Engineering electromagnetics and waves 2nd edition inan solutions manual

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6

T a his th nd wo o eir is rk w r sa co pro is ill le u vi pr de o rse de ot st f a s d s ec ro n an o te y y p d le d th a a ly by e rt ss fo U in o e r te f t ss th nite gr hi in e ity s w g us d S of or stu e o tat th k ( de f i es e in nt ns co w cl le tr p or ud a uc y r k an ing rnin tors igh d on g. in t la is w D no the iss tea s t p W em ch er or in ing m ld a itt W tio ed id n . e W eb )

The Static Magnetic Field

6.1 Force between two infinitely long wires.

Following a similar approach as in Example 6.1, the magnetic repulsion force per unit length experienced by each wire can be written as F=

µ0 I2 4π × 10−7 I2 = = 0.5 N-m−1 2πa 2π(0.5)

from which we find I ≃ 1118 A 6.2 Forces between two wires.

Two wires of a transmission line carry oppositely directed current and therefore experience magnetic repulsion, which can balance electrical attraction, due to opposite charges induced on the wires. The electric field produced by wire 1 at the location of wire 2 is given by (see Example 4.12 and Figure 6.1 below) E = −x̂

ρl 2πϵ0 d

The electrostatic force exerted on a unit length portion of wire 2 is thus given by Fe12 = E(ρl ) = −x̂

ρ2l 2πϵ0 d

We can express this force in terms of the applied potential difference V0 by using the per unit length capacitance of this configuration. From Example 4.29 we have Cu =

ρl πϵ0 πϵ0 ]≃ = [ √ V0 ln (d/a) ln d/(2a) + d/(2a) − 1

→

ρl =

πϵ0 V0 ln(d/a)

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2

Chapter 6/ The Static Magnetic Field

Figure 6.1 Figure for Problem 6.2. since the separation of the wires (d = 1 cm) is much larger than the wire diameter (2a = 1 mm).

The magnetic force exerted by wire 1 on a unit length portion of wire 2 is (see Example 6.1) µ0 I2 2πd

T a his th nd wo o eir is rk w r sa co pro is ill le u vi pr de o rse de ot st f a s d s ec ro n an o te y y p d le d th a a ly by e rt ss fo U in o e r te f t ss th nite gr hi in e ity s w g us d S of or stu e o tat th k ( de f i es e in nt ns co w cl le tr p or ud a uc y r k an ing rnin tors igh d on g. in t la is w D no the iss tea s t p W em ch er or in ing m ld a itt W tio ed id n . e W eb )

Fm 12 = x̂

For balance between the two forces we must have Fe12 + Fm 12 = 0

→

πϵ0 V0 µ0 I2 = 2πd ln(d/a)

→

I2 =

2π 2 dV0 ϵ0 µ0 ln(d/a)

Substituting the values yields I ≃ 27.9 mA. 6.3 Bundle clash in a transmission line.

Assuming the wires to be infinitely thin (since 50 cm ≫ 4 cm) and following a similar approach as in Example 6.1, the peak magnetic force of attraction per unit length on each wire can be calculated as Fpeak =

µ0 I2 4π × 10−7 (105 )2 = = 4000 N-m−1 !

2πa 2π(0.5)

To get a better feel about the strength of this peak force, consider this force as being the weight of an object which has a corresponding mass given by m = F/g ≃ 4000/(9.81) ≃ 408 kg. Assuming a typical person weighs about 70 to 80 kg, the peak attraction force is equivalent to the total weight of about 5 to 6 persons. That is, the peak force of attraction exerted over 1 m length of each wire can be thought of as the force that would be exerted on the wire if 5 to 6 people were to stand up over 1 m portion of this wire. Therefore, this strong peak attraction force may well make the wires in the bundle to collapse onto one another and cause the bundle to clash.

6.4 Magnetic force between current elements.

The differential force between two current elements is given by the integrand in equation (6.1), namely dF12 =

µ0 I2 dl2 × (I1 dl1 × R) 4π R2

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3 where dF12 is the force exerted by element 1 on element 2 and R̂ is the unit vector from element 1 to element 2. (a) Taking the origin to be at the location of element 1, we have

I1 dl1 = I1 dl1 (−ŷ) ; so that

I2 dl2 = I2 dl2 (x̂) ;

R=

√

a2 + b2 ;

ax̂ + bŷ R= √ a2 + b2

µ0 I2 dl2 (x̂) × [Id l1 (−ŷ) × (ax̂ + bŷ)] 4π(a2 + b2 ) µ0 µ0 aI1 I2 dl1 dl2 = I2 dl2 (x̂) × I1 dl1 a(ẑ) = (−ŷ) 2 2 4π(a + b ) 4π(a2 + b2 )

dF12 =

T a his th nd wo o eir is rk w r sa co pro is ill le u vi pr de o rse de ot st f a s d s ec ro n an o te y y p d le d th a a ly by e rt ss fo U in o e r te f t ss th nite gr hi in e ity s w g us d S of or stu e o tat th k ( de f i es e in nt ns co w cl le tr p or ud a uc y r k an ing rnin tors igh d on g. in t la is w D no the iss tea s t p W em ch er or in ing m ld a itt W tio ed id n . e W eb )

(b) For this case the problem statement and the figure in the text fail to specify the distance between the two elements. For the purposes of our solution here, we shall take this distance to be a. We then have

I1 dl1 = I1 dl1 (ŷ) ; Thus

I2 dl2 = I2 dl2 (x̂ sin ϕŷ + cos ϕ) ;

R=a;

R̂ = x̂

−ẑ

z }| { µ0 ŷ × ax̂) dF12 = [(x̂ sin ϕŷ + cos ϕ)I dl ] × (I l 2 2 1 d 4πa2 µ0 I1 I2 dl1 dl2 = [ŷ sin ϕ − x̂ cos ϕ] 4πa2 6.5 Square loop.

Consider a square loop carrying a current I. To find the total magnetic force on each side, we use (6.1) where Idl2 × (Idl1 × R) is a vector which is always perpendicular to and outward from each side of the square loop. So, the total force on each side calculated from (6.1) is perpendicular to that side and outward from it and as a result, the loop has a tendency to expand.

6.6 Force on a current-carrying wire.

Assuming the earth's magnetic field of B0 ≃ 0.5 G to be in the y direction and the current I = 1 kA to be flowing in the x direction, the force per meter on the current-carrying conductor exerted by the earth's B field can be found as ∫ 0.5 F= −0.5

(Idx)x̂ × ŷB0 = ẑIB0

∫ 0.5 −0.5

dx = ẑ(1000)(0.5 × 10−4 ) = ẑ0.05 N-m−1

6.7 Rigid rectangular loop.

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4

Chapter 6/ The Static Magnetic Field

Figure 6.2 Figure for Problem 6.7. Rectangular current-carrying loop in the presence of a magnetic field. (a) B = x̂1.5 T. (b) B = ẑ1.5 T.

T a his th nd wo o eir is rk w r sa co pro is ill le u vi pr de o rse de ot st f a s d s ec ro n an o te y y p d le d th a a ly by e rt ss fo U in o e r te f t ss th nite gr hi in e ity s w g us d S of or stu e o tat th k ( de f i es e in nt ns co w cl le tr p or ud a uc y r k an ing rnin tors igh d on g. in t la is w D no the iss tea s t p W em ch er or in ing m ld a itt W tio ed id n . e W eb )

(a) We assume the current in the rectangular loop to be flowing in the clockwise direction, as shown in Figure 6.2. For I = 5 A and B = x̂B0 = x̂1.5 T, the forces on sides 1, 2, 3, and 4 can be calculated using a similar approach as in Example 6.2 as ∫ 2

F1 =

0

∫ 2

F3 =

0

(−Idy)ŷ × x̂B0 = ẑ(5)(1.5)(2) = ẑ15 N

(Idy)ŷ × x̂B0 = −ẑ(5)(1.5)(2) = −ẑ15 N

F2 = −F4 =

∫ 1 0

(−Idx)x̂ × x̂B0 = 0

(b) Following a similar approach as in part (a) with B = ẑ1.5 T, the forces on sides 1, 2, 3, and 4 can be found as F1 = −F3 = x̂15 N and F2 = −F4 = ŷ7.5 N respectively.

6.8 Two parallel wires. (a) Using the superposition principle, the B field at the origin is given by B = B1 + B2 = −ŷ

µ0 I µ0 I µ0 I + x̂ = [x̂ − ŷ] 2π(2) 2π(2) 4π

(b) The B field at point (1,1,0) can be written as µ0 I 1 µ0 I 1 √ √ − ŷ √ √ 2π( 2) 2 2π( 2) 2 µ0 I 1 µ0 I 1 √ √ + ŷ √ √ =0 + x̂ 2π( 2) 2 2π( 2) 2

B = B1 + B2 = − x̂

(c) Similarly, the B field at point (2,2,0) can be found as B = B1 + B2 = −x̂

µ0 I µ0 I µ0 I + ŷ = [−x̂ + ŷ] 2π(2) 2π(2) 4π

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5

Figure 6.3 Figure for Problem 6.8. (a) I1 and I2 in the same direction. (b) I1 and I2 in opposite directions.

T a his th nd wo o eir is rk w r sa co pro is ill le u vi pr de o rse de ot st f a s d s ec ro n an o te y y p d le d th a a ly by e rt ss fo U in o e r te f t ss th nite gr hi in e ity s w g us d S of or stu e o tat th k ( de f i es e in nt ns co w cl le tr p or ud a uc y r k an ing rnin tors igh d on g. in t la is w D no the iss tea s t p W em ch er or in ing m ld a itt W tio ed id n . e W eb )

(d) When the direction of the current on the x axis is switched, all the B1 fields also switch direction.

Thus, the B field at the origin is now given by B = B1 + B2 = ŷ

µ0 I µ0 I µ0 I + x̂ = [x̂ + ŷ] 2π(2) 2π(2) 4π

The B field at point (1,1,0) is

µ0 I 1 µ0 I 1 √ √ + ŷ √ √ 2π( 2) 2 2π( 2) 2 µ0 I 1 µ0 I 1 √ √ + ŷ √ √ + x̂ 2π( 2) 2 2π( 2) 2 µ0 I = [x̂ + ŷ] 2π Finally, the B field at point (2,2,0) follows as B = B1 + B2 = x̂

B = B1 + B2 = x̂

µ0 I µ0 I µ0 I + ŷ = [x̂ + ŷ] 2π(2) 2π(2) 4π

6.9 An infinitely long L-shaped wire. (a) We can consider the L-shaped wire as two wires, one from the origin to x = ∞ and the other from the origin to y = ∞ and apply the superposition principle. Assume the current flowing from x = ∞ to y = ∞. At point P1 (−a, 0, 0), no B field is produced due to the x portion of the L-shaped wire since the observation point P2 lies along the x axis. The B field produced by the y portion of the L-shaped wire at P1 can be found from Example 6.3 by changing z to y, z′ to y′ , r direction to x direction, z′ = −a to y′ = 0, and z′ = a to y′ = ∞, yielding µ0 I 4πa (b) Following a similar approach for point P2 (0, −a, 0), since P2 lies on the y axis, only the x portion of the L-shaped wire produced B at point P2 given by B1 = ẑ

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6

Chapter 6/ The Static Magnetic Field

µ0 I 4πa (c) At point P3 (0, 0, a), both the x and y portions of the L-shaped wire contribute to the B field resulting in B2 = ẑ

B3 = (x̂ + ŷ)

µ0 I 4πa

6.10 B-field in a cylindrical conductor. (a) Using the differential form of Ampère's law given by (6.11) as J(r) =

1 ∇×B µ

we can easily compute J by noting that B only contains a ϕ̂ϕ component that varies with r, so T a his th nd wo o eir is rk w r sa co pro is ill le u vi pr de o rse de ot st f a s d s ec ro n an o te y y p d le d th a a ly by e rt ss fo U in o e r te f t ss th nite gr hi in e ity s w g us d S of or stu e o tat th k ( de f i es e in nt ns co w cl le tr p or ud a uc y r k an ing rnin tors igh d on g. in t la is w D no the iss tea s t p W em ch er or in ing m ld a itt W tio ed id n . e W eb )

[ ] 1 1 ∂ J(r) = ẑ (rBϕ (r)) µ0 r ∂r ( ) B0 r r2 = ẑ − 2 µ0 r a a ( B0 r) = 1− ẑ µ0 a a

The total current I passing through the conductor is given by (5.1) as ∫

I=

J · ds ∫ a ∫ 2π S

J(r)rdϕdr ) ∫ ( 2πB0 a r2 = r− dr µ0 a 0 a ( ) 2πB0 a2 a2 = − µ0 a 2 3 πB0 a = 3µ0 =

0

0

(b) Applying Ampère's law given by (6.9): I B · dl = µ0 Iencl and noting the cylindrical symmetry, we can evaluate the azimuthal component of B as 2πrBϕ =

µ0 πB0 a 3µ0

→

Bϕ =

B0 a 6r

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7 whence B(r) =

B0 a ϕ̂ϕ 6r

6.11 Irregular loop.

We consider each side of the loop by itself and apply the superposition principle. Since point P lies on the axes of two of the sides of the loop, the current flowing in those sides do not contribute any B field at point P. Using the result of Example 6.7 and assuming the z direction to be out of the page, the B field produced at point P by current flowing in the closest side of the loop is given by µ0 I sin ϕ0 4πa Similarly, the B field contributed by the current flowing in the farthest side is B1 = −ẑ

µ0 I sin ϕ0 4πb Therefore, the total B field at point P due to the current in the loop follows as ( ) 1 1 µ0 I sin ϕ0 − B = B1 + B2 = ẑ 4π b a T a his th nd wo o eir is rk w r sa co pro is ill le u vi pr de o rse de ot st f a s d s ec ro n an o te y y p d le d th a a ly by e rt ss fo U in o e r te f t ss th nite gr hi in e ity s w g us d S of or stu e o tat th k ( de f i es e in nt ns co w cl le tr p or ud a uc y r k an ing rnin tors igh d on g. in t la is w D no the iss tea s t p W em ch er or in ing m ld a itt W tio ed id n . e W eb )

B2 = ẑ

6.12 An N-sided regular loop.

(a) We consider each side of the N-sided polygon-shaped loop separately and apply the superposition principle. Note that due to symmetry, the magnetic field due to the current of each side is the same. Using the result of Example 6.3 with Figure 6.4, the magnitude of the total B field produced at the center of the loop can be written as B=N where we replaced

µ0 I 2 π µ0 NI π µ Ia √0 = tan = tan 2 2 2π d N πd N 2πr r + a

√ a2 + r2 = d/2 and a/r = tan(π/N)

(b) As N → ∞, tan(π/N) ≃ π/N and as a result, we find B≃

µ0 NI π µ0 I = πd N d

which is the same result obtained in Example 6.6.

6.13 Square loop of current.

In this and some of the other problems in this chapter, it is useful to note that, as depicted in Figure 6.5a and 6.5b, the B-field at some distance r from the edge of a current carrying wire of length L is one-half of that at the same distance r from the center of a wire of length 2L, the expression for which was derived in Example 6.3. Namely, B = ϕ̂ϕ

µ I(L) √0 2πr r2 + (L)2

B field at a distance r from the center of wire of length 2L

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8

Chapter 6/ The Static Magnetic Field

Figure 6.4 Figure for Problem 6.12. An N-sided regular polygon-shaped loop.

T a his th nd wo o eir is rk w r sa co pro is ill le u vi pr de o rse de ot st f a s d s ec ro n an o te y y p d le d th a a ly by e rt ss fo U in o e r te f t ss th nite gr hi in e ity s w g us d S of or stu e o tat th k ( de f i es e in nt ns co w cl le tr p or ud a uc y r k an ing rnin tors igh d on g. in t la is w D no the iss tea s t p W em ch er or in ing m ld a itt W tio ed id n . e W eb )

On this basis, the contributions of each side of the square depicted in Figure 6.5c can simply be formulated as the superposition of the B-fields at various distances from edges of wires of lengths L = a/4 and L = 3a/4. We first note that the contributions due to segments 1 and 4 are equal, as are those from 2 and 3. Using the right-hand rule, it is clear than all segments produce a B-field at point P in the −z direction, so that the different contributions can be simply added to find the total field. We have 



 ( )  µ I(3a/4) 1  µ I(a/4) 0 0   √ √ + |BP |seg 1 =  2  2π(a/4)2 (a/4)2 + (a/4)2 2π(a/4)2 (a/4)2 + (3a/4)2   {z } | {z } | r=a/4, L=a/4

( √ ) ( ) 1 µ0 I √ 6 6 = 8+ 2 2πa 5 and

r=a/4, L=3a/4





 ( )  µ0 I(a/4) 1  µ0 I(3a/4)   √ √ |BP |seg 2 = +  2 2 2 2 2 2 2  2π(3a/4) (3a/4) + (a/4) 2π(3a/4) (3a/4) + (3a/4)   | {z } | {z } r=3a/4, L=a/4

( √ √ ) ( ) 1 µ0 I 2 10 2 2 = + 2 2πa 15 3

r=3a/4, L=3a/4

Thus, the total B-field at point P is ( √ ) √ √ √ −µ0 I 6 10 2 10 2 2 2 2+ + + ẑ BP = 2πa 5 15 3 √ ) µ0 I −2 ( √ = 2 2 + 10 ẑ 3 πa

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9

Figure 6.5 Figure for Problem 6.13.

T a his th nd wo o eir is rk w r sa co pro is ill le u vi pr de o rse de ot st f a s d s ec ro n an o te y y p d le d th a a ly by e rt ss fo U in o e r te f t ss th nite gr hi in e ity s w g us d S of or stu e o tat th k ( de f i es e in nt ns co w cl le tr p or ud a uc y r k an ing rnin tors igh d on g. in t la is w D no the iss tea s t p W em ch er or in ing m ld a itt W tio ed id n . e W eb )

6.14 A wire with two circular arcs.

We consider each portion of the loop separately and apply the superposition principle. The straight portions do not produce any B field at point P since P lies along their axis. Assuming the z direction to be out of the page, the B field at point P produced by the current in the circular-arc portion with radius a is ∫ ϕ0

B1 =

0

µ0 Iadϕ′ (−ϕ̂ϕ) × (−r̂) µ0 Iϕ0 = −ẑ 2 4π a 4πa

Similarly, the B field produced by the current in the circular arc with radius b is ∫ ϕ0

B2 =

0

µ0 Ibdϕ′ (ϕ̂ϕ) × (−r̂) µ0 Iϕ0 = −ẑ 4π b2 4πb

Therefore, the total B produced by the wire loop is

µ0 Iϕ0 B = B1 + B2 = ẑ 4π

(

1 1 − b a

)

6.15 Wire with four 90◦ bends.

With reference to Figure 6.6, and the result of Example 6.3 the B-field at point P can be determined as a superposition of the contributions from different segments 1 through 5. The two formulas used are: ( ) 1 µ I(2L) √0 B = ϕ̂ϕ 2 2πr r2 + (2L)2 ( ) 1 µ0 I B = ϕ̂ϕ 2 2πr

B field at a distance r from the edge of a semi-infinitely long wire B field at a distance r from the edge of wire of length L

Designating BiP as the contribution to the B-field at point P by the ith wire segment we have

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10

Chapter 6/ The Static Magnetic Field

Figure 6.6 Figure for Problem 6.15. ( ) 1 µ0 I (−x̂) 2 2π(1) ( ) 1 µ0 I(1) 2 √ BP = (−ŷ) 2 2π(1) 12 + 12 ( ) 1 µ0 I(1) 3 √ BP = (−ŷ) 2 2π(1) 12 + 12 ( ) µ0 I(1) 1 4 √ (−ẑ) BP = 2 2π(1) 12 + 12 ( ) ( ) 1 µ0 I 1 µ0 I(1) 5 √ BP = (ẑ) + (−ẑ) 2 2π(1) 2 2π(1) 12 + 12 T a his th nd wo o eir is rk w r sa co pro is ill le u vi pr de o rse de ot st f a s d s ec ro n an o te y y p d le d th a a ly by e rt ss fo U in o e r te f t ss th nite gr hi in e ity s w g us d S of or stu e o tat th k ( de f i es e in nt ns co w cl le tr p or ud a uc y r k an ing rnin tors igh d on g. in t la is w D no the iss tea s t p W em ch er or in ing m ld a itt W tio ed id n . e W eb )

B1P =

The total B-field at point P is then

( ) 1 1 1 1 BP = −x̂ − √ ŷ − √ ŷ − √ ẑ + ẑ − √ ẑ 4π 2 2 2 2 i [ ] √ √ −µ0 = x̂ + 2ŷ + ( 2 − 1)ẑ T 4π ∑

µ0 BiP =

6.16 Infinitely long copper cylindrical wire carrying uniform current.

Using the results of Example 6.10 with I = 1000 A and a = 1 cm, we have    ϕ̂ϕ(200r) G r ≤ 1 cm    B= ( )   200    ϕ̂ϕ G r > 1 cm r where r in the B expressions is in units of cm (e.g., at r = 1 cm, the B field is B = ϕ̂ϕ200 Gauss).

6.17 Helmholtz coils.

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11 From Example 6.6, we know that the B-field at a point along the axis of a circular current carrying loop of radius a is µ0 Ia2 ẑ B= 2(a2 + z2 )3/2 where z is the distance from the loop center, and the z-axis has chosen to be in the direction of the thumb, when the right-hand fingers move in the direction of the current flow in the loop. (a) Consider a point P(0,0,z) along the z axis at a distance of z from the center of the loop on the left in Figure 6.59 on page 524 of the text. The distance between this point P and the center of the loop on the right is then (d − z). The B-field at this point P is then given as the superposition of the contributions from the two loops, namely µ0 NIa2 µ0 NIa2 + ẑ 2(a2 + z2 )3/2 2[a2 + (d − z)2 ]3/2 [ ] 1 µ0 NIa2 1 = ẑ + 2 (a2 + z2 )3/2 [a2 + (d − z)2 ]3/2

T a his th nd wo o eir is rk w r sa co pro is ill le u vi pr de o rse de ot st f a s d s ec ro n an o te y y p d le d th a a ly by e rt ss fo U in o e r te f t ss th nite gr hi in e ity s w g us d S of or stu e o tat th k ( de f i es e in nt ns co w cl le tr p or ud a uc y r k an ing rnin tors igh d on g. in t la is w D no the iss tea s t p W em ch er or in ing m ld a itt W tio ed id n . e W eb )

BP = ẑ

where we have taken note of the fact that both of the loops have N turns. (b) We have

dBz dz

{ } 3 2 3 2 2 −5/2 2 −5/2 − (a + z ) (2z) + [a + (d − z) ] [2(d − z)] 2 2 z=d/2 z=d/2 ( ) 2 −3d 3d µ0 NIa = + =0 2 2 2[a2 + (d/2)2 ]5/2 µ0 NIa2 = 2

(c) For d = a we have

] [ µ0 NIa2 1 1 BP = ẑ + 2 (a2 + z2 )3/2 [a2 + (a − z)2 ]3/2 and ] dBz µ0 NIa2 [ = −3z(a2 + z2 )−5/2 + 3(a − z)[a2 + (a − z)2 ]−5/2 dz 2 so that d2 Bz µ0 NIa2 = dz2 2

{

− 3a(a2 + z2 )−5/2 + 15z2 (a2 + z2 )−7/2 − 3[a2 + (a − z)2 ]−5/2 2

2

2 −7/2

}

+ 15(a − z) [a + (a − z) ] at z = a/2 we have

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12

Chapter 6/ The Static Magnetic Field

[ ( )−5/2 ( )−7/2 ( 2 )−5/2 ( )−7/2 ] d2 Bz µ0 NIa2 5a2 15a2 5a2 5a 15a2 5a2 = −3 + −3 + dz2 2 4 4 4 4 4 4 ] [ ( )−5/2 ( 2 )−5/2 ( 2 )−5/2 µ0 NIa2 5a2 5a 5a =0 = −6 +3 +3 2 4 4 4 Following the same procedure as given above, and with some patience, it is a rather straightforward and useful exercise for the student to differentiate the d2 Bz /dz2 expression given above and show that d3 Bz /dz3 = 0 for z = a/2.

The natural question to ask at this point is whether the fourth derivative is also zero. It can be shown by expanding the expression Bz (z) around z = a/2 that given by −144(z − a/2)4 125a4 so that d4 Bz /dz4 ̸= 0. The fact that the first three derivatives are zero provides for an extremely smooth and uniform magnetic field, which is the basic motivation for using the Helmholtz coil configuration. (d) We have

T a his th nd wo o eir is rk w r sa co pro is ill le u vi pr de o rse de ot st f a s d s ec ro n an o te y y p d le d th a a ly by e rt ss fo U in o e r te f t ss th nite gr hi in e ity s w g us d S of or stu e o tat th k ( de f i es e in nt ns co w cl le tr p or ud a uc y r k an ing rnin tors igh d on g. in t la is w D no the iss tea s t p W em ch er or in ing m ld a itt W tio ed id n . e W eb )

Bz (z) ≃

] [ µ0 NIa2 1 1 Bz = + 2 (a2 + z2 )3/2 [a2 + (d − z)2 ]3/2 For d = a, the field at the midpoint (z = a/2) is [( )−3/2 ( 2 )−3/2 ] µ0 NIa2 5a2 5a Bz = + 2 4 d ( )−3/2 ( 2 )−3/2 5 NI NI 2 5a = µ0 ≃ 0.8992 × 10−6 = µ0 NIa d d a a (e) At the center of the coil on the left (z = 0) we have ] µ µ0 NIa2 [ −3 NI NI 0 −3/2 −3 Bz = a +2 a = (1 + 2−3/2 ) ≃ 0.8505 × 10−6 2 2 a a which is quite close to the approximate expression given for the mid-point, underscoring the purpose of the Helmholtz coil configuration, namely to establish a B-field which is very nearly uniform.

Note that by symmetry, the B-field at the center of the coil on the right (z = a) must be the same.

6.18 Helmholtz coils.

Using the midway-point Bz expression provided in Problem 6.17 with Bz = 100 µT, I = 1 A, and d = a = 0.5 m, we have Bz ≃ 0.8992 × 10−6

N(1) = 10−4 0.5

from which we solve N ≃ 56 turns.

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13

Figure 6.7 Figure for Problem 6.19.

6.19 Helmholtz coils.

Using the result of Example 6.6, we simply switch the sign of one of the magnetic fields to find: T a his th nd wo o eir is rk w r sa co pro is ill le u vi pr de o rse de ot st f a s d s ec ro n an o te y y p d le d th a a ly by e rt ss fo U in o e r te f t ss th nite gr hi in e ity s w g us d S of or stu e o tat th k ( de f i es e in nt ns co w cl le tr p or ud a uc y r k an ing rnin tors igh d on g. in t la is w D no the iss tea s t p W em ch er or in ing m ld a itt W tio ed id n . e W eb )

] [ µ0 NIa2 1 1 Bz = − 2 (a2 + z2 )3/2 [a2 + (a − z)2 ]3/2 } { µ0 NIa2 1 1 1 = − 2 a3 [1 + (z/a)2 ]3/2 [1 + (1 − z/a)2 ]3/2 which is plotted as a function of z/a in Figure 6.7b. For comparison, Figure 6.7a shows bz (z) versus z for the case when the currents in the coils flow in the direction shown in Figure 6.59. We see that Bz is quite uniform for the latter case but that this uniformity is completely destroyed when one of the currents is reserved.

6.20 Square Helmholtz coils.

The configuration of the square Helmholtz coils is shown in Figure 6.8a. To determine the B-field at any point P along the axis, we rely on the result found in Example 6.3 for the B field at a distance r from the center of a current carrying wire of length a, namely B = ϕ̂ϕ

µ I(a/2) √0 2πr r2 + (a/2)2

For the case in hand, we have r = R, and by symmetry, it is clear that only the projection of the field along the z axis (i.e., its z component) is nonzero. Thus, the z component of the B field produced by one of the lower square coil is µ I(a/2) √0 sin θ 2πR R2 + (a/2)2 µ0 I(a/2) a/2 √ √ = √ 2 2 2 2 2 2 2π z + (a/2) z + (a/2) + (a/2) z + (a/2)2

B1z = |B1 | sin θ =

µ0 Ia2 √ =√ 2π(4z2 + a2 ) 2z2 + a2

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Chapter 6/ The Static Magnetic Field

T a his th nd wo o eir is rk w r sa co pro is ill le u vi pr de o rse de ot st f a s d s ec ro n an o te y y p d le d th a a ly by e rt ss fo U in o e r te f t ss th nite gr hi in e ity s w g us d S of or stu e o tat th k ( de f i es e in nt ns co w cl le tr p or ud a uc y r k an ing rnin tors igh d on g. in t la is w D no the iss tea s t p W em ch er or in ing m ld a itt W tio ed id n . e W eb )

14

Figure 6.8 Figure for Problem 6.20. (a) Coil configuration and calculation of the B-field. (b) Normalized Bz (z) versus z for optimum spacing (d = 0.5445a). (c) Expanded view of the region near the maximum for d = 0.5a, 0.5445a, and 0.6a.

The total due to the lower loop is 4B1z and that due to the upper loop is identical except for the fact that the center of the loop is at distance (from P) of (d − z), rather than z. Thus, the total B field at point P due to both of the square loops (both assumed to be carrying currents I, flowing in the same direction so that they produce fields in the same direction at point P) is then 4µ0 NIa2 4µ0 NIa2 √ √ +√ B(z) = √ 2π(4z2 + a2 ) 2z2 + a2 2π[4(z − d2 ) + a2 ] 2(z − d)2 + a2 where we have taken into account the fact that both of the loops have N turns. Figure 6.8b shows a plot of the normalized B field (i.e., B(z)/Bmax ) as a function of z for d = 0.5445a. To demonstrate that this is indeed the optimum spacing for which the field variation is the smoothest (i.e., d2 B(z)/dz2 = 0), an expanded view of the region near the maximum is shown in Figure 6.8c for d = 0.5a, 0.544a, and 0.6a. We see clearly that d = 0.5445a case is smoother than d = 0.5a and the distribution is already double-peaked for d = 0.6a, indicating that this separation is larger than optimum.

In comparing the results of this problem with those of Problem 6.17, it is useful to note that the optimum spacing of d = a for circular coils of radius a is only slightly different from the optimum spacing of d = 0.5445a for square coils of side-length a. Physically, square coils of side length a should be compared with circular coils of radius a/2, for which the optimum separation is d = 0.5a.

6.21 A circular and a square coil.

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T a his th nd wo o eir is rk w r sa co pro is ill le u vi pr de o rse de ot st f a s d s ec ro n an o te y y p d le d th a a ly by e rt ss fo U in o e r te f t ss th nite gr hi in e ity s w g us d S of or stu e o tat th k ( de f i es e in nt ns co w cl le tr p or ud a uc y r k an ing rnin tors igh d on g. in t la is w D no the iss tea s t p W em ch er or in ing m ld a itt W tio ed id n . e W eb )

15

Figure 6.9 Figure for Problem 6.21. (a) Coil configuration and calculation of the Bfield. (b) Normalized Bz (z) versus z for the case in hand, i.e., a square and a circular coil. (c) Normalized Bz (z) versus z for two circular coils.

The configuration of the square Helmholtz coils is shown in Figure 6.9a. To determine the B-field at any point P along the axis, we reply on the result found in Example 6.3 for the B field at a distance r from the center of a current carrying wire of length L, namely B = ϕ̂ϕ

µ IL √0 2πr r2 + L2

For the case in hand, we have r = R, and L = 2a, and by symmetry, it is clear that only the projection of the field along the z axis (i.e., its z component) is nonzero. Thus, the z component of the B field produced by one side of the square coil is µ I(2a) √0 sin θ 2πR R2 + (2a)2 2a/2 µ0 Ia √ √ = √ 2π z2 + (2a/2)2 z2 + (2a/2)2 + a2 z2 + (2a/2)2

B1z = |B1 | sin θ =

=

2µ0 I(2a)2 √ π[4z2 + (2a)2 ] 4z2 + 5(2a)2

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16

Chapter 6/ The Static Magnetic Field The total field on the square loop is Bsqr = 4B1z . To determine the contribution due the circular coil, we rely on the result of Example 6.6, specifying that the B-field at a point along the axis of a circular current carrying loop of radius a is B=

µ0 Ia2 ẑ 2(a2 + z2 )3/2

where z is the distance from the loop center. Since the distance between point P in Figure 6.9a and the center of the circular loop is (d − z), the B-field at point P due to the circular coil is Bcirc =

1 µ0 Ia2 2 2 [a + (d − z)2 ]3/2

The total B-field at point P due to both the square and circular coils is then B(z) =

2µ0 NI(2a)2 1 µ0 NIa2 √ + 2 + (d − z)2 ]3/2 2 2 2 2 2 [a π[4z + (2a) ] 4z + 5(2a) T a his th nd wo o eir is rk w r sa co pro is ill le u vi pr de o rse de ot st f a s d s ec ro n an o te y y p d le d th a a ly by e rt ss fo U in o e r te f t ss th nite gr hi in e ity s w g us d S of or stu e o tat th k ( de f i es e in nt ns co w cl le tr p or ud a uc y r k an ing rnin tors igh d on g. in t la is w D no the iss tea s t p W em ch er or in ing m ld a itt W tio ed id n . e W eb )

The normalized amplitude of the B-field is plotted versus z in Figure 6.9b for d = a. For comparison, a similar plot for two circular coils of radius a are shown in Figure 6.9c. We note that the field for the two circular coils is somewhat smoother near the origin (i.e., near z = 0 where the square coil is located).

6.22 Magnetic field of a solenoid.

This is a very long solenoid since l = 40 cm ≫ 2 cm = a.

(a) Using the results of Section 6.2.1 on page 436 of the text, with N = 350 and I = 1 A, the B field at the center of the solenoid can be approximately calculated as 4π × 10−7 × 350 × 1 µ0 NI = ẑ T ≃ ẑ11 G l 0.4 (b) Similarly, the B field at the ends of solenoid can be found as Bctr ≃ ẑ

1 Bend ≃ Bctr ≃ ẑ5.5 G 2 6.23 Magnetic field of a surface current distribution.

We can use the result found in Example 6.6, for the B-field at a point along the axis of a circular loop of current of radius a, namely B = ẑ

µ0 Ia2 2(a2 + z2 )3/2

The circular disk with a surface current distribution can be thought of as being constituted by differential rings of radius r, radial thickness dr, each carrying a differential current dI = |Js (r)|2πrdr and producing a differential B-field of dB = ẑ

µ0 dIr2 µ0 (Kr)(2πrdr)r2 µ0 Kπr4 dr = ẑ = ẑ 2(r2 + z2 )3/2 2(r2 + z2 )3/2 (r2 + z2 )3/2

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17

Figure 6.10

Figure for Problem 6.24. Variation of |B| along the solenoid axis.

The total B-field can then simply be found by integrating the above between r = 0 and r = a.

We find B = ẑ 0

[ ] √ µ0 Kπr4 dr a3 + 3az2 3z2 ln(a + a2 + z2 ) 2 = ẑ √ + 3z ln(z) − 2 (r2 + z2 )3/2 2 a2 + z2 T a his th nd wo o eir is rk w r sa co pro is ill le u vi pr de o rse de ot st f a s d s ec ro n an o te y y p d le d th a a ly by e rt ss fo U in o e r te f t ss th nite gr hi in e ity s w g us d S of or stu e o tat th k ( de f i es e in nt ns co w cl le tr p or ud a uc y r k an ing rnin tors igh d on g. in t la is w D no the iss tea s t p W em ch er or in ing m ld a itt W tio ed id n . e W eb )

∫ a

where the integral can be evaluated either looking up Tables or using an analysis package such as MATHEMATICA. 6.24 B inside a solenoid.

(a) From Section 6.2.1 (page 434 √ the B field along the axis of a solenoid with N = 750, √ of the text), I = 1 A, l = 20 cm, and a = A/π = 5/π ≃ 1.26 cm is given by 

B(z) = ẑ



 z − 0.1 4π × 10−7 × 750 × 1  z + 0.1  T √ √ −   −4 2(0.2) 5 × 10−4 5 × 10 + (z + 0.1)2 + (z − 0.1)2 π π

(b) Since 20 cm ≫∼ 1.26 cm (i.e., very long solenoid), the B field at the center of the solenoid can be approximately found as (page 436 of the text) Bctr ≃ ẑ

µ0 NI 4π × 10−7 × 750 × 1 = ẑ T ≃ 47.1 G l 0.2

(c) Using the expression provided in part (a), we sketch |B| versus z as shown in Figure 6.10. The striking uniformity of the field within the solenoid is apparent, as discussed in Section 6.2.1 of the text.

6.25 B inside a solenoid. (a) From Section 6.2.1, the B field as the center of a long solenoid made of a magnetic core material with µr = 300 and having a current of I = 10 A can be written as (page 436 of the text)

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18

Chapter 6/ The Static Magnetic Field

µr µ0 NI (300)(4π × 10−7 )(10) = = 1.5 T l l from which we find the number of turns of wire needed to produce 1.5 T field at the center as N/l ≃ 398 turns/m or ∼ 4 turns/cm.

Bctr ≃

(b) Repeating the same calculation in part (a) for an air-core solenoid yields N/l ≃ 1194 turns/cm.

It is clear from these results that an air-core solenoid requires a much larger number of turns to produce the same field as produced by a solenoid of same geometry which is made of a magnetic core with a high µr .

6.26 Split washer.

Let the thickness of the washer be t. Consider a thin ring as shown at a radius r and having a width dr. Using the expression for the resistance of a block of length l and area A as R = l/(σA) or, equivalent, a conductance of G = σA/l, the conductance of the differential ring can be written as σtV0 dr σtdr → dI = V0 dG = 2πr 2πr where dI is the current flowing in this ring. The total current flowing in the washer is then ( ) ∫ b ∫ b σtV0 dr σtV0 b dI = Itotal = = ln = 100 A 2πr 2π a a a T a his th nd wo o eir is rk w r sa co pro is ill le u vi pr de o rse de ot st f a s d s ec ro n an o te y y p d le d th a a ly by e rt ss fo U in o e r te f t ss th nite gr hi in e ity s w g us d S of or stu e o tat th k ( de f i es e in nt ns co w cl le tr p or ud a uc y r k an ing rnin tors igh d on g. in t la is w D no the iss tea s t p W em ch er or in ing m ld a itt W tio ed id n . e W eb )

dG =

from which V0 can be determined. Using the result of Example 6.6, the B-field produced at the center of the washer, due to the current dI flowing in the ring is dB =

µ0 dI µ0 σtV0 dr = 2r 2r 2πr

so that the total B-field at the center is ∫ b B=

dB = a

µ0 σtV0 4π

∫ b a

dr µ0 σtV0 = r2 4π

(

1 1 − a b

)

=

µ0 (2π)Itotal ln(b/a)

(

1 1 − a b

)

Thus, we can write µ0 Itotal B= 2 ln(b/a)

(

1 1 − a b

)

(4π × 10−7 )(100) ẑ = ẑ 2 ln(50/10)

(

1 1 − 0.01 0.05

)

≃ ẑ3.12 × 10−3 T

6.27 Long wire encircled by a small loop.

Using superposition principle, the B field at a distance of 1 m from the long straight wire with I1 = 1 A current can be found using the result of Example 6.3 as µ0 I1 4π × 10−7 × 1 = ϕ̂ϕ = 2 × 10−7 T 2πr 2π × 1 To find the B field at the same point due to the current loop, a magnetic dipole approximation can be used since r = 100 cm ≫ 5 cm= a. From Section 6.6, the B field at a distance of 1 m from the center and on the plane of the 5-cm radius loop with I2 = 100 A is given by B1 = ϕ̂ϕ

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19

Figure 6.11

Figure for Problem 6.26.

µ0 I2 a2 4π × 10−7 (100)(0.05)2 = −ẑ ≃ −ẑ7.85 × 10−8 T 4r3 4 × 13 Therefore, the total B field due to both currents is T a his th nd wo o eir is rk w r sa co pro is ill le u vi pr de o rse de ot st f a s d s ec ro n an o te y y p d le d th a a ly by e rt ss fo U in o e r te f t ss th nite gr hi in e ity s w g us d S of or stu e o tat th k ( de f i es e in nt ns co w cl le tr p or ud a uc y r k an ing rnin tors igh d on g. in t la is w D no the iss tea s t p W em ch er or in ing m ld a itt W tio ed id n . e W eb )

B2 = −ẑ

B = B1 + B2 ≃ ϕ̂ϕ2 × 10−7 − ẑ7.85 × 10−8 T

6.28 Flux through a rectangular loop.

(a) The magnetic flux linking the rectangular loop due the very long straight current-carrying wire (see Example 6.3) can be found using (6.26) as ∫

Ψ1 =

S

∫ 0.5a ∫ 3a

µ0 I ϕ̂ϕ · ϕ̂ϕdrdz 2πr −0.5a a ∫ ∫ µ0 I 0.5a 3a 1 µ0 Ia = drdz = ln 3 2π −0.5a a r 2π

B · ds =

(b) Following a similar approach, the magnetic flux in this case can be evaluated as µ0 I Ψ2 = 2π

∫ a ∫ 2.5a −a

1.5a

1 µ0 Ia 5 drdz = ln r π 3

As a result, the percentage change in Ψ linking the rectangular loop from part (a) to part (b) can be calculated as ∆Ψ =

2 ln 53 − ln 3 Ψ2 − Ψ1 × 100 = × 100 ≃ −7% Ψ1 ln 3

6.29 Flux through a triangular loop. (a) We can choose a coordinate system as shown in Figure 6.12 and define the origin to be the point P, i.e., the lower left edge of the triangle. We then identify a differential strip as shown, at coordinate location y and having width dy in the y direction. The extent of this strip in the x direction is clearly given by (b−a−y) tan ϕ0 , which in turn means that its differential area is dA = (b−a−y) tan ϕ0 dy.

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20

Chapter 6/ The Static Magnetic Field

Figure 6.12

Figure for Problem 6.29.

In view of the result of Example 6.3, the B-field produced by the long wire is the same all along this strip and is given by µ0 I 2π(a + y)

T a his th nd wo o eir is rk w r sa co pro is ill le u vi pr de o rse de ot st f a s d s ec ro n an o te y y p d le d th a a ly by e rt ss fo U in o e r te f t ss th nite gr hi in e ity s w g us d S of or stu e o tat th k ( de f i es e in nt ns co w cl le tr p or ud a uc y r k an ing rnin tors igh d on g. in t la is w D no the iss tea s t p W em ch er or in ing m ld a itt W tio ed id n . e W eb )

B = ẑ

since r = a + y is the radial distance away from the wire. The flux linked by the triangular loop can be found by integrating the |B|dA as follows: ∫ b−a

µ0 I (b − a − y) tan ϕ0 dy 2π(a + y) 0 ∫ µ0 I tan ϕ0 b−a b − (a + y) = dy Let u = (a + y), du = dy 2π a+y 0 ) [ ( ) ] ∫ ( µ0 I tan ϕ0 b b µ0 I tan ϕ0 b = − 1 du = b ln − (b − a) 2π u 2π a a

Ψ=

(b) Substituting the given values we find Ψ ≃ 5.09 × 10−6 Wb. 6.30 A toroidal coil around a long straight wire.

The total magnetic flux Ψ linking the N = 200-turn rectangular toroid with radii a = 4 cm and b = 6 cm, thickness t = 3 cm, and relative permeability µr = 250 due to the long, straight wire with I = 100 A along its axis can be calculated using (6.26) as ∫ Ψ=N S

B · ds = N

∫ 0.5t ∫ b −0.5t

a

µr µ0 I ϕ̂ϕ · ϕ̂ϕdrdz 2πr

µr µ0 NIt b = ln 2π a 250 × 4π × 10−7 × 200 × 100 × 0.03 6 = ln ≃ 1.22 × 10−2 Wb 2π 4 6.31 Vector potential.

We begin with the solution to B found in Example 6.10:

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21  µ Ir 0  , r≤a  ϕ̂ϕ 2πa2 B=   ϕ̂ϕ µ0 I , r>a 2πr We need to find a vector potential A such that B = ∇ × A. Since B only contains a ϕ component, in cylindrical coordinates this relation reduces to ∂Ar ∂Az − ∂z ∂r Furthermore, in this case, B does not have a z dependence, so Bϕ =

Bϕ = −

∂Az ∂r

Hence we can write T a his th nd wo o eir is rk w r sa co pro is ill le u vi pr de o rse de ot st f a s d s ec ro n an o te y y p d le d th a a ly by e rt ss fo U in o e r te f t ss th nite gr hi in e ity s w g us d S of or stu e o tat th k ( de f i es e in nt ns co w cl le tr p or ud a uc y r k an ing rnin tors igh d on g. in t la is w D no the iss tea s t p W em ch er or in ing m ld a itt W tio ed id n . e W eb )

 µ Ir  0 , r≤a ∂Az  2πa2 − =  ∂r  µ0 I , r>a 2πr Simple integration, with assuming the constant of integration to be zero for simplicity, yields  2   −µ0 Ir ,  r≤a 4πa2 Az =    −µ0 I ln(r), ,r > a 2π that is,  −µ0 Ir2    ẑ, r≤a 4πa2 A=    −µ0 I ln(r)ẑ, r>a 2π 6.32 Infinitely long wire with a cylindrical hole. (a) We first assume that the cylindrical conductor of radius b carrying a current of density J1 = ẑJ0 = ẑ

I π(b2 − a2 )

does not have a hole and find its associated magnetic vector potential A1 . We then account for the hole by assuming a current of density J2 = −ẑJ0 which only exits in the hole region (to cancel J1 ) and find its associated vector potential A2 and apply superposition principle to find the net total A. To find A1 in the region r < b, we apply Ampère's law (6.9) as was done in Example 6.13 along with Figure 6.18 on page 443 in the text: I ∫ B1 · dl = B1ϕ (2πr1 ) = µ0 J1 · ds = µ0 J0 (πr21 ) C

S

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22

Chapter 6/ The Static Magnetic Field from which we find µ0 J0 r1 2 Note that based on (6.14) B1 = ∇ × A1 and that A1 is only in the z direction, i.e., A1 = ẑA1z .

In addition, note also that A1z does not vary with z (since the conductor is infinitely long) and ϕ (because of azimuthal symmetry). As a result, we can write B1ϕ =

B1 = ∇ × A1 = ∇ × [ẑA1z (r1 )]

→

−

µ0 J0 r1 ∂A1z = B1ϕ = ∂r1 2

or µ0 J0 r21 4 Following a similar approach for the current in the hole with density J2 = −ẑJ0 yields A1z = −

µ0 J0 r22 4

T a his th nd wo o eir is rk w r sa co pro is ill le u vi pr de o rse de ot st f a s d s ec ro n an o te y y p d le d th a a ly by e rt ss fo U in o e r te f t ss th nite gr hi in e ity s w g us d S of or stu e o tat th k ( de f i es e in nt ns co w cl le tr p or ud a uc y r k an ing rnin tors igh d on g. in t la is w D no the iss tea s t p W em ch er or in ing m ld a itt W tio ed id n . e W eb )

A2 = ẑA2z = ẑ

Therefore, we can write the total A vector as

A = A1 + A2 = ẑ

µ0 J0 2 (r2 − r21 ) 4

(b) We convert the expression for A into rectangular coordinates by making the substitutions r21 = x2 + y2 and r22 = (d − x)2 + y2 yielding µ0 J0 2 µ0 J0 d (d − 2dx) = ẑ (d − 2x) 4 4 Using B = ∇ × A in rectangular coordinates, we find A = ẑ

B = ∇ × A = ∇ × [ẑAz (x)] = −ŷ

µ0 J0 d µ0 Id ∂Az (x) = ŷ = ŷ ∂x 2 2π(b2 − a2 )

which is the expression for the B field found in Example 6.13, as expected.

6.33 Finite-length straight wire: A at an off-axis point. (a) Using Figure 6.11, the magnetic vector potential of A at an arbitrary point P(r, ϕ, z) due to the straight, current-carrying filamentary conductor is given by (6.20) as ∫ µ0 I a dz′ √ A = ẑ 4π z′ =−a r2 + (z − z′ )2 [ ] √ −(z − a) + r2 + (z − a)2 µ0 I √ = ẑ ln 4π −(z + a) + r2 + (z + a)2 (b) The B field can be found from (6.14) as B = ∇ × A = ∇ × [ẑAz (r, z)] = −ϕ̂ϕ

∂Az ∂r

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23 Performing the partial derivative yields [ ] [r2 + (z − a)2 ]−1/2 µ0 Ir [r2 + (z + a)2 ]−1/2 √ √ B = −ϕ̂ϕ − 4π [−(z − a) + r2 + (z − a)2 ] [−(z + a) + r2 + (z + a)2 ] which, after some rigorous algebraic manipulations1 results in the B field expression found in Example 6.7 given by ] [ µ0 I z+a z−a √ B = ϕ̂ϕ −√ 4πr r2 + (z + a)2 r2 + (z − a)2 6.34 Two infinitely long wires.

T a his th nd wo o eir is rk w r sa co pro is ill le u vi pr de o rse de ot st f a s d s ec ro n an o te y y p d le d th a a ly by e rt ss fo U in o e r te f t ss th nite gr hi in e ity s w g us d S of or stu e o tat th k ( de f i es e in nt ns co w cl le tr p or ud a uc y r k an ing rnin tors igh d on g. in t la is w D no the iss tea s t p W em ch er or in ing m ld a itt W tio ed id n . e W eb )

Let us first with two equal length parallel wires (with currents I and −I) oriented in the z direction and extending from z = −d to z = +d, with their symmetry axis being the z axis. We choose the observation point on the xy plane since this choice maintains its generality in the case when the wires are infinitely long, i.e., when we let d → ∞. Using the result of Example 6.22 along with the superposition principle, the magnetic vector potential at point P(x, y, 0) due to both finite wires can be written as [ ( ) ( )] √ √ µ0 I d d2 d d2 A = ẑ + + ln + 1 − ln +1 2π a a2 b b2 [ ] √ b d + d2 + a2 µ0 I √ ln = ẑ 2π a d + d2 + b2 Next, letting d → ∞, and noting that

√ d2 + a2 √ lim =1 d→∞ d + d2 + b2 the magnetic vector potential at point P due to two infinitely long parallel wires can be found as d+

A = ẑ

µ0 I b ln 2π a

6.35 Wire with oval-shaped hole.

We can use superposition and represent the hole as being produced by the cancellation of the two currents in opposite directions. Applying Ampère's law given by (6.9), the B-field at point P1 is given by BP1 = B1 + B2 = 2B1 =

2µ0 J0 π(d/2)2 µ0 J0 d ẑ = ẑ 2π(d/2) 2

The field at P2 can be calculated with reference to Figure 6.13 and by noting that the horizontal components of the two contributions cancel. We have 1 See S.V.Marchall, R.E.DuBroof, and G.G.Skitek, Electromagnetic Concepts and Applications, 4th ed., p.207, Prentice Hall, 1996.

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24

Chapter 6/ The Static Magnetic Field

Figure 6.13

Figure for Problem 6.35.

BP2 = (|B1 | + |B2 |) cos θẑ where T a his th nd wo o eir is rk w r sa co pro is ill le u vi pr de o rse de ot st f a s d s ec ro n an o te y y p d le d th a a ly by e rt ss fo U in o e r te f t ss th nite gr hi in e ity s w g us d S of or stu e o tat th k ( de f i es e in nt ns co w cl le tr p or ud a uc y r k an ing rnin tors igh d on g. in t la is w D no the iss tea s t p W em ch er or in ing m ld a itt W tio ed id n . e W eb )

√ µ0 J0 r µ0 J0 (d/2)2 + (R/2)2 µ0 J0 √ 2 |B1 | = |B2 | = = d + R2 2 2 4 √ where we have noted that r = (d/2)2 + (R/2)2 . Thus we have ( ) ] [ µ0 J0 √ 2 1 d/2 2 √ BP2 = 2 ẑ = µ0 J0 dẑ d +R 2 2 4 2 (1/2) d + R | {z } cos θ

6.36 Inductance of a short solenoid.

For l = 10 cm and 2a = 5.5 cm, we have 2a/l = 0.55. From Figure 6.41b of the text, the Nagaoka correction factor kN corresponding to 2a/l = 0.55 can be read as kN ≃ 0.8. Using these values along with N = 500, the self-inductance of the short solenoid can be calculated from the Nagaoka formula given in Section 6.7.3 as L = kN

µ 0 N2 A 4π × 10−7 (500)2 [π(2.75 × 10−2 )2 ] ≃ (0.8) H ≃ 7.46 mH l 0.1

6.37 Inductance of a finite-length solenoid. (a) From Example 6.26, the self-inductance of a long air-core solenoid with N = 780 turns, l = 1.8 m and a = 17.7 cm can be calculated as µ0 N2 A 4π × 10−7 (780)2 π(17.7 × 10−2 )2 = H ≃ 41.8 mH l 1.8 (b) Since l > 0.8a, the self-inductance of this solenoid can be calculated more accurately by using the approximate formula provided in Section 6.7.3 as L≃

L≃

10µ0 πa2 N2 10(4π × 10−7 )π(17.7 × 10−2 )2 (780)2 = H ≃ 38.4mH 9a + 10l 9(17.7 × 10−2 ) + 10(1.8)

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25 One can also use the Nagaoka formula (Section 6.7.3) with kN ≃ 9 (which is read from Figure 6.41b corresponding to 2a/l ≃ 0.2) as L = kN

4π × 10−7 (780)2 π(17.7 × 10−2 )2 µ0 n2 A ≃ (0.9) H ≃ 37.6 mH l 1.8

As expected, these results are very close to each other. They are also very close to the selfinductance found in part (a) since l ≫ a (i.e., a long solenoid approximation can be used).

6.38 Inductance of a rectangular toroid. (a) The inductance of the rectangular air-core toroid can be calculated from (Section 6.7.2) Λ N L= = I I

∫

N B · ds = I S

∫ 0.5t ∫ b −0.5t

a

B · ϕ̂ϕ drdz

T a his th nd wo o eir is rk w r sa co pro is ill le u vi pr de o rse de ot st f a s d s ec ro n an o te y y p d le d th a a ly by e rt ss fo U in o e r te f t ss th nite gr hi in e ity s w g us d S of or stu e o tat th k ( de f i es e in nt ns co w cl le tr p or ud a uc y r k an ing rnin tors igh d on g. in t la is w D no the iss tea s t p W em ch er or in ing m ld a itt W tio ed id n . e W eb )

where the B field inside an air-core toroidal coil was found in Example 6.15 as B = ϕ̂ϕ

µ0 NI 2πr

Substituting the B field expression into the inductance equation above yields L=

µ0 N 2π

∫ 0.5t

∫ b

dz

−0.5t

a

b dr = 2 × 10−7 N2 t ln H r a

where t is in m or we can write

L = 2 × 10−9 N2 t ln

where t is in cm.

b H a

(b) For a = 1.2 cm, b = 2 cm, t = 1.5 cm, and N = 1000, the inductance of the rectangular air-core toroid is L = 2 × 10−9 (1000)2 (1.5) ln

2 ≃ 1.53 × 10−3 H = 1.53 mH 1.2

(c) Substituting the value in the inductance expression found in Example 6.27 for the case when the mean radius of the toroid satisfies the condition rm = (a + b)/2 ≫ b − a yields

L=

µ0 N2 (b − a)t 4π × 10−7 (1000)2 (0.02 − 0.012)(0.015) µ0 N2 A ( a+b ) = = 2πtm 2π(0.016) 2π 2

= 1.5 × 10−3 H = 1.5 mH Note that this value is only 2% lower than the value calculated in part (b).

6.39 Inductance of a rectangular toroid.

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26

Chapter 6/ The Static Magnetic Field (a) Using the inductance expression of a rectangular toroid given in Problem 6.38 with a = 3.8 mm, b = 6.4 mm, t = 4.8 mm, and L = 1 mH, we have 6.4 = 10−3 3.8 from which we solve for the number of turns needed for the air-core toroid as N ≃ 141 turns.

L = 2 × 10−9 (0.48)N2 ln

(b) Repeating the same calculation as in part (a) for a powdered nickel-iron with µr = 200 as L = 2 × 10−9 (200)(0.48)N2 ln

6.4 = 10−3 3.8

yields N ≃ 100 turns.

6.40 Inductance of a rectangular toroid.

Using the inductance expression of a rectangular toroid provided in Problem 6.38 with a = 7.37 mm, b = 13.5 mm, t = 11.2 mm, µr = 125, and L = 50 mH, we can write 13.5 = 0.05 7.37

T a his th nd wo o eir is rk w r sa co pro is ill le u vi pr de o rse de ot st f a s d s ec ro n an o te y y p d le d th a a ly by e rt ss fo U in o e r te f t ss th nite gr hi in e ity s w g us d S of or stu e o tat th k ( de f i es e in nt ns co w cl le tr p or ud a uc y r k an ing rnin tors igh d on g. in t la is w D no the iss tea s t p W em ch er or in ing m ld a itt W tio ed id n . e W eb )

L = 2 × 10−9 (125)(0.112)N2 ln

from which we find N ≃ 1718 turns.

6.41 Inductance of a circular toroid.

The magnetic field inside a toroid was found in Example 6.15 to be given by µNI 2πr Using B given above along with the definitions shown in Figure 6.14, the inductance of a circular toroid can be written, based on Section 6.7.2, as ∫ ∫ Λ N N µNI L= = ϕ̂ϕ · ϕ̂ϕdsϕ B · ds = I I S I S 2πr ∫ ∫ µN2 2π a R dRdθ = 2π 0 0 rm − R cos θ B = ϕ̂ϕ

where rm is the mean radius of the toroid. From integral tables2 , we have ∫ π du π =√ 1 − k2 0 1 ± k cos u Using this integral in the inductance expression obtained above, we find µN2 L= 2π

∫ a[ 0

= µr µ0 N2

[

√

2π

]

] [ √ a 2 2 R dR = µr µ0 N − rm − R 2

r2m − R2 ] √ rm − r2m − a2

0

where µr is the relative permeability of the core material used.

2

H.B.Dwight, Tables of Integrals and Other Mathematical Data, 4th ed., p.218, Macmillan, 1961.

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27

Figure for Problem 6.41. (a) Top view of the toroid. (b) Cross-sectional

T a his th nd wo o eir is rk w r sa co pro is ill le u vi pr de o rse de ot st f a s d s ec ro n an o te y y p d le d th a a ly by e rt ss fo U in o e r te f t ss th nite gr hi in e ity s w g us d S of or stu e o tat th k ( de f i es e in nt ns co w cl le tr p or ud a uc y r k an ing rnin tors igh d on g. in t la is w D no the iss tea s t p W em ch er or in ing m ld a itt W tio ed id n . e W eb )

Figure 6.14 view.

6.42 Inductance of a two-wire line.

(a) Using the result of Example 6.29 with a = 1 mm and d = 6 cm (i.e., d ≫ a), the per-unit-length inductance of the two wire line can be found as Lu ≃ 0.4 ln

d 6 = 0.4 ln ≃ 1.64 µH-m−1 a 0.1

(b) Repeating the calculation in part (a) with d = 12 cm yields Lu = 0.4 ln

12 ≃ 1.91 µH-m−1 0.1

As observed, increasing the conductor spacing of the-wire line increases its inductance per unit length. Do you recall how the increase in conductor spacing affects the capacitance per unit length of the two-wire line?

6.43 Inductance of a thin circular loop of wire. (a) Since d = 0.2 cm ≪ 5 cm = a, a thin wire approximation can be used. From Section 6.7.3, the inductance of the single loop can be calculated as ( ) ( ) 8a 8×5 −7 − 2 = 4π × 10 (0.05) ln − 2 H ≃ 207 nH L ≃ µ0 a ln d 0.2 (b) For d = 0.05 cm ≪ 10 cm = a, we have L ≃ 4π × 10

−7

( ) 8 × 10 (0.1) ln − 2 H ≃ 676 nH 0.05

6.44 Mutual inductance between two solenoidal coils.

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28

Chapter 6/ The Static Magnetic Field The mutual inductance L12 between two air-core coils as described in the problem can be written based on Section 6.7.2 as ∫ Λ12 N2 B1 · ds2 L12 = = I1 I1 S2 where Λ12 is the total flux linkage through the area S2 of coil 2 (with l2 = 3 cm, 2a2 = 2.5 cm, and N2 = 100) due to the B1 field produced by the current I1 in coil 1 (with l1 = 25 cm, 2a1 = 2.5 cm, and N1 = 1000). Note that the field B1 can be assumed uniform over the length of the shorter coil (coil 2) approximately given (from Section 6.2.1) by B1 ≃ B1ctr ≃ ẑ

µ0 N1 I1 l1

T a his th nd wo o eir is rk w r sa co pro is ill le u vi pr de o rse de ot st f a s d s ec ro n an o te y y p d le d th a a ly by e rt ss fo U in o e r te f t ss th nite gr hi in e ity s w g us d S of or stu e o tat th k ( de f i es e in nt ns co w cl le tr p or ud a uc y r k an ing rnin tors igh d on g. in t la is w D no the iss tea s t p W em ch er or in ing m ld a itt W tio ed id n . e W eb )

since coil 2 is located at the center of the longer coil with l1 ≫ a1 . Therefore, substituting there values into the mutual inductance expression with ds2 = ẑds2 yields ∫ µ0 N1 I1 N2 µ0 N1 N2 (πa22 ) L12 ≃ ds2 = l1 I1 l 1 S2 =

4π × 10−7 (1000)(100) [π(0.0125)2 ] H ≃ 0.247 mH 0.25

6.45 Mutual inductance between a wire and a circular loop.

Assuming a current I to flow in the wire, the B field at a point P(r, θ) as shown in Figure 6.15 is given by BP = ϕ̂ϕ

µ0 I 2π(d + r cos ϕ)

The flux linked by the circular loop can be found by integration over the area of the loop which we sweep if we vary ϕ between 0 and 2π and r between 0 and a. We thus have µ0 I Ψ12 = 2π

∫ a ∫ 2π 0

0

µ0 I rdrdϕ = d + r cos ϕ 2π

∫ a 0

√ 2πrdr √ = µ0 I(d − d2 − a2 ) d2 − r2

so that the mutual inductance is given by L12 = µ0 (d −

√

d2 − a2 )

6.46 Long wire and loop. (a) Since d ≫ a, we can use the dipole approximation for the loop. We have µ0 I µ0 Ia2 BP1 = (−ẑ) = (µ0 I) (−ẑ) + 3 |2πd{z } | 4d {z } long wire

(

1 a2 + 3 2πd 4d

) (−ẑ)

loop

(b) In this case, we can either use the exact formula (Example 6.6) or the dipole approximation for the loop. We have

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29

Figure 6.15

BP2 =

Figure for Problem 6.45.

µ0 I µ0 Ia2 (ŷ) + (ẑ) 2πd 2(a2 + d2 )3/2

Exact

T a his th nd wo o eir is rk w r sa co pro is ill le u vi pr de o rse de ot st f a s d s ec ro n an o te y y p d le d th a a ly by e rt ss fo U in o e r te f t ss th nite gr hi in e ity s w g us d S of or stu e o tat th k ( de f i es e in nt ns co w cl le tr p or ud a uc y r k an ing rnin tors igh d on g. in t la is w D no the iss tea s t p W em ch er or in ing m ld a itt W tio ed id n . e W eb )

µ0 I µ0 Ia2 Dipole approximation (ŷ) + (ẑ) 2πd 2d3 (c) The mutual inductance is zero since the B-field produced by the long wire which links the loop is in opposite directions on both sides of the x axis, thus canceling one another.

BP2 ≃

6.47 Inductance and energy of a solenoid.

(a) Since l = 30 cm ≫ 1 cm = 2a, a very long solenoid assumption can be made. Using the expression found in Example 6.26 with N = 1000, l = 30 cm and a = 0.5 cm, the inductance of the air-core solenoid can be found as µ0 N2 A 4π × 10−7 (1000)2 π(5 × 10−3 )2 = H ≃ 0.329 mH l 0.3 (b) For I = 10 A, the energy stored in the solenoid can be evaluated using our knowledge of electric circuits as L=

1 1 Wm = LI2 ≃ (0.329 × 10−3 )(10)2 J ≃ 16.5 mJ 2 2 6.48 Two square coils. (a) For d ≫ a, we can treat the square loop as a magnetic dipole with |m| = Ia2 . From equation (6.22) we have B = r̂

µ0 |m| cos θ µ0 |m| sin θ + θ̂θ 2πr3 4πr3

For all points on the xy plane, we have θ = π/2, so that µ0 Ia2 µ0 Ia2 = −ẑ 4πr3 4πr3 which only depends on radial distance r. The flux produced by coil 1 and linked by coil 2 is then B = θ̂θ

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30

Chapter 6/ The Static Magnetic Field

−L/2

T a his th nd wo o eir is rk w r sa co pro is ill le u vi pr de o rse de ot st f a s d s ec ro n an o te y y p d le d th a a ly by e rt ss fo U in o e r te f t ss th nite gr hi in e ity s w g us d S of or stu e o tat th k ( de f i es e in nt ns co w cl le tr p or ud a uc y r k an ing rnin tors igh d on g. in t la is w D no the iss tea s t p W em ch er or in ing m ld a itt W tio ed id n . e W eb )

L/2

Figure 6.16

Figure for Problem 6.48.

∫ d+a/2

∫ µ0 Ia3 d+a/2 1 Ψ12 = B · (ẑadr) = dr 4π d−a/2 r3 d−a/2 [ ]d+a/2 [ ] µ0 Ia3 1 µ0 Ia3 −r−2 1 = = − 4π 2 d−a/2 8π (d − a/2)2 (d + a/2)2 and mutual inductance is given by [ ] Ψ12 µ0 Ia3 1 1 L12 = = − I 8π (d − a/2)2 (d + a/2)2 (b) We now have d = 4a, which means that we cannot assume d ≫ a. What we have to do in this case is to use the exact expression for the B-field produced by a fixed length wire of length L at a general point P(r, ϕ, z) at a distance r from the center of the wire and at a distance z from its axis, as shown in Figure 6.16b. The general expression for the B-field at such a point P was found in Example 6.7 as ] [ µ0 I z + L/2 z − L/2 √ BP = ϕ̂ϕ −√ 4πr r2 + (z + L/2)2 r2 + (z − L/2)2 We now separately evaluate the magnetic flux produced by each side of coil 1 and linking coil 2. We start with side A, selecting our origin (z = 0) to be at its center as shown in Figure 6.16d.

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31 Noting that the wire length is L = a, we have [ ] µ0 I z + a/2 z − a/2 √ ΨA = −√ dzdr r2 + (z + a/2)2 r2 + (z − a/2)2 0 3.5a 4πr [ ] ∫ a ∫ µ0 I 4.5a z + a/2 z − a/2 √ = −√ dzdr r2 + (z + a/2)2 r2 + (z − a/2)2 0 4πr 3.5a ] [∫ ∫ ∫ 4.5a 4.5a µ0 I a 1 z + a/2 z − a/2 √ √ = dz − dz dr 4π 0 r 3.5a r2 + (z + a/2)2 r2 + (z − a/2)2 3.5a ∫ a ∫ 4.5a

To evaluate the first integral we let u =

so that

r2 + (z + a/2)2 in which case we have

1 1 z + a/2 √ 2(z + a/2)dz = √ dz 2 r2 + (z + a/2)2 r2 + (z + a/2)2 T a his th nd wo o eir is rk w r sa co pro is ill le u vi pr de o rse de ot st f a s d s ec ro n an o te y y p d le d th a a ly by e rt ss fo U in o e r te f t ss th nite gr hi in e ity s w g us d S of or stu e o tat th k ( de f i es e in nt ns co w cl le tr p or ud a uc y r k an ing rnin tors igh d on g. in t la is w D no the iss tea s t p W em ch er or in ing m ld a itt W tio ed id n . e W eb )

du =

√

∫ √r2 +16a2 √

Similarly, we have

∫ 4.5a 3.5a

Thus,

√

du =

√

r2 +9a2

z − a/2

r2 + (z − a/2)2

r2 + 16a2 −

dz =

√

√

r2 + 9a2

r2 + 25a2 −

√

r2 + 16a2

∫ ) √ √ √ µ0 I a 1 (√ 2 ΨA = r + 16a2 − r2 + 9a2 − r2 + 25a2 + r2 + 16a2 dr 4π 0 r ∫ ) √ √ µ0 I a 1 ( √ 2 = 2 r + 16a2 − r2 + 9a2 − r2 + 25a2 dr 4π 0 r Looking up a Table of Integrals we find ∫ √2 √ √ r + ζ2 dr = ζ 2 + r2 + ζ ln r − ζ ln[ζ 2 + ζ ζ 2 + r2 ] r Thus we have ∫ a√2 {√ }a √ r + 16a2 dr = 16a2 + r2 + 4a ln r − 4a ln[16a2 + 4a 16a2 + r2 ] r 0 0 √ √ √ 2 2 = a( 17 − 16) + 4a ln a − 4a ln(0)−4a ln[16a + 4a 17] √ + 4a ln[16a2 + 4a2 16]

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32

Chapter 6/ The Static Magnetic Field Note that the ln(0) term does not cause a problem since this term cancels out when we integrate the other two terms in the expression for ΨA . Integrating the other two terms we find

] √ µ0 I { [ √ 2 a 17 − 4a + 4a ln a − 4a ln 0 − 4a ln[16a2 + a2 17] + 4a ln(32a2 ) 4π [ √ ] √ − a 10 − 3a + 3a ln a − 3a ln 0 − 3a ln[9a2 + 3a2 10] + 3a ln(18a2 ) ]} [ √ √ − a 26 − 5a + 5a ln a − 5a ln 0 − 5a ln[25a2 + 5a2 26] + 5a ln(50a2 ) √ √ √ √ µ0 I [ √ = a(2 17 − 10 − 26) − 8a ln(16a2 4 17a2 ) + 3a ln(9a2 + 3 10a2 ) 4π ] √ + 5a ln(25a2 + 5 16a2 ) + 8a ln(32a2 ) − 3a ln(18a2 ) − 5a ln(50a2 ) ( [ ( √ ) √ ) √ √ √ 4 + 17 3 + 10 µ0 I ΨA = a(2 17 − 10 − 26) − 8a ln +3a ln 4π 8 6 ( √ )] 5 + 26 + 5a ln 10 T a his th nd wo o eir is rk w r sa co pro is ill le u vi pr de o rse de ot st f a s d s ec ro n an o te y y p d le d th a a ly by e rt ss fo U in o e r te f t ss th nite gr hi in e ity s w g us d S of or stu e o tat th k ( de f i es e in nt ns co w cl le tr p or ud a uc y r k an ing rnin tors igh d on g. in t la is w D no the iss tea s t p W em ch er or in ing m ld a itt W tio ed id n . e W eb )

ΨA =

We now proceed with side B, selecting our z axis as shown in Figure 6.16e. We have ] [ µ0 I z + a/2 z − a/2 √ ΨB = −√ dzdr r2 + (z + a/2)2 r2 + (z − a/2)2 −a/2 4πr 4a ] [ ∫ ∫ µ0 I 5a 1 a/2 z + a/2 z − a/2 √ = −√ dzdr 4π 4a r −a/2 r2 + (z + a/2)2 r2 + (z − a/2)2 [ ] ∫ ∫ µ0 I 5a 2 a/2 z + a/2 z − a/2 √ = −√ dzdr 4π 4a r 0 r2 + (z + a/2)2 r2 + (z − a/2)2 ∫ 5a ∫ a/2

Using the same substitution as was used for ΨA , the two terms under the integral over z yield ∫ a/2

√

0

∫ a/2 0

√

z + a/2 r2 + (z + a/2)2 z − a/2 r2 + (z − a/2)2

dz = dz =

√

r2 + a2 −

√ r2 + (a/2)2

√ √ r2 − r2 + (a/2)2

Thus we have

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33

∫ √ ) µ0 I 5a 1 (√ 2 r + a2 − r2 dr 2π 4a r [ ]}5a √ µ0 I {√ 2 a + r2 + a ln r − a ln a2 + a a2 + r2 −a = 2π 4a ] √ √ √ µ0 I [ √ = a 26 + a ln(5a) − a ln(a2 + a2 26) − a 17 − a ln(4a) + a ln(a2 + a2 17) − a 2π [ ( √ )] ( ) √ √ µ0 I 5 1 + 26 √ ΨB = a( 26 − 17 − 1) + a ln − a ln 2π 4 1 + 17

ΨB =

T a his th nd wo o eir is rk w r sa co pro is ill le u vi pr de o rse de ot st f a s d s ec ro n an o te y y p d le d th a a ly by e rt ss fo U in o e r te f t ss th nite gr hi in e ity s w g us d S of or stu e o tat th k ( de f i es e in nt ns co w cl le tr p or ud a uc y r k an ing rnin tors igh d on g. in t la is w D no the iss tea s t p W em ch er or in ing m ld a itt W tio ed id n . e W eb )

It is clear that the contribution from side D should be quite similar to that from side B except for all distances being smaller by an amount a. We have [ ] ∫ 4a ∫ a/2 z + a/2 z − a/2 µ0 I √ −√ dzdr ΨD = − r2 + (z + a/2)2 r2 + (z − a/2)2 3a −a/2 4πr ( [ √ )] ( ) √ √ 1 + 17 −µ0 I 4 √ = − a ln a( 17 − 10 − 1) + a ln 2π 3 1 + 10 Also noting that by symmetry we have ΨC = ΨA , we are now ready to find the total magnetic flux produced by coil 1 and linked by coil 2. We have Ψ12 = ΨA + ΨB + ΨC + ΨD = 2ΨA + ΨB + ΨD Evaluating the numerical values of the various terms gives

µ0 Ia 2π where the minus sign simply indicates the fact that the net flux is into the paper. The mutual inductance is then found to be Ψ12 ≃ −0.024

|Ψ12 | µ0 a = 0.024 = 4.8 × 10−9 a H I 2π where a is the side length given in meters.

L12 =

6.49 Fixed-length piece of wire.

We can first consider the inductance of single turn coils of different shapes. The formula for the inductance of single-conductor thin-wire loops is given at the end of Section 6.7.3: ( ) µ0 l 4l L≃ 2.303 log10 − θs 2π d where l is the length of wire, d is its diameter and θs is a shape factor. It is clear from the discussion in the text that the circular shape gives the highest inductance value. Thus, it makes sense to consider multiple circular turns, or a cylindrical solenoid. The inductance of solenoidal coils of length l and radius a is given approximately by the expression given in Section 6.7.3 as

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34

Chapter 6/ The Static Magnetic Field

L≃

10µ0 πa2 N2 9a + 10l

for l > 0.8a

With a fixed length (L) piece of wire, wound tightly but in a non-overlapping manner we have 2πaN = L and for Nd ≃ l so that L≃

10µ0 π[L/(2πN)]2 N2 5µ0 L2 N = 9[L/(2πN)] + 10Nd 9L + 20πdN2

Substituting values we find L(µH) ≃

0.251N 1.8 + 0.0628N2

T a his th nd wo o eir is rk w r sa co pro is ill le u vi pr de o rse de ot st f a s d s ec ro n an o te y y p d le d th a a ly by e rt ss fo U in o e r te f t ss th nite gr hi in e ity s w g us d S of or stu e o tat th k ( de f i es e in nt ns co w cl le tr p or ud a uc y r k an ing rnin tors igh d on g. in t la is w D no the iss tea s t p W em ch er or in ing m ld a itt W tio ed id n . e W eb )

Calculations for different values of N indicate that the maximum inductance is achieved for N = 5 turns, for which L ≃ 0.3718 µH. The values for N = 4 and N = 6 are respectively L ≃ 0.3574 µH and L ≃ 0.3702 µH, so that this is a rather broad maximum. (b) The length of the 0.1 mm diameter wire can be found by equating the volume, i.e., ( ) ( ) 1 0.1 π 20 = π L → L = 2000 cm 2 2 The inductance in microhenries is now given by L(µH) ≃

2509.8N 180.25 + 0.00629N2

which maximizes at a value of L ≃ 1.18 mH for N = 169. Thus, the best arrangement is 169-turn circular cylindrical solenoid. Although the inductance is substantially higher for the thinner wire, it should be noted that the wire resistance is also higher by almost the same amount. To see this, and noting that σcopper = 5.8 × 107 S-m−1 , the resistance values for the 0.1 mm and 1 mm diameter wires can be found simply using the formula R = L/(σA), where A is the cross-sectional area. We find R1 mm wire ≃ 4.39 × 10−3 Ω

and

R0.1 mm wire ≃ 43.9 Ω

6.50 Magnetic energy.

The current density at any radius r such that a ≤ r ≤ b is uniform and is given by J0 = Iπ(b2 − a2 ) The B-field can be found using Ampère's law given by (6.9) as 2

2

2πrBϕ (r) = µ J0 [π(r − a )] | {z } | H {z } C B·dl

Ienc

→

[ ] µ0 I a2 2 r − 2 Bϕ (r) = 2π(b2 − a2 ) r

The magnetic energy stored in the cylinder per unit length is then

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35

[ ]2 ∫ b [ ]2 ∫ 2π ∫ b 1 2π µ0 I a2 2 Bϕ (r)rdrdϕ = r − 2 dr 2µ0 0 2µ0 2π(b2 − a2 ) r V a a [ 4 ( )] 4 2 b 3a b µ0 I − a2 b2 + + a4 ln = 4π(b2 − a2 )2 4 4 a

1 Wm = 2µ0

∫

B2ϕ dv =

where the integral is simply evaluated upon expansion of the term in brackets into various powers of r.

6.51 Explosion in a power transformer.

Assuming the wires connecting the transformers to the high voltage line to be infinitely long and using a similar approach as in Example 6.1, the peak magnetic repulsion force per unit length on each of the two wires which are 20 cm apart and each carrying a peak current of 100 kA can be calculated as 4π × 10−7 (105 )2 µ0 I = = 10000 N-m−1 !

2πa 2π(0.2)

T a his th nd wo o eir is rk w r sa co pro is ill le u vi pr de o rse de ot st f a s d s ec ro n an o te y y p d le d th a a ly by e rt ss fo U in o e r te f t ss th nite gr hi in e ity s w g us d S of or stu e o tat th k ( de f i es e in nt ns co w cl le tr p or ud a uc y r k an ing rnin tors igh d on g. in t la is w D no the iss tea s t p W em ch er or in ing m ld a itt W tio ed id n . e W eb )

Fpeak =

This strong peak repulsion force causes the two wires to exert a strong force on their insulation structure to move further away from each other and as a result the insulation structure suddenly breaks and explodes.

6.52 Force between a long wire and a loop.

Assuming the z direction to be out of the paper, the B field of the long wire is given by (Example 6.3) µ0 I1 2πr The force on the rigid loop with current I2 due to the field B1 can be calculated as (equation (6.2)) I I Floop = I2 dl2 × B1 = I2 dl2 × B1 B1 = ϕ̂ϕ

C2

C2

Note that this integral can be broken into four integrals, each taken over one of the four sides of the loop. For the circular arcs, we have dl2 = −ϕ̂ϕ adϕ or dl2 = +ϕ̂ϕbdϕ and as a result, dl2 × B1 = 0 since ϕ̂ϕ × ϕ̂ϕ = 0, i.e., the B1 field does not exert any force on the circular portions of the loop. For the straight side along the x axis, we have µ0 I1 I2 F1 = 2π

∫ b

1 µ0 I1 I2 b drr̂ × ϕ̂ϕ = ẑ ln r 2π a a

Similarly, for the other straight side, we have µ0 I1 I2 F2 = 2π

∫ b

1 µ0 I1 I2 b dr(−r̂) × ϕ̂ϕ = −ẑ ln r 2π a a

Therefore, the net force in the loop due to the B1 field is

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36

Chapter 6/ The Static Magnetic Field

F = F1 + F2 = 0 The loop experiences a torque around the axis which is defined by a line in the plane of the paper, at an angle ϕ0 /2 from the horizontal, i.e., the line described by y = [tan(ϕ0 /2)]x, where x and y are respectively the horizontal and the vertical axes. Based on Section 6.10.3, the torque can be found from ∫ b T= a

(r1 × F1 )dr +

∫ b a

(r2 × F2 )dr

T a his th nd wo o eir is rk w r sa co pro is ill le u vi pr de o rse de ot st f a s d s ec ro n an o te y y p d le d th a a ly by e rt ss fo U in o e r te f t ss th nite gr hi in e ity s w g us d S of or stu e o tat th k ( de f i es e in nt ns co w cl le tr p or ud a uc y r k an ing rnin tors igh d on g. in t la is w D no the iss tea s t p W em ch er or in ing m ld a itt W tio ed id n . e W eb )

where |r1 | is the arc length between the torque axis (i.e., the line y = [tan(ϕ0 /2)]x) and line 1 (i.e., x axis) with r being in the ϕ̂ϕ direction. Similarly for r2 and line 2, i.e., y = (tan ϕ0 )x.

© 2015 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.