝ ⎝ π /3 ⎠ ⎠ At θ2, sin θ2 = 2λ/d, so β / 2 =

2

π ( d / 3)(2λ / d ) = 2π / 3 . Using the same procedure as for θ1, we have I2 = λ

2

⎛ sin 2π / 3 ⎞ I 0 (1) ⎜ ⎟ = 0.171 I0 ⎝ 2π / 3 ⎠ At θ3, we get β / 2 = π , which gives I3 = 0 since sin π = 0. 2

⎛ sin 4π / 3 ⎞ At θ4, sin θ4 = 4λ/d, so β / 2 = 4π / 3 , which gives I 4 = I 0 ⎜ ⎟ = 0.0427 I0 ⎝ 4π / 3 ⎠ (c) Since d = 3a, every third interference maximum is missing. (d) In Figure 36.12c in the textbook, every fourth interference maximum at the sides is missing because d = 4a.


Diffraction

36.21.

36-7

EVALUATE: The result in this problem is different from that in Figure 36.12c because in this case d = 3a, so every third interference maximum at the sides is missing. Also the “envelope” of the intensity function decreases more rapidly here than in Figure 36.12c because the first diffraction minimum is reached sooner, and the decrease in intensity from one interference maximum to the next is faster for a = d/3 than for a = d/4. (a) IDENTIFY and SET UP: The interference fringes (maxima) are located by d sin θ = mλ , with 2

⎛ sin β / 2 ⎞ ⎛ 2π ⎞ m = 0, ± 1, ± 2, …. The intensity I in the diffraction pattern is given by I = I 0 ⎜ ⎟ , with β = ⎜ ⎟ a sin θ . β / 2 ⎝ λ ⎠ ⎝ ⎠ We want m = ±3 in the first equation to give θ that makes I = 0 in the second equation. ⎛ 2π ⎞ ⎛ 3λ ⎞ EXECUTE: d sin θ = mλ gives β = ⎜ ⎟ a ⎜ ⎟ = 2π (3a / d ). ⎝ λ ⎠ ⎝ d ⎠ sin β / 2 = 0 so β = 2π and then 2π = 2π (3a / d ) and ( d / a ) = 3. β /2 (b) IDENTIFY and SET UP: Fringes m = 0, ± 1, ± 2 are within the central diffraction maximum and the m = ±3 fringes coincide with the first diffraction minimum. Find the value of m for the fringes that coincide with the second diffraction minimum. EXECUTE: Second minimum implies β = 4π .

I = 0 says

⎛ 2π ⎞ ⎛ 2π ⎞ ⎛ mλ ⎞ β = ⎜ ⎟ a sin θ = ⎜ ⎟ a ⎜ ⎟ = 2π m(a / d ) = 2π (m / 3) ⎝ λ ⎠ ⎝ λ ⎠ ⎝ d ⎠ Then β = 4π says 4π = 2π ( m / 3) and m = 6. Therefore the m = ±4 and m = +5 fringes are contained within the

36.22.

36.23.

first diffraction maximum on one side of the central maximum; two fringes. EVALUATE: The central maximum is twice as wide as the other maxima so it contains more fringes. IDENTIFY and SET UP: Use Figure 36.14b in the textbook. There is totally destructive interference between slits whose phasors are in opposite directions. EXECUTE: By examining the diagram, we see that every fourth slit cancels each other. EVALUATE: The total electric field is zero so the phasor diagram corresponds to a point of zero intensity. The first two maxima are at φ = 0 and φ = π , so this point is not midway between two maxima. (a) IDENTIFY and SET UP: If the slits are very narrow then the central maximum of the diffraction pattern for each slit completely fills the screen and the intensity distribution is given solely by the two-slit interference. The maxima are given by d sin θ = mλ so sin θ = mλ / d . Solve for θ . λ 580 × 10−9 m EXECUTE: 1st order maximum: m = 1, so sin θ = = = 1.094 × 10−3 ; θ = 0.0627° d 0.530 × 10−3 m 2λ 2nd order maximum: m = 2, so sin θ = = 2.188 × 10−3 ; θ = 0.125° d 2

⎛ sin β / 2 ⎞ (b) IDENTIFY and SET UP: The intensity is given by Eq.(36.12): I = I 0 cos (φ / 2) ⎜ ⎟ . Calculate φ and ⎝ β /2 ⎠ β at each θ from part (a). 2

⎛ 2π d ⎞ ⎛ 2π d ⎞⎛ mλ ⎞ 2 2 ⎟ sin θ = ⎜ ⎟⎜ ⎟ = 2π m, so cos (φ / 2) = cos ( mπ ) = 1 d λ λ ⎝ ⎠ ⎝ ⎠⎝ ⎠ (Since the angular positions in part (a) correspond to interference maxima.) ⎛ 2π a ⎞ ⎛ 2π a ⎞⎛ mλ ⎞ ⎛ 0.320 mm ⎞ β =⎜ ⎟ sin θ = ⎜ ⎟⎜ ⎟ = 2π m(a / d ) = m 2π ⎜ ⎟ = m(3.794 rad) ⎝ λ ⎠ ⎝ λ ⎠⎝ d ⎠ ⎝ 0.530 mm ⎠ EXECUTE:

φ =⎜

2

⎛ sin(3.794 / 2) rad ⎞ 1st order maximum: m = 1, so I = I 0 (1) ⎜ ⎟ = 0.249 I 0 ⎝ (3.794 / 2) rad ⎠ 2

36.24.

⎛ sin 3.794 rad ⎞ 2nd order maximum: m = 2, so I = I 0 (1) ⎜ ⎟ = 0.0256 I 0 ⎝ 3.794 rad ⎠ EVALUATE: The first diffraction minimum is at an angle θ given by sin θ = λ / a so θ = 0.104°. The first order fringe is within the central maximum and the second order fringe is inside the first diffraction maximum on one side of the central maximum. The intensity here at this second fringe is much less than I 0 . IDENTIFY: A double-slit bright fringe is missing when it occurs at the same angle as a double-slit dark fringe. SET UP: Single-slit diffraction dark fringes occur when a sin θ = mλ, and double-slit interference bright fringes occur when d sin θ = m′ λ.


36-8

Chapter 36

EXECUTE: (a) The angles are the same for cancellation, so dividing the equations gives d/a = m′ /m ⇒ m′ /m = 7 ⇒ m′ = 7m

36.25.

When m = 1, m′ = 7; when m = 2, m′ = 14, and so forth, so every 7th bright fringe is missing from the double-slit interference pattern. EVALUATE: (b) The result is independent of the wavelength, so every 7th fringe will be cancelled for all wavelengths. But the bright interference fringes occur when d sin θ = mλ, so the location of the cancelled fringes does depend on the wavelength. IDENTIFY and SET UP: The phasor diagrams are similar to those in Fig.36.14. An interference minimum occurs when the phasors add to zero. EXECUTE: (a) The phasor diagram is given in Figure 36.25a

Figure 36.25a

There is destructive interference between the light through slits 1 and 3 and between 2 and 4. (b) The phasor diagram is given in Figure 36.25b.

Figure 36.25b

There is destructive interference between the light through slits 1 and 2 and between 3 and 4. (c) The phasor diagram is given in Figure 36.25c.

Figure 36.25c

36.26.

36.27.

There is destructive interference between light through slits 1 and 3 and between 2 and 4. EVALUATE: Maxima occur when φ = 0, 2π , 4π , etc. Our diagrams show that there are three minima between the maxima at φ = 0 and φ = 2π . This agrees with the general result that for N slits there are N − 1 minima between each pair of principal maxima. IDENTIFY: A double-slit bright fringe is missing when it occurs at the same angle as a double-slit dark fringe. SET UP: Single-slit diffraction dark fringes occur when a sin θ = mλ, and double-slit interference bright fringes occur when d sin θ = m′ λ. EXECUTE: (a) The angle at which the first bright fringe occurs is given by tan θ1 = (1.53 mm)/(2500 mm) ⇒ θ1 = 0.03507°. d sin θ1 = λ and d = λ/(sinθ1) = (632.8 nm)/sin(0.03507°) = 0.00103 m = 1.03 mm (b) The 7th double-slit interference bright fringe is just cancelled by the 1st diffraction dark fringe, so sinθdiff = λ/a and sinθinterf = 7λ/d The angles are equal, so λ/a = 7λ/d → a = d/7 = (1.03 mm)/7 = 0.148 mm. EVALUATE: We can generalize that if d = na, where n is a positive integer, then every nth double-slit bright fringe will be missing in the pattern. mλ IDENTIFY: The diffraction minima are located by sin θ = d and the two-slit interference maxima are located a miλ . The third bright band is missing because the first order single slit minimum occurs at the same by sinθ = d angle as the third order double slit maximum. 3 cm SET UP: The pattern is sketched in Figure 36.27. tan θ = , so θ = 1.91° . 90 cm

λ 500 nm = = 1.50 × 10 4 nm = 15.0 μ m (width) sinθ sin1.91° 3λ 3(500 nm) Double-slit bright fringe: d sin θ = 3λ and d = = = 4.50 × 104 nm = 45.0 μ m (separation) . sinθ sin1.91° EXECUTE: Single-slit dark spot: a sin θ = λ and a =


Diffraction

36-9

EVALUATE: Note that d / a = 3.0 .

Figure 36.27 36.28.

36.29.

IDENTIFY: The maxima are located by d sin θ = mλ . SET UP: The order corresponds to the values of m. EXECUTE: First-order: d sin θ1 = λ . Fourth-order: d sin θ 4 = 4λ . d sin θ 4 4λ , sinθ 4 = 4sin θ1 = 4sin8.94° and θ 4 = 38.4° . = d sin θ1 λ EVALUATE: We did not have to solve for d. IDENTIFY and SET UP: The bright bands are at angles θ given by d sin θ = mλ. Solve for d and then solve for θ for the specified order. EXECUTE: (a) θ = 78.4° for m = 3 and λ = 681 nm, so d = mλ / sin θ = 2.086 × 10−4 cm The number of slits per cm is 1/ d = 4790 slits/cm (b) 1st order: m = 1, so sin θ = λ / d = (681× 10−9 m) /(2.086 × 10−6 m) and θ = 19.1°

36.30.

2nd order: m = 2, so sin θ = 2λ / d and θ = 40.8° (c) For m = 4, sinθ = 4λ / d is greater than 1.00, so there is no 4th-order bright band. EVALUATE: The angular position of the bright bands for a particular wavelength increases as the order increases. IDENTIFY: The bright spots are located by d sin θ = mλ . SET UP: Third-order means m = 3 and second-order means m = 2 . mλ mλ mλ = d = constant , so r r = v v . EXECUTE: sin θ sin θ r sin θ v

36.31.

⎛ m ⎞⎛ λ ⎞ ⎛ 2 ⎞⎛ 400 nm ⎞ sin θ v = sin θ r ⎜ v ⎟⎜ v ⎟ = (sin 65.0°) ⎜ ⎟⎜ ⎟ = 0.345 and θ v = 20.2° . m λ 3 ⎠⎝ 700 nm ⎠ ⎝ ⎝ r ⎠⎝ r ⎠ EVALUATE: The third-order line for a particular λ occurs at a larger angle than the second-order line. In a given order, the line for violet light (400 nm) occurs at a smaller angle than the line for red light (700 nm). IDENTIFY and SET UP: Calculate d for the grating. Use Eq.(36.13) to calculate θ for the longest wavelength in the visible spectrum and verify that θ is small. Then use Eq.(36.3) to relate the linear separation of lines on the screen to the difference in wavelength. ⎛ 1 ⎞ −5 EXECUTE: (a) d = ⎜ ⎟ cm = 1.111 × 10 m ⎝ 900 ⎠ For λ = 700 nm, λ / d = 6.3 × 10−2. The first-order lines are located at sin θ = λ / d ; sin θ is small enough for sin θ ≈ θ to be an excellent approximation. (b) y = xλ / d , where x = 2.50 m. The distance on the screen between 1st order bright bands for two different wavelengths is Δy = x( Δλ ) / d , so

36.32.

Δλ = d ( Δy ) / x = (1.111 × 10 −5 m)(3.00 × 10−3 m) /(2.50 m) = 13.3 nm EVALUATE: The smaller d is (greater number of lines per cm) the smaller the Δλ that can be measured. IDENTIFY: The maxima are located by d sin θ = mλ . 1 SET UP: 350 slits mm ⇒ d = = 2.86 × 10 −6 m 3.50 × 105 m −1


36-10

Chapter 36

⎛ 4.00 × 10−7 m ⎞ ⎛λ ⎞ D m = 1: θ 400 = arcsin ⎜ ⎟ = arcsin ⎜ ⎟ = 8.05 . −6 × 2.86 10 m ⎝d⎠ ⎝ ⎠ ⎛ 7.00 × 10−7 m ⎞ ⎛λ ⎞ D = arcsin ⎜ ⎟ = arcsin ⎜ ⎟ = 14.18 . Δθ1 = 14.18° − 8.05° = 6.13°. −6 ⎝d⎠ ⎝ 2.86 × 10 m ⎠

EXECUTE:

θ 700

⎛ 3(4.00 × 10 −7 m) ⎞ ⎛ 3λ ⎞ m = 3 : θ 400 = arcsin ⎜ ⎟ = arcsin ⎜ ⎟ = 24.8° . −6 ⎝ d ⎠ ⎝ 2.86 × 10 m ⎠ ⎛ 3(7.00 × 10−7 m) ⎞ ⎛ 3λ ⎞ ⎟ = 47.3° . Δθ1 = 47.3° − 24.8° = 22.5°. ⎟ = arcsin ⎜ −6 ⎝ d ⎠ ⎝ 2.86 × 10 m ⎠ EVALUATE: Δθ is larger in third order. IDENTIFY: The maxima are located by d sin θ = mλ . SET UP: d = 1.60 × 10 −6 m

θ 700 = arcsin ⎜

36.33.

EXECUTE:

36.34.

⎛ m[6.328 × 10−7 m] ⎞ ⎛ mλ ⎞ ⎟ = arcsin([0.396]m) . For m = 1, θ1 = 23.3° . For ⎟ = arcsin ⎜ −6 ⎝ d ⎠ ⎝ 1.60 × 10 m ⎠

θ = arcsin ⎜

m = 2 , θ 2 = 52.3° . There are no other maxima. EVALUATE: The reflective surface produces the same interference pattern as a grating with slit separation d. IDENTIFY: The maxima are located by d sin θ = mλ . 1 SET UP: 5000 slits cm ⇒ d = = 2.00 × 10−6 m. 5.00 × 105 m −1 d sin θ (2.00 × 10−6 m)sin13.5D = = 4.67 × 10−7 m. m 1 ⎛ 2(4.67 × 10−7 m) ⎞ ⎛ mλ ⎞ D (b) m = 2 : θ = arcsin ⎜ ⎟ = 27.8 . ⎟ = arcsin ⎜ −6 × d 2.00 10 m ⎝ ⎠ ⎝ ⎠ EXECUTE: (a) λ =

36.35.

EVALUATE: Since the angles are fairly small, the second-order deviation is approximately twice the first-order deviation. IDENTIFY: The maxima are located by d sin θ = mλ . 1 SET UP: 350 slits mm ⇒ d = = 2.86 × 10 −6 m 3.50 × 105 m −1

⎛ m(5.20 × 10−7 m) ⎞ ⎛ mλ ⎞ ⎟ = arcsin((0.182) m) . ⎟ = arcsin ⎜ −6 ⎝ d ⎠ ⎝ 2.86 × 10 m ⎠ m = 1: θ = 10.5°; m = 2 : θ = 21.3°; m = 3 : θ = 33.1°. EVALUATE: The angles are not precisely proportional to m, and deviate more from being proportional as the angles increase. EXECUTE:

36.36.

θ = arcsin ⎜

IDENTIFY: The resolution is described by R = SET UP:

λ = Nm . Maxima are located by d sin θ = mλ . Δλ

For 500 slits/mm, d = (500 slits mm) −1 = (500,000 slits m) −1 .

EXECUTE: (a) N =

λ mΔλ

=

6.5645 × 10−7 m = 1820 slits. 2(6.5645 × 10 −7 m − 6.5627 × 10−7 m)

⎛ mλ ⎞ −1 −7 −1 (b) θ = sin −1 ⎜ ⎟ ⇒ θ1 = sin ((2)(6.5645 × 10 m)(500,000 m )) = 41.0297° and ⎝ d ⎠

θ 2 = sin −1 ((2)(6.5627 × 10−7 m)(500,000 m −1 )) = 41.0160° . Δθ = 0.0137° EVALUATE: d cosθ dθ = λ / N , so for 1820 slits the angular interval Δθ between each of these maxima and the λ 6.56 × 10 −7 m first adjacent minimum is Δθ = = = 0.0137°. This is the same as the angular Nd cosθ (1820)(2.0 × 10 −6 m)cos 41° 36.37.

separation of the maxima for the two wavelengths and 1820 slits is just sufficient to resolve these two wavelengths in second order. IDENTIFY: The resolving power depends on the line density and the width of the grating. SET UP: The resolving power is given by R = Nm = = λ/Δλ. EXECUTE: (a) R = Nm = (5000 lines/cm)(3.50 cm)(1) = 17,500 (b) The resolving power needed to resolve the sodium doublet is R = λ/Δλ = (589 nm)/(589.59 nm – 589.00 nm) = 998


Diffraction

36-11

so this grating can easily resolve the doublet. (c) (i) R = λ/Δλ. Since R = 17,500 when m = 1, R = 2 × 17,500 = 35,000 for m = 2. Therefore Δλ = λ/R = (587.8 nm)/35,000 = 0.0168 nm

λmin = λ +Δλ = 587.8002 nm + 0.0168 nm = 587.8170 nm (ii) λmax = λ − Δλ = 587.8002 nm − 0.0168 nm = 587.7834 nm EVALUATE: (iii) Therefore the range of resolvable wavelengths is 587.7834 nm < λ < 587.8170 nm. 36.38.

IDENTIFY and SET UP:

λ 587.8002 nm 587.8002 N 3302 slits . = = = 3302 slits . = = 2752 0.178 1.20 cm 1.20 cm cm mΔλ (587.9782 nm − 587.8002 nm) EVALUATE: A smaller number of slits would be needed to resolve these two lines in higher order. IDENTIFY and SET UP: The maxima occur at angles θ given by Eq.(36.16), 2d sin θ = mλ , where d is the spacing between adjacent atomic planes. Solve for d. EXECUTE: second order says m = 2. mλ 2(0.0850 × 10−9 m) d= = = 2.32 × 10−10 m = 0.232 nm 2sin θ 2sin 21.5° EVALUATE: Our result is similar to d calculated in Example 36.5. IDENTIFY: The maxima are given by 2d sin θ = mλ , m = 1 , 2, … SET UP: d = 3.50 × 10−10 m . 2d sin θ EXECUTE: (a) m = 1 and λ = = 2(3.50 × 10−10 m)sin15.0° = 1.81 × 10−10 m = 0.181 nm . This is an x ray. m ⎛ 1.81 × 10 −10 m ⎞ ⎛ λ ⎞ (b) sin θ = m ⎜ ⎟ = m(0.2586) . m = 2 : θ = 31.1° . m = 3 : θ = 50.9° . The equation ⎟ = m⎜ −10 ⎝ 2d ⎠ ⎝ 2[3.50 × 10 m] ⎠ EXECUTE:

36.39.

36.40.

λ = Nm Δλ

N=

doesn't have any solutions for m > 3 . EVALUATE: In this problem λ / d = 0.52. 36.41.

IDENTIFY: Rayleigh's criterion says sin θ = 1.22

λ

D SET UP: The best resolution is 0.3 arcseconds, which is about (8.33 × 10 −5 )° .

1.22λ 1.22(5.5 × 10−7 m) = = 0.46 m sin θ sin(8.33 × 10−5 °) EVALUATE: (b) The Keck telescopes are able to gather more light than the Hale telescope, and hence they can detect fainter objects. However, their larger size does not allow them to have greater resolution⎯atmospheric conditions limit the resolution. EXECUTE: (a) D =

36.42.

IDENTIFY: Apply sin θ = 1.22 SET UP:

36.44.

.

θ = (1/ 60)°

1.22λ 1.22(5.5 × 10−7 m) = = 2.31 × 10−3 m = 2.3 mm sin θ sin(1/ 60)D EVALUATE: The larger the diameter the smaller the angle that can be resolved. λ IDENTIFY: Apply sin θ = 1.22 . D W SET UP: θ = , where W = 28 km and h = 1200 km . θ is small, so sin θ ≈ θ . h 1.22λ h 1.2 × 106 m EXECUTE: D = = 1.22λ = 1.22(0.036 m) = 1.88 m W sin θ 2.8 × 10 4 m EVALUATE: D must be significantly larger than the wavelength, so a much larger diameter is needed for microwaves than for visible wavelengths. λ IDENTIFY: Apply sin θ = 1.22 . D SET UP: θ is small, so sin θ ≈ θ = 1.00 × 10 −8 rad . D sin θ Dθ (8.00 × 106 m)(1.00 × 10−8 ) EXECUTE: λ = ≈ = = 0.0656 m = 6.56 cm 1.22 1.22 1.22 EVALUATE: λ corresponds to microwaves. EXECUTE:

36.43.

λ D

D=


36-12

Chapter 36

36.45.

IDENTIFY and SET UP: The angular size of the first dark ring is given by sin θ1 = 1.22λ / D (Eq.36.17). Calculate

θ1 , and then the diameter of the ring on the screen is 2(4.5 m) tan θ1. ⎛ 620 × 10 −9 m ⎞ sin θ1 = 1.22 ⎜ ⎟ = 0.1022; θ1 = 0.1024 rad −6 ⎝ 7.4 × 10 m ⎠ The radius of the Airy disk (central bright spot) is r = (4.5 m) tan θ1 = 0.462 m. The diameter is 2r = 0.92 m = 92 cm. EVALUATE: λ / D = 0.084. For this small D the central diffraction maximum is broad. IDENTIFY: Rayleigh’s criterion limits the angular resolution. SET UP: Rayleigh’s criterion is sin θ ≈ θ = 1.22 λ/D. EXECUTE: (a) Using Rayleigh’s criterion sinθ ≈ θ = 1.22 λ/D = (1.22)(550 nm)/(135/4 mm) = 1.99 × 10–5 rad On the bear this angle subtends a distance x. θ = x/R and x = Rθ = (11.5 m)(1.99 × 10–5 rad) = 2.29 × 10–4 m = 0.23 mm (b) At f/22, D is 4/22 times as large as at f/4. Since θ is proportional to 1/D, and x is proportional to θ, x is 1/(4/22) = 22/4 times as large as it was at f/4. x = (0.229 mm)(22/4) = 1.3 mm EVALUATE: A wide-angle lens, such as one having a focal length of 28 mm, would have a much smaller opening at f/22 and hence would have an even less resolving ability. IDENTIFY and SET UP: Resolved by Rayleigh’s criterion means angular separation θ of the objects equals 1.22λ / D. The angular separation θ of the objects is their linear separation divided by their distance from the telescope. 250 × 103 m EXECUTE: θ = , where 5.93 × 1011 m is the distance from earth to Jupiter. Thus θ = 4.216 × 10−7. 5.93 × 1011 m λ 1.22λ 1.22(500 × 10−9 m) Then θ = 1.22 and D = = = 1.45 m θ D 4.216 × 10 −7 EVALUATE: This is a very large telescope mirror. The greater the angular resolution the greater the diameter the lens or mirror must be. EXECUTE:

36.46.

36.47.

36.48.

IDENTIFY: Rayleigh’s criterion says θ res = 1.22

.

y , where s is the distance of the object from the lens and y = 4.00 mm . s y λ yD (4.00 × 10 −3 m)(7.20 × 10−2 m) EXECUTE: = 1.22 . s = = = 429 m . s D 1.22λ 1.22(550 × 10−9 m) EVALUATE: The focal length of the lens doesn’t enter into the calculation. In practice, it is difficult to achieve resolution that is at the diffraction limit. IDENTIFY and SET UP: Let y be the separation between the two points being resolved and let s be their distance λ y from the telescope. Then the limit of resolution corresponds to 1.22 = . D s EXECUTE: (a) Let the two points being resolved be the opposite edges of the crater, so y is the diameter of the crater. For the moon, s = 3.8 × 108 m. y = 1.22λ s D . SET UP:

36.49.

λ D

D = 7.20 cm . θ res =

Hubble: D = 2.4 m and λ = 400 nm gives the maximum resolution, so y = 77 m Arecibo: D = 305 m and λ = 0.75 m; y = 1.1 × 106 m yD . Let y ≈ 0.30 (the size of a license plate). s = (0.30 m)(2.4 m) [(1.22)(400 × 10−9 m)] = 1500 km . 1.22λ EVALUATE: D / λ is much larger for the optical telescope and it has a much larger resolution even though the diameter of the radio telescope is much larger.

(b) s =

36.50.

IDENTIFY: Apply sin θ = 1.22

λ

D

.

θ is small, so sin θ ≈ θ . Smallest resolving angle is for short-wavelength light (400 nm). λ 400 × 10−9 m 10,000 mi EXECUTE: θ ≈ 1.22 = (1.22) , where R is the distance to the star. = 9.61× 10−8 rad . θ = SET UP:

R=

R D 5.08 m 16,000 km = = 1.7 × 1011 km . 9.6 × 10−8 rad This is less than a light year, so there are no stars this close.

10,000 mi

θ EVALUATE:


Diffraction

36.51.

IDENTIFY: Let y be the separation between the two points being resolved and let s be their distance from the telescope. The limit of resolution corresponds to 1.22 λ D = y s . s = 4.28 ly = 4.05 × 1016 m . Assume visible light, with λ = 400 m .

SET UP:

y = 1.22 λ s D = 1.22(400 × 10−9 m) (4.05 × 1016 m (10.0 m) = 2.0 × 109 m

EXECUTE:

36.52.

36-13

EVALUATE: The diameter of Jupiter is 1.38 × 108 m, so the resolution is insufficient, by about one order of magnitude. IDENTIFY: If the apparatus of Exercise 36.4 is placed in water, then all that changes is the wavelength

λ → λ′ =

λ

. n SET UP: For y << x , the distance between the two dark fringes on either side of the central maximum is D′ = 2 y′ . Let D = 2 y be the separation of 5.91 × 10 −3 m found in Exercise 36.4.

2 xλ ′ 2 xλ D 5.91 × 10 −3 m = = = = 4.44 × 10 −3 m = 4.44 mm. a an n 1.33 EVALUATE: The water shortens the wavelength and this decreases the width of the central maximum. EXECUTE:

2y1′ =

2

36.53.

⎛ sin β / 2 ⎞ (a) IDENTIFY and SET UP: The intensity in the diffraction pattern is given by Eq.(36.5): I = I 0 ⎜ ⎟ , where ⎝ β /2 ⎠

1 ⎛ 2π ⎞ ⎟ a sin θ . Solve for θ that gives I = I 0 . The angles θ + and θ − are shown in Figure 36.53. λ 2 ⎝ ⎠ 1 sin β / 2 1 EXECUTE: I = I 0 so = 2 β /2 2

β =⎜

sin x 1 = = 0.7071. x 2 Use trial and error to find the value of x that is a solution to this equation. x (sin x) / x 1.0 rad 0.841 Let x = β / 2; the equation for x is

1.5 rad 1.2 rad

0.665 0.777

1.4 rad 1.39 rad

0.7039 0.7077;

thus x = 1.39 rad and β = 2 x = 2.78 rad Δθ = θ + − θ − = 2θ +

sin θ + =

λβ = 2π a

λ ⎛ 2.78 rad ⎞ ⎛λ⎞ ⎜ ⎟ = 0.4425 ⎜ ⎟ a ⎝ 2π rad ⎠ ⎝a⎠ Figure 36.53

⎛1⎞ = 2, sin θ + = 0.4425 ⎜ ⎟ = 0.2212; θ + = 12.78°; Δθ = 2θ + = 25.6° ⎝ 2⎠ a ⎛1⎞ (ii) For = 5, sin θ + = 0.4425 ⎜ ⎟ = 0.0885; θ + = 5.077°; Δθ = 2θ + = 10.2° λ ⎝5⎠

(i) For

a

λ

(iii) For

⎛ 1⎞ = 10, sin θ + = 0.4425 ⎜ ⎟ = 0.04425; θ + = 2.536°; Δθ = 2θ + = 5.1° λ ⎝ 10 ⎠

a

(b) IDENTIFY and SET UP:

λ a

locates the first minimum. Solve for θ 0 .

1 = 2, sin θ 0 = ;θ 0 = 30.0°; 2θ 0 = 60.0° 2 a 1 (ii) For = 5, sin θ 0 = ;θ 0 = 11.54°; 2θ 0 = 23.1° 5 λ

EXECUTE: (i) For

a

sin θ 0 =

λ


36-14

Chapter 36

⎛1⎞ = 10, sin θ 0 = ⎜ ⎟ ;θ 0 = 5.74°; 2θ 0 = 11.5° ⎝ 10 ⎠ EVALUATE: Either definition of the width shows that the central maximum gets narrower as the slit gets wider. IDENTIFY: The two holes behave like double slits and cause the sound waves to interfere after they pass through the holes. The motion of the speakers causes a Doppler shift in the wavelength of the sound. SET UP: The wavelength of the sound that strikes the wall is λ = λ0 – vsTs, and destructive interference first occurs where sin θ = λ/2. EXECUTE: (a) First find the wavelength of the sound that strikes the openings in the wall. λ = λ0 – vsTs = v/ fs – vs/ fs = (v – vs)/ fs = (344 m/s – 80.0 m/s)/(1250 Hz) = 0.211 m Destructive interference first occurs where d sin θ = λ/2, which gives d = λ/(2 sinθ) = (0.211 m)/(2 sin 12.7°) = 0.480 m (b) λ = v/f = (344 m/s)/(1250 Hz) = 0.275 m sinθ = λ/2d = (0.275 m)/[2(0.480 m)] → θ = ±16.7° EVALUATE: The moving source produces sound of shorter wavelength than the stationary source, so the angles at which destructive interference occurs are smaller for the moving source than for the stationary source. IDENTIFY and SET UP: sin θ = λ / a locates the first dark band. In the liquid the wavelength changes and this changes the angular position of the first diffraction minimum.

(iii) For

36.54.

36.55.

a

λ

EXECUTE:

sin θ air =

λair

; sin θ liquid =

λliquid

a a ⎛ sin θ liquid ⎞ λliquid = λair ⎜ ⎟ = 0.4836 ⎝ sin θ air ⎠ λ = λair / n (Eq.33.5), so n = λair / λliquid = 1/ 0.4836 = 2.07

36.56.

36.57.

EVALUATE: Light travels faster in air and n must be > 1.00. The smaller λ in the liquid reduces θ that located the first dark band. 1 IDENTIFY: d = , so the bright fringes are located by 1 sin θ = λ N N SET UP: Red: 1 sin λR = 700 nm . Violet: 1 sin λV = 400 nm . N N sin θ R 7 sin(θ V + 15°) 7 EXECUTE: = . θ R − θ V = 15° → θ R = θ V + 15°. = . Using a trig identify from Appendix B, sin θ V 4 sin θ V 4 sin θ V cos15° + cosθ V sin15° = 7 4 . cos15° + cot θ V sin15° = 7 4 . sin θ V tan θ V = 0.330 ⇒ θ V = 18.3°and θ R = θ V + 15° = 18.3° + 15° = 33.3°. Then 1 sin θ R = 700 nm gives N sin θ R sin 33.3D = = 7.84 × 105 lines m = 7840 lines cm . The spectrum begins at 18 .3D and ends at 33.3D . N= 700 nm 700 × 10−9 m EVALUATE: As N is increased, the angular range of the visible spectrum increases. (a) IDENTIFY and SET UP: The angular position of the first minimum is given by a sin θ = mλ (Eq.36.2), with m = 1. The distance of the minimum from the center of the pattern is given by y = x tan θ .

λ

540 × 10−9 m = 1.50 × 10 −3 ; θ = 1.50 × 10−3 rad a 0.360 × 10−3 m y1 = x tan θ = (1.20 m) tan(1.50 × 10−3 rad) = 1.80 × 10−3 m = 1.80 mm. (Note that θ is small enough for θ ≈ sin θ ≈ tan θ , and Eq.(36.3) applies.) (b) IDENTIFY and SET UP: Find the phase angle β where I = I 0 / 2. Then use Eq.(36.6) to solve for θ and y = x tan θ to find the distance. sin θ =

=

1 From part (a) of Problem 36.53, I = I 0 when β = 2.78 rad. 2 2 π βλ ⎛ ⎞ β = ⎜ ⎟ a sin θ (Eq.(36.6)), so sin θ = . 2π a ⎝ λ ⎠ βλ x (2.78 rad)(540 × 10−9 m)(1.20 m) y = x tan θ ≈ x sin θ ≈ = = 7.96 × 10−4 m = 0.796 mm 2π a 2π (0.360 × 10 −3 m) EVALUATE: The point where I = I 0 / 2 is not midway between the center of the central maximum and the first minimum; see Exercise 36.15. EXECUTE:


Diffraction

36-15

2

36.58.

⎛ sin γ ⎞ I = I0 ⎜ ⎟ . The maximum intensity occurs when the derivative of the intensity function with ⎝ γ ⎠ respect to γ is zero.

IDENTIFY:

SET UP:

d ⎛1⎞ 1 d sin γ = cos γ . ⎜ ⎟=− 2 . dγ ⎝ γ ⎠ γ dγ 2

⎛ sin γ ⎞⎛ cos γ sin γ ⎞ dI d ⎛ sin γ ⎞ cos γ sin γ = I0 − 2 ⎟ =0. − 2 ⇒ γ cos γ = sin γ ⇒ γ = tan γ . ⎜ ⎟ = 2⎜ ⎟⎜ γ γ γ dγ dγ ⎝ γ ⎠ γ γ ⎝ ⎠⎝ ⎠ (b) The graph in Figure 36.58 is a plot of f( γ ) = γ − tan γ . When f (γ ) equals zero, there is an intensity maximum. Getting estimates from the graph, and then using trial and error to narrow in on the value, we find that the three smallest γ -values are γ = 4.49 rad 7.73 rad, and 10.9 rad. EVALUATE: γ = 0 is the central maximum. The three values of γ we found are the locations of the first three secondary maxima. The first four minima are at γ = 3.14 rad , 6.28 rad, 9.42 rad, and 12.6 rad. The maxima are between adjacent minima, but not precisely midway between them.

EXECUTE:

Figure 36.58 36.59.

IDENTIFY and SET UP: Relate the phase difference between adjacent slits to the sum of the phasors for all slits. The 2π d 2π dθ λφ when θ is small and sin θ ≈ θ . Thus θ = . sin θ ≈ phase difference between adjacent slits is φ = 2π d λ λ EXECUTE: A principal maximum occurs when φ = φ max = m 2π , where m is an integer, since then all the phasors

add. The first minima on either side of the mth principal maximum occur when φ = φ ±min = m 2π ± (2π / N ) and the phasor diagram for N slits forms a closed loop and the resultant phasor is zero. The angular position of a principal ⎛ λ ⎞ ⎛ λ ⎞ ± ± maximum is θ = ⎜ ⎟φ max . The angular position of the adjacent minimum is θ min = ⎜ ⎟φ min. ⎝ 2π d ⎠ ⎝ 2π d ⎠ 2π ⎞ λ ⎛ λ ⎞⎛ ⎛ λ ⎞⎛ 2π ⎞ θ +min = ⎜ ⎟⎜ φ max + ⎟ =θ +⎜ ⎟⎜ ⎟ =θ + 2 π d N 2 π d N Nd ⎝ ⎠⎝ ⎠ ⎝ ⎠⎝ ⎠ 2π ⎞ λ ⎛ λ ⎞⎛ ⎟⎜ φ max − ⎟ =θ − N ⎠ Nd ⎝ 2π d ⎠⎝

θ −min = ⎜

2λ , as was to be shown. Nd EVALUATE: The angular width of the principal maximum decreases like 1/ N as N increases. IDENTIFY: The change in wavelength of the Hα line is due to a Doppler shift in the wavelength due to the motion of the galaxy. c−v fS . SET UP: From Equation 16.30, the Doppler effect formula for light is f R = c+v EXECUTE: First find the wavelength of the light using the grating information. λ = d sin θ1 = [1/(575,800 lines/m)] sin 23.41° = 6.900 × 10–7 m = 690.0 nm The angular width of the principal maximum is θ

36.60.

+ min

−θ

− min

=

c−v fS . In this case, fR is the frequency of the 690.0-nm light that the c+v cosmologist measures, and fS is the frequency of the 656.3-nm light of the Hα line obtained in the laboratory. Using Equation 16.30, we have f R =


36-16

Chapter 36

Solving for v gives v =

1 − ( f R / fS )

1 + ( f R / fS )

2 2

c. Since fλ = c, f = c/λ, which gives fR/fS = λS/λR. Substituting this into the

equation for v, we get ⎛λ 1− ⎜ S λ v= ⎝ R ⎛λ 1+ ⎜ S ⎝ λR

36.61.

2

2 ⎞ ⎛ 656.3 nm ⎞ 1− ⎜ ⎟ ⎟ ⎝ 690.0 nm ⎠ 3.00 × 108 m/s = 1.501 × 107 m/s, ⎠ c= ) 2 ( 2 ⎞ ⎛ 656.3 nm ⎞ 1+ ⎜ ⎟ ⎟ ⎝ 690.0 nm ⎠ ⎠

which is 5.00% the speed of light. EVALUATE: Since v is positive, the galaxy is moving away from us. We can also see this because the wavelength has increased due to the motion. IDENTIFY and SET UP: Draw the specified phasor diagrams. There is totally destructive interference between two slits when their phasors are in opposite directions. EXECUTE: (a) For eight slits, the phasor diagrams must have eight vectors. The diagrams for each specified value of φ are sketched in Figure 36.61a. In each case the phasors all sum to zero. (b) The additional phasor diagrams for φ = 3π / 2 and 3π / 4 are sketched in Figure 36.61b. 3π 5π 7π 3π ,φ = , and φ = , totally destructive interference occurs between slits four apart. For φ = , For φ = 4 4 4 2 totally destructive interference occurs with every second slit. EVALUATE: At a minimum the phasors for all slits sum to zero.

Figure 36.61 36.62.

IDENTIFY: Maxima are given by 2d sin θ = mλ . SET UP: d is the separation between crystal planes. ⎛ 0.125 nm ⎞ ⎛ mλ ⎞ EXECUTE: (a) θ = arcsin ⎜ ⎟ = arcsin(0.2216m) . ⎟ = arcsin ⎜ m ⎝ 2d ⎠ ⎝ 2(0.282 nm) ⎠ For m = 1: θ = 12.8°, m = 2 : θ = 26.3°, m = 3: θ = 41.7°, and m = 4 : θ = 62.4°. No larger m values yield answers. ⎛ 2mλ ⎞ a , then θ = arcsin ⎜⎜ ⎟⎟ = arcsin(0.3134m). 2 ⎝ 2a ⎠ So for m = 1: θ = 18.3°, m = 2: θ = 38.8°, and m = 3: θ = 70.1°. No larger m values yield answers. EVALUATE: In part (b), where d is smaller, the maxima for each m are at larger θ IDENTIFY and SET UP: In each case consider the relevant phasor diagram. EXECUTE: (a) For the maxima to occur for N slits, the sum of all the phase differences between the slits must add to zero (the phasor diagram closes on itself). This requires that, adding up all the relative phase shifts, 2π m N φ = 2π m, for some integer m . Therefore φ = , for m not an integer multiple of N , which would give a N maximum. 2π m (b) The sum of N phase shifts φ = brings you full circle back to the maximum, so only the N − 1 previous N phases yield minima between each pair of principal maxima. EVALUATE: The N − 1 minima between each pair of principal maxima cause the maxima to become sharper as N increases. IDENTIFY: Set d = a in the expressions for φ and β and use the results in Eq.(36.12). SET UP: Figure 36.64 shows a pair of slits whose width and separation are equal

(b) If the separation d =

36.63.

36.64.


Diffraction

36-17

EXECUTE: Figure 36.64 shows that the two slits are equivalent to a single slit of width 2a . 2π d 2π a φ= sin θ , so β = sin θ = φ . So then the intensity is

λ

λ

⎛ sin 2 ( β /2) ⎞ (2sin( β /2)cos( β /2)) 2 sin 2 β sin 2 ( β ′/2) 2π (2a) = = = , where β ′ = I = I 0 cos 2 ( β /2) ⎜ I I I sin θ , ⎟ 0 0 0 2 2 2 2 λ β β ( β ′/2) ⎝ ( β /2) ⎠ which is Eq. (35.5) with double the slit width. EVALUATE: In Chapter 35 we considered the limit where a << d . a > d is not possible.

Figure 36.64 36.65.

36.66.

IDENTIFY and SET UP: The condition for an intensity maximum is d sin θ = mλ , m = 0, ± 1, ± 2,… Third order means m = 3. The longest observable wavelength is the one that gives θ = 90° and hence θ = 1. 1 EXECUTE: 6500 lines/cm so 6.50 × 105 lines/m and d = m = 1.538 × 10−6 m 6.50 × 105 d sin θ (1.538 × 10−6 m)(1) λ= = = 5.13 × 10−7 m = 513 nm m 3 EVALUATE: The longest wavelength that can be obtained decreases as the order increases. IDENTIFY and SET UP: As the rays first reach the slits there is already a phase difference between adjacent slits of 2π d sin θ ′ . This, added to the usual phase difference introduced after passing through the slits, yields the

λ

condition for an intensity maximum. For a maximum the total phase difference must equal 2π m . 2π d sin θ 2π d sin θ ′ EXECUTE: + = 2π m ⇒ d (sin θ + sin θ ′) = mλ

λ

(b) 600 slits mm ⇒ d =

λ

1 = 1.67 × 10−6 m. 6.00 × 105 m −1

For θ ′ = 0D , m = 0 : θ = arcsin(0) = 0. ⎛ 6.50 × 10−7 m ⎞ ⎛λ⎞ D m = 1: θ = arcsin ⎜ ⎟ = arcsin ⎜ ⎟ = 22.9 . −6 × 1.67 10 m d ⎝ ⎠ ⎝ ⎠ ⎛ 6.50 × 10−7 m ⎞ ⎛ λ⎞ D m = −1: θ = arcsin ⎜ − ⎟ = arcsin ⎜ − ⎟ = −22.9 . −6 × 1.67 10 m ⎝ d⎠ ⎝ ⎠

For θ ′ = 20.0D , m = 0 : θ = arcsin(− sin 20.0D ) = −20.0D.

36.67.

⎛ 6.50 × 10−7 m ⎞ − sin 20.0D ⎟ = 2.71D. m = 1: θ = arcsin ⎜ −6 × 1.67 10 m ⎝ ⎠ ⎛ 6.50 × 10−7 m ⎞ − sin 20.0D ⎟ = −47.0D. m = −1: θ = arcsin ⎜ − −6 ⎝ 1.67 × 10 m ⎠ EVALUATE: When θ ′ > 0 , the maxima are shifted downward on the screen, toward more negative angles. mλ IDENTIFY: The maxima are given by d sin θ = mλ . We need sin θ = ≤ 1in order for all the visible d wavelengths are to be seen. 1 SET UP: For 650 slits mm ⇒ d = = 1.53 × 10−6 m. 6.50 × 105 m −1


36-18

Chapter 36

λ1 2λ 3λ = 0.26; m = 2 : 1 = 0.52; m = 3 : 1 = 0.78. d d d λ 2 λ 3 λ λ2 = 7.00 × 10 −7 m : m = 1: 2 = 0.46; m = 2 : 2 = 0.92; m = 3 : 2 = 1.37. So, the third order does not contain the violet d d d end of the spectrum, and therefore only the first and second order diffraction patterns contain all colors of the spectrum. EVALUATE: θ for each maximum is larger for longer wavelengths. λ IDENTIFY: Apply sin θ = 1.22 . D Δx SET UP: θ is small, so sin θ ≈ , where Δx is the size of the detail and R = 7.2 ×108 ly. 1 ly = 9.41×1012 km . λ = c/f R λ Δx 1.22λ R (1.22)cR (1.22)(3.00 × 105 km s)(7.2 × 108 ly) EXECUTE: sin θ = 1.22 ≈ ⇒ Δx = = = = 2.06 ly . D R D Df (77.000 × 103 km)(1.665 × 109 Hz) (9.41 × 1012 km ly)( 2.06 ly) = 1.94 × 1013 km. Δx EVALUATE: λ = 18 cm . λ / D is very small, so is very small. Still, R is very large and Δx is many orders of R magnitude larger than the diameter of the sun. IDENTIFY and SET UP: Add the phases between adjacent sources. EXECUTE: (a) d sin θ = mλ . Place 1st maximum at ∞ or θ = 90D. d = λ . If d < λ , this puts the first maximum “beyond ∞. ” Thus, if d < λ there is only a single principal maximum. (b) At a principal maximum when δ = 0 , the phase difference due to the path difference between adjacent slits ⎛ d sin θ ⎞ is Φ path = 2π ⎜ ⎟ . This just scales 2π radians by the fraction the wavelength is of the path difference between ⎝ λ ⎠ adjacent sources. If we add a relative phase δ between sources, we still must maintain a total phase difference of zero to keep our principal maximum. 2π d sin θ ⎛ δλ ⎞ Φ path ± δ = 0 ⇒ = ±δ or θ = sin −1 ⎜ ⎟ λ ⎝ 2π d ⎠ 0.280 m (c) d = = 0.0200 m (count the number of spaces between 15 points). Let θ = 45D. Also recall f λ = c, so 14 2π (0.0200 m)(8.800 × 109 Hz)sin 45D δ max = ± = ±2.61 radians. (3.00 × 108 m s) EXECUTE:

36.68.

36.69.

36.70.

36.71.

λ1 = 4.00 × 10 −7 m : m = 1:

EVALUATE: δ must vary over a wider range in order to sweep the beam through a greater angle. IDENTIFY: The wavelength of the light is smaller under water than it is in air, which will affect the resolving power of the lens, by Rayleigh’s criterion. SET UP: The wavelength under water is λ = λ0/n, and for small angles Rayleigh’s criterion is θ = 1.22λ/D. EXECUTE: (a) In air the wavelength is λ0 = c/f = (3.00 × 108 m/s)/(6.00 × 1014 Hz) = 5.00 × 10–7 m. In water the wavelength is λ = λ0/n = (5.00 × 10–7 m)/1.33 = 3.76 × 10–7 m. With the lens open all the way, we have D = f/2.8 = (35.0 mm)/2.80 = (0.0350 m)/2.80. In the water, we have sin θ ≈ θ = 1.22 λ/D = (1.22)(3.76 × 10–7 m)/[(0.0350 m)/2.80] = 3.67 × 10–5 rad Calling w the width of the resolvable detail, we have θ = w/R → w = Rθ = (2750 mm)(3.67 × 10–5 rad) = 0.101 mm (b) θ = 1.22 λ/D = (1.22)(5.00 × 10–7 m)/[(0.0350 m)/2.80] = 4.88 × 10–5 rad w = Rθ = (2750 mm)(4.88 × 10–5 rad) = 0.134 mm EVALUATE: Due to the reduced wavelength underwater, the resolution of the lens is better under water than in air. IDENTIFY and SET UP: Resolved by Rayleigh’s criterion means the angular separation θ of the objects is given by θ = 1.22λ / D. θ = y / s, where y = 75.0 m is the distance between the two objects and s is their distance from the astronaut (her altitude). y λ = 1.22 EXECUTE: s D yD (75.0 m)(4.00 × 10−3 m) s= = = 4.92 × 105 m = 492 km 1.22λ 1.22(500 × 10−9 m) EVALUATE: In practice, this diffraction limit of resolution is not achieved. Defects of vision and distortion by the earth’s atmosphere limit the resolution more than diffraction does.


Diffraction

36.72.

IDENTIFY: Apply sin θ = 1.22

36-19

λ

. D Δx , where Δx is the size of the details and R is the distance to the earth. SET UP: θ is small, so sin θ ≈ R 1 ly = 9.41 × 1015 m . EXECUTE: (a) R =

DΔx (6.00 × 106 m)(2.50 × 105 m) = = 1.23 × 1017 m = 13.1 ly 1.22λ (1.22)(1.0 × 10−5 m)

1.22λ R (1.22)(1.0 × 10−5 m)(4.22 ly)(9.41 × 1015 m ly) = = 4.84 × 108 km . This is about 10,000 times the 1.0 m D diameter of the earth! Not enough resolution to see an earth-like planet! Δx is about 3 times the distance from the earth to the sun. (1.22)(1.0 × 10−5 m)(59 ly)(9.41× 1015 m ly) = 1.13 × 106 m = 1130 km. (c) Δx = 6.00 × 106 m (b) Δx =

Δx 1130 km = = 8.19 × 10−3 ; Δx is small compared to the size of the planet. Dplanet 1.38 × 105 km 36.73.

EVALUATE: The very large diameter of Planet Imager allows it to resolve planet-sized detail at great distances. IDENTIFY and SET UP: Follow the steps specified in the problem. EXECUTE: (a) From the segment dy′, the fraction of the amplitude of E0 that gets through is

⎛ dy′ ⎞ ⎛ dy′ ⎞ E0 ⎜ ⎟ ⇒ dE = E0 ⎜ ⎟ sin( kx − ωt ). a ⎝ ⎠ ⎝ a ⎠ (b) The path difference between each little piece is E dy′ y′ sin θ ⇒ kx = k ( D − y′ sin θ ) ⇒ dE = 0 sin(k ( D − y′ sin θ ) − ωt ). This can be rewritten as a E0 dy′ dE = (sin( kD − ωt )cos(ky′ sin θ ) + sin( ky′ sin θ )cos( kD − ωt )). a (c) So the total amplitude is given by the integral over the slit of the above. a 2 E a2 ⇒ E = ∫ dE = 0 ∫ dy′ (sin(kD − ωt ) cos(ky′ sin θ ) + sin(ky′ sin θ ) cos( kD − ωt )). −a 2 a −a 2 But the second term integrates to zero, so we have: a 2

E=

a 2 ⎡⎛ sin(ky′ sin θ ) ⎞ ⎤ E0 sin( kD − ωt ) ∫ dy′(cos( ky′ sin θ )) = E0 sin ( kD − ωt ) ⎢⎜ ⎟⎥ a − 2 a ⎣⎝ ka sin θ 2 ⎠ ⎦ − a 2

⎛ sin( ka (sin θ ) 2) ⎞ ⎛ sin(π a (sin θ ) λ ) ⎞ ⇒ E = E0 sin( kD − ωt ) ⎜ ⎟ = E0 sin( kD − ωt ) ⎜ ⎟. ka (sin ) 2 θ ⎝ ⎠ ⎝ π a (sin θ ) λ ⎠ sin [. . .] At θ = 0, = 1 ⇒ E = E0 sin(kD − ωt ). [. . .] 2

36.74.

2

⎛ sin( ka (sin θ )/2) ⎞ ⎛ sin( β 2) ⎞ 2 2 (d) Since I ∝ E 2 ⇒ I = I 0 ⎜ ⎟ = I0 ⎜ ⎟ , where we have used I 0 = E0 sin ( kx − ωt ). ka (sin )/ 2 2 θ β ⎝ ⎠ ⎝ ⎠ EVALUATE: The same result for I (θ ) is obtained as was obtained using phasors. IDENTIFY and SET UP: Follow the steps specified in the problem. EXECUTE: (a) Each source can be thought of as a traveling wave evaluated at x = R with a maximum amplitude of E0 . However, each successive source will pick up an extra phase from its respective pathlength to point ⎛ d sin θ ⎞ P . φ = 2π ⎜ ⎟ which is just 2π , the maximum phase, scaled by whatever fraction the path difference, ⎝ λ ⎠ d sin θ , is of the wavelength, λ . By adding up the contributions from each source (including the accumulating phase difference) this gives the expression provided. (b) ei ( kR −ωt + nφ ) = cos( kR − ωt + nφ ) + i sin(kR − ωt + nφ ). The real part is just cos ( kR − ωt + nφ ). So, ⎡ N −1 ⎤ N −1 Re ⎢ ∑ E0ei ( kR −ω t + nφ ) ⎥ = ∑ E0 cos( kR − ωt + nφ ). (Note: Re means “the real part of . . . .”). But this is just ⎣ n=0 ⎦ n =0 E0 cos( kR − ωt ) + E0 cos( kR − ωt + φ ) + E0 cos( kR − ωt + 2φ ) + " + E0 cos(kR − ωt + ( N − 1)φ )


36-20

Chapter 36 N −1

(c)

∑E

0

ei ( kR −ωt + nφ ) = E0

n=0

N −1

∑e

− iω t

e + ikR einφ = E0ei ( kR −ω t )

n=0

N −1

N −1

N −1

∑ e φ . ∑ e φ = ∑ (e φ ) . But recall ∑ x n=0

in

n=0

in

n =0

i

n

n=0

n

=

xN −1 . x −1

eiN φ − 1 eiNφ − 1 eiNφ / 2 (eiNφ / 2 − e −iNφ / 2 ) (eiN φ / 2 − e −iN φ / 2 ) Let x = e so ∑ (e ) = iφ (nice trick!). But iφ = iφ / 2 iφ / 2 − iφ / 2 = ei ( N −1)φ / 2 iφ / 2 −iφ / 2 . e −1 e −1 e (e − e ) (e − e ) n=0 Putting everything together: N −1 (eiN φ / 2 − e − iNφ / 2 ) E0ei ( kR −ωt + nφ ) = E0ei ( kR −ωt + ( N −1)φ / 2) iφ / 2 − iφ / 2 ∑ (e − e ) n=0 N −1

iφ n

⎡ cos Nφ /2 + i sin Nφ /2 − cos Nφ /2 + i sin Nφ /2 ⎤ = E0 [ cos( kR − ωt + ( N − 1)φ /2) + i sin( kR − ωt + ( N − 1)φ / 2) ] ⎢ ⎥ cosφ /2 + i sin φ /2 − cosφ /2 + i sin φ /2 ⎣ ⎦ sin( Nφ /2) Taking only the real part gives ⇒ E0 cos( kR − ωt + ( N − 1)φ /2) = E. sin φ /2 2

(d) I = E av = I 0

I0 ∝

sin 2 ( Nφ / 2) . (The cos 2 term goes to sin 2 (φ / 2)

in the time average and is included in the definition of I 0 .)

E02 . 2

EVALUATE: (e) N = 2. I = I 0

36.75.

1 2

sin 2 (2φ / 2) I 0 (2sin φ / 2cos φ / 2) 2 φ = = 4 I 0 cos 2 . Looking at Eq.(35.9), sin 2 φ / 2 sin 2 φ / 2 2

I 0′ ∝ 2 E02 but for us I 0 ∝

E02 I 0′ = . 2 4

IDENTIFY and SET UP:

From Problem 36.74, I = I 0

sin 2 ( Nφ / 2) . Use this result to obtain each result specified sin 2 φ / 2

in the problem. ⎛ N 2 ⎞ cos( Nφ / 2) 0 sin ( Nφ / 2) ˆ rule: lim . Use l'Hopital's = lim ⎜ = N . So lim I = N 2 I 0 . ⎟ φ →0 φ →0 φ →0 sin φ / 2 1 2 cos( φ / 2) 0 ⎝ ⎠ N 2π . The (b) The location of the first minimum is when the numerator first goes to zero at φmin = π or φmin = 2 N 1 width of the central maximum goes like 2φmin , so it is proportional to . N Nφ (c) Whenever = nπ where n is an integer, the numerator goes to zero, giving a minimum in intensity. That is, 2 2nπ . This is true assuming that the denominator doesn’t go to zero as well, which I is a minimum wherever φ = N EXECUTE: (a) lim I → φ →0

occurs when

φ 2

= mπ , where m is an integer. When both go to zero, using the result from part(a), there is a

n is an integer, there will be a maximum. N n n (d) From part (c), if is an integer we get a maximum. Thus, there will be N − 1 minima. (Places where is N N not an integer for fixed N and integer n .) For example, n = 0 will be a maximum, but n = 1, 2. . ., N − 1 will be minima with another maximum at n = N . φ ⎛ π 3π ⎞ , etc.) ⎟ and if N is odd then (e) Between maxima is a half-integer multiple of π ⎜ i.e. , 2 2 2 ⎝ ⎠ maximum. That is, if

sin 2 ( Nφ / 2) → 1, so I → I 0 . sin 2 φ / 2 EVALUATE: These results show that the principal maxima become sharper as the number of slits is increased.


37

RELATIVITY

37.1.

IDENTIFY and SET UP: Consider the distance A to O′ and B to O′ as observed by an observer on the ground (Figure 37.1).

Figure 37.1

37.2.

EXECUTE: Simultaneous to observer on train means light pulses from A′ and B′ arrive at O′ at the same time. To observer at O light from A′ has a longer distance to travel than light from B′ so O will conclude that the pulse from A( A′) started before the pulse at B ( B′). To observer at O bolt A appeared to strike first. EVALUATE: Section 37.2 shows that if they are simultaneous to the observer on the ground then an observer on the train measures that the bolt at B′ struck first. 1 (a) γ = = 2.29. t = γτ = (2.29) (2.20 × 10−6 s) = 5.05 × 10 −6 s. 1 − (0.9) 2 (b) d = vt = (0.900) (3.00 × 108 m s) (5.05 × 10−6 s) = 1.36 × 103 m = 1.36 km.

37.3.

1 IDENTIFY and SET UP: The problem asks for u such that Δt0 / Δt = . 2

37.4.

37.5.

2

Δt0

u 2 ⎛1⎞ gives u = c 1 − ( Δt0 / Δt ) = (3.00 × 108 m/s) 1 − ⎜ ⎟ = 2.60 × 108 m/s ; = 0.867 c ⎝ 2⎠ 1 − u 2 / c2 Jet planes fly at less than ten times the speed of sound, less than about 3000 m/s. Jet planes fly at much lower speeds than we calculated for u. IDENTIFY: Time dilation occurs because the rocket is moving relative to Mars. SET UP: The time dilation equation is Δt = γΔt0 , where t0 is the proper time. EXECUTE: (a) The two time measurements are made at the same place on Mars by an observer at rest there, so the observer on Mars measures the proper time. 1 (75.0 μ s) = 435 μ s (b) Δt = γΔt0 = 1 − (0.985) 2 EXECUTE:

Δt =

EVALUATE: The pulse lasts for a shorter time relative to the rocket than it does relative to the Mars observer. (a) IDENTIFY and SET UP: Δt0 = 2.60 × 10 −8 s; Δt = 4.20 × 10−7 s. In the lab frame the pion is created and decays at different points, so this time is not the proper time. 2 Δt0 u 2 ⎛ Δt ⎞ EXECUTE: Δt = says 1 − 2 = ⎜ 0 ⎟ c ⎝ Δt ⎠ 1 − u 2 / c2 2

⎛ 2.60 × 10−8 s ⎞ u ⎛ Δt ⎞ = 1− ⎜ 0 ⎟ = 1− ⎜ ⎟ = 0.998; u = 0.998c −7 c ⎝ Δt ⎠ ⎝ 4.20 × 10 s ⎠ EVALUATE: u < c, as it must be, but u/c is close to unity and the time dilation effects are large. (b) IDENTIFY and SET UP: The speed in the laboratory frame is u = 0.998c; the time measured in this frame is Δt , so the distance as measured in this frame is d = u Δt 2

EXECUTE: d = (0.998)(2.998 × 108 m/s)(4.20 × 10−7 s) = 126 m EVALUATE: The distance measured in the pion’s frame will be different because the time measured in the pion’s frame is different (shorter). 37-1


37-2

37.6.

Chapter 37 γ = 1.667

1.20 × 108 m = 0.300 s. γ γ(0.800c) (b) (0.300 s) (0.800c) = 7.20 × 107 m. (c) Δt0 = 0.300 s γ = 0.180 s. (This is what the racer measures your clock to read at that instant.) At your origin

(a) Δt0 =

Δt

=

you read the original 37.7.

37.8.

37.9.

37.10.

order of events! IDENTIFY and SET UP: A clock moving with respect to an observer appears to run more slowly than a clock at rest in the observer’s frame. The clock in the spacecraft measurers the proper time Δt0 . Δt = 365 days = 8760 hours. EXECUTE: The clock on the moving spacecraft runs slow and shows the smaller elapsed time. Δt0 = Δt 1 − u 2 / c 2 = (8760 h) 1 − (4.80 × 106 / 3.00 × 108 ) 2 = 8758.88 h . The difference in elapsed times is 8760 h − 8758.88 h = 1.12 h . IDENTIFY and SET UP: The proper time is measured in the frame where the two events occur at the same point. EXECUTE: (a) The time of 12.0 ms measured by the first officer on the craft is the proper time. Δt0 gives u = c 1 − ( Δt0 / Δt ) 2 = c 1 − (12.0 × 10−3 / 0.190) 2 = 0.998c . (b) Δt = 2 2 1− u / c EVALUATE: The observer at rest with respect to the searchlight measures a much shorter duration for the event. IDENTIFY and SET UP: l = l0 1 − u 2 / c 2 . The length measured when the spacecraft is moving is l = 74.0 m; l0 is the length measured in a frame at rest relative to the spacecraft. l 74.0 m = = 92.5 m. EXECUTE: l0 = 2 2 1− u / c 1 − (0.600c / c ) 2 EVALUATE: l0 > l. The moving spacecraft appears to an observer on the planet to be shortened along the direction of motion. IDENTIFY and SET UP: When the meterstick is at rest with respect to you, you measure its length to be 1.000 m, and that is its proper length, l0 . l = 0.3048 m . EXECUTE:

37.11.

l = l0 1 − u 2 / c 2 gives u = c 1 − (l / l0 ) 2 = c 1 − (0.3048 /1.00) 2 = 0.9524c = 2.86 × 108 m/s .

IDENTIFY and SET UP: The 2.2 μs lifetime is Δt0 and the observer on earth measures Δt. The atmosphere is moving relative to the muon so in its frame the height of the atmosphere is l and l0 is 10 km. EXECUTE: (a) The greatest speed the muon can have is c, so the greatest distance it can travel in 2.2 × 10 −6 s is d = vt = (3.00 × 108 m/s)(2.2 × 10−6 s) = 660 m = 0.66 km . (b) Δt =

37.12.

1.20 × 108m = 0.5 s. Clearly the observers (you and the racer) will not agree on the (0.800) (3 × 108 m s)

Δt0 1 − u 2 / c2

=

2.2 × 10−6 s 1 − (0.999) 2

= 4.9 × 10−5 s

d = vt = (0.999)(3.00 × 108 m/s)(4.9 × 10−5 s) = 15 km In the frame of the earth the muon can travel 15 km in the atmosphere during its lifetime. (c) l = l0 1 − u 2 / c 2 = (10 km) 1 − (0.999) 2 = 0.45 km In the frame of the muon the height of the atmosphere is less than the distance it moves during its lifetime. IDENTIFY and SET UP: The scientist at rest on the earth’s surface measures the proper length of the separation between the point where the particle is created and the surface of the earth, so l0 = 45.0 km . The transit time

measured in the particle’s frame is the proper time, Δt0 . EXECUTE: (a) t =

l0 45.0 × 103 m = = 1.51 × 10−4 s v (0.99540)(3.00 × 108 m/s)

(b) l = l0 1 − u 2 / c 2 = (45.0 km) 1 − (0.99540) 2 = 4.31 km (c) time dilation formula: Δt0 = Δt 1 − u 2 / c 2 = (1.51 × 10−4 s) 1 − (0.99540) 2 = 1.44 × 10−5 s

l 4.31 × 103 m = = 1.44 × 10−5 s v (0.99540)(3.00 × 108 m/s) The two results agree. (a) l0 = 3600 m . from Δl : t =

37.13.

l = l0 1 −

u2 (4.00 × 107 m s) 2 = l (3600 m) 1 − = (3600 m)(0.991) = 3568 m. 0 c2 (3.00 × 108 m s) 2


Relativity

37-3

l0 3600 m = = 9.00 × 10 −5 s. u 4.00 × 107 m s l 3568 m (c) Δt = = = 8.92 × 10−5 s. u 4.00 × 107 m s Multiplying the last equation of (37.21) by u and adding to the first to eliminate t gives (b) Δt0 =

37.14.

⎛ u2 ⎞ 1 x′ + ut ′ = γ x ⎜1 − 2 ⎟ = x, ⎝ c ⎠ γ and multiplying the first by

u and adding to the last to eliminate x gives c2 t′ +

37.15.

37.16.

⎛ u2 ⎞ 1 u x′ = γt ⎜1 − 2 ⎟ = t , 2 c ⎝ c ⎠ γ

so x = γ( x′ + ut ′) and t = γ(t ′ + ux′ c 2 ), which is indeed the same as Eq. (37.21) with the primed coordinates replacing the unprimed, and a change of sign of u. v′ + u 0.400c + 0.600c (a) v = = = 0.806c 1 + uv′ c 2 1 + (0.400) (0.600) v′ + u 0.900c + 0.600c (b) v = = = 0.974c 1 + uv′ c 2 1 + (0.900)(0.600) v′ + u 0.990c + 0.600c (c) v = = = 0.997c. 2 ′ 1 + uv c 1 + (0.990)(0.600) γ = 1.667( γ = 5 3 if u = (4 5)c ). (a) In Mavis’s frame the event “light on” has space-time coordinates x′ = 0 and t′ = 5.00 s, so from the result of ux′ ⎞ ⎛ Exercise 37.14 or Example 37.7, x = γ( x′ + ut′) and t = γ ⎜ t ′ + 2 ⎟ ⇒ x = γut ′ = 2.00 × 109 m, t = γt ′ = 8.33 s . c ⎠ ⎝

(b) The 5.00-s interval in Mavis’s frame is the proper time Δt0 in Eq.(37.6), so Δt = γΔt0 = 8.33 s, as in part (a). 37.17.

(c) (8.33 s) (0.800c ) = 2.00 × 109 m, which is the distance x found in part (a). IDENTIFY: The relativistic velocity addition formulas apply since the speeds are close to that of light. v −u . SET UP: The relativistic velocity addition formula is v′x = x uv 1 − 2x c EXECUTE: (a) For the pursuit ship to catch the cruiser, the distance between them must be decreasing, so the velocity of the cruiser relative to the pursuit ship must be directed toward the pursuit ship. (b) Let the unprimed frame be Tatooine and let the primed frame be the pursuit ship. We want the velocity v′ of the cruiser knowing the velocity of the primed frame u and the velocity of the cruiser v in the unprimed frame (Tatooine). v −u 0.600c − 0.800c = = −0.385c v′x = x uvx 1 − (0.600) (0.800) 1− 2 c

37.18.

The result implies that the cruiser is moving toward the pursuit ship at 0.385c. EVALUATE: The nonrelativistic formula would have given –0.200c, which is considerably different from the correct result. Let u y be the y-component of the velocity of S ′ relative to S. Following the steps used in the derivation of v′y + u y . Eq.(37.23) we get v y = 1 + u y v′y / c 2

37.19.

IDENTIFY and SET UP: Reference frames S and S ′ are shown in Figure 37.19.

Frame S is at rest in the laboratory. Frame S ′ is attached to particle 1. Figure 37.19 u is the speed of S ′ relative to S; this is the speed of particle 1 as measured in the laboratory. Thus u = +0.650c. The speed of particle 2 in S ′ is 0.950c. Also, since the two particles move in opposite directions, 2 moves in the − x′ direction and v′x = −0.950c. We want to calculate vx , the speed of particle 2 in frame S; use Eq.(37.23).


37-4

Chapter 37

v′x + u −0.950c + 0.650c −0.300c = = = −0.784c. The speed of the second particle, 2 2 ′ 1 + uvx / c 1 + (0.950c)(−0.650c) / c 1 − 0.6175 as measured in the laboratory, is 0.784c. EVALUATE: The incorrect Galilean expression for the relative velocity gives that the speed of the second particle in the lab frame is 0.300c. The correct relativistic calculation gives a result more than twice this. IDENTIFY and SET UP: Let S be the laboratory frame and let S ′ be the frame of one of the particles, as shown in Figure 37.20. Let the positive x direction for both frames be from particle 1 to particle 2. In the lab frame particle 1 is moving in the +x direction and particle 2 is moving in the − x direction. Then u = 0.9520c and v = −0.9520c . v′ is the velocity of particle 2 relative to particle 1. v−u −0.9520c − 0.9520c EXECUTE: v′ = = = −0.9988c . The speed of particle 2 relative to particle 1 1 − uv / c 2 1 − (0.9520c )( −0.9520c) / c 2 is 0.9988c . v′ < 0 shows particle 2 is moving toward particle 1. EXECUTE:

37.20.

vx =

Figure 37.20 37.21.

37.22.

37.23.

IDENTIFY: The relativistic velocity addition formulas apply since the speeds are close to that of light. v −u SET UP: The relativistic velocity addition formula is v′x = x . uv 1 − 2x c EXECUTE: In the relativistic velocity addition formula for this case, vx′ is the relative speed of particle 1 with respect to particle 2, v is the speed of particle 2 measured in the laboratory, and u is the speed of particle 1 measured in the laboratory, u = – v. v′ v − (−v) 2v v′x = = . x v 2 − 2v + vx′ = 0 and (0.890c)v 2 − 2c 2v + (0.890c 3 ) = 0 . 1 − ( − v )v c 2 1 + v 2 c 2 c 2 This is a quadratic equation with solution v = 0.611c (v must be less than c). EVALUATE: The nonrelativistic result would be 0.445c, which is considerably different from this result. IDENTIFY and SET UP: Let the starfighter’s frame be S and let the enemy spaceship’s frame be S ′ . Let the positive x direction for both frames be from the enemy spaceship toward the starfighter. Then u = +0.400c . v′ = +0.700c . v is the velocity of the missile relative to you. v′ + u 0.700c + 0.400c = = 0.859c EXECUTE: (a) v = 1 + uv′ / c 2 1 + (0.400)(0.700) (b) Use the distance it moves as measured in your frame and the speed it has in your frame to calculate the time it 8.00 × 109 m takes in your frame. t = = 31.0 s . (0.859)(3.00 × 108 m/s) IDENTIFY and SET UP: The reference frames are shown in Figure 37.23.

S = Arrakis frame S ′ = spaceship frame The object is the rocket. Figure 37.23

u is the velocity of the spaceship relative to Arrakis. vx = +0.360c; v′x = +0.920c (In each frame the rocket is moving in the positive coordinate direction.)


Relativity

Use the Lorentz velocity transformation equation, Eq.(37.22): v′x =

37-5

vx − u . 1 − uvx / c 2

vx − u ⎛ v v′ ⎞ ⎛ v v′ ⎞ so v′x − u ⎜ x 2 x ⎟ = vx − u and u ⎜ 1 − x 2 x ⎟ = vx − v′x c ⎠ 1 − uvx / c 2 ⎝ c ⎠ ⎝ vx − v′x 0.360c − 0.920c 0.560c u= = =− = −0.837c 1 − vx v′x / c 2 1 − (0.360c)(0.920c ) / c 2 0.6688

EXECUTE:

37.24.

v′x =

The speed of the spacecraft relative to Arrakis is 0.837c = 2.51 × 108 m/s. The minus sign in our result for u means that the spacecraft is moving in the − x-direction, so it is moving away from Arrakis. EVALUATE: The incorrect Galilean expression also says that the spacecraft is moving away from Arrakis, but with speed 0.920c – 0.360c = 0.560c. IDENTIFY: We need to use the relativistic Doppler shift formula. SET UP: The relativistic Doppler shift formula, Eq.(37.25), is f =

c+u f0 . c−u

c+u 2 f 0 . (c − u ) f 2 = (c + u ) f 02 . cf 2 − uf 2 = cf 02 + uf 02 . cf 2 − cf 02 = uf 2 + uf 02 and c−u c( f 2 − f 02 ) ( f / f 0 ) 2 − 1 = u= c. f 2 + f 02 ( f / f0 )2 + 1 (a) For f/f0 = 0.95, u = – 0.051c moving away from the source. (b) For f/f0 = 5.0, u = 0.923c moving towards the source. EVALUATE: Note that the speed required to achieve a 10 times greater Doppler shift is not 10 times the original speed. EXECUTE:

37.25.

f2=

IDENTIFY and SET UP: Source and observer are approaching, so use Eq.(37.25): f =

c+u f 0 . Solve for u, the c−u

speed of the light source relative to the observer. ⎛c+u⎞ 2 (a) EXECUTE: f 2 = ⎜ ⎟ f0 ⎝ c −u ⎠ (c − u ) f 2 = (c + u ) f 02 and u =

⎛ ( f / f0 )2 − 1 ⎞ c( f 2 − f 02 ) c = ⎜ ⎟ 2 f 2 + f 02 ⎝ ( f / f0 ) + 1 ⎠

λ0 = 675 nm, λ = 575 nm ⎛ (675 nm/575 nm) 2 − 1 ⎞ 8 7 u =⎜ ⎟ c = 0.159c = (0.159)(2.998 × 10 m/s) = 4.77 × 10 m/s; definitely speeding 2 (675 nm/575 nm) 1 + ⎝ ⎠ (b) 4.77 × 107 m/s = (4.77 × 107 m/s)(1 km/1000 m)(3600 s/1 h) = 1.72 × 108 km/h. Your fine would be $1.72 × 108 (172 million dollars). EVALUATE: The source and observer are approaching, so f > f 0 and λ < λ0 . Our result gives u < c, as it must. 37.26.

Using u = −0.600c = − ( 3 5 ) c in Eq.(37.25) gives 1 − ( 3 5) 25 f0 = f 0 = f 0 2. 1 + ( 3 5) 85 G G G G IDENTIFY and SET UP: If F is parallel to v then F changes the magnitude of v and not its direction. ⎞ dp d ⎛ mv = ⎜ F= ⎟ dt dt ⎝ 1 − v 2 / c 2 ⎠ d df dv Use the chain rule to evaluate the derivative: f (v (t )) = . dt dv dt m mv ⎛ dv ⎞ ⎛ 1 ⎞⎛ 2v ⎞⎛ dv ⎞ EXECUTE: (a) F = ⎜ ⎟+ ⎜ − ⎟⎜ − ⎟⎜ ⎟ (1 − v 2 / c 2 )1/ 2 ⎝ dt ⎠ (1 − v 2 / c 2 )3 / 2 ⎝ 2 ⎠⎝ c 2 ⎠⎝ dt ⎠ f =

37.27.

F=

⎛ v 2 v 2 ⎞ dv dv m m ⎜1 − + ⎟ = dt (1 − v 2 / c 2 )3 / 2 ⎝ c 2 c 2 ⎠ dt (1 − v 2 / c 2 )3 / 2

dv = a, so a = ( F / m)(1 − v 2 / c 2 )3 / 2 . dt EVALUATE: Our result agrees with Eq.(37.30).

But


37-6

37.28.

Chapter 37

G G G G (b) IDENTIFY and SET UP: If F is perpendicular to v then F changes the direction of v and not its magnitude. G G d⎛ ⎞ mv F= ⎜ ⎟. 2 2 dt ⎝ 1 − v / c ⎠ G G a = dv / dt but the magnitude of v in the denominator of Eq.(37.29) is constant. ma EXECUTE: F = and a = ( F / m)(1 − v 2 / c 2 )1/ 2 . 2 2 1− v / c EVALUATE: This result agrees with Eq.(37.33). 1 IDENTIFY and SET UP: γ = . If γ is 1.0% greater than 1 then γ = 1.010 , if γ is 10% greater than 1 1 − v2 / c2 then γ = 1.10 and if γ is 100% greater than 1 then γ = 2.00 .

EXECUTE:

v = c 1 − 1/ γ 2

(a) v = c 1 − 1/(1.010) 2 = 0.140c (b) v = c 1 − 1/(1.10) 2 = 0.417c (c) v = c 1 − 1/(2.00) 2 = 0.866c 37.29.

(a) p =

mv

= 2mv .

1 − v2 c2

1 v2 3 3 = 1 − 2 ⇒ v2 = c2 ⇒ v = c = 0.866c. 4 c 4 2 v 1 (b) F = γ 3ma = 2ma ⇒ γ3 = 2 ⇒ γ = (2)1/ 3 so = 22 / 3 ⇒ = 1 − 2−2 / 3 = 0.608 v2 c 1− 2 c The force is found from Eq.(37.32) or Eq.(37.33). (a) Indistinguishable from F = ma = 0.145 N. ⇒ 1 = 2 1 − v2 c2 ⇒

37.30.

(b) γ3ma = 1.75 N. (c) γ3ma = 51.7 N. (d) γma = 0.145 N, 0.333 N, 1.03 N. 37.31.

(a) K =

mc 2 1− v c 2

(b) K = 5mc 2 ⇒ 37.32. 37.33.

2

− mc 2 = mc 2

1 1− v c 2

2

1 1 − v2 c2

=2⇒

v2 1 3 =1− 2 ⇒ v = c = 0.866c. c 4 4

=6⇒

v2 1 35 =1− 2 ⇒ v = c = 0.986c. c 36 36

E = 2mc 2 = 2(1.67 ×10−27 kg)(3.00×108 m s) 2 = 3.01×10 −10 J = 1.88×109 eV. IDENTIFY and SET UP: Use Eqs.(37.38) and (37.39). EXECUTE: (a) E = mc 2 + K , so E = 4.00mc 2 means K = 3.00mc 2 = 4.50 × 10 −10 J

(b) E 2 = (mc 2 ) 2 + ( pc) 2 ; E = 4.00mc 2 , so 15.0( mc 2 ) 2 = ( pc ) 2

p = 15mc = 1.94 × 10−18 kg ⋅ m/s (c) E = mc 2 / 1 − v 2 / c 2

37.34.

37.35.

E = 4.00mc 2 gives 1 − v 2 / c 2 = 1/16 and v = 15/16c = 0.968c EVALUATE: The speed is close to c since the kinetic energy is greater than the rest energy. Nonrelativistic expressions relating E, K, p and v will be very inaccurate. (a) W = ΔK = ( γ f − 1) mc 2 = (4.07 × 10−3 ) mc 2 . (b) ( γ f − γ i ) mc 2 = 4.79mc 2 . (c) The result of part (b) is far larger than that of part (a). IDENTIFY: Use E = mc 2 to relate the mass increase to the energy increase. (a) SET UP: Your total energy E increases because your gravitational potential energy mgy increases.


Relativity

37-7

ΔE = mg Δy

EXECUTE:

ΔE = ( Δm)c so Δm = ΔE / c 2 = mg (Δy ) / c 2 2

Δm / m = ( g Δy ) / c 2 = (9.80 m/s 2 )(30 m)/(2.998 × 108 m/s) 2 = 3.3 × 10 −13% This increase is much, much too small to be noticed. (b) SET UP: The energy increases because potential energy is stored in the compressed spring. EXECUTE: ΔE = ΔU = 12 kx 2 = 12 (2.00 × 104 N/m)(0.060 m) 2 = 36.0 J

37.36.

Δm = ( ΔE ) / c 2 = 4.0 × 10 −16 kg Energy increases so mass increases. The mass increase is much, much too small to be noticed. EVALUATE: In both cases the energy increase corresponds to a mass increase. But since c 2 is a very large number the mass increase is very small. m0 = 2m0 . (a) E0 = m0c 2 . 2 E = mc 2 = 2m0c 2 . Therefore, m = 2m0 ⇒ 1 − v2 / c2

1 v2 v2 3 = 1 − 2 ⇒ 2 = ⇒ v = c 3 4 = 0.866c = 2.60 × 108 m s 4 c c 4 m0 (b) 10 m0c 2 = mc 2 = c2 . 1 − v2 c2 1− 37.37.

v2 1 v 2 99 99 = . v=c ⇒ 2= = 0.995c = 2.98 × 108 m s . 2 c 100 c 100 100

IDENTIFY and SET UP: The energy equivalent of mass is E = mc 2 . ρ = 7.86 g/cm3 = 7.86 × 103 kg/m3 . For a

cube, V = L3 . 1.0 × 1020 J E = = 1.11 × 103 kg 2 (3.00 × 108 m/s) 2 c m m 1.11 × 103 kg (b) ρ = so V = = = 0.141 m3 . L = V 1/ 3 = 0.521 m = 52.1 cm V ρ 7.86 × 103 kg/m3 EVALUATE: Particle/antiparticle annihilation has been observed in the laboratory, but only with small quantities of antimatter. (5.52 × 10 −27 kg)(3.00 × 108 m s) 2 = 4.97 × 10−10 J = 3105 MeV. IDENTIFY and SET UP: The total energy is given in terms of the momentum by Eq.(37.39). In terms of the total energy E, the kinetic energy K is K = E − mc 2 (from Eq.37.38). The rest energy is mc 2 . EXECUTE: (a) m =

37.38. 37.39.

EXECUTE: (a) E = (mc 2 ) 2 + ( pc) 2 =

[(6.64 × 10−27 )(2.998 × 108 ) 2 ]2 + [(2.10 × 10−18 )(2.998 × 108 )]2 J E = 8.67 × 10−10 J (b) mc 2 = (6.64 × 10 −27 kg)(2.998 × 108 m/s) 2 = 5.97 × 10 −10 J

37.40.

37.41.

K = E − mc 2 = 8.67 × 10 −10 J − 5.97 × 10−10 J = 2.70 × 10−10 J K 2.70 × 10−10 J (c) = = 0.452 mc 2 5.97 × 10−10 J EVALUATE: The incorrect nonrelativistic expressions for K and p give K = p 2 / 2m = 3.3 × 10 −10 J; the correct relativistic value is less than this. 12 ⎛ ⎛ p ⎞2 ⎞ 2 4 2 2 12 2 E = ( m c + p c ) = mc ⎜1 + ⎜ ⎜ ⎝ mc ⎟⎠ ⎟⎟ ⎝ ⎠ 2 2 ⎛ 1 p ⎞ p 1 = mc 2 + = mc 2 + mv 2 , the sum of the rest mass energy and the classical kinetic energy. E ≈ mc 2 ⎜ 1 + 2 2 ⎟ 2 m c 2 m 2 ⎝ ⎠ 1 1 (a) v = 8 × 107 m s ⇒ γ = = 1.0376 . For m = mp , K nonrel = mv 2 = 5.34 × 10−12 J . 2 1 − v2 c2 K rel = ( γ − 1) mc 2 = 5.65 × 10 −12 J.

K rel = 1.06. K nonrel

(b) v = 2.85 × 108 m s; γ = 3.203. 1 K rel = mv 2 = 6.78 × 10−11 J; K rel = (γ − 1)mc 2 = 3.31 × 10−10 J; K rel K nonrel = 4.88. 2


37-8

Chapter 37

37.42.

IDENTIFY: Since the speeds involved are close to that of light, we must use the relativistic formula for kinetic energy. ⎛ ⎞ 1 SET UP: The relativistic kinetic energy is K = (γ − 1) mc 2 = ⎜ − 1 ⎟ mc 2 . 2 2 ⎝ 1− v / c ⎠ ⎛ ⎞ ⎛ ⎞ 1 1 − 1⎟ mc 2 = (1.67 × 10−27 kg)(3.00 × 108 m s) 2 ⎜ − 1⎟ (a) K = (γ − 1)mc 2 = ⎜ 2 2 ⎜ 1 − 0.100c / c 2 ⎝ 1− v / c ⎠ ( ) ⎟⎠ ⎝

37.43.

37.44.

1 ⎛ ⎞ K = (1.50 × 10−10 J) ⎜ − 1⎟ = 7.56 × 10−13 J = 4.73 MeV ⎝ 1 − 0.0100 ⎠ ⎛ ⎞ 1 (b) K = (1.50 × 10−10 J) ⎜ − 1⎟ = 2.32 × 10−11 J = 145 MeV ⎜ 1 − (0.500) 2 ⎟ ⎝ ⎠ ⎛ ⎞ 1 (c) K = (1.50 × 10−10 J) ⎜ − 1⎟ = 1.94 × 10−10 J = 1210 MeV ⎜ 1 − (0.900) 2 ⎟ ⎝ ⎠ (d) ΔE = 2.32 × 10−11 J − 7.56 × 10−13 J = 2.24 × 10−11 J = 140 MeV (e) ΔE = 1.94 × 10−10 J − 2.32 × 10−11 J = 1.71× 10−10 J = 1070 MeV 1 (f) Without relativity, K = mv 2 . The work done in accelerating a proton from 0.100c to 0.500c in the 2 1 1 nonrelativistic limit is ΔE = m(0.500c) 2 − m(0.100c) 2 = 1.81 × 10−11 J = 113 MeV . 2 2 The work done in accelerating a proton from 0.500c to 0.900c in the nonrelativistic limit is 1 1 ΔE = m(0.900c) 2 − m(0.500c) 2 = 4.21× 10−11 J = 263 MeV . 2 2 EVALUATE: We see in the first case the nonrelativistic result is within 20% of the relativistic result. In the second case, the nonrelativistic result is very different from the relativistic result since the velocities are closer to c. IDENTIFY and SET UP: Use Eq.(23.12) and conservation of energy to relate the potential difference to the kinetic energy gained by the electron. Use Eq.(37.36) to calculate the kinetic energy from the speed. EXECUTE: (a) K = qΔV = eΔV ⎛ ⎞ 1 K = mc 2 ⎜ − 1⎟ = 4.025mc 2 = 3.295 × 10−13 J = 2.06 MeV 2 2 ⎝ 1− v / c ⎠ 6 ΔV = K / e = 2.06 × 10 V (b) From part (a), K = 3.30 × 10−13 J = 2.06 MeV EVALUATE: The speed is close to c and the kinetic energy is four times the rest mass. (a) According to Eq.(37.38) and conservation of mass-energy m 9.75 2 Mc 2 + mc 2 = γ 2 Mc 2 ⇒ γ = 1 + =1+ = 1.292. 2M 2(16.7) Note that since γ =

v 1 1 1 = 0.6331. , we have that = 1 − 2 = 1 − 2 2 γ c (1.292) 2 1− v c

(b) According to Eq.(37.36), the kinetic energy of each proton is ⎛ 1.00 MeV ⎞ K = (γ − 1) Mc 2 = (1.292 − 1)(1.67 × 10−27 kg)(3.00 × 108 m s ) 2 ⎜ ⎟ = 274 MeV. −13 ⎝ 1.60 × 10 J ⎠

⎛ 1.00 MeV ⎞ (c) The rest energy of η 0 is mc 2 = (9.75 × 10−28 kg)(3.00 × 108 m s) 2 ⎜ ⎟ = 548 MeV. −13 ⎝ 1.60 × 10 J ⎠ (d) The kinetic energy lost by the protons is the energy that produces the η 0 , 548 MeV = 2(274 MeV). 37.45.

IDENTIFY: The relativistic expression for the kinetic energy is K = (γ − 1) mc 2 , where γ =

1 The Newtonian expression for the kinetic energy is K N = mv 2 . 2 3 SET UP: Solve for v such that K = K N . 2

1 and x = v 2 / c 2 . 1− x


Relativity

37-9

2

3 1 3 1 ⎛ 3 ⎞ (γ − 1) mc 2 = mv 2 . −1 = x. = ⎜ 1 + x ⎟ . After a little algebra this becomes 4 4 1− x ⎝ 4 ⎠ 1− x 1 −15 ± (15) 2 + 4(9)(8) . The positive root is x = 0.425 . x = v 2 / c 2 , so 9 x 2 + 15 x − 8 = 0 . x = 18 v = x c = 0.652c . EVALUATE: The fractional increase of the relativistic expression above the nonrelativistic one increases as v increases. (4.0015 u) The fraction of the initial mass (a) that becomes energy is 1 − = 6.382 × 10−3 , and so the energy released 2(2.0136 u) per kilogram is (6.382 × 10 −3 )(1.00 kg)(3.00 × 108 m s) 2 = 5.74 × 1014 J. EXECUTE:

)

(

37.46.

1.0 × 1019 J = 1.7 × 104 kg. 5.74 × 1014 J kg (a) E = mc 2 , m = E c 2 = (3.8 × 1026 J) (2.998 × 108 m s ) 2 = 4.2 × 109 kg .

(b) 37.47.

1 kg is equivalent to 2.2 lbs, so m = 4.6 × 106 tons (b) The current mass of the sun is 1.99 × 1030 kg, so it would take it 37.48.

(1.99 × 1030 kg) (4.2 × 109 kg s) = 4.7 × 1020s = 1.5 × 1013 years to use up all its mass. IDENTIFY: Since the final speed is close to the speed of light, there will be a considerable difference between the relativistic and nonrelativistic results. 1 1 SET UP: The nonrelativistic work-energy theorem is F Δx = mv 2 − mv02 , and the relativistic formula for a 2 2 constant force is F Δx = (γ − 1)mc 2 . (a) Using the classical work-energy theorem and solving for Δx , we obtain

Δx =

m(v 2 − v02 ) (0.100 × 10−9 kg)[(0.900)(3.00 × 108 m s)]2 = = 3.65 m. 2F 2(1.00 × 106 N)

(b) Using the relativistic work-energy theorem for a constant force, we obtain (γ − 1) mc 2 Δx = . F 1 For the given speed, γ = = 2.29, thus 1 − 0.9002

Δx =

37.49.

(2.29 − 1)(0.100 × 10−9 kg)(3.00 × 108 m s) 2 = 11.6 m. (1.00 × 106 N)

EVALUATE: (c) The distance obtained from the relativistic treatment is greater. As we have seen, more energy is required to accelerate an object to speeds close to c, so that force must act over a greater distance. (a) IDENTIFY and SET UP: Δt0 = 2.60 × 10−8 s is the proper time, measured in the pion’s frame. The time measured in the lab must satisfy d = cΔt , where u ≈ c. Calculate Δt and then use Eq.(37.6) to calculate u. d 1.20 × 103 m = 4.003 × 10 −6 s EXECUTE: Δt = = c 2.998 × 108 m/s 2 Δt0 Δt ⎛ Δt ⎞ Δt = so (1 − u 2 / c 2 )1/ 2 = 0 and (1 − u 2 / c 2 ) = ⎜ 0 ⎟ Δt ⎝ Δt ⎠ 1 − u 2 / c2 Write u = (1 − Δ )c so that (u / c) 2 = (1 − Δ ) 2 = 1 − 2Δ + Δ 2 ≈ 1 − 2Δ since Δ is small.

⎛ Δt ⎞ Using this in the above gives 1 − (1 − 2Δ) = ⎜ 0 ⎟ ⎝ Δt ⎠ 2

2

1 ⎛ Δt ⎞ 1 ⎛ 2.60 × 10−8 s ⎞ −5 Δ= ⎜ 0⎟ = ⎜ ⎟ = 2.11 × 10 2 ⎝ Δt ⎠ 2 ⎝ 4.003 × 10 −6 s ⎠ EVALUATE: An alternative calculation is to say that the length of the tube must contract relative to the moving pion so that the pion travels that length before decaying. The contracted length must be l = cΔt0 = (2.998 × 108 m/s)(2.60 × 10−8 s) = 7.79 m. 2

⎛l ⎞ l = l0 1 − u 2 / c 2 so 1 − u 2 / c 2 = ⎜ ⎟ ⎝ l0 ⎠

2


37-10

Chapter 37 2

1 ⎛ l ⎞ 1 ⎛ 7.79 m ⎞ −5 Then u = (1 − Δ)c gives Δ = ⎜ ⎟ = ⎜ ⎟ = 2.11 × 10 , which checks. 2 ⎝ l0 ⎠ 2 ⎝ 1.20 × 103 m ⎠ 2

(b) IDENTIFY and SET UP: E = γ mc 2 (Eq.(37.38). 1 1 1 EXECUTE: γ = = = = 154 2 2 2Δ 1− u /c 2(2.11 × 10−5 )

37.50.

37.51.

E = 154(139.6 MeV) = 2.15 × 104 MeV = 21.5 GeV EVALUATE: The total energy is 154 times the rest energy. IDENTIFY and SET UP: The proper length of a side is l0 = a . The side along the direction of motion is shortened

to l = l0 1 − v 2 / c 2 . The sides in the two directions perpendicular to the motion are unaffected by the motion and still have a length a. EXECUTE: V = a 2l = a 3 1 − v 2 / c 2 IDENTIFY and SET UP: There must be a length contraction such that the length a becomes the same as b; l0 = a, l = b. l0 is the distance measured by an observer at rest relative to the spacecraft. Use Eq.(37.16) and solve for u. b l EXECUTE: = 1 − u 2 / c 2 so = 1 − u 2 / c 2 ; a l0 a = 1.40b gives b /1.40b = 1 − u 2 / c 2 and thus 1 − u 2 / c 2 = 1/(1.40) 2

37.52.

u = 1 − 1/(1.40) 2 c = 0.700c = 2.10 × 108 m/s EVALUATE: A length on the spacecraft in the direction of the motion is shortened. A length perpendicular to the motion is unchanged. IDENTIFY and SET UP: The proper time Δt0 is the time that elapses in the frame of the space probe. Δt is the time that elapses in the frame of the earth. The distance traveled is 42.2 light years, as measured in the earth frame. c ⎛ ⎞ EXECUTE: (a) Light travels 42.2 light years in 42.2 yr, so Δt = ⎜ ⎟ (42.2 yr) = 42.6 yr . 0.9910 c ⎝ ⎠

Δt0 = Δt 1 − u 2 / c 2 = (42.6 yr) 1 − (0.9910) 2 = 5.7 yr . She measures her biological age to be 19 yr + 5.7 yr = 24.7 yr. (b) Her age measured by someone on earth is 19 yr + 42.6 yr = 61.6 yr .

37.53.

(a) E = γ mc 2 and γ = 10 =

1 1 − (v c )

2

2 γ −1 99 v v = ⇒ = = 0.995. 2 γ 100 c c

⎞ ⎛ ⎛ v ⎞2 (b) ( pc) 2 = m 2v 2 γ 2c 2 , E 2 = m 2c 4 ⎜ ⎜ ⎟ γ 2 + 1⎟ ⎜⎝ c ⎠ ⎟ ⎝ ⎠ ⇒

37.54.

E 2 − ( pc) 2 = E2

1 ⎛v⎞ 1+ γ 2 ⎜ ⎟ ⎝c⎠

2

=

1 = 0.01 = 1%. 1 + (10 (0.995)) 2

IDENTIFY and SET UP: The clock on the plane measures the proper time Δt0 . Δt = 4.00 h = 4.00 h (3600 s/1 h) = 1.44 × 10 4 s. Δt0 Δt = and Δt0 = Δt 1 − u 2 / c 2 2 2 1− u / c ⎛ 1 u2 ⎞ u 1 u2 ; EXECUTE: small so 1 − u 2 / c 2 = (1 − u 2 / c 2 )1/ 2 ≈ 1 − thus Δ t = Δ t ⎜1 − 0 2 ⎟ 2 c2 c ⎝ 2c ⎠ 2

The difference in the clock readings is Δt − Δt0 =

37.55.

1 u2 1⎛ 250 m/s ⎞ −9 4 Δt = ⎜ ⎟ (1.44 × 10 s) = 5.01 × 10 s. The 2 c2 2 ⎝ 2.998 × 108 m/s ⎠

clock on the plane has the shorter elapsed time. EVALUATE: Δt0 is always less than Δt ; our results agree with this. The speed of the plane is much less than the speed of light, so the difference in the reading of the two clocks is very small. IDENTIFY: Since the speed is very close to the speed of light, we must use the relativistic formula for kinetic energy.


Relativity

37-11

⎛ ⎞ 1 SET UP: The relativistic formula for kinetic energy is K = mc 2 ⎜ − 1⎟ and the relativistic mass is ⎜ 1 − v2 c2 ⎟ ⎝ ⎠ m mrel = . 1 − v2 c2 EXECUTE: (a) K = 7 × 1012 eV = 1.12 × 10−6 J . Using this value in the relativistic kinetic energy formula and ⎛ ⎞ 1 substituting the mass of the proton for m, we get K = mc 2 ⎜ − 1⎟ ⎜ 1 − v2 c2 ⎟ ⎝ ⎠ 2 v 1 1 which gives = 7.45 × 103 and 1 − 2 = . Solving for v gives 2 2 × (7.45 103 ) 2 c 1− v c

1−

v 2 (c + v )(c − v) 2(c − v) = = , since c + v ≈ 2c. Substituting v = (1 − Δ )c , we have. c2 c2 c 1

3 1 − v 2 / c 2 ( 7.45 × 10 ) v 2 2(c − v) 2 [ c − (1 − Δ )c ] 1− 2 = = = 2Δ . Solving for Δ gives Δ = = = 9 × 10−9 , to one 2 2 c c c significant digit. 1 (b) Using the relativistic mass formula and the result that = 7.45 × 103 , we have 2 2 1− v c 2

⎛ ⎞ 1 ⎟ = (7 × 103 )m , to one significant digit. = m⎜ 2 2 ⎜ 1− v c ⎟ 1− v c ⎝ ⎠ EVALUATE: At such high speeds, the proton’s mass is over 7000 times as great as its rest mass. E ⎛ 1 ⎞ IDENTIFY and SET UP: The energy released is E = ( Δm)c 2 . Δm = ⎜ 4 ⎟ (8.00 kg) . Pav = . The change in t ⎝ 10 ⎠ gravitational potential energy is mg Δy . m

mrel =

37.56.

2

2

⎛ 1 ⎞ EXECUTE: (a) E = ( Δm)c 2 = ⎜ 4 ⎟ (8.00 kg)(3.00 × 108 m/s) 2 = 7.20 × 1013 J ⎝ 10 ⎠ E 7.20 × 1013 J = = 1.80 × 1019 W t 4.00 × 10−6 s E 7.20 × 1013 J = = 7.35 × 109 kg (c) E = ΔU = mg Δy . m = g Δy (9.80 m/s 2 )(1.00 × 103 m) (b) Pav =

37.57.

c IDENTIFY and SET UP: In crown glass the speed of light is v = . Calculate the kinetic energy of an electron that n has this speed. 2.998 × 108 m/s = 1.972 × 108 m/s. EXECUTE: v = 1.52 K = mc 2 (γ − 1) mc 2 = (9.109 × 10 −31 kg)(2.998 × 108 m/s) 2 = 8.187 × 10 −14 J(1 eV/1.602 × 10−19 J) = 0.5111 MeV

γ=

37.58.

37.59.

1 1− v /c 2

2

=

1 1 − ((1.972 × 10 m/s)/(2.998 × 108 m/s)) 2 8

= 1.328

K = mc 2 (γ − 1) = (0.5111 MeV)(1.328 − 1) = 0.168 MeV EVALUATE: No object can travel faster than the speed of light in vacuum but there is nothing that prohibits an object from traveling faster than the speed of light in some material. p ( E c) E , where the atom and the photon have the same magnitude of momentum, E c . (a) v = = = m m mc E (b) v =  c, so E  mc 2 . mc IDENTIFY and SET UP: Let S be the lab frame and S ′ be the frame of the proton that is moving in the +x direction, so u = + c / 2 . The reference frames and moving particles are shown in Figure 37.59. The other proton moves in


37-12

Chapter 37

the − x direction in the lab frame, so v = −c / 2 . A proton has rest mass mp = 1.67 × 10−27 kg and rest energy

mp c 2 = 938 MeV . EXECUTE: (a) v′ =

−c / 2 − c / 2 v −u 4c = =− 1 − uv / c 2 1 − (c / 2)( −c / 2) / c 2 5

4 c. 5 (b) In nonrelativistic mechanics the speeds just add and the speed of each relative to the other is c. mc 2 (c) K = − mc 2 1 − v2 / c2 (i) Relative to the lab frame each proton has speed v = c / 2 . The total kinetic energy of each proton is 938 MeV K= − (938 MeV) = 145 MeV . 2 ⎛1⎞ 1− ⎜ ⎟ ⎝ 2⎠ 4 (ii) In its rest frame one proton has zero speed and zero kinetic energy and the other has speed c . In this frame 5 938 MeV the kinetic energy of the moving proton is K = − (938 MeV) = 625 MeV 2 ⎛ 4⎞ 1− ⎜ ⎟ ⎝5⎠ (d) (i) Each proton has speed v = c / 2 and kinetic energy mc 2 938 MeV 1 2 ⎛1 ⎞ K = mv 2 = ⎜ m ⎟ ( c / 2 ) = = = 117 MeV 2 8 8 ⎝2 ⎠ (ii) One proton has speed v = 0 and the other has speed c. The kinetic energy of the moving proton 1 938 MeV is K = mc 2 = = 469 MeV 2 2 EVALUATE: The relativistic expression for K gives a larger value than the nonrelativistic expression. The kinetic energy of the system is different in different frames. The speed of each proton relative to the other is

Figure 37.59 37.60.

IDENTIFY and SET UP: Let S be the lab frame and let S ′ the frame of the proton that is moving in the +x direction in the lab frame, as shown in Figure 37.60. In S ′ the other proton moves in the − x′ direction with speed c / 2 , so v′ = −c / 2 . In the lab frame each proton has speed α c , where α is a constant that we need to solve for. v′ + u −0.50c + α c and EXECUTE: (a) v = with v = −α c , u = +α c and v′ = −0.50c gives −α c = 2 1 + uv′ / c 1 + (α c)(−0.50c) / c 2

−0.50 + α . α 2 − 4α + 1 = 0 and α = 0.268 or α = 3.73 . Can’t have v > c , so only α = 0.268 is physically 1 − 0.50α allowed. The speed measured by the observer in the lab is 0.268c. (b) (i) v = 0.269c . γ = 1.0380 . K = (γ − 1)mc 2 = 35.6 MeV . −α =


Relativity

37-13

(ii) v = 0.500c . γ = 1.1547 . K = (γ − 1)mc = 145 MeV . 2

37.61.

37.62.

37.63.

x′ = c t ′ ⇒ ( x − ut ) γ = c γ ( t − ux c 2

2 2

2

2

2

2

)

Figure 37.60

2 2

⎛ u⎞ 1 ⇒ x − ut = c(t − ux c 2 ) ⇒ x ⎜ 1 + ⎟ = x(u + c ) = t (u + c) ⇒ x = ct ⇒ x 2 = c 2t 2 . ⎝ c⎠ c IDENTIFY and SET UP: Let S be the lab frame and let S ′ be the frame of the nucleus. Let the +x direction be the direction the nucleus is moving. u = 0.7500c . v′ + u 0.9995c + 0.7500c EXECUTE: (a) v′ = +0.9995c . v = = = 0.999929c 1 + uv′ / c 2 1 + (0.7500)(0.9995) −0.9995c + 0.7500c (b) v′ = −0.9995c . v = = −0.9965c 1 + (0.7500)( −0.9995) (c) emitted in same direction: ⎛ ⎞ ⎛ ⎞ 1 1 − 1⎟ mc 2 = (0.511 MeV) ⎜ − 1⎟ = 42.4 MeV (i) K = ⎜ 2 2 2 ⎜ 1 − (0.999929) ⎟ ⎝ 1− v / c ⎠ ⎝ ⎠ ⎛ ⎞ ⎛ ⎞ 1 1 (ii) K ′ = ⎜ − 1⎟ mc 2 = (0.511 MeV) ⎜ − 1⎟ = 15.7 MeV 2 2 2 ⎜ ⎟ ⎝ 1− v /c ⎠ ⎝ 1 − (0.9995) ⎠ (d) emitted in opposite direction: ⎛ ⎞ ⎛ ⎞ 1 1 (i) K = ⎜ − 1⎟ mc 2 = (0.511 MeV) ⎜ − 1⎟ = 5.60 MeV 2 2 ⎜ 1 − (0.9965) 2 ⎟ ⎝ 1− v / c ⎠ ⎝ ⎠ ⎛ ⎞ ⎛ ⎞ 2 1 1 (ii) K ′ = ⎜ − 1⎟ mc = (0.511 MeV) ⎜ − 1⎟ = 15.7 MeV 2 2 2 ⎜ ⎟ ⎝ 1− v /c ⎠ ⎝ 1 − (0.9995) ⎠ IDENTIFY and SET UP: Use Eq.(37.30), with a = dv / dt , to obtain an expression for dv / dt. Separate the variables v and t and integrate to obtain an expression for v (t ). In this expression, let t → ∞. dv F EXECUTE: a = = (1 − v 2 / c 2 )3 / 2 . (One-dimensional motion is assumed, and all the F, v, and a refer to xdt m components.) dv ⎛F⎞ = ⎜ ⎟ dt (1 − v 2 / c 2 )3 / 2 ⎝ m ⎠ Integrate from t = 0, when v = 0, to time t, when the velocity is v. v t⎛ F ⎞ dv ∫ 0 (1 − v 2 / c 2 )3 / 2 = ∫ 0 ⎜⎝ m ⎟⎠ dt t⎛ F ⎞ Ft Since F is constant, ∫ ⎜ ⎟ dt = . In the velocity integral make the change of variable y = v / c; then dy = dv / c. 0 m m ⎝ ⎠ v/c

v/c ⎡ ⎤ dv dy y v ∫ 0 (1 − v 2 / c 2 )3 / 2 = c ∫ 0 (1 − y 2 )3 / 2 = c ⎢⎣ (1 − y 2 )1/ 2 ⎥⎦ = 1 − v 2 / c 2 0 v Ft Thus = . 1 − v2 / c2 m v


37-14

Chapter 37

37.64.

Solve this equation for v: 2 2 v2 ⎛ Ft ⎞ ⎛ Ft ⎞ 2 2 2 v = = and ⎜ ⎟ (1 − v / c ) ⎜ ⎟ 1 − v2 / c2 ⎝ m ⎠ ⎝m⎠ ⎛ ⎛ Ft ⎞2 ⎞ ⎛ Ft ⎞ 2 ( Ft / m) Ft so v = v 2 ⎜1 + ⎜ = =c 2 2 2 ⎜ ⎝ mc ⎟⎠ ⎟⎟ ⎜⎝ m ⎟⎠ 1 + ( Ft / mc) m c + F 2t 2 ⎝ ⎠ Ft Ft → → 1, so v → c. As t → ∞, 2 2 2 2 m c +F t F 2t 2 Ft is always less than 1, so v < c always and v approaches c only when t → ∞. EVALUATE: Note that 2 2 m c + F 2t 2 Setting x = 0 in Eq.(37.21), the first equation becomes x′ = −γ ut and the last, upon multiplication by c, becomes

37.65.

ct ′ = γ ct.Squaring and subtracting gives c 2t′2 − x′2 = γ 2 (c 2t 2 − u 2t 2 ) = c 2t 2 , or x′ = c t ′2 − t 2 = 4.53 × 108 m. (a) IDENTIFY and SET UP: Use the Lorentz coordinate transformation (Eq.37.21) for ( x1 , t1 ) and ( x2 , t2 ) : x1 − ut1 x − ut2 x1′ = , x2′ = 2 2 2 1− u /c 1 − u 2 / c2 2 t − ux1 / c t − ux2 / c 2 t1′ = 1 , t2′ = 2 1 − u 2 / c2 1 − u 2 / c2 Same point in S ′ implies x1′ = x2′ . What then is Δt ′ = t2′ − t1′ ? EXECUTE: x1′ = x′2 implies x1 − ut1 = x2 − ut2 x − x Δx u (t2 − t1 ) = x2 − x1 and u = 2 1 = t2 − t1 Δt From the time transformation equations, 1 ( Δt − uΔx / c 2 ) Δt ′ = t2′ − t1′ = 1 − u 2 / c2 Δx gives Using the result that u = Δt 1 Δt ′ = ( Δt − ( Δx) 2 /(( Δt )c 2 )) 2 1 − ( Δx ) /(( Δt ) 2 c 2 )

Δt ′ =

Δt ′ =

Δt ( Δt ) − ( Δx ) 2 / c 2 2

( Δt ) 2 − ( Δx ) 2 / c 2 ( Δt ) 2 − ( Δx ) 2 / c 2

( Δt − ( Δx) 2 /((Δt )c 2 ))

= ( Δt ) 2 − ( Δx / c ) 2 , as was to be shown.

This equation doesn’t have a physical solution (because of a negative square root) if ( Δx / c ) 2 > (Δt ) 2 or Δx ≥ cΔt. (b) IDENTIFY and SET UP: Now require that t2′ = t1′ (the two events are simultaneous in S ′ ) and use the Lorentz coordinate transformation equations. EXECUTE: t2′ = t1′ implies t1 − ux1 / c 2 = t2 − ux2 / c 2 c 2 Δt ⎛x −x ⎞ ⎛ Δx ⎞ t2 − t1 = ⎜ 2 2 1 ⎟ u so Δt = ⎜ 2 ⎟ u and u = Δx ⎝ c ⎠ ⎝c ⎠ From the Lorentz transformation equations, ⎛ ⎞ 1 Δx′ = x′2 − x1′ = ⎜ ⎟ (Δx − u Δt ). 2 2 ⎝ 1− u / c ⎠ Using the result that u = c 2 Δt / Δx gives 1 Δx′ = (Δx − c 2 ( Δt ) 2 / Δx) 2 1 − c (Δt ) 2 /( Δx) 2

Δx′ = Δx′ =

Δx ( Δx) 2 − c 2 ( Δt ) 2 (Δx) 2 − c 2 ( Δt ) 2 ( Δx ) 2 − c 2 ( Δt ) 2

(Δx − c 2 (Δt ) 2 / Δx) = (Δx) 2 − c 2 (Δt ) 2

(c) IDENTIFY and SET UP: The result from part (b) is Δx′ = ( Δx) 2 − c 2 (Δt ) 2


Relativity

37-15

Solve for Δt : ( Δx′) 2 = ( Δx) 2 − c 2 ( Δt ) 2 ( Δx) 2 − ( Δx′) 2 (5.00 m) 2 − (2.50 m) 2 = = 1.44 × 10−8 s c 2.998 × 108 m/s EVALUATE: This provides another illustration of the concept of simultaneity (Section 37.2): events observed to be simultaneous in one frame are not simultaneous in another frame that is moving relative to the first. 1 (a) 80.0 m s is non-relativistic, and K = mv 2 = 186 J. 2 (b) (γ − 1) mc 2 = 1.31 × 1015 J. Δt =

EXECUTE:

37.66.

(c) In Eq. (37.23), c) v′ = 2.20 × 108 m s, u = −1.80 × 108 m s,and so v = 7.14 × 107 m s.

20.0 m

(d) (e)

γ

20.0 m = 9.09 × 10−8 s. 2.20 × 108 m s

(f) t ′ = 37.67.

= 13.6 m.

t 13.6 m = 6.18 × 10−8 s, or t ′ = = 6.18 × 10−8 s. γ 2.20 × 108 m s

IDENTIFY and SET UP: An increase in wavelength corresponds to a decrease in frequency ( f = c / λ ), so the

atoms are moving away from the earth. Receding, so use Eq.(37.26): f =

c−u f0 c+u

⎛ 1 − ( f / f0 )2 ⎞ EXECUTE: Solve for u: ( f / f 0 ) 2 (c + u ) = c − u and u = c ⎜ 2 ⎟ ⎝ 1 + ( f / f0 ) ⎠ f = c / λ , f 0 = c / λ0 so f / f 0 = λ0 / λ

37.68.

⎛ 1 − (λ0 / λ ) 2 ⎞ ⎛ 1 − (656.3/ 953.4) 2 ⎞ u = c⎜ c = = 0.357c = 1.07 × 108 m/s ⎟ ⎜ 2 2 ⎟ ⎝ 1 + (656.3/ 953.4) ⎠ ⎝ 1 + (λ0 / λ ) ⎠ EVALUATE: The relative speed is large, 36% of c. The cosmological implication of such observations will be discussed in Section 44.6. The baseball had better be moving non-relativistically, so the Doppler shift formula (Eq.(37.25)) becomes f ≅ f 0 (1 − (u c)). In the baseball’s frame, this is the frequency with which the radar waves strike the baseball, and the baseball reradiates at f. But in the coach’s frame, the reflected waves are Doppler shifted again, so the detected frequency is f (1 − (u c)) = f 0 (1 − (u c)) 2 ≈ f 0 (1 − 2(u c)), so Δf = 2 f 0 (u c) and the fractional frequency shift is

Δf = 2(u c ). In this case, f0 u= 37.69.

Δf (2.86 × 10 −7 ) c= (3.00 × 108 m) = 42.9 m s = 154 km h = 92.5 mi h. 2 f0 2

IDENTIFY and SET UP: 500 light years = 4.73 × 1018 m . The proper distance l0 to the star is 500 light years. The energy needed is the kinetic energy of the rocket at its final speed. d 4.73 × 1018 m EXECUTE: (a) u = 0.50c . Δt = = = 3.2 × 1010 s = 1000 yr u (0.50)(3.00 × 108 m/s)

The proper time is measured by the astronauts. Δt0 = Δt 1 − u 2 / c 2 = 866 yr

⎛ ⎞ 1 − mc 2 = (1000 kg)(3.00 × 108 m/s) 2 ⎜ − 1⎟ = 1.4 × 1019 J ⎜ 1 − (0.500) 2 ⎟ 1 − v2 / c2 ⎝ ⎠ This is 140% of the U.S. yearly use of energy. d 4.73 × 1018 m (b) u = 0.99c . Δt = = = 1.6 × 1010 s = 505 yr , Δt0 = 71 yr u (0.99)(3.00 × 108 m/s) K=

mc 2

⎛ ⎞ 1 K = (9.00 × 1019 J) ⎜ − 1⎟ = 5.5 × 1020 J ⎜ 1 − (0.99) 2 ⎟ ⎝ ⎠ This is 55 times the U.S. yearly use.


37-16

Chapter 37

(c) u = 0.9999c . Δt =

37.70.

d 4.73 × 1018 m = = 1.58 × 1010 s = 501 yr , Δt0 = 7.1 yr u (0.9999)(3.00 × 108 m/s)

⎛ ⎞ 1 K = (9.00 × 1019 J) ⎜ − 1⎟ = 6.3 × 1021 J 2 ⎜ 1 − (0.9999) ⎟ ⎝ ⎠ This is 630 times the U.S. yearly use. The energy cost of accelerating a rocket to these speeds is immense. (a) As in the hint, both the sender and the receiver measure the same distance. However, in our frame, the ship has moved between emission of successive wavefronts, and we can use the time T = 1 f as the proper time, with the result that f = γ f 0 > f 0 . 1/ 2

(b) Toward: f = f 0

c+u ⎛ 1 + 0.758 ⎞ = 345 MHz ⎜ ⎟ c−u ⎝ 1 − 0.758 ⎠

= 930 MHz

f − f 0 = 930 MHz − 345 MHz = 585 MHz. 1/ 2

Away: f = f 0

37.71.

c−u ⎛ 1 − 0.758 ⎞ = 345 MHz ⎜ ⎟ c+u ⎝ 1 + 0.758 ⎠

= 128 MHz and f − f 0 = −217 MHz.

(c) γf 0 = 1.53 f 0 = 528 MHz, f − f 0 = 183 MHz. The shift is still bigger than f 0 , but not as large as the approaching frequency. The crux of this problem is the question of simultaneity. To be “in the barn at one time” for the runner is different than for a stationary observer in the barn. The diagram in Figure 37.71a shows the rod fitting into the barn at time t = 0 , according to the stationary observer. The diagram in Figure 37.71b is in the runner’s frame of reference. The front of the rod enters the barn at time t1 and leaves the back of the barn at time t2 . However, the back of the rod does not enter the front of the barn until the later time t3 .

Figure 37.71 37.72.

(c/n) + V (c/n) + V = . For V non-relativistic, this is cV 1 + (V/nc) 1+ 2 nc c ⎛ 1 ⎞ 1 ⎞ ⎛ v ≈ ((cn) + V )(1 − (V/nc )) = (nc/n) + V − (V/n 2 ) − (V 2 /nc) ≈ + ⎜1 − 2 ⎟V , so k = ⎜1 − 2 ⎟ . For water, n = 1.333 n ⎝ n ⎠ ⎝ n ⎠ and k = 0.437. In Eq.(37.23), u = V , v′ = (c n), and so v =


Relativity

37.73.

37-17

dv dv v−u u dv . dt ′ = γ ( dt − udx c 2 ) . dv′ = + 2 2 2 dt ′ (1 − uv c ) (1 − uv c ) c 2 1 dv′ v −u ⎛u⎞ = + ⎜ ⎟. dv 1 − uv c 2 (1 − uv c 2 ) 2 ⎝ c 2 ⎠

(a) a′ =

⎛ ⎛ 1 − u 2 c2 ⎞ 1 (v − u ) u c 2 ⎞ dv′ = dv ⎜ + = dv ⎜ 2 2 2 ⎟ 2 2 ⎟ (1 − uv c ) ⎠ ⎝ 1 − uv c ⎝ (1 − uv c ) ⎠ (1 − u 2 c 2 ) 1 (1 − uv c 2 ) 2 dv (1 − u 2 c 2 ) a′ = = 2 γdt − uγ dx c dt (1 − uv c 2 ) 2 γ(1 − uv c 2 )

dv

= a (1 − u 2 c 2 )3 2 (1 − uv c 2 ) −3 . −3

37.74.

⎛ uv′ ⎞ (b) Changing frames from S ′ → S just involves changing a → a′, v → − v′ ⇒ a = a′(1 − u 2 c 2 )3 2 ⎜1 + 2 ⎟ . c ⎠ ⎝ (a) The speed v′ is measured relative to the rocket, and so for the rocket and its occupant, v′ = 0. The acceleration as seen in the rocket is given to be a′ = g , and so the acceleration as measured on the earth is 32

⎛ u2 ⎞ du = g ⎜1 − 2 ⎟ . dt ⎝ c ⎠ (b) With v1 = 0 when t = 0 , t1 1 du 1 v1 du v1 dt = . ∫ dt = ∫ . t1 = . 2 2 32 0 g (1 − u c ) g 0 (1 − u 2 c 2 )3 2 g 1 − v12 c 2

a=

(c) dt ′ = γ dt = dt / 1 − u 2 c 2 , so the relation in part (b) between dt and du, expressed in terms of dt ′ and du, is 1 du 1 du dt ′ = γ dt = . = 2 2 g (1 − u 2 c 2 ) 3 2 g (1 − u 2 c 2 ) 2 1− u c

c ⎛v ⎞ arctanh ⎜ 1 ⎟ . For those who wish to g ⎝c⎠ avoid inverse hyperbolic functions, the above integral may be done by the method of partial fractions; c ⎛ c + v1 ⎞ du 1 ⎡ du du ⎤ gdt ′ = 1n ⎜ = ⎢ + ⎟. ⎥ , which integrates to t1′ = 2 g ⎝ c − v1 ⎠ (1 + u c)(1 − u c) 2 ⎣1 + u c 1 − uc ⎦ (d) Solving the expression from part (c) for v1 in terms of t1 , (v1 c ) = tanh( gt1′ c), so that Integrating as above (perhaps using the substitution z = u c ) gives t1′ =

1 − (v1 c) 2 = 1 cosh ( gt1′ c), using the appropriate indentities for hyperbolic functions. Using this in the expression gt c tanh( gt1′ c) c ⎛ gt ′ ⎞ = sinh( gt1′ c), which may be rearranged slightly as 1 = sinh ⎜ 1 ⎟ . If c g 1 cosh( gt1′ c ) g ⎝ c ⎠ gt1′/c − gt1′/c v e −e hyperbolic functions are not used, v1 in terms of t′1 is found to be 1 = gt1′/c which is the same as c e + e− gt1′/c c gt1′ c − gt1′ c (e −e ), tanh( gt1′ c ). Inserting this expression into the result of part (b) gives, after much algebra, t1 = 2g which is equivalent to the expression found using hyperbolic functions. (e) After the first acceleration period (of 5 years by Stella’s clock), the elapsed time on earth is c t1′ = sinh( gt1′ c) = 2.65 × 109 s = 84.0 yr. g

found in part (b), t1 =

37.75.

The elapsed time will be the same for each of the four parts of the voyage, so when Stella has returned, Terra has aged 336 yr and the year is 2436. (Keeping more precision than is given in the problem gives February 7 of that year.) (a) f 0 = 4.568110 × 1014 Hz; f + = 4.568910 × 1014 Hz; f − = 4.567710 × 1014 Hz c + (u + v ) ⎫ f0 ⎪ c − (u + v) ⎪ f +2 (c − (u + v)) = f 02 (c + (u + v)) ⎬⇒ 2 f − (c − (u − v)) = f 02 (c + (u − v)) c + (u − v) ⎪ f− = f0 ⎪ c − (u − v ) ⎭ f+ =


37-18

Chapter 37

where u is the velocity of the center of mass and v is the orbital velocity.

⇒ (u + v ) =

( f + f0 ) 2 − 1 ( f −2 f 02 ) − 1 and c ( u − v ) = c ( f + f0 )2 + 1 ( f −2 f 02 ) + 1

⇒ u + v = 5.25 × 10 4 m s and u − v = −2.63 × 104 m s .

This gives u = +1.31 × 104 m s (moving toward at 13.1 km s) and v = 3.94 × 10 4 m/s . (b) v = 3.94 × 104 m s; T = 11.0 days. 2π R = vt ⇒

(3.94 × 104 m s)(11.0 days)(24 hrs day)(3600 sec hr) = 5.96 × 109 m . This is about 2π 0.040 times the earth-sun distance. Also the gravitational force between them (a distance of 2R) must equal the centripetal force from the center of mass: (Gm 2 ) mv 2 4 Rv 2 4(5.96 × 109 m)(3.94 × 104 m s) 2 ⇒m= = = 5.55 × 10 29 kg = 0.279 m sun . = 2 (2 R) R G 6.672 × 10 −11 N ⋅ m 2 kg 2 For any function f = f ( x, t ) and x = x( x′, t′), t = t ( x′, t′), let F ( x′, t ′) = f ( x( x′, t ′), t ( x′, t ′)) and use the standard (but mathematically improper) notation F ( x′, t ′) = f ( x′, t ′). The chain rule is then R=

37.76.

∂f ( x′, t ′) ∂f ( x, t ) ∂x′ ∂f ( x′, t ′) ∂t′ , = + ∂x ∂x′ ∂x ∂t ′ ∂x ∂f ( x′, t ′) ∂f ( x, t ) ∂x′ ∂f ( x′, t ′) ∂t ′ . = + ∂t ∂x′ ∂t ∂t ′ ∂t In this solution, the explicit dependence of the functions on the sets of dependent variables is suppressed, and the ∂f ∂f ∂x′ ∂f ∂t′ ∂f ∂f ∂x′ ∂f ∂t ′ = + , = + . above relations are then ∂x ∂x′ ∂x ∂t′ ∂x ∂t ∂x′ ∂t ∂t ′ ∂t ∂x′ ∂x′ ∂t ′ ∂t ′ ∂E ∂E ∂2E ∂2E (a) = 1, = −v, = 0 and = 1. Then, = , and 2 = 2 . For the time derivative, ∂x ∂t ∂x ∂t ∂x ∂x′ ∂x ∂x′ ∂E ∂E ∂E = −v + . To find the second time derivative, the chain rule must be applied to both terms; that is, ∂t ∂x′ ∂t′

∂ ∂t ∂ ∂t

∂E ∂2E ∂2E = −v 2 + , ∂x′ ∂x′ ∂t ′∂x′ ∂E ∂2E ∂2E = −v + . ∂t′ ∂x′∂t ′ ∂t ′2

∂2E , collecting terms and equating the mixed partial derivatives gives ∂t 2 2 ∂2E ∂2E ∂2E ∂2E 2 ∂ E = v − 2 v + , and using this and the above expression for gives the result. ∂t 2 ∂x′2 ∂x′∂t ′ ∂t ′2 ∂x′2 ∂x′ ∂x′ ∂t′ ∂t ′ = γ, = γ v, = γ v / c 2 and = γ. (b) For the Lorentz transformation, ∂x ∂t ∂x ∂t The first partials are then Using these in

∂E ∂E v ∂E ∂E ∂E ∂E =γ −γ 2 = − γv +γ , ∂x ∂x′ c ∂t′ ∂t ∂x′ ∂t′ and the second partials are (again equating the mixed partials) 2 2 2 ∂2E ∂2E 2 ∂ E 2 v ∂ E 2 v = + − γ γ 2 γ ∂x 2 ∂x′2 c 4 ∂t ′2 c 2 ∂x′∂t ′ 2 2 2 ∂ E ∂ E ∂ E ∂2E = γ 2 v 2 2 + γ 2 2 − 2 γ 2v . 2 ∂t ∂x′ ∂t ′ ∂x′∂t′

Substituting into the wave equation and combining terms (note that the mixed partials cancel), 2 v2 ⎞ ∂2 E ∂2E 1 ∂2E 1 ⎞ ∂2E ∂2E 1 ∂2E 2⎛ 2⎛v γ 1 γ − = − + − = 2− 2 = 0. ⎜ ⎟ ⎜ 2 2 4 2 ⎟ 2 ∂x 2 c 2 ∂t 2 ∂x′ c ∂t ′2 ⎝ c ⎠ ∂x′ ⎝ c c ⎠ ∂t ′


Relativity

37.77.

37-19

(a) In the center of momentum frame, the two protons approach each other with equal velocities (since the protons have the same mass). After the collision, the two protons are at rest─but now there are kaons as well. In this situation the kinetic energy of the protons must equal the total rest energy of the two kaons ⇒ 2( γ cm − 1)mp c 2 = 2mk c 2 ⇒ γ cm = 1 + γ cm − 1

mk = 1.526. The velocity of a proton in the center of momentum frame is then mp

2

vcm = c

2 γ cm

= 0.7554c.

To get the velocity of this proton in the lab frame, we must use the Lorentz velocity transformations. This is the same as “hopping” into the proton that will be our target and asking what the velocity of the projectile proton is. Taking the lab frame to be the unprimed frame moving to the left, u = vcm and v′ = vcm (the velocity of the projectile proton in the center of momentum frame). v′ + u 2vcm 1 vlab = = = 0.9619c ⇒ γ lab = = 3.658 ⇒ K lab = ( γ lab − 1)mpc 2 = 2494 MeV. 2 2 uv′ v v 1 + 2 1 + cm2 1 − lab2 c c c

K lab 2494 MeV = = 2.526. 2mk 2(493.7 MeV) (c) The center of momentum case considered in part (a) is the same as this situation. Thus, the kinetic energy required is just twice the rest mass energy of the kaons. K cm = 2(493.7 MeV) = 987.4 MeV. This offers a substantial advantage over the fixed target experiment in part (b). It takes less energy to create two kaons in the proton center of momentum frame. (b)


PHOTONS, ELECTRONS, AND ATOMS

38.1.

38

IDENTIFY and SET UP: The stopping potential V0 is related to the frequency of the light by V0 =

h φ f − . The e e

slope of V0 versus f is h/e. The value fth of f when V0 = 0 is related to φ by φ = hf th . EXECUTE: (a) From the graph, f th = 1.25 × 1015 Hz . Therefore, with the value of h from part (b),

φ = hf th = 4.8 eV . (b) From the graph, the slope is 3.8 × 10−15 V ⋅ s . h = (e)(slope) = (1.60 × 10−16 C)(3.8 × 10−15 V ⋅ s) = 6.1 × 10 −34 J ⋅ s (c) No photoelectrons are produced for f < f th .

38.2.

(d) For a different metal fth and φ are different. The slope is h/e so would be the same, but the graph would be shifted right or left so it has a different intercept with the horizontal axis. EVALUATE: As the frequency f of the light is increased above fth the energy of the photons in the light increases and more energetic photons are produced. The work function we calculated is similar to that for gold or nickel. IDENTIFY and SET UP: c = f λ relates frequency and wavelength and E = hf relates energy and frequency for a

photon. c = 3.00 × 108 m/s . 1 eV = 1.60 × 10 −16 J . c 3.00 × 108 m/s = 5.94 × 1014 Hz EXECUTE: (a) f = = λ 505 × 10−9 m (b) E = hf = (6.626 × 10 −34 J ⋅ s)(5.94 × 1014 Hz) = 3.94 × 10−19 J = 2.46 eV (c) K = 12 mv 2 so v = 38.3.

38.4.

c 3.00 × 108 m s = = 5.77 × 1014 Hz λ 5.20 × 10−7 m h 6.63 × 10 −34 J ⋅ s = 1.28 × 10−27 kg ⋅ m s p= = λ 5.20 × 10 −7 m f =

E = pc = (1.28 × 10−27 kg ⋅ m s) (3.00 × 108 m s) = 3.84 × 10−19 J = 2.40 eV. energy hc . 1 eV = 1.60 × 10−19 J . For a photon, E = hf = . h = 6.63 × 10−34 J ⋅ s. IDENTIFY and SET UP: Pav = t λ EXECUTE: (a) energy = Pavt = (0.600 W)(20.0 × 10−3 s) = 1.20 × 10 −2 J = 7.5 × 1016 eV

(6.63 × 10−34 J ⋅ s)(3.00 × 108 m/s) = 3.05 × 10−19 J = 1.91 eV 652 × 10−9 m λ (c) The number of photons is the total energy in a pulse divided by the energy of one photon: 1.20 × 10−2 J = 3.93 × 1016 photons . 3.05 × 10 −19 J/photon EVALUATE: The number of photons in each pulse is very large. IDENTIFY and SET UP: Eq.(38.2) relates the photon energy and wavelength. c = f λ relates speed, frequency and wavelength for an electromagnetic wave. E (2.45 × 106 eV)(1.602 × 10−19 J/1 eV) = 5.92 × 1020 Hz EXECUTE: (a) E = hf so f = = 6.626 × 10−34 J ⋅ s h c 2.998 × 108 m/s = 5.06 × 10−13 m (b) c = f λ so λ = = f 5.92 × 1020 Hz (c) EVALUATE: λ is comparable to a nuclear radius. Note that in doing the calculation the energy in MeV was converted to the SI unit of Joules. (b) E =

38.5.

2K 2(3.94 × 10−19 J) = = 9.1 mm/s m 9.5 × 10−15 kg

hc

=

38-1


38-2

38.6.

Chapter 38

λth = 272 nm . c = f λ .

IDENTIFY and SET UP:

h = 4.136 × 10−15 eV ⋅ s . EXECUTE: (a) f th =

38.7.

c

λth

1 2 mvmax = hf − φ . At the threshold frequency, f th , vmax → 0. 2

3.00 × 108 m/s = 1.10 × 1015 Hz . 272 × 10−9 m eV ⋅ s)(1.10 × 1015 Hz) = 4.55 eV .

=

(b) φ = hf th = (4.136 × 10−15 1 2 (c) mvmax = hf − φ = (4.136 × 10−15 eV ⋅ s)(1.45 × 1015 Hz) − 4.55 eV = 6.00 eV − 4.55 eV = 1.45 eV 2 EVALUATE: The threshold wavelength depends on the work function for the surface. hc 1 2 IDENTIFY and SET UP: Eq.(38.3): mvmax = hf − φ = − φ . Take the work function φ from Table 38.1. Solve 2 λ for vmax . Note that we wrote f as c / λ . EXECUTE:

1 2 (6.626 × 10−34 J ⋅ s)(2.998 × 108 m/s) − (5.1 eV)(1.602 × 10−19 J/1 eV) mvmax = 2 235 × 10−9 m

1 2 mvmax = 8.453 × 10 −19 J − 8.170 × 10−19 J = 2.83 × 10−20 J 2 2(2.83 × 10 −20 J) vmax = = 2.49 × 105 m/s 9.109 × 10−31 kg

38.8.

EVALUATE: The work function in eV was converted to joules for use in Eq.(38.3). A photon with λ = 235 nm has energy greater then the work function for the surface. hc IDENTIFY and SET UP: φ = hf th = . The minimum φ corresponds to the minimum λ .

λth

EXECUTE: 38.9.

φ=

hc

λth

=

(4.136 × 10

−15

eV ⋅ s)(3.00 × 108 m/s) = 1.77 eV 700 × 10−9 m c = f λ . The source emits (0.05)(75 J) = 3.75 J of energy as visible light each second.

IDENTIFY and SET UP:

E = hf , with h = 6.63 × 10−34 J ⋅ s. EXECUTE: (a) f =

38.10.

c

λ

=

3.00 × 108 m/s = 5.00 × 1014 Hz 600 × 10 −9 m J ⋅ s)(5.00 × 1014 Hz) = 3.32 × 10−19 J . The number of photons emitted per second is

(b) E = hf = (6.63 × 10−34 3.75 J = 1.13 × 1019 photons . 3.32 × 10−19 J/photon (c) No. The frequency of the light depends on the energy of each photon. The number of photons emitted per second is proportional to the power output of the source. IDENTIFY: In the photoelectric effect, the energy of the photon is used to eject an electron from the surface, and any excess energy goes into kinetic energy of the electron. SET UP: The energy of a photon is E = hf, and the work function is given by φ = hf0, where f0 is the threshold frequency. EXECUTE: (a) From the graph, we see that Kmax = 0 when λ = 250 nm, so the threshold wavelength is 250 nm. Calling f0 the threshold frequency, we have f0 = c/λ0 = (3.00 × 108 m/s)/(250 nm) = 1.2 × 1015 Hz. (b) φ = hf0 = (4.136 × 10–15 eV ⋅ s )(1.2 × 1015 Hz) = 4.96 eV = 5.0 eV (c) The graph (see Figure 38.10) is linear for λ < λ0 (1/λ > 1/λ0), and linear graphs are easier to interpret than curves. EVALUATE: If the wavelength of the light is longer than the threshold wavelength (that is, if 1/λ < 1/λ0), the kinetic energy of the electrons is really not defined since no photoelectrons are ejected from the metal.

Figure 38.10


Photons, Electrons, and Atoms

38.11.

38-3

IDENTIFY: Protons have mass and photons are massless. (a) SET UP: For a particle with mass, K = p 2 / 2m. EXECUTE: p2 = 2 p1 means K 2 = 4 K1. (b) SET UP: For a photon, E = pc.

38.12.

EXECUTE: p2 = 2 p1 means E2 = 2 E1. EVALUATE: The relation between E and p is different for particles with mass and particles without mass. 1 2 IDENTIFY and SET UP: eV0 = mvmax , where V0 is the stopping potential. The stopping potential in volts equals 2 1 2 eV0 in electron volts. mvmax = hf − φ . 2 1 2 EXECUTE: (a) eV0 = mvmax so 2 (4.136 × 10−15 eV ⋅ s)(3.00 × 108 m/s) − 2.3 eV = 4.96 eV − 2.3 eV = 2.7 eV . The stopping potential eV0 = hf − φ = 250 × 10 −9 m is 2.7 electron volts. 1 2 (b) mvmax = 2.7 eV 2 (c) vmax =

38.13.

2(2.7 eV)(1.60 × 10 −19 J/eV) = 9.7 × 105 m/s 9.11 × 10−31 kg

First use Eq.(38.4) to find the work function φ . hc eV0 = hf − φ so φ = hf − eV0 = − eV0

(a) IDENTIFY: SET UP:

λ

(6.626 × 10 −34 J ⋅ s)(2.998 × 108 m/s) EXECUTE: φ = − (1.602 × 10−19 C)(0.181 V) 254 × 10−9 m φ = 7.821 × 10−19 J − 2.900 × 10 −20 J = 7.531 × 10−19 J(1 eV/1.602 × 10 −19 J) = 4.70 eV IDENTIFY and SET UP: The threshold frequency f th is the smallest frequency that still produces photoelectrons.

It corresponds to K max = 0 in Eq.(38.3), so hf th = φ . f =

EXECUTE:

λ

says

hc

λth

(6.626 × 10 −34 J ⋅ s)(2.998 × 108 m/s) = 2.64 × 10 −7 m = 264 nm φ 7.531 × 10 −19 J (b) EVALUATE: As calculated in part (a), φ = 4.70 eV. This is the value given in Table 38.1 for copper. IDENTIFY and SET UP: A photon has zero rest mass, so its energy and momentum are related by Eq.(37.40). Eq.(38.5) then relates its momentum and wavelength. EXECUTE: (a) E = pc = (8.24 × 10−28 kg ⋅ m/s)(2.998 × 108 m/s) = 2.47 × 10−19 J =

λth =

38.14.

c

hc

=

(2.47 × 10 −19 J)(1 eV/1.602 × 10−19 J) = 1.54 eV

h 6.626 × 10−34 J ⋅ s = = 8.04 × 10 −7 m = 804 nm λ p 8.24 × 10 −28 kg ⋅ m/s EVALUATE: This wavelength is longer than visible wavelengths; it is in the infrared region of the electromagnetic spectrum. To check our result we could verify that the same E is given by Eq.(38.2), using the λ we have calculated. 1 ⎛ 1 1 ⎞ IDENTIFY and SET UP: Balmer’s formula is = R ⎜ 2 − 2 ⎟ . For the Hγ spectral line n = 5. Once we have λ , λ ⎝2 n ⎠ calculate f from f = c / λ and E from Eq.(38.2). (b) p =

38.15.

h

so λ =

⎛1 1⎞ ⎛ 25 − 4 ⎞ ⎛ 21 ⎞ = R⎜ 2 − 2 ⎟ = R⎜ ⎟ = R⎜ ⎟. 2 5 100 ⎝ ⎠ ⎝ ⎠ ⎝ 100 ⎠ 100 100 m = 4.341 × 10 −7 m = 434.1 nm. = Thus λ = 21R 21(1.097 × 107 )

EXECUTE: (a)

(b) f =

c

λ

=

1

λ

2.998 × 108 m/s = 6.906 × 1014 Hz 4.341 × 10−7 m


38-4

Chapter 38

(c) E = hf = (6.626 × 10−34 J ⋅ s)(6.906 × 1014 Hz) = 4.576 × 10−19 J = 2.856 eV EVALUATE: Section 38.3 shows that the longest wavelength in the Balmer series (Hα ) is 656 nm and the

shortest is 365 nm. Our result for Hγ falls within this range. The photon energies for hydrogen atom transitions are 38.16.

in the eV range, and our result is of this order. IDENTIFY and SET UP: For the Lyman series the final state is n = 1 and the wavelengths are given by 1 ⎛1 1 ⎞ = R ⎜ 2 − 2 ⎟ , n = 2,3,.... For the Paschen series the final state is n = 3 and the wavelengths are given by λ ⎝1 n ⎠

⎛1 1⎞ = R ⎜ 2 − 2 ⎟ , n = 4,5,.... R = 1.097 × 107 m −1 . The longest wavelength is for the smallest n and the shortest λ ⎝3 n ⎠ wavelength is for n → ∞ . 1 4 ⎛ 1 1 ⎞ 3R EXECUTE: Lyman Longest: = R⎜ 2 − 2 ⎟ = . λ= = 121.5 nm . λ 3(1.097 × 107 m −1 ) ⎝1 2 ⎠ 4 1

Shortest:

1

λ

1 ⎛1 1 ⎞ = R⎜ 2 − 2 ⎟ = R . λ = = 91.16 nm × ∞ 1 1.097 107 m −1 ⎝ ⎠

Paschen Longest:

1 ⎞ R ⎛1 = R⎜ 2 − 2 ⎟ = . λ ⎝3 ∞ ⎠ 9 hc (6.63 × 10−34 J ⋅ s) (3.00 × 108 m s) = = 2.31 × 10 −19 J = 1.44 eV. (a) Eγ = λ 8.60 × 10−7 m So the internal energy of the atom increases by 1.44 eV to E = −6.52 eV + 1.44 eV = −5.08 eV. Shortest:

38.17.

144 ⎛ 1 1 ⎞ 7R = R⎜ 2 − 2 ⎟ = . λ= = 1875 nm . λ 7(1.097 × 107 m −1 ) ⎝ 3 4 ⎠ 144 1

1

hc (6.63 × 10 −34 J ⋅ s) (3.00 × 108 m s) = = 4.74 × 10 −19 J = 2.96 eV. λ 4.20 × 10−7 m So the final internal energy of the atom decreases to E = −2.68 eV − 2.96 eV = −5.64 eV. IDENTIFY and SET UP: The ionization threshold is at E = 0 . The energy of an absorbed photon equals the energy gained by the atom and the energy of an emitted photon equals the energy lost by the atom. EXECUTE: (a) ΔE = 0 − (−20 eV) = 20 eV (b) When the atom in the n = 1 level absorbs a 18 eV photon, the final level of the atom is n = 4 . The possible transitions from n = 4 and corresponding photon energies are n = 4 → n = 3, 3 eV ; n = 4 → n = 2, 8 eV ; n = 4 → n = 1, 18 eV . Once the atom has gone to the n = 3 level, the following transitions can occur: n = 3 → n = 2, 5 eV ; n = 3 → n = 1, 15 eV . Once the atom has gone to the n = 2 level, the following transition can occur: n = 2 → n = 1, 10 eV . The possible energies of emitted photons are: 3 eV, 5 eV, 8 eV, 10 eV, 15 eV, and 18 eV. (c) There is no energy level 8 eV higher in energy than the ground state, so the photon cannot be absorbed. (d) The photon energies for n = 3 → n = 2 and for n = 3 → n = 1 are 5 eV and 15 eV. The photon energy for n = 4 → n = 3 is 3 eV. The work function must have a value between 3 eV and 5 eV. IDENTIFY and SET UP: The wavelength of the photon is related to the transition energy Ei − Ef of the atom by (b) Eγ =

38.18.

38.19.

Ei − Ef =

hc

λ

where hc = 1.240 × 10−6 eV ⋅ m .

EXECUTE: (a) The minimum energy to ionize an atom is when the upper state in the transition has E = 0 , so 1.240 × 10 −6 eV ⋅ m = 16.79 eV . E1 = −17.50 eV . For n = 5 → n = 1 , λ = 73.86 nm and E5 − E1 = 73.86 × 10−9 m E5 = −17.50 eV + 16.79 eV = −0.71 eV . For n = 4 → n = 1 , λ = 75.63 nm and E4 = −1.10 eV . For

n = 3 → n = 1 , λ = 79.76 nm and E3 = −1.95 eV . For n = 2 → n = 1 , λ = 94.54 nm and E2 = −4.38 eV . hc 1.240 × 10−6 eV ⋅ m = = 378 nm Ei − Ef 3.28 eV EVALUATE: The n = 4 → n = 2 transition energy is smaller than the n = 4 → n = 1 transition energy so the wavelength is longer. In fact, this wavelength is longer than for any transition that ends in the n = 1 state. (b)

Ei − Ef = E4 − E2 = −1.10 eV − ( −4.38 eV) = 3.28 eV and λ =


Photons, Electrons, and Atoms

38.20.

38.21.

38-5

(a) Equating initial kinetic energy and final potential energy and solving for the separation radius r, 1 (92e) (2e) 1 (184) (1.60 × 10 −19 C) = = 5.54 × 10−14 m. r= 4π P0 4π P0 (4.78 × 106 J C) K (b) The above result may be substituted into Coulomb’s law, or, the relation between the magnitude of the force and the magnitude of the potential energy in a Coulombic field is K (4.78 × 106 eV) (1.6 × 10−19 J ev) = 13.8 N. F= = (5.54 × 10−14 m) r 1 q1q2 . (a) IDENTIFY: If the particles are treated as point charges, U = 4π P0 r SET UP: q1 = 2e (alpha particle); q2 = 82e (gold nucleus); r is given so we can solve for U. EXECUTE:

(2)(82)(1.602 × 10−19 C) 2 = 5.82 × 10 −13 J 6.50 × 10 −14 m J) = 3.63 × 106 eV = 3.63 MeV

U = (8.987 × 109 N ⋅ m 2 /C 2 )

U = 5.82 × 10 −13 J(1 eV/1.602 × 10−19

(b) IDENTIFY: Apply conservation of energy: K1 + U1 = K 2 + U 2 . SET UP: Let point 1 be the initial position of the alpha particle and point 2 be where the alpha particle momentarily comes to rest. Alpha particle is initially far from the lead nucleus implies r1 ≈ ∞ and U1 = 0. Alpha

particle stops implies K 2 = 0. EXECUTE: Conservation of energy thus says K1 = U 2 = 5.82 × 10−13 J = 3.63 MeV.

1 2K 2(5.82 × 10 −13 J) = = 1.32 × 107 m/s (c) K = mv 2 so v = 2 m 6.64 × 10−27 kg

38.22. 38.23.

38.24.

EVALUATE: v / c = 0.044, so it is ok to use the nonrelativistic expression to relate K and v. When the alpha particle stops, all its initial kinetic energy has been converted to electrostatic potential energy. h (a), (b) For either atom, the magnitude of the angular momentum is = 1.05 × 10−34 kg ⋅ m 2 s. 2π IDENTIFY and SET UP: Use the energy to calculate n for this state. Then use the Bohr equation, Eq.(38.10), to calculate L. EXECUTE: En = −(13.6 eV)/n 2 , so this state has n = 13.6 /1.51 = 3. In the Bohr model. L = nU so for this state L = 3U = 3.16 × 10−34 kg ⋅ m 2 /s. EVALUATE: We will find in Section 41.1 that the modern quantum mechanical description gives a different result. 13.6 eV hc IDENTIFY and SET UP: For a hydrogen atom En = − . ΔE = , where ΔE is the magnitude of the λ n2 energy change for the atom and λ is the wavelength of the photon that is absorbed or emitted. ⎛ 1 1⎞ EXECUTE: ΔE = E4 − E1 = −(13.6 eV) ⎜ 2 − 2 ⎟ = +12.75 eV . ⎝4 1 ⎠

hc (4.136 × 10−15 eV ⋅ s)(3.00 × 108 m/s) c = = 97.3 nm . f = = 3.08 × 1015 Hz . ΔE λ 12.75 eV 1 Ze 2 , where Z = 4 is the nuclear IDENTIFY: The force between the electron and the nucleus in Be3+ is F = 4π P0 r 2

λ=

38.25.

charge. All the equations for the hydrogen atom apply to Be3+ if we replace e 2 by Ze 2 . (a) SET UP: Modify Eq.(38.18). 1 me 4 EXECUTE: En = − (hydrogen) becomes P0 8n 2h 2 ⎛ 1 me 4 ⎞ 1 m( Ze2 ) 2 ⎛ 13.60 eV ⎞ 3+ = Z2⎜− = Z2⎜− ⎟ (for Be ) 2 2 2 2 ⎟ 2 8 n h n P0 8n h P ⎝ ⎠ ⎝ 0 ⎠ ⎛ 13.60 eV ⎞ The ground-level energy of Be3+ is E1 = 16 ⎜ − ⎟ = −218 eV. 12 ⎝ ⎠ En = −

EVALUATE: The ground-level energy of Be3+ is Z 2 = 16 times the ground-level energy of H. (b) SET UP: The ionization energy is the energy difference between the n → ∞ level energy and the n = 1 level energy.


38-6

Chapter 38

EXECUTE: The n → ∞ level energy is zero, so the ionization energy of Be3+ is 218 eV. EVALUATE: This is 16 times the ionization energy of hydrogen. ⎛ 1 1 ⎞ 1 (c) SET UP: = R ⎜ 2 − 2 ⎟ just as for hydrogen but now R has a different value. λ ⎝ n1 n2 ⎠ EXECUTE:

RH =

me4 = 1.097 × 107 m −1 for hydrogen becomes 8P0 h3c

me4 = 16(1.097 × 107 m −1 ) = 1.755 × 108 m −1 for Be3+ . 8P0 h3c 1 ⎛1 1⎞ For n = 2 to n = 1, = RBe ⎜ 2 − 2 ⎟ = 3R/4. λ ⎝1 2 ⎠ RBe = Z 2

λ = 4 /(3R ) = 4 /(3(1.755 × 108 m −1 )) = 7.60 × 10−9 m = 7.60 nm. EVALUATE: This wavelength is smaller by a factor of 16 compared to the wavelength for the corresponding transition in the hydrogen atom. n 2h 2 (d) SET UP: Modify Eq.(38.12): rn = P0 (hydrogen). π me2 2 2 nh EXECUTE: rn = P0 (Be3+ ). π m( Ze 2 )

38.26.

38.27.

EVALUATE: For a given n the orbit radius for Be3+ is smaller by a factor of Z = 4 compared to the corresponding radius for hydrogen. (a) We can find the photon’s energy from Eq. 38.8 ⎛ 1 1 ⎞ ⎛ 1 1⎞ E = hcR ⎜ 2 − 2 ⎟ = (6.63 × 10−34 J ⋅ s) (3.00 × 108 m s ) (1.097 × 107 m −1 ) ⎜ 2 − 2 ⎟ = 4.58 × 10 −19 J. The ⎝2 n ⎠ ⎝2 5 ⎠ E corresponding wavelength is λ = = 434 nm. hc (b) In the Bohr model, the angular momentum of an electron with principal quantum number n is given by h Eq. 38.10: L = n . Thus, when an electron makes a transition from n = 5 to n = 2 orbital, there is the following 2π loss in angular momentum (which we would assume is transferred to the photon): h 3(6.63 × 10 −34 J ⋅ s) ΔL = (2 − 5) =− = −3.17 × 10−34 J ⋅ s. 2π 2π However, this prediction of the Bohr model is wrong (as shown in Chapter 41). 1 e2 (1.60 × 10−19 C) 2 : n = 1 ⇒ v1 = = 2.18 × 106 m/s (a) vn = P0 2nh P0 2 (6.63 × 10−34 J ⋅ s)

h = 2 ⇒ v2 = (b) Orbital period =

v1 v = 1.09 × 106 m s. h = 3 ⇒ v3 = 1 = 7.27 × 105 m s. 2 3

2π rn 2P0 n 2h 2 me 2 4P02n3h3 = = vn 1 P0 ⋅ e2 2nh me4 n = 1 ⇒ T1 =

4P02 (6.63 × 10−34 J ⋅ s)3 = 1.53 × 10−16 s (9.11 × 10 −31 kg) (1.60 × 10−19 C) 4

n = 2 : T2 = T1 (2)3 = 1.22 × 10 −15 s. n = 3 : T3 = T1 (3)3 = 4.13 × 10 −15 s. (c) number of orbits = 38.28.

1.0 × 10−8 s = 8.2 × 106. 1.22 × 10−15 s

IDENTIFY and SET UP: EXECUTE: (a) En = −

En = −

13.6 eV n2

13.6 eV 13.6 eV and En +1 = − n2 ( n + 1) 2

⎡ 1 1⎤ n 2 − (n + 1) 2 ΔE = En +1 − En = (−13.6 eV) ⎢ − 2 ⎥ = −(13.6 eV) 2 2 n ⎦ (n )(n + 1) 2 ⎣ (n + 1) 2n 2 2n + 1 ΔE = (13.6 eV) 2 As n becomes large, ΔE → (13.6 eV) 4 = (13.6 eV) 3 n n (n )(n + 1) 2


Photons, Electrons, and Atoms

38.29.

38-7

Thus ΔE becomes small as n becomes large. (b) rn = n 2r1 so the orbits get farther apart in space as n increases. IDENTIFY and SET UP: The number of photons emitted each second is the total energy emitted divided by the energy of one photon. The energy of one photon is given by Eq.(38.2). E = Pt gives the energy emitted by the laser in time t. EXECUTE: In 1.00 s the energy emitted by the laser is (7.50 × 10−3 W)(1.00 s) = 7.50 × 10 −3 J. The energy of each photon is E =

hc

λ

=

(6.626 × 10 −34 J ⋅ s)(2.998 × 108 m/s) = 1.874 × 10−20 J. 10.6 × 10 −6 m

−3

7.50 × 10 J/s = 4.00 × 1017 photons/s 1.874 × 10−20 J/photon EVALUATE: The number of photons emitted per second is extremely large. IDENTIFY and SET UP: Visible light has wavelengths from about 400 nm to about 700 nm. The energy of each hc 1.99 × 10 −25 J ⋅ m photon is E = hf = . The power is the total energy per second and the total energy Etot is the =

Therefore 38.30.

λ

λ

number of photons N times the energy E of each photon. EXECUTE: (a) 193 nm is shorter than visible light so is in the ultraviolet. hc (b) E = = 1.03 × 10 −18 J = 6.44 eV

λ

38.31.

E NE Pt (1.50 × 10−3 W)(12.0 × 10 −9 s) (c) P = tot = so N = = = 1.75 × 107 photons t t E 1.03 × 10 −18 J EVALUATE: A very small amount of energy is delivered to the lens in each pulse, but this still corresponds to a large number of photons. n − ( E − E ) / kT IDENTIFY: Apply Eq.(38.21): 5 s = e 5 s 3 p n3 p SET UP: EXECUTE:

38.32.

From Fig.38.24a in the textbook, E5 s = 20.66 eV and E3 p = 18.70 eV

E5 s − E3 p = 20.66 eV − 18.70 eV = 1.96 eV(1.602 × 10−19 J/1 eV) = 3.140 × 10 −19 J

(a)

−19 −23 n5 s = e − (3.140×10 J)/[(1.38×10 J/K)(300 K)] = e −75.79 = 1.2 × 10−33 n3 p

(b)

−19 −23 n5 s = e− (3.140×10 J)/[(1.38×10 J/K)(600 K)] = e −37.90 = 3.5 × 10−17 n3 p

(c)

−19 −23 n5 s = e− (3.140×10 J)/[(1.38×10 J/K)(1200 K)] = e−18.95 = 5.9 × 10−9 n3 p

(d) EVALUATE: At each of these temperatures the number of atoms in the 5s excited state, the initial state for the transition that emits 632.8 nm radiation, is quite small. The ratio increases as the temperature increases. n2 P3 2 ) KT −( E −E = e 2 P3 2 2 P1 2 . n2 P1/ 2

hc (6.626 × 10−34 J)(3.000 × 108 m s) = = 3.375 × 10−19 J. λ1 5.890 × 10−7 m hc (6.626 × 10−34 J)(3.000 × 108 m s) = = = 3.371 × 10−19 J. so ΔE3 / 2 −1/ 2 = 3.375 × 10−19 J − 3.371 × 10 −19 J = λ2 5.896 × 10−7 m

From the diagram ΔE3 / 2 − g = ΔE1 2 − g

38.33.

4.00 × 10−22 J.

n2 P3 / 2

eVAC = hf max =

hc λ min

n2 P1/ 2

= e − (4.00 × 10

−22

J) (1.38 × 10−23 J / K ⋅500 K).

⇒ λ min =

= 0.944. So more atoms are in the 2 p1 2 state.

hc (6.63 × 10−34 J ⋅ s)(3.00 × 108 m s) = = 3.11× 10−10 m eVAC (1.60 × 10−19 C)(4000 V)

This is the same answer as would be obtained if electrons of this energy were used. Electron beams are much more easily produced and accelerated than proton beams.


38-8

Chapter 38

38.34.

IDENTIFY and SET UP: EXECUTE: (a) V =

hc

λ

= eV , where λ is the wavelength of the x ray and V is the accelerating voltage.

hc (6.63 × 10 −34 J ⋅ s)(3.00 × 108 m/s) = = 8.29 kV eλ (1.60 × 10−19 C)(0.150 × 10 −9 m)

hc (6.63 × 10−34 J ⋅ s)(3.00 × 108 m/s) = = 4.14 × 10−11 m = 0.0414 nm eV (1.60 × 10−19 C)(30.0 × 103 V) (c) No. A proton has the same magnitude of charge as an electron and therefore gains the same amount of kinetic energy when accelerated by the same magnitude of potential difference. IDENTIFY: The initial electrical potential energy of the accelerated electrons is converted to kinetic energy which is then given to a photon. SET UP: The electrical potential energy of an electron is eVAC, where VAC is the accelerating potential, and the energy of a photon is hf. Since the energy of the electron is all given to a photon, we have eVAC = hf. For any wave, fλ = v. EXECUTE: (a) eVAC = hfmin gives (b) λ =

38.35.

fmin = eVAC/h = (1.60 × 10–19 C)(25,000 V)/(6.626 × 10–34 J ⋅ s ) = 6.037 × 1018 Hz

38.36.

= 6.04 × 1018 Hz, rounded to three digits (b) λmin = c/fmax = (3.00 × 108 m/s)/(6.037 × 1018 Hz) = 4.97 × 10–11 m = 0.0497 nm (c) We assume that all the energy of the electron produces only one photon on impact with the screen. EVALUATE: These photons are in the x-ray and γ-ray part of the electromagnetic spectrum (see Figure 32.4 in the textbook) and would be harmful to the eyes without protective glass on the screen to absorb them. hc IDENTIFY and SET UP: The wavelength of the x rays produced by the tube is give by = eV .

λ

h h hc λ′ = λ + (1 − cosφ ) . = 2.426 × 10−12 m . The energy of the scattered x ray is . λ′ mc mc hc (6.63 × 10−34 J ⋅ s)(3.00 × 108 m/s) = = 6.91 × 10−11 m = 0.0691 nm EXECUTE: (a) λ = (1.60 × 10−19 C)(18.0 × 103 V) eV h (1 − cos φ ) = 6.91 × 10−11 m + (2.426 × 10−12 m)(1 − cos 45.0°) . mc λ ′ = 6.98 × 10 −11 m = 0.0698 nm . hc (4.136 × 10−15 eV ⋅ s)(3.00 × 108 m/s) = = 17.8 keV (c) E = 6.98 × 10 −11 m λ′ EVALUATE: The incident x ray has energy 18.0 keV. In the scattering event, the photon loses energy and its wavelength increases. h IDENTIFY: Apply Eq.(38.23): λ ′ − λ = (1 − cosφ ) = λC (1 − cos φ ) mc SET UP: Solve for λ ′ : λ ′ = λ + λC (1 − cos φ ) The largest λ ′ corresponds to φ = 180°, so cosφ = −1. (b) λ ′ = λ +

38.37.

38.38.

EXECUTE: λ ′ = λ + 2λC = 0.0665 × 10 −9 m + 2(2.426 × 10 −12 m) = 7.135 × 10 −11 m = 0.0714 nm. This wavelength occurs at a scattering angle of φ = 180°. EVALUATE: The incident photon transfers some of its energy and momentum to the electron from which it scatters. Since the photon loses energy its wavelength increases, λ ′ > λ . Δλ (a) From Eq. (38.23), cosφ = 1 − , and so Δλ = 0.0542 nm − 0.0500 nm, ( h mc)

cos φ = 1 −

0.0042 nm = −0.731, and φ = 137°. 0.002426 nm

0.0021 nm = 0.134. φ = 82.3°. 0.002426 nm (c) Δλ = 0, the photon is undeflected, cosφ = 1 and φ = 0.

(b) Δλ = 0.0521 nm − 0.0500 nm. cosφ = 1 −

38.39.

IDENTIFY and SET UP: The shift in wavelength of the photon is λ ′ − λ =

wavelength after the scattering and

h (1 − cos φ ) where λ ′ is the mc

h = λc = 2.426 × 10−12 m . The energy of a photon of wavelength λ is mc


Photons, Electrons, and Atoms

E=

38.40.

hc

λ

=

1.24 × 10

−6

eV ⋅ m

λ

38-9

. Conservation of energy applies to the collision, so the energy lost by the photon

equals the energy gained by the electron. EXECUTE: (a) λ ′ − λ = λc (1 − cos φ ) = (2.426 × 10−12 m)(1 − cos35.0°) = 4.39 × 10 −13 m = 4.39 × 10 −4 nm (b) λ ′ = λ + 4.39 × 10−4 nm = 0.04250 nm + 4.39 × 10−4 nm = 0.04294 nm hc hc (c) Eλ = = 2.918 × 104 eV and Eλ ′ = = 2.888 × 104 eV so the photon loses 300 eV of energy. λ λ′ (d) Energy conservation says the electron gains 300 eV of energy. The change in wavelength of the scattered photon is given by Eq. 38.23 Δλ = h (1 − cos φ ) ⇒ λ = h (1 − cosφ ). mcλ λ ⎛ Δλ ⎞ mc ⎜ ⎟ ⎝ λ ⎠ (6.63 × 10−34 J ⋅ s) (1 + 1) = 2.65 × 10 −14 m. (1.67 × 10−27 kg)(3.00 × 108 m/s)(0.100) The derivation of Eq.(38.23) is explicitly shown in Equations (38.24) through (38.27) with the final substitution of h p′ = h λ′ and p = h λ yielding λ′ − λ = (1 − cos φ ). mc 2.898 × 10−3 m ⋅ K c From Eq. (38.30), (a) λ m = = 0.966 mm, and f = = 3.10 × 1011 Hz. Note that a more precise 3.00 K λm value of the Wien displacement law constant has been used. (b) A factor of 100 increase in the temperature lowers λm by a factor of 100 to 9.66 μ m and raises the frequency

Thus, λ =

38.41.

38.42.

by the same factor, to 3.10 × 1013 Hz. (c) Similarly, λ m = 966 nm and f = 3.10 × 1014 Hz. 38.43.

(a) H = AeσT 4 ; A = π r 2l 14

⎛ ⎞ 100 W ⎛ H ⎞ T =⎜ ⎟ =⎜ 2 4 ⎟ −3 −8 ⎝ Aeσ ⎠ ⎝ 2π (0.20 × 10 m)(0.30 m)(0.26)(5.671 × 10 W m ⋅ K ) ⎠ 3 T = 2.06 × 10 K 14

38.44. 38.45.

(b) λ mT = 2.90 × 10−3 m ⋅ K; λ m = 1410 nm Much of the emitted radiation is in the infrared. 2.90 × 10−3 m ⋅ K 2.90 × 10−3 m ⋅ K T= = = 7.25 × 103 K. λm 400 × 10−9 m IDENTIFY and SET UP: The wavelength λm where the Planck distribution peaks is given by Eq.(38.30). 2.90 × 10−3 m ⋅ K = 1.06 × 10 −3 m = 1.06 mm. 2.728 K EVALUATE: This wavelength is in the microwave portion of the electromagnetic spectrum. This radiation is often referred to as the “microwave background” (Section 44.7). Note that in Eq.(38.30), T must be in kelvins. IDENTIFY: Since the stars radiate as blackbodies, they obey the Stefan-Boltzmann law and Wien’s displacement law. SET UP: The Stefan-Boltzmann law says that the intensity of the radiation is I = σT 4, so the total radiated power is P = σAT 4. Wien’s displacement law tells us that the peak-intensity wavelength is λm = (constant)/T. EXECUTE: (a) The hot and cool stars radiate the same total power, so the Stefan-Boltzmann law gives σAhTh4 =

EXECUTE:

38.46.

λm =

σAcTc4 ⇒ 4πRh2Th4 = 4πRc2Tc4 = 4π(3Rh)2Tc4 ⇒ Th4 = 9T 4 ⇒ Th = T 3 = 1.7T, rounded to two significant digits. (b) Using Wien’s law, we take the ratio of the wavelengths, giving λm (hot) Tc T 1 = = = = 0.58, rounded to two significant digits. λm (cool) Th T 3 3

38.47.

EVALUATE: Although the hot star has only 1/9 the surface area of the cool star, its absolute temperature has to be only 1.7 times as great to radiate the same amount of energy. (a) Let α = hc / kT . To find the maximum in the Planck distribution:

dI d ⎛ 2π hc 2 ⎞ (2π hc 2 ) 2π hc 2 ( −α λ 2 ) = ⎜ 5 αλ − 5 αλ ⎟ = 0 = −5 5 α λ dλ dλ ⎝ λ (e − 1) ⎠ λ (e − 1) λ (e − 1) 2 α hc ⇒ − 5(eα λ − 1) λ = α ⇒ − 5eα λ + 5 = α λ ⇒ Solve 5 − x = 5e x where x = = . λ λkT


38-10

Chapter 38

Its root is 4.965, so

α hc . = 4.965 ⇒ λ = λ (4.965) kT

hc (6.63 × 10−34 J ⋅ s)(3.00 × 108 m s) = = 2.90 × 10−3 m ⋅ K. (4.965)k (4.965)(1.38 × 10−23 J K) IDENTIFY: Since the stars radiate as blackbodies, they obey the Stefan-Boltzmann law. SET UP: The Stefan-Boltzmann law says that the intensity of the radiation is I = σT 4, so the total radiated power is P = σAT 4. EXECUTE: (a) I = σT 4 = (5.67 × 10–8 W/m 2 ⋅ K 4 )(24,000 K)4 = 1.9 × 1010 W/m2 (b) Wien’s law gives λm = (0.00290 m ⋅ K )/(24,000 K) = 1.2 × 10–7 m = 120 nm This is not visible since the wavelength is less than 400 nm. (c) P = AI ⇒ 4πR2 = P/I = (1.00 × 1025 W)/(1.9 × 1010 W/m2) which gives RSirius = 6.51 × 106 m = 6510 km. RSirius/Rsun = (6.51 × 106 m)/(6.96 × 109 m) = 0.0093, which gives RSirius = 0.0093 Rsun ≈ 1% Rsun (d) Using the Stefan-Boltzmann law, we have (b) λ mT =

38.48.

2

38.49.

x V1 ⇒ I ( λ) ≈ 38.50.

4

2

4 2 4 ⎛ ⎞ ⎛ 5800 K ⎞ ⎛ R ⎞ ⎛ T ⎞ Psun σ AsunTsun 4π Rsun Tsun P Rsun = = = ⎜ sun ⎟ ⎜ sun ⎟ ⋅ sun = ⎜ ⎟ ⎜ ⎟ = 39 4 2 4 PSirius σ ASiriusTSirius 4π RSiriusTSirius R T P 0.00935 R Sirius sun ⎠ ⎝ 24,000 K ⎠ ⎝ Sirius ⎠ ⎝ Sirius ⎠ ⎝ EVALUATE: Even though the absolute surface temperature of Sirius B is about 4 times that of our sun, it radiates only 1/39 times as much energy per second as our sun because it is so small. 2π hc 2 x2 but e x = 1 + x + + " ≈ 1 + x for Eq. (38.32): I ( λ) = 5 hc λkT − 1) λ (e 2 4

2π hc 2 2π ckT = = Eq. (38.31), which is Rayleigh’s distribution. λ (hc λkT ) λ4 5

2.90 × 10−3 K ⋅ m = 9.7 × 10−8 m = 97 nm (a) Wien’s law: λ m = k . λ m = T 30,000 K This peak is in the ultraviolet region, which is not visible. The star is blue because the largest part of the visible light radiated is in the blue violet part of the visible spectrum (b) P = σAT 4 (Stefan-Boltzmann law)

W ⎞ ⎛ (100, 000)(3.86 × 1026 W) = ⎜ 5.67 × 10−8 2 4 ⎟ (4π R 2 )(30,000 K) 4 mK ⎠ ⎝ 9 R = 8.2 × 10 m

Rstar Rsun =

38.51.

8.2 × 109 m = 12 6.96 × 108 m

(c) The visual luminosity is proportional to the power radiated at visible wavelengths. Much of the power is radiated nonvisible wavelengths, which does not contribute to the visible luminosity. IDENTIFY and SET UP: Use c = f λ to relate frequency and wavelength and use E = hf to relate photon energy and frequency. EXECUTE: (a) One photon dissociates one AgBr molecule, so we need to find the energy required to dissociate a single molecule. The problem states that it requires 1.00 × 105 J to dissociate one mole of AgBr, and one mole contains Avogadro’s number (6.02 × 10 23 ) of molecules, so the energy required to dissociate one AgBr is

1.00 × 105 J/mol = 1.66 × 10−19 J/molecule. 6.02 × 1023 molecules/mol The photon is to have this energy, so E = 1.66 × 10−19 J(1eV/1.602 × 10−19 J) = 1.04 eV. hc (6.626 × 10−34 J ⋅ s)(2.998 × 108 m/s) = = 1.20 × 10 −6 m = 1200 nm E λ 1.66 × 10 −19 J c 2.998 × 108 m/s = 2.50 × 1014 Hz (c) c = f λ so f = = λ 1.20 × 10−6 m (d) E = hf = (6.626 × 10−34 J ⋅ s)(100 × 106 Hz) = 6.63 × 10−26 J (b) E =

hc

so λ =

E = 6.63 × 10−26 J(1 eV/1.602 × 10−19 J) = 4.14 × 10 −7 eV


Photons, Electrons, and Atoms

38.52.

38-11

(e) EVALUATE: A photon with frequency f = 100 MHz has too little energy, by a large factor, to dissociate a AgBr molecule. The photons in the visible light from a firefly do individually have enough energy to dissociate AgBr. The huge number of 100 MHz photons can’t compensate for the fact that individually they have too little energy. h h (a) Assume a non-relativistic velocity and conserve momentum ⇒ mv = ⇒ v = . mλ λ 2 1 1 ⎛ h ⎞ h2 . = (b) K = mv 2 = m ⎜ ⎟ 2 2 ⎝ mλ ⎠ 2mλ 2 K h2 λ h = ⋅ = . Recoil becomes an important concern for small m and small λ since this ratio (c) 2 E 2mλ hc 2mcλ becomes large in those limits. hc (6.63 × 10−34 J ⋅ s)(3.00 × 108 m s ) = = 1.22 × 10−7 m = 122 nm. (d) E = 10.2 eV ⇒ λ = E (10.2 eV)(1.60 × 10 −19 J eV)

K=

(6.63 × 10 −34 J ⋅ s) 2 = 8.84 × 10−27 J = 5.53 × 10−8 eV. 2(1.67 × 10−27 kg)(1.22 × 10−7 m) 2

K 5.53 × 10−8 eV = = 5.42 × 10 −9. This is quite small so recoil can be neglected. E 10.2 eV 38.53.

IDENTIFY and SET UP:

f =

c

λ

. The ( f ,V0 ) values are: (8.20 × 1014 Hz, 1.48 V) , (7.41 × 1014 Hz, 1.15 V) ,

(6.88 × 1014 Hz, 0.93 V) , (6.10 × 1014 Hz, 0.62 V) , (5.49 × 1014 Hz, 0.36 V) , (5.18 × 1014 Hz, 0.24 V) . The graph

of V0 versus f is given in Figure 38.53. EXECUTE: (a) The threshold frequency, f th , is f where V0 = 0 . From the graph this is f th = 4.56 × 1014 Hz . c 3.00 × 108 m/s = = 658 nm f th 4.56 × 1014 Hz (c) φ = hf th = (4.136 × 10 −15 eV ⋅ s)(4.56 × 1014 Hz) = 1.89 eV (b) λth =

h ⎛h⎞ (d) eV0 = hf − φ so V0 = ⎜ ⎟ f − φ . The slope of the graph is . e ⎝e⎠ h ⎛ 1.48 V − 0.24 V ⎞ −15 =⎜ ⎟ = 4.11 × 10 V/Hz and e ⎝ 8.20 × 1014 Hz − 5.18 × 1014 Hz ⎠ h = (4.11 × 10 −15 V/Hz)(1.60 × 10 −19 C) = 6.58 × 10 −34 J ⋅ s .

Figure 38.53 38.54.

dN ( dE dt ) P (200 W)(0.10) = = = = 6.03 × 1019 photons sec. dt ( dE dN ) hf h(5.00 × 1014 Hz) (dN dt ) (b) Demand = 1.00 × 1011 photons sec ⋅ cm 2 . 4π r 2 (a)

1/ 2

38.55.

⎛ ⎞ 6.03 × 1019 photons sec Therefore, r = ⎜ ⎟ = 6930 cm = 69.3 m. 11 2 ⎝ 4π (1.00 × 10 photons sec ⋅ cm ) ⎠ (a) IDENTIFY: Apply the photoelectric effect equation, Eq.(38.4). SET UP: eV0 = hf − φ = ( hc / λ ) − φ . Call the stopping potential V01 for λ1 and V02 for λ2 . Thus eV01 = (hc / λ1 ) − φ and eV02 = (hc / λ2 ) − φ . Note that the work function φ is a property of the material and is independent of the wavelength of the light. ⎛λ −λ ⎞ EXECUTE: Subtracting one equation from the other gives e(V02 − V01 ) = hc ⎜ 1 2 ⎟ . ⎝ λ 1λ 2 ⎠


38-12

Chapter 38

(6.626 × 10−34 J ⋅ s)(2.998 × 108 m/s) ⎛ 295 × 10 −9 m − 265 × 10−9 m ⎞ ⎜ ⎟ = 0.476 V. −9 −9 1.602 × 10−19 C ⎝ (295 × 10 m)(265 × 10 m) ⎠ EVALUATE: eΔV0 , which is 0.476 eV, is the increase in photon energy from 295 nm to 265 nm. The stopping potential increases when λ deceases because the photon energy increases when the wavelength decreases. IDENTIFY: The photoelectric effect occurs, so the energy of the photon is used to eject an electron, with any excess energy going into kinetic energy of the electron. SET UP: Conservation of energy gives hf = hc/λ = Kmax + φ. EXECUTE: (a) Using hc/λ = Kmax + φ, we solve for the work function: (b) ΔV0 =

38.56.

φ = hc/λ – Kmax = (4.136 × 10–15 eV ⋅ s )(3.00 × 108 m/s)/(124 nm) – 4.16 eV = 5.85 eV

38.57.

(b) The number N of photoelectrons per second is equal to the number of photons per second that strike the metal per second. N × (energy of a photon) = 2.50 W. N(hc/λ) = 2.50 W. N = (2.50 W)(124 nm)/[(6.626 × 10–34 J ⋅ s )(3.00 × 108 m/s)] = 1.56 × 1018 electrons/s (c) N is proportional to the power, so if the power is cut in half, so is N, which gives N = (1.56 × 1018 el/s)/2 = 7.80 × 1017 el/s (d) If we cut the wavelength by half, the energy of each photon is doubled since E = hc/λ. To maintain the same power, the number of photons must be half of what they were in part (b), so N is cut in half to 7.80 × 1017 el/s. We could also see this from part (b), where N is proportional to λ. So if the wavelength is cut in half, so is N. EVALUATE: In part (c), reducing the power does not reduce the maximum kinetic energy of the photons; it only reduces the number of ejected electrons. In part (d), reducing the wavelength does change the maximum kinetic energy of the photoelectrons because we have increased the energy of each photon. IDENTIFY and SET UP: The energy added to mass m of the blood to heat it to Tf = 100 °C and to vaporize it is

Q = mc (Tf − Ti ) + mLv , with c = 4190 J/kg ⋅ K and Lv = 2.256 × 106 J/kg . The energy of one photon is

E=

38.58.

38.59.

hc

λ

=

1.99 × 10−25 J ⋅ m

λ

.

EXECUTE: (a) Q = (2.0 × 10−9 kg)(4190 J/kg ⋅ K)(100°C − 33°C) + (2.0 × 10 −9 kg)(2.256 × 106 J/kg) = 5.07 × 10−3 J The pulse must deliver 5.07 mJ of energy. energy 5.07 × 10−3 J (b) P = = = 11.3 W t 450 × 10−6 s hc 1.99 × 10−25 J ⋅ m (c) One photon has energy E = = = 3.40 × 10−19 J . The number N of photons per pulse is the 585 × 10−9 m λ 5.07 × 10−3 J energy per pulse divided by the energy of one photon: N = = 1.49 × 1016 photons 3.40 × 10−19 J/photon hc (a) λ 0 = , and the wavelengths are: cesium: 590 nm, copper: 264 nm, potassium: 539 nm, zinc: 288 nm. E b) The wavelengths of copper and zinc are in the ultraviolet, and visible light is not energetic enough to overcome the threshold energy of these metals. 207 me mp mm (a) IDENTIFY and SET UP: Apply Eq.(38.20): mr = 1 2 = m1 + m2 207 me + mp EXECUTE:

mr =

207(9.109 × 10−31 kg)(1.673 × 10−27 kg) = 1.69 × 10 −28 kg 207(9.109 × 10−31 kg) + 1.673 × 10−27 kg

We have used me to denote the electron mass. (b) IDENTIFY: In Eq.(38.18) replace m = me by mr : En = −

1 mr e 4 . P02 8n 2 h 2

⎛ m ⎞ ⎛ 1 m e4 ⎞ 1 m e4 SET UP: Write as En = ⎜ r ⎟ ⎜ − 2 H2 2 ⎟ , since we know that 2 H 2 = 13.60 eV. Here mH denotes the P0 8h ⎝ mH ⎠⎝ P0 8n h ⎠ −31 reduced mass for the hydrogen atom; mH = 0.99946(9.109 × 10 kg) = 9.104 × 10−31 kg. ⎛ m ⎞ ⎛ 13.60 eV ⎞ En = ⎜ r ⎟ ⎜ − ⎟ n2 ⎠ ⎝ mH ⎠ ⎝ 1.69 × 10−28 kg E1 = ( −13.60 eV) = 186( −13.60 eV) = −2.53 keV 9.104 × 10−31 kg

EXECUTE:


Photons, Electrons, and Atoms

38-13

⎛ m ⎞ ⎛ R ch ⎞ From part (b), En = ⎜ r ⎟ ⎜ − H 2 ⎟ , where RH = 1.097 × 107 m −1 is the Rydberg constant for the ⎝ mH ⎠ ⎝ n ⎠ hc hydrogen atom. Use this result in = Ei − E f to find an expression for 1/ λ . The initial level for the transition is (c) SET UP:

λ

the ni = 2 level and the final level is the n f = 1 level. EXECUTE:

hc

λ

=

mr ⎛ RHch ⎛ RHch ⎞ ⎞ − ⎜ − 2 ⎟⎟ ⎜− ⎟ mH ⎜⎝ ni2 ⎝ nf ⎠ ⎠

⎛ 1 1⎞ mr RH ⎜ 2 − 2 ⎟ λ mH ⎝ nf ni ⎠ 1 1.69 × 10−28 kg ⎛1 1⎞ (1.097 × 107 m −1 ) ⎜ 2 − 2 ⎟ = 1.527 × 109 m −1 = λ 9.104 × 10−31 kg ⎝1 2 ⎠ λ = 0.655 nm EVALUATE: From Example 38.6 the wavelength of the radiation emitted in this transition in hydrogen is 122 nm. m The wavelength for muonium is H = 5.39 × 10 −3 times this. The reduced mass for hydrogen is very close to the mr 1

=

electron mass because the electron mass is much less then the proton mass: mp / me = 1836. The muon mass is

38.60.

207 me = 1.886 × 10−28 kg. The proton is only about 10 times more massive than the muon, so the reduced mass is somewhat smaller than the muon mass. The muon-proton atom has much more strongly bound energy levels and much shorter wavelengths in its spectrum than for hydrogen. (a) The change in wavelength of the scattered photon is given by Eq. 38.23

h h (1 − cos φ ) ⇒ λ = λ′ − (1 − cosφ ) = mc mc (6.63 × 10−34 J ⋅ s) (0.0830 × 10−9 m) − (1 + 1) = 0.0781 nm. (9.11 × 10−31 kg)(3.00 × 108 m s) λ′ − λ =

(b) Since the collision is one-dimensional, the magnitude of the electron’s momentum must be equal to the magnitude of the change in the photon’s momentum. Thus,

1 ⎞ 9 −1 ⎛ 1 −1 ⎞ ⎛ 1 pe = h ⎜ − ⎟ = (6.63 × 10−34 J ⋅ s) ⎜ + ⎟ (10 m ) ′ λ λ 0.0781 0.0830 ⎝ ⎠ ⎝ ⎠ = 1.65 × 10−23 kg ⋅ m s ≈ 2 × 10−23 kg ⋅ m s. (c) Since the electron is non relativistic ( β = 0.06), K e = 38.61.

IDENTIFY and SET UP:

λ′ = λ +

pe2 = 1.49 × 10 −16 J ≈ 10−16 J. 2m

h (1 − cosφ ) mc

2h = 0.09485 m. Use Eq.(38.5) to calculate the momentum of the scattered photon. Apply mc conservation of energy to the collision to calculate the kinetic energy of the electron after the scattering. The energy of the photon is given by Eq.(38.2), EXECUTE: (a) p′ = h / λ ′ = 6.99 × 10 −24 kg ⋅ m/s.

φ = 180° so λ ′ = λ +

(b) E = E′ + Ee ; hc / λ = hc / λ ′ + Ee

38.62.

λ′ − λ ⎛1 1 ⎞ Ee = hc ⎜ − ⎟ = (hc) = 1.129 × 10−16 J = 705 eV λλ′ ⎝ λ λ′ ⎠ EVALUATE: The energy of the incident photon is 13.8 keV, so only about 5% of its energy is transferred to the electron. This corresponds to a fractional shift in the photon’s wavelength that is also 5%. 2h (a) φ = 180° so (1 − cosφ ) = 2 ⇒ Δλ = = 0.0049 nm, so λ′ = 0.1849 nm. mc ⎛1 1 ⎞ (b) ΔE = hc ⎜ − ⎟ = 2.93 × 10−17 J = 183 eV. This will be the kinetic energy of the electron. ⎝ λ λ′ ⎠ (c) The kinetic energy is far less than the rest mass energy, so a non-relativistic calculation is adequate; v = 2 K m = 8.02 × 106 m s.


38-14

Chapter 38

38.63.

IDENTIFY and SET UP: The Hα line in the Balmer series corresponds to the n = 3 to n = 2 transition.

En = −

13.6 eV hc . = ΔE . λ n2

⎛1 1⎞ EXECUTE: (a) The atom must be given an amount of energy E3 − E1 = −(13.6 eV) ⎜ 2 − 2 ⎟ = 12.1 eV . ⎝3 1 ⎠ hc (b) There are three possible transitions. n = 3 → n = 1 : ΔE = 12.1 eV and λ = = 103 nm ; ΔE ⎛1 1⎞ n = 3 → n = 2 : ΔE = −(13.6 eV) ⎜ 2 − 2 ⎟ = 1.89 eV and λ = 657 nm ; n = 2 → n = 1 : ⎝3 2 ⎠

38.64.

⎛ 1 1⎞ ΔE = −(13.6 eV) ⎜ 2 − 2 ⎟ = 10.2 eV and λ = 122 nm . ⎝2 1 ⎠ −( Eex − Eg ) n2 − ( E − E ) kT ⇒T = . = e ex g n1 kln ( n2 / n1 )

Eex = E2 =

−13.6 eV = −3.4 eV. Eg = −13.6 eV. Eex − Eg = 10.2 eV = 1.63 × 10 −18 J. 4

(a)

−(1.63 × 10−18 J) n2 = 10 −12. T = = 4275 K. (1.38 × 10 −23 J K ) ln(10 −12 ) n1

(b)

−(1.63 × 10−18 J) n2 = 10 −8. T = = 6412 K. (1.38 × 10 −23 J K ) ln(10−8 ) n1

−(1.63 × 10−18 J) n2 = 10 −4. T = = 12824 K. (1.38 × 10 −23 J K ) ln(10 −4 ) n1 (d) For absorption to take place in the Balmer series, hydrogen must start in the n = 2 state. From part (a), colder stars have fewer atoms in this state leading to weaker absorption lines. (a) IDENTIFY and SET UP: The photon energy is given to the electron in the atom. Some of this energy overcomes the binding energy of the atom and what is left appears as kinetic energy of the free electron. Apply hf = Ef − Ei , the energy given to the electron in the atom when a photon is absorbed. (c)

38.65.

EXECUTE: The energy of one photon is

hc

λ

= 2.323 × 10

−18

J(1 eV/1.602 × 10

−19

hc

λ

=

(6.626 × 10−34 J ⋅ s)(2.998 × 108 m/s) 85.5 × 10−9 m

J) = 14.50 eV.

The final energy of the electron is Ef = Ei + hf . In the ground state of the hydrogen atom the energy of the electron

38.66.

is Ei = −13.60 eV. Thus Ef = −13.60 eV + 14.50 eV = 0.90 eV. (b) EVALUATE: At thermal equilibrium a few atoms will be in the n = 2 excited levels, which have an energy of −13.6 eV/4 = −3.40 eV, 10.2 eV greater than the energy of the ground state. If an electron with E = −3.40 eV gains 14.5 eV from the absorbed photon, it will end up with 14.5 eV − 3.4 eV = 11.1 eV of kinetic energy. IDENTIFY: The diffraction grating allows us to determine the peak-intensity wavelength of the light. Then Wien’s displacement law allows us to calculate the temperature of the blackbody, and the Stefan-Boltzmann law allows us to calculate the rate at which it radiates energy. SET UP: The bright spots for a diffraction grating occur when d sin θ = mλ. Wien’s displacement law is 2.90 × 10 −3 m ⋅ K λpeak = , and the Stefan-Boltzmann law says that the intensity of the radiation is I = σT 4, so the T total radiated power is P = σAT 4. EXECUTE: (a) First find the wavelength of the light: λ = d sin θ = [1/(385,000 lines/m)] sin(11.6°) = 5.22 × 10–7 m Now use Wien’s law to find the temperature: T = (2.90 × 10–3 m ⋅ K )/(5.22 × 10–7 m) = 5550 K. (b) The energy radiated by the blackbody is equal to the power times the time, giving U = Pt = IAt = σAT 4t, which gives t = U/(σAT 4) = (12.0 × 106 J)/[(5.67 × 10–8 W/m 2 ⋅ K 4 )(4π)(0.0750 m)2(5550 K)4] = 3.16 s. EVALUATE: By ordinary standards, this blackbody is very hot, so it does not take long to radiate 12.0 MJ of energy.


Photons, Electrons, and Atoms

38.67.

38-15

IDENTIFY: Assuming that Betelgeuse radiates like a perfect blackbody, Wien’s displacement and the StefanBoltzmann law apply to its radiation. 2.90 × 10 −3 m ⋅ K , and the Stefan-Boltzmann law says that the SET UP: Wien’s displacement law is λpeak = T 4 intensity of the radiation is I = σT , so the total radiated power is P = σAT 4. EXECUTE: (a) First use Wien’s law to find the peak wavelength: λm = (2.90 × 10–3 m ⋅ K )/(3000 K) = 9.667 × 10–7 m Call N the number osf photons/second radiated. N × (energy per photon) = IA = σAT 4. λ σ AT 4 N (hc/λm) = σAT 4. N = m . hc (9.667 × 10−7 m)(5.67 × 10−8 W/m 2 ⋅ K 4 )(4π )(600 × 6.96 × 108 m)2 (3000 K) 4 . N= (6.626 × 10−34 J ⋅ s)(3.00 × 108 m/s) N = 5 × 1049 photons/s. 2

I B AB σ ABTB4 4π RB2TB4 ⎛ 600 RS ⎞ ⎛ 3000 K ⎞ 4 = = =⎜ ⎟ ⎜ ⎟ = 3 × 10 I S AS σ ASTS4 4π RS2TS4 ⎝ RS ⎠ ⎝ 5800 K ⎠ EVALUATE: Betelgeuse radiates 30,000 times as much energy per second as does our sun! IDENTIFY: The blackbody radiates heat into the water, but the water also radiates heat back into the blackbody. The net heat entering the water causes evaporation. Wien’s law tells us the peak wavelength radiated, but a thermophile in the water measures the wavelength and frequency of the light in the water. dQ 4 4 SET UP: By the Stefan-Boltzman law, the net power radiated by the blackbody is = σ A (Tsphere − Twater ) . Since dt dQ dm = Lv . Wien’s displacement law is this heat evaporates water, the rate at which water evaporates is dt dt 2.90 × 10−3 m ⋅ K λm = , and the wavelength in the water is λw = λ0/n. T dQ dQ dm 4 4 EXECUTE: (a) The net radiated heat is = σ A (Tsphere − Twater and the evaporation rate is = Lv , where ) dt dt dt dm 4 4 = σ A (Tsphere − Twater dm is the mass of water that evaporates in time dt. Equating these two rates gives Lv ). dt 4

(b)

38.68.

dm σ ( 4π R = dt

2

)(T

4 sphere

Lv

4 − Twater )

.

−8 2 4 2 4 4 dm ( 5.67 × 10 W/m ⋅ K ) ( 4π ) (0.120 m) ⎡⎣ (498 K) − (373 K) ⎤⎦ = = 1.92 × 10 −4 kg/s = 0.193 g/s dt 2256 × 103 J/Kg (b) (i) Wien’s law gives λm = (0.00290 m ⋅ K )/(498 K) = 5.82 × 10–6 m But this would be the wavelength in vacuum. In the water the thermophile organism would measure λw = λ0/n = (5.82 × 10–6 m)/1.333 = 4.37 × 10–6 m = 4.37 µm (ii) The frequency is the same as if the wave were in air, so f = c/λ0 = (3.00 ×108 m/s)/(5.82 × 10–6 m) = 5.15 × 1013 Hz

EVALUATE: An alternative way is to use the quantities in the water: f =

38.69.

c/n

λ0 / n

= c/λ0, which gives the same

answer for the frequency. An organism in the water would measure the light coming to it through the water, so the wavelength it would measure would be reduced by a factor of 1/n. IDENTIFY: The energy of the peak-intensity photons must be equal to the energy difference between the n = 1 and the n = 4 states. Wien’s law allows us to calculate what the temperature of the blackbody must be for it to radiate with its peak intensity at this wavelength. 13.6 eV SET UP: In the Bohr model, the energy of an electron in shell n is En = − , and Wien’s displacement law n2 2.90 × 10−3 m ⋅ K is λm = . The energy of a photon is E = hf = hc/λ. T EXECUTE: First find the energy (ΔE) that a photon would need to excite the atom. The ground state of the atom is n = 1 and the third excited state is n = 4. This energy is the difference between the two energy levels. Therefore


38-16

Chapter 38

⎛ 1 1⎞ ΔE = ( −13.6 eV ) ⎜ 2 − 2 ⎟ = 12.8 eV. Now find the wavelength of the photon having this amount of energy. ⎝4 1 ⎠ hc/λ = 12.8 eV and

λ = (4.136 × 10–15 eV ⋅ s )(3.00 × 108 m/s)/(12.8 eV) = 9.73 ×10–8 m

38.70.

38.71.

Now use Wien’s law to find the temperature. T = (0.00290 m ⋅ K )/(9.73 × 10–8 m) = 2.98 × 104 K. EVALUATE: This temperature is well above ordinary room temperatures, which is why hydrogen atoms are not in excited states during everyday conditions. IDENTIFY and SET UP: Electrical power is VI. Q = mcΔT . EXECUTE: (a) (0.010)VI = (0.010)(18.0 × 103 V)(60.0 × 10 −3 A) = 10.8 W = 10.8 J/s (b) The energy in the electron beam that isn’t converted to x rays stays in the target and appears as thermal energy. Q 1.07 × 103 J For t = 1.00 s , Q = (0.990)VI (1.00 s) = 1.07 × 103 J and ΔT = = = 32.9 K . The mc (0.250 kg)(130 J/kg ⋅ K) temperature rises at a rate of 32.9 K/s. EVALUATE: The target must be made of a material that has a high melting point. IDENTIFY: Apply conservation of energy and conservation of linear momentum to the system of atom plus photon. (a) SET UP: Let Etr be the transition energy, Eph be the energy of the photon with wavelength λ ′, and Er be

the kinetic energy of the recoiling atom. Conservation of energy gives Eph + Er = Etr . Eph =

hc hc hc so = Etr − Er and λ ′ = . λ′ λ′ Etr − Er

EXECUTE: If the recoil energy is neglected then the photon wavelength is λ = hc / Etr .

⎛ 1 ⎞ 1 ⎞ ⎛ hc ⎞⎛ 1 Δλ = λ ′ − λ = hc ⎜ − − 1⎟ ⎟=⎜ ⎟⎜ − − 1 / E E E E E E r tr ⎠ r tr ⎝ tr ⎝ tr ⎠⎝ ⎠ −1

⎛ 1 E ⎞ E E = ⎜ 1 − r ⎟ ≈ 1 + r since r V 1 1 − Er / Etr ⎝ E tr ⎠ Etr Etr (We have used the binomial theorem, Appendix B.) hc ⎛ Er ⎞ ⎛ Er ⎞ 2 Thus Δλ = ⎜ ⎟ , or since Etr = hc / λ , Δλ = ⎜ ⎟ λ . Etr ⎝ Etr ⎠ ⎝ hc ⎠ SET UP: Use conservation of linear momentum to find Er : Assuming that the atom is initially at rest, the

momentum pr of the recoiling atom must be equal in magnitude and opposite in direction to the momentum pph = h / λ of the emitted photon: h / λ = pr .

pr2 h2 , where m is the mass of the atom, so Er = . 2m 2mλ 2 ⎛ h 2 ⎞⎛ λ 2 ⎞ h ⎛E ⎞ Use this result in the above equation: Δλ = ⎜ r ⎟ λ 2 = ⎜ ; ⎟= 2 ⎟⎜ 2 m hc 2 mc λ ⎝ hc ⎠ ⎝ ⎠⎝ ⎠ EXECUTE:

Er =

note that this result for Δλ is independent of the atomic transition energy. 6.626 × 10−34 J ⋅ s h (b) For a hydrogen atom m = mp and Δλ = = = 6.61 × 10−16 m 2mpc 2(1.673 × 10−27 kg)(2.998 × 108 m/s)

38.72.

EVALUATE: The correction is independent of n. The wavelengths of photons emitted in hydrogen atom transitions are on the order of 100 nm = 10−7 m, so the recoil correction is exceedingly small. (a) Δλ1 = (h mc )(1 − cos θ1 ), Δλ 2 = (h mc)(1 − cos θ2 ), and so the overall wavelength shift is

Δλ = (h mc )(2 − cos θ1 − cos θ2 ). (b) For a single scattering through angle θ , Δλ s = ( h mc)(1 − cos θ ). For two successive scatterings through an angle of θ 2 for each scattering,

Δλ t = 2(h mc )(1 − cosθ 2). 1 − cos θ = 2(1 − cos 2 (θ 2)) and Δλ s = ( h mc)2(1 − cos 2 (θ 2)) cos(θ 2) ≤ 1 so 1 − cos 2 (θ 2) ≥ (1 − cos(θ 2)) and Δλ s ≥ Δλ t


Photons, Electrons, and Atoms

38-17

Equality holds only when θ = 180°. (c) ( h mc)2(1 − cos30.0°) = 0.268( h mc). 38.73.

(d) ( h mc)(1 − cos 60°) = 0.500( h mc), which is indeed greater than the shift found in part (c). IDENTIFY and SET UP: Find the average change in wavelength for one scattering and use that in Δλ in Eq.(38.23) to calculate the average scattering angle φ . EXECUTE: (a) The wavelength of a 1 MeV photon is hc (4.136 × 10 −15 eV ⋅ s)(2.998 × 108 m/s) = 1 × 10−12 m λ= = E 1 × 106 eV The total change in wavelength therefore is 500 × 10 −9 m − 1 × 10 −12 m = 500 × 10−9 m. If this shift is produced in 10 26 Compton scattering events, the wavelength shift in each scattering event is 500 × 10−9 m Δλ = = 5 × 10−33 m. 1 × 1026 h (b) Use this Δλ in Δλ = (1 − cos φ ) and solve for φ . We anticipate that φ will be very small, since Δλ is mc much less than h / mc, so we can use cosφ ≈ 1 − φ 2 / 2. h h 2 Δλ = (1 − (1 − φ 2 / 2)) = φ mc 2mc

φ=

2Δλ 2(5 × 10−33 m) = = 6.4 × 10 −11 rad = (4 × 10−9 )° ( h / mc ) 2.426 × 10−12 m

φ in radians is much less than 1 so the approximation we used is valid. (c) IDENTIFY and SET UP: We know the total transit time and the total number of scatterings, so we can calculate the average time between scatterings. EXECUTE: The total time to travel from the core to the surface is (106 y)(3.156 × 107 s/y) = 3.2 × 1013 s. There are 3.2 × 1013 s = 3.2 × 10−13 s. 1026 The distance light travels in this time is d = ct = (3.0 × 108 m/s)(3.2 × 10 −13 s) = 0.1 mm EVALUATE: The photons are on the average scattered through a very small angle in each scattering event. The average distance a photon travels between scatterings is very small. hc (a) The final energy of the photon is E ′ = ,and E = E ′ + K , where K is the kinetic energy of the electron after λ′ the collision. Then, hc hc hc λ′ = = = . λ= ⎤ E ′ + K ( hc λ′) + K ( hc λ′) + (γ − 1) mc 2 λ′mc ⎡ 1 −1 1+ h ⎢⎣ (1 − v 2 c 2 )1 2 ⎥⎦ 10 26 scatterings during this time, so the average time between scatterings is t =

38.74.

( K = mc 2 (γ − 1) since the relativistic expression must be used for three-figure accuracy).

(b) φ = arccos(1 − Δ λ ( h mc)). (c) γ − 1 =

(1 − (

1

)

12 1.80 2 3.00

)

− 1 = 1.25 − 1 = 0.250,

⇒λ= 1+

(5.10 × 10

−12

5.10 × 10 −3 mm = 3.34 × 10−3 nm . m)(9.11 × 10 −31 kg)(3.00 × 108 m s)(0.250) (6.63 × 10−34 J ⋅ s)

φ = arccos ⎜1 − ⎝

38.75.

h = 2.43 × 10 −12 m mc

(5.10 × 10−12 m − 3.34 × 10−12 m) ⎞ ⎟ = 74.0°. 2.43 × 10−12 m ⎠

(a) IDENTIFY and SET UP: Conservation of energy applied to the collision gives Eλ = Eλ ′ + Ee , where Ee is the

kinetic energy of the electron after the collision and Eλ and Eλ ′ are the energies of the photon before and after the collision. The energy of a photon is related to its wavelength according to Eq.(38.2).


38-18

Chapter 38

⎛1 1 ⎞ ⎛ λ′ − λ ⎞ Ee = hc ⎜ − ⎟ = hc ⎜ ⎟ ⎝ λ λ′ ⎠ ⎝ λλ ′ ⎠

EXECUTE:

⎛ ⎞ 0.0032 × 10−9 m Ee = (6.626 × 10−34 J ⋅ s)(2.998 × 108 m/s) ⎜ ⎟ −9 −9 ⎝ (0.1100 × 10 m)(0.1132 × 10 m) ⎠ Ee = 5.105 × 10−17 J = 319 eV

1 2 Ee 2(5.105 × 10 −17 J) Ee = mv 2 so v = = = 1.06 × 107 m/s 2 m 9.109 × 10 −31 kg (b) The wavelength λ of a photon with energy Ee is given by Ee = hc / λ so

hc (6.626 × 10−34 J ⋅ s)(2.998 × 108 m/s) = = 3.89 nm Ee 5.105 × 10−17 J EVALUATE: Only a small portion of the incident photon’s energy is transferred to the struck electron; this is why the wavelength calculated in part (b) is much larger than the wavelength of the incident photon in the Compton scattering. h IDENTIFY: Apply the Compton scattering formula λ ′ − λ = Δλ = (1 − cos φ ) = λc (1 − cos φ ) mc (a) SET UP: Largest Δλ is for φ = 180°. EXECUTE: For φ = 180°, Δλ = 2λc = 2(2.426 pm) = 4.85 pm.

λ=

38.76.

λ ′ − λ = λc (1 − cos φ ) Wavelength doubles implies λ ′ = 2λ so λ ′ − λ = λ . Thus λ = λC (1 − cosφ ). λ is related to E by Eq.(38.2). EXECUTE: E = hc / λ , so smallest energy photon means largest wavelength photon, so φ = 180° and (b) SET UP:

λ = 2λc = 4.85 pm. Then E =

38.77.

hc

λ

=

(6.626 × 10−34 J ⋅ s)(2.998 × 108 m/s) = 4.096 × 10−14 J(1 eV/1.602 × 10−19 J) = 4.85 × 10−12 m

0.256 MeV. EVALUATE: Any photon Compton scattered at φ = 180° has a wavelength increase of 2λc = 4.85 pm. 4.85 pm is near the short-wavelength end of the range of x-ray wavelengths. 2π hc 2 c (a) I (λ ) = 5 hc λkT but λ = f − 1) λ (e ⇒ I( f ) = (b)

=∫

∫ ∞ 0

∞ 0

2π hc 2 2π hf 5 = 3 hf kT 5 hf kT − 1) c (e − 1) (c f ) (e

0 ⎛ −c ⎞ I (λ ) d λ = ∫ I ( f ) df ⎜ 2 ⎟ ∞ ⎝ f ⎠

2π hf 3 df 2π (kT ) 4 ∞ x3 2π ( kT ) 4 1 (2π )5 (kT ) 4 2π 5k 4T 4 dx = (2π ) 4 = = = 2 3 2 3 hf kT x ∫ 0 c (e ch e −1 ch 240 240h3c 2 15c 2h3 − 1) 2

2π 5k 4T 4 = σ as shown in Eq. (38.36). Plugging in the values for the constants we get 15h3c 2 σ = 5.67 × 10−8 W m 2 ⋅ K 4 .

(c) The expression

38.78.

I = σT 4 , P = IA, and ΔE = Pt ; combining,

t= 38.79.

ΔE (100 J) = = 8.81 × 103 s = 2.45 hrs. Aσ T 4 (4.00 × 10−6 m 2 )(5.67 × 10−8 W m 2 ⋅ K 4 )(473 K) 4

(a) The period was found in Exercise 38.27b: T =

4P02n 3h3 1 me 4 and frequency is just f = = 2 3 3 . 4 me T 4P0 n h

1 me4 ⎛ 1 1 ⎞ (b) Eq. (38.6) tells us that f = ( E2 − E1 ). So f = 2 3 ⎜ 2 − 2 ⎟ (from Eq. (38.18)). h 8P0 h ⎝ n2 n1 ⎠ If n2 = n and n1 = n + 1, then =

1 1 1 1 − = − n22 n12 n 2 ( n + 1) 2

⎞ 1⎛ ⎛ 2 1⎛ 1 me 4 ⎞⎞ 2 1− ≈ 2 ⎜ 1 − ⎜1 − + " ⎟ ⎟ = 3 for large n ⇒ f ≈ 2 3 3 . 2 ⎜ 2 ⎟ n ⎝ (1 + 1 n) ⎠ n ⎝ ⎝ n 4P0 n h ⎠⎠ n


Photons, Electrons, and Atoms

38.80.

38.81.

38-19

h Each photon has momentum p = , and if the rate at which the photons strike the surface is ( dN dt ) , the force λ on the surface is ( h λ )( dN dt ), and the pressure is ( h λ )( dN dt ) A. The intensity is I = (dN dt )( E ) A = ( dN dt )(hc λ ) A , and comparison of the two expressions gives the pressure as ( I c). G G G G Momentum: p + P = p′ + P ′ ⇒ p − P = − p′ − P′ ⇒ p′ = P − ( p + P′) energy: pc + E = p′c + E′ = p′c + ( P′c ) 2 + (mc 2 ) 2 ⇒ ( pc − p′c + E ) 2 = ( P′c) 2 + ( mc 2 ) 2 = (Pc) 2 + ((p + p′)c) 2 − 2 P (p + p′)c 2 + (mc 2 ) 2 ( pc − p′c) 2 + E 2 = E 2 + ( pc + p′c) 2 − 2( Pc 2 )( p + p′) + 2 Ec( p − p′) − 4 pp′c 2 + 2 Ec( p − p′) +2( Pc 2 )( p + p′) = 0

⇒ p′( Pc 2 − 2 pc 2 − Ec) = p ( − Ec − Pc 2 ) ⇒ p′ = p

Ec + Pc 2 E + Pc =p 2 pc 2 + Ec − Pc 2 2 pc + ( E − Pc )

2hc ⎛ 2 hc λ + ( E − Pc) ⎞ ⎛ E − Pc ⎞ ⇒ λ′ = λ ⎜ ⎟ = λ⎜ ⎟+ E + Pc ⎝ ⎠ ⎝ E + Pc ⎠ E + Pc (λ ( E − Pc) + 2hc) ⇒ λ′ = E + Pc ⎛ mc 2 ⎞ If E W mc 2 , Pc = E 2 − ( mc 2 ) 2 = E 1 − ⎜ ⎟ ⎝ E ⎠ ⇒ E − Pc ≈

2

⎛ 1 ⎛ mc 2 ⎞ 2 ⎞ ≈ E ⎜1 − ⎜ + "⎟ ⎟ ⎜ 2⎝ E ⎠ ⎟ ⎝ ⎠

1 ( mc 2 ) 2 λ( mc 2 ) 2 hc hc ⎛ m 2c 4 λ ⎞ ⇒ λ1 ≈ + = ⎜1 + ⎟ 2 E 2 E (2 E ) E E⎝ 4hcE ⎠

(b) If λ = 10.6 × 10 −6 m, E = 1.00 × 1010 eV = 1.60 × 10−9 J

⇒ λ′ ≈

hc 1.60 × 10−9

⎛ (9.11 × 10−31 kg) 2c 4 (10.6 × 10−6 m) ⎞ ⎜1 + ⎟ J⎝ 4hc (1.6 × 10−9 J) ⎠

= (1.24 × 10−16 m)(1 + 56.0) = 7.08 × 10−15 m.

(c) These photons are gamma rays. We have taken infrared radiation and converted it into gamma rays! Perhaps useful in nuclear medicine, nuclear spectroscopy, or high energy physics: wherever controlled gamma ray sources might be useful.


39

THE WAVE NATURE OF PARTICLES

39.1.

IDENTIFY and SET UP: EXECUTE: (a) λ =

h h = . For an electron, m = 9.11 × 10 −31 kg . For a proton, m = 1.67 × 10 −27 kg . p mv

6.63 × 10−34 J ⋅ s = 1.55 × 10−10 m = 0.155 nm (9.11 × 10−31 kg)(4.70 × 106 m/s)

(b) λ is proportional to 39.2.

λ=

IDENTIFY and SET UP:

⎛m ⎞ ⎛ 9.11 × 10 −31 kg ⎞ 1 −14 , so λp = λe ⎜ e ⎟ = (1.55 × 10−10 m) ⎜ ⎟ = 8.46 × 10 m . −27 ⎜ mp ⎟ m 1.67 10 kg × ⎝ ⎠ ⎝ ⎠ For a photon, E =

hc

λ

. For an electron or proton, p =

h

λ

and E =

p2 h2 , so E = . 2m 2mλ 2

(4.136 × 10−15 eV ⋅ s)(3.00 × 108 m/s) EXECUTE: (a) E = = = 6.2 keV 0.20 × 10 −9 m λ hc

2

(b) E =

39.3.

⎛m ⎞ ⎛ 9.11 × 10−31 kg ⎞ (c) Ep = Ee ⎜ e ⎟ = (38 eV) ⎜ ⎟ = 0.021 eV −27 ⎜ ⎟ ⎝ 1.67 × 10 kg ⎠ ⎝ mp ⎠ EVALUATE: For a given wavelength a photon has much more energy than an electron, which in turn has more energy than a proton. h h (6.63 × 10−34 J ⋅ s) = 2.37 × 10−24 kg ⋅ m s. (a) λ = ⇒ p = = p λ (2.80 × 10 −10 m) (b) K =

39.4.

λ=

=

39.5.

⎛ 6.63 × 10−34 J ⋅ s ⎞ h2 1 = = 6.03 × 10−18 J = 38 eV ⎜ ⎟ 2mλ 2 ⎝ 0.20 × 10 −9 m ⎠ 2(9.11 × 10−31 kg)

p 2 (2.37 × 10 −24 kg ⋅ m s) 2 = = 3.08 × 10−18 J = 19.3 eV. 2m 2(9.11 × 10 −31 kg)

h h = p 2mE (6.63 × 10−34 J ⋅ s) 2(6.64 × 10 −27 kg) (4.20 × 106 eV) (1.60 × 10 −19 J e V)

= 7.02 × 10 −15 m.

h h = . In the Bohr model, mvrn = n( h / 2π ), p mv so mv = nh /(2π rn ). Combine these two expressions and obtain an equation for λ in terms of n. Then

IDENTIFY and SET UP: The de Broglie wavelength is λ =

⎛ 2π rn ⎞ 2π rn . ⎟= n ⎝ nh ⎠ EXECUTE: (a) For n = 1, λ = 2π r1 with r1 = a0 = 0.529 × 10−10 m, so λ = 2π (0.529 × 10−10 m) = 3.32 × 10−10 m

λ = h⎜

λ = 2π r1; the de Broglie wavelength equals the circumference of the orbit. (b) For n = 4, λ = 2π r4 / 4. rn = n 2a0 so r4 = 16a0 .

λ = 2π (16a0 ) / 4 = 4(2π a0 ) = 4(3.32 × 10−10 m) = 1.33 × 10 −9 m 1 1 = times the circumference of the orbit. n 4 EVALUATE: As n increases the momentum of the electron increases and its de Broglie wavelength decreases. For any n, the circumference of the orbits equals an integer number of de Broglie wavelengths.

λ = 2π r4 / 4; the de Broglie wavelength is


39-2

39.6.

39.7.

39.8.

Chapter 39

(a) For a nonrelativistic particle, K =

(b) (6.63 × 10−34 J ⋅ s) 2(800 eV)(1.60 × 10-19 J/eV)(9.11 × 10-31 kg) = 4.34 × 10 −11 m. IDENTIFY: A person walking through a door is like a particle going through a slit and hence should exhibit wave properties. SET UP: The de Broglie wavelength of the person is λ = h/mv. EXECUTE: (a) Assume m = 75 kg and v = 1.0 m/s. λ = h/mv = (6.626 × 10–34 J ⋅ s)/[(75 kg)(1.0 m/s)] = 8.8 × 10–36 m EVALUATE: (b) A typical doorway is about 1 m wide, so the person’s de Broglie wavelength is much too small to show wave behavior through a “slit” that is about 1035 times as wide as the wavelength. Hence ordinary objects do not show wave behavior in everyday life.

Combining Equations 37.38 and 37.39 gives p = mc γ 2 − 1. (a) λ =

39.9.

p2 h h , so λ = = . 2m p 2 Km

h = (h mc) p

γ 2 − 1 = 4.43 × 10−12 m. (The incorrect nonrelativistic calculation gives 5.05 × 10−12 m.)

(b) ( h mc) γ 2 − 1 = 7.07 × 10−13 m. IDENTIFY and SET UP: A photon has zero mass and its energy and wavelength are related by Eq.(38.2). An electron has mass. Its energy is related to its momentum by E = p 2 / 2m and its wavelength is related to its momentum by Eq.(39.1). EXECUTE: (a) photon hc hc (6.626 × 10−34 J ⋅ s)(2.998 × 108 m/s) E= so λ = = = 62.0 nm E (20.0 eV)(1.602 × 10 −19 J/eV) λ electron

E = p 2 /(2m) so p = 2mE = λ = h / p = 0.274 nm

2(9.109 × 10−31 kg)(20.0 eV)(1.602 × 10 −19 J/eV) = 2.416 × 10 −24 kg ⋅ m/s

(b) photon E = hc / λ = 7.946 × 10−19 J = 4.96 eV electron λ = h / p so p = h / λ = 2.650 × 10 −27 kg ⋅ m/s E = p 2 /(2m) = 3.856 × 10 −24 J = 2.41 × 10 −5 eV (c) EVALUATE: You should use a probe of wavelength approximately 250 nm. An electron with λ = 250 nm has much less energy than a photon with λ = 250 nm, so is less likely to damage the molecule. Note that λ = h / p applies to all particles, those with mass and those with zero mass. E = hf = hc / λ applies only to photons and

39.10.

E = p 2 / 2m applies only to particles with mass. IDENTIFY: Any moving particle has a de Broglie wavelength. The speed of a molecule, and hence its de Broglie wavelength, depends on the temperature of the gas. SET UP: The average kinetic energy of the molecule is Kav = 3/2 kT, and the de Broglie wavelength is λ = h/mv = h/p.

EXECUTE: (a) Combining Kav = 3/2 kT and K = p2/2m gives 3/2 kT = pav2/2m and pav = 3mkT . The de Broglie 6.626 × 10−34 J ⋅ s h h wavelength is λ = = = = 1.08 × 10−10 m . p 3mkT 3( 2 × 1.67 × 10−27 kg )(1.38 × 10 −23 J/K ) ( 273 K ) (b) For an electron, λ = h/p = h/mv gives v=

h 6.626 × 10−34 J ⋅ s = = 6.75 × 106 m/s mλ ( 9.11 × 10 −31 kg )(1.08 × 10 −10 m )

This is about 2% the speed of light, so we do not need to use relativity. (c) For photon: E = hc/λ = (6.626 × 10–34 J ⋅ s)(3.00 × 108 m/s)/(1.08 × 10–10 m) = 1.84 × 10–15 J For the H2 molecule: Kav = (3/2)kT = 3/2 (1.38 × 10–23 J/K)(273 K) = 5.65 × 10–21 J For the electron: K = ½ mv2 = ½ (9.11 × 10–31 kg)(6.73 × 106 m/s)2 = 2.06 × 10–17 J EVALUATE: The photon has about 100 times more energy than the electron and 300,000 times more energy than the H2 molecule. This shows that photons of a given wavelength will have much more energy than particles of the same wavelength.


The Wave Nature of Particles

39.11.

39.12.

39-3

IDENTIFY and SET UP: Use Eq.(39.1). h h 6.626 × 10−34 J ⋅ s EXECUTE: λ = = = = 3.90 × 10−34 m p mv (5.00 × 10−3 kg)(340 m/s) EVALUATE: This wavelength is extremely short; the bullet will not exhibit wavelike properties. (a) λ = h mv → v = h m λ

1 Energy conservation: eΔV = mv 2 2 2

⎛ h ⎞ m⎜ ⎟ mv h2 (6.626 × 10−34 J ⋅ s) 2 mλ ⎠ ΔV = = ⎝ = = = 66.9 V 2 −19 2e 2e 2emλ 2(1.60 × 10 C) (9.11× 10−31 kg) (0.15 × 10−9 m) 2 2

(b) Ephoton = hf =

hc

λ

=

(6.626 × 10−34 J ⋅ s) (3.0 × 108 m s) = 1.33 × 10 −15 J 0.15 × 10−9 m

eΔV = K = Ephoton and ΔV =

39.13.

39.14.

Ephoton e

=

1.33 × 10 −15 J = 8310 V 1.6 × 10−19 C

(a) λ = 0.10 nm . p = mv = h λ so v = h ( mλ ) = 7.3 × 106 m s .

1 (b) E = mv 2 = 150 eV 2 (c) E = hc / λ = 12 KeV (d) The electron is a better probe because for the same λ it has less energy and is less damaging to the structure being probed. IDENTIFY: The electrons behave like waves and are diffracted by the slit. SET UP: We use conservation of energy to find the speed of the electrons, and then use this speed to find their de Broglie wavelength, which is λ = h/mv. Finally we know that the first dark fringe for single-slit diffraction occurs when a sin θ = λ. EXECUTE: (a) Use energy conservation to find the speed of the electron: ½ mv2 = eV. v=

2 (1.60 × 10−19 C ) (100 V) 2eV = = 5.93 × 106 m/s 9.11 × 10 −31 kg m

which is about 2% the speed of light, so we can ignore relativity. (b) First find the de Broglie wavelength:

λ=

h 6.626 × 10−34 J ⋅ s = = 1.23 × 10–10 m = 0.123 nm mv ( 9.11 × 10−31 kg )( 5.93 × 106 m/s )

For the first single slit dark fringe, we have a sin θ = λ, which gives

a=

39.15.

39.16.

λ sin θ

=

1.23 × 10−10 m = 6.16 × 10–10 m = 0.616 nm sin(11.5°)

EVALUATE: The slit width is around 5 times the de Broglie wavelength of the electron, and both are much smaller than the wavelength of visible light. h For m =1, λ = d sin θ = . 2mE h2 (6.63 × 10 −34 J ⋅ s) 2 = = 6.91 × 10−20 J = 0.432 eV. E= 2 2 −27 2md sin θ 2(1.675 × 10 kg) (9.10 × 10−11 m) 2 sin 2 (28.6°) h h mh Intensity maxima occur when d sin θ = mλ. λ = = so d sin θ = . (Careful! Here, m is the order p 2ME 2ME of the maxima, whereas M is the mass of the incoming particle.) mh (2)(6.63 × 10 −34 J ⋅ s) (a) d = = 2 ME sin θ 2(9.11 × 10−31 kg)(188 eV)(1.60 × 10−19 J/eV) sin(60.6°) = 2.06 × 10−10 m = 0.206 nm.

(b) m = 1 also gives a maximum.

⎛ ⎞ (1) (6.63 × 10−34 J ⋅ s) ⎟ = 25.8° . θ = arcsin ⎜ ⎜ 2(9.11 × 10−31 kg) (188 eV) (1.60 × 10−19 J e V) (2.06 × 10−10 m) ⎟ ⎝ ⎠


39-4

Chapter 39

This is the only other one. If we let m ≥ 3, then there are no more maxima. (c) E =

m2h2 (1) 2 (6.63 × 10 −34 J ⋅ s) 2 = 2 Md 2 sin 2 θ 2(9.11 × 10−31 kg) (2.60 × 10−10 m) 2 sin 2 (60.6°)

= 7.49 × 10−18 J = 46.8 eV. Using this energy, if we let m = 2, then sin θ > 1. Thus, there is no m = 2 maximum in this case.

39.17.

h h ⎛ mh ⎞ = , so θ = arcsin ⎜ ⎟ . (Careful! Here, m is the order of p Mv ⎝ dMv ⎠ the maximum, whereas M is the incoming particle mass.) ⎛ h ⎞ (a) m = 1 ⇒ θ1 = arcsin ⎜ ⎟ ⎝ dMv ⎠ The condition for a maximum is d sinθ = mλ . λ =

⎛ ⎞ 6.63 × 10−34 J ⋅ s = arcsin ⎜ ⎟ = 2.07°. −6 −31 4 × × × (1.60 10 m) (9.11 10 kg) (1.26 10 m s) ⎝ ⎠ ⎛ ⎞ (2) (6.63 × 10−34 J ⋅ s) m = 2 ⇒ θ2 = arcsin ⎜ ⎟ = 4.14°. −6 −31 4 × × × (1.60 10 m) (9.11 10 kg) (1.26 10 m s) ⎝ ⎠ ⎛ π radians ⎞ (b) For small angles (in radians!) y ≅ Dθ , so y1 ≈ (50.0 cm) (2.07°) ⎜ ⎟ = 1.81 cm , ⎝ 180° ⎠

39.18.

⎛ π radians ⎞ y2 ≈ (50.0 cm) (4.14°) ⎜ ⎟ = 3.61 cm and y2 − y1 = 3.61 cm − 1.81 cm = 1.81 cm. ⎝ 180° ⎠ IDENTIFY: Since we know only that the mosquito is somewhere in the room, there is an uncertainty in its position. The Heisenberg uncertainty principle tells us that there is an uncertainty in its momentum. SET UP: The uncertainty principle is Δ xΔpx ≥ = . EXECUTE: (a) You know the mosquito is somewhere in the room, so the maximum uncertainty in its horizontal position is Δ x = 5.0 m. (b) The uncertainty principle gives Δ xΔpx ≥ = , and Δpx = mΔvx since we know the mosquito’s mass. This gives

Δ x mΔvx ≥ = , which we can solve for Δvx to get the minimum uncertainty in vx. Δvx =

39.19.

which is hardly a serious impediment! EVALUATE: For something as “large” as a mosquito, the uncertainty principle places a negligible limitation on our ability to measure its speed. (a) IDENTIFY and SET UP: Use Δ xΔpx ≥ h / 2π to calculate Δ x and obtain Δvx from this. EXECUTE: Δvx =

39.20.

= 1.055 × 10−34 J ⋅ s = 1.4 × 10–29 m/s = mΔx (1.5 × 10-6 kg)(5.0 m)

Δpx ≥

h 6.626 × 10−34 J ⋅ s = = 1.055 × 10−28 kg ⋅ m/s 2πΔ x 2π (1.00 × 10 −6 m)

Δp x 1.055 × 10−28 kg ⋅ m/s = = 8.79 × 10−32 m/s m 1200 kg

(b) EVALUATE: Even for this very small Δ x the minimum Δvx required by the Heisenberg uncertainty principle is very small. The uncertainty principle does not impose any practical limit on the simultaneous measurements of the positions and velocities of ordinary objects. IDENTIFY: Since we know that the marble is somewhere on the table, there is an uncertainty in its position. The Heisenberg uncertainty principle tells us that there is therefore an uncertainty in its momentum. SET UP: The uncertainty principle is Δ xΔpx ≥ = . EXECUTE: (a) Since the marble is somewhere on the table, the maximum uncertainty in its horizontal position is Δ x = 1.75 m. (b) Following the same procedure as in part (b) of problem 39.18, the minimum uncertainty in the horizontal velocity of the marble is Δvx =

= 1.055 × 10 −34 J ⋅ s = 6.03 × 10–33 m/s = mΔx ( 0.0100 kg ) (1.75 m)

(c) The uncertainty principle tells us that we cannot know that the marble’s horizontal velocity is exactly zero, so the smallest we could measure it to be is 6.03 × 10–33 m/s, from part (b). The longest time it could remain on the


The Wave Nature of Particles

table is the time to travel the full width of the table (1.75 m), so t = x/vx = (1.75 m)/(6.03 × 10 s = 9.20 × 1024 years Since the universe is about 14 × 109 years old, this time is about

–33

39-5

m/s) = 2.90 × 1032

9.0 × 1024 yr ≈ 6 × 1014 times the age of the universe! Don’t hold your breath! 14 × 109 yr

39.21.

39.22.

EVALUATE: For household objects, the uncertainty principle places a negligible limitation on our ability to measure their speed. h Heisenberg’s Uncertainty Principles tells us that Δ xΔpx ≥ . We can treat the standard deviation as a direct 2π h −10 = 1.05 × 10−34 J ⋅ s measure of uncertainty. Here Δ xΔpx = (1.2 × 10 m) (3.0 × 10−25 kg ⋅ m s) = 3.6 × 10−35 J ⋅ s but 2π h Therefore Δ xΔpx < so the claim is not valid . 2π (a) ( Δ x) (mΔvx ) ≥ h 2π , and setting Δvx = (0.010)vx and the product of the uncertainties equal to h / 2π (for the minimum uncertainty) gives vx = h (2πm(0.010)Δ x ) = 57.9 m s. (b) Repeating with the proton mass gives 31.6 mm s.

h (6.63 × 10−34 J ⋅ s) = = 2.03 × 10−32 J = 1.27 × 10−13 eV. 2π Δt 2π (5.2 × 10−3 s)

39.23.

ΔE >

39.24.

IDENTIFY and SET UP: The Heisenberg Uncertainty Principle says Δ xΔpx ≥

Δ xΔpx is h / 2π . Δpx = mΔvx .

h . The minimum allowed 2π

h h 6.63 × 10−34 J ⋅ s . Δvx = = = 3.2 × 104 m/s 2π mΔ x 2π (1.67 × 10−27 kg)(2.0 × 10−12 m) 2π h 6.63 × 10−34 J ⋅ s (b) Δ x = = = 4.6 × 10−4 m 2π mΔvx 2π (9.11× 10−31 kg)(0.250 m/s) EXECUTE: (a) mΔ xΔvx =

39.25.

ΔE Δt =

h (6.63 × 10−34 J ⋅ s) h = = 1.39 × 10−14 J = 8.69 × 104 eV = 0.0869 MeV. . ΔE = 2π Δt 2π (7.6 × 10−21 s) 2π ΔE 0.0869 MeV c 2 = = 2.81× 10−5. E 3097 MeV c 2

h . ΔE = Δmc 2 . Δm = 2.06 × 109 eV c 2 = 3.30 × 10−10 J c 2 . 2π h 6.63 × 10−34 J ⋅ s Δt = = = 3.20 × 10−25 s. 2π Δmc 2 2π (3.30 × 10−10 J)

39.26.

ΔE Δt =

39.27.

IDENTIFY and SET UP:

For a photon Eph =

39.29.

λ

=

1.99 × 10−25 J ⋅ m

λ

2

. For an electron Ee =

p2 1 ⎛h⎞ h2 = . ⎜ ⎟ = 2m 2m ⎝ λ ⎠ 2mλ 2

−25

1.99 × 10 J ⋅ m = 1.99 × 10−17 J 10.0 × 10−9 m (6.63 × 10−34 J ⋅ s) 2 electron Ee = = 2.41× 10−21 J 2(9.11 × 10−31 kg)(10.0 × 10−9 m) 2 Eph 1.99 × 10−17 J = = 8.26 × 103 Ee 2.41 × 10−21 J (b) The electron has much less energy so would be less damaging. EVALUATE: For a particle with mass, such as an electron, E ~ λ −2 . For a massless photon E ~ λ −1 . p 2 (h λ )2 (h λ )2 , so V = = = 419 V. (a) eV = K = 2m 2m 2me 9.11× 10−31 kg (b) The voltage is reduced by the ratio of the particle masses, (419 V) = 0.229 V. 1.67 × 10−27 kg EXECUTE: (a) photon Eph =

39.28.

hc

ψ ( x ) = A sin kx. The position probability density is given by ψ ( x ) = A2 sin 2 kx. EXECUTE: (a) The probability is highest where sin kx = 1 so kx = 2π x / λ = nπ / 2, n = 1, 3, 5,… x = nλ / 4, n = 1, 3, 5,… so x = λ / 4, 3λ / 4, 5λ /4,… IDENTIFY and SET UP:

2


39-6

Chapter 39

39.30.

(b) The probability of finding the particle is zero where ψ = 0, which occurs where sin kx = 0 and kx = 2π x / λ = nπ , n = 0, 1, 2,… x = nλ / 2, n = 0, 1, 2,… so x = 0, λ / 2, λ , 3λ / 2,… EVALUATE: The situation is analogous to a standing wave, with the probability analogous to the square of the amplitude of the standing wave. 2 2 2 Ψ ∗ = ψ ∗ sin ωt , so Ψ = Ψ *Ψ = ψ *ψ sin 2 ωt = ψ sin 2 ωt . Ψ is not time-independent, so Ψ is not the 2

39.31.

wavefunction for a stationary state. IDENTIFY: To describe a real situation, a wave function must be normalizable. SET UP: |ψ |2 dV is the probability that the particle is found in volume dV. Since the particle must be somewhere, ψ must have the property that ∫|ψ |2 dV = 1 when the integral is taken over all space. EXECUTE: (a) In one dimension, as we have here, the integral discussed above is of the form (b) Using the result from part (a), we have

∫ ( e ) dx = ∫ ∞

ax 2

−∞

∞ −∞

e2 ax dx =

e 2 ax 2a

−∞

| ψ ( x) |2 dx = 1.

= ∞ . Hence this wave function cannot −∞

be normalized and therefore cannot be a valid wave function. (c) We only need to integrate this wave function of 0 to ∞ because it is zero for x < 0. For normalization we have ∞

1 = ∫ |ψ |2 dx = −∞

39.32.

∫ ( Ae ) dx = ∫ ∞

− bx 2

0

0

A2e−2bx dx =

A2e −2bx −2b

= 0

A2 A2 , which gives = 1, so A = 2b . 2b 2b

EVALUATE: If b were positive, the given wave function could not be normalized, so it would not be allowable. (a) The uncertainty in the particle position is proportional to the width of ψ ( x ) , and is inversely proportional to

α . This can be seen by either plotting the function for different values of α , finding the expectation value x 2 = ∫ ψ 2 x 2 dx for the normalized wave function or by finding the full width at half-maximum. The particle’s uncertainty in position decreases with increasing α . The dependence of the expectation value 〈 x 2 〉 on α may be found by considering ∞

〈 x2 〉 =

∫xe

2 −2α x 2

−∞ ∞

∫e

dx

=−

−2α x 2

dx

∞ 1 ∂ ⎡ −2α x2 ⎤ 1 ∂ ⎡ 1 dx ⎥ = − ln ⎢ ∫ e ln ⎢ 2 ∂α ⎣ −∞ 2 ∂α ⎣ 2α ⎦

∫e

−∞

−u 2

⎤ 1 du ⎥ = , ⎦ 4α

−∞

39.33. 39.34.

where the substitution u = α x has been made. (b) Since the uncertainty in position decreases, the uncertainty in momentum must increase. ⎛ x − iy ⎞ ⎛ x + iy ⎞ ⎛ x − iy ⎞ ⎛ x + iy ⎞ 2 * * f ( x, y ) = ⎜ ⎟ ⇒ f = f f =⎜ ⎟ and f ( x, y ) = ⎜ ⎟⋅⎜ ⎟ = 1. + − x iy x iy ⎠ ⎝ ⎠ ⎝ ⎝ x + iy ⎠ ⎝ x − iy ⎠ 2

The same. ψ ( x, y , z ) = ψ * ( x, y , z )ψ ( x, y, z ) 2

ψ ( x, y, z )eiφ = (ψ * ( x, y, z )e− iφ )(ψ ( x, y, z )e + iφ ) = ψ * ( x, y, z )ψ ( x, y, z ). 39.35.

The complex conjugate means convert all i’s to–i’s and vice-versa. eiφ ⋅ e −iφ = 1. IDENTIFY: To describe a real situation, a wave function must be normalizable. SET UP: |ψ |2 dV is the probability that the particle is found in volume dV. Since the particle must be somewhere, ψ must have the property that ∫|ψ |2 dV = 1 when the integral is taken over all space. EXECUTE: (a) For normalization of the one-dimensional wave function, we have ∞

1 = ∫ |ψ |2 dx = −∞

∫ ( Ae ) dx + ∫ ( Ae ) dx = ∫ 0

bx 2

−∞

− bx 2

0

0 −∞

A2e 2bx dx + ∫ A2e −2bx dx . 0

∞ ⎧⎪ e2bx 0 e −2bx ⎫⎪ A2 –1/2 −1 1 = A2 ⎨ + ⎬ = , which gives A = b = 2.00 m = 1.41 m b b b 2 2 − ⎪ −∞ 0 ⎭ ⎩⎪ (b) The graph of the wavefunction versus x is given in Figure 39.35.

(c) (i) P = ∫

+ 5.00 m − 0.500 m

|ψ |2 dx = 2 ∫

+ 5.00 m 0

A2e −2bx dx , where we have used the fact that the wave function is an even

function of x. Evaluating the integral gives − A2 −2b (0.500 m) −(2.00 m −1 ) −2.00 e − 1) = ( ( e − 1) = 0.865 b 2.00 m −1 There is a little more than an 86% probability that the particle will be found within 50 cm of the origin. P=


The Wave Nature of Particles

(ii) P =

∫ ( Ae ) dx = ∫

(iii) P =

2

39-7

−1

A 2.00 m 1 = = = 0.500 −1 2b 2(2.00 m ) 2 There is a 50-50 chance that the particle will be found to the left of the origin, which agrees with the fact that the wave function is symmetric about the y-axis. 0

bx 2

−∞

1.00 m 0.500 m

0

−∞

A2e 2bx dx =

A2e −2bx dx

(

)

A2 −2(2.00 m-1 )(1.00 m) −2(2.00 m-1 )(0.500 m) 1 e −e = − ( e −4 − e−2 ) = 0.0585 2 −2b EVALUATE: There is little chance of finding the particle in regions where the wave function is small. =

39.36.

Figure 39.35 −= 2 d 2ψ + Uψ = Eψ . Let ψ = Aψ1 + Bψ 2 Eq. (39.18): 2m dx 2 −= 2 d 2 ⇒ ( Aψ1 + Bψ 2 ) + U ( Aψ1 + Bψ 2 ) = E ( Aψ1 + Bψ 2 ) 2m dx 2 ⎛ = 2 d 2ψ1 ⎞ ⎛ = 2 d 2ψ 2 ⎞ ⇒ A⎜ − + Uψ − Eψ + B + Uψ 2 − Eψ 2 ⎟ = 0. ⎟ ⎜− 1 1 2 2 ⎝ 2m dx ⎠ ⎝ 2m dx ⎠

But each of ψ1 and ψ2 satisfy Schrödinger’s equation separately so the equation still holds true, for any A or B. 39.37.

39.38.

= 2 d 2ψ + Uψ = BE1ψ1 + CE2ψ 2 . If ψ were a solution with energy E, then BE1ψ1 + CE2ψ 2 = BEψ1 + CEψ2 or 2m dx 2 B( E1 − E )ψ1 = C ( E − E2 )ψ 2 . This would mean that ψ1 is a constant multiple of ψ 2 , and ψ1 and ψ 2 would be wave functions with the same energy. However, E1 ≠ E2 , so this is not possible, and ψ cannot be a solution to Eq. (39.18). −

(a) λ =

h (6.63 × 10 −34 J ⋅ s) = 1.94 × 10−10 m. = −31 −19 2mK 2(9.11 × 10 kg)(40 eV)(1.60 × 10 J eV)

(2.5 m)(9.11 × 10 −31 kg)1 2 R R = = = 6.67 × 10−7 s. v 2E m 2(40 eV)(1.6 × 10−19 J eV) λ (c) The width w is w = 2R ' and w = Δv yt = Δp yt m, where t is the time found in part (b) and a is the slit width. a 2mλ R Combining the expressions for w, Δp y = = 2.65 × 10 −28 kg ⋅ m s. at h = 0.40 μm, which is the same order of magnitude. (d) Δy = 2π Δp y

(b)

39.39.

(a) E = hc λ = 12 eV (b) Find E for an electron with λ = 0.10 × 10 −6 m. λ = h p so p = h λ = 6.626 × 10 −27 kg ⋅ m s . E = p 2 (2m) = 1.5 × 10−4 eV . E = qΔV so ΔV = 1.5 × 10−4 V v = p m = (6.626 × 10 −27 kg ⋅ m s) (9.109 × 10 −31 kg) = 7.3 × 103 m s

(c) Same λ so same p. E = p 2 /(2m) but now m = 1.673 × 10−27 kg so E = 8.2 × 10−8 eV and ΔV = 8.2 × 10−8 V. v = p m = (6.626 × 10 −27 kg ⋅ m s) (1.673 × 10 −27 kg) = 4.0 m s

39.40.

(a) Single slit diffraction: a sin θ = mλ . λ = a sin θ = (150 × 10−9 m)sin20° = 5.13 × 10−8 m

λ = h mv → v = h mλ . v = (b) a sin θ2 = 2λ . sin θ2 = ±2

6.626 × 10−34 J ⋅ s = 1.42 × 104 m s (9.11× 10−31 kg)(5.13 × 10−8 m)

⎛ 5.13 × 10−8 m ⎞ = ±2 ⎜ ⎟ = ±0.684 . θ2 = ±43.2° −9 a ⎝ 150 × 10 m ⎠

λ


39-8

Chapter 39

39.41.

IDENTIFY: The electrons behave like waves and produce a double-slit interference pattern after passing through the slits. SET UP: The first angle at which destructive interference occurs is given by d sin θ = λ/2. The de Broglie wavelength of each of the electrons is λ = h/mv. EXECUTE: (a) First find the wavelength of the electrons. For the first dark fringe, we have d sin θ = λ/2, which gives (1.25 nm)(sin 18.0°) = λ/2 , and λ = 0.7725 nm. Now solve the de Broglie wavelength equation for the speed of the electron:

v=

39.42.

h 6.626 × 10−34 J ⋅ s = = 9.42 × 105 m/s mλ (9.11× 10−31 kg)(0.7725 × 10−9 m)

which is about 0.3% the speed of light, so they are nonrelativistic. (b) Energy conservation gives eV = ½ mv2 and V = mv2/2e = (9.11 × 10–31 kg)(9.42 × 105 m)2/[2(1.60 × 10–19 C)] = 2.52 V EVALUATE: The hole must be much smaller than the wavelength of visible light for the electrons to show diffraction. IDENTIFY: The alpha particles and protons behave as waves and exhibit circular-aperture diffraction after passing through the hole. SET UP: For a round hole, the first dark ring occurs at the angle θ for which sinθ = 1.22λ /D, where D is the diameter of the hole. The de Broglie wavelength for a particle is λ = h/p = h/mv. EXECUTE: Taking the ratio of the sines for the alpha particle and proton gives sin θα 1.22λα λα = = sin θ p 1.22λp λp The de Broglie wavelength gives λp = h/pp and λα = h/pα, so

sin θα h / pα pp . Using K = p2/2m, we have = = sin θ p h / pp pα

p = 2mK . Since the alpha particle has twice the charge of the proton and both are accelerated through the same potential difference, Kα = 2Kp. Therefore pp = 2mp K p and pα = 2mα Kα = 2mα (2 K p ) = 4mα K p . Substituting these quantities into the ratio of the sines gives 2mp K p mp sin θα pp = = = sin θ p pα 2 mα 4mα K p Solving for sin θα gives sin θα =

39.43.

1.67 × 10−27 kg sin15.0° and θα = 5.3°. 2(6.64 × 10 −27 kg)

EVALUATE: Since sin θ is inversely proportional to the mass of the particle, the larger-mass alpha particles form their first dark ring at a smaller angle than the ring for the lighter protons. IDENTIFY: Both the electrons and photons behave like waves and exhibit single-slit diffraction after passing through their respective slits. SET UP: The energy of the photon is E = hc/λ and the de Broglie wavelength of the electron is λ = h/mv = h/p. Destructive interference for a single slit first occurs when a sin θ = λ. EXECUTE: (a) For the photon: λ = hc/E and a sinθ = λ. Since the a and θ are the same for the photons and electrons, they must both have the same wavelength. Equating these two expressions for λ gives a sin θ = hc/E. h h and a sin θ = λ. Equating these two expressions for λ gives a sin θ = . For the electron, λ = h/p = 2mK 2mK h , which gives E = c 2mK = (4.05 × 10−7 J1/2 ) K Equating the two expressions for asinθ gives hc/E = 2mK

E c 2mK 2mc 2 = = . Since v << c, mc2 > K, so the square root is >1. Therefore E/K > 1, meaning that the K K K photon has more energy than the electron. EVALUATE: As we have seen in Problem 39.10, when a photon and a particle have the same wavelength, the photon has more energy than the particle. d sin θ (40.0 × 10−6 m)sin(0.0300 rad) According to Eq.(35.4) λ = = = 600 nm. The velocity of an electron with m 2 this wavelength is given by Eq.(39.1) (b)

39.44.

v=

p h (6.63 × 10−34 J ⋅ s) = = = 1.21 × 103 m s. m mλ (9.11 × 10−31 kg)(600 × 10 −9 m)


The Wave Nature of Particles

39-9

Since this velocity is much smaller than c we can calculate the energy of the electron classically

39.45.

1 1 K = mv 2 = (9.11 × 10−31 kg)(1.21 × 103 m s) 2 = 6.70 × 10 −25 J = 4.19 μeV. 2 2 The de Broglie wavelength of the blood cell is λ=

h (6.63 × 10 −34 J ⋅ s) = = 1.66 × 10 −17 m. mv (1.00 × 10−14 kg)(4.00 × 10−3 m s)

We need not be concerned about wave behavior. 12

39.46.

⎛ v2 ⎞ h ⎜1 − 2 ⎟ c ⎠ h (a) λ = = ⎝ p mv

⎛ v2 ⎞ h 2v 2 v2 ⇒ λ 2 m 2 v 2 = h 2 ⎜ 1 − 2 ⎟ = h 2 − 2 ⇒ λ 2 m 2v 2 + h 2 2 = h 2 c c ⎝ c ⎠ ⇒ v2 =

(b) v =

c 12

⎛ ⎛ λ ⎞2 ⎞ ⎜1 + ⎜ ⎟ ⎜ ⎝ ( h mc) ⎟⎠ ⎟ ⎝ ⎠

h2 ⎛ 2 2 h2 ⎞ ⎜λ m + 2 ⎟ c ⎠ ⎝

=

c c2 ⇒v= . 12 2 2 2 ⎛λ m c ⎞ ⎛ ⎛ mcλ ⎞ 2 ⎞ 1 + ⎜ ⎟ 2 ⎜⎜1 + ⎜ ⎟ ⎟⎟ ⎝ h ⎠ ⎝ ⎝ h ⎠ ⎠

⎛ 1 ⎛ mcλ ⎞2 ⎞ m 2 c 2λ 2 ≈ c ⎜1 − ⎜ = − Δ (1 ) c . Δ = . ⎟ ⎜ 2 ⎝ h ⎟⎠ ⎟ 2h 2 ⎝ ⎠

h . mc (9.11 × 10−31 kg) 2 (3.00 × 108 m s) 2 (1.00 × 10−15m) 2 Δ= = 8.50 × 10−8 2(6.63 × 10−34 J ⋅ s) 2 ⇒ v = (1 − Δ )c = (1 − 8.50 × 10 −8 )c.

(c) λ = 1.00 × 10 −15 m <<

39.47.

(a) Recall λ =

λ=

h h h = = . So for an electron: p 2mE 2mqΔV 6.63 × 10−34 J ⋅ s

2(9.11 × 10

−31

kg)(1.60 × 10

(b) For an alpha particle: λ = 39.48.

−19

C)(125 V)

⇒ λ = 1.10 × 10−10 m.

6.63 × 10 −34 J ⋅ s 2(6.64 × 10

−27

kg)2(1.60 × 10−19 C)(125 V)

IDENTIFY and SET UP: The minimum uncertainty product is Δ xΔpx =

= 9.10 × 10 −13 m.

h . Δ x = r1 , where r1 is the radius of the 2π

h h . and p1 = mv1 = 2π 2π r1 h h 6.63 × 10−34 J ⋅ s = = = 2.0 × 10−24 kg ⋅ m/s . This is the same as the magnitude of EXECUTE: Δpx = 2πΔ x 2π r1 2π (0.529 × 10−10 m) the momentum of the electron in the n = 1 Bohr orbit. EVALUATE: Since the momentum is the same order of magnitude as the uncertainty in the momentum, the uncertainty principle plays a large role in the structure of atoms. hc IDENTIFY and SET UP: Combining the two equations in the hint gives PC = K ( K + 2mc 2 ) and λ = . K ( K + 2mc 2 ) hc h EXECUTE: (a) With K = 3mc 2 this becomes λ = = . 2 2 2 15 mc 3mc (3mc + 2mc ) n = 1 Bohr orbit. In the n = 1 Bohr orbit, mv1r1 =

39.49.

(b) (i) K = 3mc 2 = 3(9.109 × 10 −31 kg)(2.998 × 108 m/s) 2 = 2.456 × 10 −13 J = 1.53 MeV h 6.626 × 10 −34 J ⋅ s = = 6.26 × 10−13 m 15mc 15(9.109 × 10−31 kg)(2.998 × 108 m/s) (ii) K is proportional to m, so for a proton K = (mp / me )(1.53 MeV) = 1836(1.53 MeV) = 2810 MeV

λ=

λ is proportional to 1/m, so for a proton λ = (me / mp )(6.26 × 10 −13 m) = (1/1836)(6.626 × 10−13 m) = 3.41 × 10 −16 m EVALUATE: The proton has a larger rest mass energy so its kinetic energy is larger when K = 3mc 2 . The proton also has larger momentum so has a smaller λ .


39-10

Chapter 39

39.50.

(a)

(6.626 × 10−34 J ⋅ s) = 2.1 × 10−20 kg ⋅ m s. 2π (5.0 × 10−15 m)

(b) K = ( pc ) 2 + (mc 2 ) 2 − mc 2 = 1.3 × 10−13 J = 0.82 MeV.

39.51.

(c) The result of part (b), about 1 MeV = 1 × 106 eV , is many orders of magnitude larger than the potential energy of an electron in a hydrogen atom. (a) IDENTIFY and SET UP: Δ xΔpx ≥ h / 2π

Estimate Δ x as Δ x ≈ 5.0 × 10 −15 m. EXECUTE: Then the minimum allowed Δpx is Δpx ≈

h 6.626 × 10−34 J ⋅ s = = 2.1× 10−20 kg ⋅ m/s 2πΔx 2π (5.0 × 10−15 m)

(b) IDENTIFY and SET UP: Assume p ≈ 2.1 × 10−20 kg ⋅ m/s. Use Eq.(37.39) to calculate E, and then K = E − mc 2 . EXECUTE:

E = (mc 2 ) 2 + ( pc) 2

mc 2 = (9.109 × 10 −31 kg)(2.998 × 108 m/s) 2 = 8.187 × 10−14 J pc = (2.1 × 10−20 kg ⋅ m/s)(2.998 × 108 m/s) = 6.296 × 10−12 J

E = (8.187 × 10−14 J) 2 + (6.296 × 10−12 J) 2 = 6.297 × 10−12 J

39.52.

39.53.

39.54.

K = E − mc 2 = 6.297 × 10−12 J − 8.187 × 10−14 J = 6.215 × 10−12 J(1 eV/1.602 × 10 −19 J) = 39 MeV (c) IDENTIFY and SET UP: The Coulomb potential energy for a pair of point charges is given by Eq.(23.9). The proton has charge +e and the electron has charge –e. ke 2 (8.988 × 109 N ⋅ m 2 / C2 )(1.602 × 10 −19 C) 2 =− = −4.6 × 10−14 J = −0.29 MeV EXECUTE: U = − 5.0 × 10−15 m r EVALUATE: The kinetic energy of the electron required by the uncertainty principle would be much larger than the magnitude of the negative Coulomb potential energy. The total energy of the electron would be large and positive and the electron could not be bound within the nucleus. (a) Take the direction of the electron beam to be the x-direction and the direction of motion perpendicular to the Δp h 6.626 × 10−34 J ⋅ s = = 0.23 m/s beam to be the y-direction. Δv y = y = 2π mΔy 2π (9.11 × 10−31 kg)(0.50 × 10 −3 m) m (b) The uncertainty Δr in the position of the point where the electrons strike the screen is Δp x h x Δr = Δv yt = y = = 9.56 × 10 −10 m, m vx 2πmΔy 2K m

(c) This is far too small to affect the clarity of the picture. h h IDENTIFY and SET UP: ΔE Δt ≥ . Take the minimum uncertainty product, so ΔE = , with 2π 2πΔt ΔE Δt = 8.4 × 10 −17 s . m = 264me . Δm = 2 . c 6.63 × 10 −34 J ⋅ s 1.26 × 10 −18 J EXECUTE: ΔE = = 1.26 × 10−18 J . Δm = = 1.4 × 10−35 kg . −17 2π (8.4 × 10 s) (3.00 × 108 m/s) 2 Δm 1.4 × 10 −35 kg = = 5.8 × 10−8 m (264)(9.11 × 10 −31 kg) IDENTIFY: The insect behaves like a wave as it passes through the hole in the screen. SET UP: (a) For wave behavior to show up, the wavelength of the insect must be of the order of the diameter of the hole. The de Broglie wavelength is λ = h/mv. EXECUTE: The de Broglie wavelength of the insect must be of the order of the diameter of the hole in the screen, so λ ≈ 5.00 mm. The de Broglie wavelength gives

v=

39.55.

h 6.626 × 10−34 J ⋅ s = = 1.33 × 10–25 m/s mλ (1.25 × 10−6 kg ) ( 0.00400 m )

(b) t = x/v = (0.000500 m)/(1.33 × 10–25 m/s) = 3.77 × 1021 s = 1.4 × 1010 yr The universe is about 14 billion years old (1.4 × 1010 yr), so this time would be about 85,000 times the age of the universe. EVALUATE: Don’t expect to see a diffracting insect! Wave behavior of particles occurs only at the very small scale. IDENTIFY and SET UP: Use Eq.(39.1) to relate your wavelength and speed. h h 6.626 × 10−34 J ⋅ s , so v = = = 1.1× 10−35 m/s EXECUTE: (a) λ = mv mλ (60.0 kg)(1.0 m)


The Wave Nature of Particles

39-11

distance 0.80 m = = 7.3 × 1034 s(1 y/3.156 × 107 s) = 2.3 × 1027 y velocity 1.1 × 10−35 m/s Since you walk through doorways much more quickly than this, you will not experience diffraction effects. EVALUATE: A 1 kg object moving at 1 m/s has a de Broglie wavelength λ = 6.6 × 10 −34 m, which is exceedingly small. An object like you has a very, very small λ at ordinary speeds and does not exhibit wavelike properties. hc (a) E = 2.58 eV = 4.13 × 10−19 J, with a wavelength of λ = = 4.82 × 10−7 m = 482 nm E h (6.63 × 10−34 J ⋅ s) = = 6.43 × 10−28 J = 4.02 × 10−9 eV. (b) ΔE = 2π Δt 2π (1.64 × 10−7 s) (c) λ E = hc, so ( Δλ ) E + λΔE = 0, and ΔE E = Δλ λ , so (b) t =

39.56.

39.57.

39.58.

39.59.

⎛ 6.43 × 10−28 J ⎞ −16 −7 Δλ = λ ΔE E = (4.82 × 10 −7 m) ⎜ ⎟ = 7.50 × 10 m = 7.50 × 10 nm. −19 ⎝ 4.13 × 10 J ⎠ IDENTIFY: The electrons behave as waves whose wavelength is equal to the de Broglie wavelength. SET UP: The de Broglie wavelength is λ = h/mv, and the energy of a photon is E = hf = hc/λ. EXECUTE: (a) Use the de Broglie wavelength to find the speed of the electron. h 6.626 × 10−34 J ⋅ s v= = = 7.27 × 105 m/s mλ ( 9.11 × 10 −31 kg )(1.00 × 10 −9 m ) which is much less than the speed of light, so it is nonrelativistic. (b) Energy conservation gives eV = ½ mv2. V = mv2/2e = (9.11 × 10–31 kg)(7.27 × 105 m/s)2/[2(1.60 × 10–19 C)] = 1.51 V (c) K = eV = e(1.51 V) = 1.51 eV, which is about ¼ the potential energy of the NaCl crystal, so the electron would not be too damaging. (d) E = hc/λ = (4.136 × 10–15 eV s)(3.00 × 108 m/s)/(1.00 × 10–9 m) = 1240 eV which would certainly destroy the molecules under study. EVALUATE: As we have seen in Problems 39.10 and 39.43, when a particle and a photon have the same wavelength, the photon has much more energy. h λ′ ⎛ ⎞ sin θ ′ = sin θ , and λ ′ = ( h p′) = (h 2mE ′ ), and so θ ′ = arcsin ⎜ sin θ ⎟ . λ ⎝ λ 2mE ′ ⎠ − 34 ⎛ ⎞ (6.63 × 10 J ⋅ s)sin 35.8° ⎟ = 20.9° θ ′ = arcsin ⎜ ⎜ (3.00 × 10−11 m) 2(9.11 × 10−31 kg)(4.50 × 10+3 )(1.60 × 10−19 J eV) ⎟ ⎝ ⎠ (a) The maxima occur when 2d sin θ = mλ as described in Section 38.7. (b) λ =

(6.63 × 10−34 J ⋅ s) h h = . λ= = 1.46 × 10−10 m = 0.146 nm . −37 −19 p 2mE 2(9.11 × 10 kg)(71.0 eV) (1.60 × 10 J/eV )

⎛ mλ ⎞ θ = sin −1 ⎜ ⎟ (Note: This m is the order of the maximum, not the mass.) ⎝ 2d ⎠ ⎛ (1)(1.46 × 10−10 m) ⎞ ⇒ sin −1 ⎜ ⎟ = 53.3°. −11 ⎝ 2(9.10 × 10 m) ⎠

39.60.

39.61.

(c) The work function of the metal acts like an attractive potential increasing the kinetic energy of incoming electrons by eφ . An increase in kinetic energy is an increase in momentum that leads to a smaller wavelength. A smaller wavelength gives a smaller angle θ (see part (b)). 1 (a) Using the given approximation, E = ( (h x) 2 m + kx 2 ) , ( dE dx) = kx − ( h 2 mx 3 ), and the minimum energy 2 h . The minimum energy is then h k m. occurs when kx = ( h 2 mx3 ), or x 2 = mk (b) They are the same. (a) IDENTIFY and SET UP: U = A x . Eq.(7.17) relates force and potential. The slope of the function A x is not

continuous at x = 0 so we must consider the regions x > 0 and x < 0 separately. d ( Ax) EXECUTE: For x > 0, x = x so U = Ax and F = − = − A. For x < 0, x = − x so U = − Ax and dx d (− Ax) F =− = + A. We can write this result as F = − A x / x, valid for all x except for x = 0. dx


39-12

Chapter 39

(b) IDENTIFY and SET UP: Use the uncertainty principle, expressed as ΔpΔ x ≈ h, and as in Problem 39.50 estimate Δp by p and Δ x by x. Use this to write the energy E of the particle as a function of x. Find the value of x that gives the minimum E and then find the minimum E. p2 EXECUTE: E = K + U = +Ax 2m px ≈ h, so p ≈ h / x

h2 +Ax. 2mx 2 h2 + Ax. For x > 0, E = 2mx 2 Then E ≈

To find the value of x that gives minimum E set 0=

dE = 0. dx

−2h 2 +A 2mx 3 1/ 3

⎛ h2 ⎞ h2 x = and x = ⎜ ⎟ mA ⎝ mA ⎠ With this x the minimum E is 3

h2 ⎛ mA ⎞ E= ⎜ ⎟ 2m ⎝ h 2 ⎠

2/3

1/ 3

⎛ h2 ⎞ + A⎜ ⎟ ⎝ mA ⎠

1 = h 2 / 3m −1/ 3 A2 / 3 + h 2 / 3m −1/ 3 A2 / 3 2

1/ 3

39.62.

⎛ h 2 A2 ⎞ E = 32 ⎜ ⎟ ⎝ m ⎠ EVALUATE: The potential well is shaped like a V. The larger A is the steeper the slope of U and the smaller the region to which the particle is confined and the greater is its energy. Note that for the x that minimizes E, 2K = U. For this wave function, Ψ ∗ = ψ1∗eiω1t + ψ 2∗eiω2 t , so Ψ 2 = Ψ ∗Ψ = (ψ1∗eiω1t + ψ2∗eiω2 t )(ψ1e −iω1t + ψ 2e −iω2 t ) = ψ1∗ψ1 + ψ 2∗ψ2 + ψ1*ψ 2ei (ω1 − ω2 )t + ψ 2∗ψ1ei (ω2 − ω1 )t .

The frequencies ω1 and ω2 are given as not being the same, so Ψ 39.63.

2

is not time-independent, and Ψ is not the

wave function for a stationary state. The time-dependent equation, with the separated form for Ψ ( x, t ) as given becomes ⎛ = 2 d 2ψ ⎞ i=ψ (−iω) = ⎜ − + U ( x)ψ ⎟ . 2 2 m dx ⎝ ⎠

Since ψ is a solution of the time-independent solution with energy E , the term in parenthesis is Eψ , and so ω= = E , and ω = ( E =). 39.64.

p 2 (=k ) 2 =k 2 = ⇒ω= . 2m 2m 2m = 2 ∂ 2ψ ( x, t ) (b) From Problem 39.63 the time-dependent Schrödinger’s equation is − + 2m ∂x 2 ∂ψ ( x, t ) ∂ 2ψ ( x, t ) 2mi ∂ψ ( x, t ) U ( x )ψ ( x, t ) = i= . U ( x) = 0 for a free particle, so . =− ∂t ∂x 2 ∂t = Try ψ ( x, t ) = cos( kx − ωt ) : (a) ω = 2π f =

2π E E 2 π 2π p p= . = . k= = h h = λ =

=ω = E = K =

∂ψ ( x, t ) = Aω sin(kx − ωt ) ∂t ∂ψ ( x, t ) ∂ 2ψ = − Ak sin(kx − ωt ) and 2 = Ak 2 cos( kx − ωt ). ∂x ∂x ⎛ 2mi ⎞ Putting this into the Schrödinger’s equation, Ak 2 cos(kx − ωt ) = − ⎜ ⎟ Aω sin(kx − ωt ). ⎝ = ⎠ This is not generally true for all x and t so is not a solution.


The Wave Nature of Particles

39-13

(c) Try ψ ( x, t ) = A sin( kx − ωt ) :

∂ψ ( x, t ) = − Aω cos( kx − ωt ) ∂t ∂ψ ( x, t ) ∂ 2ψ ( x, t ) = Ak cos(kx − ωt ) and = − Ak 2 sin( kx − ωt ). ∂x ∂x 2 ⎛ 2mi ⎞ Again, − Ak 2 sin( kx − ωt ) = − ⎜ ⎟ Aω cos(kx − ωt ) is not generally true for all x and t so is not a solution. ⎝ = ⎠ (d) Try ψ ( x, t ) = A cos( kx − ωt ) + B sin(kx − ωt ) :

39.65.

∂ψ ( x, t ) = + Aω sin( kx − ωt ) − Bω cos(kx − ωt ) ∂t ∂ψ ( x, t ) ∂ 2ψ ( x, t ) = − Ak sin(kx − ωt ) + Bk cos(kx − ωt ) and = − Ak 2 cos((kx − ωt ) − Bk 2 sin( kx − ωt ). ∂x ∂x 2 Putting this into the Schrödinger’s equation, 2mi − Ak 2 cos(kx − ωt ) − Bk 2 sin( kx − ωt ) = − ( + Aω sin(kx − ωt ) − Bω cos(kx − ωt )). = =k 2 . Collect sin and cos terms. Recall that ω = 2m ( A + iB ) k 2 cos( kx − ωt ) + (iA − B) k 2 sin (kx − ωt ) = 0. This is only true if B = iA. (a) IDENTIFY and SET UP: Let the y-direction be from the thrower to the catcher, and let the x-direction be horizontal and perpendicular to the y-direction. A cube with volume V = 125 cm3 = 0.125 × 10−3 m3 has side length l = V 1/ 3 = (0.125 × 10−3 m3 )1/ 3 = 0.050 m. Thus estimate Δ x as Δ x ≈ 0.050 m. Use the uncertainty principle to estimate Δpx .

h 0.0663 J ⋅ s = = 0.21 kg ⋅ m/s 2πΔ x 2π (0.050 m) (The value of h in this other universe has been used.) (b) IDENTIFY and SET UP: Δ x = (Δvx )t is the uncertainty in the x-coordinate of the ball when it reaches the EXECUTE:

Δ xΔpx ≥ h / 2π then gives Δpx ≈

catcher, where t is the time it takes the ball to reach the second student. Obtain Δvx from Δ px . EXECUTE: The uncertainty in the ball’s horizontal velocity is Δvx =

Δpx 0.21 kg ⋅ m/s = = 0.84 m/s 0.25 kg m

12 m = 2.0 s. The uncertainty in the x-coordinate 6.0 m/s of the ball when it reaches the second student that is introduced by Δvx is Δ x = (Δvx )t = (0.84 m/s)(2.0 s) = 1.7 m. The ball could miss the second student by about 1.7 m. EVALUATE: A game of catch would be very different in this universe. We don’t notice the effects of the uncertainty principle in everyday life because h is so small. The time it takes the ball to travel to the second student is t =

39.66.

(a) ψ 2 = A2 x 2e−2( αx

2

+ βy 2 + γz 2 )

. To save some algebra, let u = x 2 , so that ψ = ue−2α u f ( y, z ) . 2

∂ 2 1 1 2 , x0 = ± . ψ = (1 − 2α u ) ψ ; the maximum occurs at u0 = ∂u 2α 2α (b) ψ vanishes at x = 0, so the probability of finding the particle in the x = 0 plane is zero. The wave function vanishes for x = ±∞. 39.67.

(a) IDENTIFY and SET UP: The probability is P = ψ dV with dV = 4π r 2 dr 2

EXECUTE:

ψ = A2e −2α r so P = 4π A2 r 2e −2α r dr 2

2

2

(b) IDENTIFY and SET UP: P is maximum where EXECUTE:

dP =0 dr

d 2 −2α r 2 (r e )=0 dr

2re −2α r − 4α r 3e −2α r = 0 and this reduces to 2r − 4α r 3 = 0 r = 0 is a solution of the equation but corresponds to a minimum not a maximum. Seek r not equal to 0 so divide by r and get 2 − 4α r 2 = 0 2

2


39-14

Chapter 39

1 (We took the positive square root since r must be positive.) 2α

This gives r =

EVALUATE: This is different from the value of r, r = 0, where ψ

2

is a maximum. At r = 0, ψ

2

has a

maximum but the volume element dV = 4π r dr is zero here so P does not have a maximum at r = 0. 2

39.68.

(a) B (k ) = e −α

2 2

k

B (0) = Bmax = 1 B (kh ) =

2 2 1 1 = e −α kh ⇒ ln(1 2) = −α 2 kh2 ⇒ kh = ln(2) = ωk . 2 α

(b) Using integral tables: ψ ( x) = ∫ e −α 0

2 2

k

cos kxdk =

π − x2 / 4α 2 (e ). ψ ( x) is a maximum when x = 0. 2α

1 − x2 ⇒ h2 = ln(1/2) ⇒ xh = 2α ln2 = ωx 4α 2 4α h ω h 1 h h ln 2 ⎛ ⎞ ⎛ ⎞ (d) ω p ωx = ⎜ k ⎟ ωx = ln2 ⎟ 2α ln2 = (2ln2) = . ⎜ π 2π ⎝ α 2π ⎝ 2π ⎠ ⎠ (c) ψ ( xh ) =

π

when e − xh / 4α = 2

2

(

39.69.

)

k0 ⎛ 1 ⎞ ∞ sin kx (a) ψ ( x) = ∫ B ( k )cos kxdk = ∫ ⎜ ⎟ cos kxdk = 0 0 k0 x ⎝ k0 ⎠

k0

= 0

sin k0 x k0 x

(b) ψ ( x) has a maximum value at the origin x = 0. ψ ( x0 ) = 0 when k0 x0 = π so x0 =

π . Thus the width of this k0

2π 2π . If k0 = , wx = L. B( k ) versus k is graphed in Figure 39.69a. The graph of ψ ( x) versus L k0 x is in Figure 39.69b.

function wx = 2 x0 =

(c) If k0 =

π L

wx = 2 L.

hk h ⎛ hw ⎞ ⎛ 2π ⎞ hw (d) wp wx = ⎜ k ⎟ ⎜ ⎟ = k = 0 = h. The uncertainty principle states that wp wx ≥ . For us, no matter what 2 2 π k k k π ⎝ ⎠⎝ 0 ⎠ 0 0 h k0 is, wp wx = h, which is greater than . 2π

Figure 39.69 39.70.

p 2 (h λ )2 n 2h 2 = = . 2m 2m 8mL2 m, E1 = 2.15 × 10−17 J = 134 eV.

(a) For a standing wave, nλ = 2 L, and En = (b) With L = a0 = 0.5292 × 10−10


The Wave Nature of Particles

39.71.

Time of flight of the marble, from a free-fall kinematic equation is just t =

2y 2(25.0 m) = = 2.26 s . 9.81 m s 2 g

ht ⎛ Δp ⎞ Δ x f = Δ xi + ( Δvx )t = Δ xi + ⎜ x ⎟ t = + Δ xi 2πΔ xi m ⎝ m ⎠ d (Δ x f ) − ht =0= +1 To minimize Δ x f with respect to Δ xi , 2πm( Δ xi ) 2 d ( Δ xi ) ⎛ ht ⎞ ⇒ Δxi (min) = ⎜ ⎟ ⎝ 2πm ⎠ ⇒ Δx f (min) =

ht ht 2ht 2(6.63 × 10−34 J ⋅ s)(2.26 s) + = = = 2.18 × 10−16 m = 2.18 × 10−7 nm. 2πm 2πm πm π (0.0200 kg)

39-15


40

QUANTUM MECHANICS

40.1.

IDENTIFY and SET UP: The energy levels for a particle in a box are given by En = EXECUTE: (a) The lowest level is for n = 1, and E1 =

n2h 2 . 8mL2

(1)(6.626 × 10−34 J ⋅ s) 2 = 1.2 × 10−67 J. 8(0.20 kg)(1.5 m) 2

1 2E 2(1.2 × 10−67 J) (b) E = mv 2 so v = = = 1.1 × 10−33 m/s. If the ball has this speed the time it would take it 2 m 0.20 kg to travel from one side of the table to the other is t =

1.5 m = 1.4 × 1033 s. 1.1× 10−33 m/s

h2 , E2 = 4 E1 , so ΔE = E2 − E1 = 3E1 = 3(1.2 × 10−67 J) = 3.6 × 10 −67 J 8mL2 (d) EVALUATE: No, quantum mechanical effects are not important for the game of billiards. The discrete, quantized nature of the energy levels is completely unobservable. h L= 8mE1 (c) E1 =

40.2.

L=

40.3.

8(1.673 × 10

−27

kg)(5.0 × 10 eV)(1.602 × 10 6

−19

J eV )

= 6.4 × 10−15 m.

IDENTIFY: An electron in the lowest energy state in this box must have the same energy as it would in the ground state of hydrogen. nh 2 SET UP: The energy of the nth level of an electron in a box is En = . 8mL2 EXECUTE: An electron in the ground state of hydrogen has an energy of −13.6 eV, so find the width corresponding to an energy of E1 = 13.6 eV. Solving for L gives L=

40.4.

(6.626 × 10−34 J ⋅ s)

h (6.626 × 10−34 J ⋅ s) = = 1.66 × 10 −10 m. 8mE1 8(9.11 × 10−31 kg)(13.6 eV)(1.602 × 10−19 J eV )

EVALUATE: This width is of the same order of magnitude as the diameter of a Bohr atom with the electron in the K shell. (a) The energy of the given photon is E = hf = h

c

λ

= (6.63 × 10−34 J ⋅ s)

(3.00 × 103 m/s) = 1.63 × 10−18 J. (122 × 10−9 m)

The energy levels of a particle in a box are given by Eq.40.9 h2 h 2 ( n12 − n2 ) (6.63 × 10−34 J ⋅ s) 2 (2 2 − 12 ) ( n 2 − n2 ). L = = = 3.33 × 10 −10 m. 2 8mL 8mΔ E 8(9.11 × 10−31 kg)(1.63 × 10−20 J) (b) The ground state energy for an electron in a box of the calculated dimensions is h2 (6.63 × 10−34 J ⋅ s) 2 E= = = 5.43 × 10−19 J = 3.40 eV (one-third of the original photon energy), 2 8mL 8(9.11 × 10−31 kg)(3.33 × 10 −10 m) 2 which does not correspond to the −13.6 eV ground state energy of the hydrogen atom. Note that the energy levels for a particle in a box are proportional to n 2 , whereas the energy levels for the hydrogen atom are proportional to − 12 . n

ΔE =

40-1


40-2

40.5.

Chapter 40

IDENTIFY and SET UP: Eq.(40.9) gives the energy levels. Use this to obtain an expression for E2 − E1 and use the value given for this energy difference to solve for L. 4h 2 h2 . The energy separation EXECUTE: Ground state energy is E1 = ; first excited state energy is E2 = 2 8mL 8mL2 3h2 3 = . This gives L = h between these two levels is Δ E = E2 − E1 = 8mΔ E 8mL2 3 = 6.1 × 10−10 m = 0.61 nm. 8(9.109 × 10 −31 kg)(3.0 eV)(1.602 × 10−19 J/1 eV) EVALUATE: This energy difference is typical for an atom and L is comparable to the size of an atom. (a) The wave function for n = 1 vanishes only at x = 0 and x = L in the range 0 ≤ x ≤ L. (b) In the range for x, the sine term is a maximum only at the middle of the box, x = L / 2. (c) The answers to parts (a) and (b) are consistent with the figure. IDENTIFY and SET UP: For the n = 2 first excited state the normalized wave function is given by Eq.(40.13). 2 2 ⎛ 2π x ⎞ 2 2 2 ⎛ 2π x ⎞ sin ⎜ ψ 2 ( x) = ⎟ . ψ 2 ( x) dx = sin ⎜ ⎟ dx. Examine ψ 2 ( x) dx and find where it is zero and where it is L L L L ⎝ ⎠ ⎝ ⎠ maximum. 2 ⎛ 2π x ⎞ EXECUTE: (a) ψ 2 dx = 0 implies sin ⎜ ⎟=0 ⎝ L ⎠ 2π x = mπ , m = 0, 1, 2, . . . ; x = m( L/2) L For m = 0, x = 0; for m = 1, x = L/2; for m = 2, x = L The probability of finding the particle is zero at x = 0, L/2, and L. 2 ⎛ 2π x ⎞ (b) ψ 2 dx is maximum when sin ⎜ ⎟ = ±1 ⎝ L ⎠ 2π x = m(π /2), m = 1, 3, 5, . . . ; x = m( L/4) L For m = 1, x = L/4; for m = 3, x = 3L/4 The probability of finding the particle is largest at x = L/4 and 3L/4. L = 6.626 × 10 −34 J ⋅ s

40.6.

40.7.

(c) EVALUATE:

The answers to part (a) correspond to the zeros of ψ

answers to part (b) correspond to the two values of x where ψ

2

2

shown in Fig.40.5 in the textbook and the

in the figure is maximum.

dψ 8π 2 m 2m 2 2 = − k ψ , and for ψ to be a solution of Eq.(40.3), k = E =E 2 . 2 2 dx h = (b) The wave function must vanish at the rigid walls; the given function will vanish at x = 0 for any k , but to vanish at x = L, kL = nπ for integer n. 2

40.8.

40.9.

(a) IDENTIFY and SET UP: satisfied. EXECUTE: Eq.(40.3): −

ψ = A cos kx. Calculate dψ 2 /dx 2 and substitute into Eq.(40.3) to see if this equation is

h2 d 2ψ = Eψ 8π 2 m dx 2

dψ = A( −k sin kx) = − Ak sin kx dx 2 dψ = − Ak ( k cos kx) = − Ak 2 cos kx dx 2 h2 Thus Eq.(40.3) requires − 2 (− Ak 2 cos kx ) = E ( A cos kx). 8π m 2 2 hk 2mE 2mE = This says − 2 = E ; k = 8π m ( h/2π ) = 2mE . = (b) EVALUATE: The wave function for a particle in a box with rigid walls at x = 0 and x = L must satisfy the boundary conditions ψ = 0 at x = 0 and ψ = 0 at x = L. ψ (0) = A cos0 = A, since cos 0 = 1. Thus ψ is not 0 at x = 0 and this wave function isn't acceptable because it doesn't satisfy the required boundary condition, even though it is a solution to the Schrödinger equation.

ψ = A cos kx is a solution to Eq.(40.3) if k =


Quantum Mechanics

40.10.

(a) The third excited state is n = 4, so

Δ E = (42 − 1) (b) λ = 40.11.

40-3

h2 15(6.626 × 10 −34 J ⋅ s) 2 = = 5.78 × 10 −17 J = 361 eV. 2 8mL 8(9.11 × 10 −31 kg)(0.125 × 10 −9 m) 2

hc (6.63 × 10−34 J ⋅ s)(3.0 × 108 m/s) = = 3.44 nm ΔE 5.78 × 10−17 J

Recall λ = (a) E1 =

h h = . p 2mE

h2 h ⇒ λ1 = = 2 L = 2(3.0 × 10−10 m) = 6.0 × 10 −10 m. The wavelength is twice the width of 2 8mL 2mh 2 /8mL2

the box. p1 =

h

λ1

=

(6.63 × 10−34 J ⋅ s) = 1.1 × 10−24 kg ⋅ m/s 6.0 × 10−10 m

2

(b) E2 =

p2 =

h

λ2

4h ⇒ λ2 = L = 3.0 × 10 −10 m. The wavelength is the same as the width of the box. 8mL2 = 2 p1 = 2.2 × 10−24 kg ⋅ m/s.

9h 2 2 ⇒ λ3 = L = 2.0 × 10−10 m. The wavelength is two-thirds the width of the box. 8mL2 3 −24 p3 = 3 p1 = 3.3 × 10 kg ⋅ m/s. IDENTIFY: If the given wave function is a solution to the Schrödinger equation, we will get an identity when we substitute that wave function into the Schrödinger equation.

(c) E3 =

40.12.

SET UP: We must substitute the equation Ψ ( x, t ) =

equation −

2 ⎛ nπ x ⎞ −iEn t/= sin ⎜ into the one-dimensional Schrödinger ⎟e L ⎝ L ⎠

= 2 d 2ψ ( x) + U ( x)ψ ( x) = Eψ ( x). 2m dx 2 2

EXECUTE: Taking the second derivative of Ψ ( x, t ) with respect to x gives

d 2 Ψ (x, t ) ⎛ nπ ⎞ = −⎜ ⎟ Ψ (x, t ) dx 2 ⎝ L ⎠

= 2 d 2ψ (x) = 2 ⎛ nπ ⎞ + = we get U ( x ) ψ ( x ) E ψ ( x ), ⎜ ⎟ Ψ (x, t ) = E Ψ (x, t ) which 2m dx 2 2m ⎝ L ⎠ 2

Substituting this result into − 2

= 2 ⎛ nπ ⎞ ⎜ ⎟ , the energies of a particle in a box. 2m ⎝ L ⎠ EVALUATE: Since this process gives us the energies of a particle in a box, the given wave function is a solution to the Schrödinger equation. −= 2 d 2ψ (a) Eq.(40.1): + Uψ = Eψ. 2m dx 2 ⎛ = 2k 2 ⎞ −= 2 d 2 = 2k 2 ( sin ) sin sin sin Left-hand side: A kx + U A kx = A kx + U A kx = + U 0 ⎟ ψ. ⎜ 0 0 2 2m dx 2m ⎝ 2m ⎠ gives En =

40.13.

= 2k 2 = 2k 2 + U 0 > U 0 > E for constant k . But + U 0 should equal E ⇒ no solution. 2m 2m = 2k 2 (b) If E > U 0 , then + U 0 = E is consistent and so ψ = A sin kx is a solution of Eq.(40.1) for this case. 2m According to Eq.(40.17), the wavelength of the electron inside of the square well is given by h 2mE k= . By an analysis similar to that used to derive Eq.40.17, we can show that outside ⇒ λin = = 2m(3U 0 )

But

40.14.

the box

λout =

Thus, the ratio of the wavelengths is

h h = . 2 m( E − U 0 ) 2m(2U 0 )

2m(3U 0 ) λout 3 = = . 2 λin 2m(2U 0 )


40-4

Chapter 40

40.15.

E1 = 0.625 E∞ = 0.625

π 2= 2 ; E1 = 2.00 eV = 3.20 × 10−19 J 2mL2 1/ 2

40.16.

40.17.

⎛ ⎞ 0.625 −10 L = π= ⎜ ⎟ = 3.43 × 10 m −31 −19 2(9.109 10 kg)(3.20 10 J) × × ⎝ ⎠ Since U 0 = 6 E∞ we can use the result E1 = 0.625 E∞ from Section 40.3, so U 0 − E1 = 5.375 E∞ and the maximum wavelength of the photon would be

Eq.(40.16): ψ = Asin

λ=

hc hc 8mL2c = = 2 2 U 0 − E1 (5.375)(h /8mL ) (5.375) h

λ=

8(9.11 × 10−31 kg)(1.50 × 10−9 m) 2 (3.00 × 108 m/s) = 1.38 × 10−6 m. (5.375)(6.63 × 10−34 J ⋅ s)

2mE 2mE x + B cos x = =

d 2ψ 2mE 2mE −2mE ⎛ 2mE ⎞ ⎛ 2mE ⎞ = − A ⎜ 2 ⎟ sin x − B ⎜ 2 ⎟ cos x= (ψ ) = Eq.(40.15). 2 dx = = =2 ⎝ = ⎠ ⎝ = ⎠ 40.18.

40.19.

dψ d 2ψ = κ (Ceκ x − De −κ x ), = κ 2 (Ceκ x + De −κ x ) = κ 2ψ for all constants C and D. Hence ψ is a solution to dx dx 2 =2 2 Eq.(40.1) for − κ + U 0 = E , or κ = [2m(U 0 − E )]1/ 2 =, and κ is real for E < U 0 . 2m IDENTIFY: Find the transition energy Δ E and set it equal to the energy of the absorbed photon. Use E = hc/λ to find the wavelength of the photon. π 2= 2 SET UP: U 0 = 6 E∞ , as in Fig.40.8 in the textbook, so E1 = 0.625 E∞ and E3 = 5.09 E∞ with E∞ = . In this 2mL2 problem the particle bound in the well is a proton, so m = 1.673 × 10 −27 kg. EXECUTE:

E∞ =

π 2= 2 2

2mL

=

π 2 (1.055 × 10−34 J ⋅ s) 2 2(1.673 × 10−27 kg)(4.0 × 10−15 m) 2

= 2.052 × 10−12 J. The transition energy is

Δ E = E3 − E1 = (5.09 − 0.625) E∞ = 4.465 E∞ . Δ E = 4.465(2.052 × 10 −12 J) = 9.162 × 10−12 J The wavelength of the photon that is absorbed is related to the transition energy by Δ E = hc/λ , so hc (6.626 × 10−34 J ⋅ s)(2.998 × 108 m/s) = = 2.2 × 10−14 m = 22 fm. ΔE 9.162 × 10−12 J EVALUATE: The wavelength of the photon is comparable to the size of the box. IDENTIFY: The longest wavelength corresponds to the smallest energy change. h2 SET UP: The ground level energy level of the infinite well is E∞ = , and the energy of the photon must be 8mL2 equal to the energy difference between the two shells. EXECUTE: The 400.0 nm photon must correspond to the n = 1 to n = 2 transition. Since U 0 = 6 E∞ , we have

λ=

40.20.

E2 = 2.43E∞ and E1 = 0.625 E∞ . The energy of the photon is equal to the energy difference between the two levels, and E∞ =

h2 hc 1.805 h 2 , which gives = − ⇒ = − = E E E (2.43 0.625) E 2 1 γ ∞ 8mL2 8mL2 λ

(1.805) hλ (1.805)(6.626 × 10 −34 J ⋅ s)(4.00 × 10−7 m) = = 4.68 × 10 −10 m = 0.468 nm. 8mc 8(9.11 × 10 −31 kg)(3.00 × 108 m s) EVALUATE: This width is approximately half that of a Bohr hydrogen atom. E⎛ E ⎞ −2 L 2 m (U 0 − E )/= E 6.0 eV = and E − U 0 = 5 eV = 8.0 × 10−19 J. T = 16 ⎜1 − . ⎟e U0 ⎝ U0 ⎠ U 0 11.0 eV

Solving for L gives L =

40.21.

⎛ 6.0 eV ⎞ ⎛ 6.0 ev ⎞ −2(0.80×10−9 m) (a) L = 0.80 × 10−9 m: T = 16 ⎜ ⎟ ⎜1 − ⎟e 11.0 eV 11.0 eV ⎠ ⎝ ⎠⎝ (b) L = 0.40 × 10−9 m: T = 4.2 × 10−4.

2(9.11×10−31 kg)(8.0×10−19 J) /1.055×10−34 J ⋅s

= 4.4 × 10 −8


Quantum Mechanics

40.22.

The transmission coefficient is T = 16

E⎛ E ⎞ −2 ⎜1 − ⎟e U0 ⎝ U0 ⎠

2 m (U 0 − E ) L/=

40-5

, with E = 5.0 eV, L = 0.60 × 10−9 m, and

m = 9.11 × 10 −31 kg

(a) U 0 = 7.0 eV ⇒ T = 5.5 × 10 −4 . (b) U 0 = 9.0 eV ⇒ T = 1.8 × 10−5 (c) U 0 = 13.0 eV ⇒ T = 1.1 × 10−7. 40.23.

IDENTIFY and SET UP: Use Eq.(39.1), where K = p 2 /2m and E = K + U .

λ = h/p = h/ 2mK , so λ K is constant

EXECUTE:

λ1 K1 = λ2 K 2 ; λ1 and K1 are for x > L where K1 = 2U 0 and λ2 and K 2 are for 0 < x < L where K2 = E − U0 = U 0

λ1 K2 U0 1 = = = K1 2U 0 λ2 2 40.24.

EVALUATE: When the particle is passing over the barrier its kinetic energy is less and its wavelength is larger. IDENTIFY: The probability of tunneling depends on the energy of the particle and the width of the barrier. E⎛ E ⎞ SET UP: The probability of tunneling is approximately T = Ge−2κ L , where G = 16 ⎜ 1 − ⎟ and U0 ⎝ U0 ⎠

κ=

2 m (U 0 − E ) =

EXECUTE:

κ=

.

G = 16

E⎛ E ⎞ 50.0 eV ⎛ 50.0 eV ⎞ ⎜1 − ⎟ = 16 ⎜1 − ⎟ = 3.27. U0 ⎝ U0 ⎠ 70.0 eV ⎝ 70.0 eV ⎠

2m(U 0 − E ) 2(1.67 × 10 −27 kg)(70.0 eV − 50.0 eV)(1.60 × 10 −19 J/eV) = = 9.8 × 1011 m −1 (6.63 × 10−34 J ⋅ s) 2π =

1 1 ⎛ 3.27 ⎞ −12 ln(G / T ) = ln ⎜ ⎟ = 3.6 × 10 m = 3.6 pm 2κ 2(9.8 × 1011 m −1 ) ⎝ 0.0030 ⎠ If the proton were replaced with an electron, the electron’s mass is much smaller so L would be larger. EVALUATE: An electron can tunnel through a much wider barrier than a proton of the same energy. Solving T = Ge−2κ L for L gives L =

40.25.

IDENTIFY and SET UP: The probability is T = Ae−2κ L , with A = 16

2m(U 0 − E ) E⎛ E ⎞ . ⎜1 − ⎟ and κ = U0 ⎝ U0 ⎠ =

E = 32 eV, U 0 = 41 eV, L = 0.25 × 10 −9 m. Calculate T.

EXECUTE: (a) A = 16

E⎛ E ⎞ 32 ⎛ 32 ⎞ ⎜1 − ⎟ = 16 ⎜ 1 − ⎟ = 2.741. U0 ⎝ U0 ⎠ 41 ⎝ 41 ⎠

2m(U 0 − E ) = 2(9.109 × 10−31 kg)(41 eV − 32 eV)(1.602 × 10 −19 J/eV) = 1.536 × 1010 m −1 κ= 1.055 × 10−34 J ⋅ s

κ=

−1

−9

T = Ae−2κ L = (2.741)e −2(1.536×10 m )(0.25×10 m) = 2.741e −7.68 = 0.0013 (b) The only change in the mass m, which appears in κ . 2m(U 0 − E ) κ= = 2(1.673 × 10 −27 kg)(41 eV − 32 eV)(1.602 × 10−19 J/eV) = 6.584 × 1011 m −1 κ= 1.055 × 10−34 J ⋅ s 11 -1 −9 Then T = Ae−2κ L = (2.741)e −2(6.584×10 m )(0.25×10 m) = 2.741e −392.2 = 10−143 EVALUATE: The more massive proton has a much smaller probability of tunneling than the electron does. 10


40-6

Chapter 40

40.26.

T = Ge −2κ L with G = 16

−2 2m(U 0 − E ) E⎛ E ⎞ E⎛ E ⎞ , so T = 16 ⎜ 1 − ⎜1 − ⎟ and κ = ⎟e U0 ⎝ U0 ⎠ = U0 ⎝ U0 ⎠

2 m (U 0 − E ) =

L

.

(a) If U 0 = 30.0 × 106 eV, L = 2.0 × 10 −15 m, m = 6.64 × 10 −27 kg and U 0 − E = 1.0 × 106 eV (E = 29.0 × 106 eV), T = 0.090.

(b) If U 0 − E = 10.0 × 106 eV (E = 20.0 × 106 eV), T = 0.014. 40.27.

IDENTIFY and SET UP: The energy levels are given by Eq.(40.26), where ω =

k′ 110 N/m = = 21.0 rad/s m 0.250 kg The ground state energy is given by Eq.(40.26): 1 1 E0 = =ω = (1.055 × 10−34 J ⋅ s)(21.0 rad/s) = 1.11 × 10 −33 J(1 eV/1.602 × 10 −19 J) = 6.93 × 10−15 eV 2 2 1⎞ 1⎞ ⎛ ⎛ En = ⎜ n + ⎟ =ω ; E( n +1) = ⎜ n + 1 + ⎟ =ω 2 2⎠ ⎝ ⎠ ⎝ The energy separation between these adjacent levels is ΔE = En +1 − En = =ω = 2 E0 = 2(1.11 × 10−33 J) = 2.22 × 10−33 J = 1.39 × 10−14 eV EVALUATE: These energies are extremely small; quantum effects are not important for this oscillator. dψ d 2ψ = (4 x 2δ 2 − 2δ)ψ , and ψ is a solution of Eq.(40.21) if Let mk ′ 2= = δ , and so = −2 xδ ψ and dx dx 2 =2 1 1 E = δ = = k ′/m = =ω. 2 2 m IDENTIFY: We can model the molecule as a harmonic oscillator. The energy of the photon is equal to the energy difference between the two levels of the oscillator. SET UP: The energy of a photon is Eγ = hf = hc/λ , and the energy levels of a harmonic oscillator are given by EXECUTE:

40.28.

40.29.

k′ . m

ω=

1 ⎞ k′ ⎛ 1⎞ ⎛ En = ⎜ n + ⎟ = = ⎜ n + ⎟ =ω . 2⎠ m ⎝ 2⎠ ⎝ (6.63 × 10−34 J ⋅ s)(3.00 × 108 m/s) = 0.21 eV λ 5.8 × 10−6 m k′ 2π =c k′ (b) The transition energy is Δ E = En +1 − En = =ω = = , which gives . Solving for k ′, we get == λ m m 4π 2c 2 m 4π 2 (3.00 × 108 m s ) 2 (5.6 × 10−26 kg) = = 5,900 N/m. k′ = λ2 (5.8 × 10−6 m) 2 EVALUATE: This would be a rather strong spring in the physics lab. According to Eq.(40.26), the energy released during the transition between two adjacent levels is twice the ground state energy E3 − E2 = =ω = 2 E0 = 11.2 eV. For a photon of energy E c hc (6.63 × 10−34 J ⋅ s)(3.00 × 108 m s ) E = hf ⇒ λ = = = = 111 nm. f E (11.2 eV)(1.60 × 10−19 J/eV) EXECUTE: (a) The photon’s energy is Eγ =

40.30.

40.31.

hc

=

IDENTIFY and SET UP: Use the energies given in Eq.(40.26) to solve for the amplitude A and maximum speed vmax of the oscillator. Use these to estimate Δ x and Δpx and compute the uncertainty product Δ xΔpx . EXECUTE: The total energy of a Newtonian oscillator is given by E = 12 k ′A2 where k ′ is the force constant and A

is the amplitude of the oscillator. Set this equal to the energy E = (n + 12 ) =ω of an excited level that has quantum number n, where ω =

k′ , and solve for A: m

1 2

k ′A2 = ( n + 12 ) =ω

(2n + 1)=ω k′ 2 . Set this equal to E = (n + 12 ) =ω and The total energy of the Newtonian oscillator can also be written as E = 12 mvmax A=

solve for vmax : vmax =

1 2

2 mvmax = ( n + 12 ) =ω

(2n + 1)=ω m


Quantum Mechanics

40-7

Thus the maximum linear momentum of the oscillator is pmax = mvmax = (2n + 1)=mω . Assume that A represents the uncertainty Δ x in position and that pmax is the corresponding uncertainty Δ px in momentum. Then the (2n + 1)=ω m ⎛1⎞ (2n + 1)=mω = (2n + 1)=ω = (2n + 1)=ω ⎜ ⎟ = (2n + 1)=. k′ k′ ⎝ω ⎠ EVALUATE: For n = 1 this gives Δ xΔ px = 3=, in agreement with the result derived in Section 40.4. The uncertainty product Δ xΔ px increases with n. uncertainty product is Δ xΔ px =

40.32.

⎛ ω⎞ mk ′ 2 ⎞ ⎛ A ⎟⎟ = exp ⎜ − mk ′ ⎟ = e −1 = 0.368. = exp ⎜⎜ − = k′ ⎠ ψ (0) ⎝ ⎝ ⎠ This is consistent with what is shown in Figure 40.20 in the textbook. (a)

2

2

⎛ ⎞ mk ′ ω⎞ ⎛ = exp ⎜⎜ − (2 A) 2 ⎟⎟ = exp ⎜ − mk ′ 4 ⎟ = e −4 = 1.83 × 10−2. = k′ ⎠ ⎝ ψ (0) ⎝ ⎠ This figure cannot be read this precisely, but the qualitative decrease in amplitude with distance is clear. IDENTIFY: We model the atomic vibration in the crystal as a harmonic oscillator. 1 ⎞ k′ ⎛ 1⎞ ⎛ SET UP: The energy levels of a harmonic oscillator are given by En = ⎜ n + ⎟ = = ⎜ n + ⎟ =ω . 2 m 2⎠ ⎝ ⎠ ⎝ EXECUTE: (a) The ground state energy of a simple harmonic oscillator is (b)

40.33.

ψ ( A)

ψ (2 A) 2 2

1 1 k ′ (1.055 × 10 −34 J ⋅ s) 12.2 N/m E0 = =ω = = = = 9.43 × 10−22 J = 5.89 × 10−3 eV 2 2 m 2 3.82 × 10−26 kg (b) E4 − E3 = =ω = 2 E0 = 0.0118 eV, so λ =

40.34.

hc (6.63 × 10 −34 J ⋅ s)(3.00 × 108 m/s) = = 106 μ m E 1.88 × 10 −21 J

(c) En +1 − En = =ω = 2 E0 = 0.0118 eV EVALUATE: These energy differences are much smaller than those due to electron transitions in the hydrogen atom. IDENTIFY: If the given wave function is a solution to the Schrödinger equation, we will get an identity when we substitute that wave function into the Schrödinger equation. SET UP: The given function is ψ ( x) = Aeikx , and the one-dimensional Schrödinger equation is

= d 2ψ ( x) + U ( x)ψ ( x) = Eψ ( x). 2m dx 2

EXECUTE: Start with the given function and take the indicated derivatives: ψ ( x) = Aeikx .

dψ ( x ) = Aikeikx . dx

= d 2ψ ( x) = 2 2 d 2ψ ( x) d 2ψ ( x) = Ai 2k 2eikx = − Ak 2eikx . = − k 2ψ ( x). − = k ψ ( x). Substituting these results into the 2 2 dx dx 2m dx 2 2m = 2k 2 ψ ( x) + U 0ψ ( x) = E ψ ( x). one-dimensional Schrödinger equation gives 2m = 2k 2 EVALUATE: ψ ( x ) = A eikx is a solution to the one-dimensional Schrödinger equation if E − U 0 = or 2m 2 m( E − U 0 ) . (Since U 0 < E was given, k is the square root of a positive quantity.) In terms of the particle’s =2 momentum p: k = p/=, and in terms of the particle’s de Broglie wavelength λ : k = 2π /λ . k=

40.35.

IDENTIFY: Let I refer to the region x < 0 and let II refer to the region x > 0, so ψ I ( x) = Aeik1 x + Be− ik1 x and

ψ II ( x) = Ceik x . Set ψ I (0) = ψ II (0) and 2

SET UP:

dψ I dψ II = at x = 0. dx dx

d ikx (e ) = ikeikx . dx

dψ I dψ II = at x = 0 gives ik1 A − ik1B = ik2C. Solving this pair of dx dx ⎛k −k ⎞ ⎛ 2k 2 ⎞ equations for B and C gives B = ⎜ 1 2 ⎟ A and C = ⎜ ⎟ A. ⎝ k1 + k2 ⎠ ⎝ k1 + k2 ⎠

EXECUTE:

ψ I (0) = ψ II (0) gives A + B = C.


40-8

Chapter 40

EVALUATE: The probability of reflection is R =

B 2 (k1 − k2 ) 2 = . The probability of transmission is A2 ( k1 + k2 ) 2

C2 4k12 = . Note that R + T = 1. A2 ( k1 + k2 ) 2 (n + 1) 2 − n 2 2n + 1 2 1 (a) Rn = = = + 2 . This is never larger than it is for n = 1, and R1 = 3. n2 n2 n n (b) R approaches zero; in the classical limit, there is no quantization, and the spacing of successive levels is vanishingly small compared to the energy levels. n2h 2 IDENTIFY and SET UP: The energy levels are given by Eq.(40.9): En = . Calculate Δ E for the transition 8mL2 and set Δ E = hc/λ , the energy of the photon. T=

40.36.

40.37.

EXECUTE: (a) Ground level, n = 1, E1 =

First excited level, n = 2, E2 =

4h 2 8mL2

The transition energy is ΔE = E2 − E1 =

3h 2 . Set the transition energy equal to the energy hc/λ of the emitted 8mL2

3h 2 . λ 8mL2 2 −31 8mcL 8(9.109 × 10 kg)(2.998 × 108 m/s)(4.18 × 10−9 m) 2 λ= = 3h 3(6.626 × 10 −34 J ⋅ s) −5 λ = 1.92 × 10 m = 19.2 μ m. 9h 2 9h 2 4h 2 5h 2 (b) Second excited level has n = 3 and E3 = . The transition energy is Δ E = E3 − E2 = − = . 2 2 2 8mL 8mL 8mL 8mL2 hc 5h 2 8mcL2 3 = so λ = = (19.2 μ m) = 11.5 μ m. 2 5h 5 λ 8mL EVALUATE: The energy spacing between adjacent levels increases with n, and this corresponds to a shorter wavelength and more energetic photon in part (b) than in part (a). L 4 2 L/4 πx 2 L/ 4 1 ⎛ 2πx ⎞ 1⎛ L 2πx ⎞ 1 1 (a) ∫ = − , which is about 0.0908. sin 2 dx = ∫ 1 − cos dx = ⎜ x − sin ⎜ ⎟ ⎟ 0 0 L L L 2⎝ L ⎠ L⎝ 2π L ⎠0 4 2π photon. This gives

40.38.

h2 8mL2

hc

=

L2

(b) Repeating with limits of L 4 and L 2 gives

1⎛ L 2πx ⎞ 1 1 , ⎜ x − sin ⎟ = + L ⎠ L 4 4 2π L⎝ 2π

about 0.409. (c) The particle is much likely to be nearer the middle of the box than the edge. (d) The results sum to exactly 1/2, which means that the particle is as likely to be between x = 0 and L 2 as it is to be between x = L 2 and x = L. (e) These results are represented in Figure 40.5b in the textbook. 40.39.

IDENTIFY: The probability of the particle being between x1 and x2 is

x2 x1

2

ψ dx, where ψ is the normalized

wave function for the particle. 2 ⎛πx⎞ sin ⎜ ⎟. L ⎝ L ⎠ EXECUTE: The probability P of the particle being between x = L / 4 and x = 3L / 4 is 3L / 4 2 3L / 4 2 ⎛ π x ⎞ 2 P=∫ ψ 1 dx = ∫ sin ⎜ ⎟ dx. Let y = π x / L; dx = ( L / π ) dy and the integration limits become π / 4 and L/4 L L/4 ⎝ L ⎠ 3π / 4. 3π / 4 2 ⎛ L ⎞ 3π / 4 2 2 ⎡1 1 ⎤ P = ⎜ ⎟∫ sin y dy = ⎢ y − sin 2 y ⎥ L ⎝ π ⎠ π /4 4 π ⎣2 ⎦ π /4 (a) SET UP: The normalized wave function for the ground state is ψ 1 =

2 ⎡ 3π π 1 ⎛ 3π ⎞ 1 ⎛ π ⎞ ⎤ − − sin + sin π ⎢⎣ 8 8 4 ⎜⎝ 2 ⎟⎠ 4 ⎜⎝ 2 ⎟⎠ ⎥⎦ 2 ⎛π 1 1 ⎞ 1 1 1 1 P = ⎜ − ( −1) + (1) ⎟ = + = 0.818. (Note: The integral formula ∫ sin 2 y dy = y − sin 2 y was used.) 4 ⎠ 2 π 2 4 π⎝4 4 P=


Quantum Mechanics

(b) SET UP: The normalized wave function for the first excited state is ψ 2 = EXECUTE:

P=∫

3L / 4 L/4

2

ψ 2 dx =

40-9

2 ⎛ 2π x ⎞ sin ⎜ ⎟ L ⎝ L ⎠

2 3 L / 4 2 ⎛ 2π x ⎞ sin ⎜ ⎟ dx. Let y = 2π x / L; dx = ( L / 2π ) dy and the integration limits L ∫L/ 4 ⎝ L ⎠

become π / 2 and 3π / 2. 3π / 2 2 ⎛ L ⎞ 3π / 2 2 1 ⎡1 1 1 ⎛ 3π π ⎞ ⎤ P= ⎜ y dy y y sin sin 2 = − = ⎜ − ⎟ = 0.500 ⎟ ⎥ L ⎝ 2π ⎠ ∫ π / 2 4 π ⎣⎢ 2 ⎦ π /2 π ⎝ 4 4 ⎠ (c) EVALUATE:

These results are consistent with Fig.40.4b in the textbook. That figure shows that ψ

2

is more

concentrated near the center of the box for the ground state than for the first excited state; this is consistent with the answer to part (a) being larger than the answer to part (b). Also, this figure shows that for the first excited state half the area under ψ 40.40.

2

curve lies between L/4 and 3L/4, consistent with our answer to part (b).

Using the normalized wave function ψ1 = 2 L sin(πx L) , the probabilities | ψ |2 dx are (a) (2 L) sin 2 ( π 4)dx = dx / L (b) (2 L) sin 2 (π / 2) dx = 2dx / L (c) (2 L )sin 2 (3π 4) = dx L .

40.41.

IDENTIFY and SET UP: The normalized wave function for the n = 2 first excited level is ψ 2 =

2 ⎛ 2π x ⎞ sin ⎜ ⎟. L ⎝ L ⎠

P = ψ ( x) dx is the probability that the particle will be found in the interval x to x + dx. 2

EXECUTE: (a) x = L/4

2 ⎛ ⎛ 2π ⎞⎛ L ⎞ ⎞ sin ⎜ ⎜ ⎟⎜ ⎟ ⎟ = L ⎝ ⎝ L ⎠⎝ 4 ⎠ ⎠ P = (2/L) dx (b) x = L/2

2 ⎛π ⎞ sin ⎜ ⎟ = L ⎝2⎠

2 ⎛ ⎛ 2π ⎞⎛ L ⎞ ⎞ sin ⎜ ⎜ ⎟⎜ ⎟ ⎟ = L ⎝ ⎝ L ⎠⎝ 2 ⎠ ⎠

2 sin(π ) = 0 L

ψ ( x) =

ψ ( x) =

2 . L

P=0 (c) x = 3L/4 2 ⎛ ⎛ 2π ⎞⎛ 3L ⎞ ⎞ sin ⎜ ⎜ ⎟⎜ ⎟ ⎟ = L ⎝ ⎝ L ⎠⎝ 4 ⎠ ⎠ P = (2/L) dx

ψ ( x) =

2 2 ⎛ 3π ⎞ sin ⎜ ⎟ = − . L L ⎝ 2 ⎠

EVALUATE: Our results are consistent with the n = 2 part of Fig.40.5 in the textbook. ψ 40.42.

40.43.

2

is zero at the center

of the box and is symmetric about this point. G G hn G G G =nπ hn Δ p = pfinal − pinitial . p = =k = = . At x = 0 the initial momentum at the wall is pinitial = − iˆ and the final L 2L 2L G G hn ˆ hn ˆ ⎛ hn ˆ ⎞ hn ˆ momentum, after turning around, is pfinal = + i . So, Δp = + i − ⎜ − i ⎟ = + i . At x = L the initial 2L 2L ⎝ 2L ⎠ L G G hn hn momentum is pinitial = + iˆ and the final momentum, after turning around, is pfinal = − iˆ. So, 2L 2L G hn hn ˆ hn Δ p = − iˆ − i = − iˆ 2L 2L L d 2ψ ( x ) 2m  (a) For a free particle, U ( x) = 0 so Schrodinger's = − 2 Eψ ( x ). The graph is given in equation becomes 2 dx h Figure 40.43. = 2κ 2 dψ ( x) d 2ψ ( x ) 2m . (b) For x < 0: ψ ( x) = e +κ x . = κ e +κ x . = κ 2e +κ x . So κ 2 = − 2 E ⇒ E = − dx dx 2m = dψ ( x) d 2ψ ( x) (c) For x > 0: ψ ( x ) = e −κ x . = − ke −κ x . = κ 2e−κ x dx dx


40-10

Chapter 40

So again κ 2 = −

2m −= 2κ 2  E E . Parts (c) and (d) show ψ ( x) satisfies the Schrodinger's ⇒ = equation, provided 2m =2

−= 2κ 2 . 2m dψ ( x) (d) Note is discontinuous at x = 0. (That is, negative for x > 0 and positive for x < 0.) dx

E=

40.44.

Figure 40.43 IDENTIFY: We start with the penetration distance formula given in the problem. = SET UP: The given formula is η = . 2m(U 0 − E ) EXECUTE: (a) Substitute the given numbers into the formula: = 1.055 × 10−34 J ⋅ s η= = = 7.4 × 10 −11 m 2m (U 0 − E ) 2(9.11 × 10 −31 kg)(20 eV − 13 eV)(1.602 × 10−19 J/eV)

(b) η = 40.45.

1.055 × 10−34 J ⋅ s 2(1.67 × 10

−27

kg)(30 MeV − 20 MeV)(1.602 × 10

−13

J/MeV)

= 1.44 × 10−15 m

EVALUATE: The penetration depth varies widely depending on the mass and energy of the particle. (a) We set the solutions for inside and outside the well equal to each other at the well boundaries, x = 0 and L. x = 0 : A sin(0) + B = C ⇒ B = C , since we must have D = 0 for x < 0.

2mEL 2mEL + B cos = + De−κ L since C = 0 for x > L. = = 2mE . This gives A sin kL + B cos kL = De −κ L , where k = = (b) Requiring continuous derivatives at the boundaries yields dψ x = 0: = kA cos(k ⋅ 0) − kB sin(k ⋅ 0) = kA = κ Ce k ⋅0 ⇒ kA = κ C dx x = L: kA cos kL − kB sin kL = −κ De −κ L . 2m(U 0 − E ) 1 ⎛T ⎞ E⎛ E ⎞ T = Ge−2κ L with G = 16 ⎜ 1 − ⇒ L = − ln ⎜ ⎟ . ⎟ and κ = = 2κ ⎝ G ⎠ U0 ⎝ U0 ⎠ x = L: A sin

40.46.

If E = 5.5 eV, U 0 = 10.0 eV, m = 9.11 × 10−31 kg, and T = 0.0010. Then

κ=

2(9.11× 10−31 kg)(4.5 eV)(1.60 × 10−19 J eV) 5.5 eV ⎛ 5.5 eV ⎞ = 1.09 × 1010 m −1 and G = 16 ⎜1 − ⎟ = 3.96 −34 10.0 eV ⎝ 10.0 eV ⎠ (1.054 × 10 J ⋅ s)

so L = − 40.47.

1 ⎛ 0.0010 ⎞ −10 ln ⎜ ⎟ = 3.8 × 10 m = 0.38 nm. 2(1.09 × 1010 m −1 ) ⎝ 3.96 ⎠

IDENTIFY and SET UP: When κ L is large, then eκ L is large and e −κ L is small. When κ L is small, sinh κ L → κ L. Consider both κ L large and κ L small limits.

⎡ (U sinh κ L) 2 ⎤ EXECUTE: (a) T = ⎢1 + 0 ⎥ 4E (U 0 − E ) ⎦ ⎣

−1


Quantum Mechanics

sinh κ L =

e

κL

−e 2

40-11

−κ L

−1

⎡ eκ L U 02e 2κ L ⎤ 16 E (U 0 − E ) and T → ⎢1 + ⎥ = − 2 16 E ( U E ) 16 E ( U 0 − E ) + U 02e 2κ L ⎣ ⎦ 0 For κ L W1, 16 E (U 0 − E ) + U 02e2κ L → U 02e 2κ L

For κ L W1, sinhκ L →

T→

⎛ E ⎞⎛ 16 E (U 0 − E ) E ⎞ − 2κ L = 16 ⎜ ⎟⎜ 1 − ⎟ e , which is Eq.(40.21). U 02e 2κ L U U 0 ⎠ ⎝ 0 ⎠⎝

L 2m(U 0 − E ) . So κ L W1 when L is large (barrier is wide) or U 0 − E is large. (E is small compared to U 0 . ) = 2m(U 0 − E ) ; κ becomes small as E approaches U 0 . For κ small, sinh κ L → κ L and (c) κ = = −1 −1 ⎡ ⎡ U 02 2m(U 0 − E ) L2 ⎤ U 02κ 2 L2 ⎤ = + T → ⎢1 + 1 ⎥ ⎢ ⎥ (using the definition of κ ) = 2 4 E (U 0 − E ) ⎦ ⎣ 4 E (U 0 − E ) ⎦ ⎣ (b) κ L =

⎡ 2U 2 L2 m ⎤ Thus T → ⎢1 + 0 2 ⎥ 4 E= ⎦ ⎣ U 0 → E so

−1

⎡ 2 EL2 m ⎤ U 02 → E and T → ⎢1 + ⎥ E 4= 2 ⎦ ⎣

−1

−1

⎡ ⎛ kL ⎞ 2 ⎤ 2mE But k = 2 , so T → ⎢1 + ⎜ ⎟ ⎥ , as was to be shown. = ⎢⎣ ⎝ 2 ⎠ ⎥⎦ EVALUATE: When κ L is large Eq.(40.20) applies and T is small. When E → U 0 , T does not approach unity. 1 (a) E = mv 2 = (n + (1 2))=ω = ( n + (1 2)) hf , and solving for n, 2 1 2 mv 1 (1/2)(0.020 kg)(0.360 m/s) 2 1 n= 2 − = − = 1.3 × 1030. hf 2 (6.63 × 10−34 J ⋅ s)(1.50 Hz) 2 2

40.48.

40.49.

40.50.

(b) The difference between energies is =ω = hf = (6.63 × 10−34 J ⋅ s)(1.50 Hz) = 9.95 × 10 −34 J. This energy is too small to be detected with current technology IDENTIFY and SET UP: Calculate the angular frequency ω of the pendulum and apply Eq.(40.26) for the energy levels. 2π 2π EXECUTE: ω = = = 4π s −1 T 0.500 s 1 1 The ground-state energy is E0 = =ω = (1.055 × 10−34 J ⋅ s)(4π s −1 ) = 6.63 × 10−34 J. 2 2 E0 = 6.63 × 10−34 J(1 eV/1.602 × 10−19 J) = 4.14 × 10 −15 eV

1⎞ ⎛ En = ⎜ n + ⎟ =ω 2⎠ ⎝ 1⎞ ⎛ En +1 = ⎜ n + 1 + ⎟ =ω 2⎠ ⎝ The energy difference between the adjacent energy levels is Δ E = En +1 − En = =ω = 2 E0 = 1.33 × 10−33 J = 8.30 × 10−15 eV EVALUATE: These energies are much too small to detect. Quantum effects are not important for ordinary size objects. IDENTIFY: We model the electrons in the lattice as a particle in a box. The energy of the photon is equal to the energy difference between the two energy states in the box. n2h 2 SET UP: The energy of an electron in the nth level is En = . We do not know the initial or final levels, but 8mL2 we do know they differ by 1. The energy of the photon, hc/λ , is equal to the energy difference between the two states. hc (6.63 × 10−34 J ⋅ s)(3.00 × 108 m/s) = = EXECUTE: The energy difference between the levels is Δ E = λ 1.649 × 10−7 m −18 1.206 × 10 J. Using the formula for the energy levels in a box, this energy difference is equal to h2 h2 Δ E = ⎡⎣ n2 − (n − 1) 2 ⎤⎦ = (2n − 1) . 2 8mL 8mL2


40-12

40.51.

Chapter 40

⎛ Δ E 8mL2 ⎞ 1 ⎛ (1.206 × 10−18 J)8(9.11 × 10−31 kg)(0.500 × 10−9 m) 2 ⎞ + 1⎟ = ⎜ + 1⎟ = 3. Solving for n gives n = ⎜ 2 (6.626 × 10−34 J ⋅ s) 2 ⎝ h ⎠ 2⎝ ⎠ The transition is from n = 3 to n = 2. EVALUATE: We know the transition is not from the n = 4 to the n = 3 state because we let n be the higher state and n − 1 the lower state. IDENTIFY: If the given wave function is a solution to the Schrödinger equation, we will get an identity when we substitute that wave function into the Schrödinger equation. 2 2 SET UP: The given wave function is ψ 0 ( x ) = A0 e −α x / 2 and the Schrödinger equation is = d 2ψ ( x) k ′x 2 − + ψ ( x) = E ψ ( x). 2m dx 2 2 2 2 2 2 dψ 0 ( x ) EXECUTE: (a) Start by taking the derivatives: ψ 0 ( x ) = A0e −α x / 2 . = −α 2 xA0 e −α x / 2 . dx 2 d 2ψ 0 ( x) −α 2 x 2 / 2 d ψ 0 ( x ) 2 −α 2 x 2 / 2 2 2 2 2 2 2 2 . = − A0α e + (α ) x A0 e = [ −α + (α ) x ] ψ 0 ( x). dx 2 dx 2 = d 2ψ 0 ( x) =2 = d 2ψ ( x) k ′x 2 [ −α 2 + (α 2 ) 2 x 2 ] ψ 0 ( x). Equation (40.22) is − + − =− ψ ( x) = E ψ ( x). Substituting 2 2m dx 2m 2m dx 2 2 2 2 = k ′x mω [ −α 2 + (α 2 ) 2 x 2 ] ψ 0 ( x) + ψ 0 ( x ) = E ψ 0 ( x ). Since α 2 = the above result into that equation gives − and = 2m 2 2 =2 2 2 k′ = 2 ⎛ mω ⎞ mω 2 k′ = 0. (α ) + = − ω= , the coefficient of x2 is − ⎜ ⎟ + 2m 2 2m ⎝ = ⎠ 2 m

⎛ mω ⎞ (b) A0 = ⎜ ⎟ ⎝ =π ⎠

1/ 4

= 2 . The probability density function ψ is ωm 1/ 2 1/ 2 2 ⎛ mω ⎞ ⎛ mω ⎞ − mω x 2 /= =⎜ = . x = 0, At e ψ 0 ⎟ ⎜ ⎟ . ⎝ =π ⎠ ⎝ =π ⎠

(c) The classical turning points are at A = ±

ψ 0 ( x ) = A02e −α 2

x

d ψ 0 ( x) d ψ 0 ( x) mω ⎛ mω ⎞ ⎛ mω ⎞ −α 2 x 2 −α 2 x 2 2 =⎜ = −2 = 0. . At x = 0, ⎟ (−α 2 x)e ⎜ ⎟ xe dx = ⎝ =π ⎠ dx ⎝ =π ⎠ 2 2 1/ 2 d 2 ψ 0 ( x) d 2 ψ 0 ( x) mω ⎛ mω ⎞ −α 2 x 2 2 2 < 0. Therefore, at x = 0, the first derivative is = −2 . At x = 0, ⎜ ⎟ [1 − 2α x ]e 2 2 dx dx = ⎝ =π ⎠ zero and the second derivative is negative. Therefore, the probability density function has a maximum at x = 0. 2 2 =2 EVALUATE: ψ 0 ( x) = A0e−α x / 2 is a solution to equation (40.22) if − (−α 2 )ψ 0 ( x ) = E ψ 0 ( x) or 2m = 2α 2 =ω =ω E= = . E0 = corresponds to n = 0 in Equation (40.26). 2 2m 2 IDENTIFY: If the given wave function is a solution to the Schrödinger equation, we will get an identity when we substitute that wave function into the Schrödinger equation. 2 2 SET UP: The given wave function is ψ 1 ( x ) = A1 2 xe −α x / 2 and the Schrödinger equation is = d 2ψ ( x ) k ′x 2 − + ψ ( x ) = E ψ ( x ). 2m dx 2 2 2 2 EXECUTE: (a) Start by taking the indicated derivatives: ψ 1 ( x ) = A1 2 xe−α x / 2 . 2

40.52.

2 2

1/ 2

1/ 2

2

2 2 2 2 2 2 2 2 2 2 dψ 1 ( x ) d 2ψ 1 ( x) = −2α 2 x 2 A1e −α x / 2 + 2 A1e −α x / 2 . = −2 A1α 2 2 xe−α x / 2 − 2 A1α 2 x 2 (−α 2 x )e −α x / 2 + 2 A1 ( −α 2 x)e−α x / 2 . 2 dx dx d 2ψ 1 ( x) 2 2 2 2 2 = [−2α + (α ) x − α ] ψ 1 ( x ) = [ −3α 2 + (α 2 ) 2 x 2 ] ψ 1 ( x) dx 2 = d 2ψ 1 ( x) =2 [ −3α 2 + (α 2 ) 2 x 2 ] ψ 1 ( x ) − = − 2m dx 2 2m


Quantum Mechanics

40-13

= d ψ ( x) k ′x ψ ( x) = E ψ ( x). Substituting the above result into that equation gives + 2m dx 2 2 mω k ′x 2 =2 k′ [−3α 2 + (α 2 ) 2 x 2 ] ψ 1 ( x) + ψ 1 ( x) = E ψ 1 ( x). Since α 2 = and ω = , the coefficient of x2 is − 2m 2 = m 2 =2 2 2 k′ = 2 ⎛ mω ⎞ mω 2 − =0 (α ) + = − ⎜ ⎟ + 2m 2 2m ⎝ = ⎠ 2 2

2

Equation (40.22) is −

1 ⎛ mω ⎞ ⎜ ⎟ 2 ⎝ =π ⎠

1/ 4

(b) A1 =

1 ⎛ mω ⎞ = ⎜ ⎟ 2 ⎝ =π ⎠

1/ 2

(c) The probability density function ψ is ψ 1 ( x) = A12 4 x 2 e−α 2

2

d ψ 1 ( x)

At x = 0, ψ 1 = 0. 2

d ψ 1 ( x)

At x = 0, d 2 ψ 1 ( x) 2

x

4 x 2e

mω x 2 =

2

= A12 8 xe−α

dx

2

= 0. At x = ±

dx

2 2

1

α

,

2 2

x

+ A12 4 x 2 (−α 2 2 x )e −α

d ψ 1 ( x)

2 2

x

= A12 8 xe −α

2 2

x

− A12 8 x3α 2e −α

2 2

x

2

dx

= 0.

2

dx 2 d 2 ψ 1 ( x)

= A12 8e −α

2 2

+ A12 8 x( −α 2 2 x)e−α

= A12 8e −α

2 2

− A1216 x 2α 2e −α

x

2 2

2 2

x

− A12 8(3 x 2 )α 2e −α

− A12 24 x 2α 2e−α

2 2

2 2

x

− A12 8 x3α 2 ( −α 2 2 x)e−α

2 2

x

.

+ A1216 x 4 (α 2 ) 2 e −α x . At x = 0, 2 2

d 2 ψ 1 ( x)

2

> 0. So at dx 2 dx 2 x = 0, the first derivative is zero and the second derivative is positive. Therefore, the probability density function x

has a minimum at x = 0. At x = ±

1

α

,

x

d 2 ψ 1 ( x) dx 2

x

2

< 0. So at x = ±

1

α

, the first derivative is zero and the second

derivative is negative. Therefore, the probability density function has maxima at x = ± classical turning points for n = 0 as found in the previous question. EVALUATE:

ψ 1 ( x ) = A1 2 xe −α

2 2

x /2

is a solution to equation (40.22) if −

, corresponding to the

=2 (−3α 2 )ψ 1 ( x) = E ψ 1 ( x) or 2m

3= 2α 2 3=ω 3=ω . E1 = = corresponds to n = 1 in Equation (40.26). 2 2m 2 IDENTIFY and SET UP: Evaluate ∂ 2ψ / ∂x 2 , ∂ 2ψ / ∂y 2 , and ∂ 2ψ / ∂z 2 for the proposed ψ and put Eq.(40.29). Use E=

40.53.

1

α

that ψ nx , ψ ny , and ψ nz are each solutions to Eq.(40.22). ⎞ ⎟ + Uψ = Eψ ⎠ 2 = 2 d ψ nx 1 2 ψ nx , ψ ny , ψ nz are each solutions of Eq.(40.22), so − + k ′x ψ nx = Enxψ nx 2m dx 2 2 2 = 2 d ψ ny 1 2 − + k ′y ψ ny = Enyψ ny 2m dy 2 2 2 = 2 d ψ nz 1 2 − + k ′z ψ nz = Enzψ nz 2m dz 2 2 1 1 1 ψ = ψ nx ( x)ψ ny ( y )ψ nz ( z ), U = k ′x 2 + k ′y 2 + k ′z 2 2 2 2 2 2 2 2 2 ⎛ ⎞ d ψ ⎛ ⎞ d ψ ∂ψ ∂ψ ∂ 2ψ ⎛ d ψ nz ⎞ ny nx ⎜ ⎟ ψ ψ , ψ ψ , = = = ⎜ ⎟ ⎜ ⎟ψ n ψ n . n n n n ∂x 2 ⎜⎝ dx 2 ⎟⎠ y z ∂y 2 ⎜⎝ dy 2 ⎟⎠ x z ∂z 2 ⎜⎝ dz 2 ⎟⎠ x y ⎛ = 2 d 2ψ nx 1 2 ⎞ = 2 ⎛ ∂ 2ψ ∂ 2ψ ∂ 2ψ ⎞ + + + = − + k ′x ψ nx ⎟ψ nyψ nz U So − ψ ⎜ ⎜ ⎟ 2 ⎜ ⎟ 2m ⎝ ∂x 2 ∂y 2 ∂z 2 ⎠ 2 ⎝ 2m dx ⎠

EXECUTE: (a) −

= 2 ⎛ ∂ 2ψ ∂ 2ψ ∂ 2ψ + + ⎜ 2m ⎝ ∂x 2 ∂y 2 ∂z 2

⎛ = 2 d 2ψ ny 1 2 +⎜ − + k ′y ψ ny ⎜ 2m dy 2 2 ⎝ −

= 2 ⎛ ∂ 2ψ ∂ 2ψ ∂ 2ψ + + ⎜ 2m ⎝ ∂x 2 ∂y 2 ∂z 2

⎞ ⎛ = 2 d 2ψ nz 1 2 ⎞ ⎟ψ nxψ nz + ⎜ − + k ′z ψ nz ⎟ψ nxψ ny ⎜ 2m dz 2 ⎟ ⎟ 2 ⎝ ⎠ ⎠

⎞ ⎟ + Uψ = ( Enx + Eny + Enz )ψ ⎠


40-14

Chapter 40

Therefore, we have shown that this ψ is a solution to Eq.(40.29), with energy 3⎞ ⎛ Enx ny nz = Enx + Eny + Enz = ⎜ nx + n y + nz + ⎟ =ω 2⎠ ⎝ 3 (b) and (c) The ground state has nx = n y = nz = 0, so the energy is E000 = =ω. There is only one set of nx , n y and 2 nz that give this energy. 5 First-excited state: nx = 1, ny = nz = 0 or ny = 1, nx = nz = 0 or nz = 1, nx = ny = 0 and E100 = E010 = E001 = =ω 2 There are three different sets of nx , n y , nz quantum numbers that give this energy, so there are three different

40.54.

quantum states that have this same energy. EVALUATE: For the three-dimensional isotropic harmonic oscillator, the wave function is a product of onedimensional harmonic oscillator wavefunctions for each dimension. The energy is a sum of energies for three onedimensional oscillators. All the excited states are degenerate, with more than one state having the same energy. 1⎞ ⎛ ω1 = k1′ m , ω2 = k2′ m . Let ψnx ( x) be a solution of Eq.(40.22) with Enx = ⎜ nx + ⎟ =ω1 , ψ nx ( y ) be a similar 2⎠ ⎝ solution, ψ nz ( z ) be a solution of Eq.(40.22) but with z as the independent variable instead of x, and 1⎞ ⎛ energy Enz = ⎜ nz + ⎟ ω2. 2⎠ ⎝ (a) As in Problem 40.53, look for a solution of the form ψ ( x, y, z ) = ψ nx ( x )ψ ny ( y )ψ nz ( z ). Then, = 2 ∂ 2ψ ⎛ 1 ∂ 2ψ ∂ 2ψ ⎞ = ⎜ Enx − k1′x 2 ⎟ ψ with similar relations for and 2 . Adding, 2 2 2m ∂x 2 ∂y ∂z ⎝ ⎠ 2 2 2 2 ⎛ ⎞ = ∂ψ ∂ψ ∂ψ 1 2 1 2 1 2⎞ ⎛ − + + ⎜ ⎟ = ⎜ En + Eny + Enz − k1′x − k1′ y − k2′ z ⎟ ψ 2m ⎝ ∂x 2 ∂y 2 ∂z 2 ⎠ ⎝ x 2 2 2 ⎠

= ( Enx + Eny + Enz − U )ψ = ( E − U )ψ

40.55.

⎡ 1⎞ ⎤ ⎛ where the energy E is E = Enx + Eny + Enz = = ⎢( nx + n y + 1)ω12 + ⎜ nz + ⎟ ω22 ⎥ , with nx , n y and nz all nonnegative 2⎠ ⎦ ⎝ ⎣ integers. 1 ⎞ ⎛ (b) The ground level corresponds to nx = n y = nz = 0, and E = = ⎜ ω21 + ω 22 ⎟ . The first excited level corresponds to 2 ⎠ ⎝ 3 ⎛ ⎞ nx = ny = 0and nz = 1, since ω12 > ω 22 , and E = =ω ⎜ ω 21 + ω 22 ⎟ . There is only one set of quantum numbers for both 2 ⎠ ⎝ the ground state and the first excited state. (a) ψ ( x ) = A sin kx and ψ (− L 2) = 0 = ψ (+ L 2) 2nπ 2π ⎛ + kL ⎞ + kL ⇒ 0 = A sin ⎜ = nπ ⇒ k = = ⎟⇒ 2 L λ ⎝ 2 ⎠

L h nh p 2 n 2 h2 (2n) 2 h2 ⇒ pn = = ⇒ En = n = = , where n = 1, 2... λn L n 2m 2mL2 8mL2 (b) ψ ( x ) = A cos kx and ψ ( − L / 2) = 0 = ψ ( + L / 2) ⇒λ =

π (2n + 1)π 2π ⎛ kL ⎞ kL ⇒ 0 = A cos ⎜ ⎟ ⇒ = (2n + 1) ⇒ k = = 2 2 2 L λ ⎝ ⎠ 2L (2n + 1) h ⇒λ = ⇒ pn = (2n + 1) 2L (2n + 1) 2 h 2 n = 0, 1, 2... 8mL2 (c) The combination of all the energies in parts (a) and (b) is the same energy levels as given in Eq.(40.9), where n2 h 2 En = . 8mL2 (d) Part (a)’s wave functions are odd, and part (b)’s are even. (a) As with the particle in a box, ψ ( x) = A sin kx, where A is a constant and k 2 = 2mE = 2 . Unlike the particle in a box, however, k and hence E do not have simple forms. ⇒ En =

40.56.


Quantum Mechanics

40-15

(b) For x > L, the wave function must have the form of Eq.(40.18). For the wave function to remain finite as x → ∞, C = 0. The constant κ 2 = 2m(U 0 − E ) =, as in Eq.(14.17) and Eq.(40.18).

(c) At x = L, A sin kL = De − κL and kA cos kL = − κDe − κL . Dividing the second of these by the first gives k cot kL = − κ , a transcendental equation that must be solved numerically for different values of the length L and

the ratio E U 0 . 40.57.

p2 h h + U ( x) ⇒ p = 2m( E − U ( x)). λ = ⇒ λ ( x) = . 2m p 2m( E − U ( x)) (b) As U ( x ) gets larger (i.e., U ( x) approaches E from below—recall k ≥ 0), E − U ( x) gets smaller, so λ ( x) gets larger. (c) When E = U ( x), E − U ( x) = 0, so λ ( x ) → ∞. b dx b b dx 1 b n hn 2m( E − U ( x )) dx = ⇒ ∫ 2m( E − U ( x)) dx = . (d) ∫ =∫ = a λ ( x) a a 2 2 h 2m( E − U ( x)) h ∫ a (e) U ( x ) = 0 for 0 < x < L with classical turning points at x = 0 and x = L. So, (a) E = K + U ( x) =

b

2m( E − U ( x )) dx = ∫

a

L 0

L

2mEdx = 2mE ∫ dx = 2mEL. So, from part (d), 0

2

hn 1 ⎛ hn ⎞ hn ⇒E= ⎜ ⎟ = 2 2m ⎝ 2 L ⎠ 8mL2 . (f ) Since U ( x) = 0 in the region between the turning points at x = 0 and x = L, the results is the same as part (e). 2 2

2mE L =

The height U 0 never enters the calculation. WKB is best used with smoothly varying potentials U ( x). 40.58.

1 2 2E ⇒ xTP = ± . (a) At the turning points E = k ′xTP 2 k′ (b)

+

2 E/k ′

∫ − 2 E/k ′

1 nh ⎛ ⎞ 2m ⎜ E − k ′x 2 ⎟ dx = . To evaluate the integral, we want to get it into a form that matches the 2 2 ⎝ ⎠

standard integral given. Letting A2 = ⇒ mk ′ ∫

40.59.

b a

1 2mE 2E ⎛ ⎞ 2m ⎜ E − k ′x 2 ⎟ = 2mE − mk ′x 2 = mk ′ − x 2 = mk ′ − x2. 2 mk ′ k′ ⎝ ⎠

2E 2E 2E ,a=− ,and b = + k′ k′ k′ ⎛ x ⎞⎤ mk ′ ⎡ 2 2 2 A − x dx = 2 ⎢ x A − x + A arcsin ⎜⎜ ⎟⎟ ⎥ 2 ⎣⎢ ⎝ A ⎠ ⎥⎦ 2

b

2

0

⎡ 2E 2E 2E 2E ⎛ 2E k ′ ⎞⎤ 2E m⎛π⎞ arcsin ⎜ arcsin (1) = 2 E = mk ′ ⎢ − + ⎟⎟ ⎥ = mk ′ ⎜ ⎟. ⎜ ′ ′ ′ ′ ′ k k k k ′ k k′ ⎝ 2 ⎠ ⎝ 2 E k ⎠ ⎥⎦ ⎣⎢ hn m hn k′ h π = . Recall ω = , so E = ωn = =ωn. Using WKB, this is equal to , so E 2 k′ 2 m 2π =ω ⎛ 1 ⎞⎞ ⎛ (c) We are missing the zero-point-energy offset of ⎜ recall E = =ω ⎜ n + ⎟ ⎟ . However, our approximation isn’t 2 ⎝ 2 ⎠⎠ ⎝ bad at all! E (a) At the turning points E = A xTP ⇒ xTP = ± . A (b)

+E / A −E / A

2m( E − A x ) dx = 2∫

dy = −2mA dx when x = 2∫

E A 0

2m( E − Ax)dx = −

E/A 0

2m( E − Ax) dx. Let y = 2m( E − Ax) ⇒

E , y = 0, and when x = 0, y = 2mE. So A 1 0 12 2 32 y dy = − y ∫ 2 mE mA 3mA 23

2 hn 1 ⎛ 3mAh ⎞ 23 ⇒E= (2mE )3 2 = ⎜ ⎟ n . 3mA 2 2m ⎝ 4 ⎠

0

= 2 mE

hn 2 . So, (2mE )3 2 . Using WKB, this is equal to 2 3mA


40-16

Chapter 40

(c) The difference in energy decreases between successive levels. For example: 12 3 − 02 3 = 1, 2 2 3 − 12 3 = 0.59, 33 2 − 23 2 = 0.49,...

A sharp ∞ step gave ever-increasing level differences (~ n 2 ).

A parabola (~ x 2 ) gave evenly spaced levels (~n).

• Now, a linear potential (~ x ) gives ever-decreasing level differences (~ n 2 3 ). Roughly speaking, if the curvature of the potential (~ second derivative) is bigger than that of a parabola, then the level differences will increase. If the curvature is less than a parabola, the differences will decrease.


41

ATOMIC STRUCTURE

41.1.

IDENTIFY and SET UP:

L = l (l + 1)=. Lz = ml = . l = 0, 1, 2,..., n − 1. ml = 0, ± 1, ± 2,..., ± l . cosθ = Lz / L .

EXECUTE: (a) l = 0 : L = 0 , Lz = 0 . l = 1: L = 2= , Lz = =,0, −= . l = 2 : L = 6= , Lz = 2=, =,0, −=, −2= .

41.2.

(b) In each case cosθ = Lz / L . L = 0 : θ not defined. L = 2= : 45.0°, 90.0°, 135.0° . L = 6= : 35.3°, 65.9°, 90.0°, 114.1°, 144.7°. G EVALUATE: There is no state where L is totally aligned along the z axis. IDENTIFY and SET UP: L = l (l + 1)= . Lz = ml = . l = 0,1, 2,..., n − 1. ml = 0, ±1, ±2,..., ±l . cosθ = Lz / L . EXECUTE: (a) l = 0 : L = 0 , Lz = 0 . l = 1: L = 2= , Lz = =,0, −= . l = 2 : L = 6= , Lz = 2=, =,0, −=, −2= . l = 3 : L = 2 3= , Lz = 3=,2=, =,0, −=, −2=, −3= . l = 4 : L = 2 5= , Lz = 4=,3=,2=, =,0, −=, −2=, −3=, −4= . (b) L = 0 : θ not defined. L = 2= : 45.0°,90.0°,135.0° . L = 6= : 35.3°,65.9°,90.0°,114.1°,144.7°. L = 2 3= :

41.3.

54.7°,73.2°,90.0°,106.8°,125.3°,150.0°. L = 2 5= : 26.6°,47.9°,63.4°,77.1°,90.0°,102.9°,116.6°,132.1°,153.4° . (c) The minimum angle is 26.6° and occurs for l = 4 , ml = +4 . The maximum angle is 153.4° and occurs for l = 4 , ml = −4 . IDENTIFY and SET UP: The magnitude of the orbital angular momentum L is related to the quantum number l by Eq.(41.4): L = l (l + 1)=, 1 = 0, 1, 2,… −34 2 ⎛ L ⎞ ⎛ 4.716 × 10 kg ⋅ m /s ⎞ l (l + 1) = ⎜ ⎟ = ⎜ ⎟ = 20 −34 ⎝ = ⎠ ⎝ 1.055 × 10 J ⋅ s ⎠ And then l (l + 1) = 20 gives that l = 4. EVALUATE: l must be integer. (a) ( ml ) max = 2, so (Lz ) max = 2=. 2

EXECUTE:

41.4.

(b)

l (l + 1)= = 6= = 2.45=.

⎛m ⎞ ⎛L ⎞ (c) The angle is arccos ⎜ z ⎟ = arccos ⎜ l ⎟ , and the angles are, for ml = −2 to ml = 2, 144.7°, L ⎝ ⎠ ⎝ 6⎠ 114.1°, 90.0°, 65.9°, 35.3°. The angle corresponding to ml = l will always be larger for larger l . 41.5.

IDENTIFY and SET UP: The angular momentum L is related to the quantum number l by Eq.(41.4), L = l (l + 1)=.

The maximum l, lmax , for a given n is lmax = n − 1. EXECUTE:

For n = 2, lmax = 1 and L = 2= = 1.414=.

For n = 20, lmax = 19 and L = (19)(20)= = 19.49=.

41.6.

41.7.

For n = 200, lmax = 199 and L = (199)(200)= = 199.5=. EVALUATE: As n increases, the maximum L gets closer to the value n= postulated in the Bohr model. The (l , ml ) combinations are (0, 0), (1, 0), (1, ± 1) , (2, 0), (2, ± 1), (2, ± 2), (3, 0), (3, ± 1), (3, ± 2), (3, ± 3), (4, 0), (4, ± 1), (4, ± 2), (4, ± 3), and (4, ± 4), a total of 25. 13.60 eV = −0.544 eV. (b) Each state has the same energy (n is the same), − 25 −19 2 1 q1q2 −1 (1.60 × 10 C) = = −2.3 × 10−18 J U= 4πP0 r 4πP0 1.0 × 10−10 m U=

−2.3 × 10−18 J = −14.4 eV. 1.60 × 10−19 J eV 41-1


41-2

41.8.

Chapter 41

(a) As in Example 41.3, the probability is a 2

P=∫

a/2 0

| ψ1s |2 4πr 2 dr =

4 ⎡⎛ ar 2 a 2 r a 3 ⎞ −2 r a ⎤ 5e −1 − − − ⎟e = 0.0803 . ⎥ =1− 3 ⎢⎜ a ⎣⎝ 2 2 4⎠ 2 ⎦0

(b) The difference in the probabilities is (1 − 5e −2 ) − (1 − (5 2)e −1 ) = (5 2)(e−1 − 2e −2 ) = 0.243. 41.9.

(a) | ψ |2 = ψ*ψ =| R (r ) |2 | Θ(θ ) |2 ( Ae− imlφ )( Ae + imlφ ) = A2 | R( r ) |2 | Θ(θ ) |2 , which is independent of φ . (b)

41.10.

2π 0

En = −

| Φ (φ ) |2 dφ = A2

2π 0

dφ = 2πA2 = 1 ⇒ A =

1 . 2π

1 mr e4 E Δ E12 = E2 − E1 = 21 − E1 = −(0.75) E1. 2 2 2 (4πP0 ) 2n = 2

(a) If mr = m = 9.11 × 10 −31 kg 2 mr e 4 (9.109 × 10−31 kg)(1.602 × 10−19 C) 4 = 8.988 × 109 N ⋅ m 2 C ) = 2.177 × 10 −18 J = 13.59 eV ( 2 2 2 −34 (4πP0 ) = 2(1.055 × 10 J ⋅ s) For 2 → 1 transition, the coefficient is (0.75)(13.59 eV) = 10.19 eV. m (b) If mr = , using the result from part (a), 2

mr e 4 ⎛ m 2 ⎞ ⎛ 13.59 eV ⎞ = (13.59 eV) ⎜ ⎟=⎜ ⎟ = 6.795 eV. (4πP0 ) 2 = 2 2 ⎝ m ⎠ ⎝ ⎠ ⎛ 10.19 eV ⎞ Similarly, the 2 → 1 transition, ⇒ ⎜ ⎟ = 5.095 eV. 2 ⎝ ⎠ (c) If mr = 185.8m, using the result from part (a),

mr e4 ⎛ 185.8m ⎞ = (13.59 eV) ⎜ ⎟ = 2525 eV, 2 2 (4πP0 ) = ⎝ m ⎠

41.11.

and the 2 → 1 transition gives ⇒ (10.19 eV)(185.8) = 1893 eV. 4π P0= 2 P0 h2 . IDENTIFY and SET UP: Eq.(41.8) gives a = = mr e 2 π mr e2 EXECUTE: (a) mr = m

a=

P0 h 2 (8.854 × 10−12 C2 /N ⋅ m 2 )(6.626 × 10−34 J ⋅ s) 2 = = 0.5293 × 10 −10 m π mr e 2 π (9.109 × 10−31 kg)(1.602 × 10−19 C) 2

(b) mr = m / 2

⎛ P h2 ⎞ a = 2 ⎜ 0 2 ⎟ = 1.059 × 10−10 m ⎝ π mr e ⎠ (c) mr = 185.8m a=

41.12.

41.13.

1 ⎛ P0 h 2 ⎞ −13 ⎜ ⎟ = 2.849 × 10 m 185.8 ⎝ π mr e 2 ⎠

EVALUATE: a is the radius for the n = 1 level in the Bohr model. When the reduced mass mr increases, a decreases. For positronium and muonium the reduced mass effect is large. eiml φ = cos( mlφ ) + i sin( mlφ ), and to be periodic with period 2π , ml 2π must be an integer multiple of 2π , so ml must be an integer. a a 1 P (a ) = ∫ ψ1s 2V = ∫ e −2 r a (4πr 2dr ) . 0 πa 3 0 a

4 a 4 ⎡⎛ − ar 2 a 2 r a 2 ⎞ −2 r a ⎤ 4 ⎡⎛ − a 3 a 3 a 3 ⎞ −2 a 3 0 ⎤ P (a ) = 3 ∫ r 2e −2 r a dr = 3 ⎢⎜ − − ⎟e = 3 ⎢⎜ − − ⎟e + e ⎥ ⎥ 2 4⎠ 2 4⎠ 4 ⎦ a o a ⎣⎝ 2 ⎦ 0 a ⎣⎝ 2 ⇒ P ( a ) = 1 − 5e −2 . 41.14.

(a) ΔE = μB B = (5.79 × 10 −5 e V T)(0.400 T) = 2.32 × 10−5 eV (b) ml = −2 the lowest possible value of ml .


Atomic Structure

41-3

(c) The energy level diagram is sketched in Figure 41.14.

41.15.

Figure 41.14 IDENTIFY and SET UP: The interaction energy between an external magnetic field and the orbital angular momentum of the atom is given by Eq.(41.18). The energy depends on ml with the most negative ml value having the lowest energy. EXECUTE: (a) For the 5g level, l = 4 and there are 2l + 1 = 9 different ml states. The 5g level is split into 9 levels by the magnetic field. (b) Each ml level is shifted in energy an amount given by U = ml μ B B. Adjacent levels differ in ml by one, so ΔU = μ B B. e= (1.602 × 10−19 C)(1.055 × 10−34 J ⋅ s) = = 9.277 × 10 −24 A ⋅ m 2 2m 2(9.109 × 10 −31 kg) ΔU = μ B B = (9.277 × 10−24 A/m 2 )(0.600 T) = 5.566 × 10−24 J(1 eV/1.602 × 10−19 J) = 3.47 × 10−5 eV

μB =

(c) The level of highest energy is for the largest ml , which is ml = l = 4; U 4 = 4 μ B B. The level of lowest energy is

for the smallest ml , which is ml = −l = −4; U −4 = −4 μ B B. The separation between these two levels is

41.16.

U 4 − U −4 = 8μ B B = 8(3.47 × 10−5 eV) = 2.78 × 10−4 eV. EVALUATE: The energy separations are proportional to the magnetic field. The energy of the n = 5 level in the absence of the external magnetic field is ( −13.6 eV)/52 = −0.544 eV, so the interaction energy with the magnetic field is much less than the binding energy of the state. (a) According to Figure 41.11 in the textbook there are three different transitions that are consistent with the selection rules. The initial ml values are 0, ±1; and the final ml value is 0. (b) The transition from ml = 0 to ml = 0 produces the same wavelength (122 nm) that was seen without the magnetic field. (c) The larger wavelength (smaller energy) is produced from the ml = −1 to ml = 0 transition. (d) The shorter wavelength (greater energy) is produced from the ml = +1 to ml = 0 transition.

41.17.

3 p ⇒ n = 3, l = 1, ΔU = μB B ⇒ B =

U (2.71 × 10 −5 eV) = = 0.468 T μB (5.79 × 10 −5 e V T)

(b) Three: ml = 0, ± 1. 41.18.

41.19.

(2.00232) ⎛ e ⎞ ⎛ −= ⎞ (a) U = + (2.00232) ⎜ μB B ⎟⎜ ⎟B = − 2 ⎝ 2m ⎠ ⎝ 2 ⎠ (2.00232) U =− (5.788 × 10 −5 e V T)(0.480 T) = −2.78 × 10−5 eV. 2 (b) Since n = 1, l = 0 so there is no orbital magnetic dipole interaction. But if n ≠ 0 there could be since l < n allows for l ≠ 0. G G IDENTIFY and SET UP: The interaction energy is U = − μ ⋅ B , with μ z given by Eq.(41.22). G G EXECUTE: U = − μ ⋅ B = + μ z B, since the magnetic field is in the negative z-direction. ⎛ e ⎞ ⎛ e ⎞ ⎟ S z , so U = −(2.00232) ⎜ ⎟ Sz B 2 m ⎝ ⎠ ⎝ 2m ⎠ ⎛ e= ⎞ S z = ms =, so U = −2.00232 ⎜ ⎟ ms B ⎝ 2m ⎠ e= = μB = 5.788 × 10−5 eV/T 2m U = −2.00232 μ B ms B 1 The ms = + level has lower energy. 2 1⎞ 1⎞ ⎛ 1 ⎛ 1 ⎞⎞ ⎛ ⎛ ΔU = U ⎜ ms = − ⎟ − U ⎜ ms = + ⎟ = −2.00232 μB B ⎜ − − ⎜ + ⎟ ⎟ = +2.00232 μB B 2⎠ 2⎠ ⎝ ⎝ ⎝ 2 ⎝ 2 ⎠⎠

μ z = −(2.00232) ⎜

ΔU = +2.00232(5.788 × 10−5 eV/T)(1.45 T) = 1.68 × 10−4 eV


41-4

41.20. 41.21.

41.22.

41.23.

Chapter 41

EVALUATE: The interaction energy with the electron spin is the same order of magnitude as the interaction energy with the orbital angular momentum for states with ml ≠ 0. But a 1s state has l = 0 and ml = 0, so there is no orbital magnetic interaction. ⎛ 1⎞ ⎛ 1⎞ ⎛ 3⎞ ⎛ 3⎞ ⎛ 5⎞ The allowed (l , j ) combinations are ⎜ 0, ⎟ , ⎜1, ⎟ , ⎜1, ⎟ , ⎜ 2, ⎟ and ⎜ 2, ⎟ . 2 2 2 2 ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ 2⎠ IDENTIFY and SET UP: j can have the values l + 1/ 2 and l − 1/ 2. EXECUTE: If j takes the values 7/2 and 9/2 it must be that l − 1/ 2 = 7 / 2 and l = 8/ 2 = 4. The letter that labels this l is g. EVALUATE: l must be an integer. hc (4.136 × 10 −15 eV ⋅ s)(300 × 108 m s ) c (3.00 × 108 m s ) (a) λ = f = = 21 cm, = = = 1.4 × 109 Hz, a short radio λ ΔE (5.9 × 10−6 eV) 0.21 m wave. (b) As in Example 41.6, the effective field is B ≅ Δ E 2 μB = 5.1 × 10−2 T, for smaller than that found in the example. IDENTIFY and SET UP: For a classical particle L = I ω. For a uniform sphere with mass m and radius R, 2 ⎛2 ⎞ I = mR 2 , so L = ⎜ mR 2 ⎟ ω. Solve for ω and then use v = rω to solve for v. 5 5 ⎝ ⎠ EXECUTE: (a) L =

ω=

41.24. 41.25.

41.26. 41.27.

3 2 3 = so mR 2ω = = 4 5 4

5 3/ 4= 5 3/ 4(1.055 × 10−34 J ⋅ s) = = 2.5 × 1030 rad/s 2 2mR 2(9.109 × 10 −31 kg)(1.0 × 10 −17 m) 2

(b) v = rω = (1.0 × 10−17 m)(2.5 × 1030 rad/s) = 2.5 × 1013 m/s. EVALUATE: This is much greater than the speed of light c, so the model cannot be valid. However the number of electrons is obtained, the results must be consistent with Table (41.3); adding two more electrons to the zinc configuration gives 1s 2 2 s 2 2 p 6 3s 2 3 p 6 4 s 2 3d 10 4 p 2 . The ten lowest energy levels for electrons are in the n = 1 and n = 2 shells. 1 n = 1, l = 0, ml = 0, ms = ± : 2 states. 2 1 n = 2, l = 0, ml = 0, ms = ± : 2 states. 2 1 n = 2, l = 1, ml = 0, ± 1, ms = ± : 6 states. 2 For the outer electrons, there are more inner electrons to screen the nucleus. IDENTIFY and SET UP: The energy of an atomic level is given in terms of n and Z eff by Eq.(41.27),

⎛ Z2 ⎞ En = − ⎜ eff2 ⎟ (13.6 eV). The ionization energy for a level with energy − En is + En . ⎝ n ⎠ (2.771) 2 EXECUTE: n = 5 and Zeff = 2.771 gives E5 = − (13.6 eV) = −4.18 eV 52 The ionization energy is 4.18 eV. 2 EVALUATE: The energy of an atomic state is proportional to Z eff . 41.28.

41.29.

For the 4s state, E = −4.339 eV and Z eff = 4 ( −4.339) ( −13.6) = 2.26. Similarly, Z eff = 1.79 for the 4p state and 1.05 for the 4d state. The electrons in the states with higher l tend to be further away from the filled subshells and the screening is more complete. IDENTIFY and SET UP: Use the exclusion principle to determine the ground-state electron configuration, as in Table 41.3. Estimate the energy by estimating Z eff , taking into account the electron screening of the nucleus. EXECUTE: (a) Z = 7 for nitrogen so a nitrogen atom has 7 electrons. N 2+ has 5 electrons: 1s 2 2 s 2 2 p. (b) Z eff = 7 − 4 = 3 for the 2p level.

⎛ Z2 ⎞ 32 En = − ⎜ eff2 ⎟ (13.6 eV) = − 2 (13.6 eV) = −30.6 eV 2 ⎝ n ⎠ (c) Z = 15 for phosphorus so a phosphorus atom has 15 electrons. P 2+ has 13 electrons: 1s 2 2 s 2 2 p 6 3s 2 3 p


Atomic Structure

41-5

(d) Z eff = 15 − 12 = 3 for the 3p level.

41.30.

41.31.

⎛ Z2 ⎞ 32 En = − ⎜ eff2 ⎟ (13.6 eV) = − 2 (13.6 eV) = −13.6 eV 3 ⎝ n ⎠ EVALUATE: In these ions there is one electron outside filled subshells, so it is a reasonable approximation to assume full screening by these inner-subshell electrons. 13.6 eV 2 (a) E2 = − Z eff , so Z eff = 1.26. 4 (b) Similarly, Z eff = 2.26. (c) Z eff becomes larger going down the columns in the periodic table. IDENTIFY and SET UP: Estimate Z eff by considering electron screening and use Eq.(41.27) to calculate the energy. Z eff is calculated as in Example 41.8. EXECUTE: (a) The element Be has nuclear charge Z = 4. The ion Be + has 3 electrons. The outermost electron sees the nuclear charge screened by the other two electrons so Z eff = 4 − 2 = 2.

⎛ Z2 ⎞ 22 En = − ⎜ eff2 ⎟ (13.6 eV) so E2 = − 2 (13.6 eV) = −13.6 eV 2 ⎝ n ⎠ 22 (13.6 eV) = −3.4 eV 42 EVALUATE: For the electron in the highest l-state it is reasonable to assume full screening by the other electrons, as in Example 41.8. The highest l-states of Be + , Mg + , Ca + , etc. all have a Z eff = 2. But the energies are different because for each ion the outermost sublevel has a different n quantum number. 7.46 × 103 eV = 28.0, which corresponds to the element Nickel (Ni). Ekx ≅ ( Z − 1) 2 (10.2 eV) . Z ≈ 1 + 10.2 eV

(b) The outermost electron in Ca + sees a Z eff = 2. E4 = −

41.32. 41.33.

(a) Z = 20 : f = (2.48 × 1015 Hz)(20 − 1) 2 = 8.95 × 1017 Hz . c 3.00 × 108 m s = = 3.35 × 10 −10 m. f 8.95 × 1017 Hz (b) Z = 27: f = 1.68 × 1018 Hz. E = 6.96 keV. λ = 1.79 × 10 −10 m. E = hf = (4.14 × 10−15 eV ⋅ s) (8.95 × 1017 Hz) = 3.71 keV. λ =

41.34.

(c) Z = 48 : f = 5.48 × 1018 Hz, E = 22.7 keV, λ = 5.47 × 10 −11 m. IDENTIFY: The orbital angular momentum is limited by the shell the electron is in. SET UP: For an electron in the n shell, its orbital angular momentum quantum number l is limited by 0 ≤ l < n, and its orbital angular momentum is given by L = l (l + 1) = . The z-component of its angular momentum is

Lz = ml =, where ml = 0, ±1, … , ±l, and its spin angular momentum is S = 3/ 4 = for all electrons. Its energy in the nth shell is En = −(13.6 eV)/n 2 . EXECUTE: (a) L = l (l + 1) = = 12= ⇒ l = 3. Therefore the smallest that n can be is 4, so En = – (13.6 eV)/n2 = – (13.6 eV)/42 = –0.8500 eV. (b) For l = 3, ml = ±3, ±2, ±1, 0. Since Lz = ml =, the largest Lz can be is 3 = and the smallest it can be is –3 = .

41.35.

(c) S = 3/ 4 = for all electrons. (d) In this case, n = 3, so l = 2, 1, 0. Therefore the maximum that L can be is Lmax = 2(2 + 1) = = 6 = . The minimum L can be is zero when l = 0. EVALUATE: At the quantum level, electrons in atoms can have only certain allowed values of their angular momentum. IDENTIFY: The total energy determines what shell the electron is in, which limits its angular momentum. SET UP: The electron’s orbital angular momentum is given by L = l (l + 1) = , and its total energy in the nth shell

is En = −(13.6 eV)/n 2 . EXECUTE: (a) First find n: En = −(13.6 eV)/n 2 = − 0.5440 eV which gives n = 5, so l = 4, 3, 2, 1, 0. Therefore the

possible values of L are given by L = l (l + 1) = , giving L = 0,

2 =,

6=,

12=,

20=.

(b) E6 = – (13.6 eV)/6 = –0.3778 eV. ΔE = E6 – E5 = –0.3778 eV – (–0.5440 eV) = +0.1662 eV This must be the energy of the photon, so ΔE = hc/λ, which gives λ = hc/ΔE = (4.136 × 10–15 eV ⋅ s )(3.00 ×108 m/s)/(0.1662 eV) = 7.47 × 10–6 m = 7470 nm, which is in the infrared and hence not visible. EVALUATE: The electron can have any of the five possible values for its angular momentum, but it cannot have any others. 2


41-6

Chapter 41

41.36.

IDENTIFY: For the N shell, n = 4, which limits the values of the other quantum numbers. SET UP: In the nth shell, 0 ≤ l < n, ml = 0, ±1, … , ±l, and ms = ±1/2. The orbital angular momentum of the electron is L = l (l + 1) = and its spin angular momentum is S = 3/ 4 = . EXECUTE: (a) For l = 3 we can have ml = ±3, ±2±, ±1, 0 and ms = ±½; for l = 2 we can have ml = ±2, ±1, 0 and ms = ±½; for l = 1, we can have ml = ±1, 0 and ms = ±1/2 ; for l = 0, we can have ml = 0 and ms = ±1/2. (b) For the N shell, n = 4, and for an f-electron, l = 3, giving L = l (l + 1) = = 3(3 + 1) = = 12= . Lz =

ml = = ±3=, ±2=, ± =, 0, so the maximum value is 3= . S = 3/ 4 = for all electrons.

41.37.

(c) For a d-state electron, l = 2, giving L = 2(2 + 1) = = 6= . Lz = ml =, and the maximum value of ml is 2, so the maximum value of Lz is 2 = . The smallest angle occurs when Lz is most closely aligned along the angular L 2= 2 = and θmin = 35.3°. The largest momentum vector, which is when Lz is greatest. Therefore cosθ min = z = L 6= 6 angle occurs when Lz is as far as possible from the L-vector, which is when Lz is most negative. Therefore −2= 2 and θ max = 144.7° . cosθ max = =− 6= 6 (d) This is not possible since l = 3 for an f-electron, but in the M shell the maximum value of l is 2. EVALUATE: The fact that the angle in part (c) cannot be zero tells us that the orbital angular momentum of the electron cannot be totally aligned along any specified direction. IDENTIFY: The inner electrons shield part of the nuclear charge from the outer electron. Z2 SET UP: The electron’s energy in the nth shell, due to shielding, is En = − eff2 (13.6 eV) , where Zeff e is the n effective charge that the electron “sees” for the nucleus. 2 Z eff (4 2 )( −1.947 eV) and n = 4 for the 4s state. Solving for Z (13.6 eV) gives Z = − eff eff n2 13.6 eV = 1.51. The nucleus contains a charge of +11e, so the average number of electrons that screen this nucleus must be 11 – 1.51 = 9.49 electrons (b) (i) The charge of the nucleus is +19e, but 17.2e is screened by the electrons, so the outer electron “sees” 19e – 17.2e = 1.8e and Zeff = 1.8. Z2 (1.8) 2 (ii) En = − eff2 (13.6 eV) = − 2 (13.6 eV) = −2.75 eV n 4 EVALUATE: Sodium has 11 protons, so the inner 10 electrons shield a large portion of this charge from the outer electron. But they don’t shield 10 of the protons, since the inner electrons are not totally equivalent to a uniform spherical shell. (They are lumpy.)

EXECUTE: (a) En = −

2

41.38. 41.39.

2

See Example 41.3; r 2 ψ = Cr 2e −2 r/a ,

d (r 2 ψ )

= Ce −2 r/a (2r − (2r 2 /a)), and for a maximum, r = a, the distance of dr the electron from the nucleus in the Bohr model. (a) IDENTIFY and SET UP: The energy is given by Eq.(38.18), which is identical to Eq.(41.3). The potential energy is given by Eq.(23.9), with q = + Ze and q0 = −e. EXECUTE:

E1s = −

E1s = U ( r ) gives −

1 me 4 1 e2 ; U (r ) = − 2 2 (4π P0 ) 2= 4π P0 r

1 me 4 1 e2 =− 2 2 (4π P0 ) 2= 4π P0 r

(4π P0 )2= 2 = 2a me 2 EVALUATE: The turning point is twice the Bohr radius. (b) IDENTIFY and SET UP: For the 1s state the probability that the electron is in the classically forbidden region r=

is P (r > 2a ) = ∫ ψ 1s dV = 4π ∫ ψ 1s r 2 dr. The normalized wave function of the 1s state of hydrogen is given in 2

2a

Example 41.3: ψ 1s (r ) = EXECUTE:

2

2a

1

π a3

e − r / a . Evaluate the integral; the integrand is the same as in Example 41.3.

⎛ 1 ⎞ ∞ P (r > 2a ) = 4π ⎜ 3 ⎟ ∫ r 2e−2 r / a dr ⎝ π a ⎠ 2a


Atomic Structure

Use the integral formula

∫r e

2 −α r

41-7

⎛ r 2 2r 2 ⎞ dr = −e −α r ⎜ + 2 + 3 ⎟ , with α = 2 / a. α ⎠ ⎝α α ∞

⎛ ar 2 a 2 r a3 ⎞ ⎤ 4⎡ 4 P (r > 2a ) = − 3 ⎢ e −2 r / a ⎜ + + ⎟ ⎥ = + 3 e −4 (2a 3 + a 3 + a3 / 4) a ⎣ 2 4 ⎠⎦ 2a a ⎝ 2

41.40.

P (r > 2a ) = 4e −4 (13/ 4) = 13e−4 = 0.238. EVALUATE: These is a 23.8% probability of the electron being found in the classically forbidden region, where classically its kinetic energy would be negative. (a) For large values of n, the inner electrons will completely shield the nucleus, so Z eff = 1 and the ionization

energy would be

13.60 eV . n2

13.60 eV = 1.11× 10−4 eV, r350 = (350) 2 a0 = (350) 2 (0.529 × 10 −10 m) = 6.48 × 10−6 m . 3502 13.60 eV (c) Similarly for n = 650, = 3.22 × 10−5 eV, r650 = (650) 2 (0.529 × 10−10 m) = 2.24 × 10 −5 m. (650) 2 (b)

41.41.

ψ 2 s (r ) =

r ⎞ − r/ 2 a ⎛ ⎜ 2 − ⎟e a⎠ 32π a ⎝ 1

3

(a) IDENTIFY and SET UP: Let I = ∫ ψ 2 s dV = 4π ∫ ψ 2 s r 2 dr. If ψ 2 s is normalized then we will find that 2

0

2

0

I = 1. r⎞ 1 ∞⎛ 4r 3 r 4 ⎞ − r / a ⎛ 1 ⎞ ∞⎛ I = 4π ⎜ + 2 ⎟ e dr 2 − ⎟ e − r/a r 2 dr = 3 ∫ ⎜ 4r 2 − 3 ⎟ ∫0 ⎜ a⎠ a a ⎠ 8a 0 ⎝ ⎝ 32π a ⎠ ⎝ ∞ n! Use the integral formula ∫ x ne −α x dx = n +1 , with α = 1/ a 2

EXECUTE:

α

0

1 ⎛ 4 1 ⎞ 1 I = 3 ⎜ 4(2!)(a 3 ) − (3!)( a) 4 + 2 (4!)( a)5 ⎟ = (8 − 24 + 24) = 1; this ψ 2 s is normalized. a a 8a ⎝ ⎠ 8 (b) SET UP: For a spherically symmetric state such as the 2s, the probability that the electron will be found at 4a

4a

r < 4a is P (r < 4a) = ∫ ψ 2 s dV = 4π ∫ ψ 2 s r 2 dr. 2

0

EXECUTE:

P (r < 4a ) =

Let P (r < 4a) =

2

0

1 4 a ⎛ 2 4r 3 r 4 ⎞ − r / a + 2 ⎟ e dr ⎜ 4r − 8a 3 ∫ 0 ⎝ a a ⎠

1 ( I1 + I 2 + I 3 ). 8a 3

4a

I1 = 4 ∫ r 2e − r / a dr 0

⎛ r 2 2r 2 ⎞ Use the integral formula ∫ r 2e −α r dr = −e −α r ⎜ + 2 + 3 ⎟ with α = 1/ a. α ⎠ ⎝α α −r / a 2 2 3 4a −4 3 I1 = −4[e ( r a + 2ra + 2a )]0 = ( −104e + 8)a .

I2 = −

4 4a 3 −r / a r e dr a ∫0

⎛ r 3 3r 2 6r 6 ⎞ Use the integral formula ∫ r 3e −α r dr = −e −α r ⎜ + 2 + 3 + 4 ⎟ with α = 1/ a. a α ⎠ ⎝α α 4 −r / a 3 I 2 = [e (r a + 3r 2 a 2 + 6ra3 + 6a 4 )] 04 a = (568e −4 − 24) a3 . a 1 4a I 3 = 2 ∫ r 4e − r / a dr a 0 ⎛ r 4 4r 3 12r 2 24r 24 ⎞ Use the integral formula ∫ r 4e −α r dr = −e −α r ⎜ + 2 + 3 + 4 + 5 ⎟ with α = 1/ a. α a a ⎠ ⎝α α

I3 = −

1 −r / a 4 [e (r a + 4r 3a 2 + 12r 2 a 3 + 24ra 4 + 24a 5 )] 04 a = ( −824e −4 + 24) a3 . a2


41-8

Chapter 41

Thus P (r < 4a) =

1 1 ( I1 + I 2 + I 3 ) = 3 a3 ([8 − 24 + 24] + e −4 [−104 + 568 − 824]) 3 8a 8a

1 P (r < 4a) = (8 − 360e −4 ) = 1 − 45e −4 = 0.176. 8 EVALUATE: There is an 82.4% probability that the electron will be found at r > 4a. In the Bohr model the electron is for certain at r = 4a; this is a poor description of the radial probability distribution for this state. 41.42.

(a) Since the given ψ ( r ) is real, r 2 | ψ |2 = r 2ψ 2 . The probability density will be an extreme when

d 2 2 dψ ⎞ dψ ⎞ ⎛ ⎛ ( r ψ ) = 2 ⎜ rψ 2 + r 2ψ ⎟ = 2rψ ⎜ ψ + r ⎟ = 0. This occurs at r = 0, a minimum, and when ψ = 0, also a dr dr dr ⎠ ⎝ ⎠ ⎝ dψ = 0. Within a multiplicative constant, ψ ( r ) = (2 − r a )e− r 2 a , minimum. A maximum must correspond to ψ + r dr dψ 1 = − (2 − r 2a)e − r 2 a , and the condition for a maximum is (2 − r a ) = (r a ) (2 − r 2a ), or r 2 − 6ra + 4a 2 = 0. dr a The solutions to the quadratic are r = a(3 ± 5). The ratio of the probability densities at these radii is 3.68, with

41.43.

the larger density at r = a(3 + 5) . (b) ψ = 0 at r = 2a Parts (a) and (b) are consistent with Figure 41.5 in the textbook; note the two relative maxima, one on each side of the minimum of zero at r = 2a. L ⎛L ⎞ IDENTIFY: Use Figure 41.2 in the textbook to relate θ L to Lz and L: cosθ L = z so θ L = arccos ⎜ z ⎟ L ⎝ L⎠ (a) SET UP: The smallest angle (θ L ) min is for the state with the largest L and the largest Lz . This is the state with

l = n − 1 and ml = l = n − 1. EXECUTE:

Lz = ml = = ( n − 1)=

L = l (l + 1)= = (n − 1) n= ⎛ (n − 1)= ⎞ ⎛ ( n − 1) ⎞ ⎛ n −1 ⎞ (θ L ) min = arccos ⎜ = arccos ⎜ = arccos ⎜⎜ ⎟ ⎟ = arccos( 1 − 1/ n ). ⎜ ( n − 1)n= ⎟ ⎜ (n − 1) n ⎟⎟ n ⎠⎟ ⎝ ⎝ ⎠ ⎝ ⎠ EVALUATE: Note that (θ L ) min approaches 0° as n → ∞. (b) SET UP: The largest angle (θ L ) max is for l = n − 1 and ml = −l = −(n − 1). EXECUTE: A similar calculation to part (a) yields (θ L ) max = arccos( − 1 − 1/ n ) EVALUATE: Note that (θ L ) max approaches 180° as n → ∞. 41.44.

(a) L2x + L2y = L2 − L2z = l (l + 1)= 2 − ml2= 2 so L2x + L2y = l (l + 1) − ml2 =. (b) This is the magnitude of the component of angular momentum perpendicular to the z-axis. (c) The maximum value is

l (l + 1)= = L, when ml = 0. That is, if the electron is known to have no z-component

of angular momentum, the angular momentum must be perpendicular to the z-axis. The minimum is ml = ±l. 41.45.

l= when

4 r4 ⎛ 1 ⎞ 4 − r 2 a dP ⎛ 1 ⎞ ⎛ 3 r ⎞ − r 2 a dP 4r − ⎟ e P( r ) = ⎜ r e . =⎜ . = 0 when 4r 3 − = 0; r = 4a. In the Bohr 5 ⎟⎜ 5 ⎟ dr ⎝ 24a ⎠ ⎝ a⎠ dr a ⎝ 24a ⎠

model, rn = n 2 a so r2 = 4a, which agrees. 41.46.

The time required to transit the horizontal 50 cm region is t =

Δ x 0.500 m = = 0.952 ms. The force required to vx 525 m s

⎛ ⎞ 2(0.50 × 10 −3 m) 2Δz 0.1079 kg mol =±⎜ = ⎟ −3 2 23 2 t ⎝ 6.022 × 10 atoms mol ⎠ (0.952 × 10 s) N. According to Eq.(41.22), the value of μz is | μ z | = 9.28 × 10 −24 A ⋅ m 2 . Thus, the required

deflect each spin component by 0.50 mm is Fz = maz = ± m ±1.98 × 10−22

magnetic-field gradient is

dBz F 1.98 × 10−22 N = z = = 21.3 T m. μz 9.28 × 10−24 J T dz


Atomic Structure

41.47.

41-9

Decay from a 3d to 2 p state in hydrogen means that n = 3 → n = 2 and ml = ±2, ± 1, 0 → ml = ±1, 0. However selection rules limit the possibilities for decay. The emitted photon carries off one unit of angular momentum so l must change by 1 and hence ml must change by 0 or ±1. The shift in the transition energy from the zero field e=B ( ml3 − ml2 ), where ml3 is the 3d ml value and ml2 is the 2 p ml value. Thus 2m there are only three different energy shifts. They and the transitions that have them, labeled by the ml names, are:

value is just U = ( ml3 − ml2 ) μB B =

e=B : 2 → 1, 2m 0 :1 → 1,

41.48.

1 → 0,

0 → −1

0 → 0,

− 1 → −1

e=B − : 0 → 1, − 1 → 0, − 2 → −1 2m IDENTIFY: The presence of an external magnetic field shifts the energy levels up or down, depending upon the value of ml. SET UP: The selection rules tell us that for allowed transitions, Δl = 1 and Δml = 0 or ±1. EXECUTE: (a) E = hc/λ = (4.136 × 10–15 eV ⋅ s )(3.00 × 108 m/s)/(475.082 nm) = 2.612 eV. (b) For allowed transitions, Δl = 1 and Δml = 0 or ±1. For the 3d state, n = 3, l = 2, and ml can have the values 2, 1, 0, –1, –2. In the 2p state, n = 2, l = 1, and ml can be 1, 0, –1. Therefore the 9 allowed transitions from the 3d state in the presence of a magnetic field are: l = 1, ml = 1; l = 2 , ml = 2 l = 1, ml = 0 l = 2 , ml = 1 l = 1, ml = 1 l = 2 , ml = 1 l = 1, ml = 0 l = 2 , ml = 0 l = 1, ml = 1 l = 2 , ml = 0 l = 1, ml = –1 l = 2 , ml = 0 l = 1, ml = 0 l = 2 , ml = –1 l = 1, ml = –1 l = 2 , ml = –1 l = 1, ml = –1 l = 2 , ml = –2 (c) ΔE = µ BB = (5.788 × 10–5 eV/T)(3.500 T) = 0.000203 eV So the energies of the new states are –8.50000 eV + 0 and –8.50000 eV ± 0.000203 eV, giving energies of: –8.50020 eV, –8.50000 eV, and –8.49980 eV (d) The energy differences of the allowed transitions are equal to the energy differences if no magnetic field were present (2.61176 eV, from part (a)), and that value ±ΔE (0.000203 eV, from part (c)). Therefore we get the following. For E = 2.61176 eV: λ = 475.082 nm (which was given) For E = 2.61176 eV + 0.000203 eV = 2.611963 eV:

→ → → → → → → → →

λ = hc/E = (4.136 ×10–15 eV ⋅ s )(3.00 × 108 m/s)/(2.611963 eV) = 475.045 nm For E = 2.61176 eV – 0.000203 eV = 2.61156 eV:

λ = hc/E = (4.136 ×10–15 eV ⋅ s )(3.00 × 108 m/s)/(2.61156 eV) = 475.119 nm

41.49.

41.50.

EVALUATE: Even a strong magnetic field produces small changes in the energy levels, and hence in the wavelengths of the emitted light. IDENTIFY: The presence of an external magnetic field shifts the energy levels up or down, depending upon the value of ml. SET UP: The energy difference due to the magnetic field is ΔE = µ BB and the energy of a photon is E = hc/λ. EXECUTE: For the p state, ml = 0 or ±1, and for the s state ml = 0. Between any two adjacent lines, ΔE = µ BB. Since the change in the wavelength (Δλ) is very small, the energy change (ΔE ) is also very small, so we can use hcΔλ hcΔλ hc hcΔλ . Since ΔE = µ BB, we get µ B B = and B = . differentials. E = hc/λ . | dE | = 2 d λ and ΔE = 2 2 λ λ λ µ Bλ 2 B = (4.136 × 10–15 eV ⋅ s )(3.00 × 108 m/s)(0.0462 nm)/(5.788 × 10–5 eV/T)(575.050 nm)2 = 3.00 T EVALUATE: Even a strong magnetic field produces small changes in the energy levels, and hence in the wavelengths of the emitted light. (a) The energy shift from zero field is ΔU 0 = ml μ B B. For ml = 2, ΔU 0 = (2) (5.79 × 10−5 e V T) (1.40 T) = 1.62 × 10−4 eV. For ml = 1, ΔU 0 = (1)(5.79 × 10 −5 e V T) (1.40 T) = 8.11 × 10−5 eV.

(b) | Δλ |= λ0

|ΔE | ⎛ 36 ⎞ 1 , where E0 = (13.6 eV)((1/ 4) − (1/ 9)), λ0 = ⎜ ⎟ = 6.563 × 10−7 m E0 ⎝ 5 ⎠R


41-10

41.51.

Chapter 41

and Δ E = 1.62 × 10−4 eV − 8.11 × 10−5 eV = 8.09 × 10 −5 eV from part (a). Then, | Δλ | = 2.81 × 10 −11 m = 0.0281 nm . The wavelength corresponds to a larger energy change, and so the wavelength is smaller. n IDENTIFY: The ratio according to the Boltzmann distribution is given by Eq.(38.21): 1 = e − ( E1 − E0 ) / kT , where 1 is n0 the higher energy state and 0 is the lower energy state. ⎛ e= ⎞ SET UP: The interaction energy with the magnetic field is U = − μ z B = 2.00232 ⎜ ⎟ ms B (Example 41.5.). The ⎝ 2m ⎠ 1 1 energy of the ms = + level is increased and the energy of the ms = − level is decreased. 2 2 n1/ 2 − (U1/ 2 −U −1/ 2 ) / kT =e n−1/ 2

EXECUTE:

⎛ e= ⎞ ⎛ 1 ⎛ 1 ⎞ ⎞ ⎛ e= ⎞ U1/ 2 − U −1/ 2 = 2.00232 ⎜ ⎟ B ⎜ − ⎜ − ⎟ ⎟ = 2.00232 ⎜ ⎟ B = 2.00232μB B m 2 2 2 ⎝ ⎠ ⎝ ⎝ ⎠⎠ ⎝ 2m ⎠

n1/ 2 = e − ( 2.00232) μB B / kT n−1/ 2 (a) B = 5.00 × 10−5 T −24 2 −5 −23 n1/ 2 = e −2.00232(9.274×10 A/m )(5.00×10 T)/([1.381×10 J/K][300 K ]) n−1/ 2 −7 n1/ 2 = e −2.24×10 = 0.99999978 = 1 − 2.2 × 10 −7 n−1/ 2

(b) B = 5.00 × 10 −5 T,

−3 n1/ 2 = e −2.24×10 = 0.9978 n−1/ 2

−2 n1/ 2 = e −2.24×10 = 0.978 n−1/ 2 EVALUATE: For small fields the energy separation between the two spin states is much less than kT for T = 300 K and the states are equally populated. For B = 5.00 T the energy spacing is large enough for there to be a small excess of atoms in the lower state.

(c) B = 5.00 × 10 −5 T,

41.52.

Using Eq.(41.4), L = mvr = l (l + 1)=, and the Bohr radius from Eq.(38.15), we obtain the following value for v :

l (l + 1)= 2(6.63 × 10−34 J ⋅ s) = = 7.74 × 105 m s. The magnetic field generated by the 2 m( n a0 ) 2π (9.11 × 10−31 kg) (4) (5.29 × 10−11 m) “moving” proton at the electrons position can be calculated from Eq.(28.1): μ | q | v sin φ (1.60 × 10−19 C) (7.74 × 105 m s) sin(90°) B= 0 = (10−7 T ⋅ m A) = 0.277 T. 2 4π (4) 2 (5.29 × 10−11 m) 2 r v=

41.53.

3 1 1 3 ms can take on 4 different values: ms = − , − , + , + . Each nlml state can have 4 2 2 2 2 electrons, each with one of the four different ms values. Apply the exclusion principle to determine the electron configurations. EXECUTE: (a) For a filled n = 1 shell, the electron configuration would be 1s 4 ; four electrons and Z = 4. For a IDENTIFY and SET UP:

filled n = 2 shell, the electron configuration would be 1s 4 2 s 4 2 p12 ; twenty electrons and Z = 20.

41.54.

(b) Sodium has Z = 11; 11 electrons. The ground-state electron configuration would be 1s 4 2 s 4 2 p 3. EVALUATE: The chemical properties of each element would be very different. (a) Z 2 ( −13.6 eV) = (7) 2 (−13.6 eV) = −666 eV. (b) The negative of the result of part (a), 666 eV. (c) The radius of the ground state orbit is inversely proportional to the nuclear charge, and a = (0.529 × 10−10 m) 7 = 7.56 × 10−12 m. Z hc hc (d) λ = = , where E0 is the energy found in part (b), and λ = 2.49 nm. ΔE E 1 − 1 0 2 2 1 2

(

)


Atomic Structure

41.55.

41-11

(a) IDENTIFY and SET UP: The energy of the photon equals the transition energy of the atom: Δ E = hc / λ . The energies of the states are given by Eq.(41.3). 13.60 eV 13.60 eV 13.60 eV EXECUTE: En = − so E2 = − and E1 = − n2 4 1 1 3 ⎛ ⎞ Δ E = E2 − E1 = 13.60 eV ⎜ − + 1⎟ = (13.60 eV) = 10.20 eV = (10.20 eV)(1.602 × 10−19 J/eV) = 1.634 × 10−18 J ⎝ 4 ⎠ 4

hc (6.626 × 10−34 J ⋅ s)(2.998 × 108 m/s) = = 1.22 × 10−7 m = 122 nm ΔE 1.634 × 10 −18 J (b) IDENTIFY and SET UP: Calculate the change in ΔE due to the orbital magnetic interaction energy, Eq.(41.17), and relate this to the shift Δλ in the photon wavelength. EXECUTE: The shift of a level due to the energy of interaction with the magnetic field in the z-direction is U = ml μ B B. The ground state has ml = 0 so is unaffected by the magnetic field. The n = 2 initial state has

λ=

ml = −1 so its energy is shifted downward an amount U = ml μ B B = (−1)(9.274 × 10−24 A/m 2 )(2.20 T) = ( −2.040 × 10−23 J)(1 eV /1.602 × 10−19 J) = 1.273 × 10−4 eV Note that the shift in energy due to the magnetic field is a very small fraction of the 10.2 eV transition energy. Problem 39.56c shows that in this situation Δλ / λ = Δ E / E . This gives

⎛ 1.273 × 10 −4 eV ⎞ −3 Δλ = λ Δ E / E = 122 nm ⎜ ⎟ = 1.52 × 10 nm = 1.52 pm. 10.2 eV ⎝ ⎠ EVALUATE: The upper level in the transition is lowered in energy so the transition energy is decreased. A smaller ΔE means a larger λ ; the magnetic field increases the wavelength. The fractional shift in wavelength, Δλ / λ is 41.56.

41.57.

small, only 1.2 × 10 −5. The effective field is that which gives rise to the observed difference in the energy level transition, Δ E hc ⎛ λ1 − λ2 ⎞ 2πmc ⎛ λ1 − λ2 ⎞ −3 = B= ⎜ ⎟= ⎜ ⎟ . Substitution of numerical values gives B = 3.64 × 10 T, much smaller μB μB ⎝ λ1λ2 ⎠ e ⎝ λ1λ2 ⎠ than that for sodium. IDENTIFY: Estimate the atomic transition energy and use Eq.(38.6) to relate this to the photon wavelength. (a) SET UP: vanadium, Z = 23 minimum wavelength; corresponds to largest transition energy EXECUTE: The highest occupied shell is the N shell ( n = 4). The highest energy transition is N → K , with transition energy ΔE = EN − EK . Since the shell energies scale like 1/ n 2 neglect EN relative to EK , so Δ E = EK = ( Z − 1) 2 (13.6 eV) = (23 − 1) 2 (13.6 eV) = 6.582 × 103 eV = 1.055 × 10 −15 J. The energy of the emitted photon equals this transition energy, so the photon’s wavelength is given by Δ E = hc / λ so λ = hc / Δ E.

(6.626 × 10 −34 J ⋅ s)(2.998 × 108 m/s) = 1.88 × 10 −10 m = 0.188 nm. 1.055 × 10 −15 J SET UP: maximum wavelength; corresponds to smallest transition energy, so for the Kα transition EXECUTE: The frequency of the photon emitted in this transition is given by Moseley’s law (Eq.41.29): f = (2.48 × 1015 Hz)(Z − 1) 2 = (2.48 × 1015 Hz)(23 − 1) 2 = 1.200 × 1018 Hz

λ=

c 2.998 × 108 m/s = = 2.50 × 10−10 m = 0.250 nm f 1.200 × 1018 Hz (b) rhenium, Z = 45 Apply the analysis of part (a), just with this different value of Z. minimum wavelength Δ E = EK = ( Z − 1) 2 (13.6 eV) = (45 − 1) 2 (13.6 eV) = 2.633 × 104 eV = 4.218 × 10−15 J.

λ=

(6.626 × 10−34 J ⋅ s)(2.998 × 108 m/s) = 4.71 × 10−11 m = 0.0471 nm. 4.218 × 10 −15 J maximum wavelength f = (2.48 × 1015 Hz)(Z − 1) 2 = (2.48 × 1015 Hz)(45 − 1) 2 = 4.801× 1018 Hz

λ = hc / Δ E =

c 2.998 × 108 m/s = = 6.24 × 10−11 m = 0.0624 nm f 4.801 × 1018 Hz EVALUATE: Our calculated wavelengths have values corresponding to x rays. The transition energies increase when Z increases and the photon wavelengths decrease.

λ=


41-12

Chapter 41

41.58.

(a) Δ E = (2.00232)

41.59.

(a) To calculate the total number of states for the n th principal quantum number shell we must multiply all the possibilities. The spin states multiply everything by 2. The maximum l value is (n –1), and each l value has (2l + 1)ml values. So the total number of states is

e e= hc 2πmc ⇒B= BΔS z ≈ B = . λ λe 2m m 2π (9.11 × 10 −31 kg) (3.00 × 108 m s ) (b) B = = 0.307 T. (0.0350 m)(1.60 × 10−19 C)

n −1

n −1

n −1

l =0

l =0

l =0

N = 2∑ (2l + 1) = 2∑1 + 4∑ l = 2n + 41.60.

4( n − 1)(n) = 2n + 2n 2 − 2n = 2n 2 . 2

(b) The n = 5 shell (O-shell) has 50 states. IDENTIFY: We treat the Earth as an electron. SET UP: The intrinsic spin angular momentum of an electron is S =

3 = , and the angular momentum of the 4

spinning Earth is S = I ω , where I = 2/5 mR2. EXECUTE: (a) Using S = I ω =

41.61.

3 = and solving for ω gives 4

3 3 = (1.055 ×10−34 J ⋅ s ) 4 4 ω= = = 9.40 × 10−73 rad/s 2 2 2 2 24 6 mR ( 5.97 ×10 kg )( 6.38 × 10 m ) 5 5 (b) We could not use this approach on the electron because in quantum physics we do not view it in the classical sense as a spinning ball. EVALUATE: The angular velocity we have just calculated for the Earth would certainly be masked by its present angular spin of one revolution per day. 1 The potential U ( x ) = k ′x 2 is that of a simple harmonic oscillator. Treated quantum mechanically (see Section 40.4) 2 each energy state has energy En = =ω (n + 12 ). Since electrons obey the exclusion principle, this allows us to put two electrons (one for each ms = ± 12 ) for every value of n⎯each quantum state is then defined by the ordered pair of quantum numbers ( n, ms ). By placing two electrons in each energy level the lowest energy is then N −1 ⎛ N −1 ⎛ 1 ⎞⎞ 1⎤ ⎛ N −1 ⎞ ⎡ N −1 ⎡ ( N − 1)( N ) N ⎤ + ⎥= 2 ⎜ ∑ En ⎟ = 2 ⎜ ∑ =ω ⎜ n + ⎟ ⎟ = 2=ω ⎢ ∑ n + ∑ ⎥ = 2=ω ⎢ 2 ⎠⎠ 2 2⎦ ⎝ ⎣ n =0 2 ⎦ ⎝ n =0 ⎠ ⎣ n=0 ⎝ n=0

=ω [ N 2 − N + N ] = =ω N 2 = =N 2

41.62.

k′ . m

Here we used the hint from Problem 41.59 to do the first sum, realizing that the first value of n is zero and the last value of n is N – 1, giving us a total of N energy levels filled. (a) Apply Coulomb’s law to the orbiting electron and set it equal to the centripetal force. There is an attractive (+2e)( −e) ( −e)(−e) force with charge +2e a distance r away and a repulsive force a distance 2r away. So, + = 4π P0 r 2 4π P0 (2r ) 2 − mv 2 = . But, from the quantization of angular momentum in the first Bohr orbit, L = mvr = = ⇒ v = . r mr 2

⎛ = ⎞ −m ⎜ 2 2 2 ⎟ e −2e −mv ⎝ mr ⎠ = − = ⇒ −7 e = − 4πP0= . So + = = 4 r2 mr 3 4πP0 r 2 4πP0 (4r ) 2 r r mr 3 2

2

2

4 ⎛ 4πP0= 2 ⎞ 4 4 −10 −11 r= ⎜ ⎟ = a0 = (0.529 × 10 m) = 3.02 × 10 m. 7 ⎝ me 2 ⎠ 7 7 −= 7 = 7 (1.054 × 10 −34 J ⋅ s) = = = 3.83 × 106 m s. And v = mr 4 ma0 4 (9.11 × 10−31 kg)(0.529 × 10−10 m) ⎛1 ⎞ (b) K = 2 ⎜ mv 2 ⎟ = 9.11 × 10−31 kg (3.83 × 106 m s) 2 = 1.34 × 10 −17 J = 83.5 eV. ⎝2 ⎠


Atomic Structure

41-13

⎛ −2e2 ⎞ −4e2 −7 ⎛ e 2 ⎞ e2 e2 −17 = + = (c) U = 2 ⎜ ⎟+ ⎜ ⎟ = −2.67 × 10 J = −166.9 eV 4 π r 4 π (2 r ) 4 π r 4 πE (2 r ) 2 4 π r P P P P ⎝ ⎝ 0 ⎠ 0 0 0 0 ⎠ (d) E∞ = −[ −166.9 eV + 83.5 eV] = 83.4 eV, which is only off by about 5% from the real value of 79.0 eV. 41.63.

(a) The radius is inversely proportional to Z, so the classical turning radius is 2a Z . (b) The normalized wave function is ψ1s ( r ) =

outside the classical turning point is P = ∫

2a Z

1 πa Z 3

3

e − Z r a and the probability of the electron being found

2

ψ1s 4πr 2 dr = ∞

4 a Z3 3

2a Z

e −2 Zr a r 2dr. Making the change of variable

u = Zr a , dr = (a Z ) du changes the integral to P = 4∫ e −2uu 2du, which is independent of Z. The probability is 2

that found in Problem 41.39, 0.238, independent of Z.


42

MOLECULES AND CONDENSED MATTER

42.1.

3 2 K 2(7.9 × 10−4 eV)(1.60 × 10−19 J eV) (a) K = kT ⇒ T = = = 6.1 K 2 3k 3(1.38 × 10−23 J K) 2(4.48 eV) (1.60 × 10 −19 J eV) (b) T = = 34,600 K. 3(1.38 × 10−23 J K)

(c) The thermal energy associated with room temperature (300 K) is much greater than the bond energy of He 2 (calculated in part (a)), so the typical collision at room temperature will be more than enough to break up He2 .

42.2.

42.3.

However, the thermal energy at 300 K is much less than the bond energy of H 2 , so we would expect it to remain intact at room temperature. 1 e2 (a) U = − = −5.0 eV. 4πε0 r (b) −5.0 eV + (4.3 eV − 3.5 eV) = −4.2 eV. IDENTIFY: The energy given to the photon comes from a transition between rotational states. =2 SET UP: The rotational energy of a molecule is E = l (l + 1) and the energy of the photon is E = hc/λ. 2I EXECUTE: Use the energy formula, the energy difference between the l = 3 and l = 1 rotational levels of the =2 5= 2 molecule is ΔE = [3(3 + 1) − 1(1 + 1) ] = . Since ΔE = hc/λ, we get hc/λ = 5 = 2 /I. Solving for I gives 2I I

I=

−34 5=λ 5 (1.055 × 10 J ⋅ s ) (1.780 nm) = = 4.981× 10−52 kg ⋅ m 2 . 2π c 2π ( 3.00 × 108 m/s )

Using I = mr r02, we can solve for r0: r0 =

I ( mN + mH ) = mN mH

( 4.981×10

−52

kg ⋅ m 2 )( 2.33 × 10−26 kg + 1.67 × 10−27 kg )

( 2.33 ×10

−26

kg )(1.67 × 10−27 kg )

42.4.

r0 = 5.65 × 10–13 m EVALUATE: This separation is much smaller than the diameter of a typical atom and is not very realistic. But we are treating a hypothetical NH molecule. The energy of the emitted photon is 1.01 × 10−5 eV, and so its frequency and wavelength are

42.5.

E (1.01 × 10−5 eV)(1.60 × 10−19 J eV) c (3.00 × 108 m s) and λ = = 2.44 GHz = = = 0.123 m. This frequency h (6.63 × 10−34 J ⋅ s) f (2.44 × 109 Hz) corresponds to that given for a microwave oven. Let 1 refer to C and 2 to O. m1 = 1.993 × 10−26 kg, m2 = 2.656 × 10−26 kg, r0 = 0.1128 nm . f =

⎛ m2 ⎞ ⎛ m1 ⎞ r1 = ⎜ ⎟ r0 = 0.0644 nm (carbon) ; r2 = ⎜ ⎟ r0 = 0.0484 nm (oxygen) ⎝ m1 + m2 ⎠ ⎝ m1 + m2 ⎠ (b) I = m1r12 + m2r22 = 1.45 × 10−46 kg ⋅ m 2 ; yes, this agrees with Example 42.2. 42.6.

Each atom has a mass m and is at a distance L 2 from the center, so the moment of inertia is 2( m)( L 2) 2 = mL2 2 = 2.21 × 10−44 kg ⋅ m 2 .

42.7.

IDENTIFY and SET UP: Set K = E1 from Example 42.2. Use K = 12 I ω 2 to solve for ω and v = rω to solve for v. EXECUTE: (a) From Example 42.2, E1 = 0.479 meV = 7.674 × 10−23 J and I = 1.449 × 10 −46 kg ⋅ m 2 K = 12 I ω 2 and K = E gives ω = 2 E1 / I = 1.03 × 1012 rad/s 42-1


42-2

Chapter 42

(b) v1 = r1ω1 = (0.0644 × 10 −9 m)(1.03 × 1012 rad/s) = 66.3 m/s (carbon) v2 = r2ω2 = (0.0484 × 10−9 m)(1.03 × 1012 rad/s) = 49.8 m/s (oxygen)

(c) T = 2π / ω = 6.10 × 10−12 s EVALUATE: From the information in Example 42.3 we can calculate the vibrational period to be T = 2π / ω = 2π mr / k ′ = 1.5 × 10−14 s. The rotational motion is over an order of magnitude slower than the

42.8. 42.9.

vibrational motion. 2 hc ⎛ 2πc ⎞ ΔE = = = k ′ mr , and solving for k ′, k ′ = ⎜ ⎟ mr = 205 N m. λ ⎝ λ ⎠ IDENTIFY and SET UP: The energy of a rotational level with quantum number l is El = l (l + 1)= 2 / 2 I (Eq.(42.3)). I = mr r 2 , with the reduced mass mr given by Eq.(42.4). Calculate I and Δ E and then use Δ E = hc / λ to find λ .

EXECUTE: (a) mr =

m1m2 mLi mH (1.17 × 10−26 kg)(1.67 × 10−27 kg) = = = 1.461 × 10−27 kg m1 + m2 mLi + mH 1.17 × 10 −26 kg + 1.67 × 10−27 kg

I = mr r 2 = (1.461 × 10−27 kg)(0.159 × 10 −9 m) 2 = 3.694 × 10 −47 kg ⋅ m 2

⎛ =2 ⎞ ⎛ =2 ⎞ l = 3 : E = 3(4) ⎜ ⎟ = 6 ⎜ ⎟ ⎝ 2I ⎠ ⎝ I ⎠ ⎛ =2 ⎞ ⎛ =2 ⎞ l = 4 : E = 4(5) ⎜ ⎟ = 10 ⎜ ⎟ ⎝ 2I ⎠ ⎝ I ⎠

42.10.

⎛ =2 ⎞ ⎛ (1.055 × 10−34 J ⋅ s) 2 ⎞ Δ E = E4 − E3 = 4 ⎜ ⎟ = 4 ⎜ = 1.20 × 10−21 J = 7.49 × 10−3 eV 2 ⎟ −47 I 3.694 10 kg m × ⋅ ⎝ ⎠ ⎝ ⎠ hc (4.136 × 10−15 eV)(2.998 × 108 m/s) = = 166 μ m (b) Δ E = hc / λ so λ = ΔE 7.49 × 10−3 eV EVALUATE: LiH has a smaller reduced mass than CO and λ is somewhat smaller here than the λ calculated for CO in Example 42.2 IDENTIFY: The vibrational energy of the molecule is related to its force constant and reduced mass, while the rotational energy depends on its moment of inertia, which in turn depends on the reduced mass. 1⎞ 1⎞ k′ =2 ⎛ ⎛ SET UP: The vibrational energy is En = ⎜ n + ⎟ =ω = ⎜ n + ⎟ = and the rotational energy is El = l (l + 1) . 2I 2⎠ 2 ⎠ mr ⎝ ⎝ EXECUTE:

For a vibrational transition, we have ΔEv = =

rotational transition is ΔER = mr r02 =

k′ , so we first need to find mr. The energy for a mr

=2 2= 2 [ 2(2 + 1) − 1(1 + 1)] = . Solving for I and using the fact that I = mrr02, we have 2I I

2= 2 , which gives ΔER

mr =

2 (1.055 × 10 −34 J ⋅ s )( 6.583 × 10 −16 eV ⋅ s ) 2= 2 = 2.0014 × 10–28 kg = 2 −9 −4 r02 ΔER 0.8860 10 m 8.841 10 eV × × ( )( )

Now look at the vibrational transition to find the force constant. ΔEv = =

42.11.

k′ mr

⇒ k′ =

2 ( 2.0014 ×10−28 kg ) (0.2560 eV) 2 = 30.27 N/m mr ( ΔEv ) = 2 =2 ( 6.583 ×10−16 eV ⋅ s )

EVALUATE: This would be a rather weak spring in the laboratory. l (l + 1)= 2 l (l − 1)= 2 =2 l= 2 , El −1 = (a) El = ⇒ ΔE = (l 2 + l − l 2 + l ) = 2I 2I 2I I Δ E ΔE l= = = . (b) f = h 2π= 2πI


Molecules and Condensed Matter

42.12.

IDENTIFY:

42-3

Find Δ E for the transition and compute λ from Δ E = hc / λ .

=2 =2 , with = 0.2395 × 10−3 eV. From Example 42.3, Δ E = 0.2690 eV 2I 2I is the spacing between vibrational levels. Thus En = ( n + 12 )=ω , with =ω = 0.2690 eV. By Eq.(42.9),

SET UP:

From Example 42.2, El = l (l + 1)

=2 . 2I (a) n = 0 → n = 1 and l = 1 → l = 2

E = En + El = ( n + 12 )=ω + l (l + 1)

EXECUTE:

⎛ =2 ⎞ For n = 0, l = 1, Ei = 12 =ω + 2 ⎜ ⎟ . ⎝ 2I ⎠ ⎛ =2 ⎞ For n = 1, l = 2, E f = 32 =ω + 6 ⎜ ⎟ . ⎝ 2I ⎠ ⎛ =2 ⎞ Δ E = E f − Ei = =ω + 4 ⎜ ⎟ = 0.2690 eV + 4(0.2395 × 10−3 eV) = 0.2700 eV ⎝ 2I ⎠ hc (4.136 × 10−15 eV ⋅ s)(2.998 × 108 m/s) hc = = 4.592 × 10−6 m = 4.592 μ m = Δ E so λ = ΔE 0.2700 eV λ (b) n = 0 → n = 1 and l = 2 → l = 1 ⎛ =2 ⎞ For n = 0, l = 2, Ei = 12 =ω + 6 ⎜ ⎟ . ⎝ 2I ⎠ ⎛ =2 ⎞ For n = 1, l = 1, E f = 32 =ω + 2 ⎜ ⎟ . ⎝ 2I ⎠ ⎛ =2 ⎞ Δ E = E f − Ei = =ω − 4 ⎜ ⎟ = 0.2690 eV − 4(0.2395 × 10−3 eV) = 0.2680 eV ⎝ 2I ⎠ −15 hc (4.136 × 10 eV ⋅ s)(2.998 × 108 m/s) = = 4.627 × 10−6 m = 4.627 μ m λ= ΔE 0.2680 eV (c) n = 0 → n = 1 and l = 3 → l = 2 ⎛ =2 ⎞ For n = 0, l = 3, Ei = 12 =ω + 12 ⎜ ⎟ . ⎝ 2I ⎠ ⎛ =2 ⎞ For n = 1, l = 2, E f = 32 =ω + 6 ⎜ ⎟ . ⎝ 2I ⎠ ⎛ =2 ⎞ Δ E = E f − Ei = =ω − 6 ⎜ ⎟ = 0.2690 eV − 6(0.2395 × 10−3 eV) = 0.2676 eV ⎝ 2I ⎠ −15 hc (4.136 × 10 eV ⋅ s)(2.998 × 108 m/s) = = 4.634 × 10−6 m = 4.634 μ m λ= ΔE 0.2676 eV EVALUATE: All three transitions are for n = 0 → n = 1. The spacing between vibrational levels is larger than the spacing between rotational levels, so the difference in λ for the various rotational transitions is small. When the transition is to a larger l, Δ E > =ω and when the transition is to a smaller l, Δ E < =ω. 42.13.

(a) IDENTIFY and SET UP: Use ω = k ′ / mr and ω = 2π f to calculate k ′. The atomic masses are used in

Eq.(42.4) to calculate mr . EXECUTE: mr =

f =

ω 2π

=

1 2π

k′ , so k ′ = mr (2π f ) 2 mr

m1m2 mH mF (1.67 × 10−27 kg)(3.15 × 10−26 kg) = = = 1.586 × 10−27 kg m1 + m2 mH + mF 1.67 × 10 −27 kg + 3.15 × 10−26 kg

k ′ = mr (2π f ) 2 = (1.586 × 10−27 kg)(2π [1.24 × 1014 Hz]) 2 = 963 N/m

(b) IDENTIFY and SET UP: The energy levels are given by Eq.(42.7). En = (n + 12 )=ω = ( n + 12 )hf , since =ω = (h / 2π )ω and (ω / 2π ) = f . The energy spacing between adjacent levels is

Δ E = En +1 − En = (n + 1 + 12 − n − 12 ) hf = hf , independent of n.


42-4

Chapter 42

EXECUTE: Δ E = hf = (6.626 × 10 −34 J ⋅ s)(1.24 × 1014 Hz) = 8.22 × 10 −20 J = 0.513 eV (c) IDENTIFY and SET UP: The photon energy equals the transition energy so Δ E = hc / λ . c 2.998 × 108 m/s = = 2.42 × 10−6 m = 2.42 μ m 1.24 × 1014 Hz f EVALUATE: This photon is infrared, which is typical for vibrational transitions. For an average spacing a, the density is ρ = m a 3 , where m is the average of the ionic masses, and so EXECUTE:

42.14.

hf = hc / λ so λ =

a3 =

42.15.

−26 −25 m ( 6.49 × 10 kg + 1.33 × 10 kg ) 2 = = 3.60 × 10 −29 m 3 , (2.75 × 103 kg m 3 ) ρ

and a = 3.30 × 10−10 m = 0.330 nm . (b) The larger (higher atomic number) atoms have the larger spacing. IDENTIFY and SET UP: Find the volume occupied by each atom. The density is the average mass of Na and Cl divided by this volume. EXECUTE: Each atom occupies a cube with side length 0.282 nm. Therefore, the volume occupied by each atom is V = (0.282 × 10−9 m)3 = 2.24 × 10−29 m 3. In NaCl there are equal numbers of Na and Cl atoms, so the average mass of the atoms in the crystal is m = 12 ( mNa + mCl ) = 12 (3.82 × 10−26 kg + 5.89 × 10−26 kg) = 4.855 × 10 −26 kg m 4.855 × 10−26 kg = = 2.17 × 103 kg/m3 . 2.24 × 10−29 m3 V EVALUATE: The density of water is 1.00 × 103 kg/m 3 , so our result is reasonable.

The density then is ρ =

42.16.

−34 8 hc ( 6.63 × 10 J ⋅ s ) ( 3.00 × 10 m s ) = = 0.200 nm. E ( 6.20 × 103 eV ) (1.60 × 10−19 J eV )

(a) As a photon, λ =

(b) As a matter wave,

λ=

( 6.63 ×10−34 J ⋅ s ) h h = = = 0.200 nm p 2mE 2 ( 9.11 × 10−31 kg ) ( 37.6 eV ) (1.60 × 10−19 J eV )

(c) As a matter wave,

λ= 42.17.

( 6.63 ×10−34 J ⋅ s ) h = = 0.200 nm . 2mE 2 (1.67 × 10−27 kg ) ( 0.0205 eV ) (1.60 × 10−19 J eV )

IDENTIFY: The energy gap is the energy of the maximum-wavelength photon. SET UP: The energy difference is equal to the energy of the photon, so ΔE = hc/λ. EXECUTE: (a) Using the photon wavelength to find the energy difference gives ΔE = hc/λ = (4.136 × 10–15 eV ⋅ s )(3.00 × 108 m/s)/(1.11 × 10–6 m) = 1.12 eV

42.18.

42.19.

(b) A wavelength of 1.11 µm = 1110 nm is in the infrared, shorter than that of visible light. EVALUATE: Since visible photons have more than enough energy to excite electrons from the valence to the conduction band, visible light will be absorbed, which makes silicon opaque. hc (a) = 2.27 × 10−7 m = 227 nm , in the ultraviolet. ΔE (b) Visible light lacks enough energy to excite the electrons into the conduction band, so visible light passes through the diamond unabsorbed. (c) Impurities can lower the gap energy making it easier for the material to absorb shorter wavelength visible light. This allows longer wavelength visible light to pass through, giving the diamond color. hc (6.63 × 10−34 J ⋅ s) (3.00 × 108 m s) ΔE = = = 2.14 × 10−13 J = 1.34 × 106 eV . So the number of electrons that can be 9.31 × 10−13 m λ

excited to the conduction band is n = 42.20.

1.34 × 106 eV = 1.20 × 106 electrons 1.12 eV

1 = ∫ ψ dV 2

3 L ⎛L ⎛ n πy ⎞ ⎞ ⎛ L ⎛ n πx ⎞ ⎞ ⎛ ⎛ n πz ⎞ ⎞ ⎛ L⎞ = A2 ⎜ ∫ sin 2 ⎜ x ⎟ dx ⎟ ⎜⎜ ∫ sin 2 ⎜ y ⎟ dy ⎟⎟ ⎜ ∫ sin 2 ⎜ z ⎟ dz ⎟ = A2 ⎜ ⎟ ⎝ L ⎠ ⎠⎝0 ⎝ L ⎠ ⎠ ⎝2⎠ ⎝ L ⎠ ⎠⎝0 ⎝0

so A = ( 2 L )

32

(assuming A to be real positive).


Molecules and Condensed Matter

42.21.

42-5

Density of states: g(E) =

( 2m )

32

2π =

V

2 3

E1 2 =

(2(9.11 × 10 −31 kg))3 2 (1.0 × 10 −6 m 3 )(5.0 eV)1 2 (1.60 × 10−19 J eV)1 2 2π 2 (1.054 × 10−34 J ⋅ s)3

g ( E ) = ( 9.5 × 1040 states J ) (1.60 × 10 −19 J eV ) = 1.5 × 1022 states eV. 42.22.

42.23.

vrms = 3kT m = 1.17 × 105 m s , as found in Example 42.9. The equipartition theorem does not hold for the electrons at the Fermi energy. Although these electrons are very energetic, they cannot lose energy, unlike electrons in a free electron gas. = 2 ⎛ ∂ 2ψ ∂ 2ψ ∂ 2ψ ⎞ + + (a) IDENTIFY and SET UP: The three-dimensional Schrödinger equation is − ⎜ ⎟ + Uψ = Eψ 2m ⎝ ∂x 2 ∂y 2 ∂z 2 ⎠ (Eq.40.29). For free electrons, U = 0. Evaluate ∂ 2ψ /∂x 2 , ∂ 2ψ /∂y 2 , and ∂ 2ψ / ∂z 2 for ψ as given by Eq.(42.10). Put the results into Eq.(40.20) and see if the equation is satisfied. ∂ψ nxπ ⎛ n πx⎞ ⎛ n π y ⎞ ⎛ n πz⎞ A cos ⎜ x ⎟ sin ⎜ y ⎟ sin ⎜ z ⎟ EXECUTE: = L ∂x ⎝ L ⎠ ⎝ L ⎠ ⎝ L ⎠ ∂ 2ψ ⎛nπ ⎞ ⎛n πx⎞ ⎛n π y⎞ ⎛ n πz ⎞ ⎛nπ ⎞ = − ⎜ x ⎟ A sin ⎜ x ⎟ sin ⎜ y ⎟ sin ⎜ z ⎟ = − ⎜ x ⎟ ψ L L L L ∂x 2 ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ L ⎠ ⎠ ⎝ 2

2

2

Similarly

⎛nπ ⎞ ∂ 2ψ ∂ 2ψ ⎛nπ ⎞ = − ⎜ y ⎟ ψ and = −⎜ z ⎟ ψ . 2 2 ∂y L z ∂ ⎝ L ⎠ ⎝ ⎠ 2

(nx2 + ny2 + nz2 )π 2= 2 ⎞ =2 ⎛ π 2 ⎞ 2 2 2 ( n n n ) = + + = ψ ψ ⎟ ⎜ 2⎟ x y z 2mL2 ⎠ 2m ⎝ L ⎠ ( n 2 + n y2 + nz2 )π 2= 2 This equals Eψ , with E = x , which is Eq.(42.11). 2mL2 EVALUATE: ψ given by Eq.(42.10) is a solution to Eq.(40.29), with E as given by Eq.(42.11). (b) IDENTIFY and SET UP: Find the set of quantum numbers nx , n y , and nz that give the lowest three values of Therefore, −

= 2 ⎛ ∂ 2ψ ∂ 2ψ ∂ 2ψ + + ⎜ 2m ⎝ ∂x 2 ∂y 2 ∂z 2

E. The degeneracy is the number of sets nx , n y , nz and ms that give the same E. 3π 2= 2 . No other combination of nx , n y , and nz 2mL2 gives this same E, so the only degeneracy is the degeneracy of two due to spin. π 2= 2 6π 2= 2 = First excited level: next lower E so one n equals 2 and the others equal 1. E = (22 + 12 + 12 ) 2mL2 2mL2 There are three different sets of nx , n y , nz values that give this E:

EXECUTE: Ground level: lowest E so nx = n y = nz = 1 and E =

nx = 2, ny = 1, nz = 1; nx = 1, ny = 2, nz = 1; nx = 1, n y = 1, nz = 2 This gives a degeneracy of 3 so the total degeneracy, with the factor of 2 from spin, is 6. Second excited level: next lower E so two of nx , n y , nz equal 2 and the other equals 1. 9π 2= 2 2mL2 2mL2 There are different sets of nx , n y , nz values that give this E: E = (22 + 2 2 + 12 )

π 2= 2

=

nx = 2, ny = 2, nz = 1; nx = 2, ny = 1, nz = 2; nx = 1, ny = 2, nz = 2.

42.24. 42.25.

Thus, as for the first excited level, the total degeneracy, including spin, is 6. EVALUATE: The wavefunction for the 3-dimensional box is a product of the wavefunctions for a 1-dimensional box in the x, y, and z coordinates and the energy is the sum of energies for three 1-dimensional boxes. All levels except for the ground level have a degeneracy greater than two. Compare to the 3-dimensional isotropic harmonic oscillator treated in Problem 40.53. 12 Eq.(42.13) may be solved for nrs = ( 2mE ) ( L =π ), and substituting this into Eq. (42.12), using L3 = V , gives

Eq.(42.14). (a) IDENTIFY and SET UP: The electron contribution to the molar heat capacity at constant volume of a metal is ⎛ π 2 KT ⎞ CV = ⎜ ⎟ R. ⎝ 2 EF ⎠ π 2 (1.381 × 10 −23 J/K)(300 K) EXECUTE: CV = R = 0.0233R. 2(5.48 eV)(1.602 × 10−19 J/eV)


42-6

Chapter 42

(b) EVALUATE: The electron contribution found in part (a) is 0.0233R = 0.194 J/mol ⋅ K. This is 0.194 / 25.3 = 7.67 × 10−3 = 0.767% of the total CV . (c) Only a small fraction of CV is due to the electrons. Most of CV is due to the vibrational motion of the ions. 42.26.

3 (a) From Eq. (42.22), Eav = EF = 1.94 eV. 5 (b)

(c) 42.27.

42.28.

2E m =

2 (1.94 eV ) (1.60 × 10−19 J eV ) 9.11× 10−31 kg

= 8.25 × 105 m s.

−19 EF ( 3.23 eV ) (1.60 × 10 J eV ) = = 3.74 × 104 K. k (1.38 ×10−23 J K )

IDENTIFY: The probability is given by the Fermi-Dirac distribution. 1 SET UP: The Fermi-Dirac distribution is f ( E ) = ( E − EF ) / kT . e +1 EXECUTE: We calculate the value of f (E), where E = 8.520 eV, EF = 8.500 eV, k = 1.38 × 10–23 J/K = 8.625 × 10–5 eV/K, and T = 20°C = 293 K. The result is f (E) = 0.312 = 31.2%. EVALUATE: Since the energy is close to the Fermi energy, the probability is quite high that the state is occupied by an electron. (a) See Example 42.10: The probabilities are 1.78 × 10−7 , 2.37 × 10−6 , and 1.51 × 10 −5 . (b) The Fermi distribution, Eq.(42.17), has the property that f ( EF − E ) = 1 − f ( E ) (see Problem (42.48)), and so

42.29.

42.30.

the probability that a state at the top of the valence band is occupied is the same as the probability that a state of the bottom of the conduction band is filled (this result depends on having the Fermi energy in the middle of the gap). 1 IDENTIFY: Use Eq.(42.17), f ( E ) = ( E − EF ) / kT . Solve for E − EF . e +1 1 SET UP: e( E − EF ) / kT = −1 f (E) The problem states that f ( E ) = 4.4 × 10−4 for E at the bottom of the conduction band. 1 − 1 = 2.272 × 103. EXECUTE: e( E − EF ) / kT = 4.4 × 10−4 E − EF = kT ln(2.272 × 103 ) = (1.3807 × 10−23 J/T)(300 K)ln(2.272 × 103 ) = 3.201× 10−20 J = 0.20 eV EF = E − 0.20 eV; the Fermi level is 0.20 eV below the bottom of the conduction band. EVALUATE: The energy gap between the Fermi level and bottom of the conduction band is large compared to kT at T = 300 K and as a result f ( E ) is small. IDENTIFY: The current depends on the voltage across the diode and its temperature, so the resistance also depends on these quantities. SET UP: The current is I = IS (eeV/kT – 1) and the resistance is R = V/I. V V eV e(0.0850 V) = = EXECUTE: (a) The resistance is R = = . The exponent is I IS ( eeV / kT − 1) kT (8.625 × 10−5 eV/K ) (293 K) 3.3635, giving R =

85.0 mV = 4.06 Ω. (0.750 mA) ( e3.3635 − 1)

(b) In this case, the exponent is

which gives R =

42.31.

eV e(−0.050 V) = = −1.979 kT ( 8.625 × 10−5 eV/K ) (293 K)

−50.0 mV = 77.4 Ω (0.750 mA) ( e−1.979 − 1)

EVALUATE: Reversing the voltage can make a considerable change in the resistance of a diode. IDENTIFY and SET UP: The voltage-current relation is given by Eq.(42.23): I = Is (eeV / kT − 1). Use the current for

V = +15.0 mV to solve for the constant I s . EXECUTE: (a) Find Is : V = +15.0 × 10−3 V gives I = 9.25 × 10 −3 A eV (1.602 × 10−19 C)(15.0 × 10−3 V) = = 0.5800 kT (1.381 × 10 −23 J/K)(300 K) I 9.25 × 10 −3 A I s = eV / kT = = 1.177 × 10−2 = 11.77 mA e −1 e0.5800 − 1


Molecules and Condensed Matter

Then can calculate I for V = 10.0 mV:

−19

42-7

−3

eV (1.602 × 10 C)(10.0 × 10 V) = = 0.3867 kT (1.381 × 10 −23 J/K)(300 K)

I = Is (eeV / kT − 1) = (11.77 mA)(e0.3867 − 1) = 5.56 mA eV eV has the same magnitude as in part (a) but not V is negative so is negative. kT kT eV V = −15.0 mV : = −0.5800 and I = I s (eeV / kT − 1) = (11.77 mA)(e −0.5800 − 1) = −5.18 mA kT eV V = −10.0 mV : = −0.3867 and I = I s (eeV / kT − 1) = (11.77 mA)(e−0.3867 − 1) = −3.77 mA kT EVALUATE: There is a directional asymmetry in the current, with a forward-bias voltage producing more current than a reverse-bias voltage of the same magnitude, but the voltage is small enough for the asymmetry not be pronounced. Compare to Example 42.11, where more extreme voltages are considered. (a) Solving Eq.(42.23) for the voltage as a function of current, (b)

42.32.

V=

42.33.

⎞ kT ⎛ 40.0 mA ⎞ kT ⎛ I ln ⎜ + 1⎟ = ln ⎜ + 1⎟ = 0.0645 V. e ⎝ 3.60 mA ⎠ ⎝ IS ⎠ e

(b) From part (a), the quantity eeV kT = 12.11 , so far a reverse-bias voltage of the same magnitude, ⎛ 1 ⎞ I = I S ( e − eV kT − 1) = I S ⎜ − 1 ⎟ = −3.30 mA . ⎝ 12.11 ⎠ IDENTIFY: During the transition, the molecule emits a photon of light having energy equal to the energy difference between the two vibrational states of the molecule. 1⎞ 1⎞ k′ ⎛ ⎛ SET UP: The vibrational energy is En = ⎜ n + ⎟ =ω = ⎜ n + ⎟ = . 2⎠ 2 ⎠ mr ⎝ ⎝ EXECUTE: (a) The energy difference between two adjacent energy states is ΔE = =

k′ , and this is the energy of mr 2

⎛ ΔE ⎞ the photon, so ΔE = hc/λ. Equating these two expressions for ΔE and solving for k ′ , we have k ′ = mr ⎜ ⎟ = ⎝ = ⎠ 2 ΔE hc / λ 2π c mH mO ⎛ ΔE ⎞ = = with the appropriate numbers gives us ⎜ ⎟ , and using = = λ mH + mO ⎝ = ⎠

k′ =

ω 1 = (b) f = 2π 2π

(1.67 ×10

−27

1.67 × 10−27

k′ 1 = mr 2π

kg )( 2.656 × 10−26 kg ) ⎡ 2π ( 3.00 × 108 m/s ) ⎤ ⎢ ⎥ = 977 N/m kg + 2.656 × 10−26 kg ⎢ 2.39 × 10−6 m ⎥ ⎣ ⎦ 2

mH mO mH + mO . Substituting the appropriate numbers gives us k′

(1.67 ×10 f =

42.34. 42.35.

1 2π

−27

kg )( 2.656 × 10−26 kg )

1.67 × 10−27 kg + 2.656 × 10 −26 kg = 1.25 × 1014 Hz 977 N/m

EVALUATE: The frequency is close to, but not quite in, the visible range. 2= 2 hλ I= = = 7.14 × 10−48 kg ⋅ m 2 . ΔE 2π 2c IDENTIFY and SET UP: Eq.(21.14) gives the electric dipole moment as p = qd , where the dipole consists of charges ± q separated by distance d. EXECUTE: (a) Point charges +e and −e separated by distance d, so p = ed = (1.602 × 10−19 C)(0.24 × 10−9 m) = 3.8 × 10−29 C ⋅ m p 3.0 × 10−29 C ⋅ m = = 1.3 × 10−19 C d 0.24 × 10−9 m q 1.3 × 10−19 C = 0.81 (c) = e 1.602 × 10−19 C

(b) p = qd so q =


42-8

Chapter 42

p 1.5 × 10 −30 C ⋅ m = = 9.37 × 10 −21 C d 0.16 × 10 −9 m q 9.37 × 10−21 C = = 0.058 e 1.602 × 10−19 C EVALUATE: The fractional ionic character for the bond in HI is much less than the fractional ionic character for the bond in NaCl. The bond in HI is mostly covalent and not very ionic. 1 e2 The electrical potential energy is U = −5.13 eV, and r = − = 2.8 × 10−10 m. 4πP0 U

(d) q =

42.36. 42.37.

(a) IDENTIFY:

E (Na) + E (Cl) = E (Na + ) + E (Cl− ) + U ( r ). Solving for U ( r ) gives

U ( r ) = −[ E (Na + ) − E (Na)] + [ E (Cl) − E (Cl− )].

SET UP:

[ E (Na + ) − E (Na)] is the ionization energy of Na, the energy required to remove one electron, and is

equal to 5.1 eV. [ E (Cl) − E (Cl− )] is the electron affinity of Cl, the magnitude of the decrease in energy when an electron is attached to a neutral Cl atom, and is equal to 3.6 eV. 1 e2 EXECUTE: U = −5.1 eV + 3.6 eV = −1.5 eV = −2.4 × 10−19 J, and − = −2.4 × 10−19 J 4π P0 r −19 ⎛ 1 ⎞ C) 2 e2 9 2 2 (1.602 × 10 = × ⋅ (8.988 10 N m /C ) r =⎜ ⎟ −19 2.4 × 10 −19 J ⎝ 4π P0 ⎠ 2.4 × 10 J r = 9.6 × 10−10 m = 0.96 nm (b) ionization energy of K = 4.3 eV; electron affinity of Br = 3.5 eV

Thus U = −4.3 eV + 3.5 eV = −0.8 eV = −1.28 × 10 −19 J, and −

42.38.

1 e2 = −1.28 × 10 −19 J 4π P0 r

⎛ 1 ⎞ (1.602 × 10−19 C) 2 e2 = (8.988 × 109 N ⋅ m 2 / C2 ) r =⎜ ⎟ −19 1.28 × 10−19 J ⎝ 4π P0 ⎠ 1.28 × 10 J r = 1.8 × 10−9 m = 1.8 nm EVALUATE: K has a smaller ionization energy than Na and the electron affinities of Cl and Br are very similar, so it takes less energy to make K + + Br − from K + Br than to make Na + + Cl− from Na + Cl. Thus, the stabilization distance is larger for KBr than for NaCl. The energies corresponding to the observed wavelengths are 3.29 × 10 −21 J, 2.87 × 10−21 J, 2.47 × 10 −21 J, 2.06 × 10−21 J and 1.65 × 10−21 J. The average spacing of these energies is 0.410 × 10−21 J and these are seen to

correspond to transition from levels 8, 7, 6, 5 and 4 to the respective next lower levels. Then,

42.39.

=2 = 0.410 × 10−21 J , I

from which I = 2.71 × 10 −47 kg ⋅ m 2 . (a) IDENTIFY: The rotational energies of a molecule depend on its moment of inertia, which in turn depends on the separation between the atoms in the molecule. SET UP: Problem 42.38 gives I = 2.71 × 10 −47 kg ⋅ m 2 . I = mr r 2 . Calculate mr and solve for r. EXECUTE:

mr =

mH mCl (1.67 × 10−27 kg)(5.81 × 10−26 kg) = = 1.623 × 10−27 kg mH + mCl 1.67 × 10−27 kg + 5.81 × 10−26 kg

I 2.71 × 10 −47 kg ⋅ m 2 = = 1.29 × 10 −10 m = 0.129 nm mr 1.623 × 10−27 kg EVALUATE: This is a typical atomic separation for a diatomic molecule; see Example 42.2 for the corresponding distance for CO. (b) IDENTIFY: Each transition is from the level l to the level l − 1. The rotational energies are given by Eq.(42.3). The transition energy is related to the photon wavelength by Δ E = hc / λ . r=

⎛ =2 ⎞ ⎛ =2 ⎞ El = l (l + 1)= 2 / 2 I , so Δ E = El − El −1 = [l (l + 1) − l (l − 1)] ⎜ ⎟ = l ⎜ ⎟ . ⎝ 2I ⎠ ⎝ I ⎠ 2 ⎛ = ⎞ hc EXECUTE: l ⎜ ⎟ = ⎝ I ⎠ λ SET UP:

l=

2π cI 2π (2.998 × 108 m/s)(2.71 × 10−47 kg ⋅ m 2 ) 4.843 × 10 −4 m = = (1.055 × 10 −34 J ⋅ s)λ λ =λ


Molecules and Condensed Matter

42-9

4.843 × 10−4 m = 8. 60.4 × 10−6 m 4.843 × 10−4 m = 7. For λ = 69.0 μ m, l = 69.0 × 10−6 m 4.843 × 10−4 m = 6. For λ = 80.4 μ m, l = 80.4 × 10−6 m 4.843 × 10−4 m = 5. For λ = 96.4 μ m, l = 96.4 × 10−6 m 4.843 × 10−4 m = 4. For λ = 120.4 μ m, l = 120.4 × 10−6 m EVALUATE: In each case l is an integer, as it must be. (c) IDENTIFY and SET UP: Longest λ implies smallest Δ E , and this is for the transition from l = 1 to l = 0. For λ = 60.4 μ m, l =

⎛ =2 ⎞ (1.055 × 10−34 J ⋅ s) 2 Δ E = l ⎜ ⎟ = (1) = 4.099 × 10−22 J −47 2 2.71 10 kg m × ⋅ I ⎝ ⎠ hc (6.626 × 10−34 J ⋅ s)(2.998 × 108 m/s) λ= = = 4.85 × 10−4 m = 485 μ m. ΔE 4.099 × 10−22 J EVALUATE: This is longer than any wavelengths in part (b). (d) IDENTIFY: What changes is mr , the reduced mass of the molecule. EXECUTE:

⎛ =2 ⎞ hc 2π cI SET UP: The transition energy is Δ E = l ⎜ ⎟ and Δ E = , so λ = (part (b)). I = mr r 2 , so λ is directly l= λ ⎝ I ⎠ λ (HCl) λ (DCl) m (DCl) proportional to mr . = so λ (DCl) = λ (HCl) r mr (HCl) mr (DCl) mr (HCl) EXECUTE: The mass of a deuterium atom is approximately twice the mass of a hydrogen atom, so mD = 3.34 × 10−27 kg. mr (DCl) =

m D m Cl (3.34 × 10−27 kg)(5.81 × 10 −27 kg) = = 3.158 × 10 −27 kg m D + m Cl 3.34 × 10−27 kg + 5.81 × 10 −26 kg

⎛ 3.158 × 10 −27 kg ⎞ ⎟ = (1.946)λ (HCl) −27 ⎝ 1.623 × 10 kg ⎠ l = 8 → l = 7; λ = (60.4 μ m)(1.946) = 118 μ m l = 7 → l = 6; λ = (69.0 μ m)(1.946) = 134 μ m l = 6 → l = 5; λ = (80.4 μ m)(1.946) = 156 μ m l = 5 → l = 4; λ = (96.4 μ m)(1.946) = 188 μ m l = 4 → l = 3; λ = (120.4 μ m)(1.946) = 234 μ m EVALUATE: The moment of inertia increases when H is replaced by D, so the transition energies decrease and the wavelengths increase. The larger the rotational inertia the smaller the rotational energy for a given l (Eq.42.3). = 2l hl λ From the result of Problem 42.11, the moment inertia of the molecule is I = = = 6.43 × 10−46 kg ⋅ m 2 and ΔE 4π 2c

λ (DCl) = λ (HCl) ⎜

42.40.

from Eq.(42.6) the separation is r0 = 42.41.

I = 0.193 nm. mr

L2 = 2l (l + 1) = . Eg = 0 (l = 0), and there is an additional multiplicative factor of 2l + 1 because for each l 2I 2I 2 n state there are really (2l + 1) ml -states with the same energy. So l = (2l + 1)e − = l ( l +1) /(2 IkT ) . n0 (a) Eex =

(b) T = 300 K, I = 1.449 × 10 −46 kg ⋅ m 2 .

= 2 (1) (1 + 1) E 7.67 × 10−23 J = 7.67 × 10−23 J. l =1 = = 0.0185. 2 −46 2(1.449 × 10 kg ⋅ m ) kT (1.38 × 10−23 J K) (300 K) n (2l + 1) = 3 , so l =1 = (3)e −0.0185 = 2.95. n0

(i) El =1 =


42-10

Chapter 42

El = 2 = 2 (2) (2 + 1) = = 0.0556. 2(1.449 × 10−46 kg ⋅ m 2 ) (1.38 × 10 −23 J K) (300 K) kT n (2l + 1) = 5 , so l =1 = (5)(e −0.0556 ) = 4.73. n0

(ii)

(iii)

El =10 = 2 (10) (10 + 1) = = 1.02. kT 2(1.449 × 10−46 kg ⋅ m 2 ) (1.38 × 10−23 J K) (300 K)

(2l + 1) = 21, so (iv)

nl =10 = (21) (e −1.02 ) = 7.57. n0

El = 20 = 2 (20) (20 + 1) = = 3.89. kT 2(1.449 × 10−46 kg ⋅ m 2 ) (1.38 × 10−23 J K) (300 K)

(2l + 1) = 41 , so

(v)

nl = 20 = (41)e −3.89 = 0.838. n0

El = 50 = 2 (50) (50 + 1) = = 23.6. −46 kT 2(1.449 × 10 kg ⋅ m 2 ) (1.38 × 10−23 J K) (300 K)

nl =50 = (101)e −23.6 = 5.69 × 10−9. n0 (c) There is a competing effect between the (2l + 1) term and the decaying exponential. The 2l + 1 term dominates for small l, while the exponential term dominates for large l. (a) I CO = 1.449 × 10−46 kg ⋅ m 2 . (2l + 1) = 101 , so

42.42.

El =1 =

= 2l (l + 1) (1.054 × 10−34 J ⋅ s) 2 (1) (1 + 1) = = 7.67 × 10−23 J . El = 0 = 0. 2I 2(1.449 × 10−46 kg ⋅ m 2 )

ΔE = 7.67 × 10−23 J = 4.79 × 10 −4 eV.

hc (6.63 × 10−34 J ⋅ s) (3.00 × 108 m s) = = 2.59 × 10−3 m = 2.59 mm. ΔE (7.67 × 10−23 J) (b) Let’s compare the value of kT when T = 20 K to that of ΔE for the l = 1 → l = 0 rotational transition:

λ=

kT = (1.38 × 10−23 J K) (20 K) = 2.76 × 10−22 J. kT = 3.60. ΔE Therefore, although T is quite small, there is still plenty of energy to excite CO molecules into the first rotational level. This allows astronomers to detect the 2.59 mm wavelength radiation from such molecular clouds. IDENTIFY and SET UP: El = l (l + 1)= 2 / 2 I , so El and the transition energy Δ E depend on I. Different isotopic molecules have different I. mNa mCl (3.8176 × 10 −26 kg)(5.8068 × 10−26 kg) EXECUTE: (a) Calculate I for Na 35Cl: mr = = = 2.303 × 10 −26 kg mNa + mCl 3.8176 × 10−26 kg + 5.8068 × 10 −26 kg ΔE = 7.67 × 10−23 J (from part (a)). So

42.43.

I = mr r 2 = (2.303 × 10−26 kg)(0.2361× 10−9 m)2 = 1.284 × 10−45 kg ⋅ m 2 l = 2 → l = 1 transition ⎛ = 2 ⎞ 2= 2 2(1.055 × 10−34 J ⋅ s) 2 Δ E = E2 − E1 = (6 − 2) ⎜ ⎟ = = = 1.734 × 10−23 J −45 2 2 I I 1.284 10 kg m × ⋅ ⎝ ⎠ hc hc (6.626 × 10 −34 J ⋅ s)(2.998 × 108 m/s) ΔE = so λ = = = 1.146 × 10 −2 m = 1.146 cm λ ΔE 1.734 × 10 −23 J l = 1 → l = 0 transition ⎛ =2 ⎞ =2 1 Δ E = E1 − E0 = (2 − 0) ⎜ ⎟ = = (1.734 × 10−23 J) = 8.67 × 10−24 J 2 ⎝ 2I ⎠ I hc (6.626 × 10 −34 J ⋅ s)(2.998 × 108 m/s) = = 2.291 cm λ= ΔE 8.67 × 10−24 J mNa mCl (3.8176 × 10 −26 kg)(6.1384 × 10−26 kg) (b) Calculate I for Na 37Cl: mr = = = 2.354 × 10−26 kg mNa + mCl 3.8176 × 10 −26 kg + 6.1384 × 10−26 kg

I = mr r 2 = (2.354 × 10−26 kg)(0.2361 × 10−9 m) 2 = 1.312 × 10−45 kg ⋅ m 2


Molecules and Condensed Matter

42-11

l = 2 → l = 1 transition 2= 2 2(1.055 × 10−34 J ⋅ s) 2 = = 1.697 × 10 −23 J 1.312 × 10 −45 kg ⋅ m 2 I hc (6.626 × 10−34 J ⋅ s)(2.998 × 108 m/s) = = 1.171 × 10−2 m = 1.171 cm λ= ΔE 1.697 × 10−23 J ΔE =

l = 1 → l = 0 transition

=2 1 = (1.697 × 10−23 J) = 8.485 × 10−24 J I 2 hc (6.626 × 10−34 J ⋅ s)(2.998 × 108 m/s) = = 2.341 cm λ= ΔE 8.485 × 10−24 J The differences in the wavelengths for the two isotopes are: l = 2 → l = 1 transition: 1.171 cm − 1.146 cm = 0.025 cm l = 1 → l = 0 transition: 2.341 cm − 2.291 cm = 0.050 cm EVALUATE: Replacing 35 Cl by 37 Cl increases I, decreases Δ E and increases λ . The effect on λ is small but measurable. ΔE The vibration frequency is, from Eq.(42.8), f = = 1.12 × 1014 Hz. The force constant is h k ′ = (2πf ) 2 mr = 777 N m. ΔE =

42.44.

42.45.

1⎞ k′ 1 2k ′ ⎛ En = ⎜ n + ⎟ = ⇒ E0 = = 2⎠ mr 2 mH ⎝ 1 2(576 N m ) ⇒ E0 = (1.054 × 10−34 J ⋅ s) = 4.38 × 10−20 J = 0.274 eV. 2 1.67 × 10−27 kg

42.46.

This is much less than the H 2 bond energy. (a) The frequency is proportional to the reciprocal of the square root of the reduced mass, and in terms of the atomic masses, the frequency of the isotope with the deuterium atom is 12

⎛ m m ( mH + mF ) ⎞ f = f0 ⎜ F H ⎟ ⎝ mF mD ( mD + mF ) ⎠ 42.47.

12

⎛ 1 + ( mF mD ) ⎞ = f0 ⎜ ⎟ . ⎝ 1 + (mF mH ) ⎠

Using f 0 from Exercise 42.13 and the given masses, f = 8.99 × 1013 Hz. IDENTIFY and SET UP: Use Eq.(42.6) to calculate I. The energy levels are given by Eq.(42.9). The transition energy Δ E is related to the photon wavelength by Δ E = hc / λ . EXECUTE: (a) mr =

mH mI (1.67 × 10−27 kg)(2.11 × 10−25 kg) = = 1.657 × 10−27 kg mH + mI 1.67 × 10 −27 kg + 2.11 × 10−25 kg

I = mr r 2 = (1.657 × 10 −27 kg)(0.160 × 10−9 m) 2 = 4.24 × 10−47 kg ⋅ m 2

⎛ =2 ⎞ k′ (b) The energy levels are Enl = l (l + 1) ⎜ ⎟ + ( n + 12 )= (Eq.(42.9)) I m 2 ⎝ ⎠ r ⎛ =2 ⎞ k′ = ω = 2π f so Enl = l (l + 1) ⎜ ⎟ + ( n + 12 )hf m ⎝ 2I ⎠

(i) transition n = 1 → n = 0, l = 1 → l = 0 ⎛ =2 ⎞ =2 Δ E = (2 − 0) ⎜ ⎟ + (1 + 12 − 12 )hf = + hf I ⎝ 2I ⎠ hc hc hc c ΔE = so λ = = = λ Δ E (= 2 / I ) + hf (= / 2π I ) + f =

2π I

λ=

=

1.055 × 10−34 J ⋅ s = 3.960 × 1011 Hz 2π (4.24 × 10−47 kg ⋅ m 2 )

c 2.998 × 108 m/s = = 4.30 μ m (= / 2π I ) + f 3.960 × 1011 Hz + 6.93 × 1013 Hz


42-12

Chapter 42

(ii) transition n = 1 → n = 0, l = 2 → l = 1 ⎛ =2 ⎞ 2= 2 Δ E = (6 − 2) ⎜ ⎟ + hf = + hf I ⎝ 2I ⎠ c 2.998 × 108 m/s λ= = = 4.28 μ m 2(= / 2π I ) + f 2(3.960 × 1011 Hz) + 6.93 × 1013 Hz (iii) transition n = 2 → n = 1, l = 2 → l = 3

42.48. 42.49.

⎛ =2 ⎞ 3= 2 Δ E = (6 − 12) ⎜ ⎟ + hf = − + hf I ⎝ 2I ⎠ c 2.998 × 108 m/s = = 4.40 μ m λ= −3(= / 2π I ) + f −3(3.960 × 1011 Hz) + 6.93 × 1013 Hz EVALUATE: The vibrational energy change for the n = 1 → n = 0 transition is the same as for the n = 2 → n = 1 transition. The rotational energies are much smaller than the vibrational energies, so the wavelengths for all three transitions don’t differ much. 1 1 1 e −ΔE / kT The sum of the probabilities is f ( EF + ΔE ) + f ( EF − ΔE ) = −ΔE kT + ΔE kT = −ΔE kT + = 1. e +1 e +1 e + 1 1 + e −ΔE kT 32 3 π 4 3= 2n 2 3 Since potassium is a metal we approximate EF = EF0 . ⇒ EF = . 2m ρ 851 kg m 3 = 1.31 × 1028 electron m3 But the electron concentration n = ⇒ n = m 6.49 × 10−26 kg

32 3 π 4 3 (1.054 × 10 −34 J ⋅ s) 2 (1.31 × 1028 /m3 ) 2 3 = 3.24 × 10−19 J = 2.03 eV. 2(9.11 × 10−31 kg) IDENTIFY: The only difference between the two isotopes is their mass, which will affect their reduced mass and hence their moment of inertia. =2 SET UP: The rotational energy states are given by E = l (l + 1) and the reduced mass is given by mr = 2I m1m2/(m1 + m2). EXECUTE: (a) If we call m the mass of the H-atom, the mass of the deuterium atom is 2m and the reduced masses of the molecules are H2 (hydrogen): mr(H) = mm/(m + m) = m/2 D2 (deuterium): mr(D) = (2m)(2m)/(2m + 2m) = m Using I = mr r02, the moments of inertia are IH = mr02/2 and ID = mr02. The ratio of the rotational energies is then ⇒ EF =

42.50.

2 EH l (l + 1) ( = / 2 I H ) I D mr02 = = = = 2. ED l (l + 1) ( = 2 / 2 I D ) I H m r 2 0 2

1⎞ k′ ⎛ ⎜ n + ⎟= m 2 E mr (D) m ⎝ ⎠ r (H) = = = 2. (b) The ratio of the vibrational energies is H = ED ⎛ mr (H) m/2 k′ 1⎞ ⎜ n + ⎟= 2 ⎠ mr (D) ⎝

41.51.

EVALUATE: The electrical force is the same for both molecules since both H and D have the same charge, so it is reasonable that the force constant would be the same for both of them. IDENTIFY and SET UP: Use the description of the bcc lattice in Fig.42.12c in the textbook to calculate the number of atoms per unit cell and then the number of atoms per unit volume. EXECUTE: (a) Each unit cell has one atom at its center and 8 atoms at its corners that are each shared by 8 other unit cells. So there are 1 + 8/ 8 = 2 atoms per unit cell. 2 n = = 4.66 × 10 28 atoms/m3 V (0.35 × 10−9 m)3 (b) EF0 =

32 / 3 π 4 / 3= 2 ⎛ N ⎞ ⎜ ⎟ 2m ⎝V ⎠

2/3

In this equation N/V is the number of free electrons per m3 . But the problem says to assume one free electron per atom, so this is the same as n/V calculated in part (a). m = 9.109 × 10 −31 kg (the electron mass), so EF0 = 7.563 × 10 −19 J = 4.7 eV EVALUATE: Our result for metallic lithium is similar to that calculated for copper in Example 42.8.


Molecules and Condensed Matter

42.52.

(a)

d αe 2 1 1 8 A4π P0 U tot = − 8 A 9 . Setting this equal to zero when r = r0 gives r07 = dr 4πP0 r 2 r αe 2

and so U tot =

42.53.

42-13

αe 2 ⎛ 1 r07 ⎞ 7 αe 2 = −1.26 × 10−18 J = −7.85 eV. ⎜ − + 8 ⎟ . At r = r0 , U tot = − 4πP0 ⎝ r 8r ⎠ 32πP0r0

(b) To remove a Na + Cl− ion pair from the crystal requires 7.85 eV. When neutral Na and Cl atoms are formed from the Na + and Cl− atoms there is a net release of energy −5.14 eV + 3.61 eV = −1.53 eV, so the net energy required to remove a neutral Na, Cl pair from the crystal is 7.85 eV − 1.53 eV = 6.32 eV. dE (a) IDENTIFY and SET UP: p = − tot . Relate E tot to EF0 and evaluate the derivative. dV 3N 3 ⎛ 32 / 3 π 4 / 3= 2 ⎞ 5 / 3 −2 / 3 EF0 = ⎜ EXECUTE: Etot = NEav = ⎟N V 5 5⎝ 2m ⎠

⎛ 32 / 3 π 4 / 3= 2 ⎞ ⎛ N ⎞ dEtot 3 ⎛ 32 / 3 π 4 / 3= 2 ⎞ 5 / 3 ⎛ 2 −5 / 3 ⎞ = ⎜ ⎟N ⎜− V ⎟⎜ ⎟ ⎟ so p = ⎜ 5⎝ 2m dV 5m ⎝ 3 ⎠ ⎠ ⎝ ⎠⎝ V ⎠

5/3

, as was to be shown.

(b) N / V = 8.45 × 10 28 m −3 ⎛ 32 / 3 π 4 / 3 (1.055 × 10 −34 J ⋅ s) 2 ⎞ 28 10 5 −3 5 / 3 p=⎜ ⎟ (8.45 × 10 m ) = 3.81 × 10 Pa = 3.76 × 10 atm. −31 5(9.109 10 kg) × ⎝ ⎠

42.54.

(c) EVALUATE: Normal atmospheric pressure is about 105 Pa, so these pressures are extremely large. The electrons are held in the metal by the attractive force exerted on them by the copper ions. 53 ⎡ 5 32 3 π 4 3= 2 ⎛ N ⎞ 2 3 ⎛ − N ⎞ ⎤ 5 dp 32 3 π 4 3= 2 ⎛ N ⎞ = −V ⎢ ⋅ ⋅ ⎜ ⎟ ⎜ 2 ⎟ ⎥ = p. (a) From Problem 42.53, p = ⎜ ⎟ . B = −V dV 5m 5m ⎝ V ⎠ ⎝ V ⎠ ⎝ V ⎠ ⎦⎥ 3 ⎣⎢ 3

N 5 3 2 3 π 4 3= 2 = 8.45 × 1028 m −3 . B = ⋅ (8.45 × 10 28 m −3 )5 3 = 6.33 × 1010 Pa. V 3 5m 6.33 × 1010 Pa (c) = 0.45. The copper ions themselves make up the remaining fraction. 1.4 × 1011 Pa (b)

42.55.

(a) EF0 =

32 3 π 4 3= 2 2m

23

1 ⎛N⎞ mc 2 . ⎜ ⎟ . Let EF0 = 100 V ⎝ ⎠ 32

⎤ 2m 2c 2 23 2 m3c3 23 2 m3c 3 ⎛N⎞ ⎡ = = = = 1.67 × 1033 m −3 . ⎜ ⎟ ⎢ 23 43 2⎥ 1003 23π 2= 3 3000π 2= 3 ⎝ V ⎠ ⎣ (100)3 π = ⎦ 8.45 × 10 28 m −3 (b) = 5.06 × 10 −5. Since the real concentration of electrons in copper is less than one part in 10 −4 of the 1.67 × 1033 m −3 concentration where relativistic effects are important, it is safe to ignore relativistic effects for most applications. 6(2 × 1030 kg) (c) The number of electrons is N e = = 6.03 × 1056. The concentration is 1.99 × 10−26 kg Ne 6.03 × 1056 =4 = 6.66 × 1035 m −3 . V π (6.00 × 106 m)3 3 6.66 × 1035 m −3 ≅ 400 so relativistic effects will be very important. 1.67 × 1033 m −3 IDENTIFY: The current through the diode is related to the voltage across it. SET UP: The current through the diode is given by I = IS (eeV/kT – 1). EXECUTE: (a) The current through the resistor is (35.0 V)/(125 Ω) = 0.280 A = 280 mA, which is also the current through the diode. This current is given by I = IS (eeV/kT – 1), giving 280 mA = 0.625 mA(eeV/kT – 1) and 1 + −23 kT ln 449 (1.38 × 10 J/K ) (293 K)ln 449 (280/0.625) = 449 = eeV/kT. Solving for V at T = 293 K gives V = = = e 1.60 × 10−19 C 0.154 V (b) R = V/I = (0.154 V)/(0.280 A) = 0.551 Ω EVALUATE: At a different voltage, the diode would have different resistance.

(d) Comparing this to the result from part (a)

42.56.


42-14

Chapter 42

42.57.

(a) U =

qi q j q 2 ⎛ −1 1 1 1 1 1 1 ⎞ q2 ⎛ 2 2 1 1 ⎞ = + − − + − ⎟= − ∑ ⎟. ⎜ ⎜ − − 4πP0 i < j rij 4πP0 ⎝ d r r + d r − d r d ⎠ 4πP0 ⎝ r d r + d r − d ⎠

⎛ ⎞ 1 1 1⎜ 1 1 ⎟ 1⎛ d d2 d d 2 ⎞ 2 2d 2 But + = ⎜ + ≈ ⎜ 1 − + 2 + ⋅ ⋅⋅ + 1 + + 2 ⎟ ≈ + 3 ⎟ r + d r − d r ⎜1+ d 1− d ⎟ r ⎝ r r r r ⎠ r r ⎝ r r⎠ −2q 2 ⎛ 1 d 2 ⎞ −2 p 2 2 p2 ⇒U = − . ⎜ + 3 ⎟= 3 4πP0 ⎝ d r ⎠ 4πP0 r 4πP0 d 3

(b) U =

qi q j q 2 ⎛ −1 1 1 1 1 1 1 ⎞ q 2 ⎛ −2 2 2 2d 2 ⎞ = − + + − − ⎟= − + + 3 ⎟= ⎜ ∑ ⎜ r ⎠ 4πP0 i < j rij 4πP0 ⎝ d r r + d r − d r d ⎠ 4πP0 ⎝ d r r

−2 p 2 −2q 2 ⎛ 1 d 2 ⎞ 2 p2 + . ⎜ − 3 ⎟ ⇒U = 4πP0 d 3 4πP0 r 3 4πP0 ⎝ d r ⎠ If we ignore the potential energy involved in forming each individual molecule, which just involves a different choice for the zero of potential energy, then the answers are: −2 p 2 . The interaction is attractive. (a) U = 4πP0 r 3 (b) U = 42.58.

+2 p 2 . The interaction is repulsive. 4πP0 r 3

⎛ 1 e2 ⎞ 1 e2 1 e2 ′ and (a) Following the hint, k ′dr = − d ⎜ = = = dr = ω = 2 k m = = ⎟ 2 3 πP0 mr03 ⎝ 4πP0 r ⎠ r = r0 2πP0 r0 1.23 × 10−19 J = 0.77 eV, where ( m 2) has been used for the reduced mass.

(b) The reduced mass is doubled, and the energy is reduced by a factor of

2 to 0.54 eV.


NUCLEAR PHYSICS

43.1.

43.2.

(a)

28 14

Si has 14 protons and 14 neutrons.

(b)

85 37

Rb has 37 protons and 48 neutrons.

(c)

205 81

43

Tl has 81 protons and 124 neutrons.

(a) Using R = (1.2 fm)A1 3 , the radii are roughly 3.6 fm, 5.3 fm, and 7.1 fm. (b) Using 4π R 2 for each of the radii in part (a), the areas are 163 fm 2 , 353 fm 2 and 633 fm 2 .

4 3 π R gives 195 fm3 , 624 fm 3 and 1499 fm3 . 3 (d) The density is the same, since the volume and the mass are both proportional to A: 2.3 × 1017 kg m3 (see Example 43.1). (c)

3

43.3.

(e) Dividing the result of part (d) by the mass of a nucleon, the number density is 0.14 fm3 = 1.40 × 1044 m . IDENTIFY: Calculate the spin magnetic energy shift for each spin state of the 1s level. Calculate the energy splitting between these states and relate this to the frequency of the photons. SET UP: When the spin component is parallel to the field the interaction energy is U = − μ z B. When the spin

component is antiparallel to the field the interaction energy is U = + μ z B. The transition energy for a transition between these two states is Δ E = 2μ z B, where μ z = 2.7928μ n . The transition energy is related to the photon frequency by Δ E = hf , so 2 μ z B = hf .

hf (6.626 × 10−34 J ⋅ s)(22.7 × 106 Hz) = = 0.533 T 2μ z 2(2.7928)(5.051× 10−27 J/T) EVALUATE: This magnetic field is easily achievable. Photons of this frequency have wavelength λ = c/f = 13.2 m. These are radio waves. G G (a) As in Example 43.2, ΔE = 2(1.9130)(3.15245 × 10 −8 eV T)(2.30 T) = 2.77 × 10−7 eV. Since μ and S are in opposite directions for a neutron, the antiparallel configuration is lower energy. This result is smaller than but comparable to that found in the example for protons. ΔE c (b) f = = 66.9 MHz, λ = = 4.48 m. h f IDENTIFY: Calculate the spin magnetic energy shift for each spin component. Calculate the energy splitting between these states and relate this to the frequency of the photons. G G G (a) SET UP: From Example 43.2, when the z-component of S (and μ ) is parallel to B, U = − | μ z | B = G G G −2.7928μ n B. When the z-component of S (and μ ) is antiparallel to B , U = − μ z | B = +2.7928μ n B. The state B=

EXECUTE:

43.4.

43.5.

with the proton spin component parallel to the field lies lower in energy. The energy difference between these two states is Δ E = 2(2.7928μ n B ). Δ E 2(2.7928μ n ) B 2(2.7928)(5.051 × 10−27 J/T)(1.65 T) = = h h 6.626 × 10−34 J ⋅ s f = 7.03 × 107 Hz = 7.03 MHz

EXECUTE:

Δ E = hf so f =

And then λ = EVALUATE:

c 2.998 × 108 m/s = = 4.26 m f 7.03 × 107 Hz From Figure 32.4 in the textbook, these are radio waves.

43-1


43-2

Chapter 43

(b) SET UP: From Eqs. (27.27) and (41.22) and Fig.41.14 in the textbook, the state with the z-component of G G μ parallel to B has lower energy. But, since the charge of the electron is negative, this is the state with the G electron spin component antiparallel to B. That is, for the ms = − 12 state lies lower in energy. ⎛ e ⎞ ⎛ =⎞ ⎛ e= ⎞ 1 1 For the ms = + 12 state, U = +(2.00232) ⎜ ⎟ ⎜ + ⎟ B = + 2 (2.00232) ⎜ ⎟ B = + 2 (2.00232) μ B B. 2 m 2 ⎝ ⎠ ⎝ ⎠ ⎝ 2m ⎠ For the ms = − 12 state, U = − 12 (2.00232) μB B. The energy difference between these two states is Δ E = (2.00232) μ B B.

EXECUTE:

Δ E 2.00232 μ B B (2.00232)(9.274 × 10 −24 J/T)(1.65 T) = = = 4.62 × 1010 Hz = 46.2 GHz. And h h 6.626 × 10−34 J ⋅ s c 2.998 × 108 m/s λ= = = 6.49 × 10−3 m = 6.49 mm. f 4.62 × 1010 Hz EVALUATE: From Figure 32.4 in the textbook, these are microwaves. The interaction energy with the magnetic field is inversely proportional to the mass of the particle, so it is less for the proton than for the electron. The smaller transition energy for the proton produces a larger wavelength. (a) 146mn + 92mH − mU = 1.93 u Δ E = hf so f =

43.6.

43.7.

43.8.

43.9.

(b) 1.80 × 103 MeV (c) 7.56 MeV per nucleon (using 931.5 MeV/u and 238 nucleons). IDENTIFY and SET UP: The text calculates that the binding energy of the deuteron is 2.224 MeV. A photon that breaks the deuteron up into a proton and a neutron must have at least this much energy. hc hc E= so λ = λ E (4.136 × 10 −15 eV ⋅ s)(2.998 × 108 m/s) EXECUTE: λ = = 5.575 × 10−13 m = 0.5575 pm. 2.224 × 106 eV EVALUATE: This photon has gamma-ray wavelength. IDENTIFY: The binding energy of the nucleus is the energy of its constituent particles minus the energy of the carbon-12 nucleus. SET UP: In terms of the masses of the particles involved, the binding energy is EB = (6mH + 6mn – mC-12)c2. EXECUTE: (a) Using the values from Table 43.2, we get EB = [6(1.007825 u) + 6(1.008665 u) – 12.000000 u)](931.5 MeV/u) = 92.16 MeV (b) The binding energy per nucleon is (92.16 MeV)/(12 nucleons) = 7.680 MeV/nucleon (c) The energy of the C-12 nucleus is (12.0000 u)(931.5 MeV/u) = 11178 MeV. Therefore the percent of the mass 92.16 MeV = 0.8245% . that is binding energy is 11178 MeV EVALUATE: The binding energy of 92.16 MeV binds 12 nucleons. The binding energy per nucleon, rather than just the total binding energy, is a better indicator of the strength with which a nucleus is bound. IDENTIFY: Conservation of energy tells us that the initial energy (photon plus deuteron) is equal to the energy after the split (kinetic energy plus energy of the proton and neutron). Therefore the kinetic energy released is equal to the energy of the photon minus the binding energy of the deuteron. SET UP: The binding energy of a deuteron is 2.224 MeV and the energy of the photon is E = hc/λ. Kinetic energy is K = ½mv2. EXECUTE: (a) The energy of the photon is

Eph =

hc

λ

=

( 6.626 ×10

−34

J ⋅ s )( 3.00 × 108 m/s )

3.50 × 10−13 m

= 5.68 × 10−13 J .

The binding of the deuteron is EB = 2.224 MeV = 3.56 × 10 −13 J . Therefore the kinetic energy is K = (5.68 − 3.56) × 10−13 J = 2.12 × 10−13 J = 1.32 MeV . (b) The particles share the energy equally, so each gets half. Solving the kinetic energy for v gives v=

43.10.

2K 2(1.06 × 10−13 J) = = 1.13 × 107 m/s m 1.6605 × 10 −27 kg

EVALUATE: Considerable energy has been released, because the particle speeds are in the vicinity of the speed of light. (a) 7( mn + mH ) − mN = 0.112 u, which is 105 MeV, or 7.48 MeV per nucleon. (b) Similarly, 2( mH + mn ) − mHe = 0.03038 u = 28.3 MeV, or 7.07 MeV per nucleon, slightly lower (compare to Figure 43.2 in the textbook).


Nuclear Physics

43.11.

43-3

(a) IDENTIFY: Find the energy equivalent of the mass defect. SET UP: A 115 B atom has 5 protons, 11 − 5 = 6 neutrons, and 5 electrons. The mass defect therefore is

Δ M = 5mp + 6mn + 5me − M ( 115 B). EXECUTE:

Δ M = 5(1.0072765 u) + 6(1.0086649 u) + 5(0.0005485799 u) − 11.009305 u = 0.08181 u. The energy

equivalent is EB = (0.08181 u)(931.5 MeV/u) = 76.21 MeV. (b) IDENTIFY and SET UP: Eq.(43.11): EB = C1 A − C2 A2/3 − C3Z ( Z − 1) / A1/3 − C4 ( A − 2 Z ) 2 / A The fifth term is zero since Z is odd but N is even. A = 11 and Z = 5. EXECUTE: EB = (15.75 MeV)(11) − (17.80 MeV)(11) 2/3 − (0.7100 MeV)5(4)/111/3 − (23.69 MeV)(11 − 10) 2 /11.

EB = +173.25 MeV − 88.04 MeV − 6.38 MeV − 2.15 MeV = 76.68 MeV 76.68 MeV − 76.21 MeV = 0.6% 76.21 MeV EVALUATE: Eq.(43.11) has a greater percentage accuracy for 62 Ni. The semi-empirical mass formula is more accurate for heavier nuclei. (a) 34mn + 29mH − mCu = 34(1.008665) u + 29(1.007825) u − 62.929601 u = 0.592 u, which is 551 MeV,

The percentage difference between the calculated and measured EB is

43.12.

or 8.75 MeV per nucleon (using 931.5 MeV/u and 63 nucleons). (b) In Eq.(43.11), Z = 29 and N = 34, so the fifth term is zero. The predicted binding energy is (29)(28) (5) 2 2 EB = (15.75 MeV)(63) − (17.80 MeV)(63) 3 − (0.7100 MeV) − (23.69 MeV) . 1 3 (63) (63)

43.13.

EB = 556 MeV . The fifth term is zero since the number of neutrons is even while the number of protons is odd, making the pairing term zero. This result differs from the binding energy found from the mass deficit by 0.86%, a very good agreement comparable to that found in Example 43.4. IDENTIFY In each case determine how the decay changes A and Z of the nucleus. The β + and β − particles have charge but their nucleon number is A = 0. (a) SET UP: α -decay: Z increases by 2, A = N + Z decreases by 4 (an α particle is a 42 He nucleus) EXECUTE:

239 94

β − decay: Z increases by 1, A = N + Z remains the same (a β − particle is an electron,

(b) SET UP: EXECUTE:

Pu → 42 He + 235 92 U

24 11

e)

Na → e + Mg 0 −1

24 12

(c) SET UP β + decay: Z decreases by 1, A = N + Z remains the same (a β + particle is a positron,

0 +1

e)

EXECUTE: O→ e+ N EVALUATE: In each case the total charge and total number of nucleons for the decay products equals the charge and number of nucleons for the parent nucleus; these two quantities are conserved in the decay. (a) The energy released is the energy equivalent of mn − mp − me = 8.40 × 10−4 u, or 783 keV. 15 8

43.14.

0 −1

0 +1

15 7

(b) mn > mp , and the decay is not possible. 43.15.

IDENTIFY: The energy of the photon must be equal to the difference in energy of the two nuclear energy levels. SET UP: The energy difference is ΔE = hc/λ.

43.16.

hc

( 6.626 ×10

−34

J ⋅ s )( 3.00 × 108 m/s )

= 8.015 × 10−15 J = 0.0501 MeV 0.0248 × 10−9 J EVALUATE: Since the wavelength of this photon is much shorter than the wavelengths of visible light, its energy is much greater than visible-light photons which are frequently emitted during electron transitions in atoms. This tells us that the energy difference between the nuclear shells is much greater than the energy difference between electron shells in atoms, meaning that nuclear energies are much greater than the energies of orbiting electrons. IDENTIFY: The energy released is equal to the mass defect of the initial and final nuclei. SET UP: The mass defect is equal to the difference between the initial and final masses of the constituent particles. EXECUTE: (a) The mass defect is 238.050788 u – 234.043601 u – 4.002603 u = 0.004584 u. The energy released is (0.004584 u)(931.5 MeV/u) = 4.270 MeV. (b) Take the ratio of the two kinetic energies, using the fact that K = p2/2m: EXECUTE:

ΔE =

λ

=

2 pTh K Th 2mTh mα 4 = = = . 2 p Kα mTh 234 α 2mα


43-4

Chapter 43

The kinetic energy of the Th is 4 4 (4.270 MeV) = 0.07176 MeV = 1.148 × 10−14 J K Total = 234 + 4 238 Solving for v in the kinetic energy gives

K Th =

v=

43.17.

2K 2(1.148 × 10−14 J) = = 2.431× 105 m/s −27 m (234.043601) (1.6605 × 10 kg )

EVALUATE: As we can see by the ratio of kinetic energies in part (b) , the alpha particle will have a much higher kinetic energy than the thorium. If β − decay of 14 C is possible, then we are considering the decay 146 C → 147 N + β − .

Δm = M ( 146 C) − M ( 147 N) − me Δm = (14.003242 u − 6(0.000549 u)) − (14.003074 u − 7(0.000549 u)) − 0.0005491 u Δm = +1.68 × 10−4 u. So E = (1.68 × 10−4 u)(931.5 MeV u ) = 0.156 MeV = 156 keV 43.18.

(a) A proton changes to a neutron, so the emitted particle is a positron ( β + ). (b) The number of nucleons in the nucleus decreases by 4 and the number of protons by 2, so the emitted particle is an alpha-particle. (c) A neutron changes to a proton, so the emitted particle is an electron ( β − ).

43.19.

(a) As in the example, (0.000898 u)(931.5 MeV u) = 0.836 MeV. (b) 0.836 MeV − 0.122 MeV − 0.014 MeV = 0.700 MeV.

43.20.

(a)

Sr → β − + 90 39 X . X has 39 protons and 90 protons plus neutrons, so it must be

90 39

(b) Use base 2 because we know the half life. A = A0 2

t=− 43.21.

T1 2 log 0.01 log 2

=−

−t T 1

2

and 0.01A0 = A0 2

−t T1 2

90

Y.

.

(28 yr)log 0.01 = 190 yr . log 2

IDENTIFY and SET UP:

T1/ 2 =

ln 2

λ

The mass of a single nucleus is 124mp = 2.07 × 10−25 kg .

ΔN / Δt = 0.350 Ci = 1.30 × 1010 Bq ; ΔN / Δt = λ N EXECUTE:

T1/ 2 = 43.22.

ln 2

λ

N=

6.13 × 10 −3 kg ΔN / Δt 1.30 × 1010 Bq = 2.96 × 1022 ; λ = = = 4.39 × 10−13 s −1 −25 2.07 × 10 kg N 2.96 × 10 22

= 1.58 × 1012 s = 5.01 × 104 yr

Note that Eq.(43.17) can be written as follows: N = N 0 2

− t / T1 2

. The amount of elapsed time since the source was

created is roughly 2.5 years. Thus, we expect the current activity to be N = (5000 Ci)2− (2.5 yr)/(5.271 yr) = 3600 Ci. The source is barely usable. Alternatively, we could calculate λ =

43.23.

ln(2) = 0.132(years) −1 and use the Eq. 43.17 directly T1 2

to obtain the same answer. IDENTIFY and SET UP: As discussed in Section 43.4, the activity A = dN / dt obeys the same decay equation as Eq. (43.17): A = A0e − λ t . For 14C, T1/2 = 5730 y and λ = ln2 / T1/ 2 so A = A0e− (ln 2)t / T1/ 2 ; Calculate A at each t; A0 = 180.0 decays/min. EXECUTE: (a) t = 1000 y, A = 159 decays/min (b) t = 50,000 y, A = 0.43 decays/min

43.24.

EVALUATE: The time in part (b) is 8.73 half-lives, so the decay rate has decreased by a factor or ( 12 )8.73. IDENTIFY and SET UP: The decay rate decreases by a factor of 2 in a time of one half-life. EXECUTE: (a) 24 d is 3T1/2 so the activity is (375 Bq) /(23 ) = 46.9 Bq (b) The activity is proportional to the number of radioactive nuclei, so the percent is

17.0 Bq = 36.2% 46.9 Bq

0 131 131 (c) 131 53 I → −1 e + 54 Xe The nucleus 54 Xe is produced. EVALUATE: Both the activity and the number of radioactive nuclei present decrease by a factor of 2 in one halflife.


Nuclear Physics

43.25.

(a) 31 H → −01 e + 23 He (b) N = N 0e − λt , N = 0.100 N 0 and λ = (ln 2) T1 0.100 = e

43.26.

43-5

− t (ln 2) T1 2

2

; −t (ln 2) T1

2

= ln(0.100); t =

− ln(0.100)T1 2 ln 2

= 40.9 y

dN = 500 μ Ci = (500 × 10−6 )(3.70 × 1010 s −1 ) = 1.85 × 107 decays s dt ln 2 ln 2 ln 2 →λ= = = 6.69 × 10−7 s . T1 2 = λ T1 2 12 d(86,400s d)

(a)

dN dN dt 1.85 × 107 decays / s = λN ⇒ N = = = 2.77 × 1013 nuclei . The mass of this many 131 Ba nuclei is dt λ 6.69 × 10 −7 s −1 m = 2.77 × 1013 nuclei × (131 × 1.66 × 10−27 kg nucleus ) = 6.0 × 10−12 kg = 6.0 × 10−9 g = 6.0 ng (b) A = A0e − λ t . 1 μ Ci = (500 μ Ci) e − λ t . ln(1/500) = −λt.

⎛ 1d ⎞ ln(1 500) = 9.29 × 106 s ⎜ ⎟ = 108 days −7 −1 λ 6.69 × 10 s ⎝ 86, 400 s ⎠ (ln 2)t − t (ln 2) / T1 / 2 = ln( A A0 ) . A = A0e − λ t = A0e . − T1 2

t=− 43.27.

ln(1 500)

T1 2 = − 43.28.

=−

(ln 2)t (ln 2)(4.00 days) =− = 2.80 days ln( A A0 ) ln(3091 8318)

dN ln 2 ln 2 = λN . λ = = = 1.36 × 10−11 s −1 . dt T1 2 1620 yr ( 3.15 × 107 s/yr ) ⎛ 6.022 × 1023 atoms ⎞ 25 N =1g⎜ ⎟ = 2.665 × 10 atoms . 226 g ⎝ ⎠

dN = λ N = (2.665 × 1025 )(1.36 × 10 −11 s −1 ) = 3.62 × 1010 decays/s = 3.62 × 1010 Bq dt ⎛ ⎞ 1 Ci Convert to Ci: 3.62 × 1010 Bq ⎜ ⎟ = 0.98 Ci 10 3.70 10 Bq × ⎝ ⎠ 43.29.

IDENTIFY and SET UP: Calculate the number N of 14 C atoms in the sample and then use Eq. (43.17) to find the decay constant λ. Eq. (43.18) then gives T1/ 2 . EXECUTE: Find the total number of carbon atoms in the sample. n = m/M; N tot = nN A = mN A / M = (12.0 × 10 −3 kg)(6.022 × 1023 atoms/mol)/(12.011 × 10−3 kg/mol) N tot = 6.016 × 1023 atoms, so (1.3 × 10−12 )(6.016 × 1023 ) = 7.82 × 1011 carbon-14 atoms ΔN / Δ t = −180 decays/min = −3.00 decays/s

−ΔN / Δt = 3.836 × 10−12 s −1 N T1/ 2 = (ln 2) / λ = 1.807 × 1011 s = 5730 y EVALUATE: The value we calculated agrees with the value given in Section 43.4. 360 × 106 decays = 4.17 × 103 Bq = 1.13 × 10−7 Ci = 0.113 μCi. 86,400 s Δ N / Δ t = −λ N ;

43.30. 43.31.

(a)

λ=

dN 0.693 0.693 = = 3.75 × 10−4 s −1. = 7.56 × 1011 Bq = 7.56 × 1011 decays s . λ = dt T1 2 (30.8 min)(60 s min)

N0 =

1 dN 7.56 × 1011 decays s = = 2.02 × 1015 nuclei. 3.75 × 10−4 s −1 λ dt


43-6

Chapter 43

(b) The number of nuclei left after one half-life is

N0 = 1.01 × 1015 nuclei, and the activity is half: 2

dN = 3.78 × 1011 decays s. dt (c) After three half lives (92.4 minutes) there is an eighth of the original amount, so N = 2.53 × 1014 nuclei, and an ⎛ dN ⎞ 10 eighth of the activity: ⎜ ⎟ = 9.45 × 10 decays s. dt ⎝ ⎠ 43.32.

The activity of the sample is

3070 decays min = 102 Bq kg, while the activity of atmospheric carbon is (60 sec min) (0.500 kg)

ln (102 255) = 7573 y. 1.21 × 10−4 y IDENTIFY and SET UP: Find λ from the half-life and the number N of nuclei from the mass of one nucleus and the mass of the sample. Then use Eq.(43.16) to calculate | dN / dt |, the number of decays per second. EXECUTE: (a) | dN / dt |= λ N 0.693 0.693 λ= = = 1.715 × 10−17 s −1 9 T1/ 2 (1.28 × 10 y)(3.156 × 107 s/1 y)

255 Bq kg (see Example 43.9). The age of the sample is then t = − 43.33.

The mass of 40 K atom is approximately 40 u, so the number of 1.63 × 10−9 kg 1.63 × 10 −9 kg N= = = 2.454 × 1016. 40 u 40(1.66054 × 10−27 kg)

40

ln (102 255)

λ

=−

K nuclei in the sample is

Then | dN / dt |= λ N = (1.715 × 10−17 s −1 )(2.454 × 1016 ) = 0.421 decays/s

43.34.

(b) | dN / dt |= (0.421 decays/s)(1 Ci/(3.70 × 1010 decays/s)) = 1.14 × 10−11 Ci EVALUATE: The very small sample still contains a very large number of nuclei. But the half life is very large, so the decay rate is small. (a) rem = rad × RBE. 200 = x(10) and x = 20 rad. (b) 1 rad deposits 0.010 J kg , so 20 rad deposit 0.20 J kg . This radiation affects 25 g (0.025 kg) of tissue, so the

total energy is (0.025 kg)(0.20 J kg ) = 5.0 × 10−3 J = 5.0 mJ (c) Since RBE = 1 for β -rays, so rem = rad. Therefore 20 rad = 20 rem. 43.35. 43.36.

1 rad = 10 −2 Gy, so 1 Gy = 100 rad and the dose was 500 rad. rem = (rad)(RBE) = (500 rad)(4.0) = 2000 rem. 1 Gy = 1 J kg, so 5.0 J kg . IDENTIFY and SET UP: For x rays RBE = 1 so the equivalent dose in Sv is the same as the absorbed dose in J/kg. EXECUTE: One whole-body scan delivers (75 kg)(12 × 10−3 J/kg) = 0.90 J . One chest x ray delivers

0.90 J = 900 chest x rays to deliver the same total energy. 1.0 × 10−3 J IDENTIFY and SET UP: For x rays RBE = 1 and the equivalent dose equals the absorbed dose. EXECUTE: (a) 175 krad = 175 krem = 1.75 kGy = 1.75 kSv (5.0 kg)(0.20 × 10−3 J/kg) = 1.0 × 10−3 J . It takes

43.37.

(1.75 × 103 J/kg)(0.150 kg) = 2.62 × 102 J (b) 175 krad = 1.75 kGy ; (1.50)(175 krad) = 262 krem = 2.62 kSv

43.38.

43.39.

The energy deposited would be 2.62 × 102 J , the same as in (a). EVALUATE: The energy required to raise the temperature of 0.150 kg of water 1 C° is 628 J, and 2.62 × 102 J is less than this. The energy deposited corresponds to a very small amount of heating. (a) 5.4 Sv (100 rem Sv) = 540 rem. (b) The RBE of 1 gives an absorbed dose of 540 rad. (c) The absorbed dose is 5.4 Gy, so the total energy absorbed is (5.4 Gy) (65 kg) = 351 J. The energy required to raise the temperature of 65 kg by 0.010° C is (65 kg) (4190 J kg ⋅ K) (0.01C°) = 3 kJ. (a) We need to know how many decays per second occur.

λ=

0.693 0.693 = = 1.79 × 10−9 s −1