194

CHAPTER 7

P.E. 7.1 z

1 2

2
  5, cos 1  0, cos  2 
27
 a  a y  a x  a y
a  al  a =  x   az 
 2  2

10  2   a  a y 
H3    0   x   30.63a x  30.63a y mA/m
4 (5)  27  2 

P.E. 7.2
2  3 
(a) H   1  az  0.1458az A/m
4 (2)  13 
12
  32  42  5,  2  0, cos 1   ,
(b) 13

 3a  4a z  4a x  3a z
a  a y   x 
 5  5
2  12   4a x  3az  1
H 1     4ax  3az 
4 (5)  13  5  26
= 48.97a x  36.73a z mA/m

P.E. 7.3
(a) From Example 7.3,
Ia 2
H az
2(a 2  z 2 )3/ 2
At (0,0,-1cm), z = 2cm,
50  103  25 104
H a z  400.2a z mA/m
2(52  22 )3/ 2  106

(b) At (0,0,10cm), z = 9cm,

 195

50  103  25 104
H a z  57.3a z mA/m
2(52  92 )3/ 2  106

P.E. 7.4
NI 2  103  50  103 (cos  2  cos 1 )a z
H  cos   cos 1 z
a 
2  0.75
2
2L

100
1.5
cos  2  cos  1 a z
0.75
(a) At (0,0,0),   90 o , cos  2 
0.75 2  0.05 2
= 0.9978 1 2
100
H  0.9978  0  az
1.5

= 66.52 az A/m
(b) At (0,0,0.75),  2  90 o ,cos  1   0.9978 1 2
100
H  0  0.9978  a z
1.5
= 66.52az A/m
 0.5
(c) At (0,0,0.5), cos  1    0.995
0.5 2  0.05 2
0.25 1
cos  1   0.9806 2
0.25 2  0.05 2
100
H  0.9806  0.995 a z
1.5
= 131.7az A/m

P.E. 7.5
1
H K  an
2
1
(a) H (0, 0, 0)  50a z  (a y )  25a x mA/m
2
1
(b) H (1,5, 3)  50a z  a y  25a x mA/m
2

P.E. 7.6
 NI
 ,   a      a, 9<  11
H   2

 0, otherwise
 196

(a) At (3,-4,0),   3 2  4 2 =5cm ‹ 9cm

H 0
(b) At (6,9,0),   6 2  9 2 = 117 ‹ 11
103  100 103
H   147.1 A/m
2 117  102

P.E. 7.7
(a) B    A  ( 4 xz  0 )a x  (0  4 yz )a y  ( y 2  x 2 )a z
B (1, 2,5)  20a x  40a y  3a z Wb/m2
4 1 4 1
(b)    B.dS    ( y 2  x 2 )dxdy   y 2 dy  5 x 2 dx
y 1 x  0 1 0

1 5
 (64  1)   20 Wb
3 3

Alternatively,
1 4 0
   A.dl   x 2 (1)dx   y 2 (1)dy   x 2 (4)dx  0
0 1 1

5 65
   20 Wb
3 3

P.E. 7.8
z

R
h

k y

dS

kdS  R
H  ,
4 R 3

dS  dxdy, k  k y a y ,
R  ( x,  y, h),
 197

k  R  (ha x  xa z )k y ,
k y (ha x  xa z )dxdy
H  3
4 ( x 2  y 2  h 2 ) 2

k y ha x   dxdy k y az  
xdxdy

4   (x
 
2
 y 2  h2 )
3
2

4   (x
 
2
 y 2  h2 )
3
2

The integrand in the last term is zero because it is an odd function of x.

k y ha x 2 
 d d  k y h2 a x 
3 d ( 2 )
H     (  h )
2 2 2
4  0  0 (  2  h2 )
3
2 4 0
2

kyh  1   ky
 ax   0  ax
2  (  2  h2 ) 12  2
 
1
Similarly, for point (0,0,-h), H =  k y a x
2
Hence,
1
 k yax , z0
H 2
1
 k a , z0
2 y x

Prob. 7.1
(a) See text
(b) Let H = Hy + Hz
I
For H z  a   ( 3) 2  4 2  5
2
( 3a x  4a y ) ( 3a y  4a x )
a   a z  
5 5
20
Hz  (4a x  3a y )  0.5093a x  0.382a y
2 (25)

I
For H y  a  ,   ( 3) 2  5 2  34
2

(3a x  5a z ) 3a z  5a x
a  a y  
34 34
 198

10
Hy  (5a x  3a z )  0.234a x  0.1404a z
2 (34)
H = Hy + Hz
= 0.7433ax + 0.382ay + 0.1404az A/m

Prob. 7.2

I I 2 100
H      31.83 m
2 2 H 2 (10 10 ) 
3

Prob. 7.3

Let H  H1  H 2
where H1 and H 2 are respectively due to the lines located at (0,0) and (0,5).
I
H1  a ,   5, a  a  a  a z  a x  a y
2
10 a
H1  ay  y
2 (5) 
I
H2  a ,   5 2, a  a  a  , a  a z
2
5a  5a y a x  a y
a  x 
5 2 2
 a  a y  -a x  a y
a  a z   x 
 2  2
10  -a x  a y  1
H2    ( -a x  a y )
2 5 2  2  2
ay 1
H  H1  H 2   (-a x  a y )  0.1592a x  0.1592a y
 2
 199

Prob. 7.4

Idl  R
H  dH   H1  H 2
4 R 3
For H1 , R  (3,1, 2)  (0, 0, 0)  (3,1, 2), R  9  1  4  14
1 0 0
Idl  R  4 105  4 105 (0, 2,1)
3 1 2
4  105 (0, 2,1)
H1   (0, 0.01215, 0.006076)105
4 (14) 3/2

For H 2 , R  (3,1, 2)  (0, 0,1)  (3,1, 3), R  9  1  9  19

0 1 0
Idl  R  6 105  6  105 (3, 0, 3)
3 1 3
6  105 (3, 0, 3)
H1   (0.017, 0, 0.017)105
4 (19) 3/2

H  H1  H 2  (0.0173, 0.01215, 0.01122)105  (0.173a x  1.215a y  0.1122a z )  A/m

Prob. 7.5

Let H =H y  H z
I
For H z , Hz  a
2
where I = 20,  =  3a x  4a y ,  = 32  42  5
 3a x  4a y  0 0 1
a  a  a  a z     0.8a x  0.6a y
 5  0.6 0.8 0
20
Hz  (0.8a x  0.6a y )  0.5093a x  0.382a y
2 (5)
I
For H y , Hy  a
2
where I = 10,  =  3a x  5a z ,  = 32  52  34
1 0 1 0 1
a  a  a   (5a x  3a z )
34 3 0 5 34
10
Hy  (5a x  3a z )  0.234a x  0.1404a z
2 (34)
H =H y  H z  0.7432a x  0.382a y  0.1404a z A/m
 200

Prob. 7.6
y
1 A
2 6A
P


B x
O 1 1

I
H   cos  2  cos 1  a
4
1 2
1  135o ,  2  45o ,   2 
2 2
 a x  a y   a x  a y  1 -1 1 0
a  al  a          az
 2   2  2 -1 -1 0
H


2
6
 cos 45o  cos135o  az  3 az 
4
2
H  0, 0, 0   0.954a z A/m

Prob. 7.7
10
(a) At (5,0,0),   5, a  a y , cos 1  0, cos  2 
125
2 10
H ( )a y  28.471a y mA/m
4 (5) 125
10
(b) At (5,5,0),   5 2, cos 1  0, cos  2 
150
 a x  a y  a x  a y
a  a z   
 2  2
2 10  a x  a y 
H ( )   13(a x  a y ) mA/m
4 (5 2) 150  2 
 201

10
(c ) At (5,15,0),   250  5 10, cos 1  0, cos  2 
350
 5a x + 15a y
 5a y - 15a x
a  a z   
 5 10  5 10
2 10  15a x  5a y 
H ( )   5.1a x  1.7a y mA/m
4 (5 10) 350  5 10 
d) At (5,-15,0), by symmetry,

H  5.1a x  1.7a y mA/m

Prob. 7.8
z C (0, 0, 5)

y
1
2
x A (2, 0, 0) B (1, 1, 0)
 202

(a) Consider the figure above.

AB  1, 1, 0    2, 0, 0    1, 1, 0 
AC   0, 0, 5   2, 0, 0    2, 0, 5
AB  AC  2, i.e AB and AC are not perpendicular.
AB  AC
cos 180o  1 
2 2
   cos 1  
AB AC 2 29 29
BC   0, 0, 5   1, 1, 0    1,  1, 5 
BA  1,  1, 0 
BC  BA 1  1
cos  2    0
BC BA BC BA
i.e. BC     1,  1, 5 ,   27
a  al  a 
 1, 1, 0    1,  1, 5 
 5, 5, 2 
2 27 54
10  2   5, 5, 2  5  5, 5, 2  A/m
H2   0    
4 27  29  2 27 2 29 27
 27.37 a x  27.37a y  10.95 a z mA/m

(b) H  H1  H 2  H 3   0,  59.1, 0    27.37, 27.37, 10.95 

  30.63, 30.63, 0 
  3.26 a x  1.1 a y  10.95a z mA/m

Prob. 7.9
y

(a) Let H  H x  H y  2H x
I
Hx   cos  2  cos 1  a
4 5A 1
2
x
O
 203

where a   a x  a y   a z , 1  180o ,  2  45o

Hx 
5
4  2 
 cos 45 o
 cos 180o   a z   0.3397a z

H
 2 H x  0.6792 a z A/m
(b) H  Hx  H y
5
where H x  1  0  a , a   a x  a y  a z
4  2 
 198.9a z mA/m
H y  0 since 1   2  0
H  0.1989 a z A/m
(c ) H  Hx  H y
5
where H x  1  0   a x  a z   198.9 a y mA/m
4  2 

1  0   a y  a z 
5
Hy   198.9 a x mA/m
4  2 
H  0.1989 a x  0.1989 a y A/m.

Prob. 7.10
For the side of the loop along y-axis,
I
H1   cos  2  cos 1  a
4
2
where a   a x ,   2 tan 30o  ,  2  30o , 1  150o
3

H1 
5
4 2
3
 cos 30o  cos 150o   a x   
15
8
ax

H  3H1   1.79a x A/m

 204

Prob. 7.11

3
Let H  H1  H 2  H 3  H 4
where H n is the contribution by side n. 4 2
(a) H  2H1  H 2  H 4 since H1  H 3
I 10  6 1  1
H1   cos  2  cos 1  a  
4 4  2   40  az
2
10  2  10  1 
H2   2  az , H 4   2  az
4  6   40  4  2   2
 5  3 1  5 5
H         a z = 1.964a z A/m
 2  10 2 6 10 2 2 
(b) At  4, 2, 0  , H  2  H1  H 4 
10 8 10 4
H1  az , H 4  az
4  2  20 4  4  20
2 5  1
 1 az 
  4 
H 1.78a z A/m

(c ) At  4, 8, 0  , H  H1  2H 2  H 3
10  4  10  8 1 
H1   2  az , H 2     az
4  8   4 5  4  4   4 5 2
10  2 
H3   a z 
4  4   2 
5  1 4 4 
H   az       0.1178a z A/m
8  5 5 2

(d ) At  0, 0, 2  ,
10  8  10
H1    0   ax  az    ay
4  2   68   68
10  4   2a  8a x  5  a x  4a z 
H2    0  ay   z  
4 68  84   68  17 84
 205

10  8   2a  4a y  a y  2a z
H3    0  ax   x  
4 20  84   20   21
10  4  5a x
H4  0    a y  a z  
4 2  20   20
 5 5   1 10   20 2 
H     ax     ay     az
 34 21  20    21  68   34 21  21 
 0.3457 ax  0.3165 a y  0.1798 az A/m

Prob. 7.12

I
H  4 H1  4 (cos  2  cos 1 )a
4
  a  2cm, I  5mA,  2  45o , 1  90o  45o  135o
a  a  a   a y  (a x )  a z
I 1 1 2I 2  5  103
H (  )a z  az  a  0.1125a z
a 2 2 a   2 102 z

Prob. 7.13
 206

(a ) Consider one side of the polygon as shown. The angle

subtended by the Side At the center of the circle
360o 2

n n
The field due to this side is

H1 
I
cos  2  cos 1 
4

 
where   r , cos  2  cos(90  )  sin
n n

cos 1   sin
n
I 
H1  2 sin
4 r n
nI 
H  nH1  sin
2 r n
3I 
(b) For n  3, H  sin
2 r 3
2
r cot 30o  2  r 
3
3 5 3 45
H     1.79 A/m.
2 2 2 8
3

4I  45 1
For n  4, H  sin  
2r 4 22  2
 1.128 A/m.
(c) As n  ,
nI  nI  I
H  lim sin   
n  2r n 2r n 2r
From Example 7.3, when h  0,
I
H 
2r
which agrees.
 207

Prob. 7.14
4

 3
1
2

2
Let H  H1  H 2  H 3  H 4
I 10
H1  az  a z  62.5 a z
4a 4  4  102

2 
cos  2  cos 90o  a z ,  2  tan 1
I 4
H2  H4   2.29o
4  4 10 100
 19.88 a z
I 100
H3  2 cos  a z ,   tan 1  87.7 o
4 1 4
10
 2 cos 87.7o a z  0.06361 a z
4
H   62.5  2 19.88  0.06361 a z
 102.32 a z A/m.

Prob. 7.15

4 2
o
0 x
1 1 2
 208

Let H  H1  H 2  H 3  H 4
which correspond with sides 1, 2, 3, and 4 as shown in the figure above.
H1 and H 3 can be found using eq. (7.12). It can be shown that
H 3   H1
Idl  R
For H 4 , dH 4  , dl  1d a , R   1a 
4 R 3
I 12 d a  (a  ) Id (a z )
dH 4  
413 41

I (a z ) o I  (a )
H4 
41 0  d  o z
41
Similarly, for H 2 ,
Io (a z )
H2 
4 2
Io  1 1 
H  H1  H 2  H 3  H 4     az
4  1  2 

Prob. 7.16
From Example 7.3,
I Ia 2
H1  az , H2  az
2a 2[a 2  d 2 ]3/2
I 1 a2  10  1 32 
H = H1  H 2    2 a  
2 3/2  z 2
 2 2  z
a
2  a [a  d ]  2  3 10 [3  4 ] (10 ) 
2 3/2

1 9 
 500   a z  202.67a z A/m
 3 125 

Prob. 7.17

2
2
 209

nI
H   cos  2  cos 1 
2

cos  2  -cos 1  2

 4
1
2 2
a2  

nI  0.5  150  2 102

H    69.63 A/m
 
1
2 a  
2 2 2 2 103  42  102
4

(b)
1 2

a 4
1  90o , tan  2    0.2   2  11.31o
b 20
nI 150  0.5
H  cos  2  cos 11.31o  36.77 A/m
2 2

Prob. 7.18

y
 P (4, 3, 2)

x
 210

Let H  Hl  H p
1
Hl  a
2
   4, 3, 2   (1, -2, 2)  (3, 5, 0),     34
3a x  5a y
a  , al  a z
34
 3a  5a y  3a y  5a x
a  al  a   a z   x  
 34  34
20  5a x  3a y 
Hl    x10  (  1.47a y  0.88a y ) mA/m
-3

2  34 
Hp 
1
2
K  an 
1
2
100 103  az   -a x    0.05a y A/m
H  H l  H p  1.47a x  49.12 a y mA/m
 211

Prob. 7.19

(a) See text

(b)

I
a

For   a,  H  dl  Ienc  0  H  0
I   2  a2 
For a    b, H  2 
  b2  a 2 
I   2  a2 
H   
2  b2  a2 
I
For   b, H  2  I  H 
2
Thus,

 0,  a

 I   2  a2 
H    2 2 , a   b
 2  b a 
 I
 ,  b
 2
 212

Prob.7.20
x

-1 1

1
H   K  an
2
1 1
 (20a x )  (a y )  (20a x )  a y
2 2
 10(a z )  10(a z )
 20a z A/m

Prob. 7.21
1 1
HP  k  an  10a x  a z  5a y
2 2

I I I
HL  a  (a x  a z )  ay
2 2 (3) 6

I
H P  H L  5a y  ay  0  I  30  94.25 A

6

Prob. 7.22
(a)
From eq. (7.29),
 I
 2 a 2 a , 0    a
H=
 I
a ,  a
 2
(b)
 I
1 d  a, a
J =  H= (  H  )a z    a 2 z
 d  0a z ,  a
 213

Prob. 7.23

For 0 <  < a

 L
H dl  I enc   J dS

2 Jo
H  2     d d 
 0
0

 J o 2
H  J o

For  > a
2 a
Jo
 H dl =  J dS =  
 
=0 =0

 d d 

H  2  J o 2 a
Joa
H 

 J o , 0< <a

Hence H    J o a
  ,  >a


Prob. 7.24
  
J

(a)    H  x y z  2( x  y )a z
y2 x2 0
(b) dS  dxdya z ,
5 2 2 5
 x2 2   y2 5 
I   J dS =  (2 x  2 y )dxdy  2  dy  xdx  2  dx  ydy  2(4)    2(2)  
S 1 0 0 1  2 0  2 1
 4(4  0)  2(25  1)  16  48  32 A

Prob. 7.25

1 d 1 d 2 2k
(a) J    H  (  H  )a z  ( ko )a z  o a z
 d  d a a
(b)For >a,
 214

2
a
2ko 2ko 2 a
 H  dl  I enc   J  dS   
 0  0 a
 d  d  
a
(2 )
2 0
ko a
H  2  2 ko a 
 H 

a
H  ko   a ,  a


Prob. 7.26

  
J    H  x y z  (2 x  2 y )a z
y2 x2 0
At (1,-4,7), x =1, y = -4, z=7,
J   2(1)  2(4)  a z  10a z A/m 2

Prob. 7.27
(a)
1  1 
J   H  (  H )a z  (103  3 )a z
   
 3 103 a z A/m
2

(b)
Method 1:
2 2
I   J dS   3  d d 10  3  10 3 3

2
d   d
S 0 0

 2 3
 3  103 (2 )  16  103 A  50.265 kA
3 2
Method 2:
2
I H dl 10   d  103 (8)(2 )  50.265 kA
3 2

L 0

Prob. 7.28

1 d 1 d
J   H  (  H  )a z  (4  2 )a z  8a z
 d  d
I   J  dS JS  8( a 2 )  8 104  2.513 mA
S
 215

Prob. 7.29

o I
(a) B  a
2
At (-3,4,5), =5.
4  107  2
B a  80a nWb/m 2
2 (5)
o I d  dz 4  107  2 6 4
   B  dS    ln  z
(b) 2  2 2 0
 16 10 ln 3  1.756  Wb
7

Prob. 7.30
Let H  H1  H 2
where H1 and H 2 are due to the wires centered at x  0 and x  10cm respectively.
(a) For H1 ,   50 cm, a  al  a   a z  a x  a y
5 50
H1  ay  a
2  5  10 
2
 y
For H 2 ,   5 cm, a   a z  a x  a y , H 2  H1
100
H  2H1  ay

 31.83 a y A/m

 2a  a y  2a y  a x
(b) For H1 , a  a z   x  
 5  5
5  a x  2a y 
H1  2     3.183a x  6.366a y
2 5 5 10  5 
For H 2 , a    a z  a y  a x
5
H2  a x  15.915a x
2  5 
H  H1  H 2
 12.3 a x  6.366a y A/m
 216

Prob. 7.31

(a) I   J dS

2 2
a
2 a
3
  
 0  0
J o (1 
a 2
)  d  d  J o  d  (  
0 0
a2
)d 

 2 4  2  2 a 2 
 2 J o   2 a
0  Jo  a  
 2 4a  2  2 
1
  a2 Jo
2

(b) H dl = I enc   J dS

For  < a,
H  2   J dS
 2 4 
= 2 J o   2
 2 4a 
2  2 
H  2  2 J o  2  
4  a2 
Jo   2 
H   2  
4  a2 

For  > a,

 H dl =  J o dS = I

1
H  2   a 2 J o
2
2
a Jo
H 
4
 Jo   2 
  2  ,   a
 4  a2 
Hence H  
 aJ o
,  >a
 4
 217

Prob. 7.32
0 I
B  a
2
da b 0 I
   B  dS    d z  0 2
d dz

μ 02
I
b π

d
+ d
a
I
n


Prob.7.33
For a whole circular loop of radius a, Example 7.3 gives
Ia 2 a z
H 3/ 2
2  a 2  h 2 
Let h 
0
I
H= az
2a
For a semicircular loop, H is halfed
I
H= az
4a
o I
B  o H  az
4a

Prob. 7.34

Bx By Bz

(a)   B    0
x y z
showing that B satisfies Maxwell’s equation.

(b) dS  dydza x
4 1
y3 1 4
   B  dS    y 2 dydz  ( z )  1 Wb
z 1 y  0
3 0 1
 218

B
(c )   H = J 
 J  
o
  
  B  x y z  2 za x  2 xa y  2 ya z
y 2 z 2 x2
2
J   ( za x  xa y  ya z ) A/m 2
o

(d)
Since   B =0,
 =  B  dS     Bdv  0
S v

Prob. 7.35

h
On the slant side of the ring, z     a
6
where H1 and H 2 are due to the wires centered at x  0 and x  10cm respectively.
o I
   B.dS   2 d dz

o I a b
h
(  a ) dz d o Ih a b  a

2   a
b
z 0 

2 b  a 
1   d
 
o Ih  ab
  b  a ln  as required.
2 b  a 
If a  30 cm, b  10 cm, h  5 cm, I  10 A,
4  107 10  0.05  4
   0.1  0.3 ln 
2 10  10 
2
 3
 1.37  108 Wb

Prob. 7.36

0.2 50o 106

   B dS  o  
z 0  0 
sin 2  d dz

50o
 cos 2 
  4 107 106  0.2    
 2  0

 0.04 1  cos 100o 

 0.1475 Wb
 219

Prob. 7.37

 /4 2 2  /4
20
   B dS    sin   d  d  20  d 
2
 sin
2
 d
S   0 1
  1 0
 /4
1 1  /4
 20(1) 
0
2
(1  cos 2 )d  10(  sin 2 )
2 0
 1
 10(  )  2.854 Wb
4 2

Prob. 7.38
   B dS , dS  r 2sin d d ar
S
2  /3
2
   cos  r 2sin d d  2  d  cos  sin  d
r3 r 1 0 0

sin   / 3
 /3 2
 2(2 )  sin  d (sin  )  4  2 sin 2 ( / 3)
0
2 0
 4.7123 Wb

Prob. 7.39

o J  R
4 v R 3
B  o H  dv

Since current is the flow of charge, we can express this in terms of a charge moving with
velocity u. Jdv = dqu.

 o  qu  R 
B
4  R 3 
In our case, u and R are perpendicular. Hence,
o qu 4 107 1.6 1019  2.2 106 1.6 1020
B   
4 R 2 4 (5.3  1011 ) 2 (5.3) 2  1022
 12.53 Wb/m 2
 220

Prob. 7.40

A
(a)    ya sin ax  0
  

A
  x y z
y cos ax 0 y  e-x
 a x  e  x a y  cos axa z  0
A is neither electrostatic nor magnetostatic field

1  1 
(b) B 
 
  B  
 
 20   0
 B  0
B can be E-field in a charge-free region.
1  2
(c )  C  (r sin ) = 0
r sin  
 1 3
C 
1
r sin  
 r 2 sin 2   ar -
r r
(r sin )a  0

C is possibly H field.
Prob. 7.41
(a) D  0
  
 D  x y z
y2z 2(x  1)yz -(x  1)z 2
 2(x  1)ya x  . . .  0
D is possibly a magnetostatic field.

1    sin  
(b) E   ( z  1) cos     0
  z   
1
 E  cos  a   . . .  0
2
E could be a magnetostatic field.
1  1   sin 
(c ) F   2cos  +    0
r r
2
rsin   r 2 
1   2 sin  
 F  
r  r
 r 1 sin   
r 2 
a  0

F can be neither electrostatic nor magnetostatic field.

 221

Prob. 7.42
o Idl o ILaz
A 
4 r 4 r
This requires no integration since L << r.
1 Az A
B   A  a   z a
  
But r   2  z 2

o ILa z
A
4 (  2  z 2 )1/ 2
Az o IL   IL 1
 (  2  z 2 )1/ 2  o ( )(  2  z 2 ) 3/ 2 (2  )
 4  4 2
o IL  a  IL  a
B  o 3
4 (   z )
2 2 3/ 2
4 r

Prob. 7.43

  
B  o H    A  x y z   sin  xa y  10 cos  ya z
10sin  y 0 4  cos  x
  
H  sin  xa y  10 cos  ya z 
o  
  
   
J   H  x y z   10 sin  ya x   cos  xa z 
o o  
0 sin  x 10 cos  y
2
J 10sin  ya x  cos  xa z 
o

Prob. 7.44
(a)
1  2 1  1   2 cos   1   sin 2  
 A  (r Ar )  ( A sin  )  0  2     3 
r r
2
r sin   r r  r  r sin    r 
1 1
 2 cos   4 2sin  cos   0
r 4
r sin 
 222

(b)
1   A  1  1 Ar  
B   A   ( A sin  )    ar  r  sin    r (rA )  a
r sin       
1  A 
  (rA )  r  a
r  r  
1    sin   2sin   1  2sin  2sin  
 0 ar  0a    2    a     a  0
r  r  r  3
r  r r3 r 3 

Prob. 7.45
  
(a) B   A  x y z
2x 2 y  yz xy 2  xz 3  6 xy  2z 2 y 2
B  (  6xz  4 x 2 y  3xz 2 )a x   y  6yz-4xy 2  a y   y 2  z 3  2 x 2  z  a z Wb/m 2

(b)

   6 xz  4 x y  3 xz 2  dy dz
2 2
  2
x 1
z 0 y 0

    6 xz  dy dz
0
 4  0
x 2 y dy dz  3  0
xz 2 dy dz
2 2
2 2 2 2
  6  dz  dy  4  dz  y dy  3 dy  z 2 dz
0 0 0 0
0 0

y 2 2 z  3 2
  6(2)  2   4(2)   +3(2)    -24+16+16
 2   3 
 0  0
  8 Wb
A x A y A z
(c )  A     4xy  2xy  6 xy  0
x y z
  B   6 z  8 xy  3 z 3  6 z  8 xy  1  3z 3  1  0

As a matter of mathematical necessity,

  B    (  A)  0
 223

Prob. 7.46

1   A  1  1 Ar  
B   A   ( A sin  )    ar    (rA )  a
r sin      r  sin   r 
1 k  1 
 (sin 2  )ar  (kr 1 ) sin  a
r sin  r 
2
r r
k k sin  2k cos  k sin 
 3 2sin cos ar  a  ar  a
r sin  r 3
r 3
r3

Prob. 7.47
1 Az A
B   A  a   z a
  
15
 e   cos  a  15 e   sin  a

   1 1
B  3, , -10   5 e 3 a   15 e 3 a
 4  2 2
B 107 15 3  1 
H   e  a  a 
o 4 2 3 
H  14 a   42 a  104 A/m
15
   B  dS    e   cos   d dz

 15 z  sin   0 2 e5  150 e 5    1.011 Wb
10
0

Prob. 7.48

1   A  1  1 Ar  
B   A    ( A sin  )    ar  r  sin    r (rA )  a
r sin 
   
1  A 
  (rA )  r  a
r  r  
1 10 1 
 2sin  cos  ar  (10) sin  a  0a
r sin  r r r
20
B  2 cos  ar
r
At (4, 60o , 30o ), r = 4,  =60o
B 1  20 
H  7  2
cos 60o ar   4.974  105 ar A/m
o 4  10  4 
 224

Prob. 7.49

Applying Ampere's law gives

H   2  J o   2
Jo
H   
2 a
Jo 
B  o H   o
2
AZ
But B   A   a  . . .

AZ 1 Jo  2
   Jo  
 AZ   o
 2 4
1
or A  - o J o  2 a z
4
Prob. 7.50

  
A A
B    A  x y z  z ax  z a y
y x
0 0 Az ( x, y )
 x y  x y
 sin sin ax  cos cos ay
2 2 2 2 2 2

Prob. 7.51

1  1 
B   A  ( A sin  )ar  (rA )a
r sin   r r
1 Ao A
 (2sin  cos  )ar  o sin  (r 2 )a
r sin  r 2
r
A
 3o (2 cos  ar  sin  a )
r
 225

Prob. 7.52

  
(a) J    H  x y z  (2 yz  x 2 )a x  (2 xz  2 xy )a z
xy 2 x 2 z  y 2 z
At (2,-1,3), x=2, y=-1, z=3.
J  2a x  16a z A/m 2
 v
(b)     J  0  2x  2x  0
t

At (2,-1,3),
v
 0 C/m3s
t

Prob. 7.53

(a) B    A
 1 Az A   A A  1  A 
   a      z  a   (  A )    a z
   z   z       
A
  z a  20  a  Wb/m 2

B 20 
H  a  A/m
o o

1 
J   H  (  A )a z
 
1 40
 (40  )a z  a z  A/m 2
o  o

2
40
2
(b) I   J dS     d d  , dS =  d d  a z
o  
0 0
2
40 40  2
2
   d   d  (2 )
2

o 0 0
o 2 0

80  2  106
  400 A
4  107
 226

Prob. 7.54 H   Vm  Vm    H  dl  mmf

Ia 2
From Example 7.3, H  az
2  z2  a 2 
3
2

Ia 2  Iz
 z  a2 
3
Vm   2 2
dz   c
2  z2  a 2 
1
2 2

As z  , Vm  0 , i.e.
I I
0    c  c 
2 2
Hence,
I  z 
Vm  1  
2  z2  a 2 
Prob. 7.55
I
H  a
2
But H   Vm  J  0
I 1 Vm I
a   a  Vm     C
2    2
 I 
At 10, 60o , 7  ,   60o  , Vm  0  0     C
3 2 3
I
or C 
6
I I
Vm    
2 6

At  4, 30 ,  2  ,
o
  30o 
6
,

I  I I 12
Vm      
2 6 6 12 12
Vm  1A
 227

Prob. 7.56
For an infinite current sheet,
1 1
H  K  an  50ay  an  25ax
2 2
But H   Vm  J  0
Vm
25 ax  
an  Vm   25x  c
x
At the origin, x  0, Vm  0, c  0, i.e.
Vm   25 x
(a) At  2, 0, 5  , Vm  50A.

(b) At 10, 3, 1 , Vm  250A.

Prob. 7.57

 V 1 V V 
(a)   V     a  a  az 
    z 
 1  2V 1  2V    2V  2V 
    a     a
  z  dz    z z 
1   2V  2V 
    az  0
    d  
 1 Az A 
(b)      A        a
   z 
 A Az  1  A  
   
 
 a  
  
 A 
   az 
  
 z
1  2 Az 1   A  1  A 1  2 Az
2

      
     z   z  
 1     1 A 
 
z   
  A    
z    


 2 A 1 A  2 A 1 A
      0
z  z z  z
 228

Prob. 7.58

1       2  2
1
'               
2 2
 a a a z  x x' y y' z z'
 x ' y ' z '   
x y
R
 1 2  2
3
     
   x 
           a y and a z terms

2 2
 2 x x' a x x' y y ' z z'
 2
R

R3
1
  x  x'    y  y '   z  z '  
2 2
R  r  r'
2 2

 
1       2  2
1
             

2 2
 a a a z  x x' y y' z z'
 x y z  
x y
R
2  2
3
1
  2  x  x'  a x  x  x'    y  y'    z  z' 
   a y and az terms
2 2

2  
  x  x' a z   y  y ' a y   z  z' a z  R
   3
R3 R