Fundamentals of Electromagnetics with Engineering Applications Solutions Manual.pdf (13)

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Solution Manual

Fundamentals of Resource: Electromagnetics with Engineering Applications Edition:

1st Edition

Author(s):

Stuart M. Wentworth

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2-1 Solutions for Chapter 2 Problems 1. Vectors in the Cartesian Coordinate System P2.1: Given P(4,2,1) andAAPQ PQ=2x =a 2+4y axa++6z, 4a ayfind + 6 the az =point (Qx- Q. Px)x + (Qy-Py)y+(Qz-Pz)az Qx-Px=Qx-4=2; Qx=6 Qy-Py=Qy-2=4; Qy=6 Qz-Pz=Qz-1=6; Qz=7 Ans: Q(6,6,7)

a

a

P2.2: Given the points P(4,1,0)m and Q(1,3,0)m, fill in the table and make a sketch of the vectors found in (a) through (f).

Vector Mag Unit Vector a. Find the vector A AOP = 4 ax + 1 ay 4.12 AOP = 0.97 ax + 0.24 ay BOQ = 1 ax + 3 ay 3.16 aOQ = 0.32 ax + 0.95 ay from the origin to P CPQ = -3 ax + 2 ay 3.61 aPQ = -0.83 ax + 0.55 ay b. Find the vector B from the origin to Q A + B = 5 ax + 4 ay 6.4 a = 0.78 ax + 0.62 ay c. Find the vector C C - A = -7 ax + 1 ay 7.07 a = -0.99 ax + 0.14 ay from P to Q B - A = -3 ax + 2 ay 3.6 a = -0.83 ax + 0.55 ay d. Find A + B e. Find C – A f. Find B - A a. AOP = (4-0)ax + (1-0)ay + (0-0)az = 4 ax + 1 ay.

AOP= 42+12= 17=4.12 1 4 aOP =17 a x + a y =0.97ax+0.24ay 17 (see Figure P2.2ab) b. BOQ =(1-0)ax + (3-0)a y + (0-0)az = 1 ax + 3 ay.

BOQ= 12 + 32 = 10 = 3.16

Fig. P2.2ab

3 1 aOQ =10 a x + a y =0.32ax +0.95ay 10 (see Figure P2.2ab) c. CPQ = (1-4)ax + (3-1)ay + (0-0)az = -3ax + 2 ay.

2 2 C PQ = 3 + 2 = 13 = 3.61 −3 2 aPQ = a x + +0.55ay − 13 ay = 0.83ax 13 (see Figure P2.2cd)

Fig. P2.2cd

2-2 d. A + B = (4+1)ax + (1+3)ay + (0-0)az = 5 ax + 4 ay.

52+42= 41=6.4 A+B= 5 4 ay=0.78ax+0.62ay ax+ 41 41 (see Figure P2.2cd) a=

e. C - A = (-3-4)ax + (2-1)ay + (0-0)az = -7 ax + 1 ay.

72+12= 50=7.07

C−A=

−7 1 ay=−0.99ax+0.14ay ax+ 50 50 (see Figure P2.2ef) a=

FigP2.2ef

f. B - A = (1-4)ax + (3-1)ay + (0-0)az = -3 ax + 2 ay.

32+22= 13=3.6 −3 2 ay=−0.83ax+0.55ay a= ax+ 13 13 (see Figure P2.2ef) B−A=

P2.3: MATLAB: Write a program that will find the vector between a pair of arbitrary points in the Cartesian Coordinate System.

A program or function for this task is really overkill, as it is so easy to perform the task.

Enter points P and Q (for example, P=[1 2 3]; Q=[6 5 4]). Then, the vector from P toQ is simply given by Q-P. As a function we could have: function PQ=vector(P,Q) % Given a pair of Cartesian points % P and Q, the program determines the % vector from P to Q. PQ=Q-P; Running this function we have: >> P=[1 2 3]; >> Q=[6 5 4]; >> PQ=vector(P,Q) PQ = 5

3

1

Alternatively, we could simply perform the math in the command line window:

2-3 >> PQ=Q-P PQ = 5 3 1 >> 2. Coulomb’s Law, Electric Field Intensity, and Field Lines P2.4: Suppose Q1(0.0, -3.0m, 0.0) = 4.0nC, Q2(0.0, 3.0m, 0.0) = 4.0nC, and Q3(4.0m, 0.0, 0.0) = 1.0nC. (a) Find the total force acting on the charge Q3. (b) Repeat the problem after changing the charge of Q2 to –4.0nC. (c) Find the electric field intensity for parts (a) and (b).

Q1Q2 a13, where R13 = 4 ax + 3 ay =, R13 = 5m, a13 = 0.8 ax + 0.6 ay. (a) F13= 2 4oR13 so

C)(4ax+3ay) 5 FVNM C VC 4 (10−9 F 36m) (m) 5 −9

F13=

C ) 110 x (x (410

−9

2

y = 1.15 x10 −9 +a0.86x10−9a N. x

Similarly, F23=1.15x10− 9ax − 0.86 x10−9 = 2.3ax nN ay N, so FTOT (b) with Q2 = -4 nC,F 13 is unchanged butF23= −1.15 x10a−9+ 0.86 x10−9 a N , so x

y

FTOT=1.7ay nN.

axN−)=VC V FTOT ( 2.3 x10 = C) 2.3ax m.

-9 Q3 Nm (1x10 9

(c) Ea =

V Likewise, Eb=1.7ay . m

Fig. P2.4 P2.5: Find the force exerted by Q1(3.0m, 3.0m, 3.0m) = 1.0 C on Q2(6.0m, 9.0m, 3.0m) = 10. nC. F12 =

Q1Q2

a12,where

4R2o12 R12 = (6-3)ax + (9-3)ay + (3-3)az = 3 ax + 6 ay m 3a + 6ay R12 = 32+62 = 45m,a12= x ,and 45

2-4

C ) 3a + 6 ya FV NM C ) 10 (1x10− ( x10−9 , so F12= 0.89a + 1.8a  N .

C VC 45 F 3 645m )m ) ( 4 (10− 6

F12=

x

9

x

2

y

Fig. P2.5

P2.6: Suppose 10.0 nC point charges are located on the corners of a square of side 10.0 cm. Locating the square in the x-y plane (at z = 0.00) with one corner at the origin and one corner at P(10.0, 10.0, 0.00) cm, find the total force acting at point P. We arbitrarily label the charges as shown in Figure P2.6. Then ROP = 0.1 ax + 0.1 ay ROP = 0.141 m aOP = 0.707 ax + 0.707 ay.

= FOP

( )(ax + ay ) (10nC) 10 ( nC) 0.707 (4 10−9F 36m 0.141m)2

)(

= 32( ax + a

FTP =

FSP =

(4

yN) 

)(

( nC) ay (10nC) 10

() ) ( 36m )0.1m 10−9 F

2

( nC)a (10nC) 10

x

(4

) ( 36 ) ( 10−9 F

m 0.1m)

2

= 90ayN Fig. P2.6

= 90axN

and then the total (adjusting to 2 significant digits) is:

FTOT=120(ax+ay)N.

2-5 P2.7: 1.00 nC point charges are located at (0.00, -2.00, 0.00)m, (0.00, 2.00, 0.00)m, (0.00, 0.00, -2.00)m and (0.00, 0.00, +2.00)m. Find the total force acting on a 1.00 nC charge located at (2.00, 0.00, 0.00)m.

Figure P2.7a shows the situation, but we need only find the x-directed force from one of the charges on Qt (Figure P2.7b) and multiply this result by 4. Because of the 2ax+ 2ay QQt problem’s aR,R=2ax+2ay,R= 8 m,aR= , F1t= symmetry, 4 oRthe rest of the components cancel. 8  2 xa +2a y  110 x −9C) 110 x −9C) ( (   8 soF = −12  = 796 x10 1t (ax + ay )N −9 10 F 4 8m2) 36 The force from all charges is then x10 (− 41 )2(3.2 aaxxnN. ) n N = FT O T 796=

(

Fig. P2.7a

m)(

Fig. P2.7b

P2.8: A 20.0 nC point charge exists at P(0.00,0.00,-3.00m). Where must a 10.0 nC charge be located such that the total field is zero at the origin?

For zero field at the origin, we must cancel the +az directed field from QP by placing Q at the point Q(0,0,z) (see Figure P2.8). Then we have Etot = EP + EQ = 0.

Q C 9 (2010 x − ) FV a =20a Vm z P aR= So, EP= z 2 −9 2 4 oR 4 10 36 m

( F  )(3m) C

2-6 and

Q aR 4 oR 2

EQ = =

C)(− x 10−9 az ) (10 −90 =

(

F 4 10−9

m)(

36

z( m))

2

z2

az

So then 90 20az − z2 az =0,

90 , z = 2.12.

20 Thus, Q(0,0,2.12m). z2=

Fig. P2.8

3. The Spherical Coordinate System P2.9: Convert the following points from Cartesian to Spherical coordinates: a. P(6.0, 2.0, 6.0) b. P(0.0, -4.0, 3.0) c. P(-5.0,-1.0, -4.0)

 6 o 2  =18o =47,=tan−1 8.7 6 4 −  3 1 − (b) r= 02+ 42+ 32= 5, = cos1 3,   = 5 o  = t a n −0= − 9 0 o  5   −1 −1 (c) r= 52+12+42=6.5,=cos−1 =130o,=tan−1  =190o  − 5  6.5 P2.10: Convert the following points from Spherical to Cartesian coordinates: (a) r=

62+22+62=8.7,=cos−1

a. P(3.0, 30., 45.) b. P(5.0, /4, 3/2) c. P(10., 135, 180) (a) x = rsincos= 3sin 30o cos 45o = 1.06

y = rsinsin= 3sin 30o sin 45o = 1.06 z = r cos=3cos30o =2.6 soP(1.1,1.1,2.6). (b)

2-7

x = rsincos= 5sin 45o cos 270o =0 y = rsinsin= 5sin 45o sin 270o = −3.5 z = r cos=5cos45o =3.5 soP(0, −3.5, 3.5). (c) = −7.1 x= r sincos= 10 sin135o cos180o y = r sinsin=10sin135osin180o= 0 z = r cos= 10 cos135o = −7.1 soP(−7.1,0,−7.1).

P2.11: Given a volume defined by 1.0m ≤ r ≤ 3.0m, 0 ≤  ≤ 0, 90 ≤  ≤ 90, (a) sketch the volume, (b) perform the integration to find the volume, and (c) perform the necessary integrations to find the total surface area. (a)

Fig. P2.11 (b) 3

 2

90o

13 V=   r2sindrdd= =13.6m3.  rdrsind d= 3 1 0 0 So volume V = 14 m3. (c) There are 5 surfaces: an inner, an outer, and 3 identical sides.

Sside

2

 2

=  rdrd  =2 m2 ; S  = rdr  d 3

sides

1

= 6 m2

0

90o

 2

0

0

Souter=  r2 sin  d  d  =3sin d  d = 2



m2;STOT =11  m 2 = 34.6m2 2 So Stotal= 35 m2.

Sinner =

9 m2 2

2-8 4. Line Charges and the Cylindrical Coordinate System P2.12: Convert the following points from Cartesian to cylindrical coordinates: a. P(0.0, 4.0, 3.0) b. P(-2.0, 3.0, 2.0) c. P(4.0, -3.0, -4.0)  4  (a) = 02+42=4,=tan−1 =90o,z=3, so P(4.0,90o,3.0)  0   3  (b) = 22+32=3.6,=tan−1  =124o,z=2, so P(3.6,120o,2.0) −2 −3 (c) = 42+32=5,=tan−1  = −37o,z=−4, so P(5.0,−37o,−4.0)  4  P2.13: Convert the following points from cylindrical to Cartesian coordinates: a. P(2.83, 45.0, 2.00) b. P(6.00, 120., -3.00) c. P(10.0, -90.0, 6.00)

(a) x  cos=2.83cos45o = 2.00 = y =  sin=2.83sin45o = 2.00 z = z = 2.00

soP(2.00, 2.00, 2.00). (b) =x  cos=6.00cos120o = −3.00 =y  sin=6.00sin120o = 5.20 =z z = −3.00 soP(−3.00,5.20,−3.00). (c) =0 x =  cos = 10.0cos(−90.0o) = −10.0 y =  sin = 10.0sin(−90.0o) z = z = 6.00 soP(0,−10.0,6.00).

P2.14: A 20.0 cm long section of copper pipe has a 1.00 cm thick wall and outer diameter of 6.00 cm. a. Sketch the pipe conveniently overlaying the cylindrical coordinate system, lining up the length direction with the z-axis b. Determine the total surface area (this could actually be useful if, say, you needed to do an electroplating step on this piece of pipe) c. Determine the weight of the pipe given the density of copper is 8.96 g/cm3

2-9 (a) See Figure P2.14 (b) The top area, Stop, is equal to the bottom area. We must also find the inner area, ,S inner and the outer area, Souter.

3

2

2

0

Stop =  d d  =   d   d  = 5  bottom

= Stop.

S

2

S

cm2.

20

 dz= 3 ddz= 120cm2 outer=  d 0

0

2

20

0

0

Sinner = = d dz= 2 d dz=80 cm2 The total area, then, is 210 cm2, ortotS = 660 cm2 . (c) Determining the weight of the pipe requires the volume:  d d  dz V =  3

2

20



0

0

=  d d dz =100 cm3. g   M2pipe = 8.96 (100 cm3) cm3   = 2815g.

So Mpipe = 2820g.

Fig. P2.14 P2.15: A line charge with charge density 2.00 nC/m exists at y = -2.00 m, x = 0.00. (a) A charge Q = 8.00 nC exists somewhere along the y-axis. Where must you locate Q so that the total electric field is zero at the origin? (b) Suppose instead of the 8.00 nC charge of part (a) that you locate a charge Q at (0.00, 6.00m, 0.00). What value of Q will result in a total electric field intensity of zero at the origin?

= E L + E Q. (a) The contributions to E from the line and point charge must cancel,E or / ) (2nCm  a= −9 2  o  2 10 F 

For the line: EL= L

(

V ay=18 may

) () 2 m

36 m

2-10 and for the point charge, where the point is located a distance y along the y-axis, we Q ( 8 n C ) =( 72 − −aa y ) have: EQ= 2 ( − a y ) 2( y ) − 9= 4 oy 4 y2 y Therefore: 72 72 y2 =18, or y= 18 = 2m.

36m) (10F

So Q(0,2.0m,0) (b)

Q

=18,

4 o( 6 ) 2

Q=

(18) ( )36 = nC. 9

72

Fig. P2.15

P2.16: You are given two z-directed line charges of charge density +1 nC/m at x = 0, y = -1.0 m, and charge density –1.0 nC/m at x = 0, y = 1.0 m. Find E at P(1.0m,0,0).

The situation is represented by Figure P2.16a. A better 2-dimensional view in Figure P2.16b is useful for solving the problem.  ax + a y  L1 a, and a= 2  E 1= .

2 o  2 

E1 =

( a+ a ) FV = 9 + V , and = 9 + V C) (1x10−9 (a a )m E (-axa )m.

10−9 F C 2

(

2

36 m ) )( 2m

x

y

x

y

So E TOT = 18 ay V/m.

Fig. P2.16a

Fig. P2.16b

2

y

2-11 P2.17: MATLAB: Suppose you have a segment of line charge of length 2L centered on

the z-axis and having a charge distribution L. Compare the electric field intensity at a point on the y-axis a distance d from the origin with the electric field at that point assuming the line charge is of infinite length. The ratio of E for the segment to E for the infinite line is to be plotted versus the ratio L/d using MATLAB. This is similar to MATLAB 2.3. We have for the ideal case

E ideal =

L L a. a = 2od 2 o

For the actual2L case, we have an integration to perform (Equation (2.35) with different limits): = +L + L   Lday  L  a z dz = E actual   4 o ( 2 z2)3 2 4 od2 z2+ d2  − L

E actual=

  Lay L   .

2 od  L2 d2 +

Now we manipulate these expressions to get the following ratio: L E actual d = .

2 E ideal L d In the program, the actual to ideal field ratio is termed “Eratio” and the charged line halflength L ratioed to the distance d is termed “Lod”.

1 + )(

% % M-File: % MLP0217 % % This % program is similar to ML0203. % % It % compares the E-field from a finite length % % segment % of charge (from -L to +L on the z-axis) % % to % the E-field from an infinite length line clc of charge. The ratio (E from segment to E from clear infinite length line) is plotted versus the ratio Lod=L/d, where d is the distance along the y axis.

Wentworth, 12/19/02 Variables:

Lod the ratio L/d Eratio ratio of E from segment to E from line %clears the command window %clears variables % Initialize Lod array and calculate Eratio Lod=0.1:0.01:100;

2-12 Eratio=Lod./(sqrt(1+Lod.^2)); % Plot Eratio versus Lod semilogx(Lod,Eratio) grid on xlabel('Lod=L/d') ylabel('E ratio: segment to line') Executing the program gives Figure P2.17.

Fig. P2.17 So we see that the field from a line segment of charge appears equivalent to the field from an infinite length line if the test point is close to the line.

P2.18: A segment of line charge L =10 nC/m exists on the y-axis from the origin to y = +3.0 m. Determine E at the point (3.0, 0, 0)m.

It is clear from a sketch of the problem in Figure P2.18a that the resultant field will be directed in the x-y plane. The situation is redrawn in a temporary coordinate system in Figure P2.18b.

dz a−za

z = E a + Ezz. We have from Eqn (2.34)E=  L a 3 4 o (  2 + z 2 ) 2

For E we have:

 E = 4 L  o

dz

L  

z

3

=  3 4  z (  2 + 2 ) 2o 2 z2+  2   0

With  = 3, we then haveE= 21.2 V/m.

For Ez:

 

2-13

Fig. P2.18b

Fig. P2.18a

zdz − L  = Ez = L  + z3 2 4) 2 o  4 o ( 2 

3

 V  =−8.79m 2 2  + zz=0 1

Thus we have ETOT = 21 a – 8.8 az V/m.

Converting back to the original coordinates, we have ETOT = 21 ax – 8.8 ay V/m.

5. Surface and Volume Charge

P2.19: In free space, there is a point charge Q = 8.0 nC at (-2.0,0,0)m, a line charge L = 10 nC/m at y = -9.0m, x = 0m, and a sheet charge  =s 12. nC/m2 at z = -2.0m.

Determine E at the origin.

The situation is represented by Figure P2.19, and the total field is ETOT = EQ + EL + ES.

Q aR = EQ = 4 o R 2 4 = 18ax EL =

(8x10−C 9

) a x

m2m)) ( (10 F36 −9

2

V m

C my )a (10 10−9 L a = − 9 x 2 o 2 1 F 36 9m)

(0 m)(

= 20ay



V m

(12 x10−9 Cm2) a m) (1 36

E s = s aN = 2 o 2 0−9F

z

V = 679az m So: E tot = 18 ax + 20ay + 680az V/m.

Fig. P2.19

2-14

P2.20: An infinitely long line charge (L = 21 nC/m) lies along the z-axis. An infinite ) lies in the x-z plane at y = 10 m. Find a point on the yarea sheet charges( = 3 nC/m2 axis where the electric field intensity is zero.

We have ETOT = EL + ES.

EL = =

a a = L y 2  o  2 oy L

x 10 − 9 C/ m ) (21 378 = ay

y F 2(10 9 36m)y −

x C (310−  a Es = s N = 9 2 o 2 0− F 9



ay

m 2) ) ( − a y

( 1 36m) 

= −54 a y

Fig. P2.20

so

378

−54=0, or y=7.

y Therefore, P(0, 7m, 0).

P2.21: Sketch the following surfaces and find the total charge on each surface given a 2 surface charge density of s = 1nC/m.

Units (other than degrees) are meters. (a) –3 ≤ x ≤ 3, 0 ≤ y ≤ 4, z = 0 (b) 1 ≤ r ≤ 4, 180 ≤  ≤ 360,  = /2 (c) 1 ≤  ≤ 4, 180 ≤  ≤ 360, z = 0

Fig. P2.21b&c

Fig. P2.21a

3

4

−3

0

= 24s = 24nC dS= sdxdy (a) Q=s

2-15 2

4 15 (b) Q=s rdrd= s=24nC 1 4

2



2

(c) Q=s   d   d  = 2 4 n C 1



P2.22: Consider a circular disk in the x-y plane of radius 5.0 cm. Suppose the charge

density is a function of radius such thats  = 12 nC/cm2 (when  is in cm). Find the electric field intensity a point 20.0 cm above the origin on the z-axis.

aha L z

From section 4 for a ring of charge of radius a,E=

. Now we have

2  ( )a 2 + h 2 o

L=sd and dE =

A d 

ha

3 2

z, where  =s A nC/cm2. Now the total field is given 3

2 o ( 2 +h ) 2 2

by the integral: Ahaz  2+d  E= .

3 2o ( 2 h2)2 = , This can be solved using integration by parts, where u du = d, −1 d v= . This leads to + , and dv=  2 +h2 2 h2

 a +a2 +  h 2 Ah  − a  az.  +ln    2o  a2 + h2 h     arrive at E = 6.7 kV/cm az.

Plugging in the appropriate values we    P2.23: Suppose a ribbon of charge with density s exists in the y-z plane of infinite length in the z direction and extending from –a to +a in the y direction. Find a general expression for the electric field intensity at a point d along the x-axis.

E=

The problem is represented by Figure P2.23a. A better representation for solving the problem is shown in Figure P2.23b. LWe have dE= a, where L = sdy. Then, since 2 o dax − yay , a= d2 + y2 the integral becomes dax − yay  sdy .

E = 2  d 2 + y2 d 2+ y2 o

2-16 It may be noted that the ay component will cancel by symmetry. The ax integral is found from the appendix and we have  a E = s tan−1  d ax.    o

FigP2.23b

FigP2.23a

P2.24: Sketch the following volumes and find the total charge for each given a volume charge density ofv = 1n C/m.3 Units (other than degrees) are meters. (a) 0 ≤ x ≤ 4, 0 ≤ y ≤ 5, 0 ≤ z ≤ 6

(b) 1 ≤ r ≤ 5, 0 ≤  ≤ 60 (c) 1 ≤  ≤ 5, 0 ≤  ≤ 90, 0 ≤ z ≤ 5 4

5

6

0

0

0

(a) Q=vdv=vdxdydz=120nC (b) Q=   vdv 5

60

2



0

0

5

 2

5

0

0

=  r2drsin d d =130nC v

(c) 1 Q=   vdv

=v 



Fig. P2.24a

d d dz= 94nC

1

Fig. P2.24b

2-17

Fig. P2.24c P2.25: You have a cylinder of 4.00 inch diameter and 5.00 inch length (imagine a can of tomatoes) that has a charge distribution that varies with radius as  = (6 )where nC/in3 v  is in inches. (It may help you with the units to think of this as ) v  (nC/in3)= 6 (nC/in4

(in)) Find the total charge contained in this cylinder.

2

2

5

0

0

0

Q=  v d v =    ( 6  )  d  d  d z = 6   2 d   d   d z = 1

P2.26: MATLAB: Consider a rectangular volume with 0.00 ≤ x ≤ 4.00 m, 0.00 ≤ y ≤

5.00 m and –6.00 m ≤ z ≤ 0.00 with charge density . Find the electric v = 40.0 nC/m3 field intensity at the point P(0.00,0.00,20.0m). % % % % % % % % % % % % % % % % %

MLP0226 calculate E from a rectangular volume of charge variables xstart,xstop ystart,ystop zstart,zstop xt,yt,zt rhov Nx,Ny,Nz dx,dy,dz dQ eo dEi limits on x for vol charge (m) test point (m) vol charge density, nC/m^3 discretization points differential lengths differential charge, nC free space permittivity (F/m) differential field vector dEix,dEiy,dEiz x,y and z components of dEi dEjx,dEjy,dEjz of dEj dEkx,dEky,dEkz of dEk Etot total field vector, V/m

2-18 clc clear % initialize variables xstart=0;xstop=4; ystart=0;ystop=5; zstart=-6;zstop=0; xt=0;yt=0;zt=20; rhov=40e-9; Nx=10;Ny=10;Nz=10; eo=8.854e-12; dx=(xstop-xstart)/Nx; dy=(ystop-ystart)/Ny; dz=(zstop-zstart)/Nz; dQ=rhov*dx*dy*dz; for k=1:Nz for j=1:Ny for i=1:Nx xv=xstart+(i-0.5)*dx; yv=ystart+(j-0.5)*dy; zv=zstart+(k-0.5)*dz; R=[xt-xv yt-yv zt-zv]; magR=magvector(R); uvR=unitvector(R); dEi=(dQ/(4*pi*eo*magR^2))*uvR; dEix(i)=dEi(1); dEiy(i)=dEi(2); dEiz(i)=dEi(3); end dEjx(j)=sum(dEix); dEjy(j)=sum(dEiy); dEjz(j)=sum(dEiz); end dEkx(k)=sum(dEjx); dEky(k)=sum(dEjy); dEkz(k)=sum(dEjz); end Etotx=sum(dEkx); Etoty=sum(dEky); Etotz=sum(dEkz); Etot=[Etotx Etoty Etotz] Now to run the program: Etot =

2-19

-6.9983 -8.7104 79.7668 >> So E = -7.0 ax -8.7 ay + 80. az V/m

P2.27: MATLAB: Consider a sphere with charge density v =120 nC/m3 centered at the origin with a radius of 2.00 m. Now, remove the top half of the sphere, leaving a Find the electric hemisphere below field theintensity x-y plane. at the point P(8.00m,0.00,0.00). (Hint: see MATLAB 2.4, and consider that your answer will now have two field components.) % % % % % % % % %

% M-File: % MLP0227 % This % program modifies ML0204 to find the field % at % point P(8m,0,0) from a hemispherical % distribution % of charge given by % rhov=120 % nC/m^3 from 0 < r < 2m and % pi/2 % < theta < pi. % Wentworth, % 12/23/02 % Variables: %

clc clear

d a dV eo

y axis distance to test point (m) sphere radius (m) differential charge volume where dV=delta_r*delta_theta*delta_phi free space permittivity (F/m) r,theta,phi spherical coordinate location of center of a differential charge element cartesian coord location of charge % x,y,z element R vector from charge element to P Rmag magnitude of R aR unit vector of R dr,dtheta,dphi differential spherical elements dEi,dEj,dEk partial field values Etot total field at P resulting from charge %clears the command window %clears variables

% Initialize variables eo=8.854e-12; d=8;a=2;

2-20 delta_r=40;delta_theta=72;delta_phi=144; % Perform calculation for k=(1:delta_phi) for j=(1:delta_theta) for i=(1:delta_r) r=i*a/delta_r; theta=(pi/2)+j*pi/(2*delta_theta); phi=k*2*pi/delta_phi; x=r*sin(theta)*cos(phi); y=r*sin(theta)*sin(phi); z=r*cos(theta); R=[d-x,-y,-z]; Rmag=magvector(R); aR=R/Rmag; dr=a/delta_r; dtheta=pi/delta_theta; dphi=2*pi/delta_phi; dV=r^2*sin(theta)*dr*dtheta*dphi; dQ=120e-9*dV; dEi=dQ*aR/(4*pi*eo*Rmag^2); dEix(i)=dEi(1); dEiy(i)=dEi(2); dEiz(i)=dEi(3); end dEjx(j)=sum(dEix); dEjy(j)=sum(dEiy); dEjz(j)=sum(dEiz);

end dEkx(k)=sum(dEjx); dEky(k)=sum(dEjy); dEkz(k)=sum(dEjz);

end Etotx=sum(dEkx); Etoty=sum(dEky); Etotz=sum(dEkz); Etot=[Etotx Etoty Etotz] Now to run the program: Etot = 579.4623 0.0000 56.5317 So E = 580ax + 57az V/m.

6. Electric Flux Density

2-21 P2.28: Use the definition of dot product to find the three interior angles for the triangle bounded by the points P(-3.00, -4.00, 5.00), Q(2.00, 0.00, -4.00), and R(5.00, -1.00, 0.00).

Here we use AB=ABcos

 AB.

−)−a y PR = ( − −3)ax + ( − 1 4

+(0−5)az

PR = 8a5x + 3a y − 5az, PR = 9.9 x

PQ= 5a + 4ay − 9az,PQ =11.0

PRPQ = () 8 5 ( ) +) 4()3+ ( )−−(95) = 97 = 9.9 11.0 cos P

 97  p =cos−1  27 (9.9)(11) =   RQ= 5.1 RQ=−a3 x +1ay −4az,

Fig. P2.28

RP=−a8 x−3ay+5az,RP = 9.9 R  ,R  = 89 RQ RP = 1=(5.1)(9.9)cos

 Q =180 − 27 − 89 = 64 P2.29: Given D = 2 a+ sin  a z C/m,2 find the electric flux passing through the surface  90. ≤  ≤ 180, and z = 4.0 m. defined by 2.0 ≤  ≤  m,  =

E dS, dS=  d d a  (  + sin a ) d d =a z dsin d  =6C  2a z

4

z

2



 2

P2.30: Suppose the electric flux density is given by D =r3r a –cos  a +sin2 =  a C/m2.

Find the electric flux through both surfaces of a hemisphere of radius 2.00 m and 0.00 ≤

 ≤ 90.0˚.

2  1 = dS,Dd S1=rsindda r

 1 =( ra 3r − cos a + sin2 a) r(2 sin d d ar )  2

2





= 3 3r sin d

dS2 = 0rdrda 

0

d = 48C 

2-22

 2 = cos a rdrda  − 2

 r2 2  sin = 0 = − 2   0 0 =48C

Fig. P2.30 7. Gauss’s Law and Applications P2.31: Given a 3.00 mm radius solid wire centered on the z-axis with an evenly distributed 2.00 coulombs of charge per meter length of wire, plot the electric flux

density D versus radial distance from the z-axis over the range 0 ≤  ≤ 9 mm.

For a 1 m length, v= 

2(Cm)=70.7x103C , a=3mm=.003m ( ) 3

m a2 =2LD, where L is the length of the Gaussian Qenc=DdS= D a   d a dz

Qenc is valid for both Gaussian surfaces. surface. Note that this expression for

GS1 ( < a):



2

L

0

0

0

Qenc =  vdv= v   d dz  =dv 2  L

 v  2  L = v  for   a.

2  L 2 GS2 ( > a):  a 21 Qenc = va2L,D= v for   a .

2  so D =

This is plotted with the following Matlab routine: % M-File: MLP0231 % Gauss's Law Problem % % solid cylinder with even charge % % Variables % rhov charge density (C/m^3) radius % a of cylinder (m) radial % rho distance from z-axis rho in mm % rhomm electric flux density (C/m^3) % D

2-23 % %

N number of data points maxrad max radius for plot (m)

clc;clear; % initialize variables rhov=70.7e3; a=0.003; maxrad=.009; N=100; bndy=round(N*a/maxrad); for i=1:bndy rho(i)=i*maxrad/N; rhomm(i)=rho(i)*1000; D(i)=rhov*rho(i)/2; end for i=bndy+1:N rho(i)=i*maxrad/N; rhomm(i)=rho(i)*1000; D(i)=(rhov*a^2)/(2*rho(i)); end plot(rhomm,D) xlabel('radial distance (mm)') ylabel('elect. flux density (C/m^2)') grid on

P2.32: Given a 2.00 cm radius solid wire centered on the z-axis with a charge density v = 6 C/cm3 (when  is in cm), plot the electric flux density D versus radial distance from the z-axis over the range 0 ≤  ≤ 8 cm.

Fig. P2.31

2-24

Choose Gaussian surface length L, and as usual we have Qenc = DS=  a  d  da  z=2LD, valid for both Gaussian surfaces. d  D

In GS1 ( < a): Qenc=vdv=62dddz=4L3, so D=

4L 3 =22 for a.

2L

For GS2 ( > a): Qenc=4La3, D=

2a3



for a.

This is plotted for the problem values in the following Matlab routine. % % % % % % % % % % %

M-File: MLP0232 Gauss's Law Problem solid cylinder with radially-dependent charge Variables a radius of cylinder (cm) rho radial distance from z-axis D N electric flux density (C/cm^3) number of data points maxrad max radius for plot (cm)

clc;clear; % initialize variables a=2; maxrad=8; N=100; bndy=round(N*a/maxrad); for i=1:bndy rho(i)=i*maxrad/N; D(i)=2*rho(i)^2; end for i=bndy+1:N rho(i)=i*maxrad/N; D(i)=(2*a^3)/rho(i); end plot(rho,D) xlabel('radial distance (cm)') ylabel('elect. flux density (C/cm^2)') grid on

2-25

Fig. P2.32

P2.33: A cylindrical pipe with a 1.00 cm wall thickness and an inner radius of 4.00 cm is centered on the z-axis and has an evenly distributed 3.00 C of charge per meter length of

pipe. Plot D as a function of radial distance from the z-axis over the range 0 ≤  ≤ 10 cm.

is true for all the Gaussian surfaces.

Qenc=DdS= Daddza=2hD;this 

GS1 ( < a): sinceQenc= 0, D = 0.

GS2(a <  < b): 3h 3 = v= 2   d d dz  (b − a2 ) Qenc =  vdv 

=



2

h 3  d  d  dz  (b2− a2)  a  0 0

(  −−a ) = 3h (b a ) 2

2

2

2

So,

3h ( 2 − a )2

Fig. P2.33a

( − a2) for a = D = b.

2  h (b2 − a2 ) 2  ( b−2a2) 

GS3( > b):

3

2

2-26 3 Qenc = 3h, D= for  b.

2  A plot with the appropriate values is generated by the following Matlab routine: % M-File: MLP0233 % Gauss's Law Problem % cylindrical pipe with even charge distribution % % Variables % a b inner radius of pipe (m) outer % rho radius of pipe (m) radial % rhocm distance from z-axis (m) radial % D N distance in cm electric flux % density (C/cm^3) number of data % points % maxrad max radius for plot (m) clc;clear; % initialize variables a=.04;b=.05;maxrad=0.10;N=100; bndya=round(N*a/maxrad); bndyb=round(N*b/maxrad); for i=1:bndya rho(i)=i*maxrad/N; rhocm(i)=rho(i)*100; D(i)=0; end for i=bndya+1:bndyb rho(i)=i*maxrad/N; rhocm(i)=rho(i)*100; D(i)=(3/(2*pi*rho(i)))*((rho(i)^2-a^2)/(b^2-a^2)); end for i=bndyb+1:N rho(i)=i*maxrad/N; rhocm(i)=rho(i)*100; D(i)=3/(2*pi*rho(i)); end plot(rhocm,D) xlabel('radial distance (cm)') ylabel('elect. flux density (C/m^2)') grid on

2-27

Fig. P2.33b P2.34: An infinitesimally thin metallic cylindrical shell of radius 4.00 cm is centered on the z-axis and has an evenly distributed charge of 100. nC per meter length of shell. (a) Determine the value of the surface charge density on the conductive shell and (b) plot D

as a function of radial distance from the z-axis over the range 0 ≤  ≤ 12 cm.

 s=

Q 100nC nC 100nC = = =398 2 .

S  d dz (.04 m)(2 ) ( m 1 m )

For all Gaussian surfaces,

of height h and radius , we have: Qenc=D dS, wheredS=  d dza ,

 Q enc= 2  hD  .

GS1 ( < a): Qenc= 0 so D = 0 GS2 ( > a): Qenc =  sdS=  s d  dz= 2a h s,

D = s % % % % % % % % %

a



for   a.

M-File: MLP0234 Gauss's Law Problem

Fig. P2.34a

cylindrical shell of charge Variables a radius of cylinder (m) surface Qs charge density (nC/m^2) radial rho distance from z-axis (m)

2-28 % % % %

rhocm D N

radial distance in cm electric flux density (nC/cm^3) number of data points maxrad max radius for plot (cm)

clc;clear; % initialize variables a=.04;Qs=398;maxrad=0.12;N=100; bndy=round(N*a/maxrad); for i=1:bndy rho(i)=i*maxrad/N; rhocm(i)=rho(i)*100; D(i)=0; end for i=bndy+1:N rho(i)=i*maxrad/N; rhocm(i)=rho(i)*100; D(i)=Qs*a/rho(i); end plot(rhocm,D) xlabel('radial distance (cm)') ylabel('elect. flux density (nC/m^2)') grid on

Fig. P2.34b

2-29

P2.35: A spherical charge density is given by v = o r/a for 0 ≤ r ≤ a, and v = 0 for r > a. Derive equations for the electric flux density for all r.

2 This is valid for each Gaussian Qenc=DdS=Drarrsinddar=4r2Dr.

surface.

GS1 (r < a): Qenc =  vdv= So D= r

r



2

a0

0

0

o

 or r3drsin .  d d  = 4

a

 or4  o 2 a4r 2= 4ar for r  a.

oa3 GS2 (r> ):a Qenc =  oa, 3 Dr = 4r2 for r  a.

P2.36: A thick-walled spherical shell, with inner radius 2.00 cm and outer radius 4.00 cm, has an evenly distributed 12.0 nC charge. Plot Dr as a function of radial distance from the origin over the range 0 ≤ r ≤ 10 cm.

Here we’ll let a = inner radius and b = outer radius. Then 2 sindd ar=4rDr; Qenc=DdS= a Drrr2

This is true for each Gaussian surface.

The volume containing charge is b 2  4  v= a r2d r  s i n  3d  ( b d3 −=a 3 ) .

0

So  v =

0

Q 3Q = . v 4  ( b 3

− a3)

Now we can evaluate Qenc for each Gaussian surface.

GS1 (r < a): Qenc = 0 so Dr = 0.

4    d =  v dv= vrdrsind GS2 (a < r < b): Q enc=v (r3−a3).

3 r

2

a

Inserting our value for v, we find 3 3 Q (r − a ) for a  r  b.

Dr= 4r2 (b3 − a3 )



2

0

0

Q , for r  b.

42 r This is plotted for appropriate values using the following Matlab routine: GS3 (r >b): Qenc = Q, Dr=

% % % %

M-File: MLP0236 Gauss's Law Problem thick spherical shell with even charge

2-30 % % Variables % % % % % % a % b r inner radius of sphere (m) outer rcm D radius of sphere (m) radial clc;clear; N distance from origin (m) radial maxr distance in cm electric flux Q density (nC/cm^3) number of data points max radius for plot (m) charge (nC)

% initialize variables a=.02;b=.04; Q=12; maxrad=0.10; N=100; bndya=round(N*a/maxrad); bndyb=round(N*b/maxrad); for i=1:bndya r(i)=i*maxrad/N; rcm(i)=r(i)*100; D(i)=0; end for i=bndya+1:bndyb

Fig. P2.36

r(i)=i*maxrad/N; rcm(i)=r(i)*100; D(i)=(Q/(4*pi*r(i)^2))*(r(i)^3-a^3)/(b^3-a^3); end for i=bndyb+1:N r(i)=i*maxrad/N; rcm(i)=r(i)*100; D(i)=Q/(4*pi*r(i)^2); end plot(rcm,D) xlabel('radial distance (cm)') ylabel('elect. flux density (nC/m^2)') grid on

2-31 P2.37: Given a coaxial cable with solid inner conductor of radius a, an outer conductor that goes from radius b to c, (so c > b > a), a charge +Q that is evenly distributed throughout a meter length of the inner conductor and a charge –Q that is evenly distributed throughout a meter length of the outer conductor, derive equations for the

electric flux density for all . You may orient the cable in any way you wish.

We conveniently center the cable on the z-axis. Then, for a Gaussian surface of length L,

Qe n c

=  D valid d for S all=Gaussian 2  surfaces. LD;

GS1: ( < a):  v =

Q ; ( m1  a)2(



2

L

0

0

0

)

QL Q1 =   v d v =  v  = a2d 2; ddz  a

QL2 Q = D =  for  a22L 2a2

GS2 (a <  < b): Q2 =QL; D =

QL

=

Q

GS3 (b <  < c): Q3 = Q +  vodv, where vo =

−Q



2

L

b

0

0

for a    b.

2L2

−Q

(1m) ( c− b22 )

(c2−  2)

= Q d ) dz  ( c − b2d (c2− b )

Q3 = Q +

2

(

2

2 2 Q c − ) for b  c. so D = 2  (c 2 −) b 2

GS4 ( > c): Qenc = 0, D = 0.

8. Divergence and the Point Form of Gauss’s Law P2.38: Determine the charge density at the point P(3.0m,4.0m,0.0) if the electric flux density is given as D = xyz az C/m.2  D (xyz)  D = z = = x y v=.  z z 3 )=(3)(4)=12 C/m. (3,4,0 v a +823a xyy xy C/m2 P2.39: Given D = 3a +2 , (a) determine the charge density at the x z point P(1,1,1). Find the total flux through the surface of a cube with 0.0 ≤ x ≤ 2.0m, 0.0

2-32 ≤ y ≤ 2.0m and 0.0 ≤ z ≤ 2.0m by evaluating (b) the left side of the divergence theorem and (c) the right side of the divergence theorem.  (a) D= y

C

(2xy)=2x, v(1,1,1)=2.

Ddv=

(b) DdS=

m3

+  +  +  +  +   = 85.3C  = 8x2y3a dxdya= 8 x2dx y3dy 2 top bottom left right

front

back

2

z

z

0

top

0

−  = 8x2y3a (− dxdya) = 85.3C z

z

bottom

 = 2 x y

y=0

left

= ay (− dxdza y) 0

 = 2 x y

y=2

=16C ay dxdza y

right

 = 3a dydza=12C x

x

front

 =3ax(−dydzax)=−12C

back

 Qe n c (c) D=

=  D dS =16C.

2 2 2   y(2xy)=2x; Ddv=2xdxdydz=16C. 0

0

0

P2.40: Suppose D = 6cos a C/m2 . (a) Determine the charge density at the point (3m, 90, -2m). Find the total flux through the surface of a quartered-cylinder defined by 0 ≤ 

 ≤ 4m, 0 ≤  ≤ 90, and -4m ≤ z ≤ 0 by evaluating (b) the left side of the divergence theorem and (c) the right side of the divergence theorem.

(a) (D)cylinder=

1 D   1 =

 

(6cos) 

C . (3,90,−2)=−6 m v

3

(b) DdS=  +  + +  =0

 =90

+  ,

top bottom outside

==−6sin.

2-33 note that the top, bottom and outside integrals yield zero since there is no component of D in the these dS directions.

a (− d  d z) a  = − 1 9 2 C  =6cos  =0

=0

 = 6cos 

 =90

=90

a ddza (  )= 0

= −192C. So, DdS

(c) dv=  d d  D = 6sin, −  dz



90

4

0

0

0

−4

= −192C. D 6dsin v =d− d dz

P2.41: Suppose D = r2sinra + sincosaC/m2. (a) Determine the charge density at

the point (1.0m, 45, 90). Find the total flux through the surface of a volume defined by

0.0 ≤ r ≤ 2.0 m, 0.0 ≤  ≤ 90., and 0.0 ≤  ≤ 180 by evaluating (b) the left side of the divergence theorem and (c) the right side of the divergence theorem.

The volume is that of a quartered-sphere, as indicated in Figure P2.41. (a)

 D=

1 D sin 1  r2Dr )+ =v, =4rsin  − (  r2r r r sin 

v(1,45 ,90 )= 1.83

C m3

(b) D dS =  +  +  +  ; note that  =0

 =180

 = sin   cos 

=0

=0

 = sin   cos 

=180

 =90

r =2

 = 0 sinceD = 0.

 =90

a (−rdrda −2  ) = C

=180

=−C 2 a rdrda  2



 =  sinr  ar r sin d  dr =r a sin2 d  2

r=2

2

4

0

Q = 4( 2 Summing these terms we have (c)

90

d  =(1− cos2 8  ) d  = 4  2 C

0

– 1)C = 35.5C.

0

2-34

sin  2



   Ddv=  4rsin − r r sindrdd 2

 2



2 

2



0

0

0

0

= 4 r3dr  sin2 d  d   d  = 4 2− 4 = 35.5C.  −rdrsin d sin 0

0

Fig. P2.41

9. Electric Potential 2 sP2.42: = 100AnC/m occupies thedensity x-z plane sheet of charge  at y = 0. (a) Find the work required to move a 2.0 nC charge from P(-5.0m, 10.m, 2.0m) to M(2.0m, 3.0m, 0.0). (b)Find VMP.

(a) W=−Q

M

EdL; so we need E for the sheet charge.

P

(100 x10C ) FV =5.65x103V  E= sa N = ay ay −12 m 2 o 2(8.854x10 F m) C −9

Notice that we are only concerned with movement in the y-direction. We then have: y=3   J  V  W = −2x 10−9 C  5.65x103 may  dyay  =79J  C y=10 V 

(79 J ) CV = kV W (b) VMP= = V = 40kV. 39.5 ; so MP x − 9C ) J Q (210

P2.43: A surface is defined by the function 2x + 4y2 –ln z = 12. Use the gradient equation to find a unit vector normal to the plane at the point (3.00m,2.00m,1.00m).

2-35 Let F = 2x + 4y2 − lnz =12, then F 1 ; F=2ax+8yay− ax = z az, F At (3,2,1), 2 2 2 2+16+1=16.16,  F = 2ax+16ay−az,F=

aN = 0.124ax +0.990ay −0.062az P2.44: For the following potential distributions, use the gradient equation to find E. (a) V = x+y2z (V)

(b) V = 2sin(V) (c) V = r sin cos (V).

2 ( a ) E = −  V =x−2yza − a y − yaz

1V  V V  (b) E = − V = −  a+ a+ a  = −2sina−cosa z z     (c)

V 1 V  1 V a+ sincosa−coscosa+sina E = − V = −  a+ r a r  r sin   = −  P2.45: A 100 nC point charge is located at the origin. (a) Determine the potential difference VBA between the point A(0.0,0.0,-6.0)m and point B(0.0,2.0,0.0)m. (b) How much work would be done to move a 1.0 nC charge from point A to point B against the electric field generated by the 100 nC point charge?

A

(a) VB A = −  E d L .

A

The potential difference is only a function of radial distance from the origin. Letting ra = 6m and rb = 2m, we then have rb B A = Q− 2ardrar= Q  1r −b1  =300V. V 4 4 o  Jra  ra  or CV (b) W=Q 2VBA=(10−9C)(300 =300nJ V)

P2.46: MATLAB: Suppose you have a pair of charges Q1(0.0, -5.0m, 0.0) = 1.0 nC and Q2(0.0, 5.0m, 0.0) = 2.0 nC. Write a MATLAB routine to calculate the potential VRO moving from the origin to the point R(5.0m, 0.0, 0.0). Your numerical integration will

involve choosing a step size L and finding the field at the center of the step. You should try several different step sizes to see how much this affects the solution.

2-36

MLP0246 % % M-File: % Modify ML0207 to calculate the potential % % % going from the origin (O) to the point % % difference % R(5,0,0) given a pair of point charges % % % and Q2(0,5,0)=2nC. % % Q1(0,-5,0)=1nC % % % The % approach will be to break up the distance % % % % % % % % from O to R into k sections. The total field E will be found at the center of each section (located clc at point P) and then dot(Ep,dLv) will give the clear potential drop across the kth section. Total potential is found by summing the potential drops.

Wentworth, 1/7/03 Variables: Q1,Q2 k dL dLv x(n) R1,R2 E1,E2 Etot V(n)

the point charges, in nC number of numerical integration steps magnitude of one step vector for a step x location at center of section at P vector from Q1,Q2 to P electric fields from Q1 & Q2 at P total electric field at P portion of dot(Etot,dL) at P

%clears the command window %clears variables % Initialize variables k=64; Q1=1; Q2=2; dL=5/k; dLv=dL*[1 0 0]; % Perform calculation for n=1:k x(n)=(n-1)*dL+dL/2; R1=[x(n) 5 0]; R2=[x(n) -5 0]; Rmag1=magvector(R1); Rmag2=magvector(R2); E1=9*Q1*R1/Rmag1^3;

2-37 E2=9*Q2*R2/Rmag2^3; Etot=E1+E2; V(n)=dot(Etot,dLv); end Vtot=sum(-V) Now running the program: Vtot = -1.5817 So VRO = -1.6 V.

P2.47: For an infinite length line of charge density L = 20 nC/m on the z-axis, find the potential difference VBA between point B(0, 2m, 0) and point A(0, 1m, 0).

B

VBA = −EdL ; E =

L

A

B

L a , dL = d a , 2  o  

− L so VBA = − ln(2)= −250V a da = 2 2  o   o A P2.48: Find the electric field at point P(0.0,0.0,8.0m) resulting from a surface charge

density s = 5.0 nC/m2 existing on the z = 0 plane from  = 2.0 m to  = 6.0 m. Assume V = 0 at a point an infinite distance from the origin.

(Method 1) For a ring of charge it was previously found that E=

aha L z

2  ( )a .2 + h 2 3 2

o

We can then break up our disk into differential rings (see Figure P2.48), each contributing dE as:  h d d E =S z, where we've used L=Sd.

3 a 2 o ( 2 + h2)2 So we then have  Sha d E=  z .

3 2 2o + h

(

2)2

u =  2 + h2, thendu = 2  d, and  we have This is easy to integrate if we let b 1  Shaz − 32  Shaz − 2 − Shaz u du= E= u=  2 4o 4o  2+ h2 a o

2-38 Solving, we arrive at  − h 1 1 − .

E = So   2  b2 + h2 a2 +2 h Upon inserting the appropriate values we find E = 48 V/m az.

Fig. P2.48 (method 2) Find an expression for potential and then evaluate the gradient at the point. dQ  So d  d  V = R = 2 + h2 , d Q = S  d  ,dso V=  4  h  2 + 4   R = , o

V=



S

b

2 oa

2 b d  = S  + h 2a 2o 2 + h2

 Sb2+h2− 2 o 

2

a2+h2. 

Now we let h = z andE = − V ; 1 1      2 2 − (a2+ z2 ) 2  az E = − S   (z b+ z2 ) z 2 o 

 −1 −1 z   z   1 2 z −1 = − S + z2 ) 2 2z az = − S  − b2+ 22 (a2 ) z (  az 2 2 o   2 o  b2 z + a22 z + 2 the values we find E = 48 V/m az.

Plugging in

2

P2.49: Suppose a 6.0 m diameter ring with charge density 5.0 nC/m lies in the x-y plane with the origin at its center. Determine the potential difference Vho between the point h(0.0,0.0,4.0)m and the origin. (Hint: first find an expression for E on the z-axis as a general function of z.)

For the ring of charge, replacing h with z, we have z  aa E= L z 2o a2+ z 3 ( 2)2

2-39 h

zdz a Vh o = −  E LdL = − 3 .

2 o (a2 + z2 ) 2 0 Letting u = a2 + z2, du = 2z dz, we have  a −3  La 1 Vho = − 4L o  u 2du= 2o . u Replacing u and evaluating from 0 to h,  a  1 1 Vho = L  a2+h2−  a 2o

= −36V = −113V. Fig. P2.49 10. Conductivity and Current P2.50: A columnular beam of electrons from 0 ≤  ≤ 1 mm has a charge density v =-0.1 cos(/2) nC/mm3 (where  is in mm) and a velocity of 6 x 106 m/sec in the +az direction. Find the current.

 ocos  , where o = -0.1 nC/mm.3 Then we’ll let u = uoz, Let’s let = a where uo = v  2  6x109 mm/s. Notice we convert the units to mm. Now,

 J = vu=ouocos az,

 2  and with dS =  d d az we then have a     2 I =  J d S =  o u od  c do s .  2 0 0 This becomes a



d, where A = 2ouo. cos  2

I=A

0

Now we can integrate by parts, or 2

= uv−  vdu, where u = A, du = Ad, udv

      sin  , and dv=cos   d.   2  2 We then have 2Aa  a 4A   a  I= sin   + cos  1.   2   2  2   − and then To evaluate, we first findA = 2(-0.1x10-9)(6x109)=3.77,  I = 2.40-1.53=0.87A. v=

I = 0.87A.

2-40 P2.51: Two spherical conductive shells of radius a and b (b > a) are separated by a

material with conductivity . Find an expression for the resistance between the two spheres.

First find E for a < r < b, assuming +Q at r = a and –Q at r = b. From Gauss’s law:

Q a 4 or2 r Now find Vab: E=

a

a

Q Vab=− E dL = − ardra r b   b 4 or2 a

a

−Q dr Q 1 Q  1 1 = = =  − .  b 4ob a 4 o b r2 4 o r Now can find I: Q 1 2 I = J dS= E dS =  4 o r 2 ar r sin d  dr  a

Q



2

Q .  o 0 o 0 Vab 1 1 1  = Finally, R =  a − b I 4    =4   sind 

d =

P2.52: The typical length of each piece of jumper wire on a student’s protoboard is 5.0 cm. Assuming AWG-20 (wire diameter 0.812 mm) copper wire, (a) determine the resistance for this length of wire. (b) Determine the power dissipated in the wire for 10. mA of current.

(a)R=

1 L

=

1

0.05m

=1.67m

Sm) ( 0.406 x10−3 m )   a 2 5.8x107( 2

so R = 1.7 m

(b) P=I2R=(10x10−3A)(1.7x10−3)=170nW 2

P2.53: A densely wrapped coil of AWG-22 (0.644 mm diameter) copper magnet wire is 150 m long. The wire has a very thin insulative sheath. Determine the resistance for this length of wire.

R=

150m 1 1 L = =7.94 7 2   a 5.8x10 S m (0.322x10− 3m)2

so R = 7.9

2-41

P2.54: Determine an expression for the power dissipated per unit length in coaxial cable

of inner radius a, outer radius b, and conductivity between the conductors  if a potential difference Vab is applied.

1  b  From Eqn(2.84) we have R= ln  2L a Now for a given potential difference Vab we have Vab 2 2LV2ab P 2   V 2 P= = , so L = ab.

R ln a

( b)

ln(b a)

P2.55: Find the resistance per unit length of a stainless steel pipe of inner radius 2.5 cm and outer radius 3.0 cm.

R=

1

  (b

L

a 22

,

−)

  1 1 R 1 1  = = 6  L  b x10 S .025− ( 2− a1.12 ) m (.0302 2 so R/L = 1.0 m/m so we have

   1.05m = ) m 2m  

P2.56: A nickel wire of diameter 5.0 mm is surrounded by a 0.50 mm thick layer of silver. What is the resistance per unit length for this wire? Assuming 1.0 m of this wire carries 1.0 A of current, determine the power dissipated in the nickel portion and in the silver portion of the wire.

We can treat this wire as two resistors in parallel. We have

RNi 1 1 = 7 L 1.5x10  (2.5 10 x

RAg L

=

m =3.4x10−3

−3)

2

 1 1 1.87 x10−3 = 2 2 ( 2.5x10−3 6.2x107  m ) − −3 )   (3 x10

Rtotal RNi RAg m = =1.2 L L L m To find the power dissipated, we first find the potential difference: V=IR total =1.2mV then PNi=

V2 V2 =0.42mW,PAg= =0.77mW RNi RAg

2-42

11. Dielectrics P2.57: A material has 12.0 V/m ax field intensity with permittivity 194.5 pF/m.

Determine the electric flux density.

(

D=E=194.51012F x −

nC 12V =2.3 a ( ) ) C m m FV m 2

x

P2.58: MATLAB: A 20 nC point charge at the origin is embedded in Teflon (r = 2.1).

Find and plot the magnitudes of the polarization vector, the electric field intensity and the electric flux density at a radial distance from 0.1 cm out to 10 cm.

We use the following equations: Q E= , P = eoE,D=r  oE 4 r or 2 % % % % % % % % % % % % %

M-File: MLP0258 Plot E, P and D vs distance r from a point charge Q at the origin with a dielectric.

Variables Q eo r Chi E D P

charge (C) free space permittivity (F/m) radial distance (m) electric susceptibility electric field intensity(V/m) electric flux density (C/m^2) polarization vector (C/m^2)

% initialize variables Q=20e-9; er=2.1; eo=8.854e-12; Chi=er-1; % perform calculations r=0.001:.001:0.100; rcm=r.*100; E=Q./(4*pi*r.^2); P=Chi*eo*E; D=er*eo*E; %

plot data

2-43 subplot(2,1,1) loglog(rcm,P,'-k',rcm,D,'-k') legend('P','D') ylabel('C/m^2') grid on subplot(2,1,2) loglog(rcm,E) ylabel('V/m') xlabel('radial distance (cm)') grid on

Fig. P2.58

P2.59: Suppose the force is very carefully measured between a pair of point charges separated by a dielectric material and is found to be 20 nN. The dielectric material is removed without changing the position of the point charges, and the force has increased to 100 nN. What is the relative permittivity of the dielectric?

F1=

Q12Q , F= 2 4 r oR2

Q12Q , 4 o R2

F2 100 =5 = r = F1 20

r = 10.2 isV = 12 xy2 (V). Find E, P and D. P2.60: The potential field in a material with  (12xy2)  (12xy2) V − 2ax − 24xyay m E=−V=− ax − ay = 12y x y nC D = roE= -1.1y2ax − 2.2xyay m2 9.2  e =  −r 1 = ax−2.00xyay P =eoE = (9.2)(8.854x10−12 )E =nC -9.8y2 m2 P2.61: In a mineral oil dielectric, with breakdown voltage of 15 MV/m, the potential function is V = x3 – 6x2 –3.1x (MV). Is the dielectric likely to breakdown, and if so, where?

2-44

E = −  V = ( − 3 xMV 2+12x+3.1)a m x

dE

=−6x+12,

d2E

=−6, so from 6x – 12 = 0 we find the maximum electric field dx2 dx occurs at x = 2m.

At x = 2m, we have E = -12+24+3.1 = 15.1 MV/m, exceeding the breakdown voltage.

12. Boundary Conditions

P2.62: For y < 0, r1 = 4.0 and E1 = 3ax + 6ayz+V/m.

4a At y = 0, s= 0.25 nC/m. If r2 = 5.0 for y > 0, find E2.

E1 = 3ax + 6ay + 4az V/m (a) EN1 = 6ay (b) ET1 = 3ax + 4az (d) DN1 = r1oEN1 = 24o ay

2

(g) E2 = 3ax + 20.7ay + 4az V/m (f) EN2 = DN2/5o = 20.7ay (c) ET2 = ET1 = 3ax + 4az (e) DN2 = 0.92 ay

(e) a21(D1−D2)=s, -ay(DN1−DN2)ay=s, DN2−DN1=s

nC  1 0 − 9 nC F  nC DN2=s+DN1=0.25 +24   =0.92 m2  3 6  m  m 2m2

P2.63: For z ≤ 0, r1 = 9.0 and for z > 0, r2 = 4.0. If E1 makes a 30 angle with a normal to the surface, what angle does E2 make with a normal to the surface?

Refer to Figure P2.63.

T2 ET1=E1sin1, ET2=E2sin2, and ET1=E also DN1=r1oE1cos1, DN2=r2oE2cos2, and DN1=DN2 (since s=0)

Therefore  − r2  ET1 E T2,and after routine math we find =tan1 tan1 =  2  r1  DN1 DN2 Using this formula we obtain for this problem 2 = 14°.

Fig. P2.63

2-45

P2.64: A plane defined by 3x + 2y + z = 6 separates two dielectrics. The first dielectric,

on the side of the plane containing the origin, has r1 = 3.0 and E1 = 4.0az V/m. The other dielectric has r2 = 6.0. Find E2.

We first use gradient to find a normal to the planar surface.

Let F = 3x + 2y + z – 6 = 0.  F = 3ax + 2ay+ az, and  F = 14, F + ay + 0.267az. so aN =  F = 0.802a x 0.534 Now we can work the boundary condition problem.

E1= 4a , zE N1=(E a1 + az .

Na =) 0.857 Nax + 0.570ay 0.285

E T1 = E − N1E 1

=0.857a − x − 0.570a y +3.715az , E T2 = ET1

a and D N2=DN1 = 0.855z, DN1 = 1r  o1 E No2.571ax+1.710a+ y D D E N2 = N 2 N=2 =0.429ax+0.285ay+0.143a z 6o Finallyr2o we have E V 2=ET2+EN2=−0.43ax−0.29ay+3.8a z . m P2.65: MATLAB: Consider a dielectric-dielectric charge free boundary at the plane z =

0. Construct a program that will allow the user to enter r1 (for z < 0), r2, and E1, and will then calculate E2. (Just for fun, you may want to have the program calculate the angles that E1 and E2 make with a normal to the surface).

% % M-File: % MLP0265 % % Given % E1 at boundary between a pair of clc dielectrics with no charge at boundary, clear calculate E2. Also calculates angles.

% enter variables disp('enter vector quantities in brackets,') disp('for example: [1 2 3]') er1=input('relative permittivity in material 1: '); er2=input('relative permittivity in material 2: '); a12=input('unit vector from mtrl 1 to mtrl 2: '); E1=input('electric field intensity vector in mtrl 1: '); %

perform calculations

2-46 En1=dot(E1,a12)*a12; Et1=E1-En1; Et2=Et1; Dn1=er1*En1; %ignores eo since it will factor out Dn2=Dn1; En2=Dn2/er2; E2=Et2+En2 % calculate the angles th1=atan(magvector(Et1)/magvector(En1)); th2=atan(magvector(Et2)/magvector(En2)); th1r=th1*180/pi th2r=th2*180/pi Now run the program: enter vector quantities in brackets, for example: [1 2 3] relative permittivity in material 1: 2 relative permittivity in material 2: 5 unit vector from mtrl 1 to mtrl 2: [0 0 1] electric field intensity vector in mtrl 1: [3 4 5] E2 =

3

4

2

th1r = 45 th2r = 68.1986 P2.66: A 1.0 cm diameter conductor is sheathed with a 0.50 cm thickness of Teflon and then a 2.0 cm (inner) diameter outer conductor. (a) Use Laplace’s equations to find an

expression for the potential as a function of  in the dielectric. (b) Find E as a function of . (c) What is the maximum potential difference that can be applied across this coaxial cable without breaking down the dielectric? (a) Since V is only a function of ,

2-47

 2Vcyl =

1   V   =0      

V =A  or V = Aln + B where A and B are constants.

Now we apply boundary conditions.

BC1: 0 = ln A b+ B, B= −Alnb, so 

   V =Aln   b  

 a BC2: Va = Aln  , A =  b

Va a

Fig. P2.66

 ln( b) , V =V

ln(b)

a

ab) ln(

or V = −1.443Va ln (100  ). V 1.443Va (b) E = −  V = − a = a



(c)



1.443Va = 288.5Va = Ebr= 60 x106 , .005 60 x106 =208kV,  (Vab) =210kV so V a = max 288.5 Emax =

P2.67: A 1.0 m long carbon pipe of inner diameter 3.0 cm and outer diameter 5.0 cm is cut in half lengthwise. Determine the resistance between the inner surface and the outer surface of one of the half sections of pipe.

One approach is to consider the resistance for the half-section of pipe is twice the resistance for a complete cylindrical section, given by Eqn. (2.84). But we’ll used the LaPlace equation approach instead.

Laplace:  2V cyl=

So: 

V



1

V

  =0; here we see V only depends on   

=A; V=Aln+B;

where A and B are constants.

Now apply boundary conditions.

BC1:

2-48

Vb = 0 = Alnb + B; B = − A lnb;   V = Aln    b BC2:  a Va = ; A = Alnb 

 ln( b) V =V a

Va a

;

ln(b)

ab) ln(

V V E = −  V = − a = − a a 

1

ln(b)

a

J= E

I

= J dS = −

 Va

 L V  1  d dz = 

ab) ln( ln ( a) b=5.4.

V R=a= I

Fig. P2.67

L

a

0

0

ln(ba)

L

P2.68: For a coaxial cable of inner conductor radius a and outer conductor radius b and a dielectric r in-between, assume a charge density  v =  o  is added in the dielectric

region. Use Poisson’s equation to derive an expression for V and E. Calculate s on each plate.

V  −    = o   

 

 2 V = −v = −

1 

so

 V   o   V  o V  = o  +A, where A is a constant.  = ;  d   =  d;             V  A A  = o+ ; dV=  o d +  d ; V = o  +Aln  + B, where B is a constant.  



  V = Va at  = a and V = 0 at = b Now apply boundary conditions: 

Applying the second one gives us:

V  = o b( ) − b ) + A l n (  .

Applying the first one:

2-49

Va =

Va +

o ( b− a) 

ab) ; A= ln( ab) ln(

 oa ( − b) + A 

Therefore,

V=

Va +

o ( b− a)     ln



b) .   + o(  −  b 

ab) ln(

  V    o   E = −  V = − a = −  K ln   +   − oba        b  K   E =  − − o  a    where



K=

− a) Va + o ( b ab) ln(

,

     a +o ( b − a)    V−   − o a . so E =   a   ln     b         ( − a)    a + o b  V−  − o  sa DN = ;s DNa =  E  =a =  a   = aln    b         o − a a + V ( b )   −   o  − =  sb DNb =E  =b =    a     bln   b   P2.69: For the parallel plate capacitor given in Figure 2.51, suppose a charge density z  v=  so i n  2d is added between the plates. Use Poisson’s equation to derive a new expression for V and

E. Calculate s on each plate.

2-50

o sin2d) (

z  2V ( z) − v − = =  z2  V ( z) − o  z dz= 2o d =  z 

 z 2d) +A (  sin2d) cos( d2 2d 2  z 2d) V(z)= + Az+ B sin( 2d) dz+ Adz   cos(z  2 = o

o

Now apply the boundary conditions:

Va = 0 = B; Vd = V ( z) =

22 od 2

2od 2 2

 d 2d) + Ad; A = sin(

Vd −

2od2 2 d

 +V 2od  z 2d) −  z sin( d 2  d

   2 od2 V  sin z E = −  V = − az =  − z   z  2



    V 2o d  −  2  z a +− ( 2d)   d  

(z )

V 2+  od z  od E =  −  cs o  2d − d a d 2 

d

z

z

  

od Vd 2od  s z=0 = −  − d2 + 2 = s .  od V at z = d, DN=E, so s z=d = − d + 2 = s . d  so at z = 0, D=E,  N

13. Capacitors P2.70: A parallel plate capacitor is constructed such that the dielectric can be easily removed. With the dielectric in place, the capacitance is 48 nF. With the dielectric removed, the capacitance drops to 12 nF. Determine the relative permittivity of the dielectric.

C1=

 roA; C2= oA d

d

;

C1 C =  r =4 8 =4.0 1 2 2

P2.71: A parallel plate capacitor with a 1.0 m2 surface area for each plate, a 2.0 mm plate separation, and a dielectric with relative permittivity of 1200 has a 12. V potential difference across the plates. (a) What is the minimum allowed dielectric strength for this capacitor? Calculate (b) the capacitance, and (c) the magnitude of the charge density on one of the plates.

(a) E=

12V kV kV = 6 ; (a)E brmin=6 m 0.002m m

2-51

 A=(1200)(8.854x10 F / m 1m2 ) ) ( 12 (b) C=ro =5.3F d 0.002m ; Q=CV=(5.3x10−6F)(12V) Q C −

(c) C=

V

=64C

FV

P2.72: A conical section of material extends from 2.0 cm ≤ r ≤ 9.0 cm for 0 ≤  ≤ 30 with r = 9.0 and  = 0.020 S/m. Conductive plates are placed at each radial end of the section. Determine the resistance and capacitance of the section.

1   V  V A = A ; V = − +B, where A and B are constants.  r 2 =0; r2 r 2  r  r  r r Boundary conditions: r = a, V = 0 and r = b, V = Vb  1 1  −    V = Vb a r  −     1 1V E = − V = − ar r a b − ro −Vbar Vbar ,D = 1 E= − 1 2  1 1   a b r2  −  r b    a 2

V=

Fig. P2.72

 sb =

;  − r oVb  1 1  − b2  a b

Q = sdS; dS= r 2 sin d  d 

2-52

 r oVb

30o

2

1  b sin d d 1  0  a−  b2 0  b = 1.73x10−12 Vb. Q  ; R= 1  C= b =1.7pF; RC= =2.3k Vb C 

Qb =

2

P2.73: An inhomogeneous dielectric fills a parallel plate capacitor of surface area 50.

cm2 and thickness 1.0 cm. You are given r = 3(1 + z), where z is measured from the bottom plate in cm. Determine the capacitance.

Place +Q at z = d and –Q at z = 0.

Q Q Q az, E=−  s= ,d D=− az S S  r oS d d = − -Q Q dz Vdo 0 EdL = −0 r oS az dzaz = o S0  r

evaluating the integral: d dz d dz 1 1 1 0  =r 0 3(1+ z) = 3 ln(1+z) 0 = 3 ln2 cm m F x10 3 ( 8.854 = Q −12 2 50 ( ) cm2) 3oS  m  = = C   = 19pF Vdo ln2 cm) (ln(2) 100cm P2.74: Given E = 5 xyax + 3zaz V/m, find the electrostatic potential energy stored in a volume defined by 0 ≤x ≤ 2 m, 0 ≤ y ≤ 1 m, and 0 ≤ z ≤ 1 m. Assume  = o.

1 1  WE =  o  E E d v = o25x2y2dxdydz 9+z 2 dxdydz  2 2 1 1 2 1 1  1  2 2 z2dz WE =  25x2dx  =125pJ y dy dz+9dxdy  2  0  0 0 0 0 0 o



P2.75: Suppose a coaxial capacitor with inner radius 1.0 cm, outer radius 2.0 cm and length 1.0 m is constructed with 2 different dielectrics. When oriented along the z-axis,

r for 0 ≤  ≤ 180 is 9.0, and for 180 ≤  ≤ 360 is 4.0. (a) Calculate the

capacitance. (b) If 9.0 V is applied across the conductors, determine the electrostatic potential energy stored in each dielectric for this capacitor. (a) a coaxial line,

2-53

2 L  r o ln b a But for only half the line,  C = L ro ln ba So  L r1 o = 361pF C1 = b ln a and  L r 2 o =161pF C2 = C=

( )

( )

( )

ln(b a)

So Fig. P2.75 CTOT = C1 + C2 = 522pF 1 1 (b) WE1 = C1V 2= 14.6nJ; WE 2 = C2V 2= 6.5nJ 2 2

3-1 Solutions for Chapter 3 Problems 1. Magnetic Fields and Cross Products P3.1: FindA xBfor the following: a. A = 2x a – 3ya+ 4z,a B = 5y a - 1za b. A = a + 2a+ 4z,a B = 2 a + 6za c. A = 2ra + 5a+ 1,a B = ra+ 3a (a)

ax AB= 2 0

a y za − 3 4 = 3(− 20) ax − 1 5

− x + 2a y +10az − )(ay−+ (10) 2 az = 17a

(b)

a a az A  B = 1 2 4 =12a − (6 − 8)a + (0 − 4)az =12a + 2a − 4az 2 0 6 (c)

ar AB= 2 1

a a 5 1 =15a − ( 6 −a 1+ ()0 − 5)a =15a − 5a − 5a r r     0 3

P3.2: If a parallelogram has a short side a, a long side b, and an interior angle  (the smaller of the two interior angles), the area of the parallelogram is given by area= absin .

Determine how you would use the cross product of a pair of vectors to find the area of a parallelogram defined by the points O(0,0,0), P(6,0,0), Q(8,12,0) and R(2,12,0).

(Assume dimensions in meters)

area=absin = AB = A B sin A = 6ax, B = 2ax+ 12ay A x B = 72az, Area = 72 m2

Fig. P3.2

3-2 P3.3: Given the vertices of a triangle P(1,2,0), Q(2,5,0) and R(0,4,7), find (a) the interior angles, (b) a unit vector normal to the surface containing the triangle and (c) the area of the triangle.

(a) PQ=1a x+3ay; PQ=3.16 PR = − 1a x+2ay+7az; PR=7.348

PQPR = 21ax−7a y + 5az PQPR = 22.69 A  B = A B sin ab

sinP =

22.69

;

( 3.16 ) ( 7.348 )

 = 78 P

QP

= −1ax − 3a y ; QP = 3.16

Fig. P3.3

= −2ax −1a y + 7a z; QR =7.348 QR QPQR= 22.69;  Q = 78; R=180 − P −  =Q24

PQ  = 0.93ax − 0.31ay + 0.22az PR PQ  1PR (c) area= PQPR=11.4m2 2 (b) an =

2. Biot-Savart’s Law P3.4: A segment of conductor on the z-axis extends from z = 0 to z = h. If this segment conducts current I in the +az direction, find H(0,y,0). Compare your answer to that of Example 3.2.

We use Eqn. (3.7) and change the limits: h

 I a  z  =  H 4  2 z2+  2 0 =

Iy(− ax ) 

h



h2+ y2   −Ihax  H= y 4 y h2+ y2  Note that if2the line of current is semi-

4

 

infinite (goes  from z = 0 to z = ∞), we’d have:

Fig. P3.4

3-3 H=

−I ax 4 y

P3.5: An infinite length line with 2.0 A current in the +ax direction exists at y = -3.0 m, z = 4.0 m. A second infinite length line with 3.0 A current in the +az direction exists at x = 0, y = 3.0 m. Find H(0, 0, 0).

This situation is shown in Figure P3.5a.

Ho = H1 + H2 Referring to the figure, R = 3ay – 4az, R = 5, aR = 0.6ay – 0.8az a = ax x aR = 0.80ay + 0.60az I H1 = 1 a

2

=

2A (0.80 y +) 0.60a z

2 y+38az5m

= 51a

Fig. P3.5a

a mA m

I2 a 2  3Aax = = 159ax mA m 2 3m Ho = 159ax + 51ay +38az mA/m H2 =

Or with 2 significant digits Ho = 160ax + 51ay +38az mA/m Fig. P3.5b

P3.6: A conductive loop in the shape of an equilateral triangle of side 8.0 cm is centered in the x-y plane. It carries 20.0 mA current clockwise when viewed from the +az direction. Find H(0, 0, 16cm).

The situation is illustrated in Figure P3.6a. The sketch in Figure P3.6b is used to find the +x-axis intercept for the triangle. By simple trigonometry we have: b =tan30;and since a = 4 cm we find b = 2.31 cm. a

3-4 Now for one segment we adapt Eqn. (3.7): +a

 Ia  z  H1 = 4   z2 +  2

−a

with a = RaR,  − x − 0.143az R = −2.31ax + 1 ,6 aRz = 16.17cm, a = − a y aR = 0.99a

( 20 x10 ) ( − 0.143a)  2(4)  A −0.99a H =   m) 4 16.17 ( x10−2  42 +16.172  −3

x

z

1

H1 = -4.7ax – 0.68az mA/m Now by symmetry the total H contains only theaz component: Htot = -2.0 mA/m az.

Fig. P3.6a

Fig. P3.6b

P3.7: A square conductive loop of side 10.0 cm is centered in the x-y plane. It carries 10.0 mA current clockwise when viewed from the +az direction. Find H(0, 0, 10cm).

We find H for one section of the square by adopting Eqn. (3.7): +a  z Ia  H=   4   z2 +  2 −a  Iaa H= 2   a2 +  2 With a = RaR, we have R = -5ax+ 10a,zR || = 11.18x10-2 m aR = -0.447ax + 0.894az, a = -ay x aR = -0.894 ax – 0.447 az

Fig. P3.7

3-5

H 1=

5) −) (0.894 ) (10x10− 3 ( ax− 0.447a z=−5.2ax−2.6amA m 2 11.18 ( x 10− 25 )+11.18 2

z

2

Now by symmetry the total H contains only the az component: HTOT = -10.4az mA/m

P3.8: A conductive loop on the x-y plane is bounded by  = 2.0 cm,  = 6.0 cm,  = 0 and  = 90. 1.0 A of current flows in the loop, going in the a direction on the  = 2.0 cm arm. Determine H at the origin.

By inspection of the figure, we see that only the arc portions of the loop contribute to H. From a ring example we have: 2 Ia2az H= 3  d 

4(h2 + a2 ) 2 0

For the  = a segment of the loop: /2 2 I Iaa z H a = 4a3 d  = 8aazd  = 0

At  = b: Hb=

Ib2az 0

4b3 /2

−I 8 az; b

Fig. P3.8

I  11  1 1 1  A − az=4.2a z So H TOT=  −  a z =  m 8a b 80.02 0.06

P3.9: MATLAB: How close do you have to be to the middle of a finite length of currentcarrying line before it appears infinite in length? Consider Hf(0, a, 0) is the field for the finite line of length 2h centered on the z-axis, and that Hi(0, a, 0) is the field for an infinite length line of current on the z-axis. In both cases consider current I in the +az direction. Plot Hf/Hi vs h/a.

Adapting Eqn. (3.7), for the finite length line we have:  Ia  h   Hf = 2   h2+  2    = a For the infinite length of line: Ia Hi = 2 

3-6 The ratio we wish to plot is: h Hf h a = = 2 Hi h2 + a2 h +1 a

( ) ( )

The MATLAB routine follows. % M-File: MLP0309 %Consider the field for a finite line of length 2h %oriented on z-axis with current I in +z direction. %The field is to be found a distance a away from %the current on the y axis (point (0,a,0)). %We want to compare this field with that of an infinite %length line of current. %Plot Hf/Hi versus h/a. We expect that as h/a grows large, %the line will appear more 'infinite' to an observation %point at (0,a,0). hova=0.01:.01:100; HfovHi=hova./sqrt(1+(hova).^2); semilogx(hova,HfovHi) xlabel('h/a') ylabel('Hf/Hi') grid on

Fig. P3.9a

Fig. P3.9b

3-7 P3.10: MATLAB: For the ring of current described in MATLAB 3.2, find following points (a) (0, 0, 1m), (b) (0, 2m, 0), and (c) (1m, 1m, 0).

H at the

%M-File: MLP0310 %Find the magnetic field intensity at any observation point %resulting from a ring of radius a and current I, %in the aphi direction centered in the x-y plane. df=1; %increment in degrees a=1; %ring radius in m I=1; %current in A Ro=input('vector location of observation point: '); for j=1:df:360; Fr=j*pi/180; Rs=[a*cos(Fr) a*sin(Fr) 0]; as=unitvector(Rs); dL=a*df*(pi/180)*cross([0 0 1],as); Rso=Ro-Rs; aso=unitvector(Rso); dH=I*cross(dL,aso)/(4*pi*(magvector(Rso))^2); dHx(j)=dH(1); dHy(j)=dH(2); dHz(j)=dH(3); end H=[sum(dHx) sum(dHy) sum(dHz)] Now to run the program: >> MLP0310 vector location of observation point: [0 0 1] H = -0.0000 -0.0000

0.1768

>> MLP0310 vector location of observation point: [0 2 0] H=

0

0 -0.0431

>> MLP0310 vector location of observation point: [1 1 0] H=

0

0 -0.1907

3-8 >> MLP0310 vector location of observation point: [0 1 1] H=

0.0000

0.0910

0.0768

>> So we see: (a) H = 0.18 az A/m (b) H = -0.043 az A/m (c) H = -0.19 az A/m (extra) H = 9.1ay + 7.7 az mA/m

P3.11: A solenoid has 200 turns, is 10.0 cm long, and has a radius of 1.0 cm. Assuming 1.0 A of current, determine the magnetic field intensity at the very center of the solenoid.

How does this compare with your solution if you make the assumption that 10 cm >> 1 cm?

Eqn. (3.10):   z NI  h− z = +  a H z 2  2h  ( h z2 + a2 − + a 2 z)  

  0.05 200 (1 A) 0.1− 0.05 A  +  H= az = 1961maz  2 0.052 + 0.012 2 ( 0.1 m) (0.1− 0.05) + 0.012 

Or H = 1960 A/m az

The approximate solution, assuming 10cm >> 1cm, is NI 200(1A) A az=2000 maz H = h az= 0.1m

P3.12: MATLAB: For the solenoid of the previous problem, plot the magnitude of the field position along the axis of the solenoid. Include the axis 2 cm beyond each versus M-File: MLP0312 end of the solenoid.

Plot H vs length thru center of a solenoid % % % %

3-9 clc clear % initialize variables %number of turns N=200; %height of solenoid h=0.10; %radius of solenoid a=0.01; %current I=1; %step change in z dz=0.001; z=-.02:dz:h+.02; zcm=z.*100; A1=(h-z)./sqrt((h-z).^2+a^2); A2=z./sqrt(z.^2+a^2); Atot=A1+A2; H=N*I.*Atot/(2*h); % generate plot plot(zcm,H) xlabel('z(cm)') ylabel('H (A/m)') grid on Fig. P3.12

P3.13: A 4.0 cm wide ribbon of current is centered about the y-axis on the x-y plane and has a surface current density K = 2 ay A/m. Determine the magnetic field intensity at the point (a) P(0, 0, 2cm), (b) Q(2cm, 2cm, 2cm). (a) Because of the symmetry (Figure P3.13a), we can use a modified Eqn. (3.14): Ky − d tan1 H=  ax  a 2 t a n 2−1   A = ax =1.57ax m   2  (b) Referring to Figure P3.13b; I dH = a;

2 Ra r =  a;whereR = (d − x)ax + aaz

Fig. P3.13a

3-10

(d − x)2 + 2a, and (d −)x a+xa az . So aR = 2 d − x + a 2 ( ) aax + ( d)a x − z a = a y R a = (d − x2)+ a2 + xaz− daz)dx K y (aax H= 2   ) (d − x 2 + a2 so =

Fig. P3.13b

This is separated into 3 integrals, each one solved via numerical integration, resulting in: H=1.1083ax +0.3032az – 1.1083az; or H = 1.1ax – 0.80az A/m 3. Ampere’s Circuit Law P3.14: A pair of infinite extent current sheets exists at z = -2.0 m and at z = +2.0 m. The top sheet has a uniform current density K = 3.0 ay A/m and the bottom one has K = -3.0 ay A/m. Find H at (a) (0,0,4m), (b) (0,0,0) and (c) (0,0,-4m).

We apply H=

1 Ka N , 2 1

1 (a) H= ( 3 ya) 2 (a -z 3+ a y )  a z = 0 1 2 (b) 2 1 + a(-3a H = 2 ( 3y )a z− y ) az A = −3 axm (c) H = 0

Fig. P3.14

P3.15: An infinite extent current sheet with K = 6.0 ay A/m exists at z = 0. A conductive loop of radius 1.0 m, in the y-z plane centered at z = 2.0 m, has zero magnetic field intensity measured at its center. Determine the magnitude of the current in the loop and show its direction with a sketch.

Htot = HS + HL

3-11

1 A 1 6a N = K  = ( m 2 2 For the loop, we use Eqn. (3.10): I H = 2a az HS

ay)az =3ax

where here −I I 2 )ax= HL = ( − a x 2a (sign is chosen opposite HS).

So, I/2 = 3 and I = 6A. Fig. P3.15

P3.16: Given the field H = 3y2 ax, find the current passing through a square in the x-y plane that has one corner at the origin and the opposite corner at (2, 2, 0).

Referring to Figure P3.6, we evaluate the circulation of H around the square path. b

c

d

a

a

b

c

d

H d L = Ie n c =  +  +  + 

 b

2 3(0) ax dxax = 0



a c

y2ax dyay = 0

3

0

2 dx= −24 3(2) ax dxax =12 2

b d

 c

a



=0

d we have Ienc = 24 A. The negative So Sign indicates current is going in the -az direction.

Fig. P3.16

P3.17: Given a 3.0 mm radius solid wire centered on the z-axis with an evenly distributed 2.0 amps of current in the +az direction, plot the magnetic field intensity H versus radial

distance from the z-axis over the range 0 ≤  ≤ 9 mm.

Figure P3.17 shows the situation along with the Amperian Paths. We have:

2H=I HdL=Ienc, where H=Ha and dL=da; enc

3-12 This will be true for each Amperian path. 

2

I I I 22 AP1: Ienc=JdS, J= az,I enc= 2  d  d  = 2 a a a 0 0 So: H =

I

 a for   a 2a2

AP2: Ienc = I, H =

I

a for   a

2

Fig. P3.17a % %

Fig. P3.17b

MLP0317 generate plot for ACL problem

a=3e-3; %radius of solid wire (m) I=2; %current (A) N=30; %number of data points to plot rmax=9e-3; %max radius for plot (m) dr=rmax/N; for i=1:round(a/dr) r(i)=i*dr; H(i)=(I/(2*pi*a^2))*r(i); end for i=round(a/dr)+1:N r(i)=i*dr; H(i)=I/(2*pi*r(i)); end plot(r,H) xlabel('rho(m)') ylabel('H (A/m)') grid on

3-13

P3.18: Given a 2.0 cm radius solid wire centered on the z-axis with a current density J =

3 A/cm2 az (for  in cm) plot the magnetic field intensity H versus radial distance from the z-axis over the range 0 ≤  ≤ 8 cm.

We’ll let a = 2 cm. enc where H = H L =  d  a;  2   H  = I a andd HdL=Ienc,

AP1 ( < a): I e n JcdS== 3 a d d  =  2  3 a z and H=  2a for   a

a3 3 AP2 ( > a): Ienc = 2a, so H= afor 

 a

The MATLAB plotting routine is as follows: % MLP0318 % generate plot for ACL problem a=2; %radius of solid wire (cm) N=40; %number of data points to plot rmax=8; %max radius for plot (cm) dr=rmax/N; for i=1:round(a/dr) r(i)=i*dr; H(i)=r(i)^2; end for i=round(a/dr)+1:N r(i)=i*dr; H(i)=a^3/r(i); end plot(r,H) xlabel('rho(cm)') ylabel('H (A/cm)') grid on

Fig. P3.18

P3.19: An infinitesimally thin metallic cylindrical shell of radius 4.0 cm is centered on the z-axis and carries an evenly distributed current of 10.0 mA in the +az direction. (a) Determine the value of the surface current density on the conductive shell and (b) plot H as a function of radial distance from the z-axis over the range 0 ≤  ≤ 12 cm.

3-14

(a) K s=

10mA mA ; so K=40az =39.8 m 2 a 2 (0.04m) I

=

mA m

(b) for  < a, H = 0. For  > a we have: I H= a 2  The MATLAB routine to generate the plot is as follows: % MLP0319 % a=4;generate plot for ACL problem N=120; %radius of solid wire (cm) %number of data points to plot I=10e-3; %current (A) rmax=12; %max plot radius(cm) dr=rmax/N; for i=1:round(a/dr) r(i)=i*dr; H(i)=0; end for i=round(a/dr)+1:N r(i)=i*dr; H(i)=100*I/(2*pi*r(i)); end plot(r,H) xlabel('rho(cm)') ylabel('H (A/m)') grid on

Fig. P3.19a

Fig. P3.19b

3-15 P3.20: A cylindrical pipe with a 1.0 cm wall thickness and an inner radius of 4.0 cm is centered on the z-axis and has an evenly distributed 3.0 amps of current in the +az direction. Plot the magnetic field intensity H versus radial distance from the z-axis over

the rangeAmperian 0 ≤  ≤ 10 cm.

For each Path:

2H=I HdL=Ienc, where H=Ha and dL=da; enc

Now, for  < a, Ienc = 0 so H = 0.

= J For a <  < bI , enc = JdS, where

I

az anddS =  d z d  a

2  ( b a2 )−

I enc =

I



2

a

I

( 2 − a) 2

2

0

Fig. P3.20a % %

 2− a2

= I  d, H − a2  =2 ( b− a2)  ( b 2) − d a 2 b2



Fig. P3.20b

MLP0320 generate plot for ACL problem

a=4;

%inner radius of pipe (cm)

b=5; %outer radius of pipe(cm) N=120; %number of data points to plot I=3; %current (A) rmax=10; %max radius for plot (cm) dr=rmax/N; aoverdr=a/dr boverdr=b/dr for i=1:round(a/dr) r(i)=i*dr; H(i)=0;

a

3-16 end for i=round(a/dr)+1:round(b/dr) r(i)=i*dr; num(i)=I*(r(i)^2-a^2); den(i)=2*pi*(b^2-a^2)*r(i); H(i)=100*num(i)/den(i); end for i=round(b/dr)+1:N r(i)=i*dr; H(i)=100*I/(2*pi*r(i)); end plot(r,H) xlabel('rho(cm)') ylabel('H (A/m)') grid on P3.21: An infinite length line carries current I in the +az direction on the z-axis, and this is surrounded by an infinite length cylindrical shell (centered about the z-axis) of radius a carrying the return current I in the –az direction as a surface current. Find expressions for the magnetic field intensity everywhere. If the current is 1.0 A and the radius a is 2.0 cm, plot the magnitude of H versus radial distance from the z-axis from 0.1 cm to 4 cm.

I

a and for a, H=0. HdL=Ienc; for 0<a, 2  H= 

Fig. P3.21a

Fig. P3.21b

The MATLAB routine used to generate Figure P3.21b is as follows: % MLP0321 % generate plot for ACL problem

3-17 clc clear a=2; %inner radius of cylinder(cm) N=80; %number of data points to plot %current (A) I=1; %max radius for plot (cm) rmax=4; dr=rmax/N; for i=1:40 r(i)=.1+(i-1)*dr; H(i)=100*I/(2*pi*r(i)); end for i=40:N r(i)=i*dr; H(i)=0; end plot(r,H) xlabel('rho(cm)') ylabel('H (A/m)') grid on P3.22: Consider a pair of collinear cylindrical shells centered on the z-axis. The inner shell has radius a and carries a sheet current totaling I amps in the +az direction while the outer shell of radius b carries the return current I in the –az direction. Find expressions for the magnetic field intensity everywhere. If a = 2cm, b = 4cm and I = 4A, plot the magnitude of H versus radial distance from the z-axis from 0 to 8 cm.

HdL=I enc; for 0<a, H = 0; for a<  b, H=

I

a ; 2  and for  b, H=0.

The MATLAB routine used to generate Figure P3.22b is as follows: % MLP0322 % generate plot for ACL problem a=2;

%inner radius of coax (cm)

b=4; N=160;

%outer radius of coax(cm) %number of data points to plot

3-18

Fig. P3.22a

Fig. P3.22b

I=4; rmax=8; %current (A) dr=rmax/N; %max radius for plot (cm) aoverdr=a/dr boverdr=b/dr

for i=1:round(a/dr) r(i)=i*dr; H(i)=0; end for i=round(a/dr)+1:round(b/dr) r(i)=i*dr; H(i)=100*I/(2*pi*r(i)); end for i=round(b/dr)+1:N r(i)=i*dr; H(i)=0; end plot(r,H) xlabel('rho(cm)') ylabel('H (A/m)') grid on P3.23: Consider the toroid in Figure 3.55 that is tightly wrapped with N turns of conductive wire. For an Amperian path with radius less than a, no current is enclosed and therefore the field is zero. Likewise, for radius greater than c, the net current

3-19 enclosed is zero and again the field is zero. Use Ampere’s Circuital Law to find an expression for the magnetic field at radius b, the center of the toroid.

enc; HdL=I Within the toroid, H = H a, so

d HdL=H  aa = H. 2 b  =b

Then, Ienc by the Amperian path is: Ienc = NI. NI  H = a.

2 b 

4. Curl and the Point Form of Ampere’s Circuital Law P3.24: Find A for the following fields: a. A = 3xy2/z ax

b. A = sin2 a –2z cos a c. A = r2sin ar + r/cos  a A A − 3xy 6xy = xa − az= (a) (3xy2z)a a − a z y z 1 z x

y

2

x

2

y

z

A A  A    sin2 a −  2zcos a ) = −a +  a+  (  )A − (b)   (   az z z    

32zcos 1 =  2 cos a + 0a − az −  2sincos az   = − 2 cos a 3 zcos + 2sincos  az (c)

  r a   r2 sin ar + cos   −1   A 1  1 Ar  1  ( r A  )Ar  −  = a +  ar + a  r  r r sin      rsin    2  −1 ) c 1  r  s o (  r − ) 1  2 =  r sin a ar + ( a − rsin r   r  cos  r 

=

  2 −sin − rcos  a  ar +   sn i  cos2 cos

3-20 P3.25: FindJ at (3m, 60, 4m) forH = (z/sin)a – ( 2/cos)az A/m.

 A  A  −1 A 1 A z a +  − z a +   az  z      

 H =

 1 2  z cos + a +  az  a +  cos sin2  sin cos   Now find J by evaluating  H at the given point: A J = − 1 0 a  + 1 3z a  + 0 . 8 9 a m2 =

−sin 2

P3.26: Suppose H = y2a 2 A/m. x +xay a. Calculate

→B→C→D→A HdL around theApath

, where A(2m,0,0),

B(2m,4m,0), C(0,4m,0) and D(0,0,0). b. Divide this HdLby the area S (2m*4m = 8m2). c. Evaluate   H at the center point. d. Comment on your results for (b) & (c).

B

C

A D

+ =  + HdL=Ienc (a) Referring to the figure, we evaluate C A+ B D B

 A C

4

= x2 0 0

 = y B D

2 0

A C

42

D

0

dx= −32C

2

= x2

 

=16C dy= 4(4) x=2

y=4

dy= 0 x=0

=  y 2d y = 0 y=0

= − 1C6 So we haveHdL

Fig. P3.26

(b) dividing by S = 8m2, we have -2 C/m2 (c) Evaluating the curl of H:  A A    H =  y − x a z=(2x−2y)az, and at the center point (x = 1 and y = 2) we ha  x y   C   H center= − 2 2 az m (d) In this particular case,  H = HdLS,even though S is of appreciable size.

3-21 P3.27: For the coaxial cable example 3.8, we found:

for a,

H=

for a<b,

H=

for b<c,

H=

I a , 2 a 2 I 2 

a , I

 c2 −  2  a , − b2   

2 (c2

)

and for c<, H =0. a. Evaluate the curl in all 4 regions. b. Calculate the current density in the conductive regions by dividing the current by the area. Are these results the same as what you found in (a)?

1   I  2 

I  az = az for   a     20for a a2a 2 1  I   H =  b  az =   2  2 1   I c( −  2)  −I  a   H =  z= az for b<c     2(  c )2 − b2( c 2 ) −b2   H=0 for c I a= a S z  a2 az −I for b<  c, S =  ( c− b22), J= az

(a)   H =



(b) for a, S=2,) J=(I

 (c−b2) 2

Comment: H=J is confirmed.

P3.28: Suppose you have the field H = r cos  a A/m. Now consider the cone specified by  = /4, with a height a as shown in Figure 3.56. The circular top of the cone has a radius a. a. Evaluate the right side of Stoke’s theorem through the dS = dSa surface. b. Evaluate the left side of Stoke’s theorem by integrating around the loop.

1   1  ( raH  )  (a) H= rsin siH ( n  ) r a − r r   ar derivative:

(cos2 − sin2 )  1    (rsincos)=r(cos2−sin2);  ( s i n H  ) a r = a r  sin r sin   

3-22

Fig. P3.28 a derivative: cos ) −1  ( 2 a  = − 2 a c o s r r cos2 −sin2 ) ( So,   H = ar −2cos a  sin Now we must integrate this over the surface: a 2 2 (cos   − sin  )   r −   HdS= 2cos rsindrd (−a )  a a   sin     2a 2 = cos ) =  rdr d = 2 a2 2rsin cosdrd= 2 ( sin 4

0

0

2 (b) HdL=rr=a 2 cosaada=acos

2



=  d= 2a 

2

4 0

Clearly in this case the circulation of H is the easiest approach.

5. Magnetic Flux Density P3.29: An infinite length line of 3.0 A current in the +ay direction lies on the y-axis. Find the magnetic flux density at P(7.0m,0,0) in (a) Teslas, (b) Wb/m2, and (c) Gauss.

H=

3A mA I a= (− az ) = − 6a z8 m 2 (7m) 2 

H   A  Wb Wb  86 x10−9 − x10−9  = − Taz B = oH = 4 x10−7   68 m2 az = 86 − a  z m  m  HA  G G 10, 000 x10−6 x10−9 86 860 = − B = ( − Taz ) az T

3-23 P3.30: Suppose an infinite extent sheet of current with K = 12ax A/m lies on the x-y plane at z = 0. Find B for any point above the sheet. Find the magnetic flux passing through a 2m2 area in the x-z plane for z > 0.

1 H = K  a N; 2



(4 x10−7Hm)  A 12a 

 Wb − x10−6 m2a y  x m az  = 7.54 2 2  This is valid at any point above the sheet.

B=

o

K  aN =

 

y(2m2(−ay))=15Wb

Wb 

Now, = BdS=BS=-7.54x10-6 a m2



P3.31: An infinite length coaxial cable exists along the z-axis, with an inner shell of radius a carrying current I in the +az direction and outer shell of radius b carrying the return current. Find the magnetic flux passing through an area of length h along the zaxis bounded by radius between a and b.

H

I = a , B

2 BdS

For a <  < b,  = 

=

=

 I = a, o 2

b  oI a   oI d dza = ln( )a h   2 2

oIh  b  ln Wb 2    a

6. Magnetic Forces P3.32: A 1.0 nC charge with velocity 100. m/sec in the y direction enters a region where the electric field intensity is 100. V/m za and the magnetic flux density is 5.0 Wb/m2 ax.

Determine the force vector acting on the charge. m Wb Wb F = q +( uE B); uB=100 a y  5 2 =−500 a s sm max  WbVs mN V  F =10−9C 100 az+−500 =−400nNa z az m smWb VC 

P3.33: A 10. nC charge with velocity 100. m/sec in the z direction enters a region where 2 the electric field intensity is 800. V/m ax and the magnetic flux density 12.0 Wb/m ay.

Determine the force vector acting on the charge.

3-24



V m

m s

Wb  ma y 

F = q ( E + u  B ) = 1 0 x az12 1 0 −2 9 C= −4Na 8 0x 0 ax+100 

P3.34: A 10. nC charged particle has a velocity v = 3.0ax + 4.0ay + 5.0az m/sec as it enters a magnetic field B = 1000. Ty a(recall that a tesla T = Wb/m2 ). Calculate the force vector on the charge.

Wb  1000 F = q ( u  B )x+4ay+5az) = 1 0m x 1 0 −a 9 C ( 3 a s m  

The cross-product: ax a y az    4 5  = −5000ax + 3000az 3  0 1000 0    Evaluating we find: F = -50ax + 30az N

2

y



P3.35: What electric field is required so that the velocity of the charged particle in the previous problem remains constant?

dv F = m a =m =0 (constant velocity) dt 0; F = qE ( u + B) = 

E = u−B= 3a(

x

Wb V V m − maz + 5000max + 4ay + 5az )s  1000m2a y = 3000

kV E = 5ax− 3az m

-31kg and charge q = -1.6 x 10-19 C) has a P3.36: An electron (with rest mass e= M 9.11x10 velocity of 1.0 km/sec as it enters a 1.0 nT magnetic field. The field is oriented normal to the velocity of the electron. Determine the magnitude of the acceleration on the electron caused by its encounter with the magnetic field.

F=ma=q(uB); q a = m( u  B ) ;

3-25

a=

quB = m

) m2) (−1.6x10−19C(1000m)(10−9Wb s =175x103m kg) (9.11x10−31

s2

P3.37: Suppose you have a surface current K = 20. ax A/m along the z = 0 plane. About a meter or so above this plane, a 5.0 nC charged particle is moving along with velocity v = -10.ax m/sec. Determine the force vector on this particle.

1 1 A KaN = 2 0 a x  a z = − 1 0 a y 2 2 m o B =  H = ) −y 1m2 0=( −440 xxWb 1y1 00−m2 −7 a7 W ab m   − F = q u  b (5x109C) = − 10ax s  −40 x107 ay m 2  =0.63pNa z  − H=

Wb

P3.38: A meter or so above the surface current of the previous problem there is an infinite length line conducting 1.0 A of current in the –ax direction. Determine the force per unit length acting on this line of current.

0 Wb − 2 F12 =  2I d ay 2 L1  B 2 = I  d x a x  −40x107 m L = I( F12 2 − L) ( − 4 0z ); x 1 0 − 7 a

N F12 =12.6 az = 40x10−7a z m L P3.39: Recall that the gravitational force on a mass mFis= mg, where, at the earth’s surface, g = 9.8 m/s2 (-az). A line of 2.0 A current with 100. g mass per meter length is horizontal with the earth’s surface and is directed from west to east. What magnitude and direction of uniform magnetic flux density would be required to levitate this line?

 Ns2 kg   m F g=mg; for 1 m Fg=(100g)9.8 2( −az)  = −0.98Naz   s   kgm1000g  1m

+ Naz F = IdL  B = − gF = 0.98 0

By inspection,B = Bo(-ax)

3-26 L

Idya  Bo −( ax ) = ILBoaz y

o

Wb = (2A) (m1 o) ) Bo 2  = 2 B(N  m  The unit conversion to arrive at Newtons is as follows: Am WbVsWJNm = N m2 Wb VA Ws J , and So we have Bo = 0.490 Wb/m2 B = 0.490 Wb/m2 (-ax) (directed north) Fig. P3.39

P3.40: Suppose you have a pair of parallel lines each with a mass per unit length of 0.10 kg/m. One line sits on the ground and conducts 200. A in the +ax direction, and the other one, 1.0 cm above the first (and parallel), has sufficient current to levitate. Determine the current and its direction for line 2.

4x10− 7Hm 200A 2az=4x10−I2a F12 II 3 o 12 z = y= Here we will use I a 2 0.01m 2 y L Ns2 F mg N ms2)kgm= 0.98m = L = ( 0.10 kgm)(9.8 L So solving for I2: 0.98 =245A in the -a x d irection.

I 2= 410 x −3

P3.41: In Figure 3.57, a 2.0 A line of current is shown on the z-axis with the current in the +az direction. A current loop exists on the x-y plane (z = 0) that has 4 wires (labeled 1 through 4) and carries 1.0 mA as shown. Find the force on each arm and the total force acting on the loop from the field of the 2.0 A line.

I

o1 F12=I2dL2B; a 1 1 2B=  A→B:dL

2=5da, aa=0

C→D:aa=0 3

B → C : F12= I da 2 5

o12 I II  3   2o1 a  = 2 ln  a z  5 

3-27

Fig. P3.41

12

F =

4  10 x −H7m  3 ln a z = − 2 0 4 p N a z (2A)(10−3 A) 2  5

So for B to C: F12 = -0.20 nN az Likewise, from D to A: F12 = +0.20 nN az P3.42: MATLAB: Modify MATLAB 3.4 to find the differential force acting from each individual differential segment on the loop. Plot this force against the phi location of the segment.

%MLP0342 %modify ML0304 to find dF acting from the field %

of each segment of current; plot vs phi

clear clc %current in A I=1; %loop radius, in m a=1; %free space permeability mu=pi*(4e-7); az=[0 0 1]; %unit vector in z direction DL1=a*2*(pi/180)*[0 1 0]; %Assume 2 degree increments %DL1 is the test element vector %F is the angle phi in radians %xi,yi is location of ith element on the loop %Ai & ai = vector and unit vector from origin % to xi,yi %DLi is the ith element vector %Ri1 & ri1 = vector and unit vector from ith % point to test point for i=1:179

3-28 phi(i)=i*2; F=2*i*pi/180; xi=a*cos(F); yi=a*sin(F); Ai= [xi yi 0]; ai=unitvector(Ai); DLi=(pi*a/90)*cross(az,ai); Ri1=[a-xi -yi 0]; ri1=unitvector(Ri1); num=mu*I*cross(DLi,ri1); den=4*pi*(magvector(Ri1)^2); B=num/den; dFvect=I*cross(DL1,B); dF(i)=dFvect(1);

end plot(phi,dF) xlabel('angle in degrees') ylabel('the differential force, N')

Fig. P3.42

P3.43: MATLAB: Consider a circular conducting loop of radius 4.0 cm in the y-z plane centered at (0,6cm,0). The loop conducts 1.0 mA current clockwise as viewed from the +x-axis. An infinite length line on the z-axis conducts 10. A current in the +az direction.

Find the net force on the loop.

The following MATLAB routine shows the force as a function of radial position around the loop. Notice that while there is a net force in the -y direction, the forces in the zdirection cancel.

3-29

% % %

MLP0343 find total force and torque on a loop of current next to a line of current

% % % % % % % % % % % % % %

variables I1,I2 current in the line and loop (A) yo center of loop on y axis (m) uo free space permeability (H/m) N number of segements on loop a loop radius (m) dalpha differential loop element dL length of differential section DL diff section vector B1 I1's mag flux vector (Wb/m^2) Rv vector from center of loop to the diff segment unit vector for Rv ar the location of the diff segment y,z

clc clear % initialize variables I1=10; I2=1e-3; yo=.06; uo=pi*4e-7; N=180; a=0.04; dalpha=360/N; dL=a*dalpha*pi/180; ax=[1 0 0];

% perform calculations for i=1:N dalpha=360/N;

Fig. P3.43

alpha=(i-1)*dalpha; phi(i)=alpha; z=a*sin(alpha*pi/180); y=yo+a*cos(alpha*pi/180); B1=-(uo*I1/(2*pi*y))*[1 0 0];

3-30 Rv=[0 y-.06 z]; ar=unitvector(Rv); aL=cross(ar,ax); DL=dL*aL; dF=cross(I2*DL,B1); dFx(i)=dF(1); dFy(i)=dF(2); dFz(i)=dF(3); end plot(phi,dFy,phi,dFz,'--k') legend('dFy','dFz') Fnet=sum(dFy) Running the program we get: Fnet = -4.2932e-009 >> So Fnet = -4.3 nN ay P3.44: MATLAB: A square loop of 1.0 A current of side 4.0 cm is centered on the x-y plane. Assume 1 mm diameter wire, and estimate the force vector on one arm resulting from the field of the other 3 arms. % MLP0344 V2 % % Square loop of current is centered on x-y plane.

Viewed % %from the +z axis, let current go clockwise. We want to the find the force on the arem at x = +2 cm resulting from % current in arms at y = -2 cm, x = -2 cm and y = +2 cm. %

Wentworth, 12/3/03

% % % % % % % %

Variables a side length (m) b wire radius (m) I current in loop (A) uo free space permeability (H/m) N number of segments for each arm xi,yi location of test arm segment (at x = +2 cm) xj,yj location of source arm segment (at y = -2 cm)

3-31 % % % % % %

xk,yk location xL,yL of source arm segment (at x = -2 cm) dLi dLj,location dLk, of source arm segment (at y = +2 cm) dLL differential test segment vector diff vectors on sources Rji vector from source point j to test point i aji unit vector of Rji

clc;clear; a=0.04; b=.0005; I=1; uo=pi*4e-7; N=80; for i=1:N xi=(a/2)+b; yi=-(a/2)+(i-0.5)*a/N; ypos(i)=yi; dLi=(a/N)*[0 -1 0]; for j=1:N xj=-(a/2)+(j-0.5)*a/N; yj=-a/2-b; dLj=(a/N)*[-1 0 0]; Rji=[xi-xj yi-yj 0]; aji=unitvector(Rji); num=I*CROSS(dLj,aji); den=4*pi*(magvector(Rji))^2; H=num/den; dHj(j)=H(3); end for k=1:N yk=(-a/2)+(k-0.5)*a/N; xk=-(a/2)-b; dLk=(a/N)*[0 1 0]; Rki=[xi-xk yi-yk 0]; aki=unitvector(Rki); num=I*CROSS(dLk,aki); den=4*pi*(magvector(Rki))^2; H=num/den; dHk(k)=H(3); end for L=1:N

3-32 xL=(-a/2)+(L-0.5)*a/N; yL=(a/2)+b; dLL=(a/N)*[1 0 0]; RLi=[xi-xL yi-yL 0]; aLi=unitvector(RLi); num=I*CROSS(dLL,aLi); den=4*pi*(magvector(RLi))^2; H=num/den; dHL(L)=H(3); end H=sum(dHj)+sum(dHk)+sum(dHL); B=uo*H*[0 0 1]; F=I*CROSS(dLi,B); dF(i)=F(1); end Ftot=sum(dF) plot(ypos,dF) Running the program: Ftot = 7.4448e-007 Fig. P3.44

So Ftot = 740 nN

P3.45: A current sheet K = 100ax A/m exists at z = 2.0 cm. A 2.0 cm diameter loop centered in the x-y plane at z = 0 conducts 1.0 mA current in the +a direction. Find the torque on this loop.

 = m  B;

m =IS aN =

−9 a − 3 a = 314) x10 ( A1)( 0(0.01m) z Am2 2

z

A 1 A K  aN =  1ax0 0  ( z − ) = a50ay m; m 2 2 Wb B = 50oa y 2 ; m  m =  B = -20pNma x B = oH ; H =

1

3-33 P3.46: 10 turns of insulated wire in a 4.0 cm diameter coil are centered in the x-y plane. Each strand of the coil conducts 2.0 A of current in the a direction. (a) What is the 2 (b) what is the torque on the coil? field B = 6.0ax + 3.0ay + 6.0a Wb/m, magnetic dipole moment this coil? Now suppose this coil is in a uniform z of magnetic

x=(10)(2A)((0.02m))=25.1mAm2a z 2

(a) m=NISa (b) =mB=25.1mAa

z  ( 6 a x +Wb 3=0.151a a y +y−0.075ax 6 a z )Nm mNm   = − 7x+151ay 5a

m2

P3.47: A square conducting loop of side 2.0 cm is free to rotate about one side that is fixed on the z-axis. There is 1.0 A current in the loop, flowing in the –az direction on the

fixed side. A uniform B-field exists such that when the loop is positioned at  = 90, no torque acts on the loop, and when the loop is positioned at  = 180 a maximum torque of 8.0 N-m az occurs. Determine the magnetic flux density.

At  = 90°, m=ISa N=0.0004Am2a. x Also, since  = m  B = 0, B is in direction of m, and therefore B = ±Boax. −6 At  = 180°, m=ISa N = 0.0004Am2 ay, and =mB=8x10azNm. mBo Therefore, B = -Baand = 8x10,-6 so ox 8x10−6 mWb mWb , and B=−20 Bo= 0.0004=20 a.x m2 m2

7. Magnetic Materials P3.48: A solid nickel wire of diameter 2.0 mm evenly conducts 1.0 amp of current.

Determine the magnitude of the magnetic flux density B as a function of radial distance from the center of the wire. Plot to a radius of 2 mm.

J =

I

az =

1A

kA az = 31.8m2az

 1(x10−3m) I = JdS HdL= a2

2

enc

I

I

= d 2 2 H =   a 2  d  a2 I  roI for   a H = a a; B= 2a2 2a2

3-34 for   a H = % %

I

a; B =

2

 oI a 2

MLP0348 generate plot for ACL problem

a=2e-3; %radius of solid wire (m) I=1; %current (A) N=30; %number of data points to plot rmax=4e-3; %max radius for plot (m) dr=rmax/N; uo=pi*4e-7; ur=600; for i=1:round(a/dr) r(i)=i*dr; B(i)=(ur*uo*I/(2*pi*a^2))*r(i); end for i=round(a/dr)+1:N r(i)=i*dr; B(i)=uo*I/(2*pi*r(i)); End rmm=r*1000; plot(rmm,B) xlabel('rho(cm)') ylabel('B (Wb/m^2)') grid on Fig. P3.48 8. Boundary Conditions P3.49: A planar interface separates two magnetic media. The magnetic field in media 1

(with r1) makes an angle 1 with a normal to the interface. (a) Find an equation for 2, the angle the field in media 2 (that has r2) makes with a normal to the interface, in terms of 1 and the relative permeabilities in the two media. (b) Suppose media 1 is nickel and media 2 is air, and that the magnetic field in the nickel makes an 89 angle with a normal

to the surface. Find 2.

 H 1a H 1 = H1NNa + H1TaT; H2TTa = H1TaT; B1NNa =  r1o N N  B2NaN = B1N aN =  r1o 1NHN a =  r2o H2NaN ;  H2 N =r1 H1N  r2

3-35

tan 2=

H 2T H 1T = ; H 2N  r1 H 1N

 r2

 r2 H 1T  r2 = tan 1  r1 H1N  r 1   1  2= tan− r 2 tan1   r1  1  = tan 89 5.5 2 tan−1 600   = tan  2=

Fig. P3.49

P3.50: MATLAB: Suppose the z = 0 plane separates two magnetic media, and that no

surface current exists at the interface. Construct a program that prompts the user for r1 (for z < 0), r2 (for z > 0), and one of the fields, either H1 or H2. The program is to calculate the unknown H. Verify the program using Example 3.11.

% % M-File: % MLP0350 % % % clc Given H1 at boundary between a pair of clear materials with no surface current at boundary, calculate H2.

% enter variables disp('enter vectors quantities in brackets,') disp('for example: [1 2 3]') ur1=input('relative permeability in material 1: '); ur2=input('relative permeability in material 2: '); a12=input('unit vector from mtrl 1 to mtrl 2: '); F=input('material where field is known (1 or 2): '); Ha=input('known magnetic field intensity vector: '); if F==1 ura=ur1; urb=ur2; a=a12; else ura=ur2; urb=ur1;

3-36 a=-a12; end % perform calculations Hna=dot(Ha,a)*a; Hta=Ha-Hna; Htb=Hta; %ignores uo since it will factor out Bna=ura*Hna; Bnb=Bna; Hnb=Bnb/urb; display('The magnetic field in the other medium is: ') Hb=Htb+Hnb Now run the program (for Example 3.11): enter vectors quantities in brackets, for example: [1 2 3] relative permeability in material 1: 6000 relative permeability in material 2: 3000 unit vector from mtrl 1 to mtrl 2: [0 0 1] material where field is known (1 or 2): 1 known magnetic field intensity vector: [6 2 3] ans = The magnetic field in the other medium is: Hb = 6 2 6 For a second test, run the program for problem P3.52(a). enter vectors quantities in brackets, for example: [1 2 3] relative permeability in material 1: 4 relative permeability in material 2: 1 unit vector from mtrl 1 to mtrl 2: [0 0 -1] material where field is known (1 or 2): 1 known magnetic field intensity vector: [3 0 4] ans = The magnetic field in the other medium is: Hb = 3 0 16

3-37

P3.51: The plane y = 0 separates two magnetic media. Media 1 (y < 0) has r1 = 3.0 and media 2 (y > 0) has r2 = 9.0. A sheet current K = (1/o) ax A/m exists at the interface, and B1= 4.0a y+ 6.0a Wb/m2 . (a) Find B2. (b) What angles do B1 and B2 make with a z

normal to the surface?

(a) B N1 = 4ay N2 = BN1 = 4ay (b) B (c) BT1 = 6az

(d) H T 1 (e) H T2=

2 B1 =T = a z

 r1 o 3

o

az (see below)

o (f) B T 2 =  r  o H T 2 = 2 7 a z (g) B

Wb m2 2 = 4ay + 27az

Fig. P3.51

Now for step (e):

a2 1  ( H 1 − H 2 ) = K ; 1

ax; H T2−HT1= − ( T1−HT2)ax= H o



1

− a yo ax( H T 1 a z − H T 2 a z ) = 1 ; HT2=

o

1

+

2

=

3

o o o

 BT1   BT2   =82 tan−1  =  56;  2 =  BN1   BN2

Angles: 1=tan−1

P3.52: Above the x-y plane (z > 0), there exists a magnetic material with r1 = 4.0 and a field H1 = 3.0ax + 4.0az A/m. Below the plane (z < 0) is free space. (a) Find H2, assuming the boundary is free of surface current. What angle does H2 make with a normal to the surface? (b) Find H2, assuming the boundary has a surface current K = 5.0 ax A/m. (a)

B N2 (1) H N1=4az, (2) BN1=16a =16az oz , (3) B N = B N =16 oaz ,(4) H N2= 2

1

A (5) H T 1 = 3ax , (6) H T 2 = H T 1 = 3ax , (7) H 2 = 3ax +16az m

o

3-38

 HT 2   =10.6  HN2 

2= tan−1 

(b) Now step (6) becomesa21  ( H 1

=HK , shere ) a21= az. − 2

Let’s let H2=Aax+Ba y +16az , then a21  (3ax + 4az− Aax − Ba y −16az )= 5ax Solve for A and B:  ax a y az   0   1  = Bax + ( A− 3)a y = 5ax; so A = 3 and B = 5 0  3 − A −B −12 

Finally, H

A 2=3ax+5ay+16a z m.

P3.53: The x-z plane separates magnetic material with r1 = 2.0 (for y < 0) from magnetic material with r2 = 4.0 (for y > 0). In medium 1, there is a field H1 = 2.0ax + 4.0ay + 6.0az A/m. Find H2 assuming the boundary has a surface current K = 2.0ax – 2.0az A/m.

8 (1)H1N=4ay,(2)B1N=8oay,(3)B2N=B1N=8oay,(4)H2N= ay=2ay o 4 = K ,H (5)H1T= 2ax + 6az,(6)a21  ( H 1 − let2HT2)= Hxax + Hz az , o − a

so − ay  ( 2 Hx)ax + (6 − Hz)az ) = 2 x − 2az , ay + ax az 1 − 0 0 = (Hz− 6)a (2 − Hx)a = 2a − 2az, soHz= 8, H x= 4 x z x 0 2 −Hx 6 − Hz (7)H2T=4ax+8az, H2=4ax+2ay+8az

A m

P3.54: An infinite length line of 2 A current in the +az direction exists on the z-axis.

This is surrounded by air for  ≤ 50 cm, at which point the magnetic media has r2 = 9.0 for  > 50 cm. If the field in media 2 at  = 1.0 m is H = 5.0a A/m, find the sheet current density vector at  = 50. cm, if any.

Method 1:

From just the line of current we would have H= I1 a  = 1 a  .

1

21 ()

3-39 Now, since H TOT = 5a = H1 + H2, thenH 2 = 4a is the contribution from the sheet current.

8 I I2 A = 8 maz H2 = 2 a = 4a, so I 2 = 8 A, then K = 2a az = 2 ( 1 ) 2(0.5)

A  K = 8 az m Method 2: From I1 at boundary we have H 1=

2a

=2a, but 5a at  = 1.0m corresponds to 10a  at =0.5m since H varies

2(0.5)

as 1/. So

− a  10 ( 2 a , − a ()−=8Ka −) a= 8 a z = K 



A  K = 8 maz

9. Inductance and Magnetic Energy P3.55: Consider a long pair of straight parallel wires, each of radius a, with their centers separated by a distance d. Assuming d >> a, find the inductance per unit length for this pair of wires.

Each wire can be solved using the energy approach: L   = 2 = h wires 8 4 The fields are not confined to a volume, so we must use the flux linkage approach to find inductance outside the wires.

I I a, H 2 = H1 = a ,

2

2 ( d −)

1 = BdS ,  TOT = 2B dS,

Fig. P3.55

d−a

2I d I  I d − a  = =  TOT = 2 a ddza  h  ln  dz  2 a  0 2   a  h

3-40

 d − a  1 L  = +ln  a  .

L=TOT , we arrive at h TOT  4 I     

Finally, with

 

P3.56: In problem P3.23 the task was to find the field at the center (radius b) of an N-turn toroid. If the radius of the toroid is large compared to the diameter of the coil (that is, if b >> c-a), then the field is approximately constant from radius a to radius c. (a) Obtain an expression for the toroid’s inductance. (b) Find L if there are 600 turns around a 99.8% iron core with a = 8.0 cm and c = 9.0 cm.

From P3.23 we found: H=

 NI  =  B d S =adSa 2 b 

NI a at radius b for the toroid.

2 b 

and with area (b−a), we have = 2

 NI 2b

(b−a).

2

N2I ( b − 2a ) =N= 2b

L=

 N 2 I

=

2b

( b −2 a )

(5000)(4x10− )(600) =0.33H (.085−.080) Plugging in our numbers we have: L= 7

2

2

2 ( 0.085)

P3.57: MATLAB: Consider that a solid wire of radius a = 1.0 mm is bent into a circular loop radius 10. cm. Neglecting internal inductance of the wire, write a program to find of M-File: MLP0357 the inductance for this loop.

Inductance inside a conductive loop % % % % % % % % % % % % This modifies ML0302 to calculate inductance of a conductive loop. It does this by calculating the mag field at discrete points along a pie wedge, then calculates flux through each portion of the wedge. Then it multiplies by the number of wedges in the 'pie'.

Wentworth, 1/19/03

3-41 % % % % % % % % % % % % % % % % % % % % % % %

Variables: I a b Ndeg current(A) in +phi direction on ring f df ring dL radius (m) wire radius (m) number dLmag dLuv of increments for phi angle of phi in [xL,yL,0] radians differential change in phi differential length vector on the ring magnitude of dL unit vector in direction of dL

clc clear

Ntest Rsuv R Ruv Rmag dH dHmag

location of source point number of test points unit vector from origin to source point vector from the source to test point unit vector for R magnitude of R differential contribution to H magnitude of dH

radius radial distance from origin Hz Bz total magnetic field at test point flux total mag flux density at test point flux through each differential segment %clears the command window %clears variables

% Initialize Variables a=0.1; b=1e-3; I=1; Ndeg=180; Ntest=60; uo=pi*4e-7; df=360/Ndeg; dLmag=(df*pi/180)*a; dr=(a-b)/Ntest; % Calculate flux thru each segment of pie wedge for j=1:Ntest x=(j-0.5)*dr; for i=(df/2):df:360 f=i*pi/180;

3-42 xL=a*cos(f); yL=a*sin(f); Rsuv=[xL yL 0]/a; dLuv=cross([0 0 1],Rsuv); dL=dLmag*dLuv; R=[x-xL -yL 0]; Rmag=magvector(R); Ruv=R/Rmag; dH=I*cross(dL,Ruv)/(4*pi*Rmag^2); dHmag(i)=magvector(dH);

end Hz(j)=sum(dHmag); Bz(j)=uo*Hz(j); dSz(j)=x*df*(pi/180)*dr; flux(j)=Bz(j)*dSz(j); end fluxwedge=sum(flux); Inductance=Ndeg*fluxwedge Now run the program: Inductance = 5.5410e-007 or L = 550 nH

P3.58: Find the mutual inductance between an infinitely long wire and a rectangular wire with dimensions shown in Figure 3.58.

H=

I1

 oI1 a a, B1=

2

2

B1dS2= o+a d oI  I b  + a o1ln  12=   dz= 2  o  2  o  b

1

o

0

12 b  + a M12= I1 =  oln  o  2   o  P3.59: Consider a pair of concentric conductive loops, centered in the same plane, with radii a and b. Determine the mutual inductance between these loops if b >> a.

3-43 In this case will drive the b-radius loop with current I. Here, at the center of the b-radius loop we have from Eqn. (3.10):  I B1= o az 2b a

2 2  oI1  2  oaI 1 = Then, 12= B1 dS 2 = 2b   d  d2b =2 2b  =0

oI1

   a2 So M 12= 12 = o I1 2b

P3.60: A 4.0 cm diameter solid nickel wire, centered on the z-axis, conducts current with

a density J = 4 A/cm2 az (where  is in cm). Find the internal inductance per unit length for the wire with this current distribution. 

2

8

HdL=Ie n c , 2   H  = 34   d   d  = 2

0

H =

8   34 2 4 =  , B = 2 6   3 42 3

1 4 2 3 = Wm 2  3 a

3

0

a

2

h

116   5 d  0 ddz= 8  a 6  h a  d d dz = 29 0 27 0

1 Wm= LI2; 2 Now we solve for I: I = J dS =

L=

1 h 12 

8a3 3

, and then solve forL:

4x10−7  H 1  1 L =L= = 600 =20 . h m 12 12  10. Magnetic Circuits P3.61: Referring to Figure 3.48(a), suppose 2.0 Amps flows through 80 turns of a toroid that has a core cross sectional area of 2.0 cm2 and a mean radius of 80. cm. The core is 99.8% pure iron. (a) How much magnetic flux exists in the toroid? (b) How much energy is stored in the magnetic field contained by the toroid?

3-44

( 5000 ( ) 2() = 0.200Wb ,  = 40Wb ( 4)x10−  ) 80  NI A; B = m2 2 (0.8 2 o ) 7

(a)  = BA=

1 1 B2 2oA=3.2mJ (b) Wm = BHdv= 2 2

P3.62: In Figure 3.59, a 2.0 cm diameter toroidal core with r1 = 10,000 is wrapped with a 1.0 cm thick layer of r2 = 3000. The toroid has a 1.0 m mean radius. For 20. A of current driven through 50 loops of wire, find the magnetic field intensity in each material of the toroid.

This toroid can be modeled with the circuit of Figure P3.62.

Fig. P3.62 Vm = NI = (50)(20A)=1000 A-turns 2 o l =1.592x106;R2= R=  A ; R1= 2

2  o

=1.768x106;

2  r 2  2 oa )( b − −6 B =159A V 628 x10 = BA; B = = H = 2; 1 = m = 628 x10 −6 m 1  r1o R1 (0.01 )2

 r1oa

2=

Vm

R2

H2 =

= 566 x10−6 = BA; B =

B

A =159m

 r2o

566 x10−6

= 0.6001

2  (0.02) ) ( 2 −(0.01)

3-45 P3.63: Referring to Figure 3.49(a), the 2.0 cm diameter core is characterized by the magnetization curve of Figure 3.60. The toroid has a mean radius of 60. cm. For 10. A of current driven through 100 loops of wire, find the magnetic field intensity in the 1.0 mm gap.

Referring to the model of Figure 3.49(b), Vm = NI = (100)(10A)=1000 A-turns; A = a2 = 314x10-6

R=G G

l 6 =2.53x10; oA

Approach: (1) Assume B: B = 0.4 (2) Read H from chart: H = 200

(3) Evaluate RC =

2  o 12x103H : Rc =6x106 = B A

6 6 (4) R TOT = R c + 2.53x10: R TOT =8.53x10 (5)  = Vm/ RTOT:  =117x10-6 (6) B = /A: B = 0.373

Insert this value of B into step (1) and repeat.

After 4 or 5 iterations the routine converges to a solution B = 0.382 Wb/m2 and H = 190 kA/m. Then, in the gap, we have H= B =300 kA. m o P3.64: Referring to Figure 3.52, suppose the cross sectional area of the bar is 3.0 cm2, while that of the electromagnet core is 2.0 cm2. Also, the bar has a relative permittivity of 3000, while that of the magnetic core is 10,000. The dimensions for h and w are 12. cm and 16. cm, respectively. If the mass of the bar is 20. kg, how much current must be driven through 24 loops to hold up the bar against gravity?

We will follow an approach similar to Example 3.18, but realize that Ac ≠ Ab.

l Rc=  A = c

2 2(.12)+ 0.16 100 cm =159x103   10, 000   o (2cm2 ) m 2

0.16 100cm l Rb= =   =141x103 cm2 )  m   Ab 3, 000 o (3

Vm =NI,  =

=

NI

RTOT RTOT

Ro TOT

Vm

=

NI

c; H=  oHA

Ac

3-46 2

2

  NI   NI F =  oHAc = o  Ac =   RTOToAc   R TOT  2

Solve for I: I=

1 =mg oAc

o c R TOT mgA =2.8A

N

P3.65: Consider a 1.0 mm air gap in Figure 3.49(a). The toroid mean radius and cross

sectional area are 50. cm and 2.0 cm2, respectively. If the magnetic core has a r = 6000, and 4.0 A is being driven through 30 loops, determine the force pulling the gap closed.

2

F =H o A, RTOT= V

=

m

l 2 o ( A turns )( ) + =6.06x106 Wb r  A  oA o

=

(30)(4 ) =19.7910−6= x BA

RTOT 6.06x106

= H = 78.76x103A/m In the gap,B = 19.79x10-6/A = 98.97x10-3 oH, 2

2  1  F =  oH 2A =( 4  x )−78.76 x103) (2 )   = 1.6 (  100 107 F = 1.6N

4-1 Solutions for Chapter 4 Problems 1. Current Continuity and Relaxation Time P4.1: How long does it take for charge density to drop to 1% of its initial value in polystyrene? r = 2.56 and = 10-17 S/m. − (/ ) t  v=  oe  has  Polystyrene



−10−17



t=e−441x10−9t o

  2( .56)( 8.854x10 −12) 0.01=exp o

o

Solving for t we get t = 10.4x106 sec = 120 days

P4.2: At a particular point in a slab of silver, a charge density of 109 C/m3 is introduced.

Plot v versus time for a duration of 10 relaxation times.

For silver we have  = 1 and  = 6.2x107 S/m C −(/)t e−x18t = 109 710 m3  v =  oe

 The relaxation time is: = =142x10−21sec 

The plot is obtained with the following MATLAB routine. % % % %

MLP0402 Plot rhov vs time for a charge placed in silver

rhovo=1e9; %initial charge density, C/m^3 er=1; %relative permittivity of silver sig=6.2e7; %conductivity, S/m eo=8.854e-12; tau=er*eo/sig; t=tau/10:tau/10:10*tau; rho=rhovo*exp(-t./tau); subplot(2,1,1) plot(t,rho) xlabel('t (s)') ylabel('rhov (C/m^3)') grid on

4-2

subplot(2,1,2) loglog(t,rho) xlabel('t (s)') ylabel('rhov (C/m^3)') grid on

Fig. P4.02

P4.3: A current density is given by J = e-.01ta  A/m.2 Find the charge density after 10 seconds if it has an initial value of zero.

  1 v= (2e−0.01t)=2e−0.01t t t +C dv=−2e−0.01tdt;v=  200e−0.01t At t = 0, v = 0 = 200 + C, C = -200.

C e−0.01(10) − 1 ) = − 1 9 So we have  v= 200( m3 J=−

P4.4: At t = 0 seconds, 60.0 C is evenly distributed throughout a 2.00 cm diameter pure silicon sphere. (a) Find the initial charge density. (b) How long does it take to drop to 10% of its initial value? (c) What will be the final surface charge density? (a) Volume v=

Q C 4 3 6 3 v =14.3m3 (0.01m)=4.19x10−m; vo= 3

(b) =0.10=e−t/ v vo vo 

For pure silicon, =  =

(11.8)(8.854x10−12)

=237.4x10−9sec

(4.4x10− 4 )

4-3

ln0.10 = (c)  = s

−t



= −2.303; t=(2.303)=547ns

60C Q − =47.8mC ; area=4r2=1.257x103m2; s= m2 1.257x10−3m2 area

2. Wave Fundamentals P4.5: A propagating electric field is given by

 V  .01 Ezt=100.e−z c o s   7 1 0 t +  z −. (,)  4m (a) Determine the attenuation constant, the wave frequency, the wavelength, the propagation velocity and the phase shift. (b) How far must the wave travel before its amplitude is reduced to 1.0 V/m?

1107m −  =−45 ; = 5 z;  = 2; (a)  = 0.01Npm; f =MH m u p=x s 4  1  −0.01z =1; −0.01z=ln ; z=460m (b) Ez=100e ()  100  P4.6: A 10.0 MHz magnetic field travels in a fluid for which the propagation velocity is 1.0x108 m/sec. Initially, we have H(0,0)=2.0 ax A/m. The amplitude drops to 1.0 A/m after the wave travels 5.0 meters in the y direction. Find the general expression for this wave.

A yt=H oe−  yc o s (  t −  y +  ) a x The general expression for the wave is: H(,) m 106rad = 20 x  = 2f=  2(10x106) s  up rad 2 up=1x108ms; = =10m;  = =0.2 m f  A H (0, 0)= 2.0ax m;  Ho= 2.0, = 0 y = 2.0e− (5) ; solving we get = 0.14 H ( y = 5) = 1 =He−o Finally, A H(y,t)=2.0e−0.14ycos(20x106t−0.2y)a x m

P4.7: MATLAB: Modify the simple wave program in MATLAB 4.1 to include attenuation. Generate a plot for the case where the amplitude is 4 V/m, the attenuation

4-4 constant is 0.001 Np/m, and the frequency is 1 MHz. Take your snapshot in time at 0 seconds, and let your phase shift be 0. % % % % % % % % % % % % % % % % % % % % %

M-File: MLP0407 This is a modification of ML0401 that includes attenuation. It plots a wave (in vac) versus position (in z direction) for a fixed time.

Wentworth, 1/19/03 Variables: atten Eo f omega t phi phir c

clc clear

attenuation constant wave amplitude (V/m) frequency (Hz) angular frequency (rad/sec) time snapshot (sec) phase constant (degrees) phase constant (radians) speed of light in vacuum (m/s)

lambda wavelength (m) B phase constant (1/m) E electric field intensity z position %clears the command window %clears variables

% Initialize Variables atten=0.001; Eo=4; f=1e6; t=0; phi=0; phir=phi*pi/180; c=2.998e8; lambda=c/f; B=2*pi/lambda; omega=2*pi*f; % Perform Calculation z=0:4*lambda/100:4*lambda; E=Eo*exp(-atten.*z).*cos(omega*t-B*z+phir);

4-5

% Generate the Plot plot(z,E) axis('tight') %sets axes min & max data values grid xlabel('z(m)') ylabel('E(V/m)')

Fig. P4.7

P4.8: MATLAB: Modify the traveling wave program in MATLAB 4.2 to include attenuation. Use the parameters from problem P4.7, except for the fixed time of course. % % % % % % % % % % % % % % % %

M-File: MLP0408 This program illustrates a traveling wave including attenuation. It modifies ML0402.

Wentworth, 1/19/03 Variables: atten Eo f omega t phi phir c

attenuation (Np/m) wave amplitude (V/m) frequency (Hz) angular frequency (rad/sec) time snapshot (sec) phase constant (degrees) phase constant (radians) speed of light in vacuum (m/s)

4-6 % % % %

lambda wavelength (m) B phase constant (1/m) E electric field intensity z position

clc clear

%clears the command window %clears variables

% Initialize Variables atten=0.001 Eo=4; f=1e6; t=1; phi=0; phir=phi*pi/180; c=2.998e8; lambda=c/f; B=2*pi/lambda; omega=2*pi*f; % Perform Calculation z=0:4*lambda/100:4*lambda; E=Eo.*exp(-atten.*z).*cos(omega*t-B*z+phir); % Generate a Reference Frame plot(z,E) axis([0 4*lambda -2*Eo 2*Eo]) grid xlabel('z(m)') ylabel('E(V/m)') pause % Make the Movie t=0:1/(40*f):1/f; for n=1:40; E=Eo.*exp(-atten.*z).*cos(omega*t(n)-B*z+phir); plot(z,E) axis([0 4*lambda -2*Eo 2*Eo]) grid title('General Wave Equation'); xlabel('z(m)'); ylabel('E(V/m)');

4-7 M(:,1)=getframe; end

Fig. P4.8 (at end of animation)

3. Faraday’s Law and Transformer EMF P4.9: The magnetic flux density increases at the rate of 10 (Wb/m2)/sec in the z direction. A 10 cm x 10 cm square conducting loop, centered at the origin in the x-y plane, has 10 ohms of distributed resistance. Determine the direction (with a sketch) and magnitude of the induced current in the conducting loop.

dB =10Wb az dt m2s Wb Vemf =  −dB d S = − 2a1 zdxdyaz 0 dt ms WbVs Vemf=−0.1 =−0.1V sWb 0.1V I=I= =10mA 10 I=10mA clockwise (when viewed From +z)

Fig. P4.9

P4.10: A bar magnet is dropped through a conductive ring. Indicate in a sketch the direction of the induced current when the falling magnet is just above the plane of the ring and when it is just below the plane of the ring, as shown in Figure 4.22.

Refer to Figure P4.10.

4-8 When the north pole first goes through the loop, flux is increasing and the current induced to oppose this change in flux is as shown.

When the south pole is exiting the loop, flux is decreasing and the current induced acts to oppose this change in flux.

Fig. P4.10

P4.11: Considering Figure 4.7, suppose the area of a single loop of the pair is 100 cm2, and the magnetic flux density is constant over the area of the loops but changes with time 2 and  = 0.30 Np/sec. Determine VR at 1, 10, az, where Bo = 4.0 mWb/m o as B=Be−t and 100 seconds.

B dB = − o B etaz; − V Vemf=−N dS; t dt

emf

= 2 B oe − S t 2

 1m  e0.30t− 6 .30 =24x10−e−0t VR= 2  Bo S et − 20.30 (= ) 410 (x − 3) 100 ( cm2)  100cm at t = 1 sec, VR = 17.8 V at t = 10 sec, VR = 1.20 V at t = 100 sec, V V R = 2.25x10-18



P4.12: Sometimes a transformer is used as an impedance converter, where impedance is given by v/i. Find an expression for the impedance Z1 seen by the primary side of the transformer in Figure 4.11 that has a load impedance Z2 terminating the secondary.

N N1 We have i= 2 N2i1 and v2= 2v1 N1

4-9

 N2   N1v1 v1 v2 =   Z1 = , Z2 = N1  i1 i2   N2 i1   2

N  Z1 =  12  Z2 N 

2

N  =   Z1 2  N 1

P4.13: A 1.0 mm diameter copper wire is shaped into a square loop of side 4.0 cm. It is placed in a plane normal to a magnetic field increasing with time as B = 1.0 t Wb/m2 az, where t is in seconds. (a) Find the magnitude of the induced current and indicate its direction in a sketch. (b) Calculate the magnetic flux density at the center of the loop resulting from the induced current, and compare this with the original magnetic flux density that generated the induced current at t = 1.0 sec.

We find the distributed resistance of the loop and work the problem assuming this resistance is lumped in one spot as shown in the figure. (a) The induced current is Vemf divided by the distributed resistance of the wire loop.

) ) 1 m 4(0.04m 1l R=  = =3.5m 7 A 5.8 x10 (0.0005m)2 dB Wb0.04 0.04 Wb dB = − 1 = 1m2s dS dx  dy=−1.6mV az ; Vemf = − dt dt m2 s  0 0

I ind =

1.6mV = 0.46 A(note that this answer has no time dependence) 3.5m

Fig. P4.13 (b) The field at the center of the loop from a single arm of the loop is found from Eqn. (3.7): +a   (0.46) 1 z I I 1 A H=   ( −z a) = ( ( a z ) = − a z ) = − 2; . 5 9 a z m 2 2a 4   2 z2 +  2 2 0.02 2 ) ( −a



4-10

 Wb So B=4 o H = − 1 32 a z. m

P4.14: The mean length around a nickel core of a transformer like the one shown in Figure 4.11 is 16 cm, and its cross sectional area is 1 cm2. There are 30 turns on the primary side and 45 on the secondary side. If the current on the primary side is 1.0 sin20x106t mA, (a) calculate the amplitude of the magnetic flux in the core in the absence of the output windings. (b) With the output windings in place, calculate i2.

V l (a) = m; V m=N1i1; R=  ro A R

) 1(mA) ( ) ( x10−7) (cm2) 1 m2 = 14x10−9 N1i1  roA (30 = Wb 2 l cm) (0.16m) (100 600 4 N 1 30 = 2 1 ) (b) i2 = 2 i 45 1sin 20 3 sin (20  x 1 0 6 t ) m A N =

( ( x106t )mA =

P4.15: A triangular wire loop has its vertices at the points (2, 0, 0), (0, 3, 0) and (0, 0, 4), with dimensions in meters. A time-varying magnetic field is given by B = 4ty a Wb/m2 (t in seconds). If the wire has a total distributed resistance of 2 , calculate the induced current and indicate its direction in a carefully drawn sketch.

B Ve m f = dS, t Wb B =4a y m2s t

−

MN 1 S = MN 2 MN M = −x+3ay,N=−2ax+4az 2a

S=6a+4a+3a m2z x y Vemf = -(4)(4)=-16V V 16V I = emf R = 2 = 8 A

Fig. P4.15

4. Faraday’s Law and Motional EMF P4.16: Referring to Figure 4.23, suppose a conductive bar of length h = 2.0 cm moves with velocity u = -1.0 m/s a towards an infinite length line of current I = 4.0 A. Find an expression for the voltage from one end of the bar to the other when  reaches 10 cm and indicate which end is positive.

4-11 In Figure P4.16, an imaginary circuit has been chosen. For the chosen circulation direction, we have the sign for Vemf as shown. Then,  1o Ih  o I   d z a z = m  − dz, Ve m f =  ( u  B-)1daLa=  2      s  20 Vemf=−160nV. Therefore, the bottom of the bar is positive.

Fig. P4.16

P4.17: Suppose we have a conductive bar moving along a pair of conductive rails as in Figure 4.12, only now the magnetic flux density is B = 4.0a azWb/m2. If R = 10. x + 3.0 , w = 20. cm, and uy = 3.0 m/s, calculate the current induced and indicate its direction.

 Wb m Vemf=−Buydxaz=−3   3 (0.2m )=−1.8V 2  m   s 1.8V I= =0.18A (clockwise when viewed from the +z axis) 10 P4.18: The radius r of a perfectly conducting metal loop in free space, situated in the x-y

plane, increases at the rate of (r)-1 m/sec. A break in the loop has a small 2.0 ohm

resistor across it. Meanwhile, there exists a magnetic field B = 1.0 az T. Determine the current induced in the loop, and show in a sketch the direction of flow.

Here we’ve assumed dS = -dSaz to get iind and Vemf as shown. Our approach will be to

find , then Vemf = -d/dt.

4-12

( −  d ) d  a z

= B  dS = 1az r

= − 2  =d − r 2

 0

d = − dr  1  2 r = − 2  r  = − 2 V dt dt r

Vemf = +2V I=

2V = 1,Aclockwise as shown 2 Fig. P4.18

P4.19: Rederive Vemf for the rectangular loop of Figure 4.16 if the magnetic field is now B = Boaz.

We see in Figure P4.19a that ( uBdL=0 ) for the For the 2→3 section we have:

dl

3

1 → 2 and 3→ 4 line sections.

d  a

d L== da  a  , u =  = (B u  B ) d L ; B = B o adtz =, dt 2

0

(u  ) dL =  Bo  d =  a a

−Ba2 o 2

+ B a2 , so for the 2 → 3section, the contributions to Vemf 2 0 cancel. This will also be the case for the 4 → 1 section, and therefore Vemf = 0; no current is induced.

 o u  B) d L =  oB  d = ( 

Fig. P4.19a

Fig. P4.19b

4-13 a and P4.20: In Figure 4.16, replace the rectangular loop with a circular one of radius rederive Vemf.

2 2  = BdS=BoaydS=Boaya(−sinax+cosay)=B oacos d flux = − Bo sin ;  a2

dt

d flux Vemf = − =  Bo a2 sin dt P4.21: A conductive rod, of length 6.0 cm, has one end fixed on a grounded origin and is free to rotate in the x-y plane. It rotates at 60 revolutions per second in a magnetic field B = 100. mT az. Find the voltage at the end of the bar.

rad  rev 2 rad  = 60    =120 s   u   rev  = (  BdL) emf

s

V

u= 

d a=   a dt l

Vemf =  (  Boa az ) d a = 0

1

 Bol 2 2

2  rad  Wb  2 Vs Ve m f =120   0.1 (0.06 m) Wb 2 s  m = 68mV

Fig. P4.21

We can confirm the sign by observing that a positive charge placed in the middle of the bar would move to the ungrounded end by the Lorentz force equation.

P4.22: Consider the rotating conductor shown in Figure 4.24. The center of the 2a diameter bar is fixed at the origin, and can rotate in x-y plane with B = Boaz. The outer ends of the bar make conductive contact with a ring to make one end of the electrical 2 contact to R; the other contact is made to the center of the bar. Given o =B100. mWb/m, a = 6.0 cm, and R = 50. , determine I if the bar rotates at 1.0 revolution per second.

d

= ;  au = , d Ve m f =  ( u  B ) ad L L = d  a  dt Figure P4.22 indicates one of the paths for the circulation integral.

4-14 a

a  Boa2  Boaz ) d a=  Bo  d  =  (   a     Ba2 2 0 0   Vemf rad Wb 1 VsA 1  rev o =  1  2 I= R = ) 2    100 x10−3 ( 0.06m rev  m2  2 s 2R 50  WbV I=  22.6 A 

Vemf =

Fig. P4.22

P4.23: A Faraday Disk Generator is similar to the rotating conductor of P4.22, only now the rotating element is a disk instead of a bar. Derive an expression of the Vemf produced by a Faraday Disk Generator, and using the parameters given in problem 4.22, find I. Worked exactly as P4.22.

P4.24: Consider a sliding rail problem where the conductive rails expand as they progress in the y direction as shown in Figure 4.25. If w = 10. cm and the distance between the rails increases at the rate of 1.0 cm in the x direction per 1.0 cm in the y direction, and uy = 2.0 m/sec, find the Vemf across a 100.  resistor at the instant when y = 10. cm if the field is Bo = 100. mT. First we modify the figure so that the top rail is horizontal and all the spreading occurs

via the bottom rail. As before, our approach will be to find  and then d /dt. We have:  = BdS=Boazdxdya z Now, notice that x and y are not independent and are in fact related: x=y+w So we have y y y+w 1 2   = Bo  dxdy=B o(y+w)dy=Bo  y+wy  2  y=0 x=0 0

4-15

d Wb dy   m + 0.1m) 2s  Vemf = − = −Bo ( y + wdt) = −Bo ( y + w y)=u 0.100 ( 0.1m −  m2  dt   Vemf = 40mV Alternate Method: Vemf =  u( B =)(uyd a y LBoaz )dxa x 1 −2 y

Vemf = uy Bo  dx= −uyBo(w+ y) 1 w+ 2 y

Fig. P4.24

5. Displacement Current P4.25: Suppose a vector field is given as A = 3x2yz3 ax .

Verify that the divergence of the curl of this vector field is equal to zero.

verify that A=0

ax ay az  =  (3x2yz3   A=  )ay −  y(3x2yz3)az x y z z 3x2yz3 0 0

= 9x2yz2 a y −3x2 z3 az =     A  y(9x2yz2 ) − z(3x2z3)= 9x2z2− 9x2z2= 0

4-16 P4.26: Suppose a vector field is given by A=  2cos az.

Verify that the divergence of the curl of this vector field is equal to zero.

 1   (2cos)a ( 2c o s  ) a  = −  s i n  a  − 2  c o s   −     1  1  +i n     A= 2) )cos ) (  ( −  s (−     1 = − 2sin 2sin=0 + A=



P4.27: A pair of 60 cm2 area plates are separated by a 2.0 mm thick layer of ideal

dielectric characterized by r = 9.0. If a voltage v(t) = 1.0 sin (2x103t) V is placed across the plates, determine the displacement current.

id=C

dv ; C= dt

S=(9)(8.854x10 12 F m)(60cm2 ) 10mm 1m =239pF −

d

(2mm)

cm 100cm

dv

V =2x103cos(2x103t) dtd =1.5cos(2x103t)A s

i

P4.28: Plot the loss tangent of seawater ( = 4 S/m and r = 81) versus log of frequency from 1 Hz to 1 GHz. At what frequency is the magnitude of the displacement current density equal to the magnitude of the conduction current density?



tan =   =

889x 106 (4S m) x −12 = ) f ( ( 8.854 10 F m) ) 81 ( 2) (fHz

(Plot this in Figure P4.28)

)(

Solving for f when tan = 1: f = 890MHz % MLP0428 clear clc for i=1:9 for j=1:10 m=(i-1)*10+j; f(m)=j*10^(i-1); tand(m)=889e6/f(m); end

4-17 end loglog(f,tand) xlabel('frequency (Hz)') ylabel('loss tangent') grid on

Fig. P4.28

P4.29: A 1.0 m long coaxial cable of inner conductor diameter 2.0 mm and outer

conductor diameter 6.0 mm is filled with an ideal dielectric with r = 10.2. A voltage v(t) = 10.cos(6x106 t) mV is placed on the inner conductor and the outer conductor is grounded. Neglecting fringing fields at the ends of the coax, find the displacement current between the inner and outer conductors.

C=

Q

, so Q=Cv(t)

V 2  l oQ= for coax (from chapter 2): C= s ln (ba )

DdS so D=

=

2   ol cos V t ln (ba )

 = 2 lD = Q D d adza

l cos t  Vocos t 2  Vo  Vocos t = and D = a .  ln(ba) 2lln( ba) ln(ba) osin t  D −  V = a = Jd  ln(b a) t

id = J d dS =

− Vosin t 1

ln(ba) 

=   d dz 2

l

0

0

−2   lV osin t ln(b a)

4-18 Now evaluate idwith the given parameters: −2 ( 6x106 rads ) 10.2  t) C As 1m) 0.010Vsin )(8.854x10−12 Fm)( ( ( id = FVC ln(0.006 0.002)

t) A id = −97 sinx106 ( 6

7. Lossless TEM Waves P4.30: Suppose in free space that E(z,t) = 5.0e-2zt ax V/m. Is the wave lossless? Find H(z,t).

Since the wave has an attenuation term (e-2zt) it is clearly not lossless.

ax ay az H  2zt  =  (5e−2zt )ay = 10  =   E= − o − te− ay x y z z t 5e −2 zt 0 0 −10t 2zt 2zt te− e− dta y , H= 10 dH = dta y   − o o

(

)

= uv− vdu This integral is solved by parts udv where we letu = t and dv= e−2ztdt.

We arrive at:   −10t −2zt 10 A e−2 tz ay H=  e − 2 m 4oz   2 oz P4.31: An electric field propagating in a lossless non-magnetic media is characterized by V m .

E( y, t ) = 100. cos ( 4106 t − 0.1257 y ) a z

Find the wave amplitude, frequency, propagation velocity, wavelength, and the relative permittivity of the media. (b) Find H(y,t).

V f Eo=100m, =2MHz,

 =

2



, =50m, up= 

  E=  Eo sin(  t −  ayx =) − o

H t

m 1  ; r=9 =100x106 =  s  

 Eo E = t o cos( ) t  s i n ( −  t −y  a)yx d o o E H (y,t) =  o c o st (  ax −  y ) H= −

o

Inserting the given values we find: H ( y,t ) = 0.796cos ( 4x106t − 0.126 y)ax Am

4-19 P4.32: A magnetic field propagating in free space is given by A H(z,t)=20.sin(10t+z)a m.

8

x

Find f, , , and E(z,t).

1

10 50MHz  2=  f =  x10,8f= 2 x8= x 8 2  rad c 310 = = 6 m,  = = 8 3 m  f 0.5x10 D   H= (no J c) t

=

ax   H = x t +(  Ho s i n

ay az  =  z(Ho sin   (t + azy ) ) y z   0z ) 0

E = Ho cos(  t + z) a y=o t dE=E= Ho H ( t + z)dtay = o sin( t + azy ) cos

 o V  Ho (z,t)= so E sin ( t + z)ay  m

 o

o

  kV  inserting the given values we find:E (z,t)= 7.5 sin x108t+ zay m 3   P4.33: Find the instantaneous expression for E for the magnetic field of problem P4.6.

A a. We can write this wave From P4.6 we have H=2.0e0.14cos(20x10t−0.2y) m −

y

6

x

A − ax, and can reinsert the values at the end. as H= Hoeycos(t−y) m E c . Assuming a nonmagnetic medium, we havepu= First we have H= r , t

or r = 9. Now we evaluate the curl of H:  ax ay az         H =  = − Hoe− y cos(  t −  ayz. )  z  y  y   x    He− y cos(  t −  y 0) 0  o 

4-20

 Hoe− y cos(  t −  yaz) = H−o  − e −y cos(  t −  +y)e −y sin( t −  yaz). y   So H= Ho e −y cos(  t −  ayz −) Ho e −y sin( t − y)az =  E . t Solving for E : H o−  y H  y ( t − y)dtaz z − o e − sin e cos ( t − y ) dta E= −

  Ho  y H e− sin ( t − y )z a o+e − y cos(  t−  y )z.a E=  

Now we can evaluate the variables: (2) 0.14 ( ) Ho = 56, = 6   (20 ) 9()(8.854x10−12) x10(2) 0.2 ( ) Ho = 251. = 6 −12   (20x10 ) 9()(8.854x10 ) So

V .

E = 56e−0.14y sin (20x10−6t − 0.2y)a t−0.2y)az + 251e−0.14y cos (20 m z

We can use a trigonometric identity to simplify this answer. Given that sinxsiny, cos(xy)=cosxcosy

we can let E=e−ya zA(bcos(t−y)+csin(t−y)), where b=cosy, c=siny, and tany=c/b. Now, since b/c = 251/56, then y=12.6° and A=257.

Thus E(,)=257− V . yt e0.14ycos(20x106t−0.2y−12.6)a z m This result is easier to obtain using the phasor approach of the next section.

P4.34: Given, at some point distant from a point source at the origin in free space, V m,

E(r,t)=8.0cos(9106t−r)a

find frequency, phase constant and H(r,t).

=2f,f=4.5MHz,= H B   E = − = − o t t

rad 2 c =0.094 =66.7m,= f  m

4-21

1   E = r (rE)a , E = Eocos( t−r), r  cos( t − r )= )Eocos( (Eor  t−r ) +oEr sin(t −  r ) r E  E = o   c( o ts  −  r ) +  r s ian (  t −  r ) r a + E sin(t −r)a Eocos(  t−  r) = o   r Assume we can ignore the first term in the far-field for large r, then we have H   E =o  E s i n (  t − or)t a  = − 

 Eo

Eo

 d H = H = −s i n (  t −  roc)odst( a t −= r ) a  o

Inserting the given values we find mA m H(r,t)=21cos(t−r)a

P4.35: In a lossless, non-magnetic media, the magnetic field at some point distant from a source at the origin is given by A , m find the relative permittivity of the media, the frequency and phase constant of the wave,

H(,t)=6.0sin(3108t+10)a

and E(,t).

2f=3x108;f=48MHz m c  up= =0.3x108 = r ; r=100 s   

rad m c D E H= =  t t   H =1  ( H o  s i n (  t +1 1 0  )t )+a)z + =oHcos( ( H ot s+i anz(  )   In the far-field, we will ignore the first term (has)1/leaving us with

 =

r=10

  H =  Ho c o s (  t +   ) a z H Now, d E = =E ocos(  t +   ) d t a z   Ho E =   sin(  t + az. )

)

4-22

V m

Plugging in the given values we arrive at:E(,t)=226sin(3x108t+10)a . z P4.36: Suppose, in a non-magnetic medium of relative permittivity 3, that

V  x10(7 t −  y)ax + 9.0 cos  x107(t −  y)az m.

E (y,t)= 4.0 sin Determine and H( y,t).

f = 5MHz , pu=

c



r

up rad 2 m = 1.73x108 = 34.6m, =  = 0.181 s , = m f

 t −  ayx +) 4 cos(  t −  ayz )= −o   E = 9 sin( H=

9 

−4 cos(  t −  ayx )+  sin( t −  ayz )

H t

 o

o

Inserting the given values we arrive at:

mA  t( −  ayx −) 18 sin  t (−  ayz )m H ( y,t ) = 41cos

8. Time-Harmonic Fields and Phasors P4.37: Show that  B sejt  − t

(Re

jB ( )e  .  ) Re= − s

j t

We first show that :    t B sejt  − Re  B sej   t  t Expanding the left side we have:   jt  −t B s cos( t )+ jB ssin( t )) (Re t  = − B scos(  t )=  Bs sin(t) t Now expanding the right side we have:    −Re  Bsejt  = −Re  jB sejt  = Re −  jB (s cos  t)(+ j sin (t))   t 

(Re

)=−

( R e  ) B= s−e

jB

= − R e  s cos(  t) +j2 sin(  t ) = Bs  we have : So since

sin (t )

4-23

−

 t

  B e   ) Re B e  = −Re  j B e  , = − (Re t  jt

jt

s

then  − t

jt

s

s

B e  ) Re =B−t e j  = −Re  j B e  (Re jt

jt

t

s

s

s

P4.38: Derive the differential phasor form of (a) Gauss’s Law, and (b) Ampere’s Circuit Law.

 D = v,  D(x,y,z,t)=v(x,y,z,t)  Re  sejt  =   Dvsejt e j  t   D  = R e vs   e j  t   ; s R e  Re   D s= vs



D E   H =  t = t



  H e )=Re ( j e ) E s E e ); Re(   RH eejt() =tRe( s

jt

jt

s

s

   H s = j   sE

P4.39: Find y,t H( ) in problem P4.31(b) using phasors.

E s= Eoe− j  yaz ,   Es = − j   H s −jy   Es = j− Eoe ax = − j Hs jy s = E E o − H e ax and H(y,t) =  o cos(  t − axy. )



o

o

Plugging in the values from P4.31, we find A  x106 t(−4 0.126 y)ax m.

H ( y,t ) = 0.796 cos

P4.40: Find E(z,t) in problem P4.32 using phasors.

H= Ho ejz ax,   Hs=joEs a s ay az x   = jHoejz a    Hs =  y x  y  z z jHoejz 0 0

jt