1
CHAPTER 1
P. E. 1.1
A  B  1,0,3  5,2,6   6,2,3
(a)
A  B  36  4  9  7
(b)
5 A  B  5,0,15  5,2,6   0,2,21
(c)
The component of A along ay is
(d)
Ay = 0
3 A  B  3,0,9   5,2,6   8,2,3
A unit vector parallel to this vector is
8,2,3
a11 
64  4  9
  0.9117a x  0.2279a y  0.3419a z 
P. E. 1.2 (a) rp  a x  3a y  5a z
rR  3a y  8a z
(b)
The distance vector is
rQR  rR  rQ  (0,3,8)  (2, 4, 6)  2a x  a y  2a z
(c)
The distance between Q and R is
| rQR | 4  1  4  3
Consider the figure shown on the next page:
40
uZ  uP  uW  350a x 
 a x  a y 
2
 378.28a x  28.28a y km/hr
P. E. 1.3
or
uz  379.3175.72 km/hr
Where up = velocity of the airplane in the absence of wind
uw = wind velocity
uz = observed velocity
15
CHAPTER 2
P. E. 2.1
(a) At P(1,3,5),

x = 1,
x  y
2
2
=
y = 3,
10 ,
z =5,
z = 5,
  tan 1 y / x  tan 1 3  71.6o
P(  ,  , z)  P( 10 , tan 1 3,5)  P(3.162,716
. o ,5)
Spherical system:
r
x2  y2  z2 
35  5.916
  tan 1 x 2  y 2 z  tan 1 10 5  tan 1 0.6325  32.31
P (r , ,  )  P (5.916,32.31, 71.57)
At T(0,-4,3),
x=0
y =-4,
z =3;
  x  y  4, z  3,   tan y / x  tan  4 / 0  270
T ( ,  , z)  T (4,270 ,3).
2
1
2
1
Spherical system:
r
x 2  y 2  z 2  5,  tan 1  / z  tan 1 4 / 3  5313
. .
T (r , ,  )  T (5,5313
.  ,270 ).
At S(-3-4-10),
x =-3, y =-4, z =-10;
 4 
  x 2  y 2  5,   tan 1    233.1
 3 
S (  ,  , z )  S (5, 233.1, 10).
Spherical system:
r  x 2  y 2  z 2  5 5  11.18.
  tan 1  z  tan 1
5
 153.43;
10
S (r ,  ,  )  S (11.18,153.43, 233.1).
(b)
In Cylindrical system,
Qx 

  z2
2
;
  x2  y 2 ;
yz  z  sin  ,
z  sin 
Qy  0;
Qz  
 2  z2
;
15
16
 Q   cos 
  
 Q     sin 
 Qz   0
Q  Qx cos  
 cos 
 z
2
2
sin 
cos 
0
0  Qx 
 
0  0  ;
1  Qz 
,
Q  Qx sin  
  sin 
 2  z2
Hence,
Q

 2  z2
(cos  a   sin  a   z sin  a z ).
In Spherical coordinates:
r sin 
 sin  ;
Qx 
r
1
Qz  r sin  sin  r cos   r sin  cos sin  .
r
Qr   sin  cos  sin  sin  cos  Qx 
  
 
Q   cos cos  cos sin   sin    0  ;
 
cos 
0  Qz 
Q    sin 
Qr  Qx sin  cos   Qz cos  sin 2  cos   r sin  cos2  sin  .
Q  Qx cos cos   Qz sin   sin  cos cos   r sin 2  cos sin  .
Q  Qx sin    sin  sin .
 Q  sin   sin  cos   r cos2  sin   a
r
 sin  cos (cos   r sin  sin  )a  sin  sin  a .
At T :
4
12
a x  a z  0.8a x  2.4a z ;
5
5
4
Q (  ,  , z )  (cos 270 a   sin 270 a  3sin 270a z
5
 0.8a  2.4 a z ;
Q ( x, y , z ) 
4
45
4 3
20
4
Q (r ,  ,  )  (0  (1))ar  ( )(0  (1))a  (1)a
5
25
5 5
5
5
36
48
4
 a r  a  a  1.44ar  1.92a  0.8 a ;
25
25
5
16
17
Note, that the magnitude of vector Q = 2.53 in all 3 cases above.
P.E. 2.2 (a)
 Ax  cos sin 0  z sin 
 A   sin cos 0  3 cos 
 y 


 Az   0
0
1  cos sin
A  (z cos sin 3 cos sin) ax  (z sin2  3 cos2 ) ay   cos sin az .
y
x
y
But   x2  y2 , tan  , cos 
, sin 
;
2
2
2
2
x
x y
x y
Substituting all this yields:
1
A
[(xyz 3xy)ax  (zy2  3x2 ) ay  xy az ].
2
2
x y
 Bx  sin  cos 
 B    sin  sin 
 y 
 Bz   cos 
cos  cos 
cos  sin 
 sin 
Since r  x 2  y 2  z 2 , tan  
and sin  
and sin  
x2  y 2
x y z
2
2
y
x y
2
2
,
2
 sin    r 2 


cos    0 
0  sin  
x2  y2
y
, tan   ;
z
z
z
, cos  
cos  
x  y2  z2
2
x
x  y2
2
;
;
y
1
 ( r 2 x  y ).
r
r
x
1
B y  r 2 sin  sin   sin  cos   ry   (r 2 y  x ).
r
r
1
Bz  r 2 cos   r z  ( r 2 z ).
r
Bx  r 2 sin  cos   sin  sin   rx 
Hence,
B
1
x y z
2
2
2
[{x ( x 2  y 2  z 2 )  y} a x  { y ( x 2  y 2  z 2 )  x} a y  z ( x 2  y 2  z 2 )a z ].
17
18
P.E.2.3 (a) At:
(1,  / 3, 0),
H  (0, 0.06767,1)
1
a x  cos  a   sin  a  (a   3 a )
2
H  a x  0.0586.
(b)
At:
(1,  / 3, 0),
a  cos  a  sin  a z   a z .
H
a
 az  0
0
(c)
a
az
0.06767 1   0.06767 a  .
0
1
( H  a  ) a   0 a .
H az 
(d)
a
a
az
0
0.06767
1
0
0
1
 0.06767 a  .
H  a z  0.06767
A
P.E. 2.4
(a)
B  (3, 2,  6)   4, 0,3   6.
A B 
B
A
(b)
3 2 6
 6 ar  33a  8a .
4 0 3
Thus the magnitude of

 34.48.
(c )
At (1,  / 3, 5 / 4),    / 3,
a z  cos  a r  sin  a 
1
3
ar 
a .
2
2
3 
3
 1
( A a z )a z    3   ar 
a   0.116ar  0.201a
2
2
 2

18
19
Prob. 2.1
(a)
  x 2  y 2  4  25  5.3852,
  tan 1
y
 tan 1 2.5  68.2o
x
x2  y 2
5.3852
 tan 1
 79.48o
1
z
o
o
P (r ,  ,  )  P(5.477, 79.48 , 68.2 )
r  x 2  y 2  z 2  4  25  1  5.477,
P (  ,  , z )  P(5.3852, 68.2o ,1),
  tan 1
(b)
  x 2  y 2  9  16  5,
r  x 2  y 2  z 2  5,
  tan 1
y
4
 tan 1
 360o  53.123o  306.88o
3
x
x2  y 2
 tan 1   90o
z
P(r ,  ,  )  P(5,90o ,306.88o )
  tan 1
Q(  ,  , z )  Q(5,306.88o , 0),
(c )
  x 2  y 2  36  4  6.325,
  tan 1
y
2
 tan 1  18.43o
x
6
r  x 2  y 2  z 2  36  4  16  7.483,
x2  y 2
6.325
 tan 1
 180o  57.69o  122.31o
4
z
o
R(  ,  , z )  R(6.325,18.43 , 4),
R(r ,  ,  )  R(7.483,122.31o ,18.43o )
  tan 1
Prob. 2.2
(a)
x   cos   2 cos 30  1.732;
y   sin   2sin 30  1;
z  5;
P1 ( x, y, z )  P1 (1.732,1, 5).
(b)
x  1cos 90  0;
y  1sin 90 1;
z   3.
P2 ( x, y, z )  P2 (0, 1,  3).
19
20
(c)
(d)
x  r sin  cos   10sin( / 4) cos( / 3)  3.535;
y  r sin  sin   10sin( / 4) sin( / 3)  6.124;
z  r cos 
 10 cos( / 4)
 7.0711
P3 ( x, y, z )  P3 (3.535, 6.124, 7.0711).
x  4sin 30 cos 60  1
y  4sin 30 sin 60 1.7321
z  r cos   4 cos 30  3.464
P4 ( x, y, z )  P4 (1,1.7321,3.464).
Prob. 2.3
  x 2  y 2  4  36  6.324
y
6
 tan 1  71.56o
x
2
o
P is (6.324, 71.56 , 4)
(a)   tan 1
r  x 2  y 2  z 2  4  36  16  7.485
x2  y2
6.324
4
(b)   tan
 tan 1
 90o  tan 1
 122.3o
z
4
6.324
o
o
P is (7.483,122.3 , 71.56 )
1
Prob. 2.4
(a)
x   cos   5cos120o  2.5
y   sin   5sin120o  4.33
z 1
Hence Q  (2.5, 4.33,1)
20
21
(b)
r  x 2  y 2  z 2   2  z 2  25  1  5.099
  tan 1
x2  y2
5

 tan 1  tan 1  78.69o
z
z
1
  120o
Hence Q  (5.099, 78.69o ,120o )
Prob. 2.5
T (r , ,  )


r  10,   60o ,   30o
x  r sin  cos   10sin 60o cos 30o  7.5
y  r sin  sin   10sin 60o sin 30o  4.33
z  r cos   10 cos 60o  5
T ( x, y, z )  (7.5, 4.33,5)
  r sin   10sin 60o  8.66
T (  ,  , z )  (8.66,30o ,5)
Prob. 2.6
(a)
x   cos  ,
y   sin  ,
V   z cos    2 sin  cos    z sin 
(b)
U  x2  y 2  z 2  y 2  2 z 2
 r 2  r 2 sin 2  sin 2   2r 2 cos 2 
 r 2 [1  sin 2  sin 2   2 cos 2  ]
21
22
Prob. 2.7
(a)


x
 2

   z2 
sin  0  

y



cos  0 
  2  z2 
0
1  

4


 2
2 
  z 
 F   cos 
 F     sin 
  
 Fz   0
F 
F 
Fz 
_
F 
1
 z
2
[  cos 2    sin 2  ] 
2
1
 2  z2
4
  z2
2
1
  z2
2

  z2
2
;
[  cos  sin    cos  sin  ]  0;
;
(  a  4 az )
In Spherical:
 x
 
 Fr 
sin cos sin sin cos   r 
 

  y
F   cos cos cos sin sin   r 
F 
 sin
cos
0   
 
4
 r 
r
r
4
4
Fr  sin2  cos2   sin2  sin2   cos
 sin2   cos;
r
r
r
r
4
4
F sin cos cos2   sin cos sin2   sin  sin cos  sin ;
r
r
F  sin cos sin  sin sin cos  0;
_
4
4
 F  (sin2   cos ) ar  sin (cos  )a
r
r
22
23
(b)
 x 2 
 2

   z2 
sin  0  

y 2 


cos  0 
  2  z2 
0
1  

 z 2 
 2

2
   z 
G   cos 
 G     sin 
  
 Gz   0
2
G 
[  cos 2    sin 2  ] 
 2  z2
3
 2  z2
;
G  0;
z 2
Gz 
 2  z2
2
G
 2  z2
;
(  a  z az )
Spherical :
G
2
( xa x  ya y  za z ) 
r
r 2 sin 2 
rar  r 2 sin 2  ar
r
Prob. 2.8
B   ax 
y

a y  za z
 B   cos 
 B     sin 
  
 Bz   0
B   cos  
sin 
cos 
0
y

B    sin  
0   
0   y /  
1   z 
sin 
y

cos 
Bz  z
But y   sin 
B   cos   sin 2  , B    sin   sin  cos 
Hence,
B  (  cos   sin 2  )a   sin  (cos    )a  za z
23
24
Prob. 2.9
 Ax  cos 
 A    sin 
 y 
 Az   0
 sin 
cos 
0
At P,   2,
   / 2,
0 2
0   3 
1   4 
z  1
Ax  2 cos   3sin   2 cos 90o  3sin 90o  3
Ay  2sin   3cos   2sin 90o  3cos 90o  2
Az  4
Hence, A  3a x  2a y  4a z
Prob. 2.10
(a)
 Ax  cos 
  
 Ay    sin 
 Ay   0
 
 sin 
cos 
0
0    sin  
0    cos  
1   2 z 
Ax   sin  cos    cos  sin   0
Ay   sin 2    cos 2     x 2  y 2
Az  2 z
Hence,
A  x 2  y 2 a y  2 za z
(b)
 Bx 
B  
 y
 Bz 
sin cos  cos  cos 
 sin sin cos  sin

 cos 
 sin
 sin 
cos  
0 
4 r cos  


r


 0 
Bx  4 r sin cos2   r cos  cos 
By  4 r sin sin cos   r cos  sin
Bz  4 r cos  cos   r sin
But
2
2
sin 
2
r  x y z ,
sin 
y
2
x  y2
,
cos  
x2  y 2
,
r
x
cos  
z
r
x2  y 2
24
25
Bx  4 x2  y 2
x2

x2  y 2
By  4 x 2  y 2
xy

x  y2
x  y2
zy
2
x
Bz  4 z
B
zx
2
2
x y
1
2
x y
2
2
x2  y 2
 x2  y 2
 x(4 x  z)ax  y(4 x  z)ay  (4 xz  x 2  y 2 )az 
Prob. 2.11
Method 1:
 Fx  sin  cos 
 F    sin  sin 
 y 
 Fz   cos 
cos  cos 
cos  sin 
 sin 
4
sin  cos  ,
r2
Fy 
r 2  x2  y2  z 2 ,
sin  
Fx 
sin  
Fx 
y
x2  y2
4
2
x  y2  z2
4
Fy  2
x  y2  z2
,
 sin    4 / r 2 


cos    0 
0   0 
4
sin  sin  ,
r2
cos  
x2  y 2
x y z
2
2
x
x2  y2
x2  y 2
x y z
cos  
,
z
x  y2  z2
2
x2  y 2
x2  y2  z 2
2
2
x
x2  y 2
2
4
Fz  2 cos 
r
y
2
x y
2
2

4x
( x  y 2  z 2 )3/2

4y
( x  y 2  z 2 )3/ 2
2
2
z
4
4z
 2
2
2
2
2
2
x  y  z ( x  y  z ) ( x  y 2  z 2 )3/2
Thus,
4
 xa x  ya y  za z 
F 2
2
( x  y  z 2 )3/2 
Fx 
2
25
26
Method 2:
4a r 4ra
F  2r .  3 r
r r
r
4
 xa x  ya y  za z 
F 2
( x  y 2  z 2 )3/2 
Prob. 2.12
   / 2,   3 / 2
B  2sin( / 2)ar  4 cos(3 / 2)a  2ar
r  2,
(a)
(b)
 Bx   sin  cos  cos  cos   sin    r sin  
 B    cos  cos  cos  sin  cos   

0
 y 


 Bz   cos 
 sin 
0    r 2 cos  
Bx  r sin 2  cos   r 2 sin  cos  ,
By  r sin  cos  cos   r 2 cos 2 
Bz  r sin  cos 
But

z
r  x  y  z , cos   ,sin   
r
r
2
cos  
x

2
x

x2  y2
Bx  x 2  y 2  z 2

,
sin  
x2  y 2
x2  y 2  z 2
y


x2  y2
x2  y 2  z 2
y
x2  y 2
x
x2  y 2
 ( x2  y 2  z 2 )
xy
x  y2
 ( x2  y2  z 2 )
x2
x2  y2
2
x x2  y 2
xy ( x 2  y 2  z 2 )

x2  y2
x2  y 2  z 2
By  x 2  y 2  z 2

2
xz
x2  y 2  z 2

z x2  y2
x2  y 2  z 2
2
x2  y 2
x2 ( x2  y2  z 2 )
x2  y 2
z x2  y 2
Bz  x  y  z 2

x  y2  z2
2
x
2
z x2  y 2
x2  y2  z 2
B  Bx a x  By a y  Bz a z
Prob. 2.13
26
27
x   cos 
(a)
B   cos  a z
x  r sin  cos 
(b) B  r sin  cos  a z ,
Bx  0  By , Bz  r sin  cos 
 Br   sin  cos  sin  sin  cos   
0

  



0
 B   cos  cos  cos  sin   sin   

 B    sin 
cos 
0   r sin  cos  
 
Br  r sin  cos  cos   0.5r sin(2 ) cos 
B  r sin 2  cos  ,
B  0
B  0.5r sin(2 ) cos  ar  r sin 2  cos  a
Prob. 2.14
(a)
a x  a   (cos  a   sin  a )  a   cos 
a x  a  (cos  a   sin  a )  a   sin 
a y  a   (sin  a   cos  a )  a   sin 
_
_
a y  a  (sin  a   sin  a )  a  cos 
(b) and (c)
In spherical system :
a x  sin  cos  ar  cos  cos  a  sin  a .
a y  sin  sin  a r  cos  sin  a  cos  a .
a z  cos  a x  sin  a .
27
28
Hence,
a x  a r  sin  cos  ;
a x  a  cos  cos  ;
a y  a r  sin  sin  ;
a y  a  cos  sin  ;
_
_
_
_
a z  a r  cos  ;
a z  a   sin  ;
Prob. 2.15
(a)
a   cos  a x  sin  a y ,
a   sin  a x  cos  a y
a   a 
cos 
 sin 
sin 
cos 
az  a 
0
0
cos  sin 
1
  sin  a x  cos  a y  a
0
a  a z 
 sin 
0
0
 cos  a x  sin  a y  a 
1
0
 (cos 2   sin 2  )a z  a z
0
cos 
0
(b)
ar  sin  cos  a x  sin  sin  a y  cos  a z
a  cos  cos  a x  cos  sin  a y  sin  a z
a   sin  a x  cos  a y
ar  a 
sin  cos 
cos  cos 
sin  sin 
cos  sin 
cos 
 sin 
 ( sin 2  sin   cos 2  sin  )a x  (cos 2  cos   sin 2  cos  )a y
 (sin  cos  sin  cos   sin  cos  sin  cos  )a z
  sin  a x  cos  a y  a
28
29
a  ar 
 sin 
cos 
sin  cos  sin  sin 
0
cos 
 cos  cos  a x  cos  sin  a y  ( sin  sin 2   sin  cos 2  )a z
 cos  cos  a x  cos  sin  a y  sin  a z  a
a  a 
cos  cos 
 sin 
cos  sin 
cos 
 sin 
0
 sin  cos  a x  sin  sin  a y  (cos  cos 2   cos  sin 2  )a z
 sin  cos  a x  sin  sin  a y  cos  a z  ar
Prob. 2.16
(a)
r
 2  z2 .
x2  y2  z2 

  tan 1 ;
z
  .
or

x 2  y 2  r 2 sin 2  cos2   r 2 sin 2  sin 2  .
 r sin  ;
z  r cos ;
  .
(b) From the figures below,
cos a
z
z
a
a
az
ar
-az

sin a

sin  (  az )
cos  a z


29
30
a r  sin a   cos  a z ;
a  cos  a   sin a z ;
a  a .
Hence,
 
 ar   sin
  
 a   cos 
   0
 a  
 
 
a
cos    
 
0  sin   a 
1
0   
a z 
 
From the figures below,
0
a   cos  a  sin a r ; a z  cos  a r  sin a ; a  a .
z
sin  ar
z
az
sin  (  a )
cos a
a
a
cos ar
ar


a
ar

 
a    sin 
  
 a    0
   cos 
 az  
 

 
ar
0  
 
1  a 
0   
 az 
 
cos 
0
 sin 
Prob. 2.17
At P(2, 0, 1),
  0,


 1 
  cos 1 
 116.56o

2
2
2
 x y z 
 5


  cos 1 
z
(a) a   a x  cos   1
(b) a  a y  cos   1
(c) ar  a z  cos   0.4472
30
31
Prob. 2.18
If A and B are perpendicular to each other, A B = 0
A B   2 sin 2    2 cos 2  - 2
= 2 (sin 2   cos 2  )- 2
= 2   2
=0
As expected.
Prob. 2.19
(a) A  B  8a   2a  7az
(b) A B = 15 + 0 - 8 = 7
(c ) A  B =
3 2
1
5 0 8
=-16a   (5  24)a  10a z
=-16a   29a  10a z
A B
7
7


AB
9  4  1 25  64
14 89
=0.19831
(d ) cos AB 
 AB =78.56o
Prob. 2.20
 Gx  cos   sin  0  G 
G    sin  cos  0   G 
 y 
  
 Gz   0
0
1   Gz 
Gx  G cos   G sin   3 cos    cos  sin 
 3x  x sin   3(3)  (3) sin(306.87o )  11.4
Gx  Gx a x  11.4a x
31
32
Prob. 2.21
G   cos  sin  0   yz 
 G     sin  cos  0   xz 
  
 
 Gz   0
0
1   xy 
G  yz cos   xz sin 
x   cos  , y   sin  ,
yz   z sin  , xz   z cos 
G   z sin  cos    z cos  sin   2  z sin  cos    z sin 2
G   yz sin   xz cos    z (cos 2   sin 2  )   z cos 2
Gz  xy   2 cos  sin   0.5 2 sin 2
G   z sin 2 a y   z cos 2 a  0.5 2 sin 2 a z
Prob. 2.22
 Ax 
 cos   sin  0   A 
 

  
 Ay    sin  cos  0   A 
 Az 
 0
0
1  Az 
x
y



0
 2
2
2
2
x y
  A 
 x y
y
x
  

  2
0   A 
2
2
2
x y
 A 
 x y
0
0
1  z 



 Ax 
sin cos  cos  cos 
 A    sin sin
cos  sin
 y

 Az 
 cos 
 sin

x

2
 x  y 2  z2


y

2
2
2

 x y z

z

 x2  y 2  z2

 sin   Ar 
 
cos    A 
0   A 
xz
x2  y 2 x2  y 2  z2
yz
2
x y

2
x2  y 2  z2
x2  y 2
x2  y 2  z 2


2
2 
x y
  Ar 
 
x
A 
2
2  
x  y  A 
  

0


y
32
33
Prob. 2.23 (a) Using the results in Prob.2.14,
A  z sin   r 2 sin  cos sin 
A  3 cos   3r sin  cos 
Az   cos  sin   r sin  cos  sin 
Hence,
 Ar   sin 
  
 A    cos 
A   0
  
0 cos    r 2 sin  cos  sin  


0  sin    3r sin  cos  
1
0   r sin  cos  sin  


A(r , , )  r sin sin cos   r sin  cos   ar  sin r cos2   sin cos  a  3 cos  a 


At (10 ,  / 2,3 / 4),
r  10 ,    / 2,   3  / 4
A  10(0ar  0.5a 
(b)
Br  r 2  ( 2  z 2 ),
3
a )  5a  21.21a
2
B  0 ,
 B   sin 
  
 B    0
 Bz   cos 
B  sin  
cos 
0
 sin 

 2  z2
0   Br 
 
1  B 
0   B 



z

B(  ,  , z )   2  z 2   a  2
a
a


z
  z2


At (2,  / 6 ,1),
  2,    / 6 , z  1
B  5(2a   0.4a  a z )  4.472a   0.8944a  2.236a z
Prob. 2.24
(a) d 
(b)
(6  2) 2  (  1  1) 2  (2  5) 2 
29  5.385
d 2  32  52  2(3)(5) cos   (  1  5) 2  100
d  100  10
33
34
(c)

 3
sin cos(7 
)
4
6
4
6
4 4
 125  100(0.7071)(0.866)  100(0.7071)(0.5)(0.2334)
 125  61.23  35.33  99.118
d2  102  52  2(10)(5)cos

cos

 2(10)(5)sin

d  99.118  9.956.
Prob. 2.25
Using eq. (2.33),
d 2  r12  r22  2r1r2 cos 1 cos  2  2r1r2 sin 1 sin  2 cos(2  1 )
 16  36  2(4)(6) cos 30o cos 90o  2(4)(6) sin 30o sin 90o cos(180o )
 16  36  0  48(0.5)(1)(1)  52  24  76
d  8.718
Prob. 2.26
 a   cos 
 a     sin 
  
 a z   0
sin  0   a x 
cos  0  a y 
0
1   a z 
a   cos  a x  sin  a y , a   sin  a x  cos  a y
At (0, 4, -1),   90o
a   sin 90o a y  a y
a   sin 90o a x  a x
Prob. 2.27
At (1, 60o , 1),
  1,   60o , z  1,
(a) A  (2  sin 60o )a  (4  2 cos 60o )a  3(1)(1)a z
 2.866 a   5a  3a z
B  1cos 60o a   sin 60o a  a z  0.5a  0.866a  a z
A B  1.433  4.33  3  5.897
AB  2.8662  26  9 0.25  1  0.8662  9.1885
cos  AB 
A B 5.897

 0.6419
AB 9.1885

  AB  50.07o
34
35
Let D = A  B. At (1,90o , 0),
  1,   90o , z  0
(b) A   sin 90o a   4a  a   4a
B  1cos 90o a   sin 90o a  a z  a  a z
a a
D  A  B  1 4
0 1
aD 
az
0  4a   a  a z
1
D
(4,1, 1)

 0.9428a   0.2357a  0.2357a z
D
16  1  1
Prob.2.28
  90;
 Bx   cos   sin  0   B 
 B    sin  cos  0   B 
 y 
  
At P(0, 2, 5),
 Bz 
 0
0
1   Bz 
0 1 0   5
 1 0 0   1 
0 0 1   3
B  a x  5a y  3a z
(a) A  B  (2, 4,10)  (1, 5, 3)
 ax  a y  7 az .
(b) cos  AB 
A B

AB
 AB  cos1(
(c) AB  A  a B 
52
4200
52
4200
)  143.36.
A B
52

 8.789.
B
35
Prob. 2.29
35
36
B  a x  Bx
 Bx   cos   sin  0   B 
 B    sin  cos  0   B 
 y 
  
 Bz   0
0
1   Bz 
Bx  B cos   B sin    2 sin  cos   ( z  1) cos  sin 
 16(0.5)  (2)(0.5)  8  1  9
Prob. 2.30
_
G  cos2  ax 
_
2 r cos  sin _
ay  (1 cos2  )az
r sin
_
_
_
 cos2  ax  2 cot  sin ay  sin2  az
 Gr 
 
 G  
G 
 
 sin cos  sin sin

cos  cos  cos  sin
  sin
cos 
cos    cos2  


 sin  2 cot  sin 
0   sin2  
Gr  sin cos3   2 cos  sin2   cos  sin2 
 sin cos3   3 cos  sin2 
G  cos  cos3   2 cot  cos  sin2   sin sin2 
G   sin cos2   2 cot  sin cos 
_
G  [sin cos3   3 cos  sin2  ]ar
 [cos  cos3   2 cot  cos  sin2   sin sin2  ]a
 sin cos (2 cot   cos  )a
Prob. 2.31
(a)
An infinite line parallel to the z-axis.
(b)
Point (2,-1,10).
(c)
A circle of radius r sin   5 , i.e. the intersection of a cone and a sphere.
(d)
An infinite line
parallel to the z-axis.
(e) A semi-infinite line parallel to the x-y plane.
(f) A semi-circle of radius 5 in the y-z plane.
36
37
Prob. 2.32
(a) J z  ( J  a z )a z .
At (2,  / 2, 3 / 2), a z  cos  ar  sin  a   a .
J z   cos 2 sin  a   cos  sin(3 / 2) a   a .


(b) J  tan ln r a  tan ln 2 a  ln 2a  0.6931a .
2
4
(c) J t  J  J n  J  J r   a  ln 2 a   a  0.6931a
(d )
J P  ( J  a x )a x
a x  sin  cos  ar  cos  cos  a  sin  a  a .
At (2,  / 2, 3 / 2),
J P  ln 2a .
Prob. 2.33
H ax  H x
 H x   cos 
 H    sin 
 y 
 H z   0
 sin 
cos 
0
0    2 cos  


0     sin  
1   0 
H x   2 cos 2    sin 2 
At P,   2,   60o , z  1
H x  4(1/ 4)  2(3 / 4)  1  1.5  2.5
Prob. 2.34
(a) 5  r  ax  r  ay  x  y
(b)
10  rxaz 
a plane
x y z
|yax  xay | x 2  y 2  
0 0 1
a cylinder of infinite length
37
38
CHAPTER 3
P. E. 3.1
  60
(a) DH 



 r sin  d r  3,90  3(1)[ 3  4 ]  4  0.7854.

o
 45
  90 
(b) FG 
 rd
r5
  60 
 
5
 5(  ) 
 2.618.
2 3
6
(c)
  90    60 

AEHD 
 r sin  d d
2
  60    45 
r3
90 
  60 
 9 (  cos  )|  60
  |  45 
1 
3
 9 ( )( ) 
 1178
. .
2 12
8
(d)
r  5  90
r2 r  5  
4
ABCD    rd dr 
(  )
 4.189.
2 r 3 2 3
3
r  3   60
(e)
r 5
Volume  
r 3

  60


 45
 90
2
 r sin  dr d d 
  60
r3
3
r 5
r 3
( cos  )
 90
  60

  60
  45
49
 4.276 .
36
P.E. 3.2
y
3
2
60o
1
x

1
1 
(98)( )
3
2 12
39
 A dl  (     ) A dl  C  C  C
1
1
L
2
2
3
3
2
Along (1), C1   A dl    cos  d  | 0 
0
2
20
 2.
Along (2), dl   d a , A dl  0, C2  0
0
Along (3), C3    cos  d  60  
2
2
2
0
2
1
( )  1
2
 A dl  C  C  C  2  0  1  1
1
2
3
l
P.E. 3.3
U 
(a)
U
U
U
ax 
ay
az
x
 y
z
 y (2 x  z ) a x  x( x  z ) a y  xy a z
V 
(b)
V
V
1 V
a 
a 
az


z
 ( z sin   2  ) a   ( z cos  
z2

sin 2 ) a  (  s in  2 z cos 2  ) a z
(c)
f 
f
1 f
1 f
ar 
a 
a
r
r 
r sin  
cos  sin 
sin  sin  ln r
(cos  cos  ln r  r 2 )
 2r )ar 
a 
a
r
r
r sin 
sin  sin  ln r
 cos  sin 

 cot  cos  ln r


 2r  a r 
 r cos ec  a
a 
r
r
r




(
P.E. 3.4
  ( y  z ) a x  ( x  z ) a y  ( x  y ) a z
At (1, 2,3),   (5, 4,3)
(2, 2,1) 21

 7,
3
3
where (2, 2,1)  (3, 4, 4)  (1, 2,3)
  a1  (5, 4,3) 
40
P.E. 3.5
g  x log z  y 2  4,
Let f  x y  z  3,
2
 f  2 xy a x  x 2 a y  a z ,
 g  log z a x  2 y a y 
x
az
z
At P (  1 , 2,1),
f
(4 a x  a y  a z )
g
(4 a y  a z )

,

ng  
| f |
| g |
18
17
(  5)
cos   n f . n g  
18  17
T ake positive value to get acute angle.
5
  cos  1
 73.39 
17.49 3
nf  
P.E. 3.6
(a)   A 
At
 Ax  Ay  Az


 0  4 x  0  4 x.
x
y
z
(1, 2,3),   A  4.
(b)
B 

1 
 
1

(  B ) 
2  z sin  
1

1  B
 

 Bz

3 z 2 sin   2 z sin   3 z 2 sin 
 (2  3 z ) z sin  .
At (5,
(c )

2
,1) ,
  B  (2  3)(1)  1.
1  2
1 
1  C
( r Cr ) 
(C sin  ) 
C  2
r r
r sin  
r sin  
1
 2 6r 2 cos  cos 
r
 6 cos  cos 
41
At (1,
 
  C  6 cos
, ),
6 3

6
cos

3
 2.598.
P.E. 3.7 This is similar to Example 3.7.
   D  dS   t   b   c
S
 t  0   b since D has no z-component
  2
z 1
 0
z 0
 c    2 cos 2  d dz   3  cos 2  d  dz
 4
 (4)3  (1)  64
  0  0  64  64
By the divergence theorem,
 D  dS     Ddv
S
V
 D 
1 
1 
(  3 cos 2  ) 
 
 3 cos 2  
z

 
V
2
4
 Az
dz
cos  .
     Ddv   (3 cos 2  
V
z sin  
z

cos  )  d dzd 
1
4
2
1
0
0
0
0
 3  d   cos  d  dz   d   cos  d  zdz
2
0
 3(
2
0
3
4
)  (1)  64 .
3
P.E. 3.8
(a)
  A  a x (1  0)  a y ( y  0)  a z (4 y  z )
 a x  y a y  (4 y  z ) a z
_
At (1, 2,3) ,   A  a x  2 a y 11a z
(b)
42
  B  a  (0  6  z cos  )  a (  sin   0)  a z
1

(6  z 2 cos    z cos  )
 6  z cos  a    sin  a  (6 z  1) z cos  a z
At (5,

2
,  1) ,   B  5 a
(c)
_
a
1
2r cos  sin  3 1/ 2 a 
 r )  (0  2r sin  cos  )
(r 1/ 2 cos   0) 
(
r sin 
r
sin 
2
r
3
 r 1/ 2 cot  a r  (2 cot  sin   r 1/ 2 ) a  2sin  cos  a
2
  C  ar
 
, ),   C  1.732 a r  4.5 a  0.5 a
6 3
At (1,
P.E. 3.9
 A  dl   (  A)  dS
L
S
But (  A)  sin  a z 
z cos 

a  and
d S   d d  a z
 (  A)  dS    sin  d d 
S

2 2
60
| ( cos  )|
2
0
0
1
 2(  1)  1.
2
P.E. 3.10
  V 
ax
ay
az

x
V
x

y
V
y


z
V
z
 2V
 2V
 2V
 2V
 2V
 2V
) ax  (
)ay (
)az 0
(



 y z  y z
 x z  z x
 x y  y x
43
P.E. 3.11
(a)
 2U 

 2

(2 xy  yz ) 
( x  xz ) 
( xy )
x
y
z
 2 y.
(b)
 2V 

1 
 
1

( z sin   2 ) 
( z sin   4 ) 
 4  2 cos2  
(c)
2z 2
2
1
2
1

2
(  z sin   2 z 2


sin  cos  )  (  sin   2 z cos2  )

z
( z sin   2 z 2 cos 2 )  2 cos2 .
cos 2.
1  21
1

[r cos  sin   2r 3 ]  2
[ sin 2  sin  ln r ]
2
r r
r
r sin   
1
 2 2 [ cos  sin  ln r ]
r sin 
1
 2 cos  sin  (1  2 ln r  csc 2  ln r )  6
r
2 f 
P.E. 3.12
If B is conservative ,   B  0 must be satisfied.
ax

 B 
x
y  z cos xz
ay

y
x
az

z
x cos xz
 0 a x  (cos xz  xz sin xz  cos xz  xz sin xz ) a y  (1  1) a z  0
Hence B is a conservative field.
44
Prob. 3.1
(a)
dl   d ;
3

2
 
3
L   dl  3 d  3(  ) 
 2.356
2 4
4

4
(b)
dl  r sin  d;
r  1,   30 ;

3

L   dl  r sin   d  (1) sin 30 [( )  0] 
3
0
(c)
0.5236.
dl  rd

2
 
4
 4.189
L   d l  r  d  4(  ) 
2 6
3

6
Prob. 3.2
(a)
d S   d dz
5

2


S   d S    d dz  2  dz  d  2(5)[  ] 
2 3

0
10
 5.236
6
3
(b)
In cylindrical, dS   d  d

3
4
1
0
S   d S    d   d 
 2 3
( )  3.142
4 2 1
d S  r 2 sin  d d
(c) In spherical,
2
3
2

0
2
3
S   d S  100  sin  d  d  100 (2 )( cos  ) |  200 (0.5  0.7071)  758.4
4
(d)

4
45
d S  r dr d

2
r2 4  
8
S   dS   rdr  d  |0 (  ) 
 4.189
2 2 3
6

0
4
3
Prob.3.3
(a ) dV  dxdydz
1
2
3
0
1
3
V   dxdydz   dx  dy  dz  (1) (2  1)(3   3)  6
(b) dV   d d  dz
5
4

2 5

2
V    d   dz  d 
1
2

1
3
(c) dV  r 2 sin  drd d

3

2
2
6
2
 /3 
r3 3

V   r dr  sin   d  |1 (  cos  ) | / 2 (  )
3
2 6


1
3
2

1 
26 
1
(27  1)( )( ) 
 4.538
2 3
18
3
Prob.3.4
 /6
L   dl    d
 0
L
 4
 4( / 6)  2.094
Prob. 3.5
dS  r 2 sin  d d
 /2
 /4
0
0
S  r 2  d  sin  d

r 5
2
| (4  1)(  3 )  2 (25  4)(5)( 3 )  35  110
2
 25( / 2)( cos  )
 /4
0
25
( cos( / 4)  1)  11.502
2
Prob. 3.6
dv   d  d dz
10
30
5
  2 5  5
(25  4)  54.98
V   dz  d   d   10( / 6) 

 2 2 6
z 0
 0
 2
46
Prob. 3.7
dl  dxa x  dya y
I   H dl =  ( xy 2 dx  x 2 ydy )
L
But on L, x  y 2

dx  2 ydy
 2 y6 y4  4
I   y (2 ydy )  y dy   (2 y dy  y dy )  
 
4 1
 6
y 1
1
2 1
 (4096)  64    1428.75
3
6 4
4
4
3
5
3
Prob. 3.8
The line joining P and Q is y  x  2, dy  dx
3
I   (2 x 2  4 xy )dx  (3 xy  2 x 2 y )dy    2 x 2  4 x( x  2) dx  3 x( x  2)  2 x 2 ( x  2) dx
x 1
L
 3
x4  3
2
  3x  2 x  2 x dx   x  x    27  9  81/ 2  4.5
4 1

x 1
3
2
3
Prob. 3.9
(a)
x2
1
 F  dl   ( x  z )dy|
2
y 0
2
x  0, z  0
  2 xydx |
x 0
x2 2
z3 3

3(2)
|
|
2 0
3 0
 0  4  54   50
 0  2(1)
(b)
Let x  2t. y  t , z  3t
dx  2dt , dy  dt , dz  3dt ;
1
2
2
3
 F  dl   (8t  5t  162 t ) dt
0
1
 (t 3  40.5t 4 )  39.5
0
z 3
y 1, z  0
  (3 xz 2 )dz|
z 0
x  2, y 1
47
Prob.3.10
 /4
3
W   F  dl   z  d 
z  0,   2
 0
L
  cos  dz   /4
z 0
 0  cos( / 4)(3)  3cos 45o  2.121 J
Prob. 3.11
0
 H  dl   ( x  y)dx
x 1
1
y  0, z  0
  ( x 2  zy )dy  5yzdz
0
2
1
0
  xdx   ( y 
  5 yzdz
z 0
x  0, y  0
x  0, z  1  y / 2
0
y2
)dy   (10 z  10 z 2 )dz
2
1
 1.5
Prob. 3.12
Method 1:
1
1
y 0
z 0
 B  dl    yzdy z  0   xzdz x  1   ( yzdy  xzdz ) x  1
L
But z  y
 B  dl

 dz  dy on the last segment (or integral).
 0
L

0
z2 1
1
y3 y 2 0
  ( y 2  y )dy   (  )
2 0 y 1
2
3
2 1
1 1 1 1
    0.333
2 3 2 3
Method 2:
 B  dl     B d S
L
S

  B = x
xy

y
-yz
y
1
y 0 z 0
0
1
   B dS  
S

z  ya x  za y  xa z ,
xz
2
 ydzdy   y dy 
dS  dydza x
y3 1 1
  0.333
3 0 3
48
Prob. 3.13
1
2
z2 2
   A dS    zdxdz   dx  zdz  (1)
2
2 0
0
0
S
Prob. 3.14
   D  dS     Ddv
S
v
D 
Dx Dy Dz


 2 xz  3 y 2  2 yz
x
y
z
     Ddv   (2 xz  3 y 2  2 yz )dxdydz
v
1
4
3
1
4
3
1
4
3
1
0
1
 2  xdx  dy  zdz  3  dx  y dy  dz  2  dx  ydy  zdz
2
1
0
1
1
0
1
 y 4
 y 4 z 4
 0  3(2) 
 (3)  2(2) 

  6(64)  16(9  1)  384  128
 3 0
 2 0 2 0
  512
3
2
2
Prob. 3.15
   A  dS     Adv
S
v

1  2
1
1
3cos 
5

(r r ) 
(3sin  ) 
(5)  3 
2
r r
r sin  
r sin
r sin  r sin 
2
dv  r sin  d drd
 A 
  3 r 2 sin  d  d  dr  3 r cos  d  d  dr   5d  d  dr
4
 /2
3 /2
4
 /2
0
0
0
0
3 /2
4
 /2
3 /2
0
0
0
0
 3 r 2 dr  sin  d  d 3 rdr  cos  d  d 5 dr  d  d
0
 r 4
 r 4
 /2
 /2
 3
(3 / 2)  3 
(3 / 2)  5(4)( / 2)(3 / 2)
 ( cos  )
 (sin  )
0
0
 3 0
 2 0
 96  36   15 2  336.54
3
Prob. 3.16
(a) dv = dxdydz
2
49
2
1
1
1
1
z
0
0
0
 xydv     xydxdydz   xdx  ydy  dz
z 0 y 0 z 0
v

2
2
x 1 y 1 2
z  (1/ 2)(1/ 2)(2)  0.5
2 0 2 0 0
(b)
dv   d  d dz

3
3
2

  0 z  0  1
1
0
0
2
2
  zdv      z  d  ddz    d   zdz  d
v

 3 3 z2 2
1
( )  (9  )(2 )  54.45
3 1 2 0
3
Prob. 3.17
V1
V
V
a x  1 a y  1 az
x
y
z
 (6 y  2 z )a x  6 xa y  (1  2 x)a z
(a ) V1 
V3 
(c )
V2
V
1 V2
a 
a  2 a z

z
 
 (10 cos   z )a   10sin  a   a z
(b) V2 
V3
1 V3
1 V3
ar 
a 
a
r 
r sin  
r
2
1  2

  2 cos  ar  0 
  sin   a
r
r sin   r

2
2sin 
  2 cos  ar  2
a
r
r sin 
Prob. 3.18
(a)
V  (10 yz  4 xz )a x  10 xza y  (10 xy  2 x 2 )a z
At P,
x  1, y  4, z  3. Hence,
V  (120  12)a x  30a y  (40  2)a z  132a x  30a y  42a z
(b)
50
U  (2sin   z )a  2 cos  a   a z
At Q,   2,   90o , z  1
U  (2  1)a  0a  2a z  a   2a z
(c )
8
1 4 cos  cos 
1  4sin  sin  
W   3 sin  cos  ar 
a 

 a
2
r
r
r
r sin  
r2

At R, r  1,    / 6,    / 2
4
W   3 a  4a
(1)
Prob. 3.19
r  x2  y 2  z 2 ,
r n  ( x 2  y 2  z 2 )n / 2
Method 1:
r n 
r n
r n
r n
n
ax 
ay 
a z  ( x 2  y 2  z 2 ) n / 21 (2 x)a x  
x
y
z
2
 n( x 2  y 2  z 2 )
n2
2
( xa x  ya y  za z )  nr n  2 r
Method 2:
r n
r
r n 
ar  nr n 1  nr n  2 r
r
r
 T  2x a x  2 y a y 
At
a
Prob. 3.20
z
1,1, 2  ,  T  (2, 2, 1). The mosquito should move in the direction of
2ax  2a y  az
Prob. 3.21
 F  a x  2a y  a z
an 
a x  2a y  a z
F

 0.4082a x  0.8165a y  0.4082a z
| F |
1 4 1
Prob. 3.22
51
Method 1:
T
1 T
1 T
ar 
a 
a
r
r 
r sin  
 sin  cos  ar  cos  cos  a  sin  a
T 
At P, r  2,  60o ,   30o
T  sin 60o cos 30o ar  cos 60o cos 30o a  sin 30o a
 0.75ar - 0.433a  0.5a
| T | 0.752  0.4332  0.52  1
The magnitude of T is 1 and its direction is along T.
Method 2:
T  r sin  cos   x
T  a x
| T | 1
Prob. 3.23
f
f
f
f  a x  a y  a z  (2 xy  2 y 2 )a x  ( x 2  4 xy )a y  3 z 2 a z
x
y
z
At point (2,4,-3), x  2, y  4, z  3
f  (16  32)a x  (4  32)a y  27a z  16a x  28a y  27a z
a x  2a y  a z
1
(1, 2, 1)
1+ 4 +1
6
The directional derivative is
1
99
f a  (16, 28, 27)
(1, 2, 1)  
 40.42
6
6
a

Prob. 3.24
(a) Let f = ax + by + cz – d = 0
52
f  aa x  ba y  ca z
an1 
aa  ba y  ca z
f
 x
| f |
a 2  b2  c2
Let g   x   y   z  
g  a x   a y   a z

an 2 
| g |
2  2  2
cos   an1 an 2 
  cos 1
a  b  c
a 2  b2  c 2  2   2   2
a  b  c
(a  b 2  c 2 )( 2   2   2 )
2
a  1, b  2, c  3
(b)   1,   1,   0
  cos 1
1 2  0
(12  22  32 )(12  12  02 )
 cos 1
3
 cos 1 0.5669  55.46o
28
Prob. 3.25
V
V
V
ax 
ay 
a z  4 ye z a x  4 xe z a y  4 xye z a z
x
y
z
At (3,1,-2), x  3, y  1, z  - 2
V 
V  4e 2 a x  12e 2 a y  12e 2 a z  0.5413a x  1.624a y  1.624a z
This is the direction. The maximum rate of change is
| V | e 2 42  122  122  0.1353 17.44  2.36
Prob. 3.26
(a)



UV  (UV )a x  (UV )a y  (UV )a z
x
y
z
 V
U 
U 
U 
 V
 V
 U
V
V
V
 a y  U
 ax  U
 az
x 
y 
z 
 x
 z
 y
 U V  V U
(b)
53
V  10 xya x  (5 x 2  2 z )a y  2 ya z
U V  30 x 2 y 2 za x  (15 x 3 yz  6 xyz 2 )a y  6 xy 2 za z
V U  (15 x 2 y 2 z
2
z
2
y
6
+
U  3 yza x  3 xza y  3xya z
︶ a x  (15 x3 yz  6 xyz 2 )a y  (15 x3 y  6 xy 2 z )a z
U V  V U  (45 x 2 y 2 z  6 y 2 z 2 )a x  (30 x 3 yz  12 xyz 2 )a y  (15 x3 y  12 xy 2 z )a z  (1)
UV  15 x 3 y 2 z  6 xy 2 z 2
(UV )  (45 x 2 y 2 z  6 y 2 z 2 )a x  (30 x 3 yz  12 xyz 2 )a y  (15 x 3 y  12 xy 2 z )a z  (2)
From (1) and (2), the formula is verified.
Prob. 3.27
Ax Ay A z


 3y  x
x
y
z
1
1
  B  2  z 2   2 sin cos   2  sin2 
(a)   A 

(b)

 2 z  sin2  2  sin2 
2
(c )   C 
1 2
3r  0  3
r2
Prob. 3.28
Ax Ay Az


 2 xy  0  2 y  2 y (1  x)
x
y
z
(a) At (3, 4, 2), x  3, y  4
 A  2(4)(1  3)  16
 A
 B
1 
1 B Bz 1 


3 2 sin    0  8 z cos 2 
 B  


 
  z  
 6sin   8 z cos 2 
(b)
At
(5,30o ,1), z  1,   30o
 B  6sin 30o  8(1) cos 2 30o  3  6  9
54
1  2
1 C 1  4
( r Cr )  0 
(r cos  )  0  4r cos 

2
r r
r sin   r 2 r
(c ) At (2,  / 3,  / 2), r  2,    / 3
 C = 4(2)cos( /3) = 4
C
Prob. 3.29
2
 H  k   T  k  T
 2T  2T
x
 y 2 2
 T

 50sin
 ) 0
cos h
(
 x2  y
2
2
4
4
2
2
Hence,  H  0
Prob. 3.30
(a)



(V Ax ) 
(V Ay ) 
(V Az )
x
 y
z
 Ay
 Ax
 Az
V
V
V
 ( Ax
V
)  ( Ay
V
)  ( Az
V
)
x
x
 y
 y
z
z
 Ax  Ay  Az
V
V
V
 V(


)  Ax
 Ay
 Az
x  y z
x
 y
z
  (V A) 
 V   A  A  V
(b)
A
  A  2  3  4  1;
 V  yz a x  xz a y  xy a z
  (V A)  V    A V
 xyz  2 xyz  3 xyz  4 xyz  2 x y z
Prob. 3.31
(a)
( r )T  3T  6 yz a y  3 xy 2 a y  3 x 2 yz a z
(b)
55
x
T
T
T
y
z
 x (y 2 ay  2 xyz az )  y(2 z ax  2 xy ay  x 2 z a z )
x
y
z
 z(2y ax  x2 y az )
 4yz ax  3 xy 2 ay  4 x 2 yz a z
(c)
  r (r  T )  3 (2 xyz  xy 3  x 2 yz 2 )
 6 xyz  3xy 3  3 x 2 yz 2
(d)
(r  ) r 2  (x



y
 z )(x 2  y 2  z 2 )
x
y
z
 x(2 x)  y(2y)  z(2 z)
 2(x 2  y 2  z 2 )  2 r 2
Prob. 3.32
We convert A to cylindrical coordinates; only the -component is needed.
A  Ax cos   Ay sin   2 x cos   z 2 sin 
But x   cos  ,
A  2  cos 2   z 2 sin 
   A  dS   A  d dz    2  2 cos 2    z 2 sin  d dz
S
 /2
1
1
 /2
1
 2(2)  (1  cos 2 )d  dz  2 z 2 dz  sin  d
2
0
0
0
0
2
 /2
 /2
1
z3 1
( cos  )
 4(  sin 2 )
2
 2  2 / 3  5.6165
0
0
2
3 0
56
Prob. 3.33
z
Z=1
y
Z=1
x
(a)
 D  dS  [      ]D  dS
z 1
z 1
5
    2 cos2  d d    2 cos2  d d   2  2 z 2 d dz|
 5
2
3
1
z
 2(5)2  d  z 2 dz   50(2 )( |11 )
3
0
1
200 
 209.44
3
1 
(2  2 z 2 )  4 z 2
(b)   D 

 
1
5
2
0
0
   Ddv   4 z  d d dz  4  z dz   d  d
2
1
 4x
Prob. 3.34
2
z 1  5
200
(2 ) 
 209.44
3 1 2 0
3
3
2
57
 / 2 2
 H dS    10 cos r sin  d d r  1
2
0 0
S
2
 /2
 /2
0
0
 10(1)  d  sin  cos  d  10(2 )  sin  d (sin  )
2
0
 sin    / 2
 20 
 10  31.416

 2  0
2
Prob. 3.35
 H  dS     Hdv
S
v
 H  dS   2 xydydz x  0   2 xydydz x  1  (x  z )dxdz y  1
2
2
S
  (x 2  z 2 )dxdz
y2
  2yzdxdy
 2yzdxdy
z  1 
z3
2
3
1
2
1
2
1
1
0
1
0
1
 0  2  ydy  dz  2  dx  ydy  6 dx  ydy
 12  3  9  24
H 
Hx Hy Hz


 2y  0  2y  4y
x
y
z
1
2
3
0
1
1
   Hdv   4ydxdydz  4 dx  ydy  dz
V
 4(1)
y 2
(3  1)  24
2 1
2
Prob. 36
   H  dS     Hdv
S
To find
v
 H  dS , let
S
 = t + b   s
wher  t ,  b , and  s are the fluxes from the top, bottom, and side of the cylinder.
58
dS   d  d a z ,
For  t ,
 t  2 z   d  d
z 3
10
2
0
0
 6   d   d  12
 2 10
2 0
 600
dS   d  d (a z ) ,
For  t ,
 b  2 z   d  d
z0
0
dS   d dza ,
For  s ,
 t  4   d dz
3
  10
3
2
0
0
 4000 dz  d  4000(3)(2 )  24000
  600  0  24000  23400  73,513.27
   Hdv,
To get
dv   d d  dz
v
1 
(4  3 )  2  12   2
 
 H 
10
2
3
10
2
3
0
0
0
0
0
0
   (12   2) d d  dz  12   2 d   d  dz  2   d   d  dz
(1000)
(100)
(2 )(3)  2
(2 )(3)  24000  600  73,513.27
3
2
 12
Prob. 3.37
Let
 =  A dS =   Adv
S
v
Ax Ay Az


 2( x  y  z )
x
y
z
 A = 2(  cos    sin   z ),
dv   d  d dz
 A=
   2(  cos    sin   z )  d  d dz
1
4
2
0
2
0
 0  0  2  d   zdz  d 
2 2 1 z 2 4
1
(2 )  (16  4)(2 )
2 0 2 2
2
 12  37.7
Prob. 3.38
1 
1 
(r 4 ) 
(r sin 2  cos  )
2
r r
r sin  
 4 r  2 cos  cos 
 A 
59
   Adv   4r sin  d d dr   2r sin  cos  cos  d d dr
3
2

 /2 
r4 3
2r 3 3 cos 2   / 2
2
(

cos

)
(
)

(

)
sin

|
|
|
|
|
0
0
0
0
0
4
2
3
2
1

 81(1)( )  18(0  )(1  0)
2
2
81

 9  136.23
2
4
z
y
x
 A  dS  [       ] A  dS
0
 /2
r 3
 /2
Since A has no   component, the first two integrals on the right hand side vanish.
 /2
 A  dS  
 0

 /2
4
 r sin  d d|
 0
 /2
3
r 3


r 0
 /2
 81 ( ) ( cos  )|  9(1) sin |
0
0
2
81

 9  136.23
2
 /2
 r sin  cos  drd| 

2
0
2
 /2
60
Prob. 3.39
Let    F  dS   t   b   o  i
where  t ,  b ,  o ,  i are the fluxes through the top surface, bottom surface,
outer surface (   3), and inner surface respectively.
For the top surface, dS   d d  a z ,
z  5;
F  dS   2 z d dz. Hence:
2
3
t  
2
 2
0
For the bottom surface,
z 5

190 
 198.97
3
z  0, dS   d d  (
a
  z d dz|

z
)
F  dS    z d d   0. Hence,  b  0.
2
a
For the outer curved surface,   3, dS   d dz

F  dS =  sin  d dz. Hence,
2
2
5
a 
3
 dz   sin  d|
 0
z 0
 3
0
For the inner curved surface,   2, d S   d dz ( a  )
F  dS    3 sin  d dz. Hence,
2
5
 a    dz 
z 0
 
3
 sin  d|  0
 0
2
190 
190 
 00 0
 198.97
3
3
   F  dS     FdV
F 
1 

 3  sin  
z

(  3 sin  ) 
sin   
1 
 
( z cos  )  
61
z
   Fdv   (3 sin    sin    )  d d  dz
V
5
2
3
0
0
2
 0  0   dz  d   2 d 

190 
 198.97
3
Prob. 3.40
ax

(a) xA 
x
xy
ay

y
y2
az

 zay  xaz
z
 xz
1

1
xB   2  z2 sin cos   0  a  (2  z  2 z sin2  )a 
2  sin2   0 a



 4 z sin cos  a  2( z  z sin2  )a  2 sin2  az
(b)

 2 z sin2 a  2 z(  sin2  )a  2 sin2  az
xC 
cos2 
r
(2 cos  )( sin )sin  cos  (cos2  ) ar 
(2 r )a
r sin
r
(cos3   2 sin2  cos  )

ar  2 cos2  a
sin

(c )
az




x y
z
x 2 y y 2 z 2 xz
 A  0
y 2
a

ay
a
A
Prob. 3.41
(a)
ax

1  
1 


(r cos2  sin  ar   (r 2 cos2   a

r sin  
r  r


x
z
 2 z ay  x 2

62

A
(b)
(
1  Az
 

 A
 A  Az
1  (  A )  A

) a  (  
) a  (
)az
z
z




 (0  0) a   (  2  3 z 2 ) a 
1

(4  3  0) a z
 (  2  3z 2 ) a  4  2 a z
A

0
(c )
1  sin  
1  1 cos 
1   cos  ) 

0  2  ar  
 0  a   (
 0  a
2

r sin  
r 
r  sin  r
r  r
r


sin 
cos 
cos 
 3
ar  3
a  3 a
r sin 
r sin 
r
 A 
 sin 
sin 
0 4
0
4
r sin 
r sin 
 A 
    A 0
Prob. 3.42
  H  0a   1a 
1

(2  cos    cos  )a z  a  cos  a z
 1

1
1
1
    H    sin   0  a   0a  (1  0)a z  - sin  a   a z



 

63
Prob. 3.43
Method 1: We can express A in spherical coordinates.
a
r
A  3 ar  2r ,
r
r
2
a 
1
  A     2r     2   ar  3 ar  ar  0
r
r 
r 
Method 2:
x
y
z
a  3 a y  3 az
3 x
r
r
r



3
x y z  3

 A 
  z ( x 2  y 2  z 2 ) 5/ 2 (2 y )   y ( x 2  y 2  z 2 ) 5 / 2 (2 z )  a x  ...
2
x
y
z
 2

3
3
3
r
r
r
0
A
Prob. 3.44
y
1
1
2
3
0
(a)
1
2
x
64
 F  d l  (     ) F  d l
L
1
2
3
_
_
_
For 1, y  x dy  dx, dl  dx a x  dy a y ,
1
3
 F  d l   x dx  xdx  
1
0
1
4
_
_
_
For 2, y   x  2, dy   dx, dl  dx a x  dy a y ,
2
3
2
 F  d l   ( x  2 x  x  2)dx 
2
1
17
12
For 3,
0
2
 F  d l   x ydx|
3
y 0
0
2
1
17
7
 F  d l   4  12  0  6
L
(b)
  F   x2 a z ;
S
 (  F)  d
dS  dxdy( a z )
   ( x 2 )dxdy 
1

x
2
 x y|dx 
0
0
1 x
2
 x2
0 0
1
y 0
2
  x dydx  
2
 x 2
2
 x y| dx 
1
0
 x dydx
2
2
x1
7
  x 2( x  2)dx 
|
4 0
6
1
(c) Yes
Prob. 3.45
1
 /2
2
0
 A  dl    sin d   0     d   1   sin d   90     d   2
2
o
2


0
1

 (4  1)  8( )  9.4956
2 2
2
1
 /2
3
65
Prob. 3.46
y
45o
0
2
2
 F  dl   2  zd
0
 2
2
0
 (6cos  )
 xF 
1
 /4
0
x
 /4
 3 z sin  d

z  1 0
  2, z  1
 2
0
2
0
 2  zd z  1
2
 (4  0)  6( cos  / 4  1)  (0  4)  1.757
3 z sin  0  az  ...

2
 /4
0
0
3z
 (xF) dS     sin  d d z 
 3(2)( cos  )
 /4
0
 6( cos   1)  1.757
Prob. 3.47
  A  8 xe y  8 xe y  16 xe y
(  A)  16e y ax  16 xe y ay
ax
ay


x(  A) 
x
y
y
16 xe y
16e
Should be expected since xV  0 .
Prob. 3.48
(a) V  
az

 (16e y  16e y )az  0
z
0
sin cos 
cos  cos 
sin
ar 
a  2 a
2
2
r
r
r
66
(b) xV  0
(c)
1 
1

cos  cos 
1
sin cos 
( sin cos  )  2
(sin
)  2 2 (
)
2
r
r
r r
r sin 
r sin 
cos 
cos 
0 3
(1 2 sin2  )  3
r sin
r sin
2 sin cos 

r3
  V   2V 
Prob.3.49
r
Q
r sin  [(cos   sin  ) a x  (cos   sin  ) a y ]
r sin 
 r (cos   sin  ) a x  r (cos   sin  ) a y
 Qr   sin  cos 
  
Q    cos  cos 
Q    sin 
 
sin  sin 
cos  sin 
cos 
cos    Qx 
 sin   Qy 
0   Qz 
Q  r sin  a r  r cos  a  r a 
(a)
dl   d a ,
1
2
  r sin 30  2( )  1
z  r cos 30  3
Q  r 
2
 Q  dl  
 2  z2
 2  z 2  d  2(1)(2 )  4
0
(b)
  Q  cot  a r  2 a  cos  a
For S1 , dS  r 2 sin  d d a r
 (  Q)  dS   r sin  cot  d d|
2
r 2
S1
2
30
0
0
 4  d  cos  d  4
(c)
67
dS  r sin  d dr a
For S2 ,
 (  Q)  dS   2 r sin  d dr|
 30
S2
2
2
0
0
 2sin 30 rdr  d
  4
(d)
For S1, dS  r 2 sin d d a r
 Q  d S  r  sin  d d|
3
2
r 2
S1
2
30
0
0

 8  d  sin2  d
 4 [

3

3
]  2.2767
2
(e)
_
For S2 , dS  r sin d dr a
 Q  dS   r sin cos d dr|
2
S2

(f)
 30
4 3
 7.2552
3
1 

r
(r 3 sin  ) 
(sin  cos  )  0
2
r r
r sin   
 2sin   cos  cot 
Q 
   Qdv   (2 sin  cos cot )r sin d d dr
2
30
r3 2

(2 )  (1 sin2  )d
3 0
0

4
3
( 
)  9.532
3
2
68
Check :    Qdv  (
S1
 4 [


3
 QdS
S2

3
3

]
2
3
4
3
[ 
]
3
2
(It checks!)
Prob. 3.50
Since u    r ,
  u    (  r). From Appendix A.10,
  (A B)  A(  B) B(  A) (B  ) A (A )B
  u    (  r )
_
  (  r )   (  r )  r (   )  (r  )   (  ) r
  (3)    2 
or  
1
  u.
2
Alternatively, let
u
x  r cos  t ,
y  r sin  t
x
y
ax 
ay
t
t
   r sin  t a x   r cos t a y
  y ax   x a y


u   x  y
 y  x
i.e.,  

 z  2 a z  2
0
1
u
2
Note that we have used the fact that     0,
(r  )  0 ,
(  )r  
69
Prob. 3.51
(a)   H 
H x H y H z


 2 z  5x  8
x
y
z



z
(b)   H  x y
 8a x  2 xa y  5 ya z
2 xy 5 xy 8( y  z )
Prob. 3.52
1  2
1
1 B

(r Br ) 
( B sin  ) 
2
r r
r sin  
r sin  

1 
1
 2 (r 4 ) 
(4 rsin  cos 2 )  0
r r
r sin  
1
 4r 
 4r cos  cos 2  4r sin  (2sin 2 )
sin 
  B  4r  4r cot  cos 2  8r sin 2
B 
 B =
B 

1  
1  1 Br 
1 
B 
( B sin  )    ar  
 (rB )  a   (rB )  r  a

r sin   
 
r  sin   r
r  r
 

All terms are zero except one.
1 
1 

(4r 2 cos 2 )a  8cos 2 a
  B = 0 ar  0a   (rB )  0  a 
r  r
r r

Prob. 3.53
(a)
 V
V
V 
ax  V
ay V
az 
 (VV)=  V
y
z 
 x

  V    V    V 
 V
V
 V

x  x  y  y  z  z 
2
 2V
 2V
 2V  V   V   V 
 V 2 V 2 V 2  
 
 

x
y
z
 x   y   z 
2
 V  2V  | V |2
2
70
(b)



  VA  x y z
VAx VAy VAz









  (VAz )  (VAy )  a x   (VAx )  (VAz )  a y   (VAy )  (VAx )  a z
z
x
y
 z

 y

 x

Ay 
 V
A
V
  Az
 V z  Ay
V
 ax
y
z
z 
 y
A
V
A 
 V
  Ax
 V x  Az
V z  ay
z
x
x 
 z
A
 V
A 
V
  Ay
 V y  Ax
 V x  az
x
y
y 
 x
 A A 
 A A  
 A A 
  VA  V  z  y  a x   x  z  a y   y  x  a z 
z 
x 
y  
 z
 x
 y
 V
 V
V 
V 
V 
 V
  Az
 Ay
 Az
 Ax
 a x   Ax
 az
 a y   Ay
z 
z
x 
x
y 

 y

 V   A  V  A
Prob. 3.54
(a)
 B
Bx By Bz


 2 xy  1  1  2  2 xy
x
y
z
(b)



y
z  (1  0)a x  (0  0)a y  (4 x  x 2 )a z
  B  x
x 2 y (2 x 2  y ) ( z  y )
 a x  x(4  x)a z
(c )
( B)  (2  2 xy )  2 ya x  2 xa y
(d)
71


    B  x y
1 0

z
 0a x - (4 - 2x)a y  0a z
2
(4 x  x )
 2( x  2)a y
Prob. 3.55
(a)
V1  x 3  y 3  z 3
 2V1  2V1  2V1
 V1 


x 2 y 2 z 2



( 3x 2 ) 
3 y 2   (3z 2 )


x
y
x
2
 6 x  6 y  6z  6( x  y  z)
(b)
V2   z 2 sin2
1 
 2V2 

 
z
2

(
( z 2 sin2 ) 
sin2 
3 z 2

4z
2

4 z2

sin2 

(2  z sin2 )
z
sin2  2  sin2
 2  )sin2
(c)
V3  r 2(1 cos  sin )
 2V3 
1 
[2 r 3(1 cos  sin )]
r2  r
1

1
(  sin 2  sin  )r 2  2 2 r 2 (  cos sin  )
r sin 
r sin 
2 sin
cos sin 
 6(1  cos sin  ) 
cos sin  
sin
sin 2 
cos sin 
 6  4 cos sin  
sin 2 

2
72
Prob. 3.56
(a)
U  x 3 y 2 e xz


 4 2 xz
(3 x2 y 2 e xz  x3 y 2 zexz ) 
(2 x 3 y exz ) 
(x y e )
x
y
z
2U 
 6 xy 2 e xz  3 x2 yze xz  3 x 2 y 2 zexz  x 3 y 2 z 2 exz  2 x 3 exz  x5 y 2 e xz
 e xz(6 xy 2  3 x2 y 2 z  3 x2 y 2 z  x3 y 2 z 2  2 x3  x 5 y 2 )
At (1, 1,1),
 2U  e1 (6  3  3  1 2  1)  16e 43.493
(b)
V   2 z (cos   sin  )
1 
 2V 
[2  2 z (cos   sin  )]  z (cos   sin  )  0
 
 4 z (cos   sin  )  z (cos   sin  )
 3z (cos   sin  )

At (5,
6
,  2),  2V  6(0.866  0.5) 8.196
(c )
W  e  r sin  cos 
 2W 
1 
e r

2 r




(
r
e
sin
cos
)
cos 
(sin  cos  )
2
2

r r
r sin 

e  r sin  cos 
r 2 sin 2 
1
(2re  r sin  cos  )  e  r sin  cos 
r2
e  r cos 
e  r cos 
 2
(1  2sin 2  )  2
r sin 
r sin 
2 2
 2W  e  r sin  cos  (1   2 )
r r

At
(1, 60,30),
 W  e 1 sin 60 cos 30(1  2  2)  2.25e 1   0.8277
2
73
Prob. 3.57
(a) Let V  1nr  1n x 2  y 2  z 2
V 1 1
x
2 x  (x 2  y 2  z 2 )1/ 2  2


x r 2
r
V 
V
V
ax 
ox
oy
ay 
V
oz
az 
x a x  ya y  z a z
r
2

r
r2
r 1
 ar in spherical coordinates.
r2 r
1  2
1 
(r A r )  2 (r )
 2(1nr )    (1nr )   A  2
r
r
r r
1
 2
r
(b) Let V  A 
Prob. 3.58
(a)
(b)
U 
U
U
U
U
ax 
ay 
a z  y 2 z 3a x  2 xyz 3a y  3 xy 2 z 2
az
x
y
z
z
 2U 
  U    U    U 
3
2
3
2
 

 
  0  2 xz  6 xy z  2 xz  6 xy z
x  x  y  y  z  z 
V 
V
1 V
V
sin 
cos 
1
a 
a 
a z   2 a   2 a  0  2   sin  a   cos  a 

z
 



 2V 
1   V  1  2V  2V 1 
1  sin  
 2 
  1 sin    2  
0


 2
2
z
      
 
   

sin 

3
1
3
sin   0
W
1 W
1 W
ar 
a 
a
r
r 
r sin  
 2r sin  cos  ar  r cos  cos  a  r sin  a
W 
( c)

74
1   2 W 
1
 
W 
1
 2W
 W  2 r

 sin 

r r  r  r 2 sin   
  r 2 sin 2   2
1 
1
 2
1
r sin  cos  cos    2 2 (r 2 sin  cos  )
 2  2r 3 sin  cos    2

r r
r sin  
r sin 
1
cos 
 6sin  cos  
(cos 2  cos   sin 2  cos  ) 
sin 
sin 
cos 
cos 
 6sin  cos  
 2sin  cos  
 4sin  cos 
sin 
sin 
2
Prob. 3.59
V 
V
1 V
V
a 
a 
a z  2  z cos  a  -  zsin a   2 cos  a z

z
 
 2V 
1   V  1  2V  2V 1 
1
2  2 z cos    2  2 z cos   0
 2 


 2
2
z
      
 

  4  1 z cos   3z cos 
Prob.3.60
V
1 V
1 V
ar 
a 
a
r
r 
r sin  
10
5sin 
  3 cos  ar  3
a
r
r sin 
V 
(a)
1  2 V
1

V
1
 2V
 V   V  2
(r
) 2
(sin 
) 2 2
r r
r
r sin  

r sin   2
1  2 10 cos 
1
5cos 
(b)
 2
r (
)  0  2 2 (
)
3
r r
r
r sin 
r2
10 cos  5cos 
 V 
 4 2
r4
r sin 
2
(c )
  V  0, see Example 3.10.
75
Prob. 3.61
U
U
U
U 
ax 
ay 
a z  4 yz 2 a x  (4 xz 2  10 z )a y  (8 xyz  10 y )a z
x
y
z



 U  (U x )  (U y )  (U z )  0  0  8 xy  8 xy
x
y
z
 2U 
 2U  2U  2U


 0  0  8 xy  8 xy
x 2 y 2 z 2
Hence,  2U   U
Prob. 3.62
Method 1
 2G

  2G 
2 G G

 2   2
1 
2  sin 
8  sin  2  sin 
0

 2  sin   
2
 

2
2
2sin  2sin  8sin  2sin 
6sin 







 2G


  2G 



2 G G

 2   2
1 
1
4  cos  4  cos 
(4  cos  )  4  cos   0 

 

2
2
4 cos  4 cos  4 cos  4 cos 




0


 2G
z

  2 Gz 

1


1 

  ( z 2  1)   0  (2 z  )
 
z
( z 2  1)  2 
Adding the components together gives
 2G 
6sin 




1
a    2   ( z 2  1)  a z



76
Method 2:
A
 2G  ( G )    (  G )
1 
1
Let V   G =
(2  2 sin  )  (4  sin  )  2 z   2 z 
 

( G )  V  2 za   2  a z
1


1
   G   0  0  a   0  ( z 2  1)  a   (4  2 cos  )  2  cos   a z
  



2
  ( z  1)a  6 cos  a z
Let
 6


1
    G    A    sin   2 z  a   (0  0)a   (  ( z 2  1))  0  a z
  
 




6
1
  2 z  sin   a   ( z 2  1)a z




 2G   V    A


6
1
 2 za   2  a z   2 z  sin   a   ( z 2  1)a z







1
sin  a    2   ( z 2  1)  a z




6
Prob. 3.63



( xz )  ( z 2 )  ( yz )  z  y
x
y
z
( A) = a y  a z
 A
 2 A   2 Ax a x   2 Ay a y   2 Az a z  0  2a y  0  2a y
( A) -  2 A  a y  a z

  A  x
xz

y
z2

    A  x
z
(1)

z   za x  xa y
yz

y
x

z  a y  a z
0
From (1) and (2),
    A = (  A) -  2 A
(2)
77
Prob. 3.64
 A
Ax Ay Az


 111  3  0
x
y
z

  A = x
x
 B

y
y

z  0
z
1 
1 B Bz

 4 cos   4 cos   0
(  B ) 
 
  z
B 
 1 Bz B 
 B B 
1
 B = 

a      z  a   (  B )    a z

z 
 
 
  
  
 z
1
 0a  0a   8  sin   2  sin   a z  6sin  a z  0

1  2
2sin 
(r sin  )  0  0 
0
2
r r
r
 2
1  
1


2
C =



(
sin
)
0
0
(r sin  )  a

r
a
r



r sin   
r  r


1
  0  cos   a
r
cos
 2 cos  ar - 2sin a a  0
r
(a) B is solenoidal.
(b) A is irrotational.
C
Prob. 3.65 (a)
ax
ay


G 
x
y
16 xy  z 8 x 2
az

z
x
 0 a x  (1  1) a y  (16 x  16 x) a z  0
Thus, G is irrotational.
(b) Assume that  represents the net flux.
78
   G  dS     Gdv
  G  16 y  0  0  16 y
1
1
1
0
0
0
   16 ydxdydz  16 dx  dz  ydy  16(1)(1)(
(c)
y2 1
)8
2 0
y
1
0
y 1
x 1
 G  dl   (16 xy  z )dx|
x 0
L
1
y 0
z 0
 0  8(1) y|  16(1)
0
1
y 0
x 0
  8 x dy|x 1   (16 xy  z )dx| y 1   8 x 2 dy| x 0
2
y 0
z 0
x2 0
| 0
2 1
 88  0
This is expected since G is irrotational, i.e.
 G  dl   (  G )  dS  0
  T  6  0  6
Prob. 3.66
1 
1    
0
(  E ) 
 
   2 
Hence, E is solenoidal.
E
E
  E   a   a z  0
z

showing that E is conservative.
 E 
x
x 1
z 0
y 1
z 0
79
Prob. 3.67
 H  0
 H dl   (  H) dS  0
L
S
Prob. 3.68

x
 B 
2
3x z  y 2

y
2 xy

z  (0  0)a x  (3x 2  3x 2 )a y  (2 y  2 y )a z  0
x3
showing that B is conservative.
Prob. 3.69
 D 
1
(  0) 
1
(0)  0  0


We conclude that D is solenoidal.
Prob. 3.70
 E 
1
1 1
1 
2k ( sin  ) 

(0  0)ar  
0  0  a   (kr 2 sin  ) 
 a
r sin 
r  sin 
r  r
r3

1  2k sin  2k sin  
 0  

a  0
r
r3
r 3 
showing that E is conservative.
80
CHAPTER 4
P. E. 4.1
 5 109 [(1, 3, 7)  (2, 0, 4)]

 [(1, 3, 7) (2, 0, 4)]3



9


110
9


(a) F 
(
2
10
)[(1,
3,
7)
(
3,
0,5)]





 109   

3
4 
[(1, 3, 7)  (3, 0,5)] 

 36  


[
45(1, 3,3) 18(4, 3, 2)
] nN

193/ 2
293/ 2
 1.004a x  1.284a y  1.4 a z nN
(b)
E
F
 1.004a x  1.284a y  1.4a z V/m
Q
P. E. 4.2
Let q be the charge on each sphere, i.e. q=Q/3. The free body diagram below helps us to
establish the relationship between various forces.
P


T
A
F1
d/2
F2
mg
81
At point A,
T sin  cos 30  F1  F2 cos 60
1
q2
q2

2 ( )
2 
4  0 d
4  0 d 2
3 q2

8  0d 2
T cos   mg
Hence,
sin  
But
Thus,
or
tan  cos 30 
q2 
h
d

tan  
l
3l
d
3
l2 
d2
3
d
3
( )
3q 2
3 2 
8 0 d 2 mg
d2
2
l 
3
4 0 d3 mg
3 l2 
d2
3
Q
 
3
12   0 d 3 m g
but q 
Q2 
3q 2
8 0 d 2 mg
d2
l 
3
2
q2 
Q
. Hence,
9
82
P.E. 4.3
_
d2 l
eE  m 2
dt
_
_
d 2 x _ d 2 y _ d 2z _
ax  2 a y  2 az )
dt 2
dt
dt
E0  200 kV / m
_
eE0 (  2a x  a y )  m(
where
d 2z
0
 
z  ct  c2
dt 2
 2eE0 t 2
d2x
 
 c3 t  c4
m 2   2eE0
x
2m
dt
d2y
eE0 t 2
 
 c5 t  c6
m 2  eE0
y
2m
dt
At t  0, ( x, y, z )  (0, 0, 0) c1  0  c4  c6
dx dy dz
, , )  (0, 0, 0)
dt dt dt


c1  0  c3  c5
At t  0
Also, (
Hence,
( x, y ) 
eE0t 2
(2,1)
2m
i.e. 2 | y |  | x |
Thus the largest value of is
80 cm  0.8 m
P.E. 4.4
(a)
Consider an element of area dS of the disk.
The contribution due to dS   d d  is
dE 
 s dS
 s dS

2
4 0 r
4 0 (  2  h 2 )
The sum of the contribution along  gives zero.
a
 s a 2 h  d  d 
h s
 d

Ez 
2
2 3/ 2
2



4  0  0  0 (   h )
2  0  0 (   h 2 )3/ 2
a
h s
hs
2
2 3/ 2
2
2
2 1/ 2
(
)
(
)
(
2(
)
|






h
d


h
0
4  0 0
4 0
a

s
h
[1  2
]
2 0
(h  a 2 )1/ 2
83
(b)


As a
E
,
s
az
2 0
(c) Let us recall that if a/h <<1 then (1+a/h)n can be approximated by (1+na/h).
Thus the expression for Ez from (a) can be modified for a<<h as follows.


1



2 2 
2



s 
a
1
  s 1  1       2  s  a 
Ez 
1


2 

a 0 , but  sa  Q
2 o  2h 2 
2 o 
a 2  2 o   h  

1 2 


h 

  a 2 
Q
=
 s 
2 
2 o  2h  4 o h 2
This is in keeping with original Coulomb’s law.
P. E. 4.5
2
QS    S dS  
2
 12 | y| dx dy
2 2
2
 12(4)  2 y dy  192 mC
0
E
 s dS
 dS | r  r' |
a  s
2 r
4 r
4 o | r  r' |3
where r  r'  (0, 0,10)  ( x, y, z )  ( x,  y,10).
84
12 | y |103 ( x,  y,10)
  109 2 2
x 2 y 2
4 (
)( x  y  100)3/ 2
36
2
E
2
2 2
 xdx dy a x
 y | y | dy dx a y

2
2
3/ 2


( x  y  100)
( x 2  y 2  100)3/ 2
2
2 2
2
2
 108(106 )[  | y | 
2
2
2
 10 a z 
| y | dx dy
 ( x  y  100)
2
2
3/ 2
]
2 2
1
d ( y2 )
7
E  108(10 ) a z  [2  2 2 2
]dx
( x  y  100)3/ 2
2
0
2
2
2
  216(107 ) a z  [
2
1
1
]dx
 2
1/ 2
( x  104)
( x  100)1/ 2
  216 (107 ) a z ln |
2
x  x 2  104
x  x 2  100
2
|
2
2  108
2  108
)  ln(
))
2  104
2  104
  216 (107 )a z (7.6202 (103 ) )
  216 (107 )a z (ln(
E  16.46 a z MV/m
P.E. 4.6
y = -3 plane
z
line
charge
O
y
P
x
x=2 plane
85
E1 and E2 remain the same as in Example 4.6.
E3 
L
a
2 0 
This expression, which represents the field due to a line charge,
is modified as follows. To get a  , consider the z  1 plane.   2
a   a x cos 45  a y sin 45
1
(a x  a y )
2
10(109 ) 1
E3 
(a x  a y )
109 2
2 (
)
36

 90  (a x  a y ).
Hence,
E  E 1  E 2  E3
  180  a x  270  a y  90  a x  90  a y
 282.7 a x  565.5a y V/m
P.E. 4.7

Q
a  s an
2 r
4 r
2
[(0, 4,3)  (0, 0, 0)] 10  109
ay

5
2
D  DQ  D 
30 109

4 (5) 2
30
(0, 4,3)  5 a y nC / m 2

500
 5.076a y  0.0573a z nC / m 2
86
P.E. 4.8
(a) v    D  4 x
v (1, 0,3)  4 C/m3
(b)  =  D  dS   D  dS
x0
  D  dS
x 1
  D  dS
y0
  D  dS
1 1
1 1
1 1
1 1
1 1
0 0
0 0
0 0
0 0
0 0
y 1
  D  dS
    (2 y 2  z )dydz    (2 y 2  z )dydz    4 x(1)dxdz    (x)dxdz    (x)dxdz
1 1
 4   x dxdz  4(1/ 2)(1)  2C
0 0
1 1 1
(c)   Q    v dv     4 xdxdydz
0 0 0
 4(1)(1)(1/ 2)  2 C
Q2C
P.E. 4.9
Q    vdv     D  dS
For 0  r  10,
Dr ( 4 r 2 ) 
 2r (r ) sin  d dr d
2
2r 4 r
| )  2 r 4
4 0
r2
E
ar nV/m
2 0
Dr (4 r 2 )  4 (
Dr 
r2
2
E (r  2) 
For r  10,
4(109 )
ar  72 ar  226 ar V/m
9
2( 1036 )
Dr (4 r 2 )  2 r0 4 ,
4
r
Dr  0 2
2r
E (r  12) 


r0  10m
r0 4
E
ar nV/m
2 0 r 2
104 (109 )
ar  1250 ar
109
2(
)(144)
36
 3.927ar kV/m
z0
 
87
P. E. 4.10
V (r )   k 1
3
Qk
C
4 0 | r  rk |
At V ()  0,
C 0
| r  r1 | | (1,5, 2)  (2, 1,3) | 46
| r  r2 || (1,5, 2)  (0, 4, 2) | 18
| r  r3 || (1,5, 2)  (0, 0, 0) | 30
106
4
5
3
[


]
V (1,5, 2) 
9
10
46
18
30
4 (
)
36
 10.23 kV
P.E. 4.11
V
Q
4 0 r
C
If V (0, 6, 8)  V (r  10)  2;
5(109 )
C
109
4 (
)(10)
36
2


C  2.5
(a)
5(109 )
VA 
109
4 (
)|(  3,2,6)  (0,0,0)|
36
 3.929 V
 2.5
(b)
VB 
(c )
7  12  52
 2.5  2.696 V
V AB  VB  VA  2.696  3.929   1233
.
V
P.E. 4.12
(a)
45
2
88
W
  E  dl   (3x 2  y )dx  xdy
Q
1
2
  (3x  y )dx |   x dy |
2
y 5
0
5
x2
 18  12  6 kV
W  6 Q  12 mJ
(b)
dy  3 dx
2

W
 E  dl   (3x 2  5  3 x)dx  x(3)dx
Q 
0
2
=  (3x 2  6 x  5)dx  8  12  10  6
0
W  12 mJ
P.E. 4.13
(a)
(0, 0,10)
V


100 cos 0
(1012 ) 
2
4 0 (10 )
(r  10,   0,   0)
1012
 9 mV
109
4 (
)
36
100(1012 )
[2 cos 0 ar  sin 0 a ]
109 3
4 (
)10
36
1.8 ar mV/m
E
(b)
At (1,
 
, ),
3 2
100 cos

(1012 )
3
 0.45 V
109
2
4 (
)(1)
36


100 (1012 )
E
(2 cos ar  sin a )
9
10
3
3
4 (
)(1) 2
36
V
 0.9 ar  0.7794a V/m
89
P.E. 4.14
After Q1 , W1  0
After Q2 , W2  Q2V21 

Q2 Q1
4   0 |(1,0 ,0 )  (0 ,0 ,0 )|
1(  2) (10 18 )
  18 nJ
1
9
4  (10 )
36 
After Q3 ,
W3  Q3 (V31  V32 )  Q2V21

1
2
 3(9 )(10 9 ) 

  18 nJ
 |(0 ,0 , 1)  (0 ,0 ,0 )| |(0 ,0 , 1)  (1,0 ,0 )|
2
 27 (1 
)  18
2
  29.18 nJ
After Q4 ,
W4  Q4 (V41  V42  V43 )  Q3 (V31  V32 )  Q2V21


1
3
2
  4(9 )(10 9 ) 


  W3
 |(0 ,0 ,1)  (0 ,0 ,0 )| |(0 ,0 ,1)  (1,0 ,0 ) |(0 ,0 ,1)  (0 ,0 , 1) 
2 3
  36 (1 
 )  W3
2 2
  39.09  29.18 nJ   68.27 nJ
P.E. 4.15
E  V  ( y  1)ax  (1  x)ay  2az
At (1,2,3), E  3ax  2az V/m
1 1 1
1
1
W   o  E  Edv   o    ( x 2  y 2  2 x  2 y  6)dxdydz
2
2 1 1 1
1
1
1
1

1 
  o   x 2 dx  dydz   y 2 dy  dxdz  2  xdx  dydz  2  ydy  dxdz 6(2)(2)(2) 
2  1
1
1
1

 80 o
1  x3 1
(2)(2)  0  0  6(8)  
  o 2
2  3 1
3

 0.2358 nJ
90
Prob. 4.1
Q1Q2 (rQ1  r Q2 )
20(1012 )[(3, 2,1)  (4, 0, 6)]
(7, 2, 5)
FQ1 

  180
 103
9
3
10
3
688.88
4  rQ1  rQ2
4
(3, 2,1)  (4, 0, 6)
36
 1.8291a x  0.5226 a y  1.3065 a z mN
Prob. 4.2 (a)
E ( 5, 0, 6) 
[(5, 0, 6)  (4, 0, 3)]
Q2 [(5, 0, 6)  (2, 0,1)]

3
4   0 | (5, 0, 6)  (4, 0, 3) |
4   0 | (5, 0, 6)  (2, 0,1) |3
Q1
Q1 (1, 0,9)
Q2 (3, 0,5)

4   0 ( 82)3 4   0 (34)3/ 2
If Ez  0, then

9 Q1
5 Q2
1
1

 0
3/ 2
4   0 (82)
4   0 (34)3/ 2
5
82
5 82
Q1   Q2 ( )3/ 2   4 ( )3/ 2 nC
9
34
9 34
 8.3232 nC
(b)
F (5, 0, 6)  qE (5, 0, 6)
If Fx  0, then
qQ1
3qQ2

0
3/ 2
4 0 (82)
4 0 (34)3/ 2
82 3/ 2
82
)  12( )3/ 2 nC
34
34
Q1  44.945 nC
Q1  3Q2 (
91
Prob. 4.3
Q  (0, 0, 0)  (a, 0, 0) 
Q  (0, 0, 0)  ( a, 0, 0) 
Q(r  rk' )


' 3
3
4 o | (0, 0, 0)  (a, 0, 0) | 4 o | (0, 0, 0)  ( a, 0, 0) |3
k 1 4 o | r  rk |
2
E 
(a)

(b) E 

Q( a, 0, 0) Q(a, 0, 0)
Q
ax


3
3
4 o a
4 o a
2 o a 2
Q  (0, a, 0)  (a, 0, 0) 
Q  (0, a, 0)  ( a, 0, 0) 

3
4 o | (0, a, 0)  (a, 0, 0) | 4 o | (0, a, 0)  ( a, 0, 0) |3
Q( a, a, 0)
Q(a, a, 0)
Q


ax
2 3/ 2
2 3/ 2
4 o (2a )
4 o (2a )
4 2 o a 2
(c )
E
Q  (a, 0, a )  (a, 0, 0) 
4 o | (a, 0, a)  (a, 0, 0) |

3

Q  (a, 0, a )  (a, 0, 0) 
4 o | (a, 0, a)  ( a, 0, 0) |3
Q(0, 0, a ) Q(2a, 0, a )
Q
Q 
1 
a
az



1

x
3
2 3/ 2
2

4 o a
4 o (5a )
4 o a  5 5 
10 5 o a 2
Prob. 4.4
F  qE  mg

 E
mg 2  9.8

 4.9 kV/m
q
4  103
Prob. 4.5
5
(a)
5
Q    L dl   12 x 2 dx  4 x3 | mC  0.5 C
0
0
(b)
Q    S dS 
4

2
  z 2 d dz |  9(2 )
z0  0
 3
z3 4
| nC
30
 1.206  C
(c)
10
r 2 sin  d d dr
r sin 
2
4

42
 10  d  d  rdr  10(2 ) ( )
2
0
0
0
Q   V dV  
 1579.1 C
92
Prob. 4.6
0 0 9
 x 2 a  a3 o
dxdydz  (a)(a)  o 

0
2
2
a
a


2
 /2
o x
a a a
Q    v dv    
v
Prob. 4.7
Q    v dv  
1
   5 z d d  dz mC
2
 0 z 0  / 6
v
=5
 4 2 z2 1
4 0 2 0

 /2 5
10
 (16)(1)( / 2   / 6) 
 /6 8
3
Q  10.472 mC
Prob. 4.8
Q    s dS   6 xydxdy
2
 
x 0
x
4
 x4
y 0
x2
y 2
 6 xydxdy  
 6 xydxdy
y2 x
y2 x  4
6  x
dx   6x
dx
0
2 0
2
x 0
x 2
2
2
 6  x(
x 0
4
4
x2
 0)dx  3  x (4  x) 2  0  dx
2
x2
2
4
x3
 6 dx  3 (16 x  8 x 2  x 3 )dx
2
0
2
x4 2
8 x3 x 4 4
2
3
 6(8 x 
 )
4 0
3
4 2
 12  3(128  32 
 12  3(96 
Q  32 C
512 64
  64  4)
3
3
448
 60)
3
93
Prob. 4.9
y
y=3x/2, z=4 plane
3
0
2
Q    s dS   10 x 2 yzdxdy
S
x
2 3x/2
z4
mC  10(4) 
x 0 y 0
2
2
 40  x 2
x 0
Prob. 4.10
Q    v dv    4  2 z cos  d  d dz
nC
v
2
1
 /4
0
0
0
 4   3 d   zdz  cos d   4
2 z2 1
0 2 0
(sin  )
 /4
0
 (16)(0.5)(sin  / 4)  5.657 nC
Prob. 4.11
1
12 x 3 1
(a) Q    L dl   12 x dx 
 4 nC
3 0
0
L
(b) Method 1:
2
 L dl
r
4 r 3
r = (0, 0, h) - (x, 0, 0) = (-x, 0, h), r | r | h 2  x 2
12 x 2 ( x, 0, h)dx
4  (h 2  x 2 )3/ 2
Since x<<h, we can ignore the x-component.
E
1
E
h
4
az 
Method 2:
2
y 3x / 2
x5 2
9 
 9(32)  288 mC
dx  20  x 2  x 2 dx  20(9 / 4)
0
2 0
4
5


x 0
2
E
 x ydydx
12 x 2 dx
Q
az

3
4 h 2
h
94
The line charge can be regarded as a point charge because it is very far from the
observation point.
E
Q
4 r
a 
2 r
4  109
4(9)
a z  6 a z  3.6 105 a z V / m
9
10
10
4 
(1000) 2
36
Prob. 4.12
 dl
E   L 2 aR
4 o R
L
R   aa   ha z , R | R | a 2  h 2 , dl  ad
E
 L (  aa   ha z )
ad
4 o  (a 2  h 2 )3/ 2
Due to symmetry, the -component cancels
 L 2 haa z d
 L haa z

(2 )
E
2
2 3/ 2

4 o 0 (a  h )
4 o (a 2  h 2 )3/ 2
out.
4 103  12 106  4  3a z
(2 )  260.58a z N
F  QE 
109 3
4 
5
36
Prob. 4.13
 s dS
R, R =  (-a )  ha z , dS   d d 
3
R
4

o
S
s
 d d 
E
(-  a   ha z )

4 o S (  2  h 2 )3/ 2
Due to symmetry, the  -component vanishes.
E
E
2
 s ha z b
2
2 3/ 2
h
d

(


)

0 d
4 o a
Let u= 2  h 2 , du  2  d 
b
 s ha z
 s ha z 1/ 2u 1/ 2  s ha z 
1 3/ 2
1


E

(2 )  u du 
2 o 1/ 2
2 o 
4 o
2
 2  h 2 a 
E

s h 
1
1

 2
 az
4 o  a  h 2
b2  h2 
95
Prob. 4.14
(a) From eq. (4.26),
 678.58a z V/m, z>0

12  109
E  s an 
an  18 (12)an  
9
10
2 o
678.58a z V/m, z<0
2
36
(b) Similarly,
E
678.58a z V/m, z>4
s
12  109
an 
an  18 (12)an  
9
10
2 o
 678.58a z V/m, z<4
2
36
For z > 4 and z < 0, the fields cancel out. For 0 < z < 4, they add up. Thus
E  2  678.58a z V/m= 1.357a z kV/m
Prob. 4.15
y
an
x
Let f ( x, y )  x  2 y  5;
f  a x  2 a y
f
(a x  2 a y )
 
| f |
5
Since point ( 1, 0,1) is below the plane,
an  
_
an  
E
(a x  2 a y )
.
5
s
6(109 )
(a x  2 a y )
(
)
an 
9
2 0
2(10 / 36 )
5
 151.7 a x  303.5 a y V/m
96
Prob. 4. 16
 s
s
x
x=0
x=a
(a) For x < 0,
E  E1  E2 
s
(  s )
( a x ) 
( a x )  0
2 o
2 o
(b) For 0 < x < a,.

(  s )

E  s ax 
( a x )  s a x
o
2 o
2 o
(c ) For x > a,

(  s )
E  s ax 
(a x )  0
2 o
2 o
Prob. 4.17
(a) At P(5,-1,4),
 sk
10  106
20 106
30  106
ank 
a
a
(
)

(
)

( a z )
x
y
109
109
109
k 1 2 o
2
2
2
36
36
36
3
 36 (5, 10, 15)  10  565.5a x  1131a y 1696.5a z kV/m
3
E 
(b) At R(0,-2,1)
E  36 5(a x )  10(a y )  15(a z )  103  565.5a x  1131a y 1696.5a z kV/m
(c ) At Q(3,-4,10),
E  36 5a x 10(a y )  15a z  103  565.5a x  1131a y 1696.5a z kV/m
Prob. 4.18
97
Fe 
e2
4 o r 2
ar
Fe
e2
1


Fg 4 o Gm 2 4 109
 4.17  10
 1.6 1019 

31 
 6.67  1011  9.1 10 
1
36
2
42
Prob. 4.19
Let Q1 be located at the origin. At the spherical surface of radius r,
Q1   D dS   Er (4 r 2 )
Or
Q1
ar by Gauss's law
4 r 2
If a second charge Q2 is placed on the spherical surface, Q2 experiences a force
E
Q1Q2
ar
4 r 2
which is Coulomb’s law.
F  Q2 E 
Prob. 4.20
Q
R
4 R 3
For the given three point charges,
For a point charge,
D
D
1  QR1 QR2 2QR3 
 3 


4  R13
R2
R33 
R1  (0, 0)  (1, 0)  (1, 0), R1  1
R2  (0, 0)  (1, 0)  (1, 0), R2  1
R3  (0, 0)  (0,1)  (0, 1), R3  1
D
Q
Q
Q
(1, 0)  (1, 0)  2(0, 1)  (0, 2)  a y
4
2
4
98
Prob. 4.21
(a) Assume for now that the ring is placed on the z=0 plane.
z
(0,0,h)
R
y
a
 L dl
x
D 
 L dl R
, R   a a  h az
4  R3
  2
D
L
ad (  a a   h a z )

4   0
(a 2  h 2 )3/ 2
Due to symmetry, the  component vanishes.
 L a (2  h) a z
L a h a z

2
2 3/ 2
4  (a  h )
2(a 2  h 2 )3/ 2
a  2, h  3,   5  C/m
D
L
_
Since the ring is actually placed in x  0, a z becomes a x .
D
(b)
(6)(5) a x
 0.32 a x  C/m 2
2(4  9)3/ 2
99
Q [(3, 0, 0)  (0, 3, 0)]
Q [(3, 0, 0)  (0,3, 0)]

3
4  | (3, 0, 0)  (0, 3, 0) |
4  | (3, 0, 0)  (0,3, 0) |3
Q(3,3, 0) Q(3, 3, 0)
6 Q(1, 0, 0)



3/ 2
3/ 2
4 (18)
4 (18)
4  (18)3/ 2
D  DR  DQ  0
DQ 
0.32(106 ) 
6Q
0
4  (18)3/ 2
 Q   0.32(4 )(183/ 2 )106
Prob. 4.22
(a)  v    D 
1
 51.182 C
6
Dx Dy Dz


 0  2 x  4  2 x  4 nC/m3
x
y
z
(b)
   D  dS   y 2 dydz
S
x3
5
6
0
0
  dz  y 2 dy  (5)
y3 6 5 3
 (6)  360 nC
3 0 3
Prob. 4.23
   D  dS   o  E  dS ,
S

o
dS  d  dza
S
  E  dS    6  z sin  d  dz
S
  90o
  2 2   z2 5 
1
 6sin 90   d   zdz  6 

  3(4  0) (25  0)  150
2
 2 0 2 0
0
0
109
  150 o  150 
 1.326 nC
36
2
o
5
100
Prob. 4.24
(a)
1  2
1

1
(r sin  sin  ) 
(sin  cos  sin  ) 
( sin  )
2
r r
r sin  
r sin 
2
sin 
sin 
 sin  sin  
(cos 2   sin 2  ) 
sin 
r
r sin 
2
2
2
But cos   sin   1  2sin 
v    D 
2
r
v  sin  sin  
At
sin  2sin  sin  sin 


0
sin 
sin 
r
r  2,   30o ,   60
(2,30o , 60o ),
o
v  0
   D  dS ,
dS  r 2 sin  d d ar
S
60o 30o
 
 r sin  sin  d d r  2
 
2
2
0 0
60o
30o
0
0
 4  sin  d  sin 2  d
(b)
1
sin 2  (1  cos 2 ).
2
o

60o  1 30
1

 /6
  4   cos 
(1  cos 2 )d  2( cos 60o  1)    sin 2 



2
0  2 0

 0

1


 2(1/ 2  1)   / 6  sin( / 3)  0   0.5236  0.433  0.0906 nC = 90.6 pC
2


But
Prob. 4.25
Dx Dy Dz


 8 y C/m 2
x
y
z
1 
1 D Dz

(b)  v   D =
(  D ) 
z
 
 
 8sin   2sin   4 z
(a)
v   D =
 6sin   4 z C/m3
101
1  2
1

(r D r ) 
(D sin )  0
r r
r sin  
2
2cos
=  4 cos 
r
r4
= 0
 v = D = 2
(c)
Prob. 4.26
 x2 2   y2 2   z 2 2 
(a) Q    v dv     12 xyzdxdydz  12 


  12(2)(2)(2)  96 mC
 2 0 2 0 2 0
v
0 0 0
(b)  =Q= 96 mC
2 2 2
Prob. 4.27
Gaussian surface
r
a
b
Apply Gauss’s law,
 D dS  Q
enc
Dr 4 r 2   s1 4 a 2   s 2 4 b 2  8  109  4 (1) 2  (6 103 )  4 (2) 2  0.3016
Dr 
0.3016  0.3016

 0.0027
4 r 2
4 (3) 2
D  -2.7ar mC/m 2
Prob. 4.28
102
For r <a.
 D dS  Q
enc
S
  v dv
v

2
r
0
0
0
  x dx  dz 
1
C
3
Dr (4 r 2 )   5r1/ 2 r 2 sin  d d dr  5 sin  d  d  r 5/ 2 dr
 5(2)(2 )
r 40 r
r

7/2 0
7
7/2
7/2
40 r 7 / 2
10
7
Dr   o Er 
 r 3/ 2
2
4 r
7
10 3/ 2
Er 
r ,0  r  a
7 o
For r > a,
40 7 / 2
a
7
40 7 / 2
a
Dr   o Er  7
4 r 2
10a 7 / 2
Er 
,r  a
7 o r 2
Thus,
 10 3/ 2
r ar , 0  r  a

 7 o
E
7/2
 10a a , r  a
 7 o r 2 r

Dr (4 r 2 ) 
Prob. 4.29
(a)  v    D 
Dx Dy Dz


 2 y C/m3
x
y
z
(b)    D  dS   x 2 dxdz
1
1
2
y 1
0
0
1
1
1
0
0
0
(c ) Q    v dv   2 ydxdydz  2  dx  ydy  dz  1 C
v
103
Prob. 4.30
(a)
V    D 
1 
(  D ) 
1  D
 
 
V  4 ( z  1) cos   ( z  1) cos   0

 Dz
z
V  3( z  1) cos   C/m3
(b)
Qenc   V dv   3( z  1) cos   d d  dz
2
4
 /2
4
 /2
1
2  z2
 3  d   ( z  1) dz  cos  d  3   2  (  z )| (sin | )
0
0
0 2
0
0
0
2
 3(2)(8  4)(1  0)  72 C
(c)
Let    1   2  3  4  5   D  dS
where  1 ,  2 ,  3 ,  4 ,  5 respectively correspond witn surfaces S1 , S2 , S3 , S4 , S4
(in the figure below) respectively.
y
S3
S4
S5
S1
y
x
S2
104
For S1   2, dS   d dza 
4
 /2
0
0
 1   2  ( z  1) cos  dS|  2  2(2)  ( z  1)dz  cos  d
2
 8(12)(1)  96
For S2 , z  0, dS   d d  ( a z )
 /2
2
 2     cos   d d      d   cos  d
2
3
0


4
0
2
| (1)  4
4
0
For S3, z  4, dS   d d  a z ,  3  4
For S4 ,    / 2, dS  d  dza


 4     ( z  1) sin  d  dz|  /2    sin


2
2
 2
4
   d    z  1dz
20
0
2
| (12)  (2)(12)  24
0
For S5 ,   0, dS  d  dz (a ) , 5    ( z  1) sin  d  dz|
 0
  96  4  4  24  0  72 C
This is exactly the answer obtained in part (b).
Prob. 4.31
e
R
0
105
F  eE
0 
e

3
R
3
4
3e
4  R3
  0,
0rR
0,
elsewhere
V  
 D  dS  Qenc   V dV 
Er 
3 e 4 r 3
 Dr (4 r 2 )
4 R 3 3
3e r
12 0 R 3
F  eE 
e2 r
4  0 R 3
Prob. 4.32
Using Gauss’ law, when   0,
E0
Qenc   D  dS  o  E  dS  o E (2 L)
For a    b,
S
S
where L is the length of the cable. But
2
o
 d dz   o 2 L
 0 z 0 
L
Q enc    s dS   
S
 o 2 L   o E (2 L)

For  >b, Q enc  0. Thus,
 o
a  b
a ,

E   o 
 0,
otherwise

E 
o
o
106
Prob. 4.33
(a)
  Qenc
at r  2
Qenc   V dV  
2
 10 
2
10 2
r sin  d dr d
r2

  sin  d drd
r 1   0   0
 10 (1) (2 ) (2)  (40  ) mC
Thus,   125.7 mC
At r  6;
4
2

r 1
 0
 0
Qenc.  10  dr  d  sin  d
 10 (3)(2 ) (2)  120  mC
  377 mC
(b)
  Qenc
But  
At
Dr 
2
r
r
r  1,
Qenc  0
At r  5,
 D  dS  D  dS  D (4 r )

 D0
Qenc  120 
Qenc
120 

 1.2
2
4 r
4  (5) 2
D  1.2 ar mC/m 2
Prob. 4.34
v 
Q
Q
3Q


3
volume 4 a / 3 4 a 3
For r < a,
 D dS  Q
Dr 4 r 2 
enc
  v dv
3Q 4 r 3 Qr 3
 3
4 a 3 3
a


Dr 
Qr
4 a 3
107
 D dS  Q
For r > a,
Dr 4 r 2  Q


Dr 
Q
4 r 2
Hence,
 Qr
 4 a 3 ar ,
D
 Q a,
 4 r 2 r
ra
ra
Prob. 4.35

109 
2
4



VP 
9 
10  | (1, 2,3)  (1, 0,3) | | (1, 2,3)  (2,1,5) | 
4 o r1 4 o r2
4 
36
4
2

 9 
  1.325 V
994
2
Prob. 4.36
Q
V 4
r  a 2  a 2  h 2  22  22  32  17 cm
,
4 o r
Q1
V
Q2
4  8  109
 6.985 kV
109
2
4 
 17  10
36
Prob. 4.37 (a)
Q/2
V

2
Q
2
4 0 r

6
Q
4 0 r
60(10 )
 135 kV
109
4 
x4
36
Q/2
108
(b)
Q
3( )
3  135 kV
V
4 0 r
(c)
Q
2 (4)
Q
 L dl
8

V

 135 kV
4 0 r
4  0 r 4 0 r
Prob. 4.38
(a)
VP  
Qk
4 | r p  r k |
103
3(103 )
2 (103 )


| (1,1, 2)  (0, 0, 4) |
| (1,1, 2)  (2,5,1) |
| (1,1, 2)  (3, 4, 6) |
1
2
3
1
2
3
4   0 (103 ) V p 





| (1,1, 2) | | (1, 4,1) | (4,5, 4) |
6
18
57
4   oV p 
4
109
(103 ) V p  0.3542
36 
V p  3.008  106 V
(b)
VQ  
Qk
4  o | r p  r k |
103
3(103 )
2 (103 )


| (1, 2,3)  (0, 0, 4) |
| (1, 2,3)  (2,5,1) |
| (1, 2,3)  (3, 4, 6) |
1
2
3
1
2
3
4   0 (103 ) V p 





| (1, 2, 1) | | (3, 3, 2) | | (2, 6, 3) |
6
22 7
4   oVQ 
109
4
(103 ) V p  0.410
36
VQ  3.694 (106 )V
VPQ  VQ  VP  0.686 (106 )  686 kV
109
Prob. 4.39
 E  V 
V
1 V
V
a 
a 
az

z
 
 2  e  z sin  a    e  z cos  a    2 e  z sin  a z
At (4, /4,-1),  =4, = /4,z  1
 E  2(4)e1 sin( / 4)a   4e1 cos( / 4)a  16e1 sin( / 4)a z
 2(4)(2.7183)(0.7071)a   4(2.7183)(0, 7071)a  16(2.7183)(0, 7071)a z
E  15.38a   7.688a  30.75a z V/m
Prob. 4.40
(a)
V
V
V
E  (
ax 
ay 
az )
x
 y
z
  2 xy ( z  3)a x  x 2 ( z  3) a y  x 2 y a z
x  3, y  4, z  6,
At (3, 4, 6),
E   2(3)(4)(3) a x  9 (3)a y  9(4) a z
 72 a x  27 a y  36a z V/m
(b)
V    D   0  E    0 (2 y ) ( z  3)
  Qenc   V dV   2 0  y ( z  3)dx dy dz
1
1
1
0
0
0
  2 0  dx  y dy  ( z  3)dz   2 0 (1)(1/ 2)(
1
z2
 3z ) |
0
2
1
7 109
   0 (  3) 
(
)
2
2 36 
Qenc   30.95 pC
Prob. 4.41
(a)
 r2 
Q    v dv    o 1  2 r 2 sin  d d dr
 a 
v
2

a

 a 3 a 3  8 3
r4 
a o
 o  d  sin  d   r 2  2 dr   o (2 )(2)    
a 
 3 5  15
0
0
0
110
(b) Outside the nucleus, r >a,
 D dS  Q
enc


E
S
Qenc
ar
4 o r 2
8 a 3  o
2a 3  o
15
E
ar 
ar
4 o r 2
15 o r 2
V    E dl    Er dr 
V ()  0, C1  0.
Since
V
2a 3  o
 C1
15 o r
2a  o
15 o r
3
(c ) Inside the nucleus, r <a
 r3 r5 
Qenc  4o   2 
 3 5a 
Qenc
o  r r 3 
E
a

 
a
r
 o  3 5a 2  r
4 o r 2
o  r 2
r4 
V    Er dr    
C
 o  6 20a 2  2
V (r  a ) 
2a 2  o  o  a 2 a 2 

  C
15 o
 o  20 6  2
2a 2  o 7 a 2  o a 2  o
C2 


15 o
60 o
4 o
V
o  r 4
r 2  a 2 o



 o  20a 2 6  4 o
(d) E is maximum when
  1 3r 2 
dE
0 o   2 
dr
 o  3 5a 
r

 9r 2  5a
2
5
a  0.7454a
3
We are able to say maximum because
d 2E
6r
  2  0.
2
dr
5a
111
Prob. 4.42
z
h
R
b

a
x
D
s 
D
 s dSR
,
4 R 3
R    a  ha z ,
R | R |   2  h 2 , dS   d d 
Q
Q

2
S  (b  a 2 )
 s  d d  (  a   ha z )
4 
(  2  h 2 )3/ 2
Due to symmetry, the component along a vanishes.
y
112
b
 s h b 2  d  d 
s h
Dz 

(2 )  (  2  h 2 ) 3/ 2  d 
4  a 0 (  2  h 2 )3/ 2 4
a

1  b  s h 
1
1


 2



2    h 2  a
2  a2  h2
b2  h2 


Qh
1
1
D

 az
2
2 
2 (b  a )  a 2  h 2
b2  h2 

s h 
Prob. 4.43
Dx Dy Dz


 4  20 y  2 z
x
y
z
At P(1,2,3), x=1, y= 2, z=3
v   D 
v  4  20(2)  2(3)  30 C/m3
Prob. 4.44
r=3cm
r=5cm
For r < 3cm, Qenc  0
For 3 < r < 5cm,
 D dS  Q
enc

 D0
 10 nC
Dr 4 r 2  10 nC

 Dr 
10
nC/m 2
4 r 2
For r > 5 cm,
 D dS  Q
enc
 10 -5 = 5nC

 Dr 
5
nC/m 2
2
4 r
113
Thus,


0,
r  3 cm

 10
D
a nC/m 2 , 3  r  5 cm
2 r
 4 r
5

2
r  5 cm
 4 r 2 ar nC/m ,

Prob. 4.45
v   D    o E 
1    o Eo  2 


   a 

2 o Eo
,0    a
a
Prob. 4.46
Let us choose the following path of two segments.
(2,1, 1)  (5,1, 1)  (5,1, 2)
W   q  E  dl
5
W
   E  dl   2 xyzdx
q
x2
2
z  1, y  1
  x 2 ydz
z 1
x  5, y  1
2
x2 5
 2(1)(1)
 (5) 2 (1) z
 21  75  54
1
2 2
W  54q  108  J
Prob. 4.47
(a)
v    D     E
v
1 
1
  E 
(12  2 z cos  )  (6  z cos  )  0

 

 24 z cos   6 z cos   18 z cos 
At A(2,180o , 1),
  2,   180o , z  1
v  18 z cos   18(1) cos(180o ) 
109 109

 0.1592 nC/m3
36
2
114
(b)
W   Q  E  dl ,
dl   d a
L

0o 
W  Q  6  z sin   d
 Q6(2) (1)   cos 
  24Q(1  1)

  2, z  1
180o 
 180o

0o
2
 48Q  48 10  106  480  J
Prob. 4.48
(a)
From A to B, dl  rd a ,
90
WAB  Q  10 r cos  r d |   Q (10)(5) 2 (sin  )
 30
r 5
90o
 1250 nJ
30o
(b)
From A to C , dl  dr a r ,
10
WAC   Q  20 r sin  dr
r 5
|   Q(20)(sin 30 )  2 r  5  3750 nJ
o
1 2 10
 30
(c )
From A to D, dl  r sin  d a ,
WAD  Q  0(r sin  ) d  0 J
(d)
WAE  WAD  WDF  WFE
where F is (10,30o , 60o ). Hence,
90
 10

WAE   Q   20 r sin  dr |   10 r cos  r d | 
r 10 
 30
 r 5
 30o
75 100
] nJ   8750 nJ
  100[ 
2
2
o
Prob. 4.49
B
5
10
dr
2
r
1
VAB    E dl   
A
10
1
 15  10(  1)  8V
r
5
115
Prob. 4.50
Method 1:
W   Q  E dl ,
dl   d a
L
 E   cos  sin  0   Ex 
 E     sin  cos  0   E 
  
 y
 Ez   0
0
1   Ez 
E   Ex sin   E y cos   20 x sin   40 y cos 
x   cos  , y   sin 
E  20  cos  sin   40  sin  cos   20  cos  sin 
W  Q  E dl = -2 10-3  20  cos  sin  d
L
 /2
 2(20)(2)2  sin  d (sin  ) mJ  160
0
 2
sin 2   / 2
 80 mJ
0
2
Method 2:
W
   E dl =  20xdx + 40ydy
Q L
y  2  x, dy  dx
0
W
   20xdx + 40(2 - x)(-dx) =  (60x - 80)dx
Q
x=2

0
60 x 2
 80 x  40
2
2
W  40Q  80 mJ
Method 3:



y
z  0
  E  x
20 x 40 y 10 z
V    E dl = -10x 2  20 y 2  5 z 2  C
L
W  Q(V2  V1 )  Q(20  4  10  4)  40Q
W  40Q  80 mJ
116
Prob. 4.51
W   Q  E dl
L


E  s an  s a x ,
2 o
2 o
dl  dxa x
Qs
Qs
Qs
W 
dx  
(2) 

o
2 o 3
2 o
1

10 106  40  109
 400  36 106  45.24 mJ
109
36
Prob. 4.52
(a) E  V  
V
V
V
ax 
ay 
a y  4 xa x  8 ya y
x
y
z
v   D   o E   o (4  8)  12 o  106.25 pC/m3
V
1 V
V
a 
a 
az

z
 
 (20  sin   6 z )a  10  cos  a  6  a z
(b) E  V  
 1
v   D   o E   o  
 

 40  sin   6 z   10sin  


6z 
   30sin     o C/m3
 

V
1 V
1 V
ar 
a 
a
r
r 
r sin  
 10r cos  sin  ar  5r sin  sin  a  5r cot  cos  a
(c) E  V  
v   D   o E
v 1
5r sin 
5r cot  sin 
2sin  cos  
 2 (30r 2 cos  sin  ) 
r sin 
r sin 
o r
v   o (5sin  csc2  cos   20 cos  sin  ) C/m3
117
Prob. 4.53
(a)
 E  V 
V
1 V
V
a 
a 
a z  e  z sin  a   e  z cos  a   e  z sin  a z

z
 
E  e  z sin  a   e  z cos  a   e  z sin  a z
(b)
.
0
E 
 1 Ez E 
 E
1
E 
 E  

a      z  a   (  E )    a z

z 
 
 
  
  
 z
1

1
   e  z cos   e  z cos   a    e  z sin   e  z sin   a   e  z cos   e  z cos   a z  0



since each component is zero. Alternatively,
  E    V 
Prob. 4.54
V  r 3 sin  cos 
 E  V 
V
1 V
1 V
ar 
a 
a
r
r 
r sin  
r 4 sin 
( sin  )a
sin 
3
1
sin 
E
 4 sin  cos  ar  4 cos  cos  a  4 a
r
r
r
o
o
o
o
At (1,30 , 60 ), r  1,  30 ,   60
 3r 4 sin  cos  ar  r 4 cos  cos  a 
E
 3sin 30o cos 60o ar  cos 30o cos 60o a  sin 60o a
 0.75ar  0.433a  0.866a
109
(0.75ar  0.433a  0.866a )
36
 6.635ar  3.829a  7.657a pC/m 2
D  o E 
Prob. 4.55
For a < r < b, we apply Gauss’s law.
118
 D dS  Q
enc
Q
S
Dr (4 r 2 )  Q

 Er 
b
Vab    E dl  
a
Dr
o

Q
4 o r 2
1
Q 1b
Q 1 1
dr


  
2

4 o a r
4 o r a
4 o  a b 
b
Q
Prob. 4.56
 /2
 s dS
 s 2  / 2 r 2 sin  d d
s


(2 )  sin d
V 
4 o r 4 o 0  0
4 o r
r
S
0

V
 / 2
s 
  cos 

0 ra
2 o r 

s
2 o a
Prob. 4.57
 E  0


 D  0
D 
 1 Dz D 
 D D 
1
a     z  a   (  D )    a z
 D  


  
z 
 
 
  
 z
1
 0 a   0a  2  cos  a z  0

Hence D is not a genuine EM field.
 /4 1
 /4
1
 0 z 0
0
0
   D dS    2  sin  d dz  2  sin  d  dz  2
S
 2 cos 
 /4
0
(1)(1) 2  2(cos  / 4  1)  0.5858 C
 1
119
Prob. 4.58
(a)
d2y
m 2  eE ; divide by m , and integrate once, one obtains :
dt
dy
eEt
u

 c0
dt
m
e E t2
 c0t  c1
2m
"From rest" implies c1  0  c0
y 
(1)
V
or V  E d .
d
Substituting this in (1) yields :
2m d
t2 
eE
Hence :
At t  t0 ,
y  d, E 
2md
eE
u
eE
m
that is, u 
V

2 e Ed
2eV

m
m
u k V
or
(b)
k 
2e

m
2 (1.603) 1019
9.1066 (1031 )
 5.933  105
(c )
1
u m
100  2.557 k V
V

2e
2 (1.76) (1011 )
2
9(1016 )
120
Prob. 4.59
(a)
This is similar to Example 4.3.
eEt
, u x  u0
uy 
m
e E t2
, x  u0 t
y 
2m
x 10 (102 )
 10 ns
t 
107
u0
Since x 10 cm when y 1cm,
E
2m y
2 (102 )

 1.136 kV/m
1.76 (1011 ) (1016 )
et2
E   1.136 a y kV/m
(b)
u x  u0  107 ,
uy 
2000
(1.76)1011 (108 )  2(106 )
1.76
u  (a x  0.2a y ) (107 ) m/s
Prob. 4.60
p cos 
k cos 

V
2
4 0 r
r2
At (0, l nm),
  0,
r  1 nm, V  9;
k (1)
,  k  9(1018 )
1(1018 )
cos 
V  9(1018 ) 2
r
At (1,1) nm, r  2 nm,
  45,
that is,
V
9
9(1018 ) cos 45
9

 3.182 V
18
2
10 ( 2)
2 2
121
Prob. 4.61
The dipole is oriented along y  axis.
V
pr
; p  r  Q d a y  a r  Qd sin  sin 
4 0 r 2
V
Qd sin  sin 
4 0 r 2
E   V  

1V
1 V
V
ar 
a 
a
r 
r sin   
r

Q d  2sin  sin 
cos  sin 
cos 
ar 
a  3 a  

3
3
r
r
r
4 0 


_
E
Qd
4  0r 3
(2sin  sin  a r  cos  sin  a  cos  a )
Prob. 4.62
Using eq. (4.81),
p  (r  r ')
V
4 o | r  r ' |3
r  r '  (4, 0,1)  (2,3, 1)  (2, 3, 2)
| r  r ' | 4  9  4  17
p  (r  r ')  (2, 6, 4)  (2, 3, 2)  4  18  8  22
V
22  106
198

kV= 2.825 kV
9
10
70.093
3/ 2
4 
17
36
Prob. 4.63
E  k (2 cos  ar  sin  a )
 E   sin  cos  0   2k cos  
E    0
0
1   k sin  
  
 Ez  cos   sin  0   0 
Ez  2k cos 2   k sin 2   2k cos 2   k (1  cos 2  )  3k cos 2   k
Setting this to zero gives
3cos 2  1

cos   
1
 0.5773
3

  54.74o , 125.26o
122
Prob. 4.64
W  W1  W2  0  Q2V21  Q2
Q1
4 o | (2, 0, 0)  (0, 0,1) |
40  109  (50) 109 40  9  (50)  109


109
4 1
4 
| (2, 0, 1) |
36
 8.05  J
Prob. 4.65
E  V  
V
V
ax 
a y  4 xa x  12 ya y V/m
x
y
1
1
1
1
1
W   o  | E |2 dv   o    (16 x 2  144 y 2 )dxdydz
2 v
2 z 1 y 1 x 1
x3 1
y 3 1  1 109
1 
1
  o 16(4)
 144(4)
(160)(4) (1  1)

2 
3 1
3 1 2 36
3
 1.886 nJ
Prob. 4.66
Given that E  2r sin  cos  ar  r cos  cos  a  r sin  a
E 2  4r 2 sin 2  cos 2   r 2 cos 2  cos 2   r 2 sin 2 
 r 2 cos 2   4sin 2   cos 2    r 2 sin 2 
r 2 cos 2   3r 2 cos 2  sin 2   r 2 sin 2 

 r 2 (1  3cos 2  sin 2  )
W


E r sin  drd d
2 
2 2
2 4
2 0
 
r dr   (1  3cos 2  sin 2  ) sin  d d
 0 


16
3
( sin  
sin 2 )d

5 0
2

16 109
16
x
(4 ) 
nJ= 0.36 nJ
5 36
45
123
Prob. 4.67
Method 1:
W
1
V
1
 sVdS    s dS  QV

2S
2S
2
V
But
W
Q
Q
4 o a
2
8 o a
Method 2:
1
1
W   D Edv   o  E 2 dv
2v
2 v
2
 Q  2
1
  o  
r sin  d drd
2 
2
 4 o r 
2


o
Q2
Q2
1

  o  d  sin  d 
dr

(2
)(2)
2 0
16 2 o2 r 2
2
16 2 o2 a
r a
0
W
Q2
8 o a
Prob. 68
W
2W
o
1
1
1
D  Edv   o  | E |2 dv   o   y 4  4 x 2 y 2  16 z 2  dxdydz

2v
2 v
2 v
2
4
1
2
4
1
2
4
1
  dx  dz  y dy  4 x dx  dz  y dy  16 dx  z dz  dy
4
0
0
1
2
0
2
0
2
1
0
0
1
 y 1
 x 2  y 1 
 z3 4 
 2(4)  2


4(4)
2
16(2)(2)






 5 0
 3 0  3 0 
 3 0
16 256 64  64
 

 1396.98
5
9
3
1 109
1396.98  6.176 nJ
W 
2 36
5
3
3
124
Prob. 4.69
(a)
 V
V 
1 V
E  V   
a 
a 
az 
 
z 
 
 e  z sin  a   e  z cos  a   e  z sin  a z
E E | E |2  e2 z sin 2   e2 z cos 2    2 e 2 z sin 2   e2 z   2 e2 z sin 2 
1
W   o  | E |2 dv
2 v
2W
o
  | E |2 dv    e2 z   2 e2 z sin 2   d  d dz
v
1
(b)
2
2
0
0
1
2
2
0
0
   d   d  e2 z dz    3 d   sin 2  d  e 2 z dz
0
0
 e 2   1   sin 2  2  e2 z 2 
(2 ) 



 

2 0
4  0  2 0 
 2 0  4 0  2
1
5
  (1/ 2)(e4  1)  (  0)(1/ 2)(e 4  1) 
(1  e 4 )  1.9275
4
8
9
1
1 10
(1.9275)  8.512 pJ
W   o (1.9275)  
2
2 36

 1
2
2 z
4
125
CHAPTER 5
dS =  d dza
P. E. 5.1
I

2
J  dS 
5
  10 z sin dzd|
2
 2
=
 0 z1
z2
1
1
1
 2
(1  cos 2 )d  10(52  12 )    sin 2 
 10(24)  240

2 1 0 2
2
2
 0
5 2
10(2)
I = 754 A
P. E. 5.2
I   s wu  0.5 106  0.110  0.5 A
V  IR  0.5  106 1014  50 MV
P. E. 5.3   5.8 107 S/m
J
8  106
J E


E 
 0.138 V/m
 5.8 107
8  106
J  vu

 u

 4.42 104 m/s
10
v 1.8110
P. E. 5.4 The composite bar can be modeled as a parallel combination of resistors as
shown below.
J
RL
Rc
126
l
,
 L SL
RL 
For the lead,
RL 
l  4m, S L  d 2   r 2  9 
4
 
(5 10 )  9   104
4


4
cm 2 ,  L  5 106 S/m
 973.8
6
Rc 
l
,
 c Sc
Rc 
For copper,
4
5.8 10 

4
Sc   r 2 

4
,  c  5.8 107 S/m cm2
 878.5 
10
4
RR
973.8  878.5
 461.8
R L c 
RL  Rc 973.8  878.5
7
Ps  P  as  ax2  b
P. E. 5.5
 ps x 0  P  ( ax ) x 0  (ax 2  b)
 ps x  L  P  ax
xL
 (ax 2  b)
x0
xL
b
 aL2  b

Qs   ps dS  bA  (aL2  b) A  AaL2
 pv    P  
 pv

x0
 0,
d
(ax 2  b)  2ax
dx
 pv
x L
  2aL
L

Qv   pv dv  ( 2ax ) Adx   AaL2
0
Hence,
QT  Qv  Qs   AaL2  AaL2  0
P. E. 5.6
127
V
103
ax 
ax  500ax kV/m
d
2 x10 3
10 9
P   e o E  (2.55  1)x
x0.5 x106 ax  6.853ax  C /m2
36
E
 ps  P  ax  6.853 C/m2
P. E. 5.7 (a) Since P   o  e E ,
e 
Px   o  e E x
Px
3 x10 9 1

x36 x109  2.16
10 5
 o Ex
36 109 1

(3, 1, 4)109  5ax  1.67 a y  6.67 az V/m
(b) E 
2.16 10
 e o
(c )
 P 3.16  1 
nC / m2  139.7 a x  46 .6 a y  186 .3az
D   o r E  r 

 (3, 1,4)
2.16  10  
e
P
P. E. 5.8 From Example 5.8,
F
But
s2 S
2 o


s   o E   o
s2 
s2 
Vd
.
d
2 o F  o 2Vd 2

S
d2
2 o F
S
Hence


Vd 2 
2 Fd 2
oS
i.e.
Vd  V1  V2 
2 Fd 2
oS
as required.
P. E. 5.9 (a) Since
D1n  12a x ,
E 2 t  E1t
an  a x ,
D1t  10 a x  4a z ,


D2 t 
D2 n  D1n  12a x
 2 D1t
1

( 10 a y  4a z )  4a y  16
. az
2.5
1
D2  D2 n  D2 t  12ax  4ay  1.6az nC/m2.
pC / m2
128
D
tan  2  2 t 
D2 n
(  4) 2  (16
. )2
 0.359
12
 
 2  19.75 o
(b ) E1t  E 2t  E2 sin  2  12 sin 60 o  10.392
E2
E 2t
2
E2n
E1n 
r2
1
12 cos 60 o  2.4
E2n 
2.5
 r1
E1 
tan 1 
x
E1t 2  E1n 2  10.67
E1t
E2t
 E

2.5
tan 60o  4.33

 r1 2t  r1 tan  2 
1
E1n ( r1 /  r 2 ) E2 n  r 2 E2 n  r 2


Note that 1   2 .
P. E. 5.10
10 9
D   oE 
(60 ,20 , 30 ) x10 3  0.531a x  0.177 a y  0.265az pC/m2
36 
10 9
 s  Dn  | D| 
(10 ) 36  4  9 (10 3 )  0.619 pC/m2
36 
1  77o
129
Prob. 5.1
I   J  dS ,
dS = dydza x
I   e  x cos(4 y )dydz
x2
 /3
4
0
0
 e 2  cos(4 y )dy  dz
 sin 4 y  / 3  2 
4

 4e 2 
  e  sin( )  0   0.1172 A
0 
3


 4
Prob. 5.2
Method 1:
I   J  dS = 
10 103 t 2
e r sin d d
t  2ms, r  4m
r
 10(4)e 10 210
3
3

2
 0
 0
2
2
 sin  d  d  40e (2)(2 )  160 e
 68.03 A
Method 2:
3
10 103 t
10
e dS  e 10 t (4 r 2 )
r
r
since r is constant on the surface.
I   J  dS = 
I=40 re 2  160 e 2
 68.03 A
Prob. 5.3
I   J dS  
10

 10(5)( cos  )

0

0
0
 100 A
Prob. 5.4
I   J  dS
5
sin  d dz  10  dz  sin  d
2
2
a
 0  0
0
 0
a
5 
10 
10 
 e  d d   5  d   e d 
10 
e
 a 10 10 a
 5(2 ) 
(10   1)  
 e (10a  1)  1(0  1)  ,
 100
 0 100



e 0.04 (0.04  1)  1  (0.00078)  244.7  A
10
10
a  0.004
130
Prob. 5.5
I   J  dS ,
dS  r 2 sin  d d ar
S
2
 /2
2
 /2
20 cos  2
20(9)
r sin  d d

d cos  sin  d

r 3
6 0 /4
  0   / 4 r  3
I 
 /2
 30(2 )  cos  d ( cos  )  60
 /4
cos 2   / 2
 60 (0  cos 2 ( / 4))  60 (1/ 2)  30
2  /4
I  94.2 A
Prob. 5.6
l
l
2  102
R

 
 6
 3.978  104 S/m
3 2
S
RS 10 ( )(4 10 )
8 102
8
l
Prob. 5.7 (a) R 


 33.95m
4
6
 S 3 10  (25)10
75
(b) I  V / R  9 
75
 265.1 A
8
(c ) P = IV = 2.386 kW
Prob. 5.8
(a) E 
V
9

 90 mV/m
l 100
(b) R 
V
9

 30 
I 0.3
l
S

 
R
l
100

 2.653 105 S/m
6
2
RS 30(  2 )10
131
Prob. 5.9
If R and S are the same,




 1   2 1
R1  1  R2  2
 1S
 2S
2
If 1 corresponds to copper and 2 to silver,
 1  5.8 107 S/m,  2  6.1107 S/m
5.8
 0.951 2
6.1
That is, the copper wire is shorter than silver wire or the silver wire is longer.
1   2
Prob. 5.10

V
I

 
,
S   r2
SV
S I
7
2(5)
10


 6.635  104 S/m
6
2
 (2 10 )(12) 48
R
2
Prob. 5.11 (a) Si   ri   (1.5)2 x10 4  7.068 x10 4
So   (ro 2  ri 2 )   (4  2.25) 104  5.498 104
RI 
Ro 
I l
SI

o l
So
11.8  10 8  10
 16.69  10 4
4
7.068 X10

1.77  10 8  10
 3.219  10 4
4
5.498  10
Ri Ro
16.69  3.219 104
R  Ri // Ro 

 0.27m
Ri  Ro
16.69  3.219
(b)
V  I i Ri  I o Ro


Ii
R
0.3219
 o 
 0.1929
1669
.
Io
Ri
I i  I o  11929
.
I o  60 A
I o  50.3 A
(copper),
I i  9.7 A
(steel)
Alternatively, using the principle of current division,
132
I o  60
Ri
 50.3 A
Ri  Ro
I i  60
Ro
 9.7 A
Ri  Ro
(c)
10  1.77 108
 0.141m
R
 (22 )  104
Prob. 5.12
From eq. (5.16),
 
R1  1  1 ,
S1
ab
R2 
2
S2

2
ac
1  2
1  2  2
R1 R2
R  R1 R2 
 ab ac 
R1  R2 1   2  ac 1  ab 2 
ab ac
1  2 
R
a (c 1  b 2 )
Prob. 5.13


V
 2 
S r  I
 r 2 12 (0.84 103 ) 2  6.1107
I V

 130.86 A
12.4

R
Prob. 5.14
| P | n | p | nQd  2ned   e o E
(Q  2e)
2ned 2  5  1025  1.602  1019 1018
e 

 0.000182
109
oE
4
10
36
 r  1   e  1000182
.
133
Prob. 5.15
N
P
 qi di
i 1
v
N
p
 i 1
i
v
N
| p | 2 1019 1.8 1027  3.6 108
v
| P |
P |P|ax  3.6  10 8 ax C/m2
But
P   e o E
or
P
3.6  108
e 

 0.0407
 o E (109 / 36 )105
 r  1   e  10407
.
Prob. 5.16
Q
E
ar
4 o r r 2
P   e o E 
 eQ
3(10)103

a
ar  596.8ar  C/m 2
r
4 r r 2
4 (4)12
Prob. 5.17
P   e o E


E
P
 e o

100  109
a  2.261a  kV/m
109
2.5
(2)
36
109
D   o r E = 3.5 
2.261103 a   70a  nC/m 2
36
Prob. 5.18
1 
( p  2 )  2 po
  o
The surface polarization charge is
 pv   P  
 ps  P a 
Prob. 5.19
(a)
 a
 po a
134
Qs1   P dS , dS  r 2 sin d d (ar )
S
   4r r 2 sin  d d
2
r  1.2cm

 4(1.2) (106 )  d  sin  d (1012 )
3
0
 6.912(2 )(2)  10
0
18
 86.86  1018 C
(b)
Qs 2   P dS , dS  r 2 sin d d (ar )
S
   4r r 2 sin  d d
r  2.6cm
2
 4(2.6) (10 )  d  sin  d (1012 )
6
3
0
 4(2.6) (2 )(2) 1018  883.5 1018 C
3
(c )
1 
(4r 3 ) pC/m3  12pC/m3
r r
 pv   P   2

2
2.6
0
0
1.2
Qv    pv dv  12  dv  12  sin  d  d  r 2 dr (1018 )
v
 12(2)(2 )
3
r 2.6 18
(10 )  16 (2.63  1.23 )(1018 )
3 1.2
 796.611018 C
Prob. 5.20
109
(6,12, 20)  0.1114a x  0.2228a y  0.3714a z nC/m 2
36
109
P   e o E  1.1x
(6,12, 20)  0.0584a x  0.1167a y  0.1945a z nC/m 2
36
D   o r E  2.1x
Prob. 5.21
At P  2,5,3 ,
x   2, y  5, z  3
V  4  4  5  27   2.16 kV
135
 V
V
V 
E  V   
ax 
ay 
a z     8 xyz 3a x  4 x 2 z 3a y  12 x 2 yz 2 a z 
y
z 
 x
At P,
E  (16)(5)(27)a x  4(4)(27)a y  12(4)(5)(9)a z  2.16a x  0.432a y  2.16a z kV/m
P   e o E 
7  109
(2160, 432, 2160)  133.69a x  26.74a y  133.69a z nC/m 2
36
Prob. 5.22
 V
V
V 
(a) E  V   
ax 
ay 
a z   20 xyza x  10 x 2 za y  10( x 2 y  z )a z V/m
x
y
z





(b) D   E  5 o E  0.8842 xyza x  0.4421x 2 za y  0.4421( x 2 y  z )a z nC/m 2
(c ) P   e o E  4 o E  0.7073xyza x  0.3537 x 2 za y  0.3537( x 2 y  z )a z nC/m 2
(d)  v   2V



 2V  (20 xyz )  (10 x 2 z )  (10 x 2 y  10 z )  20 yz  10
x
y
z
 v  5 o10(2 yz  1)  0.8854 yz  0.4427 nC/m3
136
Prob. 5.23
Using Gauss' law,
Qenc   D  dS
For r < a, Qenc  0

E 0 DP
For a < r <b, Qenc  4C
4  Dr (4 r 2 )
E
D



D
a
4
a  r2
2 r
4 r
r
ar
D

2 o 2 o r 2
ar
2 r 2
For b < r < c, Qenc  4  6  2C
P   e o E  (1) o E =
2  Dr (4 r 2 )
E
D



D
a
2
a  r 2
2 r
4 r
2 r
ar
D

5 o
10 o r 2
2ar
4 ar
2
10 r
5 r 2
For r >c, Qenc  4  6  10  8C
P   e o E =(4) o E = -
8  Dr (4 r 2 )
E
D


D
8
2a
a  r2
2 r
r
4 r
2ar
 o r 2
o
P   e o E = (0) o E = 0
Prob. 5.24 (a) Applying Coulomb’s law, we obtain the electric field intensity due to a
point charge as
Q
 Dr
   4 r 2 , r  b
o
Er   o
 Dr  Q , a  r  b
 
4 r 2
P
Hence
 r 1
D
r
( D   O E )
137
Pr 
r  1 Q
.
,
 r 4 r 2
a rb
1 d 2
(r Pr )  0
r dr
 pv    P   2
(b)
(c )
 ps  P  (ar )  
 ps  P  (ar )  
Q r 1
(
),
4 a2  r
Q r 1
(
),
4 b2  r
r a
r b
Prob. 5.25
F1 
Q1Q2
Q1Q2
 2.6 nN,
 1.5 nN
F2 
2
4 o d
4 o r d 2
F1 2.6

  r  1.733
F2 1.5
Prob. 5.26
(a) By Gauss’s law,
 D dS  Q
enc

 Dr 
S
Er 
Dr


Q
4 r 2
Q
4 r 2
1
W    | E |2 dv,
2
v
2  
dv  r 2 sin  drd d
1
Q2
Q2
2

sin



W    
r
drd
d
2  0  r  a 16 2 2 r 4
8 a
138
(b) Dr remains the same but
D
Q
Q

Er  r 
2
2

4 o  r  a 
 a
4 r 2 o 1  
 r
2


1
1
Q 2 r 2 sin  drd d  r  a 
2
o 
W    | E | dv    

2
2  0  0 r  a 16 2 2  r  a 4
 r 
v

1   Q2 1
Q2
dr
Q2 

(4
)




a  r  a 2 8 o  r  a a   8 o 2a
32 2 o
Q2
W
16a o
Prob. 5.27
(a)
 o ,
v  
 0,
0 r a
ra
For r < a, Er (4r 2 )  o
4 r 3
3
V    E  dl  
For r > a,  o E r (4r 2 )  o

At r = a,

o r
3
o a 3
3 o r


Er 
o a 3
3 o r 2
 c2
V  0 and c2  0
V(a+) = V(a-)
oa2
 a2
 c1  o
6 o r
3 o
V(r=0) = c1 
(b)
Er 
o r 2
 c1
6
4a 3
3
V   E  dl 
 ,
As r 


 oa 2
V (r  a ) 
3 o


c1 
oa2(2 r  1)
6 o r
oa2
(2 r  1)
6 o r
2
139
Prob. 5.28
 Dx 
4 1 1  1 
 D    E 1 3 1   1 
o o 
 y
 
 Dz 
1 1 2   1
Dx   o Eo (4  1  1)  4 o Eo
Dy   o Eo (1  3  1)  3 o Eo
Dz   o Eo (1  1  2)  0
D   o Eo (4a x  3a y ) C/m 2
Prob. 5.29
Since
v
 0,
t
  J  0 must hold.
(a)
  J  6 x2 y  0  6 x2 y  0
(b)
  J  y  ( z  1)  0
(c)
J 
(d)
J 
1 
 




( z 2 )  cos   0
1 
(sin  )  0
r2  r
This is possible.
This is not possible.




This is not possible.
This is possible.
Prob. 5.30
J x J y J z



 2e 2 y cos 2 x  2e 2 y cos 2 x  1  1   v
x
y
z
t
 v
 1 C/m3 s
Hence,
t
J
Prob. 5.31
(a)   J 

1  100
100
(
) 3
 
v
100
 J   3
t





 v 100
 3 C/m3 .s
t

140
(b)
I   J  dS  
100

 d dz  2 
2
2
1
100
d  dz  100  314.16 A
2 0
0
Prob. 5.32
  vo e t /T
r
where vo is the initial value (t=0). When t  80  s,
6
1
vo  vo e (8010 )/Tr

2
80 106
Tr 
 115.42  s
ln 2
Tr 





Tr

(80  106 ) / Tr  ln 2
109
ln 2
36
 5.746 107 S/m
6
80  10
7.5 
Prob. 5.33
From the continuity equation,

J   v
(1)
t
But
J  v u
Applying the vector identity
 • (VA)  V   A  A  V
  J   v (  u)  u   v
(2)
Substituting (2) into (1) gives

(u  ) v   v (  u)  v  0
t
as required.
Prob. 5.34

v
J
   J  x  0.5 cos  x
t
x
At P(2,4,-3), x= 2
v
 0.5 cos(2 )  0.5  1.571 C/m 2 s
t
1
2
  vo . Then,
141
Prob. 5.35
(a)



3.1
109
36  2.741 104 s
1015
109

36  5.305  104 s


1015
6
(b)



(c)
109
36  7.07  s
104
80 
Prob. 5.36
Tr 

2.5  109

 4.42  s
 5 106  36
vo 
Q
1

 29.84 kC / m3
V 4  106  8
3
v  vo e t / T  29.84e2 / 4.42  18.98 kC/m3
r
Prob. 5.37
The normal component of a solenoidal vector field is continuous across an interface.
Hence, since   J =0,
J 1n  J 2 n
Similarly, the tangential component of a curl-free vector field is continuous.
Hence, since   (J / )=0,
J1t  1

J 2t  2
Alternatively,
E1n  E2 n
as required.
Prob. 5.38
(a)

J1t
1

J 2t
2
or
J1t  1

J 2t  2
142
P1   e1 o E1  3 
109
(60, 100, 40)  1.591a x  2.6526a y  1.061a z nC/m 2
36
(b)
E2t  E1t  60a x  100a y
D2 n  D1n
E2 n 

 2 E2 n  1 E1n
1
4
(40a z )  21.33a z
E1n 
2
7.5
E2  60a x  100a y  21.33a z
109
(60, 100, 21.33)  3.979a x  6.631a y  1.414a z nC/m 2
D2   o r 2 E2  7.5 
36
Prob. 5.39
(a)
E2t  E1t  10a y  8a z
D2 n  D1n
E1n 

 2 E2 n  1 E1n
2
2
E2 n  o (6a x )  3a x
4 o
1
E1  3a x  10a y  8a z
109
P1   e1 o E1  3 
(3, 10,8)  79.6a x  265.3a y  212.2a z nC/m 2
36
109
(6, 10,8)  53.05a x  88.42a y  70.74a z nC/m 2
P2   e 2 o E2  1
36
(b)
1
1
w1  1 E12  (4 o )(9  100  64)  346 o  3.0593 nJ/m 2
2
2
1
1
w2   2 E22  (2 o )(36  100  64)  200 o  1.7684 nJ/m 2
2
2
Prob. 5.40
f(x,y)= 4x +3y –10=0
(4a x  3a y )
f

 0.8a x  0.6a y
| f |
5
The minus sign is chosen for an because it is directed toward the origin.
f  4a x  3a y


an  
D1n  ( D1  an )an  (1.6  2.4)an  0.64a x  0.48a y
D1t  D1  D1n  2.64a x  3.52a y  6.5a z
D2 n  D1n  0.64a x  0.48a y
E2t  E1t


D2t
2

D1t
2
143
D2t 
2
2.5
(2.64, 3.52, 6.5)  (6.6, 8.8,16.25)
D1t 
1
1
D2  D2 n  D2t  5.96a x  9.28a y  16.25a z nC/m 2
 2  cos 1
D2 an
 87.66o
| D2 |
Prob. 5.41
(a)
Let f ( x, y, z )  x  2 y  z  1  0
f
1

( a x  2a y  a z )
| f |
6
1
1
E1n  ( E1  an )an 
(20  20  40)
(a x  2a y  a z )  6.667a x  13.33a y  6.667a z V/m
6
6
E1t  E1  E1n  13.3a x  23.3a y  33.3a z V/m
 f  a x  2a y  a z

an 
(b)
E2t  E1t  13.3a x  23.3a y  33.3a z
D2 n  D1n
E2 n 

 2 E2 n  1 E1n
1
2
E1n  (6.667,13.33, 6.667)  2.7a x  5.3a y  33.3a z
5
2
E2  E2t + E 2 n  16a x  18a y  36a z V/m
Prob. 5.42
109
(10, 6,12)  0.1768a x  0.1061a y  0.2122a z nC/m 2
(a) P1   o  e1 E1  2 
36
(b)
E1n  6a y ,
D2 n  D1n
or
E2 n 
E2t  E1t  10a x  12a z


 2 E2 n  1 E1n
3 o
1
(6a z )  4a y
E1n 
4.5 o
2
E2  10a x  4a y  12a z V/m
E
tan  2  2 t 
E2 n
10 2  12 2
 3.905
4
 
 2  75.64 o
144
(c ) wE 
wE1 
1
1
DE   | E | 2
2
2
1
1
10 9
(102  62  122 )  3.7136 nJ/m3
1|E1|2  x3 x
2
2
36
1
1
10 9
2
(102  4 2  122 )  5.1725 nJ/m3
wE 2   2 | E2 |  x4.5 x
2
2
36
Prob. 5.43 (a) D2 n  12a   D1n ,
E 2t  E 2t
D1t 


D1t
1

D2t  6a  9a z
D2t
2
3.5 o
1
(6a  9a z )  14a  21a z
D2t 
1.5 o
2
D1  12a   14a  21a z nC/m 2
(12, 14, 21)  109
 387.8a   452.4a  678.6a z
109
3.5 
36
0.5 o
D
(12, 6,9)  4a  2a  3a z nC/m 2
(b) P2   o  e 2 E2  0.5 o 2 
 2 1.5 o
E1  D1 / 1 
v 2    P2  0
(c)
wE1 
wE 2 
1
1 D1  D1 1 (122  142  212 )x10 18

 12.62  J/m2
D1  E1 
9
2
2  o r1
2
10
3.5 x
36
1 D2  D2 1 (122  62  92 )x10 18

 9.839  J/m2
9
2  o r 2
2
10
1.5 x
36
145
Prob. 5.44
.
Gaussian surface
.
.
Q   D.dS  1 Er
4 r 2
4 r 2
  2 Er
 2 r 2 (1   2 ) Er
2
2
Q

,
ra

E r   2 (1   2 )r 2

0,
ra

Prob. 5.45
E1t  E2t  5cos  a
D1n  D2 n
1 E1n   2 E2 n

6
2
E2 n  o (10sin  )a r  30sin  ar
1
2 o
E1t  E1t + E1n  30sin  ar  5cos  a
E1n 
D1  1 E1  2 o E1   o (60sin  ar  10 cos  a )
Prob. 5.46
(a) The two interfaces are shown below
oil
1
glass
2
oil-glass
glass
air
2
3
glass-air
146
E1n  2000 ,
E1t  0  E2 t  E 3 t
D1n  D2 n  D3 n


 1 E1n   2 E 2 n   3 E 3 n
E2 n 
1
3.0
E1 n 
(2000 )  705.9 V / m,  2  0 o
2
8.5
E3 n 
1
3.0
(2000 )  6000 V / m,  3  0 o
E1 n 
3
10
.
(b)
oil
glass
1
2
2
glass
air
2
3
 1  75 o
E1n  2000 cos75 o  517 .63,
E1t  2000 sin75 o  E 2 t  E 3 t  1931.85
E2n 
1
3
E1n 
(517 .63)  1827
. ,
2
8.5
E3n 
1
3
E1n  (517 .63)  1552.89
3
1
E2 
E2 n 2  E2 t 2  1940.5,
 2  tan 1
E2 t
 84.6 o ,
E2 n
E3 
E3 n 2  E3 t 2  2478.6 ,
 3  tan 1
E3 t
 51.2 o
E3 n
Prob. 5.47
109
2900
302  402  202  103 
pC/m 2
36
36
2
 0.476 pC/m
 s  Dn   o E 
Prob. 5.48 (a)  s  Dn   o En 
(b)
Dn   s  20 nC/m 2
10 9
15 2  8 2  0.1503 nC / m2
36 
147
D  Dn an  (  20 nC)(-a y )  20 a y nC / m2
Prob. 5.49
At the interface between  o and 2 o ,
E1n  Eo cos 30o ,
E1t  Eo sin 30o
E2t  E1t  0.5 Eo
D2 n  D1n
E2 n 


1 E1n   2 E2 n

1
E1n  o (0.866 Eo )  0.433Eo
2
2 o
The angle E makes with the z-axis is
E
0.5
1  tan 1 2t  tan 1
 49.11o
E2 n
0.433
At the interface between 2 o and 3 o ,
E3t  E2t  0.5 Eo
D3n  D2 n


E3n 
2
2
E2 n  o (0.433Eo )  0.2887 Eo
3
3 o
The angle E makes with the z-axis is
E
0.5
 60o
 2  tan 1 3t  tan 1
E3n
0.2887
At the interface between 3 o and  o ,
E4t  E3t  0.5 Eo
D4 n  D3n


E4 n 
3
3
E3n  o (0.2887 Eo )  0.866 Eo
4
o
The angle E makes with the z-axis is
E
0.5
 30o
3  tan 1 4t  tan 1
E4 n
0.866
148
CHAPTER 6
P. E. 6.1
 2V  


 x
d 2V
 o
2
dx
a


 o x3
V
 Ax  B
6a
  x2

dV
E
ax   o  A  ax
dx
 2 a

If E = 0 at x =0, then
0 0 A


A0
If V= 0 at x =a, then
0
o a 3
B
6 a


B
o a 2
6
Thus
o
(a 3  x 3 ),
6 a
V
V1  A1 x  B1 ,
P. E. 6.2
V1 ( x  d )  Vo  A1d  B1
V1 ( x  0 )  0  0  B2


A1a  Vo  A1d 
1
aA1
2
B1  Vo  A1d
B2  0


 1 A1   2 A2


o x 2
ax
2 a
V2  A2 x  B2




V1(x  a)  V2(x  a)
D1n  D2 n
E
aA1  B1  A2 a


A2 
1
A1
2

 
Vo  A1   a  d  1 a
2 

or
A1 
A2 
1
 1Vo
A1
2
 2 d   2 a   1a
Voax
,
d  a  1a /  2
E2   A2 ax 
Vo
,
d  a   1a /  2
Hence
E1   A1ax 
Vo ax
a   2 d / 1   2 a / 1
149
P. E. 6.3 From Example 6.3,
V
E   o a , D   o E
o
 s  Dn (  0 )  
Vo 
 o
The charge on the plate   0 is

Q   s dS  
C
Vo 
o
L
b
Vo 
  dzd    L ln(b / a)
1
o
z0  a
| Q| L b
ln

Vo  o a
4mm
a
45o
a
a sin
C
45 o
2
2


a
2
sin 22.5 o
 5.226 mm
109
36 5ln 1000  444 pF

5.226
4
1.5 
Q  CVo  444  10 12  50 C  22.2 nC
P. E. 6.4 From Example 6.4,
Vo  50,
2  45o ,
1  90o ,
r  32  42  22  29 ,
5

   68.2o ;
tan45o  1
2
50ln(tan34.1o )
V
 22.125 V,
ln(tan22.5o )
50a
E
 11.35a V/m
29 sin68.2o ln(tan22.5o )
tan1
  tan1

z

150
P. E. 6.5
From Example 6.5,
V(x, y) 
sin(n x / b)sinh[n y / b]
n sinh(n a / b)
 nodd
4Vo

E  V  


V
V
ax 
ay
x
y
4Vo 
1
cos(n x/b)sinh (n y/b) a x  sin(n x/b)cosh(n y/b)a y 

b n odd sinh n a/b 
(a) At (x,y) = (a, a/2),
V
400

(0.3775  0.0313  0.00394  0.000585  ...)  44.51 V
E  0ax  (115.12  19.127  3.9411 0.8192  0.1703  0.035  0.0074  ...)ay
 99.25ay V/m
(b) At (x,y) = (3a/2, a/4),
V
400

(0.1238  0.006226  0.00383  0.0000264  ...)  16.50 V
E  (24.757  3.7358  0.3834  0.0369  0.00351 0.00033  ...)ax
(66.25  4.518  0.3988  0.03722  0.00352  0.000333  ...)ay
 20.68ax  70.34ay V/m
P. E. 6.6

V ( y  a )  Vo sin(7 x / b) 
 c sin(nx / b) sinh(na / b)
n
n1
By equating coefficients, we notice that cn = 0 for n  7 . For n=7,
Vo sin(7 x / b)  c7 sin(7 x / b) sinh(7 a / b)


c7 
Hence
V ( x, y) 
Vo
sin(7  x / b) sinh(7  y / b)
sinh(7  a / b)
Vo
sinh(7 a / b)
151
P. E. 6.7 Let V (r ,  ,  )  R(r )F ( ) ( ).
Substituting this in Laplace’s equation in spherical coordinates gives
F d  2 dR 
R d 
dF 
RF d 2 
sin

r


0




d  r 2 sin 2  d 2
r 2 dr  dr  r 2 sin  d 
Dividing by RF / r 2 sin 2  gives
sin 2  d 2
sin  d
1 d 2
 sin  F '  
r R' 
 2

R dr
F d
 d 2
 ''  2  0
d
1 d 2
1
 sin  F '   2 / sin 2 
r R' 

R dr
F sin  d
1 d 2
2
1
d
 sin  F '   2
r R' 


2
R dr
sin  F sin  d
2 rR ' r 2 R ''   2 R
or
R''
2
2
R ' 2 R  0
r
r
sin  d
 sin  F '   2   2 sin 2   0
F d
or
F '' cos  F ' (  2 sin    2 cs c ) F  0
P. E. 6.8 (a) This is similar to Example 6.8(a) except that here 0    2  instead of
0     / 2 . Hence
b
Vo ln a
2 tVo 
I
and
R

I 2 t
ln(b / a )
(b) This to similar to Example 6.8(b) except that here 0    2  . Hence
b 2
Vo 
Vo   (b 2  a 2 )
 d d 
I
t a 0
t
152
and R 
Vo
t

2
I   (b  a 2 )
P. E. 6.9
J1 
From Example 6.9,
 1Vo
,
b
 ln
a
J2 
 2Vo
b
 ln
a
2
L  

Vl
I   J  dS     J1 d   J 2 d  dz  o   1    2
b

z0 

  0
ln
a
b
ln
Vo
a
R

I
l  1   2


P. E. 6.10 (a)

C
4 
,
1 1

a b
C1 and C2 are in series.



109 
2.5
C1  4 x
 3
  5 / 3 pF,
3
36  10 10 



3 
 2
C




10 9 
3.5
C2  4 x
 3
  7 / 9 pF
3
36  10 10 



2 
 1
(5 / 3)(7 / 9 )
C1C2

 0.53 pF
C1  C2 (5 / 3)  (7 / 9 )
2 
, C1 and C2 are in parallel.
1 1

a b




109  2.5 
109  3.5 
C1  2 

  5 / 24 pF, C2  2 

  7 / 24 pF
36  103 103 
36  103 103 






3 
3 
 1
 1
(b)
C
C  C1 C2  0.5 pF
P. E. 6.11 As in Example 6.8, the solution of Laplace’s equation yields
V (  )  A ln   B
153
Using the boundary conditions V (  a )  0 ,
0 = Alna + B
Vo = Aln b + B
Solving this yields
V  Vo
ln  / a
,
ln b / a
E  V  
V
Q  E  dS  o
ln b / a

C
L 2
V (  b)  Vo ,
Vo
a
 ln b / a 
Vo 2 L
  dzd  ln b / a
1
z0  0
Q 2  L

Vo ln b / a
P. E. 6.12
(a) Let C1 and C2 be capacitances per unit length of each section and CT be the total
capacitance of 10m length. C1 and C2 are in series.
C1 
2 r1 o 2 x2.5 10 9

 342.54 pF/m,
ln b / c ln3 / 2 36
C2 
2 r 2 o 2 x3.5 10 9

 280.52 pF/m
ln c / a
ln2 36
C1C2
342.54 x280.52

 154.22 pF
C1  C2 342.54  280.52
CT  Cl  1.54 nF
C
(b) C1 and C2 are in parallel.
C  C1  C2 
 r1 o
ln b / a
CT  Cl  1.52 nF
 r 2 o

ln b / a

 ( r1   r 2 ) o
ln b / a
P. E. 6.13 Instead of Eq. (6.31), we now have
a
V 
 4r
b
Qdr

2
a
 4 10 r   40  ln b / a
b
Qdr
o
r
Q
2
o
6 10 9

 151.7 pF/m
ln3 36
154
40  10 9
Q
C

 113
. nF
|V | ln 4 / 1.5 36 
P. E. 6.14 Let
F  F1  F2  F3  F4  F5
i  1,2,...,5
where Fi ,
are shown on in the figure below.
Q2(ax sin30o  ay cos 30o ) Q2(ax cos 30o  ay sin30o )

F
ay 
4 o r 2
4 o(2 r cos 30o )2
4 o(2 r )2
Q2
2
o
o
Q (ax cos 30  ay sin30 )
Q2 ax


o 2
4 o(2 r cos 30 )
4 o r 2


3 ay  1  3ax ay  1
3ax ay 
1  ax



a
a







 


y
x
3 2
2  4  2
2  3
2
2
4 o r 2 




Q2
155
 1 5 3
 5
3 
 9x10 5  ax  
 ay 


   52.4279ax  30.27ay N



2
8
8
6


 
 
|F| 60.54 N
Note that the force tends to pull Q toward the origin.
Prob. 6.1
(a)
V
V 
 V
E  V   
ax 
ay 
az 
z
z 
 x
 (15 x 2 y 2 za x  10 x 3 yza y  5 x 3 y 2 a z )
At P, x=-3, y=1, z=2,
E  15(9)(1)(2)a x  10(27)(1)(2)a y  5(27)(1)a z  270a x  540a y  135a z V/m
v    D
(b)
 2V 
or
v   2V
 2V  2V  2V 


 2  2  (15 x 2 y 2 z )  (10 x3 yz )  (5 x3 y 2 )
2
x
y
y
x
y
z
 30 xy 2 z  10 x3 z
At P,
v   2V  2.25 
109
30(3)(1)(2)  10(27)(2)  14.324 nC/m3
36
Prob. 6.2
(a)
20
10
1 10
cos  sin  ar  3 sin  sin  a +
cos  cos  a
3
r
r
rsin r 2
At P(1,60o ,30o ), r  1,   60o ,   30o
 E  V  
20
10
1 10 cos 60o cos 30o
o
o
o
o
cos
60
sin
30
sin
60
sin
30
a
a
a


r

13
13
sin60o
13
 5ar  4.33a  5a V/m
E
156
(b)
 2V 
1   20 cos  sin  
1
  10sin 2  sin  




 2
r 2 r 
r
r2
 r sin   

1
10 cos  sin 
2
r sin 
r2
20 cos  sin  20sin  cos  sin  10 cos  sin 



r4
r 4 sin 
r 4 sin 2 
10 cos  sin 

r 4 sin 2 

10 o cos  sin 
 2V   v

  v    2V 
r 4 sin 2 

At P, r =1,  =60o ,  =30o

 v =10 
2
10-9 cos60o sin 30o
=29.47 pC/m3
o
4
2
36 1 sin 60
Prob. 6.3
 2V  
v

d 2V
y 109
y 109






 2.25 y
dy 2
4 4 o
4 109
4
36
2
dV
y
 2.25  B
dy
2
3
V  0.375 y  By  C
V (1)  0  0.375  B  C
V (3)  50  10.125  3B  C
From (1) and (2), B=29.875 and C=-29.5
V  0.375 y 3  29.875 y  29.5
V (2)  27.25 V
Prob. 6.4
(1)
(2)
157
E  V  
 2V 
dV
a   (0.8)(10)  0.2 a   8(0.6) 0.2 a   8.861a 
d
v 1 d
1 d  dV 
1
8 0.8   (8)0.8 0.2  6.4  1.2




 d  d 
o  d 

v   o 2V  6.4 
109 1.2
6.4(0.6) 1.2
nC/m3  0.1044 nC/m3
 
36
36
Prob. 6.5
 2V 
where
k
d 2V
v
50(1 y 2 )x10 6




 k(1 y 2 )
2


dx
50 106
 600 103
109
3
36
dV
  k ( y  y 3 / 3)  A
dy
 y2 y4 
V  k 
   Ay  B  50 .10 3 y 4  300 .10 3 y 2  Ay  B
2
12 

When y=2cm,
V=30X103,
30 103  50 103 16  106  300 103  4 104  Ay  B
or
30,376.77  0.02A  B
When y=-2cm,
(1)
V=30x103,
30,376.77  0.02A  B
From (1) and (2), A=0, B=30,376.77.
(2)
Thus,
V  157.08 y 4  942.5 y 2  30.377 kV
158
Prob. 6.6
v
 z
d 2V


 o
2

d
dz
2
 z
dV
 o A
dz
2 d
 z3
V   o  Az  B
6 d
 2V  
z  0,V  0
z  d ,V  Vo




0  0  B,
Vo  
Vo  o d

d
6
Hence,
A
o z 3  Vo o d 
V 
z


6 d  d
6 
i.e. B  0
o d
 Ad
6
2
159
Prob. 6.7
 2V  
v



10

3.6 
Let   0.1 .
 2V  
 
 1012
9
10
36

0.1

 1 d  dV 


  d   d  
d  dV 


d  d 
dV
   A
d
dV
A
  
d

V    A ln   B

At  =2, V=0

 0  2  A ln 2  B

 60  5  A ln 5  B
At  =5, V=60
Subtracting (1) from (2),
60  3
60  3  A ln 5 / 2

 A
 66.51
ln 2.5
From (1),
B  2  A ln 2  45.473
dV
A
66.51
)a 
E
a   (  )a   (0.3142 
d


(1)
(2)
160
Prob. 6.8
o
1 d  dV 


 d  d 

o 
d  dV 



d  d 
Integrating gives
 2
dV
 o
A
 2
d
Integrating again,

V 


dV
A
 o 

d
2
o  2
 A ln   B
4
where A and B are integration constants.
Prob. 6.9
v
1 d  2 dV 
10 109
60
r







9
2
10
r dr  dr 
r

6r 
36
d  2 dV 
dV

r2
 30 r 2  A
r
  60 r
dr  dr 
dr
dV
A
A
 30  2
 V  30 r   B
dr
r
r
V (r  1)  0  0  30  A  B
(1)
 2V 
V (r  4)  50  50  120  A / 4  B
Solving (1) and (2) yields A  443.66, B  537.91
Thus,
443.66
V  30 r 
 537.91
r
443.66
V (r  2)  60 
 537.91  127.58 V
2
Prob. 6.10
(a)
 2V1  2V1  2V1


 002  0
x 2 y 2 z 2
It does not satisfy Laplace's equation.
 2V1 
(2)
161
(b)
 2V2 
1 
1  10sin   10 2
10sin 
  (  210sin  )   2  
0
  (  ) sin  
 
 
  
3
It does satisfy Laplace's equation.
( c)
1  2
1
 
(5cos  ) 
 r ( r 2 5sin  )   2
sin 
2

r r
r sin   
r 2 
5
 0 3
(1  2sin 2  )  0
r sin 
It does not satisfy Laplace's equation.
 2V3 
Prob. 6.11
 2U  2U  2U
 U  2  2  2  6 xy  0  2c  0
x
y
z
c  3xy
2
Prob. 6.12
(a)
V
 4 xyz,
x
 2V
x 2
V
 2 x 2 z  3 y 2 z,
y
V
 2x2 y  y3 ,
z
 4 yz
 2V
y 2
 2V
z 2
 6 yz
0
 2V  4 yz  6 yz  0  2 yz
 2V  0 ,
V does not satisfy Laplace’s equation.
(b)
 2V  
v
 2 yz

1 1 1


v  2 yz
Q    v dv     (2yz )dxdydz  2 (1)
0 0 0
Q = 8.854 pC
y 2 1 z2 1
  / 2  2 o / 2   o
2 0 2 0
162
Prob. 6.13
 2V 
1 d  dV 

0
 d   d  

V  A ln   B
Let a = 1 cm, b = 1.5 cm, Vo  50V
V (   b)  0

0  A ln b  B or B   A ln b
V (   a)  Vo

Vo  A lna  Alnb  Aln
V  A ln   A lnb  Aln
V
a
or A  o
a
b
ln
b

b
A
V
dV
a   a   o a
a

d
 ln
b
9
10
 o rVo 50(4) 36
400(50)
 2

nC/m 2  436.14 nC/m 2
 s  Dn   En 
b 10 ln1.5 36 ln1.5
a ln
a
E  V  
Prob. 6.14
 2V 
d 2V
dz 2
0


V  Az  B
When z=0, V = 0
B=0
When z=d, V = Vo
Vo=Ad or A = Vo/d
Hence,
V
Vo z
d
V
dV
az   o az
dz
d
V
D   E   o r o a z
d
Since Vo = 50 V and d = 2mm,
E  V  
V = 25z kV, E = - 25az kV/m
D
109
(1.5)25 103 a z  332a z nC/m 2
36
163
 s  Dn   332 nC / m2
The surface charge density is positive on the plate at z=d and negative on the plate at
z=0.
Prob. 6.15 From Example 6.8, solving  2V  0 when V  V () leads to
Vo ln  / a
ln(a /  )
 Vo
ln b / a
ln(a / b)
Vo
Vo
E  V  
a 
a ,
 ln b / a
 ln a / b
V
 s  Dn  
D  E  
 o r Vo
a
 ln b / a
 o  rVo
 ln b / a   a ,b
In this case, Vo=100 V, b=5mm, a=15mm,  r  2. Hence at  = 10mm,
V
100ln(10 /15)
 36.91 V
ln(5/15)
100
a  9.102a kV/m
10 x10 3 ln3
109
2a  161a nC/m2
D  9.102 x103 x
36
E
 s (  5 mm) 
10 9
10 5
( 2)
 322 nC / m2
36 
5 ln 3
 s (  15 mm)  
10 9
10 5
( 2)
  107 .3 nC / m2
36 
15 ln 3
Prob. 6.16
1 d 2V
0
 d 2
V  A  B
0  0 B


d 2V
0
d 2



 B0
100
50  A / 2


E  V  
A
1 dV
100
a   a  
a
 d

 
A

dV
A
d
164
Prob. 6.17
(a)
V
a2
 Vo (1  2 ) sin 



V
a2
 Vo (   ) sin 


a2
  V 

(1
) sin 
V



 o
2
   
1   V 
1 a2

 V (  ) sin 
     o   3
a2
 2V

V
(
) sin 



o

 2
1  2V
1 a2
V
(


 ) sin 
o
 2  2
 3
 2V 
1   V   2V


0
      2
(b)
2
a 2 , then a
If  2
E  V  
2
1 and V
Vo  sin 
V
1 V
a 
a  Vo sin  a  Vo cos  a
  

Prob. 6.18
d 2V
 2V  2  0


V  Ax  B
dx
At x  20 mm  0.02 m, V  0

 (1)
0  0.02 A  B
E = -100 a x 
From (1)
Then
At x  0
dV
ax
dx

 A  110 
 (2)
B  0.02 A  2.2
V  110 x  2.2
V  2.2V
At x  50 mm  0.05 m,
V  110  0.05  2.2  3.3V
165
Prob. 6.19
 2V  0


V  A/r  B
At r=0.5, V=-50
-50 = -A/0.5 + B
Or
-50 = -2A + B
At r = 1, V =50
(1)
50 = -A + B
(2)
From (1) and (2), A = 100, B = 150, and
V 
100
 150
r
A
100
E  V   2 ar   2 ar V/m
r
r
Prob. 6.20 From Example 6.4,
V
Vo  100 ,
 tan  / 2 
Vo ln

 tan  1 / 2 
 tan  2 / 2 
ln

 tan  1 / 2 
 1  30 o ,
 2  120 o ,
r  3 2  0 2  4 2  5,
  tan 1  / z  tan 1 3 / 4  36 .87 o
 tan 18.435 o 
ln

 tan 15 o 
V  100
 117
. V
 tan 60 o 
ln

 tan 15 o 
sec 2  / 2
Vo a
Vo a
1 V
tan 1 / 2
E
a 

r 
 tan  2  tan  / 2
 tan  2 
r ln 
 tan  / 2 r ln 
 2sin( / 2) cos( / 2)
1
 tan 1 
 tan 1 
166
E
Vo a
100a

 17.86a V/m
 tan  2 / 2  5sin 36.87 o ln 6.464
r sin  ln 

 tan 1 / 2 
Prob. 6.21
(a)
 2V 
1   V 

 0
    
V (  b )  0


V (   a)  Vo
V 
V  A ln   B


0  A ln b  B




Vo  A ln a / b
B   A ln b


A
Vo
ln b / a
Vo
V ln b / 
ln  / b  o
ln b / a
ln b / a
V (  15 mm) = 70
ln2
 12.4 V
ln50
(b) As the electron decelerates, potential energy gained = K.E. loss
e[70  12.4] 
1
m[(107 ) 2  u 2 ]
2
u 2  1014 


1014  u 2 
2e
 57.6
m
2  1.6  1019
 57.6  1012 (100  20.25)
31
9.1 10
u  8.93 106 m/s
Prob. 6.22 This is similar to case 1 of Example 6.5.
X  c1 x  c2 ,
But
X (0 )  0
Y  c3 y  c4


0  c2 ,
Y (0 )  0


0  c4
Hence,
V ( x , y )  XY  ao xy ,
Also,
V ( xy  4)  20
ao  c1c3


20  4ao


ao  5
Thus,
V ( x, y )  5 xy and E  V  5 ya x  5 xa y
167
At (x,y) = (1,2),
V  10 V, E  10a x  5a y V/m
Prob. 6.23 (a) As in Example 6.5, X ( x )  A sin(nx / b)
For Y,
Y ( y )  c1 cosh(ny / b)  c2 sinh(ny / b)
Y (a )  0


0  c1 cosh(na / b)  c2 sinh(na / b)


c1  c2 tanh(na / b)

V
 a sin(nx / b) sinh(ny / b)  tanh(na / b) cosh(ny / b)
n
n1

V(x, y  0)  Vo   an tanh(n a / b)sin(n x / b)
n1
 4Vo
b
, n  odd
2

an tanh(n a / b)   Vo sin(n x / b)dx   n
b0
 0, n  even
Hence,
4Vo
V

4Vo


 sinh(n y / b)

 sin(n x / b) ntanh(n a / b) 
nodd

cosh(n y / b)

n

sin(n x / b)

 n sinh(n a / b) sinh(n y / b)cosh(n a / b)  cosh(n y / b)sinh(n a / b)
nodd

sin(n x / b)sinh[n (a  y)/ b]
 nodd
n sinh(n a / b)
4Vo


Alternatively, for Y
Y ( y )  c1 sinh n( y  c2 ) / b
Y (a )  0



V
0  c1 sinh[n(a  c2 ) / b]
 b sin(nx / b) sinh[n( y  a) / b]
n
n1


c2  a
168
where
4Vo


, n  odd
bn   n sinh(n a / b)

n  even
0,
(b) This is the same as Example 6.5 except that we exchange y and x. Hence
V(x, y) 
sin(n y / a)sinh(n x / a)]
 nodd
n sinh(n b / a)
4Vo


(c) This is the same as part (a) except that we must exchange x and y. Hence
V(x, y) 
sin(n y / a)sinh[n (b  x)/ a]
 nodd
n sinh(n b / a)
4Vo


Prob. 6.24 (a) X(x) is the same as in Example 6.5. Hence

V ( x, y) 
 sin(nx / b)a sinh(ny / b)  b cosh(ny / b)
n
n
n1
At y=0, V = V1

V1 
 b sin(nx / b)
n


n1
 4V1
 n , n  odd

bn  
 0, n = even


At y=a, V = V2

V2 
 sin(nx / b)a sinh(na / b)  b cosh(na / b)
n
n
n1
 4V2
 n , n  odd

an sinh(na / b)  bn cosh(na / b)  
 0, n = even


or
4V2

 n sinh(na / b) V2  V1 cosh(na / b) ,

an  

0, n = even


n  odd
Alternatively, we may apply superposition principle.
y
0
V2
0
V2
0 0
0 0
0
169
V
VA
V1
VB
0
x
V1
i.e. V  V A  VB
VA is exactly the same as Example 6.5 with Vo  V2 , while VB is exactly the same
as Prob. 6.19(a). Hence
4  sin(n x / b)
V
V sinh[n (a  y ) / b]  V2 sinh(n y / b)

 n  odd n sinh(n a / b) 1


(b)
V ( x , y )  (a1e   x  a2 e   x )(a3 sin y  a4 cos y )
lim V ( x , y )  0


a2  0
V ( x, y  0)  0


a4  0
V ( x, y  a)  0


  n / a ,
x
 
n  1,2,3,...
Hence,

V ( x, y) 
ae
n
 n x / a
sin(ny / a )
n1

V ( x  0 , y )  Vo 

n1
an sin(n y / a )
 
 4Vo

, n  odd
an   n
 0 ,
n  even

V ( x, y) 
4Vo
sin(n y / a )
exp(  n x / a )
n
 n  odd

(c ) The problem is easily solved using superposition theorem, as illustrated below.
170
y
V3
a
V2
V
V4
x
0
0
b
V1
V2
0
0
VI
VII
0
0
V1
0
V3
0
VIII
VIV
0
0
V4
0
Therefore,
V  VI  VII  VIII  VIV
 sin(n x / b)
V1 sinh(n (a  y ) / b)  V3 sinh(n y / b) 

4
1  sinh(n a / b)

  

 n odd n  sin(n x / a)
sinh(
/
)
sinh(
(
)
/
)



V
n
y
a

V
n
b

x
a


4
 sinh(n b / a) 2


where

VI 
4V1
sin(nx / b) sinh[n(a  y ) / b]
n sinh(na / b)
 n  odd
VII 
4V2
sin(nx / a ) sinh(ny / a )
 n  odd
n sinh(nb / a )



0
171

VIII 
4V3
sin(nx / b) sinh(ny / b)
n sinh(na / b)
 n  odd
VIV 
4V4
sin(ny / a ) sinh[n(b  x) / a ]
n sinh(nb / a )
 n  odd



Prob. 6.25
E  V  
V
V
ax 
ay
x
y
n sin(n y / a)
exp( n x / a)
n
n  odd
4V  n cos(n y / a)
Ey   o 
exp(n x / a )
n
 n odd a
4V 
E  o  exp( n x / a ) sin(n y / a )a x  cos(n y / a)a y 
a n odd
Ex 
4Vo

 a

Prob. 6.26
This is similar to Example 6.5 except that we must exchange x and y. Going through the
same arguments, we have
 n x   n y 
V ( x, y )   cn sinh 
 sin 

 b   b 
Applying the condition at x=a, we get
 y 
 n a   n y 
Vo sin 
   cn sinh 
 sin 

 b 
 b   b 
This yields
 n a  Vo ,
cn sinh 

 b   0,
Hence,
n 1
n 1
x   y 
sinh 

 sin 
b   b 

V ( x, y )  Vo
a 
sinh 

 b 
Prob. 6.27
 2V 
1   V  1  2V
0


      2  2
172
If we let V (, )  R()(),
 
1
(R')  2 R''  0
 

or
 
''
(R')  

R 

Hence
''   0
and

R
(R') 
0


or
R''
R' R

0
 2
Prob. 6.28
 2V 
1   2 V 
1

V
(sin  )  0
r
 2
2

r r  r  r sin  
If V (r , )  R(r ) F (),
F
r  0,
d 2
R d
(r R') 
(sin F ')  0
dr
sin  d
Dividing through by RF gives
1 d 2
1
d
(r R')  
(sin F ')  
R dr
F sin  d
Hence,
sin F '' cos F ' F sin   0
or
F '' cot F ' F  0
Also,
d 2
(r R')  R  0
dr
173
or
R''
2 R' 
 2 R0
r
r
Prob. 6.29 If the centers at   0 and    / 2 are maintained at a potential difference
of Vo, from Example 6.3,
E 
2Vo
,

J  E
Hence,
I

2V 
J  dS  o

b
t
  ddz 
1
 a z0
2Vo t
ln(b / a )

and
R
Vo


I
2t ln(b / a )
Prob. 6.30 If V (r  a )  0 ,
E
Vo
r 2 (1 / a  1 / b)
V (r  b)  Vo , from Example 6.9,
J  E
,
Hence,
I

Vo 
J  dS 
1/ a 1/b

2
2Vo 
  r r sin dd  1 / a  1 / b ( cos )|
1
2
2
 0  0
1 1

Vo
a b
R

I
2  (1  cos  )
Prob. 6.31
This is the same as Problem 6.30 except that  =  . Hence,
R
1
1 1 1
1 1
  
  
2 (1  cos  )  a b  4  a b 
Prob. 6.32 For a spherical capacitor, from Eq. (6.38),
1 1

R a b
4
0

174
For the hemisphere, R'  2 R since the sphere consists of two hemispheres in parallel. As
b


,
 1 1
2  
1
a b

R '  lim 
b  
4
2 a
G  1 / R'  2 a
C  4a.
Alternatively, for an isolated sphere,
RC 


R'  2 R 


R
1
4a
1
2 a
or
G  2 a
But
Prob. 6.33
(a) For the parallel-plate capacitor,
V
E   o ax
d
From Example 6.11,
C
V 
1
2
| E |2 dv 
o
1
Vo2
Vo2

2
2
S
 d dv  d Sd  d

(b) For the cylindrical capacitor,
Vo
a
 ln b / a
From Example 6.8,
E
C
1
Vo2
Vo2
b
2 L
d
2 L
  ln b / a dddz   ln b / a    ln b / a
2
2
a
(c )For the spherical capacitor,
E
Vo
ar
r (1/ a  1/ b)
2
From Example 6.10,
C
1
Vo2
Vo2
b

4
 r 1 / a  1 / b r sin ddrd  1 / a  1 / b 4  r  1  1
4
2
2
2
dr
2
a
a
b
175
Prob. 6.34
Assume V( =a) = 0 and V( =b)=Vo . Following Example 6.8,
Vo

ln
b a
ln
a
dS   d dza 
J   E  V ,
V  A ln   B 
2
L
 Vo
 0 z 0
 ln
I   J  dS   
S
b
a
 d dz 
 Vo
ln
b
a
(2 L)
b
ln
Vo
a
R 
I 2 L
I 2 L
G 
b
Vo
ln
a
The conductance per unit length is
G 2
G'  
L ln b
a
Prob. 6.35
From eq. (6.37) or from previous problem
b
a
R
2 L
V 2 2 LV 2
P  VI 

b
R
ln
a
The power loss per unit length is
ln
P' 
P 2 V 2

b
L
ln
a
Prob. 6.36
176
C
S


d
S
Cd
 o r

2 109  106 2
m  0.5655 cm 2
9
4  10 / 36
Prob. 6.37
This can be regarded as three capacitors in parallel.
C1
C2
C3
 S
C  C1  C2  C3   o rk k
dk
o
3  15  102  20 102  5 15 102  20 102  8 15 102  20 102 
2 103 
109 15  102  20  102
[3  5  8]  2.122 nF


36
2  103

Prob. 6.38
The structure may be treated as consisting of three capacitors in series.
C1 
o A
C2 
,
 r o A
C3 
,
o A
a
a
a
a
a
a
1 1
1
1
 




C C1 C2 C3  o A  o r A  o A
A
2
1
2  1
 
 r
aC  o  o r
 o r
C
 o r A
a(1  2 r )
Prob. 6.39
A
C1
C1
d
C3
C3
177
C2
C2
From the figure above,
C
C1C2
 C3
C1  C2
here
C1 
o A / 2 o A
,

d /2
d
C2 
or A
,
d
C3 
o A
2d
 o 2 r A 2 / d 2  o A  o A  1
 r  109 10 104  1 6 


C
 

    6 pF
 o ( r  1) A / d 2d
d  2  r  1  36 2 103  2 7 
Prob. 6.40
C
oS
d

S 
Cd
o
11103
 36 106
109 / 36
S  1.131 108 m 2
S
Prob. 6.41
Fdx  dWE
WE 


F
dWE
dx
 2| E | dv  2   E xad  2  E da(1  x)
1
1
2
1
2
o r
2
o
where E  Vo / d .
dWE 1 Vo 2
  o 2 ( r  1)da
dx
2
d
Alternatively,
WE 
C  C1  C2 
2


1
CVo 2 , where
2
F
 o r ax  o r ( L  x)
d

dWE 1 Vo a
 o
( r  1)
dx
d
2
d
 o ( r  1)Vo 2 a
2d
178
F
 o ( r  1)Vo 2 a
2d
Prob. 6.42
(a)
oS
C
d

109 200  104
 59 pF
36 3  103
(b)  s  Dn  10 6 nC / m2 . But
Dn  E n 
or
Vo 
 oVo
 s
d
s d
 106  3 103  36 109  339.3 V
o
(c )
Q2
 S 1012  200 104  36 109
F
 s 
 1.131 mN
2S o
2 o
2
2
Prob. 6.43
C1 
 o r S
d
C2 
,
C1
 r
C2
oS
d
r 


56 F
 1.75
32 F
Prob. 6.44
(a) C 
S
d

109
 0.5
36
 7.515 nF
4 103
6.8 
Q
Q
, C

 Q  CV
S
V
(b)
CV
7.515 109  9

 135.27 nC/m 2
s  
S
0.5
s 
179
Prob. 6.45
Co 
S
S
 Co / 3
d
3d
Q
Co  o
 Qo  CoV
V
Q  CV  (Co / 3)V  Qo / 3
Eo 
C
,
V
,
d
E
E
V
 o
3d
3
1
1
W  CV 2  (Co / 3)V 2  Wo / 3
2
2
This indicates that two-thirds of the energy stored is lost in the connecting wires
and source resistance.
Prob. 6.46
(a)
C1 
 o r 1 S
d
,
C2 
 o r 1 S  o  r 2 S
 o r 2 S
d
 o r1 r 2 S
109 40  104 4  6 4.8
C1C2
d
d
d




nF=42.44 pF
C
C 1  C2  o  r 1 S   o  r 2 S
 r1   r 2 36 2 103 4  6 36
d
d
Q
(b) C 
 Q  CV  509.3 pC
V
C1
+
V1
C2
-
+ V2
-
180
(c) C1 and C2 are in series as shown above
Q  C1V1  C2V2 ,
V1  V2  V  12
Solving these gives
C2
6
V  (12)  7.2
V1 
C1  C2
10
V2 
C1
4
V  (12)  4.8
C1  C2
10
V1
 3.6 kV/m
d
V
E 2  2  2.4 kV/m
d
D  D1  D2  1 E1  1.2732 107 C/m 2
E1 
P1   e1  o E1  9.549 108 C/m 2
P2   e 2 o E2  1.061 108 C/m 2
Prob.6.47
(a)
109
4
36  25 pF

C
1 1
1
1


2
a b 5 x10
10 x102
4  2.25 
(b)
s 
Q = C Vo= 25x80 pC
Q
25  80
pC/m2  63.66 nC/m2

2
4 r
4  25  10 4
181
Prob. 6.48
C1
b
c
C2
d
C3
a
1
1
1
1



C C1 C2 C3
4 3
,
where C1 
1 1

b a
C2 
4 2
,
1 1

c b
C3 
41
,
1 1

d c
4 1/ b  1/ a 1/ c  1/ b 1/ d  1/ c



C
3
2
1
4
C
1


 2  3
1 1 1 1 1 1



d c c b b a
Prob. 6.49
We may place a charge Q on the inner conductor. The negative charge –Q is on the outer
surface of the shell. Within the shell, E = 0, i.e. between r=c and r=b. Otherwise,
182
E
Q
4 o r 2
ar
The potential at r=a is
a
c
b

c
a
Va    E dl    Er dr   Er dr   Er dr


Q
c
Q
dr
Q 1
Q
1
0


  
4  r
4  r
4 c 4  a b 
2
o 
C
dr
b
a
Q

Va
2
o b
o
o
1
1
1 1 1

  
4 o c 4 o  a b 
Prob. 6.50
We can regard this as having two cylindrical capacitors in series.
2 o r1 L
2 o r 2 L
C1 
,
C2 
c
b
ln
ln
a
c
2 o r1 L 2 o r 2 L
c
b
ln
ln
2 o r1 r 2 L
CC
a
c

C 1 2 
C1  C2 2 o r1 L  2 o r 2 L  ln b   ln c
r1
r2
c
b
c
a
ln
ln
a
c
Prob. 6.51
109
2  2.5 
 3 103
2 L
36
C

 0.8665  F
ln(b / a)
ln(8 / 5)
Prob. 6.52
Let the plate at =0 be 0, i.e. V(0)=0 and let the plate at =/4 be Vo , i.e. V(/4)=Vo.
183
 2V 
1 d 2V
0
 2 d 2
V (0)  0



 0  0 B
V ( / 4)  Vo
dV
A
d

 V  A  B

 B0

 Vo  A / 4


A
4Vo

4V
A
1 dV
a   a   o a
 d


4Vo
D E  
a
E  V  

4 Vo
 s  Dn  

b
4 Vo
L
Q    s dS   
4 Vo
  d  dz    L ln(b / a)
  a z 0
C
| Q | 4 L

ln(b / a)

Vo
Prob. 6.53
Since V  V ( ),
 2V  0
1 d 2V
0
 2 d 2

d 2V
dV
0 
 A  V  A  B
2
d
d
B0
For  =0, V=0  0=0+B

V
A o
For  = , V=Vo
 Vo =A

  0,
But

A
1 dV
a   a
 d

A
dS  d  dz
 s  Dn   E   ,
E  V  

2
Q    s dS  
S
C
L
A
2
Vo
2
   d  dz  A L ln     L ln 
 
 1 z 0
Q  L 2

ln
1
Vo 
1
1
184
Prob. 6.54
C
2 o L

ln(b / a)
V  Q/C 
2 
109
100 106
36
 1.633 1015 F
ln(600 / 20)
50 1015
 30.62 V
1.633 1015
Prob. 6.55
21
2 2
C1 
, C2 
ln(b / a )
ln(c / b)
Since the capacitance are in series, the total capacitance per unit length is
CC
21 2
C 1 2 
C1  C2  2 ln(b / a )  1 ln(c / b)
Prob. 6.56
(a) This is similar to Example 6.10.
1 1
Vo   
r b
 2V  0  V  
1 1

a b
Vo
dV
E  V  
ar 
a
1 r
dr
21
r   
a b
 o rVo
At r=a,  s  Dn   Er 
1 1
a2   
a b
1 1
1 1 1
 s a 2    400 109 (4 104 )    2
a b 
 2 4  10  16(36 )102 (1/ 4)  4.524
r 
9
10
 oV
(100)
36
(b)
109
4 
4.524
4 o r
452.4
36
C


 201.1 nF
1 1
1 1 1
9(1/
4)


   2
a b
 2 4  10
185
Prob. 6.57
Each half has capacitance given by
2
2 ab
C

1 1
ba

a b
The two halves may be regarded as capacitors in parallel. Hence,
21ab 2 2 ab 2 (1   2 )ab
C  C1  C2 


ba
ba
ba
Prob. 6.58
E
Q
4 r 2
ar
Q
Vo
0
W
1
Q2
2
2

E

|
|
dv
 32 2 2 r 4  r sin d d dr
2
b
Q2
dr Q2  1 1 
(2
)(2)


c r 2  8  c  b 
32 2
Q 2 (b  c)
W
8   bc
Prob. 6.59
s
(a x ) , where  s is to be determined.

d
d
s

1 d
Vo    E  dl   
dx   s 
dx  s d ln(x  d)

 d x

0
0 o
(a) Method 1:
E
186
Vo   s d ln
E
2d
d
 
s 
Vo  o
d ln 2
s
Vo
ax  
ax
(x  d)ln2

Method 2: We solve Laplace’s equation
  (  V ) 
d
dV
(
) 0
dx dx
 

dV
 A
dx
dV A
Ad
c
 
 1
dx   o ( x  d ) x  d
V  c1 ln( x  d )  c2
V ( x  0)  0
 
0  c1 ln d  c2
V ( x  d )  Vo
 
Vo  c1 ln 2d  c1 ln d  c1 ln 2
c1 
c2   c1 ln d
Vo
ln 2
V  c1 ln
E
(b)
 
x  d Vo
x d

ln
d
ln 2
d
dV
Vo
ax  
ax
dx
(x  d)ln2
 o xVo
 x  d   oVo
P  ( r  1) oE   
 1
ax  
ax
d(x  d)ln2
 d
 (x  d)ln2
(c )
x=d
187
x=0
 ps |x 0  P  (ax )|x 0  0
 ps |x d  P  ax |x d  
E
(d)

 oVo
2d ln2
s
Q
Q
ax 
ax 
ax
x

S
 (1  ) S
o
d
Q
dx
Q
V    E dl  
d ln 2


 o S a (1  x )  o S
d
 S
Q
C  o
V d ln 2
d
Prob. 6.60
We solve Laplace’s equation for an inhomogeneous medium.
188
 (V ) 
d  dV 

0
dx  dx 

 
dV
A
dx
2
dV A
A  x 
1
 

   
dx  2 o   d  
V
A
x3
(x  2 )  B
2 o
3d
When x=d, V=Vo ,
A
d
(d  )  B
2 o
3


Vo 
When x = -d, V=0,
A
d
0
(d  )  B
2 o
3


0
Adding (1) and (2),
Vo  2 B
Vo 
2 Ad
B
3 o
(1)
2 Ad
B
3 o
(2)

 B  Vo / 2
From (2),
B
2 Ad Vo

3 o
2


A
3 oVo
4d
  x 2 
1    
3 oVo   d  
dV
A
E  V  
ax   ax  

dx
4d
 s  D an   E a x
xd
Q    s dS   s S  
S
 A  
2 o
2
3Vo   x  
ax  
1     a x
8d   d  
3 oVo
4d
3S  oVo
4d
| Q | 3 o S

C
Vo
4d
Prob. 6.61
Method 1: Using Gauss’s law,
Q   D  dS  4 r 2 Dr
E  D/ 
Q
4 o k


ar
D
Q
ar ,
4 r 2

 ok
r2
189
V    E  dl  
C
Q
a
Q
o
b
o
dr  
(b  a)
4 k 
4 k
Q 4  o k

|V | b  a
Method 2: Using the inhomogeneous Laplace’s equation,
dV
 A'
dr
V (r  a )  0
 
 
1 d   o k 2 dV 
r

 0
r 2 dr  r 2
dr 
dV
 A or V  Ar  B
dr
0  Aa  B
 
B   Aa
V (r  b)  Vo
 
Vo  Ab  B  A(b  a )
  (  V )  0
 
 ok
E
V
dV
ar   Aar   o ar
dr
ba
 s  Dn  
Vo  o k
|
b  a r 2 r  a ,b
Q    s dS  
C
1 2
Vo  o k
Vo  o k
4
2 r sin  d d  

b a
r
b a
| Q| 4  o k

Vo
b a
Prob. 6.62
C  4 o a  4 
109
 6.37 106  0.708 mF
36
 
A
Vo
b a
190
Prob.6.63
Q
C
V
Q
D
a
2 L
D
Q
E= 
 2 L o (3)(1   )
V    E dl 
Q
d

6 L o a  (1   )
b
A
B
1
 
 (1   )  1  
Using partial fractions
A=1, B= -1
Let
b d b d 



6 0 L  a  a 1   
Q

 ln   ln(1   ) ba
6 0 L
Q
V

b
a 

ln
 ln

6 0 L  1  b
1  a 
Q
If a=1 mm, and b=5 mm
Q
C
|V |
6 o

b
a
 ln
ln
1 b
1 a
9
10
6 
36

5
1
ln  ln
6
2
1
 109
1
6

  1.9591 nF
ln 0.8333  ln 0.5 6
C  0.326 nF
191
Prob. 6.64
Da D
  1  11
a
a
109

(4)
4
36
C
nF/m = 46.34 pF/m

ln11
36 ln11
Prob. 6.65
(a) From eq. (6.46),
Qh
10  109 (10)
107


 12.107 pF/m 2
2 [ x 2  y 2  h 2 ]3/2 2 [4  16  100]3/2 2 (120)3/2
(b) Qin  Q  10 nC
s 
Prob. 6.66
4nC
-2
4
3nC
-3nC
-1
0
3
1
2
-4nC
2
1
(a) Qi = -(3nC – 4nC) = 1nC
(b) The force of attraction between the charges and the plates is
F  F13  F14  F23  F24
| F |
1018
 9 2(12) 16 
 2  2   5.25 nN
9
4  10 / 36  22
3
4 
Prob. 6.67
We have 7 images as follows: -Q at (-1,1,1), -Q at (1,-1,1), -Q at (1,1,-1),
-Q at (-1,-1,-1), Q at (1,-1,-1), Q at (-1,-1,1), and Q at (-1,1,-1). Hence,
(2ax  2ay  2az ) (2ay  2az )
 2
2
2
 3 ax  3 ay  3 az 



Q
2
2
2
123 / 2
83 / 2


F
4 o  (2ax  2ay ) (2ax  2az )




3/2
3/2
8
8


2
192
 1
1
1 
 0.9(ax  ay  az )  

  0.1092(ax  ay  az ) N
 4 12 3 4 2 
Prob. 6.68
 360 o 
N
 1  7
 45 o

Prob. 6.69
(a)
E  E  E 
 L  a  1 a  2  16 109  (2, 2,3)  (3, 2, 4)
(2, 2,3)  (3, 2, 4) 




9 
2
10  | (2, 2,3)  (3, 2, 4) | | (2, 2,3)  (3, 2, 4) |2 
2 o  1  2 
2 
36
(1,0, 1) (1,0,7)
 18 x16 

 138.2ax  184.3ay V/m
2
50 

(b)  s  Dn
 L  a1 a 2  16 x10 9  (5, 6,0)  (3, 6,4)
(5, 6,0)  (3, 6, 4) 
D  D  D 





2
2  1
2 |(5, 6,0)  (3, 6,4)| |(5, 6,0)  (3, 6, 4)|2 
2 

8  (2, 0, 4) (2, 0, 4) 

nC/m 2  1.018a z nC/m 2


  20
20 
193
 s   1018
.
nC / m2
Prob. 6.70
o
y=2 y=4
y=8
y=-4
y=0
y=2
y=4
y=8
At P(0,0,0), E=0 since E does not exist for y<2.
At Q(-4,6,2), y=6 and
s
109
E
an 
(30ay  20ay  20ay  30ay )  18 (60)ay
2 o
2 x10 9 / 36
 3.4ay kV/m
194
CHAPTER 7
P.E. 7.1
z
1
2
2
27
a x  a y
 a  a y 
a  al  a =  x
  az 
2 
2

  5, cos 1  0, cos  2 
H3 
  a  a y 
10  2
 0   x

  30.63a x  30.63a y mA/m
4 (5)  27
2



P.E. 7.2
2 
3 
1

 az  0.1458az A/m
4 (2) 
13 
12
  32  42  5,  2  0, cos 1   ,
(b)
13
(a) H 
 3a  4a z  4a x  3a z
a  a y   x

5
5


2  12   4a x  3az 
1
H
 4ax  3az 
1  

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
H
50  103  25 104
a z  57.3a z mA/m
2(52  92 )3/ 2  106
P.E. 7.4
2  103  50  103 (cos  2  cos 1 )a z
NI
cos

cos
a




2
1 z
2L
2  0.75
100

cos  2  cos  1 a z
1.5
0.75
(a) At (0,0,0),   90 o , cos  2 
0.75 2  0.05 2
1
= 0.9978
100
H
 0.9978  0  az
1.5
H
= 66.52 az A/m
(b) At (0,0,0.75),  2  90 o ,cos  1   0.9978
100
 0  0.9978  a z
H
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
cos  1 
 0.9806
0.25 2  0.05 2
100
 0.9806  0.995 a z
H
1.5
= 131.7az A/m
P.E. 7.5
H
(a)
(b)
1
K  an
2
1
H (0, 0, 0)  50a z  (a y )  25a x mA/m
2
1
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

2
1
2
1
2
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
H 
103  100 103
 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
y 1 x  0
1
0
(b)    B.dS  
2
2
2
2
 ( y  x )dxdy   y dy  5 x dx
1
5
 (64  1)   20 Wb
3
3
Alternatively,
1
4
0
0
1
1
   A.dl   x 2 (1)dx   y 2 (1)dy   x 2 (4)dx  0
5 65
 
 20 Wb
3 3
P.E. 7.8
z
R
h
y
k
dS
x
H 
kdS  R
,
4 R 3
dS  dxdy, k  k y a y ,
R  ( x,  y, h),
197
k  R  (ha x  xa z )k y ,
H 

k y (ha x  xa z )dxdy
3
4 ( x 2  y 2  h 2 ) 2
k y ha x  
dxdy
4
  (x  y  h )
2
 
2
2
3

k y az  
xdxdy
4   ( x  y  h )
2
2
 
2
2
3
2
The integrand in the last term is zero because it is an odd function of x.
H

k y ha x 2 
4
 d d 
  (  h )
2
 0  0
2
3

k y h2 a x 
2
4
2
2
 (  h )
0

  ky
1
  a
ax 
 (  2  h2 ) 12  0 2 x
2


kyh
1
Similarly, for point (0,0,-h), H =  k y a x
2
Hence,
1
z0
 k yax ,
H 2
1
z0
 k a ,
2 y x
Prob. 7.1
(a) See text
(b) Let H = Hy + Hz
For H z 
a   a z 
Hz 
I
2
a
  ( 3) 2  4 2  5
( 3a x  4a y )
5

( 3a y  4a x )
5
20
(4a x  3a y )  0.5093a x  0.382a y
2 (25)
For H y 
I
a  ,   ( 3) 2  5 2  34
2
a  a y 
(3a x  5a z ) 3a z  5a x

34
34
3
2
d ( 2 )
2
198
Hy 
10
(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
H
I
2


I
2 H

2
100

 31.83 m
3
2 (10 10 ) 
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).
H1 
I
2
a ,
  5,
a  a  a  a z  a x  a y
a
10
ay  y
2 (5)

I
  5 2, a  a  a  , a  a z
H2 
a ,
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
( -a x  a y )
H2 


2 5 2 
2  2
H1 
H  H1  H 2 
ay


1
(-a x  a y )  0.1592a x  0.1592a y
2
199
Prob. 7.4
H  dH 
For H1 ,
Idl  R
 H1  H 2
4 R 3
R  (3,1, 2)  (0, 0, 0)  (3,1, 2),
Idl  R  4 105
R  9  1  4  14
1 0 0
 4 105 (0, 2,1)
3 1 2
4  105 (0, 2,1)
 (0, 0.01215, 0.006076)105
H1 
3/2
4 (14)
R  (3,1, 2)  (0, 0,1)  (3,1, 3),
For H 2 ,
Idl  R  6 105
H1 
R  9  1  9  19
0 1 0
 6  105 (3, 0, 3)
3 1 3
6  105 (3, 0, 3)
 (0.017, 0, 0.017)105
3/2
4 (19)
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
For H z ,
Hz 
I
2
a
where I = 20,  =  3a x  4a y ,  = 32  42  5
0
0 1
 3a x  4a y 
a  a  a  a z  
 0.8a x  0.6a y

5

 0.6 0.8 0
20
(0.8a x  0.6a y )  0.5093a x  0.382a y
Hz 
2 (5)
I
For H y ,
Hy 
a
2
where I = 10,  =  3a x  5a z ,  = 32  52  34
a  a  a 
1 0 1 0
1
(5a x  3a z )

34 3 0 5
34
10
(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
Hy 
200
Prob. 7.6
y
1 A
2
6A
P

B
1
O
H 
I
4
x
1
 cos  2  cos 1  a
1
2
2 
2
2
 a x  a y 
 a x  a y 
1 -1 1 0
 


 az



2 -1 -1 0
2
2




 cos 45o  cos135o  az  3 az
1  135o ,  2  45o ,  
a  al  a 
H
6


2
2
H  0, 0, 0   0.954a z A/m
4
Prob. 7.7
(a) At (5,0,0),   5,
H
a  a y ,
cos 1  0,
cos  2 
10
125
2
10
(
)a y  28.471a y mA/m
4 (5) 125
(b) At (5,5,0),   5 2,
cos 1  0,
cos  2 
10
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
(c ) At (5,15,0),   250  5 10,
cos 1  0,
cos  2 
10
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
1
x
A (2, 0, 0)
C (0, 0, 5)
y
2
B (1, 1, 0)
202
(a)
Consider the figure above.
AB

AC

1, 1, 0    2, 0, 0 
 0, 0, 5   2, 0, 0 
AB  AC


BA 
i.e.
H2 

(b)
BC  BA
BC BA

BC   
a 
AB  AC
AB AC

 0, 0, 5   1, 1, 0 
1,  1, 0 
cos  2

2, i.e AB and AC are not perpendicular.
cos 180o  1 
BC
 1, 1, 0 
 2, 0, 5

al  a


2

2 29

cos 1  
2
29
 1,  1, 5
1  1
BC BA

0
 1,  1, 5 ,   27
 1, 1, 0    1,  1, 5


2
27
 5, 5, 2 
54

 5, 5, 2  A/m
2   5, 5, 2 
5


 0 

27
29 
2 27
2 29

27.37 a x  27.37a y  10.95 a z mA/m
10
4 27
 0,  59.1, 0    27.37, 27.37, 10.95
  30.63, 30.63, 0 
H  H1  H 2  H 3 

 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
Hx 
I
4
 cos  2  cos 1  a
2
O
5A
1
x
203
where a   a x  a y   a z , 1  180o ,  2  45o
 cos 45  cos 180   a   0.3397a
4  2 
5
Hx 
o
o
z
z
H
 2 H x  0.6792 a z A/m
(b)
H  Hx  H y
where H x 
5
1  0  a , a
4  2 
  a x  a y  a z
 198.9a z mA/m
H y  0 since 1   2  0
H  0.1989 a z A/m
(c )
H  Hx  H y
where H x 
Hy 
5
1  0   a x  a z   198.9 a y mA/m
5
1  0   a y  a z   198.9 a x mA/m
4  2 
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
where a   a x ,   2 tan 30o 
2
,  2  30o , 1  150o
3
5
3
15
cos 30o  cos 150o   a x   
ax

4 2
8
H  3H1   1.79a x A/m
H1 
204
Prob. 7.11
3
Let H  H1  H 2  H 3  H 4
4
where H n is the contribution by side n.
H  2H1  H 2  H 4 since H1  H 3
(a)
H1 
H2 
I
4
 cos  2  cos 1  a
10  6
1  1

 az
4  2   40
2

10 
2 
10 
1 
2
2
az , H 4 



 az
4  6  
4  2  
40 
2
 5  3
1 
5
  



2
6 10
 2  10
At  4, 2, 0  , H  2  H1  H 4 
H
(b)
H1 

H
(c )
10
4  2 

8
10
az , H 4 
4  4 
20
2 5  1
az 
1
  4 

 a z = 1.964a z A/m
2 2 
5
4
az
20
1.78a z A/m
At  4, 8, 0  , H  H1  2H 2  H 3
H1 
10 
4 
10  8
1 
2

az , H 2 



 az
4  8   4 5 
4  4   4 5
2
H3 
10  2 
 a z 
4  4   2 
H
(d )
2

5
8
4
4 
 1



5
2
 5
 az  

0.1178a z A/m
At  0, 0, 2  ,
H1 
10  8

 0   ax  az 

4  2   68

H2 
10
 4

 2a  8a x 
 0  ay   z


4 68  84
68 




10
ay
 68

5  a x  4a z 
17 84
205
H3 
 2a  4a y 
10
 8

 0  ax   x


4 20 
84
20 



a y  2a z
 21
5a x
4 

0 
  a y  a z  
 20
4 2 
20 
5
5 
10 
2 

 1
 20



H  
 az
 ax  
 ay  
 34 21  21 
 34 21  20 
  21  68 
 0.3457 ax  0.3165 a y  0.1798 az A/m
10
H4 
Prob. 7.12
H  4 H1  4
I
4
(cos  2  cos 1 )a
  a  2cm, I  5mA,  2  45o , 1  90o  45o  135o
a  a  a   a y  (a x )  a z
H
I 1
1
2I
2  5  103
(

)a z 
az 
a  0.1125a z
a 2
a
  2 102 z
2
Prob. 7.13
206
Consider one side of the polygon as shown. The angle
(a )
subtended by the Side At the center of the circle
360o
2

n
n
The field due to this side is
H1 
where   r ,
I
cos  2  cos 1 
4


cos  2  cos(90  )  sin
n
n
cos 1   sin
H1 
I

n
2 sin
4 r

n
nI

sin
n
2 r
3I

For n  3, H 
sin
2 r
3
2
r cot 30o  2  r 
3
H  nH1 
(b)
3 5
2 2
H 
For n  4, H 

3
3
2

45
8
45 1
4I

sin


2r
4
22  2
 1.128 A/m.
(c)
As n  ,
H  lim
n 
nI

sin

2r
n
nI 
I


2r n
2r
From Example 7.3, when h  0,
H 
I
2r
which agrees.
 1.79 A/m.
207
Prob. 7.14
4

1
2
2
Let H 
H1  H 2  H 3  H 4
I
10
az 
a z  62.5 a z
4a
4  4  102
I
4
cos  2  cos 90o  a z ,  2  tan 1
 H4 
2 
4  4 10
100
 19.88 a z
H1 
H2
H3 
I
4 1
2 cos  a z ,   tan 1
100
4
 87.7 o
10
2 cos 87.7o a z  0.06361 a z
4
  62.5  2 19.88  0.06361 a z

H
 102.32 a z A/m.
Prob. 7.15
y
3
I
o
0
1
4
2
1
2
x
 2.29o
3
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
,
dl  1d a , R   1a 
4 R 3
I 12 d a  (a  ) Id (a z )
dH 4 

413
41
dH 4 
For H 4 ,

H4 
I  (a )
I (a z ) o
d  o z

41 0
41
Similarly, for H 2 ,
H2 
Io (a z )
4 2
H  H1  H 2  H 3  H 4 
Io  1 1 
   az
4  1  2 
Prob. 7.16
From Example 7.3,
H1 
I
az ,
2a
H2 
Ia 2
az
2[a 2  d 2 ]3/2


10  1
32
I 1
a2
 2
H = H1  H 2    2
a  
a
2 3/2  z
2 3/2
2
2  z
2  a [a  d ] 
2  3 10
[3  4 ] (10 ) 
1 9 
a z  202.67a z A/m
 500  
 3 125 
Prob. 7.17
2
2
209
H

nI
 cos  2  cos 1 
2

cos  2  -cos 1 
H


nI 
2
2 a  
2
4


1
2
a2  

2
2
4
1
2
0.5  150  2 102
2 103  42  102
 69.63 A/m
(b)
1
2
a
4

 0.2   2  11.31o
b
20
150  0.5
cos 11.31o  36.77 A/m

2
1  90o , tan  2 
H

nI
cos  2
2
Prob. 7.18
e
y
 P (4, 3, 2)
x
210
Let
H  Hl  H p
Hl 
 
a 
Hl
Hp
H
1
2
a
 4, 3, 2   (1, -2, 2)  (3, 5, 0),
3a x  5a y
34
 


34
, al  a z
3a y  5a x
 3a  5a y 
a  al  a   a z   x
 
34
34 

20  5a x  3a y 
-3


 x10  (  1.47a y  0.88a y ) mA/m
2 
34

1
1
K  an 

100 103  az   -a x    0.05a y A/m
2
2
 H l  H p  1.47a x  49.12 a y mA/m
211
Prob. 7.19
(a) See text
(b)
I
a
b
For   a,
 H  dl  I
For a    b,
H  2 
H 
For   b,
enc
 0  H  0
I   2  a2 
  b2  a 2 
  2  a2 


2  b2  a2 
I
H  2  I 
H 
I
2
Thus,
H

 0,
 a

 I   2  a2 
 
 2 2 ,
2

 b a 

 I
,
 b

 2
a   b
212
Prob.7.20
x
y
-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
HP 
HL 
I
2
a 
1
1
k  an  10a x  a z  5a y
2
2
I
I
(a x  a z ) 
ay
2 (3)
6
H P  H L  5a y 
I
ay  0
6

 I  30  94.25 A
Prob. 7.22
(a)
From eq. (7.29),
 I
 2 a 2 a , 0    a
H=
I

 a
a ,
 2
(b)
 I
a,
a
1 d

(  H  )a z    a 2 z
J =  H=
 d
 0a z ,
 a
213
Prob. 7.23
For 0 <  < a
 H dl  I
L
H  2  
  J dS
enc
2
 0

Jo
   d d 

0
 J o 2
H  J o
For  > a
2
a
=0
=0
Jo
 H dl =  J dS =     d d 
H  2  J o 2 a
H 
Joa

 J o , 0< <a

Hence H    J o a
  ,  >a

Prob. 7.24
(b)
dS  dxdya z ,
J
(a)

   H  x
y2

y
x2

z  2( x  y )a z
0
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)




 2 0
 2 1
S
1
0
0
1
 4(4  0)  2(25  1)  16  48  32 A
Prob. 7.25
(a) J    H 
(b)For >a,
2k
1 d
1 d
2
(  H  )a z 
( ko
)a z  o a z
a
a
 d
 d
214
2
2ko
2ko
2 a
d
d

(2
)





a
2 0
 0  0 a
a
 H  dl  I enc   J  dS  
H  2  2 ko a

 H 
a
H  ko   a ,

 a
ko a

Prob. 7.26

J    H  x
y2

y
x2

z  (2 x  2 y )a z
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)
J   H 
1 
1 
(  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   d   d
3
3
2
0
S
 3  103 (2 )
Method 2:
 2
3
3 2
0
 16  103 A  50.265 kA
2
I   H dl 10   2  d  103 (8)(2 )  50.265 kA
3
L
0
Prob. 7.28
J   H 
1 d
1 d
(  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
(a) B 
o I
a
2
At (-3,4,5), =5.
B
(b)
4  107  2
a  80a nWb/m 2
2 (5)
   B  dS 
6 4
o I d  dz 4 107  2

ln  z

2 0

2
2
 16 107 ln 3  1.756  Wb
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
H1 
5
50
ay 
a
2
 y
2  5  10 
For H 2 ,   5 cm, a   a z  a x  a y , H 2  H1
H  2H1 
100

ay
 31.83 a y A/m
(b)
2a y  a x
 2a  a y 
For H1 , a  a z   x
 
5 
5

 a x  2a y 
5
H1 
   3.183a x  6.366a y
2 
2 5 5 10 
5

For H 2 , a    a z  a y  a x
H2 
5
2  5 
a x  15.915a x
H  H1  H 2
 12.3 a x  6.366a y A/m
216
Prob. 7.31
(a) I   J dS
2
 
2
a
 J o (1 
a
 0  0
(b)
 H dl = I
a
0
0
 J o  d  (  
)  d  d
2
 2 4  a
 2 J o 
 20
 2 4a 
1
  a2 Jo
2
2

enc
2  2 a 2 
Jo  a  
2
2 

  J dS
For  < a,
H  2   J dS
 2 4 
= 2 J o 
 2
 2 4a 
H  2  2 J o
H 
2 
2 
2



4 
a2 
Jo  
2 
2



4 
a2 
For  > a,
 H dl =  J dS = I
o
1
H  2   a 2 J o
2
2
a Jo
H 
4
 Jo  
2 
2



,   a
a2 
 4 
Hence H  
aJ o

,  >a

4
3
a2
)d 
217
Prob. 7.32
0 I
a
2
 
 B  dS 
da

0 I
d dz
z  0 2
 
d
a
+ d
d
n
I
b π
I
μ 02
B 
b
Prob.7.33
For a whole circular loop of radius a, Example 7.3 gives
H
Ia 2 a z
2  a 2  h 2 
3/ 2
0
Let h 
I
az
2a
For a semicircular loop, H is halfed
H=
H=
I
az
4a
B  o H 
o I
4a
az
Prob. 7.34
Bx By Bz


0
x
y
z
showing that B satisfies Maxwell’s equation.
(a)   B 
(b)
dS  dydza x
4
1
   B  dS    y 2 dydz 
z 1 y  0
4
y3 1
( z )  1 Wb
1
3 0
218
(c )   H = J


J  
B
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
   a
6
where H1 and H 2 are due to the wires centered at x  0 and x  10cm respectively.
On the slant side of the ring, z 
 
 B.dS

o I
 2 d dz

o I a  b bh (   a ) dz d
o Ih a  b  a 

1   d


2  a z 0
2 b  a   


o Ih 
ab
 b  a ln
 as required.
a 
2 b 
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
3
2 10  10 

 1.37  108 Wb
Prob. 7.36
 
 B dS
50o
106
z 0  0

 o 
0.2

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
   B dS   
20
  0  1
S

2
 /4
1
0
sin   d  d  20  d   sin 2  d
2
 /4
1
1
(1  cos 2 )d  10(  sin 2 )
0
2
2
0
 /4
 20(1) 
 1
 10(  )  2.854 Wb
4 2
Prob. 7.38
   B dS , dS  r 2sin d d ar
S
2
r
   3 cos  r 2sin d d
 /3
r 1
 2(2 )  sin  d (sin  )  4
0
2
 /3
0
0
 2  d  cos  sin  d
sin   / 3
 2 sin 2 ( / 3)
0
2
2
 4.7123 Wb
Prob. 7.39
B  o H 
o J  R
dv
4 v R 3
Since current is the flow of charge, we can express this in terms of a charge moving with
velocity u. Jdv = dqu.
B
 o  qu  R 
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 cos ax

y
0

z
y  e-x
 a x  e  x a y  cos axa z  0
A is neither electrostatic nor magnetostatic field
1 
1 
 B  
 20   0

 
 
 B  0
B can be E-field in a charge-free region.
(b)
B 
(c )
 C 
 2
1
(r sin ) = 0
r sin  
1

1 3
C 
r 2 sin 2   ar (r sin )a  0

r sin  
r r
C is possibly H field.
Prob. 7.41
(a)
D  0

x
 D 
y2z

y

z
2(x  1)yz -(x  1)z 2
 2(x  1)ya x  . . .  0
D is possibly a magnetostatic field.
(b)
E 
1 
  sin  
 ( z  1) cos    
0
 
z   
 E 
1
2
cos  a   . . .  0
E could be a magnetostatic field.
(c )
F 
1 
1   sin 
 2cos  +

  0
2
rsin   r 2 
r r
1  
2 sin  
r 1 sin   
a  0


r  r
r 2 
F can be neither electrostatic nor magnetostatic field.
 F 
221
Prob. 7.42
A
o Idl o ILaz

4 r
4 r
This requires no integration since L << r.
A
1 Az
B   A 
a   z a
 

But r   2  z 2
o ILa z
4 (  2  z 2 )1/ 2
 IL 1
Az o IL 

(  2  z 2 )1/ 2  o ( )(  2  z 2 ) 3/ 2 (2  )

4 
4
2
o IL  a
 IL  a
B
 o 3
2
2 3/ 2
4 (   z )
4 r
A
Prob. 7.43

x
B  o H    A 
10sin  y
H

z
  sin  xa y  10 cos  ya z
4  cos  x

 
sin  xa y  10 cos  ya z 

o 



x
J   H 
o
0
J

y
0



 
y
z
10 sin  ya x   cos  xa z 


o 

sin  x 10 cos  y
2
10sin  ya x  cos  xa z 
o
Prob. 7.44
(a)
 A 
1  2
1
1   2 cos  
1

  sin 2  
(
)
(
sin

)
0




r
A
A


r



r 2 r
r sin  
r 2 r  r  r sin    r 3 
1
1
2sin  cos   0
  4 2 cos   4
r
r sin 
222
(b)
B   A 
A 

1  
1  1 Ar 
( A sin  )    ar  
 (rA )  a

r sin   
 
r  sin   r

A 
1 
  (rA )  r  a
 
r  r
1    sin   2sin  
1  2sin  2sin  
 0 ar  0a    2  

a   
a  0

3
r  r  r 
r 
r
r3
r 3 
Prob. 7.45

x
B   A 
(a)

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  3xz  dy dz
2
 
2
2
2
z 0 y 0

x 1
   6 xz  dy dz  4  x y dy dz  3  xz dy dz
2
0
2
0
2
2
2
2
0
0
0
0
0
2
2
  6  dz  dy  4  dz  y dy  3 dy  z 2 dz
0
0
y 
z 
 +3(2) 
 
  6(2)  2   4(2) 
 2 0
 3 0




  8 Wb
2 2
(c )
 A 
A x
x

A y
y

3 2
A z
z

4xy  2xy  6 xy  0
  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
-24+16+16
223
Prob. 7.46
B   A 
A 

1  
1  1 Ar 
( A sin  )    ar  
 (rA )  a

r sin   
 
r  sin   r

1 k 
1 
(sin 2  )ar 
(kr 1 ) sin  a
2
r sin  r 
r r
2k cos 
k
k sin 
k sin 
2sin cos ar 
 3
a 
ar 
a
3
3
r sin 
r
r
r3

Prob. 7.47
B   A 

1 Az
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
2
2
 4

107 15 3  1
B


e  a  a 
H 
o
4
2
3

H 
14 a  42 a  10 A/m
 
 B  dS    e


15
 15 z 0
10

4

cos   d dz
 sin   0 2 e5  150 e5

  1.011 Wb
Prob. 7.48
B   A 
A 

1  
1  1 Ar 
( A sin  )    ar  
 (rA )  a

r sin   
 
r  sin   r

A 
1 
  (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
H
B
o

1
 20

cos 60o ar   4.974  105 ar A/m
7  2
4  10  4

224
Prob. 7.49
Applying Ampere's law gives
H   2  J o   2
Jo

2
H 

B  o H   o
But B


 A
AZ




Jo 
2
AZ
a  . . .

1
 Jo 
2


AZ 
1
or A  - o J o  2 a z
4
Prob. 7.50

B    A  x
0


2
sin

y
0
x
2
sin

A
A
z
 z ax  z a y
y
x
Az ( x, y )
y
2
ax 

2
cos
x
2
cos
Prob. 7.51
1

1 
( A sin  )ar 
(rA )a
r sin  
r r
A
1 Ao

(2sin  cos  )ar  o sin  (r 2 )a
2
r sin  r
r
A
 3o (2 cos  ar  sin  a )
r
B   A 
y
2
ay
 o
Jo  2
4
a
225
Prob. 7.52



z  (2 yz  x 2 )a x  (2 xz  2 xy )a z
(a) J    H  x y
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
(b) 
 v
   J  0  2x  2x  0
t
At (2,-1,3),
v
 0 C/m3s
t
Prob. 7.53
(a) B    A
A 
 1 Az A 
 A A 
1 
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 
(  A )a z
 
1
40

(40  )a z 
a z  A/m 2
J   H 
o 
o
I   J dS 
(b)

2
2
o 0
0
40
40
2
2
o 0 0
  d   d 
 d d  , dS =  d d  a z
40  2 2
(2 )
o 2 0
80  2  106

 400 A
4  107
226
Prob. 7.54
H 
 Vm
  H  dl  mmf

Vm 
From Example 7.3, H 
Ia 2
2  z2  a 2  2
3
az
3
Ia 2
 Iz
2
2  2
dz 

z

a
 c


1

2
2  z2  a 2  2
Vm 
As z  , Vm  0 , i.e.
0  
I
 c
2

Vm 
I 
1 
2 


z2  a 2 
c 
I
2
Hence,
z
Prob. 7.55
H 
I
2
a
 J  0
H   Vm
But
I
2
a 

1 Vm
a
  
At 10, 60o , 7  ,   60o 
or
C 

3
 Vm  
I
  C
2
, Vm  0 
0  
I
6
Vm  
I
I
 
2
6
At  4, 30o ,  2  ,   30o 

6
,
I 
I


2 6
6
Vm  1 A
Vm  

I
12

12
12
I 

 C
2 3
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
an  Vm   25x  c
x
At the origin, x  0, Vm  0, c  0, i.e.
25 ax


Vm 
 25 x
(a)
At  2, 0, 5  , Vm  50A.
(b)
At 10, 3, 1 , Vm  250A.
Prob. 7.57
(a)
 V
V 
1 V
  V    
a 
a 
az 
z 
 
 

 1  2V
  2V
1  2V 

a




 dz  
  z
 z
1   2V


  
(b)




A 
 a
z 
Az 
1 
 a 

 
  

 1 
 A   


z   

 2 A
A  
 az 
  
 A  

2
1  2 Az
1   A 
1  A





   z 
 z
 

 2V 
 a
z 
 2V 

 az  0
 d  
 1 Az
     A     
  
 A
  
 z


1  2 Az
 
  1 A 


z    
 2 A
1 A



z
z
 z

1 A
 z
 0
228
Prob. 7.58
'
1
R
1
 
 


2
2
2  2








x
x'
y
'
z
z'
y
a
a
a







x
y
z 

y '
z '  
 x '

3
2
2
2  2
 1


2
x
x'
x
x'
y
'
z
z'
y








a
 x 
 
 
   a y and a z terms

   
 2
R
R3


1
R 

1
R
2
2
2
2
  x  x'    y  y '   z  z '  


1
 

  
2
2
2  2

a
a
a
x
x'
y
y
'
z
z'








 
 

x
y
z 

y
z  
 x
r  r'



3
1
2
2
2  2


 2  x  x'  a x  x  x'    y  y'    z  z' 
 a y and az terms


2
  x  x' a z   y  y ' a y   z  z' a z 
R
  3
R3
R
229
CHAPTER 8
P.E. 8.1
u
 QE  6a z N
t
u

(b)
 6a z  (u x , u y , u z ) 
t
t
u x
 0  ux  A
t
u y
 0  uy  B
t
u z
 6  u z  6t  C
t
A=B=C=0
Since u  t  0   0 ,
(a) F  m
ux = 0 = uy, uz = 6t
x
ux 
0 x A
t
y
uy 
0 y B
t
z
uz 
 6t  z  3t 2  C1
t
At t = 0, (x,y,z) = (0,0,0)  A1 = 0 = B1 = C1
Hence , (x,y,z) = (0,0,3t2),
u  6ta z at any time. At P(0,0,12), z = 12 =3t2  t =2s
t =2s
(c) u  6ta z  12a z m/s .
u
a
 6a z m 2
s
t
(d) K .E 
1
1
2
m u  1144   72 J
2
2
P.E. 8.2
(a)
ma  eu  B = (eBouy, -eBoux, 0)
d 2 x eBo dy
dy


2
dt
m dt
dt
(1)
230
d2y
eBo dx
dx

 
2
dt
m dt
dt
(2)
d 2z
dz
 0; 
 C1
2
dt
dt
(3)
From (1) and (2),
d 3x
d2y
dx


  2
3
2
dt
dt
dt
(D2 + w2 D)x = 0  Dx = (0, j)x
x = c2 + c3cost +c4sint
dy 1 d 2 x

 c3 cos t  c4 sin t
dt  dt 2
At t = 0, u  ( , 0,  ) . Hence,
c1   , c3  0, c4 


dx
dy
dz
  cos t ,   sin t ,  
dt
dt
dt
(b)
Solving these yields
a

x  sin t , y  cos t , z   t


The starting point of the particle is (0,
(c)
x2 + y2 =

,0)

2
, z=t
2
showing that the particles move along a helix of radius 
 placed along the z-axis.
P.E. 8.3
(a)
From Example 8.3, QuB = QE regardless of the sign of the charge.
E = uB = 8 x 106 x 0.5 x 10-3 = 4 kV/m
(b)
Yes, since QuB = QE holds for any Q and m.
231
P.E. 8.4
By Newton’s 3rd law, F12  F21 , the force on the infinitely long wire is:
IIb 1
1
)a
Fl   F  o 1 2 ( 
2
o o  a 
4  107  50  3  1 1 

   a   5a   N
2
 2 3
P.E. 8.5
m  ISan  10  104  50
(2, 6, 3)
7
= 7.143 x 10-3 (2, 6, -3)
 (1.429a x  4.286a y  2.143a z )  102 A-m 2
P.E. 8.6
T  mB 
(a)
10  104  50 2 6 3
6 4 5
7  10
 0.03a x  0.02a y  0.02a z N-m
(b)
T  ISB sin  
| T |max =
P.E. 8.7
(a)
r 
(b)
H
(c)
T max  ISB
50  10 -3
| 6a x  4a y  5a z | 0.04387 Nm
10

 4.6,  m   r  1  3.6
o
10  103 e  y
a A / m  1730e  y a z A/m
 4 107  4.6 z
M   m H  6228e  y a z A/m
B

P.E. 8.8
3a  4a y 6a x  8a y
an  x

5
10
(6  32)(6a x  8a y )
B1n  ( B1  an )an 
1000
232
 0.228a x  0.304a y  B 2 n
B1t  B1  B1n  0.128a x  0.096a y  0.2a z
B2t 
2
B  10 B1t  1.28a x  0.96a y  2a z
1 1t
B2  B2 n  B2t  1.052a x  1.264a y  2a z Wb/m2
P.E. 8.9
(a)
B1n  B2 n  1 H1n  z 2 H 2 n
or 1 H1  an 21  2 H 2  an 21
(6 H 2 x  10  12)
(60  2  36)
o
 2o
7
7
35  6 H 2 x
H 2 x  5.833 A/m
(b)
K  ( H1  H 2 )  an12  an 21  ( H1  H 2 )
= an 21   (10,1,12)  (35 , 5, 4) 
6


=
2 3
1 6
7 25 6 6 8
K  4.86a x  8.64a y  3.95a z A/m
(c)
Since B   H , B1 and H1 are parallel, i.e. they make the same angle with the
normal to the interface.
H a
26
 0.2373
cos 1  1 n 21 
H1
7 100  1  144
1  76.27 o
cos  2 
H 2  an 21
13

 0.2144
H2
7 (5.833) 2  25  16
 2  77.62o
P.E. 8.10
(a)
L'  o  r n 2 S  4  107  1000  16  106  4  104
= 8.042 H/m
233
(b)
Wm '  1 L' I 2  1 (8.042)(0.5 2 )  1.005 J/m
2
2
From Example 8.11,
P.E. 8.11
Lin =
o l
8
Lext =
2 wm 1
I 2

 d  d dz
I2
I 2  4 2  2
=
l
1
2
b
2 o
d
4    (1   ) 
dz d
2
0
0
a
1
2o
1 
d
 2 l   
2
4
 (1   ) 
a 
b
=
1 b 
 ol  b
ln  ln

  a
1  a 
 l  l b
1 b
L = Lin + Lext = o  o ln  ln
8
  a
1  a 
=
P.E. 8.12
(a)
L’in =
4  107
o
=
= 0.05 H/m
8
8
L’ext = L’ – L’in = 1.2 – 0.05 = 1.15 H/m
(b)
L’ =
ln
d  a
o  1
 ln

a 
2  4
d  a 2 L '
2 1.2 106

 0.25 
 0.25
a
o
4 x107
 6  0.25  5.75
d a
 e 5.75  314.19
a
d  a  314.19a  314.19 
d  407.9mm  40.79cm
2.588  10 3
 406.6mm
2
234
P.E. 8.13
This is similar to Example 8.13. In this case, however, h=0 so that
o I1a 2b
A1 
a
4b3
 I a2
 I a 2
12  o 12  2b  o 1
2b
4b
o a
12
m12 

2b
I1
= 2.632 H
2
4 107    4

23
P.E. 8.14
Lin =
4 107  10  102
o
 2 o
l= o
=
8
8
4
= 31.42 nH
P.E. 8.15
(a) From Example 7.6,
Bave 
o NI
l
  Bave  S 
or I 

o NI
2o
o NI
 a 2
2 o
2  o 2  10  102  0.5  103

a 2 N
4  10 7  10 4  103
= 795.77A
Alternatively, using circuit approach
l
2 o
2o
R


S o S oa 2
 2  o
  NI 

, as obtained before.
N
a 2 N

2 o
2  10  102

 1.591 109
7
4
2
a
4  10  10
   = 0.5x10 x1.591x10 =7.9577x10
-3
I
(b)
9

 795.77 A as obtained before.
N
If =500o,
5
235
I
795.77
 1.592 A
500
P.E. 8.16

B 2 a S (1.5) 2  10  104 22500


= 895.25N
2o
8
2  4  10 7
Prob. 8.1
At P, x = 2, y = 5, z = -3
E  2(2)(5)(3)a x  (2)2 (3)a y  (2)2 (5)a z  60a x  12a y  20a z
B  (5) 2 a x  (3) 2 a y  22 a z  25a x  9a y  4a z
F  Q( E  u  B )
u B 
1.4 3.2 1
 3.8a x  30.6a y + 92.6a z
25
9
4
E  u  B  (60, 12, 20)  (3.8, 30.6,92.6)  (63.8, 42.6,112.6)
F  Q( E  u  B )  4( E  u  B ) mN
= 255.2a x  170.4a y  450.4a z mN
Prob. 8.2
F  m 2 r  9.111031  (2 1016 ) 2 (0.4 1010 )  14.576 nN
Prob. 8.3
(a)
F  Q(u  B )  103
10 2
6
0
25
0
 103 (50a x  250a y )
= 0.05a x  0.25a y N
(b) Constant velocity implies that acceleration a = 0.
F = ma  0  Q( E  u  B )
E = -u  B 
50a x + 250 a y V/m
236
Prob. 8.4
Fe  qE,
Fm  qu  B
Fe
20  103
E


 80
Fm uB 0.5 108  5  106
Prob. 8.5
ma  Qu  B
103 a  2  103
ux
0
uy
6
uz
0
d
(u x , u y , u z )  (12u z ,0,12u x )
dt
i.e.
du x
 12u z
dt
du y
 0  u y  A1
dt
du z
 12u x
dt
(1)
(2)
(3)
From (1) and (3),
ux  12u z  144u x
or
ux  144u x  0  u x  c1 cos12t  c 2 sin 12t
From (1), uz= - c1sin12t + c2cos12t
At t=0,
ux=5, uy=0, uz=0  A1=0=c2, c1=5
Hence,
u  (5cos12t , 0, 5sin12t )
u(t  10s )  (5cos120, 0, 5sin120) = 4.071a x  2.903a z m/s
dx
 5 cos12t  x  5 sin 12t  B1
12
dt
dy
uy 
 0  y  B2
dt
dz
uz 
 5 sin 12t  z  5 cos12t  B3
12
dt
ux 
237
At t=0, (x, y, z) = (0, 1, 2)  B1=0, B2=1, B3=
19
12
5
19 
 5
( x , y , z)   sin 12t ,1, cos 12t  
 12
12
12 
(4)
At t=10s,
5
19 
 5
( x , y , z)   sin 120 ,1, cos 120   = (0.2419, 1, 1.923)
 12
12
12 
By eliminating t from (4),
x 2  ( z  19 ) 2  ( 5 ) 2 , y  1 which is a circle in the y=1 plane with center at
12
12
(0,1,19/12). The particle gyrates.
Prob. 8.6
(a)
ma  e(u  B )

u
m d
(u x , u y , u z )  x
0
e dt
uy
0
uz


 u y Bo ax  Bo u x a y
Bo
du z
 0  uz  c  0
dt
du x
Be
Be
 u y o  u y w , where w = o
m
dt
m
du y
 ux w
dt
Hence,
ux   wu y   w2u x
or ux  w2u x  0  u x  A cos wt  B sin wt
uy  
u x
 A sin wt  B cos wt
w
At t=0, ux = uo, uy = 0  A = uo, B=0
Hence,
u
dx
 x  o sin wt  c1
dt
w
dy
u
u y  uo sin wt 
 y   o cos wt  c2
dt
w
u x  uo cos wt 
238
u
At t=0, x = 0 = y  c1=0, c2= o . Hence,
w
uo
uo
x  sin wt , y  (1  cos wt )
w
w
2
u 2o
u
u 
(cos2 wt  sin 2 wt )   o   x 2  ( y  o ) 2
2
w
w
w
showing that the electron would move in a circle centered at (0,
uo
). But since the field
w
does not exist throughout the circular region, the electron passes through a semi-circle
and leaves the field horizontally.
(b)
d = twice the radius of the semi-circle
=
2u o 2u o m
=
w
Bo e
Prob. 8.7
qE
o
mg
mg  qE

 q
mg 0.4  103  9.81

 26.67 nC
E
1.5 105
Prob. 8.8
F   Idl  B  IL  B
ax
ay
az
 4.5(0.2)a x  (2.5)(a y  a z )10  4.5(0.5) 1
0
0
1
0 mN
1
3
F  2.25(a y  a z ) mN
Prob. 8.9
qE  quB

B
E 12  103 1200


 85.714 Wb/m 2
u
140
14
239
Prob. 8.10
 I I a a
F
 I1al  B2  o 1 2 l 
2
L
7
a  (a y )4 10 (100)(200)
F21  z
 4a x mN/m (repulsive)
2
  IL  B   
(a)
(b)
(c)
(d)
F12   F21  4a x mN/m (repulsive)
4
3
3
4
al  a  a z  ( a x  a y )   a x  a y ,   5
5
5
5
5
7
4
4 10 (3  10 )  3
4 
F31 
  ax  a y 
2 (5)
5 
 5
 0.72a x  0.96a y mN/m (attractive)
F3  F31  F32
4 107  6  104 )
 az  a y   4ax mN/m(attractive)
2 (3)
F3  3.28a x  0.96a y mN/m
F32 
(attractive due to L2 and repulsive due to L1)
Prob. 8.11
F
o I1 I 2 4 107 (10)10

 100  N
2
2 (20  102 )
Prob. 8.12
F   Ldl  B  3(2a z )  cos 
W    F  dl ,
F  6 cos 
2
3

a N
W    6 cos 
0
= -1.8sin
3
3
o d  6 o  3sin  3 2 0 J
2
= -1.559 J
3
Prob. 8.13
o I1 I 2
4  107
d  a   a 
(2)(5) ln 6 a z
2


2
2
 2
6
(a)
F1  
 2 ln 3a z  N = 2.197a z  N
a
240
(b)
F2   I 2 dl2  B1
o I1 I 2 1
 d  a  dza z   a
2   
II 1
 o 1 2   d  a z  dza 

2

But  = z+2, dz=d
4 107
1
(5)(2)   d  a z  dza  
F2 
2
 4 
2
2 ln 2 (a z  a )  N  1.386a  1.386a z a z  N
4
II 1
F3  o 1 2   d  a z  dza  

2
But z = - + 6, dz = -d
4
4  107
1
(5)(2)   d  a z  dza  
F3 
2
 6 
2 ln 4 (a z  a )  N  0.8109a   0.8109a z  N
6
F  F1  F2  F3
 a  (ln 4  ln 4  ln 9)  a z (ln 9  ln 4  ln 4  ln 9)
 0.575a   N
Prob. 8.14
A
fBC
From Prob. 8.7,
C
f 
o I1 I 2
a
2
f  f AC  f BC
60o
B
4 107  75  150
 1.125 103
2  2
o
f  2  1.125cos 30 a x mN/m
| f AC || f BC |
 1.949a x mN/m
30o
fAC
241
Prob. 8.15
The field due to the current sheet is
B

2
K  an 
o
2
10a x  (a z )  5 o a y
L
F  I 2  dl2  B  2.5 dxa x  (5o a y )  2.5L  5o (a z )
0
F
 12.5  4  107 (a z )  15.71a z  N/m
L
Prob. 8.16
F   Idl  B  IL  B  5(2a z )  40a x 103  0.4a y N
Prob. 8.17
m  ISan  10(2  6)(a x )  120a x
T  m  B  120a x  4.5(a y  a z )  540
Prob. 8.18
f ( x, y, z )  x  2 y  5 z  12  0
an 
1 0
0 1 1


(a x  2a y  5a z )
30
Prob. 8.19
I
 540(a y +a z ) N.m
f  a x  2a y  5a z
a x  2a y  5a z
f

| f |
30
m  NISan  2  60  8  104
m  IS
0

 I
m
m
 2
S r
8  1022
 6.275  108  627.5 MA
 (6370  103 ) 2
 17.53a x  35.05a y  87.64a z mAm
242
Prob. 8.20
Let F  F1  F2  F3
0
F1   Idl  B   2dxa x  30a z mN
5
0
 300a y mN
5
=-60a y x
5
F2   2dya y  30a z mN
0
=60a x y
5
0
 300a x mN
5
F3   2(dxa x  dza z )  30a z mN
0
5
 300a y mN
0
F  F1  F2  F3  300a y +300a x -300a y mN=300a x mN
=60(-a y ) x
1
T  m  B  ISan  B  2( )(5)(5)a y  30a z 103  0.75a x N.m
2
Prob. 8.21
For each turn, T = m  B, m = ISan
For N turns,
T  NISB  50  4  12 104 100 103  24 mNm
Prob. 8.22
F   Idl  B

 F  IB  520  0.4  103  30  103
F  6.24 mN
Prob. 8.23
M  m H  m
B
o r

m B
o (1   m )
243
Prob. 8.24
M  mH  m
(a)
B
o 
4999
1.5

5000
4  10 7
M 

1.193  106 A/m
N
M 
(b)
m
k
k 1
v
If we assume that all mk align with the applied B field,
M 
Nmk
v
mk 
1.404  10 23 A  m2

mk 
Prob. 8.25
r   m  1  6.5  1  7.5
24 y 2

 H

az
m
6.5
M
M  m H
At y = 2cm,
24  4  104
a z  1.477a z mA/m
6.5



x y
z
48 y
J   H 

ax
2
6.5
24 y
0
0
6.5
At y=2cm,
H
J
48  2 102
a x  0.1477a x A/m 2
6.5
Prob. 8.26
(a)
(b)
(c )
  80 o

 m  r  1  79
H
B


r  80
20 xa y 103
80(4  107 )
 198.9 xa y A/m
M
N
v

1.193  106
8.5  1028
244
(d )
( e)
M   m H  15.713 xa y kA/m


y
J b    M  x
0 15.713x

z  15.713a z kA/m
0
Prob. 8.27
When H = 250,
2H
2(250)
B

 1.4286 mWb/m 2
100  H 100  250
But B=o r H
r 
B
1.4286  103

 4.54
o H 4 107  250
Prob. 8.28
 H  dl
 Ienc
H  2 
 2
I
 a2
M  m H 
Jb    M 
I
2 a 2
 H 
I
a
2 a 2
 r  1
1 
 
  M  a     1  Ia a

z
r
2
z
Prob. 8.29
(a)
From H1t – H2t = K and M = mH, we obtain:
M1t
 m1

M2t
 m2
K
Also from B1n – B2n = 0 and B = H = (/m)M, we get:
1 M 1n  2 M 2 n

 m1
 m2
(b)
From B1cos1 = B1n = B2n = B2cos2
B sin1
B sin 2
 H1t  K  H2 t  K  2
and 1
1
Dividing (2) by (1) gives
2
(1)
(2)
245
tan1
1
i.e.

k
tan 2 tan 2 
k 2 

1 


B2 cos 2
2
 2  B2 sin  2 
tan  1 1 
k 2 
1 


tan  2  2  B2 sin  2 
Prob. 8.30
B2 n  B1n  1.8a z
H 2t  H1t
B2t 


B2t
2

B1t
1
4 o
2
B1t 
(6a x  4.2a y )  9.6a x  6.72a y
1
2.5o
B2  B2 n  B2t  9.6a x  6.72a y  1.8a z mWb/m 2
H2 
B2
2

103 (9.6, 6.72,1.8)
4  4 107
 1,909.86 a x  1,336.9a y  358.1a z A/m
z
B2n
2
B2t
B2 n
1.8

 0.1536
B2t
9.62  6.722
 2  8.73o
tan  2 
246
Prob. 8.31
B2 n  B1n  12a z
H 2t  H1t
B2t 


B2t
2

B1t
1
2
2
B1t  o (4a x  10a y )  1.6a x  4a y
1
5 o
B2  B2 n  B2t  1.6a x  4a y  12a z mWb/m 2
B23 (1.62  42  122 )  106
1
3
w2  2 H 2 

 32.34 J/m3
7
2
2 2
2(2)(4 10 )
Prob. 8.32
1
1
2
2
tan 1 
But
H1t
,
H 1n
tan  2 
H 2t
H 2n
H1t  H 2t
B1n  B2t

1 H1n  2 H 2 n

H 2n 
1
H
 2 1n
H 2t
H1t
6.5H1t


1
H 2n
H1n
H 1n
6.5
If 1  42o ,
tan 42o 
H1t
H 1n
tan  2  6.5(0.9004)  5.832


H1t
 0.9004
H1n
 2  80.3o
247
Prob. 8.33
x <0
K
x>0
( H1  H 2 )  an12  K

H1  a x  K  H 2  a x
H1  a x  (10a x  6a z )  a x  6a y
H 2  ax  ( H 2 x , H 2 y , H 2 z )  ax  H 2 z a y  H 2 y az
6a y  12a y  H 2 z a y  H 2 y a z
Equating components,
6  12  H 2 z

H 2 z  6,
H2 y  0

1 H1n  2 H 2 n
Also,
B1n  B2 n
H 2n 
2
1
H1n  o (10a x )  5a x
4o
2
H 2  5a x  6a z A/m
Prob. 8.34
H 2t  H1t   a x   a z
B2 n  B1n

 2 H 2 n  1 H1n
1

H1n  r1  a y
2
r 2

H   a x  r1  a y   a z
r 2
H 2n 
248
Prob. 8.35
B1n  B 2n  15a
(a)
H1t  H 2t 
B1t
1

B2t
2
1
2
B2t 
10a  20az   4a  8az
5
2
B1t 
Hence,
B1  4a   15a  8a z mWb/m 2
(b)
w m1
 4  15  8  10

2
B12
1

B1  H1 
2
21
2
2
6
2  2  4  107
w m1  60.68 J / m3
B22

2 2
w m2
Prob. 8.36
10  15  20  10

2
2
2
2  5  4  10
7
6
 57.7 J / m3
f ( x, y , z )  x  y  2 z
 f  a x  a y  2a z
an 
f
1

( a x  a y  2a z )
| f |
6
(a)
H1n  ( H1 an )an  (40  20  60)
(a x  a y  2a z )
6
 6.667a x  6.667a y  13.333a z A/m
(b)
H 2  H 2 n  H 2t
B 2 n  B1n
2 H 2 n  1 H1n


B2  2 H 2  2 H 2 n  2 H 2t  1 H1n  2 H 2t  o (2 H1n  5 H 2t )
But
 4  107  (13.333,13.333, 26.667)  (233.333, 66.666, 83.333
 4  107 (220,80, 110)
 276.5a x + 100.5a y  138.2a z  Wb/m 2
249
Prob. 8.37
an  a 
B2 n  B1n  22 o a
H 2t  H1t
B2t 


B2t
2

B1t
1
o
2
B1t 
(45o a )  0.05625o a
800 o
1
B2  o (22a   0.05625a ) Wb/m 2
Prob. 8.38
H1n  3az ,
H1t  10a x  15ay
H 2t  H1t  10a x  15a y
H 2n 
1
1
H1n 
(3a z )  0.015a z
200
2
H 2  10a x  15a y  0.015a z
B2  2 H 2  200  4  107 (10,15, 0.015)
B2  2.51a x  3.77a y  0.0037a z mWb/m2
tan  
B2 n
B2t
or   tan 1
0.0037
2.51  3.77
2
2
= 0.047o
Prob. 8.39
(a)
H  1 K  an  1 (30  40)a x  (a z )  5a y A/m
2
2
B  o H  4  107 (5a y )  6.28a y  Wb/m2
(b)
H  1 (30  40)a y  35a y A/m
2
B  o r H  4  107 (2.5)(35a y )  110a y  Wb/m 2
250
(c)
H  1 (30  40)a y  5a y
2
B  o H  6.283a y  Wb/m2
Prob. 8.40
r = a is the interface between the two media.
B2 n  B1n


Bo1 (1  1.6) cos  ar  Bo 2 cos  ar
2.6 Bo1  Bo 2
(1)
H 2t  H1t


B2t
2

B1t
1
2 B1t  1 B2t
2 Bo1 (0.2) sin  a  o Bo 2 ( sin  )a
B
2  o o 2
(2)
0.2 Bo1
Substituting (1) into (2) gives
2 
o
0.2
(2.6)  13o
Prob. 8.41
(a)
The square cross-section of the toroid is shown below. Let (u,v) be the local
coordinates and  o =mean radius. Using Ampere’s law around a circle passing
through P, we get
v
u
(0,  o )
H (2 )(  o  v)  NI
The flux per turn is


H
NI
2 ( o  v)
251
a/2


o NIa
a/2
 o  a / 2 
 Bdudv  2 ln    a / 2 
u  a / 2 v  a / 2
L
o
N  o N 2 a  2 o  a 

ln 

I
2
 2 o  a 
The circular cross-section of the toroid is shown below. Let (r,) be the local
coordinates. Consider a point P( r cos  ,  o  r sin  ) and apply Ampere’s law
around a circle that passes through P.
(b)
H (2 )(  o  r sin  )  NI


H
NI
NI  r sin  

1 

2 (  o  r sin  ) 2o 
o 
r

(0,  o )
a 2
 NI  r sin  
 NI a 2
Flux per turn    
(2 )
1 
rdrd 
2 o 
2 o 2
o 
r 0 
L
N   N 2a2

I
2 o
Or from Example 8.10,
L  L' l 
o N 2lS
l2

o N 2a 2 o N 2a 2

2 o
2 o
252
Prob. 8.42
1
2
a  2 cm
o  (3  5)  4cm
L
o N 2 a  2  o  a 
ln 

2
 2 o  a 
N2 
2 L
2 (45  106 )

 22, 023.17
 2 o  a 
8 2
7
2
o a ln 
 4 10 (2 10 ) ln  8  2 
 2 o  a 
N  148.4 or 148
Prob. 8.43
L
o  4 107 (40)

 20  107  2  H
8
8
Prob. 8.44
Lin 
o 
,
8
o 
ln(b / a )
2
o  o 
ln(b / a )




8
Lext 
If Lin = 2Lext
1
b
 e1/ 8  1.1331
8
a
b  1.1331a  7.365 mm
ln(b / a ) 
Prob. 8.45
o   2  4 107 (10)  2 10

ln  1 
ln
 1  2  106 (ln1000  1)
L
2


2  a
2

 2  10

 2(5.908)  H  11.82  H
Prob. 8.46
o  a
 12   B1  dS  
  o
b
o I
o Ib
a  o
 2 dzd   2 ln 
z 0
o
253
M 12 
For N = 1,
M 12 

N 12 N o b a   o

ln
I
2
o
 12
I1

o b a   o
ln
2
o
4  107
(1) ln 2 = 0.1386 H
2
Prob. 8.47
We may approximate the longer solenoid as infinite so that B1 
the second solenoid is:
 2  N 2 B1S1 
M 
2
I1

o N1 I1
o N1N 2
l1
o N1I1
l1
  r1 N 2
2
l1
 r1
2
Here we assume air-core solenoids.
Prob. 8.48
For a straight infinitely long conductor,
B 
o I
2
h a b  w

o I  h b  w 1
1
   B  dS 
dzd  
   dzd    

2  z 0  b 
S
z 0   a b


bw
a  b  w o Ih  b  w
o Ih 
a  b  w
ln
 ln 
 ln
ln 


b
a  b  2 
2 
b
a  b 
 
o Ih  (a  b)(b  w) 
ln 

2
 b(a  b  w) 
Prob. 8.49
H
I
a
2
1
1
I2
wm   | H |2   2 2
2
2 4 
1
I2
1
W   wm dv    2 2  d d  dz 
 I 2 L ln(b / a)
2 4 
4
. The flux linking
254

1
 4  4  107 (625  106 )3ln(18 /12)  304.1 pJ
4
Alternatively,
W
1 2 1 L b 2 I 2L b
LI 
ln  I 
ln
a
2
2 2 a
4
Prob. 8.50
r   m  1  20
1
1
B1  H1 
H  H
2
2
1
  25x 4 y 2 z 2  100x 2 y 4 z 2  225x 2 y 2 z 4 

2
Wm   wm dv
wm 
1
2
2
1
2
2
1 
  25 x 4 dx  y 2 dy  z 2 dz  100 x 2 dx  y 4 dy  z 2 dz
0
1
1
0
0
2  0
1
2
2
 225 x 2 dx  y 2 dy  zdz 

0
0
1
1
2
2
1
2
2
25  x 5
y3
z3
x3
y5 z 3

 4

2  5 0 3 0 3 1
3 0 5 0 3 1

1
2
2
x3
y3 z 5 
 9

3 0 3 0 5 1 



25  1 8 9
4 32 9
9 8 33 
         
2 5 3 3
3 3 3
3 3 5

25
3600
 4  10 7  20 
2
45
Wm  25.13 mJ
Prob. 8.51
4
3
2
1
| B |2
1
W   B  Hdv  
dv 
(42  122 )106 dxdydz
7   
2v
2
2(15)4 10 z 0 y 0 x 0
v

106
320
 101.86 J
(16  144)(2)(3)(4) 
7

(30)4  10
255
Prob. 8.52
NI  Hl 
Bl

l
o
1.5  0.6
N

o r I 4 107  600 12
 313 turns
Bl
N
Prob. 8.53
F = NI = 400 x 0.5 = 200 A.t
Ra 
100
MAt/Wb,
4
Fa 
R 1  R2 
6
MAt/Wb,
4
R3 
1.8
MAt/Wb
4
Ra F
 190.8 A.t
Ra  R3  R1 // R2
Ha 
Fa
190.8

 19080 A/m
l a 1  10  2
Prob. 8.54
Total F = NI = 2000 x 10 = 20,000 A.t
lc
(24  20  0.6)  10 2
= 0.115 x 107 A.t/m

Rc 
7
4
 o  r S 4  10  1500  2  10
la
Ra 

0.6  102
= 2.387 x 107 A.t/m
7
4
4  10 (1)  2  10
o  r S
R = Ra + Rc = 2.502 x 107 A.t/m


20,000
a c 
= 8 x 10-4 Wb/m2
R
2.502  107
Ra
2.387  20,000

 19,081 A.t
R a  Rc
2.502
Rc
0.115  20,000
c 

 919 A.t
R a  Rc
2.502
a 
256
Prob. 8.55
Rc

F
F = NI = 500 x 0.2 = 100 A.t
Rc 
lc
42  102
42  106


 S 4 107 103  4 104
16
Ra 
la
103
108


o S 4 107  4 104 16
1.42  108
Ra  Rc 
16

Ba 
F
16 100 16


 Wb
Ra  Rc 1.42  108 1.42

S

16  106
 88.5 mWb/m 2
4
1.42  4 10
Ra
257
Prob. 8.56
The equivalent circuit is shown below.
R1
R2
R3
N1 I1
R4
N2 I2
R5
Prob. 8.57

4.4 103
0.2282  106


 7.2643  104 A.t/Wb
R
7
2
o S 4 10 (4.82 10 )

Prob. 8.58
F
B2S
2
4  106


 53.05 kN
2 o 2o S 2  4  107  0.3 104
Prob. 8.59
(a)
F = NI = 200 x 10-3 x 750 = 150 A.t.
10
la

 3.183  107
6
o S 25  10 o
lt
2  0.1
= 6.7 x 107
Rt 

6
o  r S o  300  25  10
Ra 


3

150
 7
 15.23 10 7
Ra  Rt 10 (3.183  20 / 3)
Rt

Ra
258
F
B2S
2
2.32  1012


2 o 2o S 2  4 107  25  106
= 37 mN
(b)If t  , Rt  0,  

150

Ra 3.183  107
F2  I 2 dl2  B1  I 2 dl2
1
S

2  103  5  103  150
3.183  107  25  10 6
F2 = 1.885 N
Prob. 8.60
2
2
1
1

Ra


Ra
Ra
Ra
Ra/2
 1  2 2 , 1 

3 R
2 a


2
2 
3Ra
3Ra
2
  2  
3 1
2
  2 2   1 

2
 2o S  2 o S 4o S 3Ra o S
o S 2 4 107  200 104  9 106


2
3 106
3la
 24  10 3  mg  m 
24  10 3
 7694 kg
9.8
Prob. 8.61

Rs
 = NI
Ra
Rs

Ra
Rs/2
Since    for the core (see Figure) , Rc = 0.
a
R   ( 2  x)

  NI    Ra  s  
o S
2 

259

 (2 x  a)
2o S
N 2 I 2 4o2 S 2
B2S
1
1
2




2 o
2 o S 2 o S
(a  2 x)2

2 N 2 I 2 o S
(a  2 x)2
F   Fa x since the force is attractive, i.e.
 2 N 2 I 2  o Sa x
F
(a  2 x) 2
260
CHAPTER 9
P.E. 9.1
(a) Vemf    u  B   dl  uBl  8  0.5  0.1  0.4 V
(b) I 
Vemf
R

0.4
 20 mA
20
(c) Fm  Il  B  0.02  0.1a y  0.5a z   a x mN
(d) P  FU  I 2 R  8 mW
P
or
Vemf
R

0.42  8 mW
20
P.E. 9.2
(a) Vemf    u  B   dl
where B  Bo a y  Bo  sin  a   cos  a  , Bo  0.05 Wb/m2
 u  B   dl    Bo sin  dz  0.2 sin t   2  dz
0.03
Vemf    u  B   dl  6 cos 100 t  mV
0
At t = 1ms,
Vemf  6 cos 0.1   17.93 mV
Vemf
 60 cos100 .t  mA
R
At t = 3ms, i  60 cos 0.3  110.8 mA
i
(b) Method 1:
 o zo
   B  dS   Bot  cos  a   sin  a   d  dza     Bo t sin  d  dz   Bo o zo t sin 
0 0
where Bo  0.02 ,  o  0.04 , zo  0.03
  t   2
   Bo o zot cos t
Vemf  

 Bo  o zo cos t  Bo o zo t sin t
t
261
  0.02  0.04  0.03  cos t  t sin t 
 24  cos t  t sin t  V
Method 2:
B
 dS   (u  B ).dl
t
B  Bo ta x  Bot (cos  a   sin  a ),   t  
Vemf   
2
B
 Bo (cos  a  sin  a )
t
Note that only explicit dependence of B on time is accounted for, i.e. we make 
= constant because it is transformer (stationary) emf. Thus,
o zo
0
0 0
zo
Vemf   Bo   (cos  a   sin  a ) d  dza    o Bot cos  dz
 Bo o zo (sin   t cos  ),   t  
2
 Bo o zo (cos t  t sin t ) as obtained earlier.
At t = 1ms,
Vemf  24[cos18o  100  103 sin 18o ]V
= 20.5V
At t = 3ms,
i  240[cos 54o  .03 sin 54o ]mA
= -41.93 mA
P.E. 9.3
d
d
, V2   N 2
dt
dt
V2 N 2
N
300  120

 V2  2 V1 
 72V
V1 N 1
N1
500
V1   N 1
P.E. 9.4
(a)
Jd 
D
 20 o sin(t  50 x)a y A / m 2
t
262
(b)
H z
a y  20 o sin(t  50 x)a y
x
20 o
or H 
cos(t  50 x)a z
50
  H  Jd  
 0.4 o cos(t  50 x)a z A/m
(c)
  E   o
E y
H
a z  0.4 o 2 o sin(t  50 x)a z

t
x
1000  0.4o o 2  0.4
2
or  = 1.5 x 1010 rad/s
c2
P.E. 9.5

2

245o 
   j 
j 
  j 2 143.13o
o
5
 5  26.56 
2 j
3 1  j 
(a)
2

= 0.24 + j0.32
(b)
630o  j 5  3  e j 45  5.196  j 3  j 5  3  0.7071(1  j )
o
= 2.903 + j8.707
P.E. 9.6


P  2sin(10t  x   )a y  2 cos 10t  x     a y , w  10
4
4
2

 Re 2e
i.e. Ps  2e
j ( x  3 )
4
j ( x 3 )
4

a y e jwt  Re  Ps e jwt 
ay
Q  Re  Qs e jwt   Re  e j ( x  wt ) (a x  a z )  sin  y
 sin  y cos( wt  x)(a x  a z )
P.E. 9.7

H
1

1 
  E 
( E sin  )ar 
(rE )a
r sin  
r r
t
263
2 cos 

cos(t   r )ar  sin  sin(t   r )a
2
r
r
2 cos 

H 
sin(t   r )ar 
sin  cos(t   r )a
 r 2
 r
=


c

6 107
 0.2 rad/m
3 108
1
1
H 
cos  sin(6  107  0.2r )ar 
sin  cos(6 107  0.2r )a
2
12 r
120 r
P.E. 9.8
3

E


3c
r r
1

9  10 8
10
 2.846  10 8 rad/s
6
  Hdt  
cos(t  3 y )a


x
6
cos(t  3 y )a x
9  10 109

(5)
10 36
E  476.86 cos(2.846 108 t  3 y )a x V/m
=
Prob. 9.1
V 
8


B
   B  dS  
S
t
t
t
= 3770 sin377t x (0.2)2 x 10-3
= 0.4738 sin377t V
264
Prob. 9.2

,    B  dS  BS
t
B
Vemf  
S
t
V
(4  20sin 20t )(2 104 ) 160
i (t )  emf 

sin 20t (104 )
R
20  30
50
i (t )  0.32sin 20t mA
Vemf  
Prob.9.3
  B S  (0.2) 2  40 103 sin104 t

V 
 16 cos104 t
t
V 16
cos104 t
i 
4
R
 12.57 cos104 t A
Prob.9.4
Measuring the induced emf in the clockwise direction,
Vemf   (u  B ) dl
0
1.2
=  (5a x  0.2a z ) dya y   (15a x  0.2a z )dya x
1.2
0
1.2
0
0
1.2
= -  (1) dy   (3)dy
 1.2  1.2  3  1.2  3.6
 2.4 V
Prob. 9.5
1.6
1.6
y 0
y 0
Vemf   (u  B )  dl   (2a x 10 cos  ya z )  dya y    20 cos  ydy

20sin  y 1.6
20sin1.6

0


265
Prob. 9.6
B
o I
( a x )
2 y
 a
o I a
dzdy o Ia   a
ln




2 z 0 y   y
2
   B  dS 
Vemf  

 
 Ia d


  o uo
[ln(   a )  ln  ]
t
 t
2
d
 o Ia  1
 o a 2 Iu o
1

uo 
 
2
   a   2 (   a)
where   o  u o t
Prob. 9.7
 a
Vemf   3a z 

o I
3 I   a
a  d  a    o ln

2
2
4  10 7
60

 15  3 ln
 9.888V
2
20
Thus the induced emf = 9.888V, point A at higher potential.
Prob. 9.8
V   ( u  B ) dl
u   a ,
B  Bo a z

  Bo  2
1
V    Bo d    Bo  2 
0
2
2
 0

30
 60  103 (8  102 ) 2  5.76 mV
2
Prob. 9.9
Vemf   N


dS
  N  B dS   NB
t
t
dt
d
d
(  )   NB
  NB
dt
dt
 50(0.2)(30  104 )(60)  1.8V
  NB
266
Prob. 9.10
Method 1:
We assume that the sliding rode is on   y  
  x / 3  5t / 3

 25t 2 
2
Vemf   (u  B ) dl   5a x  0.6a z  dya y  3 x  dy  3 x(2)  6  
  86.6025t
 3 

Method 2:
The flux linkage is given by
5t
x. 3
 

0.6 xdxdy  0.6 
x  o y  x / 3
Vemf  
2
125t 3 / 3  28,8675t 3
3
d
 86.602t 2
dt
Prob. 9.11
u
B
B
u

Vemf   (u  B )  dl  uBl cos 
 120 103

m / s   4.3  105  1.6  cos 65o

 3600

o
 2.293cos 65  0.97 mV
Prob. 9.12
Vemf  uB  410  0.4  106  36  5.904 mV
267
Prob. 9.13
d = 0.64 – 0.45 = 0.19, dt = 0.02
d
 0.19 
 10 
  95V
dt
 0.02 
Vemf  N
I
 95 
    6.33 A
R
 15 
Vemf
Using Lenz’s law, the direction of the induced current is counterclockwise.
Prob. 9.14
V   (u  B )  dl , where u   a , B  Bo a z
2
V    Bo d  
1
V
 Bo
2
(  2 2   21 )
60 15
 103 (100  4)  104  4.32 mV
2
Prob. 9.15
J ds  jDs  J ds max  E s  

Vs
d
109 2  20  106  50

36
0.2  10 3
 277.8 A/m2
I ds  J ds  S 
1000
 2.8  10  4  77.78 mA
3.6
Prob. 9.16
Jc   E,
D
E

t
t
| J d |  | E |
Jd 
| J c |  | E |,
If I c  I d , then | J c || J d |
  2 f 
f 


2


4
109
2  9 
36
 8 GHz

   
268
Jc
E



J d E 
Prob. 9.17

(a)


(b)




(c)


2  103
2 109  81
9
10
36
25
109
2 10  81
36
 0.444  103
 5.555
9
2  104
2 109  5 
 7.2  104
9
10
36
Prob. 9.18
Jc   E
Jd  

E
t
| J cS | 


| J ds | 
J cs   Es
 J ds  j Es
4
2  107  81
9
10
36

400(18)
 88.89
81
Prob. 9.19
J d  E 


1
J
E

2 f  12  105

 



104

 12 105
9
10

3
36
f  600 kHz
269
Prob. 9.20
J c   E  0.4 cos(2 108 t )
E
0.4

cos(2  108 t )
E
0.4

(2 108 ) sin(2  108 t )
t

109
0.4  4.5 
36 (2 108 ) sin(2  108 t )

4
10
 100sin(2 108 t ) A/m 2
Jd  
Prob. 9.21
D
 J  Jd
t
Since the regin is source-free, J = 0.
 H  J 

J d    H  x
H x ( z)

y
0

H x
z 
a y   H o (  ) sin(t   z )a y
z
0
  H o sin(t   z )a y
Prob. 9.22
Jd 
D
E

t
t
E
109
25 103 cos103 t (2  5)  2.2105 108 cos103 t
I   J d  dS  J d S 
S
t
36
S
I
 22.1cos103 t nA
Prob. 9.23
(a)
  Es 
s
 ,  Hs  0
  E s  j H s ,
  H s  (  j ) E s
270
(b)
Dx Dy Dz


 v
x
y
z
B By Bz
B  0  x 

0
x
y
z
B
E E y
B
 E  
 z
 x
t
y
z
t
  D  v 
By
Ex Ez


x
t
z
E y E x
Bz


t
y
x
D
H z H y
D


 Jx  x
t
y
z
t
D y
H x H z

 Jy 
z
x
t
H y H x
D z

 Jz 
x
y
t
 H  J 
(1)
(2)
(3)
(4)
(5)
(6)
(7)
(8)
Prob. 9.24
If J  0  v , then
B  0
  D  v
B
 E  
t
D
 H  J 
t
(1)
(2)
(3)
(4)
Since     A  0 for any vector field A ,

B  0
t

 H    D  0
t
showing that (1) and (2) are incorporated in (3) and (4). Thus Maxwell’s equations can be
 E  
reduced to (3) and (4), i.e.
 E  
B
D
,  H 
t
t
271
Prob. 9.25
 E 0


(1)
 H 0


(2)
  E  
H
t

 x

 E  
0



E y
x
H 



y
(3)

z 


0 

E y ( x, t )
a z   Eo sin x cos ta z
Eo
1
  Edt 
sin sin ta


z
o
 H  
E
t

 x

 H = 
0


E



y
0
(4)

z




H z ( x, t ) 


E
H z
a y   o cos x sin ta y
o
x
1
Eo
=
  Hdt 
cos x cos ta


y
o
which is off the given E by a factor. Thus, Maxwell’s equations (1) to (3) are satisfied,
but (4) is not. The only way (4) is satisfied is for o  1 which is not true.
Prob. 9.26
 E  
 E  
B
t


J
2 E
  B     H  
  2
t
t
t
t
272
But
    E  (  E )   2 E
(  E )   2 E   
J
2 E
  2 ,
t
t
J  E
In a source-free region,   E   v /   0 . Thus,
 2 E  
E
2 E
  2
t
t
Prob. 9.27
  J  (0  0  3 z 2 ) sin 10 4 t  
v
t
v      Jdt    3z 2 sin104 tdt 
3z 2
cos104 t  Co
4
10
If  v |z 0  0, then Co  0 and
v  0.3z 2 cos104 t mC/m3
Prob. 9.28
  D    E  v
 v    E =  o
Ez
  o Eo sin z cos t
z
Prob. 9.29
 =108
Let
1
D
E
 0

E     Hdt
t
t




y
z  40  cos(t   x)a z
  H  x
0 40sin(t   x) 0
1
40
sin(t   x)a z
E     Hdt 
 H  J 

 o
273
B
H
 
t
t



x y
z
40 2
cos(t   x)a y
 E 

40
 o
0
0
sin(t   x)
 E  
(1)
 o
H
 40o cos(t   x)a y
t
Equating (1) and (2) gives
 o
40o 
40 2
 2   o  o 2

 o
(2)
    o o  10
m
/
d
a
r
3
3
3
.
0
109 1
4 10 
 
36 3
7
8
Prob. 9.30
D
E 50 o
4.421 102
Jd 
 o

(108 ) sin(108 t  kz )a   
sin(108 t  kz )a  A/m
t
t


  E   o
 E 
H 
E
z
H
t
a 
50k

sin(108 t  kz )a
1
1
50k
cos(10 t  kz )a
  Edt 
4  10 10 
 
7
8
8
o
H
2.5k
cos(108 t  kz )a A/m
2
 H  
  H  Jd
k2 
H
z


2
 4.421 102
2.5
a  

2.5k 2
sin(108 t  kz )a
2
4.421x102



2.5k 2
sin(10 t  kz )a  
sin(108 t  kz )a 
2
8
k  0.333
2
274
Prob. 9.31
  E   o
H
t


H 
1
o 
  Edt
1 
1 
(rE )a 
10sin  cos(t   r ) a
r r
r r
10

sin  cos(t   r )a
r
10 
sin   sin(t   r )dta
H 
r
10

sin  cos(t   r )a
o r
 E =
Prob. 9.32
D


D   J d dt
dt
60 103
D=
cos(109 t   z )a x  60  1012 cos(109 t   z )a x C/m 2
109
H
D
H
 E  

    
t
t




x y z
D 1
1
 
 (60)(1)  1012 sin(109 t   z )a x
 

Dx 0
0
(a) J d 

H 
1
60

 1012 sin(109 t   z )a y

D
dt  
1
(1)
60  1012
 9 cos(109 t   z )a y

10



60 

 1021 cos(109 t   z )a y A/m

275
(b)
  H  J  J  0  J d

x

y

z
Jd    H 

0
Hy
(  )(1)60 
0

 (1021 ) sin(109 t   z )a x
Equating this with the given J d
3
60 10 
60  2

 1021
109 2000
   10  2  4 10 10 

36
9
  14.907 rad/m
2
7
18
Prob. 9.33
  D  v ,
 H  J 
 H   J 
Or
J = -
D
t
  D
0
t

J 
 v
t
 v
0
t
Prob. 9.34
From Maxwell's equations,
  D   v  0,
  B  0,

 H

t
  E =0,
J   E,
    E  
But
 2 E   
 2 E  
D
t

D 
(  E )   2 E     J 

t 
t 
D E
 E  
B
,
t

 E 
E
2 E




E





t 
t 
t
t 2
E
2 E
  2  0
t
t
 H  J 
276
Prob. 9.35
(a)   A  0

x

y

z
 A 

0
E z ( x, t )
0
 E z ( x, t )
ay  0
x
Yes, A is a possible EM field.
(b)
 B  0
 B 
1 
 
10 cos(t  2  ) az  0
Yes, B is a possible EM field.
(c)
C 
C 
1 
 
1 
sin  sin t
3 cot  sin t  
0

 

3
2
 cos  sin t  a z  3 2

(cot  sin t )a z  0

No, C cannot be an EM field.
1

sin(t  5r ) (sin 2  )  0
(d)   D  2
r sin 

 D  
 D
1 
1
(rD )a  sin  (5) sin(t  5r )a  0
ar 
r r
r

No, D cannot be an EM field.
Prob. 9.36
From Maxwell’s equations,
B
 E  
t
D
 H  J 
t

Dotting both sides of (2) with E gives:
(1)
(2)
277
D
(3)

t


But for any arbitrary vectors A and B ,
  ( A  B )  B  (  A)  A  (  B )
Applying this on the left-hand side of (3) by letting A  H and B  E , we get

H  (  E )    ( H  E )  E  J  1
( D  E ) (4)
2 t
From (1),
 B  1 
H  (  E )  H   
  2 (B  H )
t
 t 
Substituting this in (4) gives:


1
(B  H )    (E  H )  J  E  1
(D  E)
2 t
2 t
Rearranging terms and then taking the volume integral of both sides:
E  (  H )  E  J  E 

   ( E  H )dv   t 1 2  ( E  D  H  B)dv   J  Edv
v
v
Using the divergence theorem, we get
W
 ( E  H )  dS   t   J  Edv
s
or
v
W
   ( E  H )  dS   E  Jdv as required.
t
s
v
v
278
Prob. 9.37
  E   o
H
t


  E  x
0

y
0

E
E
z
 z ax  z a y
y
x
E z ( x, y )
H 
1
o 
  Edt
  cos(12 x) cos(1011 t   y )a x  12 sin(12 x) sin(1011 t   y )a y
H 

1
o 
  Edt
1 
 cos(12 x)  cos(1011 t   y )dta x  12 sin(12 x)  sin(1011 t   y )dta y 


o


o10
But
11
cos(12 x) sin(1011 t   y )a x 
 H 
12
sin(12 x) cos(1011 t   y )a y
10
o10
D
E
 o
t
t
E
  o (1011 ) cos(12 x) cos(1011 t   t )a z
t



 H y H x 
x
y
z  
 H 

 az
x
y 

H x ( x , y ) H y ( x, y ) 0
o
(1)
 (12 ) 2

2
11
s(10
t
y
)
cos(12 x) cos(1011 t   y )  a z



x
cos(12

)
co

11
11
o10
 o 10

Equating (1) and (2),
 o (1011 ) 
2
(12 ) 2

o1011 o1011

o o (1022 )  (12 ) 2   2
109
106
 144 2   2
   331.2
(1022 ) 
36
9
12
12 1011

331.2  1011
3



10
,

 3  103
2.636
11
7
11
7
o10
4  10
o10
4  10
4  107 
Thus,
H  2.636 cos(12 x) sin(1011 t   y )a x  3sin(12 x) cos(1011 t   y )a y mA/m
(2)
279
Prob. 9.38
  E   o
H
t

  E  x
Ex ( z )


y
0
H 
1
o 
  Edt

E
z  x a y    Eo cos(1200 t   z )a y
z
0
E
H
  o o (1200 ) cos(1200 t   z )a y

t
Setting (1) and (2) equal,
 o
 Eo 
o Eo
(1200 )

E
t

y
H y ( z)


(1)
(2)
1200o
(3)

But   H   o

  H  x
0

H y
 Eo
z  
cos(1200 t   z )a x
ax 

z
0
E
  o (1200 ) Eo cos(1200 t   z )a x
t
Setting (4) and (5) equal,
 = o (1200 )
(6)
o
From (3) and (6),
1200o
  o (1200 )



2 
(5)
o
o
o
4 107

 120  377
109
o
36
From (3),

1200o

1200  4 107

 40  107  1.257 105 rad/m
120
Prob. 9.39 Using Maxwell’s equations,
 H   E 
But
E
t
(  0)


E
1

  Hdt
(4)
280
 H  
E
12sin 

1  H
1 
12sin 
ar 
(rH )a 
 sin(2 108 t   r )a
r sin  
r r
r
  sin(2 108 t   r )dta
o
12sin 
 cos(t   r )a ,
 o r
  2 108
Prob. 9.40
With the given A, we need to prove that
2 A
2
 A   2
t
2
 A   ( j )( j ) A   2  A
Let  2   2  , then  2 A    2 A is to be proved. We recognize that
A
o jt  j r
e e az
4 r

e j r
,
A  o e jt  a z
4
r
1  2
  1   2   j  1   j r 
 2  2
(r sin 
) 
(r ) 
 2 e


r sin   r
dr  r 2  r
r 
 r

Assume

 j r
1
2
 j r
2 e
r
j
j
e









   2


r2
r
2 A   2 A
Therefore,
We can find V using Lorentz gauge.
1
1
  Adt 
 A
V

 o o
jo o


  o  j  r jt 
1
  j  1   j r jt
e e 
 2  e e cos 


jo o r  4 r
r 
 j o (4 )  r
V
1
cos  
1  j (t   r )
 j   e
j 4 o r 
r
Prob. 9.41
Take the curl of both sides of the equation.
281

 A
t
But   V  0 and B =  A. Hence,
  E    V 
B
t
which is Faraday's law.
 E = 
Prob. 9.42
 A 
(a)
Hence,
Az x
 ,
z c
V
  xc,
t
V
  A =  o o
t
E  V 
(b)
  o o
x
x
V
 2c
c
t c
A
V 
 V
 
ax 
a z   xa z  ( za x  xa z )  xa z
t
z 
 x
  za x
E
Prob. 9.43
 A  0   
V
t

 V  constant
A
 0  Ao cos(t   z )a x
t
  Ao cos(t   z )a x
(a) E  V 
(b) Using Maxwell’s equations, we can show that
    o o
Prob. 9.44
(a)
z  430o  1050o  3.464  2 j  6.427  j 7.66  2.963  j 5.66
 6.389  117.64o
z1/ 2  2.5277  58.82o
(b)
1  j2
2.236  63.43 o
2.236  63.43 o


6  j 8  7  15 o 6  j 8  7 .761  j1.812 9.841 265.57 o
 0.2272  202.1o
282
(c)
(5 53.13 o ) 2
25 106 .26 o
z

12  j7  6  j10 18.028   70.56 o
 1.387  176 .8 o
(d)
1.897   100 o
o
o
o  0.0349   68
(576
.  90 )(9.434  122 )
Prob. 9.45
(a) H  Re  H s e jt  ,
  106
H  Re  10e j (t  /3) a x 

H s  10e j /3a x
(b) E s  4 cos(4 y )e  j 2 x a z
(c) sinA  cos(A  90o )
D  5cos(t   / 2   / 3)a x  8cos(t   / 4)a y
Ds  5e  j /6 a x  8e  j / 4 a y
Prob. 9.46
(a)
As  10e j /2 a x  20e  j /2 a y
A  Re[ As e jt ]  Re 10e j (t  /2) a x  20e j (t  /2) a y 
 10 cos(t   / 2)a x  20 cos(t   / 2)a y
A  10sin ta x  20sin ta y
(b)
B  Re[ Bs e jt ]  Re  4e j (t  2 x  /2) a x  6e j (t  2 x ) a y   4 cos(t  2 x   / 2)a x  6 cos(t  2 x)a y
 4sin(t  2 x)a x  6sin(t  2 x)a y
(c)Cs  2e j /2 e 10 z e  j /4 a z
C  Re[C s e jt ]  2 Re[e j (t  / 2 /4) e 20 z a z ]  2 e 20 z cos(t   / 4   / 2)a z
 2 e 20 z sin(t   / 4)a z
283
Prob. 9.47
1
e  j 3 z a
(a)
Hs 
(b)
  H  o

E
t
H 

E
1
o 
  Hdt
1 
3
(  H  )a z   sin(t  3 z )a   0
 

z
1 3
3
E    sin(t  3z )a  dt 
cos(t  3 z )a  V/m
 H  
But
a 
o 
 o
 E 
(c )
E
z
a 
9
 o
sin(t  3 z )a
B
H o
  o

sin(t  3 z )a
t
t


B
9
 E  

 o
t
 o


2 
9

 o o

3
 o o
 3  3 108  9 108 rad/s
Prob. 9.48
We can use Maxwell’s equations or borrow ideas from chapter 10.

1 120

 o


r
9
Ho 
Eo


10  9
 0.2387
120
    

c
r 
2 109
81  60  188.5 rad/m
3  108
284
Prob. 9.49
  H s  j o E s

  H s  x
0


y
z   j12 e j x a y
0 12e j x
Es 
  Hs
12 j x

e ay
j o
 o
But
  E s   jo H s

  E s  x
0
Hs  

y
Esy ( x)

E
12 2 j x
z  sy a z   j
e az
x
 o
0
  Es
12 2 j x
 2
e az
jo  o o
Equating this with the given H,
12=
12 2
 2 o o

 2   2  o o

 2  (109 ) 2  4 107 
109
1016
 (109 ) 2 
36
9
10
 3.333
3
12 j x
12(10 / 3) j10 x /3
e ay  
e
Es  
a y  40(36 )e j10 x /3a y  4.533e j 3.33 x a y kV/m
9
10
 o
109 
36

Prob. 9.50
( j ) 2 Y  4 jY  Y  20o ,
 3
Y ( 2  4 j  1)  2
Y
2
2
2


 0.0769  j 0.1154
  4 j  1 9  j12  1 8  j12
2
 0.1387  123.7 o
y (t )  Re(Ye jt )  0.1387 cos(3t  123.7 o )
Prob. 9.51 We begin with Maxwell’s equations:
  D  v /   0,
B
 E  
,
t
B  0
 H  J 
D
t
285
We write these in phasor form and in terms of Es and Hs only.
  Es  0
(1)
  Hs  0
(2)
  E s   j H s
(3)
  H s  (  j ) E s
(4)
Taking the curl of (3),
    E s   j  H s
(  E s )   2 E s   j (  j ) E s
 2 E s  ( 2   j ) E s  0


 2 Es   2 Es  0
Similarly, by taking the curl of (4),
    H s  (  j )  E s
(  H s )   2 H s   j (  j ) H s
 2 H s  ( 2   j ) H s  0


2 H s   2 H s  0
286
CHAPTER 10
P. E. 10.1 (a)
T
2


2
 31.42 ns,
2  108
  uT  3 108  31.42 109  9.425 m
k    2  /   0.6667 rad/m
(b) t1 = T/8 = 3.927 ns
(c )
H (t  t1 )  0.1cos(2  108

8  108
 2 x / 3)a y  0.1cos(2 x / 3   / 4)a y
as sketched below.
P. E. 10.2 Let xo 
1  ( /   ) 2 , then

 o o

 r  r ( xo  1) 
2
c
or
xo  1 
xo 2 
81
 1  ( /   ) 2
64
16
xo  1
2
 c 1/ 3  3  108
1


8
 8
10 8
8
xo  9 / 8

 0.5154

287
tan 2   0.5154
   13.63 o



xo  1

xo  1
(a)
   17 
17
17
 1.374 rad/m
3

 0.5154

 /  120  2 / 8
(c ) |  | 

 177 .72
xo
9/8
(b)
  177 .72 13.63 o 
(d)
u
 108

 7.278 107 m/s
 1.374
(e) a H  ak  a E
H


az  ax  aH


aH  a y
0.5  z / 3
e sin(108 t   z  13.63o )a y  2.817e  z / 3 sin(108 t   z  13.63o )a y mA/m
177.5
P. E. 10.3 (a) Along -z direction
(b)  
2
 2  / 2  3.142 m

10 8

f 

 15.92 MHz
2
2
        o o  r  r 

c
or  r   c /  
3 108  2
6
108
(c )    0 ,|  | 
 / 
ak  a E  a H
(1)  r
 o /  o 1/  r 
a z  a y  a H
 r  36
120 
 20 
6
aH  ax
288
H
50
sin(t   z )a x  795.8sin(108 t  2 z )a x mA/m
20
P. E. 10.4 (a)



 
 
102
109
10   4 
36
 0.09
9
 
2
 
1  

109 

r  r
(2)(0.09)  0.9425 Np/m
1  
  1 
2  2   
 2  3 108
 2c
 
2
 109 
1  


1
1
2[2  0.5(0.09) 2 ]  20.965 rad/m




8
2  2   
 3  10
E  30e 0.9425 y cos(109  t  20.96 y   / 4)a z
At t = 2ns, y = 1m,
E  30e 0.9425 cos(2  20.96   / 4)a z  2.844a z V/m
(b)  y  10 o 
10 
rad
180
or
y
 1


 8.325 mm
18  18  20.965
(c ) 30(0.6) = 30 e y
y
1
1
1
ln(1 / 0.6 ) 
ln
 542 mm

0.9425 0.6
(d)
|  |
 /
60

 188.11 
1
1.002
2
[1  (0.09) ]
4
289
2   tan 1 0.09
   2.571o
a H  ak  a E  a y  a z  a x
H
30 0.9425 y
e
cos(109  t  20.96 y   / 4  2.571o )a x
188.11
At y = 2m, t = 2ns,
H  (0.1595)(0.1518) cos(34.8963rad )a x  22.83a x mA/m
P. E. 10.5
w
w

0 0
0
0
I s    J xs dydz  J xs (0)  dy  e  z (1 j ) /  dz 
| Is|
J xs (0) w
1 j
J xs (0 ) w
2
P. E. 10.6 (a)
Rac
1.3  103
a a
 f  
 107  4 107  3.5 107 = 24.16


2
Rdc 2 2
(b)
Rac 1.3  103
  2 109  4 107  3.5 107  341.7

2
Rdc
P. E. 10.7
E  Re[ E s e jt ]  Re  Eo e jt e  j  z a x  Eo e  j / 2 e jt e  j z a y 
 Eo cos(t   z )a x  Eo cos(t   z   / 2)a y
 Eo cos(t   z )a x  Eo sin(t   z )a y
At z = 0, Ex  Eo cos t , E y  Eo sin t
2
2
 E   Ey 
  x    1
cos t  sin t  1 
 Eo   Eo 
which describes a circle. Hence the polarization is circular.
2
2
290
P. E. 10.8
1
2
Pave   H o a x
2
(a) Let f(x,z) = x + y –1 = 0
an 
ax  a y
f
,

| f |
2
Pt   P.dS =P.San =
1
=
2 2
dS = dSan
a a
1
 H o 2ax . x y
2
2
(120  )(0.2) 2 (0.1) 2  53.31 mW
(b) dS = dydzax , Pt   P.dS =
Pt 
1
(120  )(0.2) 2  (0.05) 2  59.22 mW
2
P. E. 10.9

1
 Ho 2 S
2
1  o  120 , 2 
 o


2
2 2
  1
 2 / 3,   2
 1/ 3
2  1
2  1
Ero   Eio  
Ers  
10
3
10 j 1z
e a x V/m
3
where  1   / c  100  / 3 .
20
Eto   Eio 
3
Ets 
20  j 2 z
e
a x V/m
3
where  2    r / c  2 1  200  / 3 .
291
P. E. 10.10
1  0, 1 
2

 2

c
r  r 
0.1
109
7.5  10  4 
36
2
 5 
   5c / 2  7.5  108
c
 1.2
8
2 

c
4
2
 1  1.44  1  6.021
2 

c
4
2
 1  1.44  1  7.826
| 2 | 4
2
60 
1  1.44  2
2
 95.445,  1  120   r 1  754
tan 2  2  1.2      2  37 .57 o
 2  95.445 37 .57 o
(a)
 
 2   1 95.445 37 .57 o  754

 0.8186  17108
. o
 2   1 95.445 37 .57 o  754
  1    0.2295 33.56 o
s
(b)
1 |  | 1  0.8186

 10.025
1 |  | 1  0.8186
Ei  50sin(t  5 x)a y  Im( Eis e jt ) , where Eis  50 e  j 5 x a y .
Ero   Eio  0.8186 e j 171.08 (50 )  40.93e j 171.08
o
o
Ers  40.93e j 5 x  j171.08 a y
o
Er  Im( Ers e jt )  40.93sin(t  5 x  171.1o )a y V/m
a H  a k  a E  a x  a y   a z
292
Hr  -
40.93
sin(t  5 x  171.1o )a z  0.0543sin(t  5 x  171.1o )a z A/m
754
(c )
o
o
Eto   Eio  0.229 e j 33.56 (50 )  11.475e j 33.56
Ets  11.475e  j 2 x  j 33.56 e  2 x a y
o
Et  Im( Ets e jt )  11.475e 6.021x sin(t  7.826 x  33.56o )a y V/m
a H  ak  a E  a x  a y  a z
Ht 
11.495 6.021x
sin(t  7.826 x  33.56o  37.57o )a z
e
95.445
 0.1202e 6.021x sin(t  7.826 x  4.01o )a z A/m
(d)
2
2
E
E
1
[502 a x  40.932 a x ]  0.5469a x W/m2
P1ave  io a x  ro (a x ) 
21
21
2(240 )
2
E
(11.475) 2
cos 37.57 o e 2(6.021) x a x  0.5469e 12.04 x a x W/m2
P2ave  to e 2 2 x cos 2 a x 
2 | 2 |
2(95.445)
P. E. 10.11 (a)
k  2a y  4a z 
 k  22  42  20
  kc  3 108 20  1.342 109 rad/s ,
  2 / k  1.405m
(b) H 
ak  E
o

(2a y  4a z )
20(120 )
 (10a y  5a z ) cos(t  k.r )
 29.66 cos(1.342 x109 t  2 y  4 z )a x mA/m
293
(c ) Pave 
| Eo |2
125 (2a y  4a z )
ak 
 74.15a y  148.9a z
2o
2(120 )
20
mW/m2
P. E. 10.12 (a)
y
ki
i
kt
r
z
kr
tan  i 
sin  t 
kiy
kiz

2
   i  26 .56   r
4
1 1
2 2
sin  i 
(b)  1   o ,  2   o / 2
1
sin 26 .56 o    t  12.92 o
2
E is parallel to the plane of incidence. Since  1   2   o ,
we may use the result of Prob. 10.42, i.e.
 \\ 
tan( t   i ) tan(  13.64 o )

  0.2946
tan( t   i ) tan(39.48 o )
 \\ 
2 cos 26 .56 o sin 12.92 o
 0.6474
sin 39.48 o cos(  13.64 o )
(c) kr   1 sin  r a y  1 cos  r a z .
Once kr is known, Er is chosen such that
kr .Er  0 or .Er  0. Let
Er   Eor ( cos  r a y  sin  r a z ) cos(t  1 sin  r y  1 cos  r z )
Only the positive sign will satisfy the boundary conditions. It is evident that
294
Ei  Eoi (cos i a y  sin i a z ) cos(t  2 y  4 z )
Since  r   i ,
Eor cos  r   / / Eoi cos  i  10  / /   2.946
Eor sin  r   / / Eoi sin  i  5 / /   1.473
 1 sin  r  2,  1 cos  r  4
i.e.
Er   (2.946 a y  1.473az ) cos( t  2 y  4 z)
E1  Ei  Er  (10a y  5a z ) cos(t  2 y  4 z )  (2.946a y  1.473a z ) cos(t  2 y  4 z )
V/m
(d) kt    2 sin t a y   2 cos t a z .
Since k r  Er  0 , let
Et  Eot (cos t a y  sin t a z ) cos(t   2 y sin t   2 z cos t )
 2    2  2   1  r 2  2 20
sin  t 
1
1
sin  i 
,
2
2 5
 2 cos  t  2 20
cos  t 
9
20
19
 8.718
20
Eot cos  t   / / Eoi cos  t  0.6474 125
19
 7 .055
20
Eot sin  t   / / Eoi sin  t  0.6474 125
1
 16185
.
20
Hence
E2  Et  (7.055a y  1.6185a z ) cos(t  2 y  8.718 z ) V/m
(d) tan  B / / 
2
 2    B / /  63.43 o
1
295
P.E. 10.13
Si =
1  0.4 14
.

= 2.333
1  0.4 0.6
So =
1  0.2 12
.

= 1.5
1  0.2 0.8
Prob. 10.1 (a) Wave propagates along +ax.
(b)
2
 1 s
 2 106
2 2


 1.047m

6
T
u
2

 2 106

 1.047 106 m/s

6
(c ) At t=0, E z  25 sin(  6 x )   25 sin 6 x
At t=T/8, E z  25 sin(

2 T
 6 x )  25 sin(  6 x )
T 8
4
At t=T/4, E z  25 sin(
At t=T/2, E z  25 sin(
2 T
 6 x )  25 sin(  6 x  90 o )  25 cos 6 x
T 4
2 T
 6 x )  25 sin(  6 x   )  25 sin 6 x
T 2
These are sketched below.
296
Prob. 10.2
c 3  108
 5  106 m
(a)   
f
60
3  108
 150 m
2  106
3  108
 2.5 m
(c )  
120  106
3  108
 0.125 m
(d)  
2.4  109
(b)  
Prob. 10.3
(a)   108 rad/s
(b)  
(c )  


c
2

108
 0.333 rad/m
3 108
 6  18.85 m
(d) Along -ay
At y=1, t=10ms,
1
(e) H  0.5cos(108 t 10 109   3)  0.5cos(1  1)
3
 0.1665 A/m
297
Prob. 10.4
(a)
E
  sin( x  t )  sin( x  t )
x
2 E
  cos( x  t )  cos( x  t )   E
x 2
E
  sin( x  t )   sin( x  t )
t
2 E
  2 cos( x  t )   2 cos( x  t )   2 E
t 2
2
2 E
2  E
u
  2 E  u 2 E  0
2
2
t
x
2
if   u 2 and hence, eq. (10.1) is satisfied.
(b) u  
Prob. 10.5 If
 2  j  (  j  )    2    j  
| 2 |
i.e.
(  2   2 )  4 2  2 
and     j , then
( 2   2 ) 2   2   2
 2   2    ( 2   2  2 )
(1)
Re(  )         
2
2
2
2
 2   2   2 
(2)
Subtracting and adding (1) and (2) lead respectively to

2

 
  
 1     1
 
2 



2

 
  
 1     1
 
2 


(b) From eq. (10.25), E s ( z )  Eo e z a x .
298
  E   j H s
Hs 
But H s ( z )  H o e  z a y , hence Ho 

j

  Es 
j

( Eo e z a y )
Eo
j

E

 o
j 

(c) From (b),
j 

j  (  j  )

 /
||
4
  
1  
 
j 

  j 
 
, tan 2    
  
2
 /
1 j
1





Prob. 10.6 (a)
From eq. (10.18),
 = j (  j ) 
Assuming that


j
j 

j j (1 
)  j  1 


  
1, we include up to the second power in
neglect higher-order terms.



2

  2 
2 
1 
  j  1  2 2     j 
   2 
 8  
Thus,

2 
 =  1  2 2 
 8  
Prob. 10.7
(a)



8  102
109
2  50 10  3.6 
36
6
1/2
8

and

299
 
 
 
2

2  50 106
 


1 
1
=

2 
3  108

  


2.1 3.6
[ 65  1]  5.41
2
 
2

 

 1 
 = 6.129

1

2 

  


    j  5.41  j6 .129 /m
2
2

 1025
m
.

6 .129
(b)

(c)
 2  50 106
u 
 5.125 107 m/s

6.129
(d) |  | 


  
4 1 

 
tan 2  
2.1
3.6
 101.4
4
65
120 
2


 8      41.44 o

  101.41 41.44 o 
(e)
H s  ak 
Es

o
6
6
 a x  e  z a z   e z a y  59.16e j 41.44 e z a y mA/m


Prob. 10.8
(a) tan  



102
 1.5
109
2 12  10 10 
36
4
10
 3.75  102
(b) tan  
9
10
2 12  106  4 
36
6
300
(c ) tan  
4
109
2  12  10  81
36
 74.07
6
Prob. 10.9
(a)
 
2
 2  15  109 1 9.6
 

 1  9  108  1
 1 

1


8


2 

3
10
2






 
1

 100 4.8   9 108   0.146
2

1
   6.85 m

(b) A     0.146  5  103  0.73  103 Np
Prob. 10.10
If  = , the loss tangent is
 
But

 
2

 
 
 
 1  
 

1
1  12  1  
2  1


2 
 
2 
2 




2 2 c
2 c

 
o 

2 c

o
2
 

1


2 1 
o
2  3  108
12  102
 
1
109
 4 107  4 
(0.6436)  47.66 Np/m
2
36
2
 2 c 
 
 1  
  1 
o
2
2 

  


2 1
2  3  108 1
109
7




4

10
4
(1.5538)
12  102
2
36
 115.06 rad/m

 2 c
2  3  108

 1.3652  108 m/s
u 
2
 o  12 10 115.06
Prob. 10.11
For silver, the loss tangent is
301



6.1 107
 6.1 18  108
9
10
2 108 
36
Hence, silver is a good conductor
For rubber,



1015

1
18
 1014
3.1
1
10
2 10  3.1
36
Hence, rubber is a poor conductor or a good insulator.
9
8
Prob. 10.12

4

 9, 000  1
5
 2 10  80 109 / 36
  

2
2  105
 5 105 m/s
0.4
(a)
u / 
(b)
  2 /  
(c )   1 /  
2  105
 4  107  4  0.4
2

2
5m
0.4 
1
 0.796 m
0.4 
(d)   |  |    ,    45 o


|  |
4
  0.444345o
Prob. 10.13 (a)
 
1 

  


2
 
4 107  2  105

 0.4443
 
4
302
T  1 / f  2 /  
x
(b) Let
  x  1


  x  1
But
  
1  
 


 x  1


 x  1
2
1/ 2

c
x 1 
2
 20 ns
 x10 8
 r r
2
x 1
c
0.1 3  108

 0.06752 
 x  1.0046
r  r
 108 2
2
1/ 2
  2 /  
 2.0046 


 0.0046 
1/ 2
0.1  2.088
2
 3m
2.088
 /
||
x
  
1     10046
.
 
x

377
 188.1
2 10046
.
(c )
2

 0.096  tan 2       274
. o

  188.1 274
. o

Eo   H o  12  188.1  2257.2
a E  a H  ak 
 a E  a x  a y 
 aE  az
E  2.257e 0.1 y sin(  108 t  2.088 y  2.74o )a z kV/m
303
(d) The phase difference is 2.74o.
Prob. 10.14
This is a lossy medium in which =o.
 
Let x  

  
o
j j 2  109  4


 35.3126.57

100  j 200
Eo  0.05  35.31  1.765
2
a E  a H  a k  a z
Thus, we obtain
E = -1.765cos (2 109 t  200 x  26.57o )a z V/m
 r (1/ 3) 
r
 c 100  3 108 15


2 109


1
 4.776
3
tan 2 


 4

 3
 r  14.32


  26.57 o
377
 14.32  77.175
|  | 4
1 x
5/3
Eo |  | H o  77.175  50 103  3.858
 /
a E  ( a k  a H )   (a x  a y )   a z
E  3.858e 100 x cos(2 109 t  200 x  26.57 o )a z V/m
304
Prob. 10.15

=

1
109
2  10  4 
36
=4.5
9
 
 
2

 
 1  

1

 
2 




 2  10
9
4
109 
7
1  4.52  1
 10  4  9 

2
36 
 20 2[ 21.25  1]  168.8 Np/m
 
 
2

 
 1  
  20 2[ 21.25  1]  210.5 rad/m
1

2 
 




tan 2 
|  |

 4.5

   38.73o

 /
120 9 / 4
 
4 1


  
2

4
 263.38
1  4.52
  263.3838.73o 
u
 2 109

 2.985 107 m/s

210.5
Prob. 10.16
  6.5   o o 
(a)

c
   c  6.5  3 10  1.95 109 rad/s
8
2
 0.9666 m
 6.5
(b) For z=0, Ez  0.2 cos t

2

2 
)  0.2 cos t
 2
The two waves are sketched below.
For z= /2, Ez  0.2 cos(t 
305
300
z = 0
z = /2
Amplitude (mV / m)
200
100
0
-100
-200
-300
-3
-2
-1
0
1
Time t (ns)
(c) H  H o cos(t  6.5 z )a H
Ho 
Eo
o

0.2
 5.305  104
377
a E  a H  ak

 ax  aH  az

 aH  a y
H  0.5305cos(t  6.5 z )a y mA/m
Prob. 10.17

2


2
 
 o  o r
o

 r

o o
r 
o 6.4

 2.286
 2.8

 r  5.224
2
3
306
Prob. 10.18 (a) Along -x direction.
(b)   6,
    
  2 108 ,

c
 r r
6  3 108
9
r  c /  
2  108
   o r 


 r  81
109
 81  7.162 1010 F/m
36
(c )    /   o /  o r /  r 
120
 41.89 
9
Eo  H o  25 103  41.88  1.047
a E  a H  ak 
 a E  a y  a x 
 aE  az
E  1.047 sin(2  108 t  6 x)a z V/m
Prob. 10.19 (a)



106
109
2 107  5 
36
 3.6 104  1
Thus, the material is lossless at this frequency.
(b)     

2 107
5  750  12.83 rad/m
3 108
2
2

 0.49 m

12.83
(c ) Phase difference =  l  25.66 rad
(d)    /   120
r
750
 120
 4.62 k
r
5
307
Prob. 10.20
(a)

108 

  1.0472 rad/m
     
8
c 3 10
3
(b)
E 0

 sin(108  to   xo )  0  sin(n ), n  1, 2,3,...
108  to   xo  
108   5 103 

3
xo  


xo
5 105 m
(c )
H  H o sin(108  t   x)a H
50 103
Ho 

 132.63  A/m

120
a H  a k  a E  a x  a z  a y
Eo
H  132.63sin(108  t  1.0472 x)a y  A/m
Prob. 10.21
This is a lossless material.


 377 r  105

r

u
(1)
c




 7.6 107
  
r  r
(2)
From (1),
r 105

 0.2785
 r 377
(1)a
From (2),
1
r  r

7.6  107
 0.2533
3  108
(2)a
Multiplying (1)a by (2)a,
1
r
 0.2785  0.2533  0.07054
Dividing (1)a by (2)a,
r 
0.2785
 1.0995
0.2533

  r  14.175
308
Prob. 10.22
ax
ay


 E 
x
y
(a)
Ex ( z, t ) E y ( z, t )
az
E
E

  y ax  x a y
z
z
z
0
 6  cos(t   z )a x  8 sin(t   z )a y
But   E   
H
6

H
t

 H 
sin(t   z )a x 
8

1

  Edt
cos(t   z )a y
2 f
2  40 106
4.5 
4.5  1.777 rad/m
c
3  108
2
2

 3.536 m

 1.777
(b)     

 120

 177.72 

4.5
u
1
c
3 108


 1.4142 108 m/s
4.5
4.5

Prob. 10.23
(a) E  Re[ E s e jt ]  (5a x  12a y )e 0.2 z cos(t  3.4 z )
At z = 4m, t = T/8,  t 
2 T 

T 8 4
E  (5a x  12a y )e 0.8 cos( / 4  13.6)
| E | 13e 0.8 | cos( / 4  13.6) | 5.662 V/m
(b) loss    z  0.2(3)  0.6 Np. Since 1 Np = 8.686 dB,
loss = 0.6 x 8.686 = 5.212 dB
309
(c ) Let
  
1  
 
x
  x  1


  x  1
1/ 2
 0.2 / 3.4 
x 1
 1 / 289
x1
 
    / 2 x  1 
r
2

c
 x 1
|  |
2

o 1
.
o r
x

c
1
17
.
x  100694
r /2 x 1
0.2  3 108
 7.2
108 0.00694



 r  103.68
120
 36.896
103.68  1.00694

 x 2  1  0.118

  36.8963.365o 
tan 2  
 
   3.365 o
Prob. 10.24
  

2
 |  |  
|  |
2 2
2 122

 

 36.48
 2 106  4 107

45o


2 106  4  107

 0.4652

36.48
Eo |  | H o  0.4652  20  103  9.305  103
a E  a H  a k  a y  (  a z )  a x
E  Eo e  z sin(t   z )a E
 9.305e 12 z sin(2  106 t  12 z  45o )a x mV/m
310
Prob. 10.25 For a good conductor,
(a)



102
109
2  8  10 15 
36

  1,

 1.5
say

 100



lossy
6
No, not conducting.
(b)
(c )



0.025



25
109
2  8  10  16 
36
No, not conducting.
 3.515


lossy
 694.4


conducting
6
109
2  8  10  81
36
Yes, conducting.
6
Prob. 10.26
  
But
1

u

   0.02  103  2  100 106  4 103  1.256 104 m/s

Prob. 10.27 (a)
Rdc 
l
l
600


 2.287 
2
7
 S  a 5.8 10    (1.2) 2 106
l
.
 2  a
(see Table 10.2).
(b) Rac 
Rac 
(c )
At 100 MHz,   6.6 103 mm =6.6  10-6 m mm for copper
600
 207.61 
5.8  10  2  (1.2 103 )  6.6 106
Rac
a

1
Rdc 2
7


66.1103
  a/2
f
311
f 
66.1 2 103 66.1 2

a
1.2


f  12.137 kHz
Prob. 10.28
(a) Copper is a good conductor.
   f    1010  4 107  5.8 107  2 105 5.8
  1.513 106 Np/m
(b)
(c )
1
 6.609 107 m
6
 1.513 10
1 j
1 j


 26.09(1  j ) 103 
7
7
 5.8 10  6.609 10

Prob. 10.29
1

 f 
f 
1



f 
1
 
2
1
 1.038 kHz
4  10    4  107  6.1 107
6
Prob. 10.30

1
1
1


 8.531 mm
 f 
  5.8 107 (60)4 107 2 5.8(60)
312
Prob. 10.31
This is a good conductor.

45o


2  12 106  4 107
24.6 


16 2 (0.6)

16 (0.6)
 0.1566 S/m
24.62
1
1
   
 8.333 rad/m
 0.12
2

 2 (0.12)  0.754 m
2



u     2 12 106  0.12  9.05 106 m/s

Prob. 10.32



0.12
109
2  2.42  10  5.5 
36
This is a lossy material.
 0.1623
9
 
2

 

 1 

1

 
2 




 
 2  2.42 109
4
109 
1  0.16232  1
 107  5.5 


2
36
 15.21(109 )(108 ) 0.01308
 17.39

1

 57.5 mm
Prob. 10.33
t  5 
5
5

 2.94 106 m
9
7
7
 f 
 12  10  4 10  6.110
313
Prob. 10.34



4
109
2  2  10  24 
36
 1.5
9
 

 
 1  
  2  2  109

1


2 





 130.01 Np/m
 
2
109
4  10  24 
36  1  1.5 2  1


2
7
105 Eo  Eo e  d
Taking the log of both sides gives
-5ln10   d

 d
5ln10


5ln10
 0.0886 m
130.01
Prob. 10.35
(a) Linearly polarized along az
(b)   2 f  2  107


      o o  r 
(c )
Ho 

c
 c 3  3 108

 14.32
2  107

H  H o sin(t  3 y )a H
r 
Let
f  107  10 MHz
Eo

,

r

  r  205.18
 o 120


 26.33

 r 14.32
12
 0.456
26.33
a H  ak  a E  a y  a z  a x
(d) H o 
H  0.456sin(2 107 t  3 y )a x A/m
Prob. 10.36
E  (2a y  5a z ) sin(t   x)
The ratio E y / Ez remains the same as t changes. Hence the wave is linearly polarized
314
Prob. 10.37
(a)
Ex  Eo cos(t   y ),
E y  Eo sin(t   y )
Ex (0, t )  Eo cos t

 cos t 
Ex (0, t )
Eo
E y (0, t )  Eo sin t

 sin t 
E y (0, t )
Eo
2
2
 E   Ey 
cos t  sin t  1 
  x    1
 Eo   Eo 
Hence, we have circular polarization.
2
2
(b)
Ex  Eo cos(t   y ),
E y  3Eo sin(t   y )
In the y=0 plane,
Ex (0, t )  Eo cos t

 cos t 
E y (0, t )  Eo sin t

 sin t 
2
Ex (0, t )
Eo
 E y (0, t )
3Eo
2
 E  1 E 
cos t  sin t  1 
  x    y  1
 Eo  9  Eo 
Hence, we have elliptical polarization.
2
2
Prob. 10.38
(a) We can write
E  Re( Es e jt )  (40a x  60a y ) cos(t  10 z )
Since E x / E y does not change with time, the wave is linearly polarized.
(b) This is elliptically polarized.
315
Prob. 10.39
(a) The wave is elliptically polarized.
(b)
Let E  E1  E2 ,
where E1  40 cos(t   z )a x ,
E2  60sin(t   z )a y
H1  H o1 cos(t   z )a H 1
H o1 
40
o

40
 0.106
120
a H 1  ak  a E  a z  a x  a y
H1  0.106 cos(t   z )a y
H 2  H o 2 sin(t   z )a H 2
H o1 
60
o

60
 0.1592
120
a H 2  ak  a E  a z  a y  a x
H 2  0.1592sin(t   z )a x
H  H1  H 2  159.2sin(t   z )a x  106 cos(t   z )a y mA/m
Prob. 10.40
We can write Es as
E s  E1 ( z )  E2 ( z )
where
1
Eo (a x  ja y )e  j  z
2
1
E2 ( z )  Eo (a x  ja y )e  j z
2
We recognize that E1 and E2 are circularly polarized waves. The problem is therefore proved.
E1 ( z ) 
Prob. 10.41
(a)
When   0,
E ( y, t )  ( Eo1a x  Eo 2 a z ) cos(t   y )
The two components are in phase and the wave is linearly polarized.
(b)
When    / 2,
Ez  Eo 2 cos(t   y   / 2)   Eo 2 sin(t   y )
We can combine Ex and Ez to show that the wave is elliptically polarized.
316
(c )
When    ,
E ( y, t )  Eo1 cos(t   y )a x  Eo 2 cos(t   y   )a z
 ( Eo1 a x  Eo 2 a y ) cos(t   y )
Thus, the wave is linearly polarized.
Prob. 10.42
Let E s  Er  jEi
H s  H r  jH i
and
E  Re(E s e jt )  E r cos t  Ei sin t
Similarly,
H  H r cos t  H i sin t
1
P  E  H  Er  H r cos 2 t  Ei  H i sin 2 t  ( Er  H i  Ei  H r ) sin 2t
2
T
T
T
T
1
1
1
1
Pave   Pdt   cos 2  dt ( Er  H r )   sin 2  dt ( Ei  H i ) 
sin 2 dt ( Ei  H i  Ei  H r )
T 0
T 0
T 0
2T 0

1
1
( Er  H r  Ei  H i )  Re[( Er  jEi )  ( H r  jH i )]
2
2
Pave 
1
*
Re(E s  H s )
2
as required.
Prob. 10.43
(a)
      o o  r 
r 

c
r
 c 8  3 108

 2.4

109
 r  5.76
(b)  
o 1
377



 157.1 

 o  r 2.4
(c ) u 
 109

 1.25 108 m/s

8
(d)
317
H  H o cos(109 t  8 x)aH
Let
Eo
150
 0.955
 157.1
a H  ak  a E  a x  a z  a y
Ho 

H  0.955cos(109 t  8 x)a y A/m
(e)
P  E  H = -150(0.955)cos 2 (109 t  8 x)a x
 -143.25cos 2 (109 t  8 x)a x W/m 2
Prob. 10.44
P  EH 
Eo2

cos 2 (t  10 z )a z
T
Pave 
Eo2
1

dt
az
P
2
T 0
P   Pave  dS 
S
Eo2 S (40) 2   (1.5) 2 (60) 2


 15 W
2
2  120
240
Prob. 10.45
(a)
H
Let H s  o sin  e  j 3r a H
r
E
10
1

Ho  o 
o 120 12
a H  ak  a E  ar  a  a
Hs 
1
sin  e  j 3r a A/m
12 r
(b)
1
2
Pave  Re( E s  H s ) 
Pave   Pave dS ,
10
sin 2  ar
2 12 r 2
dS  r 2 sin  d dar
S
  /6
10
5 5 3
 
 0.007145
Pave 
r 2 sin 3  d


r  2 8 32
24  0  0
 7.145 mW
318
Prob. 10.46
(a) Pave 
|E |
1
1
82 0.2 z
Re( Es H s* )  Re( s ) 
e
2
2
| |
2 | |
 
 
 
2

 
 1  

1

 
2 




 
2

 

 1 

1

 
2 




 
Let x = 1  

  
2

x 1

 0.1/ 0.3  1/ 3

x 1
x 1 1

x 1 9

 x  5/ 4
5
 
 1 

4
  
2




|  |


 3/ 4

120 / 81
 37.4657
5
4
 
1 

  
64
Pave 
e 0.2 z  0.8541e 0.2 z W/m 2
2(37.4657)
2
4
(b) 20dB  10 log
P1
P2


P1
 100
P2
P2
1
 e 0.2 z 

 e0.2 z  100
100
P1
z  5log100  23 m
319
Prob. 10.47
(a) u   / 



c
4.5

2  3  108
 2.828  108 rad/s
4.5
120 
 177 .7 
4.5

40
(b) P  E  H 
9
H  ak 
E


Pave =
(c )
  u 
az

4.5
2


2
sin(t  2 z )a  
0.225

sin(t  2 z )a A/m
sin 2 (t  2 z )a z W/m 2
a z , dS =  d d az
Pave =  Pave  dS = 4.5
3mm
d
2
   d = 4.5ln(3/2)(2 ) = 11.46 W
2mm
0
Prob. 10.48
P= EH 
Eo2
sin 2  sin 2  (t  r / c)ar
r2
T
Pave 
Eo2
1
sin 2  ar
P
dt

2

2 r
T 0
Prob. 10.49

E

c
1



   c  40(3 108 )  12 109 rad/s
  Hdt



y
z
  H = x
0 10sin(t  40 x) 20sin(t  40 x)
 800 cos(t  40 x)a y  400 cos(t  40 x)a z
320
E
1
800
400
  Hdt  
sin(t  40 x)a 
sin(t  40 x)a



y
800
400
sin(t  40 x)a z
9
10
9 10
12 10 
12 10 
36
36
 7.539sin(t  40 x)a y  3.77 sin(t  40 x)a z kV/m

9
sin(t  40 x)a y 
z
9
P  EH 
0 Ey
0 Hy
Ez
 ( E y H z  E z H y )a x
Hz
  20(7.537) sin 2 (t  40 x)  37.7 sin 2 (t  40 x)  a x 103
1
 20(7.537)  37.7 a x 103  94. 23 a x kW/m2
2
Pave 
Prob. 10.50
P
Eo2
2o

 Eo2  2o P  2(120 )10  103  7.539
Eo  2.746 V/m
Prob. 10.51
Let T  t   z.

B

   E  x
t
cos T


y
sin T

z
0
H
  cos Ta x   sin Ta y
t
H 



cos Ta x  sin Ta y dt  
sin Ta x 
cos Ta y




cosT
P  EH =



sin T

sinT
0


(cos 2 T  sin 2 T )a z

cos T 0 



az 
a

 z
which is constant everywhere.
321
Prob. 10.52
Eo2
P
2o
Eo2 S (2.4 103 ) 2  450 104

 343.8 W
P= PS 
2o
2  377
Prob. 10.53
P  EH 
Pave 
(a)
Vo I o
sin 2 (t   z )a z
2 ln(b / a)
2
T
T
Vo I o
Vo I o
1
1
1
az
dt
P

sin 2 (t   z )dta z 
2
2


2 ln(b / a) T 0
2 ln(b / a ) 2
T 0

Vo I o
az
4 ln(b / a)
2
(b)
Pave   Pave dS ,
dS   d  d a z
S
2 b
Vo I o
Vo I o
1

 d  d 
(2 ) ln(b / a )
2


4 ln(b / a )  0   a 
4 ln(b / a )
1
 Vo I o
2
Prob. 10.54
E 2
(a) Pi ,ave  io ,
2 1
Pr ,ave 
Ero 2
,
2 1
Pt ,ave 
Pr ,ave Ero 2
   1
2

R

 2
2  
Pi ,ave Eio
 2  1 



R 


o

2
o

2
Eto 2
2 2
2
2
o 
2

  o 1   o 2 
1 

 
o 
  o 1   o 2 

1 
Since n1  c  1 1  c  o  1 ,
n2  c  o  2 ,
322
 n n 
R 1 2 
 n1  n2 
Pt , ave
T
Pi ,ave
2
1 Eto 1 2 1
4n1n2
   (1  ) 2 
2
(n1  n2 ) 2
 2 Eio 2
2
2

(b) If Pr ,ave  Pt ,ave   RPi ,ave  TPi .ave   R  T
i.e. (n1  n2 ) 2  4n1n2
 
n12  6 n1n2  n2 2  0
2
n 
n 
or  1   6  1   1  0, so
 n2 
 n2 
n1
 3  8  5.828
or
0.1716
n2
(Note that these values are mutual reciprocals, reflecting the inherent symmetry of the
problem.)
Prob. 10.55
1 

2 o o
1

 ,
1
8 o
2
2 
1  2 o / 4  o / 2
1

 ,
1   2 o / 4  o / 2
3
o

2

 o
2
16 o
4
   1 
2
3
1
1
Er   (60) sin(t  10 z )a x  (30) sin(t  10 z   / 6)a y
3
3
 20sin(t  10 z )a x  10sin(t  10 z   / 6)a y V/m
2
2
Et  (60) sin(t  10 z )a x  (30) sin(t  10 z   / 6)a y
3
3
 40sin(t  10 z )a x  20sin(t  10 z   / 6)a y V/m
323
Prob. 10.56
1  o  120 ,

2 
o o



 40

9 o
3
 2  1 o / 3  o
1

 ,
2  1 o / 3  o
2
Er  Eo   Eo / 2,
Piave 
2
  1  
1
2
Et   Eo  Eo / 2
2
o
| Eo |
E

21
2o
1 2
Eo
| Et |2
E2 3
4
Ptave 

 o
22
2(o / 3) 2o 4
Ptave 3
  0.75
Piave 4
Prob. 10.57
1  o
(a)
Ei  Eio sin(t  5 x)a E
Eio  H ioo  120  4  480
a E  a H  ak 
 a E  a y  a x 
 a E  a z
Ei  480 sin(t  5 x)a z
2 
 
o 120

 60
4 o
4
 2   1 60   120 

  1 / 3,
 2   1 60   120 
  1   2 / 3
Ero  Eio  (1/ 3)(480 )  160
Er  160 sin(t  5 x)a z
E1  Ei  Er  1.508sin(t  5 x)a z  0.503sin(t  5 x)a z kV/m
324
(b) Eto   Eio  (2 / 3)(480  )  320 
E
(320 ) 2
P  to a x 
a x  2.68a x kW/m 2
2 2
2(60 )
1 |  | 1  1 / 3

2
(c) s 
1 |  | 1  1 / 3
2
Prob. 10.58  1 
 
2  1
 1 / 3,
2  1
1
  o / 2,
1
  1   4 / 3
Eot   Eio  20 / 3
Eor  Eio  (1/ 3)(5)  5 / 3,

(a)

c
108
r  r 
4  2/3
3  108
5
Er  cos(108 t  2 y / 3)a z
3
E1  Ei  Er  5cos(108 t 
2
(b)
2  o
Pave1 
2
5
2
y )a z  cos(108 t  y )a z V/m
3
3
3
2
Eio
E
25
1
(a y )  ro (a y ) 
(1  )(a y )  0.0589a y W/m 2
21
21
2(60 )
9
2
E
400
(c ) Pave2  to (a y ) 
(a y )  0.0589a y W/m 2
2 2
9(2)(120 )
325
Prob. 10.59
1   11   o o r1 r1 
 2   o o 

16 
c
90  109 (4)
 300(4)  1200
3 108
90 10
 300
3  108
o
1

1
o
1 

9
 r1  o

 30 ,
 r1 4
 2  o  120
 2  1 o  o / 4 3

 ,
2  1 o  o / 4 5
  1  
8
5
P  EH
P1  Pi  Pr 
602
1
3
(60  ) 2
5 cos 2 (t   z )a
cos 2 (t  1 z )a x 
1
x
1
602
602

 38.197,
1 30
3
(60  ) 2
5  13.75
1
P1  38.197 cos (t  1 z )a x  13.75cos 2 (t  1 z )a x W/m 2 , where 1  1200
2
8
(60  ) 2
5 cos 2 (t   z )a
P P 
2
o
t
1
x
8
(60  ) 2
2
5  96  24.46
120
o
P2  24.46 cos 2 (t   2 z )a x W/m 2 , where  2  300
Prob. 10.60
(a) In air,  1  1,  1  2  /  1  2   6 .283 m
  1c  3 108 rad/s
In the dielectric medium,  is the same .
  3 108 rad/s
2 

c
 r2   1  r2 
3
326
2 2

 3.6276 m
2
3
E
10
(b) Ho  o 
 0.0265
 o 120 
a H  a k  a E  a z  a y  a x
2 
H i  26.5cos(t  z )a x mA/m
(c ) 1  o ,
 
 2  o / 3
 2   1 (1 / 3 )  1

  0.268 ,
 2   1 (1 / 3 )  1
(d) Eto   Eio  7 .32,
  1    0.732
Ero   Eio   2.68
E1  Ei  Er  10 cos(t  z )a y  2.68cos(t  z )a y V/m
E2  Et  7.32 cos(t  z )a y V/m
Pave1 
1
1
2
2
(a z )[ Eio  Ero ] 
(a z )(102  2.682 )  0.1231a z W/m 2
21
2(120 )
2
E
3
(7.32) 2 (a z )  0.1231a z W/m 2
Pave2  to (a z ) 
2 2
2 120
Prob. 10.61
1  o  120
For seawater (lossy medium),
2 

jo

  j
j 2 108  4
109
4  j 2 10  81
36
8
2  1
 0.9461177.16
2  1
|  |2  0.8952,
Pr
 89.51%,
Pi
1 |  | 0.1084
Pt
 10.84%,
Pi
 10.44  j 9.333
327

2  1 7.92443.975  377

 0.9702178.2o
2  1 7.92443.975  377
The fraction of the incident power reflected is
Pr
|  |2  0.97022  0.9413
Pi
The transmitted fraction is
Pt
 1 |  |2  1  0.97022  0.0587
Pi
Prob. 10.62
(a)
 120
1  1 
 188.5,
1
4

2 
2 120

 210.75
2
3.2
 2  1 210.75  188.5

 0.0557,
2  1 210.75  188.5

c
4
2 2
2  210.75

 1.0557
2  1 210.75  188.5
Eto   Eio  1.0557(12)  12.668
Ero  Eio  (0.0557)(12)  0.6684
1   11 


 
1 c
2

40 (3  108 )
 6 109 rad/s
2
(b)

6 109 3.2
 2    2 2 
3.2 
 112.4
c
3 108
Er  Ero cos(t  40 x)a z  0.6684 cos(6  109 t  40 x)a z V/m
Et  Eto cos(t   2 x)a z  12.668cos(6 109 t  112.4 x)a z V/m
Prob. 10.63 (a)    c  3  3  108  9 108 rad/s
(b)   2 /   2 / 3  2.094 m
(c )

4

 2  6.288
8
 9  10  80 109 / 36
tan 2  
| 2 |

 6 .288

2 /2
 
   40.47 o
377 / 80

 16 .71
2
2
4


1
4
 2 
4 1 

 2
328
 2  16 .71 40.47 o 
(d)
 2  1 16.7140.47o  377


 0.935179.7 o
o
2  1 16.7140.47  377
Eor  Eoi  9.35179.7o
Er  9.35sin(t  3 z  179.7)a x V/m
2 

2

 2 
9 108

1 
 1 

 3  108
 2 



  r 2 r 2 
2
c
2 
9  108
3 108
80 
1  4 2  1  43.94 Np/m


2
80 
1  4 2  1  51.48 rad/m


2
22
2 16.7140.47o

 0.085738.89o

o
2  1 16.7140.47  377
Eot   Eo  0.85738.89o
Et  0.857e43.94 z sin(9 108 t  51.48 z  38.89o )a x V/m
Prob. 10.64
Induced Currents on the surface
= 0 Standing waves of H
 = 
Zero fields
z
Curve 0 is at t = 0; curve 1 is at t = T/8; curve 2 is at t = T/4; curve 3 is at t = 3T/8, etc.
329
Prob. 10.65 Since  o   1   2 ,
sin  t 1  sin  i
 o sin 45 o

 0.3333
1
4.5
 
 t 1  19.47 o
sin t 2  sin t1
1 1 4.5

 0.4714
 2 3 2.25


t 2  28.13o
 j ( kx x  k y y )
 az

Prob. 10.66
20(e jkx x  e jk x x ) (e y  e
j2
2
jk y
Es 
  j 5 e
j ( kx x  k y y )
e
j ( kx x k y y )
 jk y y
e
)
az
 j ( kx x k y y )
e
which consists of four plane waves.
  E s   jo H s
Hs  
j 20


Hs 
j
o
  Es 
 k y sin(k x x) sin(k y y )a x  k x cos(k x x) cos(k y y )a y 
o 
Prob. 10.67
1  o  377
For  2 ,
2

 2
4
109
2 1.2 10  50 
36
 1.2
9
tan 22 
2
 1.2
 2
 /
| 2 |
4
j   Ez
 Ez 
ax 
ay

o   y
 x 
 
1  2 
  2 
2


 2  25.1o
120 1/ 50
4
1  1.22
 42.658
 2  42.65825.1o
   42.65825.1o  377
 2 1 
 0.8146174.4o
o
2  1 42.65825.1  377
330
Prob. 10.68
(a)
Pt  (1 |  |2 ) Pi
s
1 |  |
1 |  |

 |  |
s 1
s 1
Pt
4s
 s 1 
 1 
 
2
Pi
 s  1  ( s  1)
2
(b) Pi  Pr  Pt


P  s 1 
Pr
 1 t  

Pi
Pi  s  1 
2
Prob. 10.69
If A is a uniform vector and  (r ) is a scalar,
  (A)    A  (  A)    A
since   A  0.
 E  (



j ( k x  k y  k z t )
ax 
ay 
a z )  Eo e
 j (k x a x  k y a y  k z a z )e j ( k r t )  Eo
x
y
z
x
y
z
 jk  Eo e j ( k r t )  jk  E

Also,
B
 j H .
t
Hence   E  
B
becomes k  E   H
t
ak  a E  a H
From this,
k
Prob. 10.70
k | | 1242  1242  2632  316.1
2

 19.88 mm
k
2 f
kc 316.1 3 108
k    

 f 

 15.093 GHz
c
2
2
124

 cos  x 

  x  66.9o   y
k  a x  k cos  x
316.1
 z  cos 1
263
 33.69o
361.1
Thus,
 x   y =66.9o ,  z  33.69o
331
Prob. 10.71
k  3.4a x  4.2a y
k E 0

 0  3.4 Eo  4.2
4.2
 1.235
3.4
Eo 
k | k |   (3.4) 2  (4.2) 2  5.403
2


f 
c

Hs 



2
 1.162
5.403
3 108
 258 MHz
1.162
1
k  Es 

1
k  Es
 kc
0
3.4 4.2
1
A
8
1 3  j4 o
4 10  5.403  3 10 Eo
7
Ao  e  j 3.4 x  4.2 y
where
H s  4.91Ao  104  4.2(3  j 4)a x  3.4(3  j 4)a y  (3.4  4.2 Eo )a z 
 0.491 (12.6  j16.8)a x  (10.2  j13.6)a y  8.59a z  e  j 3.4 x  4.2 y mA/m
Prob.10.72
 E  (



j ( k x  k y  k z t )
ax 
ay 
a z )  Eo e
 j (k x a x  k y a y  k z a z )e j ( k r t )  Eo
x
y
z
x
 jk  Eo e j ( k r t )  jk  E  0
y
z


Similarly,
  H  jk  H  0


kH 0
It has been shown in the previous problem that
B
t
k  E   H
D
t
kxH    E
 E  
Similarly,
 H 
kE 0
332
From k  E   H ,
ak  a E  a H
From k  H   E ,
a k  a H  a E
Prob. 10.73
 o   1   2, 1 
If
 \\ 
1
cos  t 
 r2
1
cos  t 
 r2
 r 1 sin  i 
and
o
o
,2 
 r1
 r2
1
cos  i
 r1
1
cos  i
 r1
 r 2 sin  t
 
 r2
 r1

sin  i
sin  t
sin  i
cos  i
sin  t cos  t  sin  i cos  i
sin  t
 \\ 

sin  i
sin  t cos  t  sin  i cos  i
cos  t 
cos  i
sin  t
sin 2t  sin 2i sin(t  i ) cos(t  i ) tan(t  i )



sin 2t  sin 2i cos(t  i ) sin(t  i ) tan(t  i )
cos  t 
Similarly,
 \\ 
2
cos  i
 r2
2 cos  i

1
1
sin  i
cos  t 
cos  i cos  t 
cos  i
sin  t
 r2
 r1
2 cos  i sin  t
sin  t cos  t (sin  i  cos2  i )  sin  i cos  i (sin 2  t  cos2  t )
2 cos  i sin  t

(sin  i cos  t  sin  t cos  i )(cos  i cos  t  sin  i sin  t )


2
2 cos  i sin  t
sin( i   t ) cos( i   t )
333
1
sin  i
cos  t cos  i 
cos  t
 r1
sin( t   i )
sin  t


1
sin  i
sin( t   i )
cos  t cos  i 
cos  t
sin  t
 r1
 
1
cos  i 
 r2
1
cos  i 
 r2
 
2
cos  i
 r2
2 cos  i
2 cos  i sin  i


1
1
sin  i
sin( t   i )
cos  i 
cos  t cos  i 
cos  t
sin  t
 r2
 r1
Prob. 10.74
(a) n1  1,
n2  c 2 2  c 6.4 o  o  6.4  2.5298
sin t 
n1
1
sin i 
sin12o  0.082185
n2
2.5298

 t  4.714o
1
 47.43
6.4
Ero
 cos t  1 cos i 47.43 cos 4.714o  120 cos12o
 2

 2 cos t  1 cos i 47.43 cos 4.714o  120 cos12o
Eio
1  120 ,
2  120
47.27  117.38
 0.4258
47.27  117.38
Eto
2 2 cos i
2 x 47.43cos12o 92.787
 


 0.5635
Eio
 2 cos t  1 cos i 47.27  117.33 164.65

Prob. 10.75
(a) ki  4a y  3az
ki  an  ki cos i
cos i  4 / 5




i  36.87o
(b)
E
1
( 82  62 ) 2 (4a y  3a z )
*
Re( E s  H s )  o ak 
 106.1a y  79.58a z mW/m 2
2
2
2  120
5
o
(c )  r   i  36 .87 . Let
2
Pave 
Er  ( Ery a x  Erz a z ) sin(t  kr  r )
334
z
kt
Er
kr
Et
r
ki
t
i
y
Ei
From the figure, kr  krz a z  kry a y . But
k rz  k r sin  r  5(3 / 5)  3,
Hence,
sin  t 
k r  ki  5
k ry  k r cos  r  5(4 / 5)  4,
kr  4a y  3a z
c  1 1
n1
3/5
sin  i 
sin  i 
 0.3
n2
c  2 2
4
t  17.46, cos t  0.9539,
1 o  120 , 2  o / 2  60

o
(0.9539)  o (0.8)
Ero 2 cos t  1 cos i
 // 

 2
 0.253
Eio  2 cos t  1 cos i o (0.9539)   (0.8)
o
2
Ero   / / Eio   0.253(10 )   2.53
But
3
4
( Ery a y  Erz a z )  Ero (sin  r a y  cos  r a z )  2.53( a y  a z )
5
5
Er  (1.518a y  2.024a z ) sin(t  4 y  3 z ) V/m
Similarly, let
Et  ( Ety a y  Etz a z ) sin(t  kt  r )
335
k t   2    2  2   4 o  o
But
ki   1    o  o
kt
2
ki
 
k t  2 ki  10
k ty  k t cos  t  9.539 ,
k tz  kt sin  t  3,
k t  9.539 a y  3az
Note that kiz  k rz  k tz  3
 \\ 
Eto
2  2 cos  i
 o (0.8 )


 0.6265

Eio  2 cos  t   1 cos  i
o
(0.9539 )   o (0.8 )
2
Eto   \\ Eio  6.265
But
( Ety a y  Etz a z )  Eto (sin t a y  cos t a z )  6.256(0.3a y  0.9539a z )
Hence,
Et  (1.879a y  5.968a z ) sin(t  9.539 y  3 z ) V/m
Prob. 10.76
(a)
k
1
tan  i  ix 
kiz
8
sin  t  sin  i
(b) 1 
(c )

c
 r1 1
 ( 3)  1
 r2 3
 r1 
  2 /  ,
 
 i   r  19.47 o
 
 t  90 o
109
 3  10  k 1  8  3k
3  108


 1  2  /  1  2  / 10  0.6283 m
 2   / c  10 / 3,
2  2 /  2  2  3 /10  1.885 m
k  3.333
336
(d)
(a x  8a z )
3
 (23.6954a x  8.3776a z ) cos(109 t  kx  k 8 z ) V/m
Ei  1 H x  ak  40 (0.2) cos(t  k  r )a y 
(e)  / / 
2 cos  i sin  t
2 cos 19.47 o sin 90 o

6
sin( i   t ) cos( t   i ) sin 19.47 o cos 19.47 o
cot 19.47 o
//  
 1
cot 19.47 o
Et   Eio (cos t a x  sin t a z ) cos(109 t   2 x sin t   2 z cos t )
Let
where
Et   Eio (cos i a x  sin i a z ) cos(109 t  1 x sin i  1 z cos i )
sin  t  1,
cos  t  0 ,
 2 sin  t  10 / 3
Eto sin  t   \\ Eio  6 (24  )(3)(1)  1357 .2
Hence,
Et  1357 cos(109 t  3.333 x)a z V/m
Since    1,
r  i
Er  (213.3a x  75.4a z ) cos(109 t  kx  k 8 z ) V/m
(f) tan  B / / 
2

1
o
 1/ 3
9 o
 
Prob. 10.77
(a) Ei   5cos(t  0.5 x  0.866 z )a y
Ei  (4a x  3a z ) cos(t  0.5 x  0.866 z )
(b) Comparing Ei with eq. (10.115a),
4a x  3a z  (cos i a x  sin i a z ) Eio
tan i 
sin i 3

cos i 4

i  36.87o
 B / /  18.43 o
337
Prob. 10.78
tan  B 

1
 o r  r
2
o
 r  tan 2  B  tan 2 68  6.126
Prob. 10.79
c
(a) n   r  r  2.11  1.45
u
(b) n  r  r  1 81  9
(c ) n   r  2.7  1.643
Prob.10.80
Microwave is used:
(1) For surveying land with a piece of equipment called the tellurometer. This radar
system can precisely measure the distance between two points.
(2) For guidance. The guidance of missiles, the launching and homing guidance of
space vehicles, and the control of ships are performed with the aid of microwaves.
(3) In semiconductor devices. A large number of new microwave semiconductor
devices have been developed for the purpose of microwave oscillator,
amplification, mixing/detection, frequency multiplication, and switching.
Without such achievement, the majority of today’s microwave systems could not
exist.
Prob.10.81
(a) In terms of the S-parameters, the T-parameters are given by
T11 = 1/S21, T12 = -S22/S21, T21 = S11/S21, T22 = S12 - S11 S22/S21
(b)
T11 = 1/0.4 = 2.5, T12 = -0.2/0.4,
T21 = 0.2/0.4, T22 = 0.4 - 0.2 x 0.2/0.4 = 0.3
Hence,
 2.5 0.5
T= 

 0.5 0.3 
Prob. 10.82
Since ZL = Zo ,  L = 0.
338
 i = S11 = 0.33 – j0.15
 g = (Zg - Zo)/ (Zg + Zo) = (2 –1)/(2 + 1) = 1/3
 o = S22 + S12S21  g /(1 - S11  g )
= 0.44 – j0.62 + 0.56x0.56 x(1/3)/[1 – (0.11 – j0.05)]
= 0.5571 - j0.6266
Prob. 10.83 The microwave wavelengths are of the same magnitude as the circuit
components. The wavelength in air at a microwave frequency of 300 GHz, for example,
is 1 mm. The physical dimension of the lumped element must be in this range to avoid
interference. Also, the leads connecting the lumped element probably have much more
inductance and capacitance than is needed.
Prob. 10.84
 = c/f =
3  108
8.4  109
= 35.71 mm
339
CHAPTER 11
P.E. 11.1
Since Zo is real and   0, this is a distortionless line.
R
G
(1)
L C

R G
(2)
  RG
(3)
Zo 
or
  L
G L

R Zo
(4)
(1)  (3)  R   Z o  0.04  80  3.2 / m ,
(3)  (1)  G 


0.04
 5 104 S / m
80
Zo
1.5  80
 Zo
L

 38.2 nH / m
2  5  10 8

C
0.04
1
LG 12
  10 8 

 5.97 pF/m

80 0.04  80
R
P.E. 11.2
(a) Zo 
R  j L

G  j C
0.03  j 2  0.1  10 3
0  j 2   0.02  10 6
 70.73  j1688
.
 70.75  1.367 o 
 R  j LG  j C  0.03  j0.2  j0.4  10  
4
(b)  
 2.121  10 4  j 8.888  10 3 / m
(c) u 

2  103

 7.069  105 m/s
3
 8.888  10
P.E. 11.3
(a) Zo  Zl  Zin  Zo  30  j60 
340
(b) Vin  Vo 
I in  I o 
Vg
Zin
Vg 
 7 .5 0 o Vrms
Zin  Zo
2
Vg
Z g  Z in

Vg
2Z o

150o
2  30  j 60 
 0.2236  63.43o A
assuming that Zg = 0.
(c ) Since Zo = Zr,   0  Vo   0 ,Vo   Vo
The load voltage is VL  Vs  z  l   Vo  e  l
e
 l
Vo 
7 .5 0 o


1.5 48 o
VL 5  48 o
e  l e jl  1.548 o
1
1
e  l  1.5    ln 1.5 
ln 1.5  0.0101
40
l
o
1 48 o
e jl  e j 48   
rad  0.02094
l 180 o
  0.0101  j 0.02094 /m
assuming that Zg =0.
P.E. 11.4
(a) Using the Smith chart, locate S at s = 1.6. Draw a circle of radius OS. Locate P
where   = 300o . At P,
 
OP 2.1cm

 0.228
OQ 9.2cm
  0.228 300 o
Also at P, zL =1.15-j0.48,
O
S
R 
Q
ZL =Zo zL =70(1.15-j0.48) = 80.5-j33.6
 =0.6  0.6  720o =432o =360o +73o
From P, move 432o to R. At R, zin  0.68  j025
341
Zin  Zo Zin  70 (0.68  j0.25)  47 .6  j17 .5
(b) The minimum voltage (the only one) occurs at    180 o ; its distance from the
180  60

load is
   0.1667 
720
6
Values obtained using formulas are as follows:
s 1

300o  0.2308300o
s 1
Z L  80.5755  j 34.018 
Z in  48.655  j17.63 
These are pretty close.
P.E. 11.5
Z  Zo 60  j60  60
j
(a)   L


 0.4472 63.43 o
Z L  Zo 60  j60  60 2  j
s
1 
1  0.4472

 2.618
1 
1  0.4472
Let x  tan l   tan
2 l

 Z  jZ o tan l  
Zin  Z o  L

 Zo  jZ L tan l  
 60  j 60  j 60 x 
120  j 60  60 

 60  j  60  j 60  x 
Or 2  j 
1  j1  x
1  x  jx
 1  x  j  2 x  2  0
Or x  1  tan l 

2l
 n 

4
i.e
l

 1  4n , n  0 ,1,2,3...
8
342
(b) z L 
Z L 60  j 60
 1 j
Zo
60
Locate the load point P on the Smith chart.
 
OP 4.1cm

 0.4457 ,   62 o
OQ 9.2cm
  0.4457 62 o
Locate the point S on the Smith chart. At S, r = s = 2.6
Zin 120  j60

 2  j , which is located at R on the chart. The angle between OP
60
Zo
90  
 .
and OR is 64o-(-25o) = 90o which is equivalent to
Q
720 8
64º
P

 
Hence l   n   1  4n , n  0 ,1,2........
8
2 8
Zin 
 Zin  max  sZo  2.61860  157.08
S
O
( Zin ) min  Zo / s  60 / 2.618  22.92 
l
62 o
720 o
R
-26º
  0.0851
P.E. 11.6
ZL 57.6 º
Vmax
S = 1.8

 37 .5  25  12.5cm or   25cm
2
2
l  37 .5  35.5  2cm 
25
l  0.08   57 .6 o
z L  1.184  j 0.622
Z L  Z o z L  50 1.184  j 0.622 
 59.22  j 31.11
343
See the Smith chart
P.E. 11.7
100  j80
 1.33  j1.067
75
132 o  65
lA 
  0.093
72
132 o  64 o
lB 
 0.272
720 o
91º
zL 
dA 
A
yL
lB
zL
dB
A´ B
91
  0.126 
720
d B  0.5  d A
Ys  
lA
B´
1 + j0.97
Y = 1 + jb
dA
-91º
 0.374
1 - j0.97
j0.95
  j12.67 mS
75
P.E. 11.8
(a)  G 
1
Z  Zo
lim
,  L  Z L   0 L
 1
3
Z L  Zo
Vg
V
ZL
12
lim
Vg  0,
I   zL 
0
 g 
 120mA
ZL  Zg
Z g  Z L Z g 100
Thus the bounce diagrams for current and waves are as shown below.
lim
V  zL 
0
L = –1
G = 13
4V
V=0
V=4
–4V
t1
2t1
– 43
V=0
V = –1.33
4t1
V = 0.444
6t1
3t1
4
3
4
9
V=0
 94
5t1
 274
V=0
(Voltage)
L = 1
G=  13
80mA
I = 80
t1
– 803
I = 160
2t1
I = 133.3
 803
4t1
I = 115.6
6t1
I=0
80
3t1
80
9
80
9
 80
27
(Current)
I = 106.27
5t1
I = 124.4
344
V (l,t)
0V
t (s)
V(0,t)
4V
4V
4
3
4
9
t (s)
0
2
4
6
– 43
–4
I(l,t)
160mA
124.45
106.67
80
80
9
0
2
6
4
t (s)
 803
I(0,t)
133.33
115.5
80
80mA
80
9
0
2
4
6
t (s)
345
(b)  G 
1
Z  Zo
lim
,  L  Z L    L
1
3
Z L  Zo
V  z L   
lim
ZL
V  Vg  12V ,
Z L  Zg g
I   zL   
lim
Vg
ZL  Zg
0
The bounce diagrams for current and voltage waves are as shown below.
L = 1
G = 13
4
V=4
t1
4
3
4
3
V = 9.33
4t1
2t1
3t1
I = 26.67
–80
80
3
 803
4t1
I = 8.89
I=0
t1
I=0
3t1
80
9
 809
I=0
5t1
6t1
V = 11.55
4
27
V = 11.7
V=8
5t1
4
9
6t1
I = 80
V = 10.67
4
9
V = 11.11
80mA
V=0
4
2t1
L =- 1
G=  13
(Voltage)
(Current)
V(l,t)
10.67
11.55
12V
8V
4V
4
3
0
2
4
4
9
t (s)
6
I (l,t)
0A
2
4
6
t (s)
346
12V
11.11
9.333
V(0,t)
4V
4V
4
3
0
2
4
9
8
6
4
t (s)
I(0,t)
80
80mA
80
80
80
3
2
0
9
3
80
9
t (s)
6
4
 80 3
–80
P.E. 11.9
1
1
G   ,  L  , t1  2 s
2
7
V 

 100 mA
 I o  max  Z max
100
Z
10
g
g
o
The bounce diagrams for maximum current are as shown below.
 =  17
 = 12
100mA
– 100
t1
7
2t1
 50 7
50
3t1
49
4t1
25
49
 25 343
6t1
5t1
 12.5 343
7t1
347
I (0,t)mA
100
1.521
0
2
6
4
8
10
8
10
t (s)
–21.43
I (l,t)mA
85.71
0
2
4
6
–6.122
P.E. 11.10
(a) For w / h  0.8 ,
(b) Zo 
4.8 2.8 
12 
 eff 

1

2
2  0.8 
P.E. 11.11
f o

c
2
 275
.
60
 8 0.8 
ln

  36 .18 ln 10.2  84.03
 0.8 4 
.
275
3  10 8
(c)   10
 18.09 mm
.
10 275
Rs 
1
  20  10 9  4   10 7
5.8  107
 3.69  10  2
R
8.686  3.69  10 2
 c  8.685 s 
wZ o
2.5  10 3  50
 2.564 dB / m
t (s)
348
Prob. 11.1
1
1

 f  c
  500 106  4 107  7 107

  2.6902 106
R
2
w c

2
 0.0354 / m
0.3  2.6902  106  7 107
 o d 4   10 7  1.2  10 2
L

 50.26 nH / m
0.3
w
C
 o w 10 9
0.3


 221 pF / m
d
36  1.2  10 2
Since   0 for air,
G
w
d
0
Prob. 11.2

1
1

 7.744 106
6
7
7
 f c c
  80 10  4 10  5.28 10
1
1



3
3
3 

1 1 1


0.8
10
2.6
10

  10 (1.25  0.3836)  0.6359 /m


R
2 c  a b  2  7.744  106  5.28 107
2569.09
 b 4 107 2.6
L
ln 
ln
 2.357 107 H/m
2 a
2
0.8
G
2 2 105

 5.33 105 S/m
b
2.6
ln
ln
a
0.8
109
2  3.5 
2
36  1.65 1010 F/m
C

b
2.6
ln
ln
a
0.8
349
Prob. 11.3
Method 1:
Assume a charge per unit length Q on the surface of the inner conductor and –Q on the
surface of the outer conductor. Using Gauss’s law,
Q
E 
a b
,
2
b
Q
ln(b / a)
2
V    E dl 
a
J E 
Q
2
2
Q
Q
(1)  d 

  0 2
I   J dS  
S
Q

I
2


Q
V
ln(b / a) ln(b / a)
2
Method 2:
G
Consider a section of unit length, Assume that a total current of I flows from inner
conductor to outer conductor. At any radius  between a and b,
I
J
a ,
a  b
2
J
I
E= 
a
 2 
b
V = -  E dl 
a
G
I
2
2
I

V ln(b / a )
ln(b / a )
350
Prob. 11.4
(a)
R

1
2
w c
 f  c c

1
  200 106  5.8 107  4 107

103
 4.67  106
2 200(5.8)
2
20

 0.2461  /m
6
7
30 10  4.67 10  5.8 10
3(4.67)5.8
R
3
d
4  107  2 103 4 107

 83.77 nH/m
w
30 103
15
 w 103  30 103
G

 15 mS/m
d
2  103
109
 30  103
4
w
60  109

36


 0.5305 nF/m
C
2 103
36
d
L

  ( R  j L)(G  jC )
(b)
 2  (0.2461  j 2  200 106  83.77 109 )(15 103  j 2  200 106  0.5305 109 )
 (0.2461  j105.3)(15 103  j 0.6667)
  0.104  j8.379 /m
Z
( R  j L)
(0.2461  j105.3)

 12.565  j 0.1266 
(G  jC )
(15  103  j 0.6667)
Prob. 11.5
C
 l
l

cosh (d / 2a ) ln(d / a )
1
since (d/2a)2 = 11.11 >> 1.
C


109
 16  103
36
 0.2342 pF
ln(2 / 0.3)
1
1

 2.09 105 m << a
7
7
7
 f  c
 10  4 10  5.8 10
351
Rac 
l
 a c

16 103
 1.4  102 
3
5
7
  0.3 10  2.09 10  5.8 10
Prob. 11.6
We use the two-wire line formulas in Table 11.1
2
d 15mm
 d 

 12.5,    39.1 1
a
1.2
 2a 
d
d
Hence, cosh 1
ln  ln12.5  2.526
a
2a
d

 d 4 107  2.526
L  cosh 1
 ln 
 10.103  107  1.01  H/m


2a  a
109



 44.105 pF/m
C
4


d 2.526

36
1 d
cosh
ln
a
2a
L
1.01 106
Zo 

 151.16 
C
44.105 1012
Prob. 11.7
V ( z  z , t )  V ( z , t )  Lz
I ( z , t )
t
or
I ( z , t )
V ( z  z , t )  V ( z , t )
 L
z
t
V ( z  z , t )
I( z  z , t )  I ( z , t )  C z
t
or
V ( z  z , t )
I( z  z , t )  I ( z , t )
 C
z
t
As z  0, we obtain
V
I
 L
z
t
I
V
 C
z
t
(1)
(2)
352
Prob.11.8
(a)
  ( R  j L)(G  jC )  j LC (1 
 j LC 1 
R
j L
)(1 
G
)
jC
RG
R
G


2
 LC j L jC
As R<< L and G<<C, dropping the 2 term gives

R
G
R
G 
  j LC 1 

 j LC 1 


j L jC
 2 j L 2 jC 
R C G L 
 

  j LC
2 L 2 C
(b)
1
R
1/ 2
R  j L
L
L
R 
j L
Zo 


1


G  jC
C 1 G
C
j L 
jC

G 
1 

jC 

1/ 2



L
R
G
L
R
G

 ...  1 
 ...  
j
 ... 
1 
1  j
C  2 j L
j 2C
C
2 L
2C




L
R 
 G
1 j 



C
 2C 2 L  
Prob. 11.9
(a) R  j L  0.2  j 2  12  106  40  106  0.2  j 24 (40)  0.2  j 3015.93
G  jC  4  103  j 2  12 106  25 106  4 103  j 50 (12)
 4 103  j1884.96
  ( R  j L)(G  jC )  (0.2  j 3015.93)(4 103  j1884.96)
 8.159 102  j 2.384  103 /m
(b) Since R<<  L and G<< L, we may use the result in Problem 11.8(a).
L 40 106

 1.6,
LC  40(25) 1012  109
C 25  106
R C G L 
 0.1 4 103

  


1.6   j 2 12  106  40  106  25 106
  j LC  
2
 1.6

2 L 2 C
3
2
 8.159  10  j 2.384  10 /m
353
Prob. 11.10
I1 (t   z ) travels along +z direction.
(a)
(b) From eq. (11.18),
Vo
Vo
Zo     
Io
Io
Hence,
V ( z , t )  Z o I1 (t   z )  Z o I 2 (t   z )
Prob. 11.11
R  j L
G  jC
Zo 
(1)
    j   ( R  j L)(G  jC )
  0.04 dB/m 
(2)
0.04
Np/m = 0.00461 Np/m
8.686
Multiplying (1) and (2),
Z o (  j  )  R  j L

 50(0.00461  j 2.5)  R  j L
R  50  0.00461  0.2305 /m
L
50  2.5
 0.3316  H/m
2  60 106
Dividing (2) by (1),
  j
 G  jC
Zo
G
C

Zo

0.00461
 92.2 S/m
50

2.5

 0.1326 nF/m
 Z o 2  60 106  50
Prob. 11.12
L
d 
d d

.

c
w w w 
d
Z o  o  78
w
Zo 
354
Z o '  o
d
 75
w'
78 w'
  w'  1.04 w
75 w
i.e. the width must be increased by 4%.
Prob. 11.13
R  j L
6.8  j 2 103  3.4 103
(a) Z o 

G  jC
0.42 106  j 2 103  8.4 109
 103
6.8  j 21.36
 644.3  j 97 
0.42  j 52.78
  ( R  j L)(G  jC )  103 (6.8  j 21.36)(0.42  j 52.78)
 (5.415  j 33.96)  103 /mi

2  103

 1.85  105 mi/s
3
 33.96 10
2
2
(c)  =

 185.02 mi
 33.96 103
(b)
u
Prob. 11.14
Using eq. (11.42a),
Z in   jZ o cot  
c
3  108

 0.75 m
f 400 106
2
2  0.1
 

 48o

0.75
Z in   j (250) cot 48o   j 225.1 
Z o  250,

Prob. 11.15
Assume that the line is lossless.
355
Zo 
L
C
L
From Table 11.1,
L  1
b
ln 
 
C   2 a 
Zo 
ln
 b
ln ,
2 a
C
2
b
ln
a
2
o
1
L
b
b

ln 
ln


C 2 a
 2  r a
Z
75
b
 2  r o  2 2.25
 1.875
120
a
o
b
 e1.875
a

 a  be 1.875  3e 1.875 mm = 0.46 mm
Prob. 11.16
(a) For a lossless line, R = 0 = G.
  j LC
u

 c

 
   LC    o co 
1
LC
(b) For lossless line, R = 0 =G
L

d
cosh 1
,C 

2a
Zo 


cosh 1
d
2a
L
 1
d 120
d
. cosh 1
cosh 1


C
 
2a   r
2a
120
d
cosh 1
2a
r
Yes, true for other lossless lines.
Prob. 11.17

c
356
L

d
0.32
cosh 1
 4  10 7 cosh 1
2a

0.12
L  0.655  H / m
10 9
 3.5


36
C

Cosh 1 2.667
1 d
Cosh
2a

C  59.4 pF / m
L
0.655  106
Zo 

 105
C
59.4  1012
or
120
cosh 1 2.667  105
Zo 
3.5
Prob. 11.18
For a distortionless cable,
R G



RC  LG
L C
L
 60
Zo 
C

1
4

 
u 
6

LC to 80  10
   0.24dB 
(1)
(2)
(3)
0.24
Np  0.0276
8.686
  RG  0.00069
(4)
From (2) and (3),
1
60  4

C 80 106
From (2),

 C
8 105
 333.3 nF/m
240
L  (60) 2 C  3600  333.3  109  1.20 mH/m
357
From (1) and (4),
C
LG
0.000692
2
G




G 0.000692
602
0.00069
G
 11.51 S/m
60
From (4),
R=
0.000692
 0.00069  60  0.0414 /m
G
Prob. 11.19
R G
R
20  63  10 12
(a)   G  C 
L C
L
0.3  10 6
G  4.2  10  3 S / m
  RG  20  4.2  10  3  0.2898
   LC  2  120  10 6 0.3  10 6  63  10 12
  0.2898  j 3.278 / m
u
 2 120 106

 2.3  108 m/s

3.278
L
0.3 106

 69 
C
63  1012
Zo 
(b) Let Vo be its original magnitude
Vo e   z  0.2Vo  e  z  5
z
1
ln 5  5.554 m

(c)  l  45o  
4
l 

4

 0.3051 m
4 4  3.278
Prob. 11.20
(a)   0.0025 Np/m,   2 rad/m,
 3.278
358
 10 8
u 
 5  107 m / s

2
V
60 1

(b)   o 
Vo
120 2
Z L  Zo
1 300  Zo
 
 Zo  100 
Z L  Zo
2 300  Zo
'
120 0.0025l '
60
cos 108  2l '   e 0.0025l cos 108 t  2l ' 
I l '  
e
Zo
Zo
But  
 1.2e0.0025l cos 108  2l '   0.6e 0.0025l cos 108 t  2l '  A
'
'
Prob. 11.21
  103 ,   0.01
    j   0.001  j 0.01  (1  j10) 103 /m
u
 2 104

 6.283  106 m/s

0.01
Prob. 11.22
L
0.6  106

 85.54 
C
82  1012
RC
RC  LG

 G
L
RC
R 10  103


 1.169  104 Np/m
  RG  R
L
Zo
85.54
Zo 
   LC  2  80 106 0.6 106  82 1012
 3.5258 rad/m
  1.169  104  j 3.5258 /m
Prob. 11.23
R  j L  6 .5  j 2   2  10 6  3.4  10 6  6 .5  j 4273
.
G  jC  8.4 103  j 2  2  106  21.5  1012   8.4  j 0.27  103
Zo 
R  j L

G  j C
6 .5  j 4273
.
8.4  j0.27   10 3
359
Zo  7171
.  39.75 o  55.12  j 45.85
 
 R  j LG  j C

43.1981.34 8.4  10 1.84 
o
3
= 0.45 + j0.4 /m


t

l
, but u  ,
u

t
l
0.39  5.6

 0.1783 s
 2  2 106
Prob. 11.24
(a) For a lossy line,
 Z  Z o tanh   
Z in  Z o  L

 Z 0  Z L tanh   
For a short-circuit, Z L  0.
Z sc  Z in
tanh   
ZL  0
 Z o tanh  
Z sc 30  j12

 0.168  j 0.276
Z o 80  j 60
      j    tanh 1 (0.168  j 0.276)  0.1571  j 0.2762
0.1571
 0.0748 Np/m
2.1
0.2762
 0.1316 rad/m

2.1

(b)
 (40  j 30)  (80  j 60)(0.168  j 0.276) 
Z in  (80  j 60) 

 (80  j 60)  (40  j 30)(0.168  j 0.276) 
 61.46  j 24.43 
o
360
Prob. 11.25
 Z  Z o tanh   
Z in  Z o  L

 Z 0  Z L tanh   
      j    1.4  0.5  j 2.6  0.5  0.7  j1.3
tanh    1.4716  j 0.3984
 200  (75  j 60)(1.4716  j 0.3984) 
Z in  (75  j 60) 

 (75  j 60)  200(1.4716  j 0.3984) 
 57.44  j 48.82 
Prob. 11.26
VL
ZL IL
2Z L I L



1 V  Z I  Z L I L  Z o I L
Vo
o L
2 L
2Z L

Z L  Zo
(a) TL 
1 L  1
(b) (i)  L 
2Z L
Z L  Zo

Z L  Z o Z L  Zo
2nZ o
2n

nZ o  Z o n  1
lim
(ii)  L YL 
0 
lim
0 
(iii)  L  Z L 
(iv)  L 
2
ZL
2Z L
0
Z L  Zo
2Z o
1
2Z o
Prob. 11.27
Z  Zo
(a)   L
Z L  Zo
Hence,
1
2
Zo
ZL 
 Z L  Z o  Z L  Z o
1 
Zo
1 

Z L (  1)  (  1) Z o
361
1  0.645o
1.424  j 0.424
(50) 
(50)  (1.252  j1.6586)(50)
o
1  0.645
0.5751  j 0.424
 62.6  j83.03 
ZL 
(b)
Prob. 11.28
L 
s
Z L  Z o 200  j 240  120 80  j 240


 0.52  j 0.36  0.6325  34.7o
Z L  Z o 200  j 240  120 320  j 240
1 |  L | 1  0.6325

 4.4415
1 |  L | 1  0.6325
Prob. 11.29
  ( R  j L)(G  jC )  (3.5  j 2  400  106  2  106 )(0  j 2  400 106 120 1012 )
 (3.5  j 5026.55)( j 0.3016)  0.0136  j 38.94
  0.0136 Np/m,
u
Zo 
  38.94 rad/m
 2  400  106

 6.452  107 m/s

38.94
R  j L
3.5  j 5026.55

 129.1  j 0.045 
G  jC
j 0.3016
Prob. 11.30
From eq. (11.33)
Z sc  Z in Z L 0  Z o tanh  l
Zoc  Zin Z L   
Zo
 Zo coth  l 
tanh  l
For lossless line,   j , tan  l   tanh j l   j tan  l 
Z sc  jZo tan  l  , Zoc   jZo cot  l 
Z sc  jZ o tan   l   (65  j 38) tanh[(0.7  j 2.5)0.8]  (65  j 38) tanh(0.56  j 2)  75.2530.3o
 7.86 60.3o  3.89  j 6.83
exp(0.56 
exp(0.56 
362
Prob. 11.31

Z L  Z o (1  j 2)  1
190
j



 0.7071  45o

Z L  Z o (1  j 2)  1 1  j
245
Prob. 11.32
Z  Z o 120  50
(a)   L

 0.4118
170
Z L  Zo
For resistive load, s 
(b) Z in  Z o
l 
ZL
 2.4
Zo
Z L  jZ o tan l 
Z o  jZ L tan l 
2 
.  60 o
 6
 
 
120  j 50 tan 60 o 
 34.63  40.65 o 
Z in  50 
o 
j
50
120
tan
60



Prob. 11.33
1
 a   l  100  25 rad  1432.4o  352.4o
4
 j 40  j 60 tan 352.4o 
Z in  60 
 j 29.375
o 
 60  40 tan 352.4 
V ( z  0)  Vo 

Z in
j 29.375(100o
Vg 
Z in  Z g
j 29.375  50  j 40
293.7590o
 5.75102o
51.116  12o
(b) Z in  Z L  j 40.
Vo 
Vg
(e
j l
 e  j  l )
(l is from the load)
363
VL 
Vg (1  )
j l
 j l
 12.620o V
( e  e )
1
 c   l '   4  1rad  57.3o
4
 j 40  j 60 tan 57.3o 
Z in  60 
  j 3471.88.
o 
 60  40 tan 57.3 
V
Vg (e j  e  j )
 j 25
 22.740o V
( e  e )
(d) 3m from the source is the same as 97m from the load., i.e.
j 25
1
4
 j 40  j 60 tan 309.42o 
Z in  60 
  j18.2
o 
 60  40 tan 309.42 
l '  100  3  97 m,
V
 l '   97  24.25rad  309.42o
Vg (e j 97 / 4  e  j 97 / 4 )
(e j 25  e  j 25 )
 6.607180o V
Prob. 11.34
V1  Vs ( z  0)  Vo  Vo
V e

o

o
V2  Vs ( z  l )  V e
I1  I s ( z  0) 
(1)
  l
o
 l
o
V
V

Zo Zo
Vo  l Vo  l
e
I2  Is (z  l)  
e 
Zo
Zo
(2)
(3)
(4)
1
(1)  (3)  Vo  (V1  Z o I1 )
2
1
(1) - (3)  Vo  (V1  Z o I1 )
2

Substituting Vo and Vo in (2) gives
1
1
V2  (V1  Z o I1 )e  l  (V1  Z o I1 )e l
2
2
1
1
 (e l  e  l )V1  Z o (e  l  e l ) I1
2
2
V2  cosh  lV1  Z o sinh  lI1
(5)
364
Substituting Vo and Vo in (4),
I2  

I2 
1
1
(V1  Z o I1 )e  l 
(V1  Z o I1 )e l
2Z o
2Z o
1
1
(e l  e  l )V1  (e l  e  l ) I1
2Zo
2
1
sinh  lV1  cosh  lI1
Zo
(6)
From (5) and (6)
 cosh  l
 V2  
  I     1 sinh  l
 2  Z
 o
But
 cosh  l
 1
  sinh  l
 Z o
Thus
 Z o sinh  l 
 V1 
cosh  l   I1 

1
 Z o sinh  l 
 cosh  l
  1
 sinh  l
cosh  l 

 Z o
 cosh  l
V1  
 I    1 sinh  l
 1 Z
 o
Z o sinh  l 

cosh  l 

Z o sinh  l 
  V2 
cosh  l    I 2 

Prob. 11.35
(a)
za 
Z a 80

 1.6
Z o 50
(b)
zb 
Z b 60  j 40

 1.2  j 0.8
50
Zo
(c)
zc 
Z c 30  j120

 0.6  j 2.4
50
Zo
The three loads are located on the Smith chart, as A, B, and C as shown next.
365
Prob. 11.36
zL 
Z L 210

 2.1  s
Z O 100
Or  
s
Z L  Z O 110

,
Z L  Z O 310
1 
1 
But s 
 2.1
Vmax
 Vmax  sVmin
Vmin


720 o
long, 
 120 o
Since the line is
4
4
4
Hence the sending end will be Vmin , while the receiving end at Vmax
Vmin  Vmax / s  80 / 2.1  38.09
Vsending  38.0990o
366
Prob. 11.37
ZL
 1 j2
Zo
We locate z L on the Smith chart.
Z L  (1  j 2) Z o


 zL 
720o
 180o
4
4
o
We move 180 toward the generator and locate point Q at which
z  0.2  j 0.4
Z  zZ o  (0.2  j 0.4) Z o


Prob. 11.38
(a) Method 1:
At Y,
 Z  jZ o tan   
Z in  Z o  L

 Z o  jZ L tan   
2 
tan   0
 
,
 2
Z
Z in  Z o L  Z L  150 
Zo
At X,
Z L '  150
2 
  / 2,
tan  / 2  
 4
ZL ' 

jZ

2
o
2

tan    Z o (75)
Z in  lim Z o 

 37.5

tan   
 jZ ' Z o  Z L ' 150
 L tan   
 
367
Method 2: Using the Smith chart,
Z ' 150
3
zL  L 
Zo
50
Since = /2, we must move 360o toward the generator. We arrive
at the same point. Hence,
Zin  Z L  150
zL ' 


150
2
75
 180o
4
We move 180o toward the generator.
zin  0.5
Z in  75(0.6)  37.5
(b) From the Smith chart,
s = 3 for section XY
s = 2 for section YZ
Z  Z o 150  50
(c)  = L

 0.5
ZL  Z o 150  50
Prob. 11.39
zin 

Z in 100  j120

 2  j 2.4
Zo
50
u 0.8  3 108

 0.4 m,
f
6 108
If   720 ,
o
then
  0.1 m =
 / 4  180

4
o
We move 180o counterclockwise from zin to get z L on the Smith chart as shown below.
z L  0.2  j 0.25

From the Smith chart,
Z L  Z o z L  50(0.2  j 0.25)  10  j12.5 
s  5.19
368
zL
s
zin
Prob. 11.40
Z
40  j 25
zL  L 
 0.8  j 0.5
Zo
50
We locate this at point P on the Smith chart shown below
OP 2.4 cm

 0.3,    96o
| L |
OQ
8 cm
 L  0.3  96o
At S,
s = r = 1.81
  0.27


0.27  720o  194.4o
From P, we move 194.4o toward the generator to G. At G,
z in  1.0425  j 0.6133
Z in  Z o z in  50(1.0425  j 0.6133)  52.13  j 30.66 
369
69.323o
G
O
S=1.81
P
-96.277o
Prob. 11.41
  720o implies that 0.2  144o . The minimum voltage is located at s on the Smith
chart. We move 144o from there to locate the load z L as shown on the Smith chart below.
At z L , we obtain
z L  0.46  j 0.26,
 |  |   

Z L  100(0.46  j 0.26)  46  j 26 
OP
3.5cm
144o 
144o  0.4167144o
OQ
8.4cm
370
zL
s=2.4
Prob. 11.42
(a)
0.12

 0.12  720o  86.4o
We draw the s=4 circle and locate Vmin . We move from that
location 86.4o toward the load.
371
0.37
Min
O
Max
s=4
P
-93.6o ZL =22.294 -j41.719 
At P, z L  0.45  j 0.83
Z  50(0.45  j 0.83)  22.3  j 41.72 
(b) the load is capacitive.
(c) Vmin and Vmax are  /4 apart. Hence the first maximum
occurs at
0.12 +0.25  0.37
Prob. 11.43
(a)
zL 
Z L 75  j 60

 1.5  j1.2
Zo
50
OP 3.8cm

 0.475,
OQ
8cm
  0.475420
|  |
(Exact value = 0.4688  41.76o )
(b) s=2.8
(Exact value = 2.765)
   42o
372
(c )
0.2  0.2 x720o  144o
zin  0.55  j 0.65
Z in  Z o zin  50(0.55  j 0.65)  27.5  j 32.5 
(d) Since    42o , Vmin occurs at
42
  0.05833
720
(e) same as in (d), i.e.. 0.05833
Prob. 11.44
If   720o , then

6
 120o
zin  0.35  j 0.24
Prob. 11.45
Z 100  j 60

 2  j1.2
z
50
Zo
We locate z on the Smith chart. We move 180o toward the generator
to reach point Q. At Q, y = 0.36-j0.22
Y  yYo 
1
(0.36  j 0.22)  7.4  j 4.4 mS
50
373
Prob. 11.46
u 0.5  3 108

 0.9375 m
f
160 106
Z
50  j 30
zL  L 
 1  j 0.6
Zo
50

We locate z at P on the Smith chart. We draw a circle that passes through P.
We locate point Q as the point where the circle crosses the  r  axis. At Q,
zin  1.8
Z in  Z o zin  50(1.8)  90 
The angular distance between P and Q is    73.3o.

If
  720o ,

73.3o
 0.9375  0.0954 m
720o
73.3o 
720o
73.3o
374
Prob. 11.47
60
 0.6
100
If   720o , then
zL 
 / 2  360o . This means that the input impedance is
located on the same spot as z L on the Smith chart.
Zin'  Z o z L  60

 / 4  180o ,
zin'  60 / 50  1.2
zin  0.84
Z in  50(0.84)  42 
Prob. 11.48
Z  Z o 0.5  j  1
(a)   L

 0.0769  j 0.6154  0.6202  82.87 o
Z L  Z o 0.5  j  1
375
(b)

zL  1
zL  1

 zL 
1   1  0.425o

1   1  0.425o
z L  1.931  j 0.7771
Z L  (1.931  j 0.7771) Z o
Prob. 11.49
(a )
Z in 
Z in
1 0 0  j1 2 0

 1 .2 5  j1 .5
Zo
80
u
0 .8  3  1 0 8

 20m
12  106
f
22
 1 .1   7 2 0   7 2 
l1  2 2 m 
20
28
 1 .4   7 2 0   7 2   2 1 6 
l2  2 8 m 
20
T o lo c a te P (th e lo a d ), w e m o v e 2 re v o lu tio n s
p lu s 7 2  to w a rd th e lo a d . A t P ,
OP
5 .1c m
L 

 0 .5 5 4 3
OQ
9 .2 c m
   72  47  25
 
 L  0 .5 5 4 3  2 5 
( E x a c t v a lu e = 0 .5 6 2 4  2 5 .1 5 o )
Z in , m a x  s Z o  3 .7 (8 0 )  2 9 6 
( E x a c t v a lu e = 2 8 5 .5 9  )
Zo
80

 2 1 .6 2 2 
3 .7
s
( E x a c t v a lu e = 2 2 .4 1  )
Z in , m in 
376
Z L  2.3  j1.55
(b) Also, at P,
Z L  80(2.3  j1.55)  184  j124
(Exact value = 183.45+j128.25 )
s  3.7
At S,
To Locate Z , we move 216 from Zin toward the geneator.
'
in
At Zin' ,
zin'  0.48  j 0.76
Z in'  80(0.48  j 0.76)  38.4  j 60.8
(Exact=37.56+j61.304 )
(c) Between ZL and Zin , we move 2 revolutions and 72. During
the movement, we pass through Zin, max 3 times and Zin,min twice.
Thus there are:
3 Z in ,max and 2 Z in ,min
Prob. 11.50
377
(a)

2
 120cm    2.4m
3 108
u f  f  
 125MHz

2.4
40 
720
(b) 40cm 
 
 120
240 6
6
Z L  Z o z L  150(0.48  j 0.48)
u
 72  j 72 
(Exact value = 73.308+j70.324 )
s  1 1.6
(c)  

 0.444,
s  1 3.9
  0.444120
Prob. 11.51
(a)
Z
Z L j 60
j 40

 j 0.75,
zin  in 
 j 0.5
80
80
Zo
Zo
The two loads fall on the r=0 circle, the outermost resistance circle. The shortest
distance between them is
zL 
106.26o
126.87o
360o  (126.87o  106.26o )
 0.4714
720o
 L  1106.26o
(b) s  ,
378
Prob. 11.52
Z
40  j 25
 0.8  j 0.5
zL  L 
50
Zo
Locate this load at point P on the Smith chart. Draw a circle that passes through
P. Locate point Q where the negative  r  axis crosses the circle.
 P  97 o. The angular distance between P and Q is
 =180   P  277 o.
720o  
277 o   

720o
277o  0.3847
379
Prob. 11.53
Z o22
(a) From Eq. (11.43), Z in 2 
ZL
Z in1 
Z2
Z2
Z o21
 Z o , i.e. Z in 2  o1  o 2
Zo
ZL
Z in 2
Z o1  Z o 2
Zo
50
 30
 24.5.
75
ZL
(b) Also,
Z Z
Z o  Z o2 
  Z o 2  o L
 
Z o1
Z o1  Z L 
(1)
2
Z 
Z
3
Also, o1   o 2   Z o 2   Z o1 Z L2
Z o2  Z L 
Z o3 Z L3
From (1) and (2), ( Z o 2 )  Z o1Z  3
Z o1
3
2
L
(2)
(3)
or Z o1  4 Z o3 Z L  4 (50)3 (75)  55.33
From (3), Z o 2  3 Z o1Z L2  3 (55.33)(75) 2  67.74.
Prob. 11.54

74
1
 1.48,
 0.6756
zL
4
50
This acts as the load to the left line. But there are two such loads in parallel due to
 180, z L 
the two lines on the right. Thus
 1 
 
Z
'
Z L  50  L   25(0.6756)  16.892
2
16.892
1
z L' 
 0.3378, z in  '  2.96
zL
50
Z in  50(2.96)  148.
380
Prob. 11.55
From the previous problem, Z in  148
I in 
Vg
Z g  Z in

120
 0.5263 A
80  148
1
1
2
I in Rin  (0.5263) 2 (148)  20.5W
2
2
Since the lines are lossless, the average power delivered to either antenna is 10.25W
Pave 
Prob. 11.56
2  
(a)  l 
.  ,
4 4 2
tan  l  
 ZL

 jZ o 

 Z  jZ o tan  l 
 tan  l

Z in  Z o  L
  Zo
 Zo

 Z o  jZ L tan  l 
 tan  l  jZ L 


As tan  l  ,
Z in 
Z o2 (50) 2

 12.5
ZL
200
(b) If Z L  0,
Z o2

(open)
0
25  
25

 25
(c) Z L  25 //  
25   1  25

2
(50)
Z in 
 200
12.5
Z in 
381
Prob. 11.57

Zo2
Z
l1   Z in1 
or yin1  L2
4
ZL
Zo
yin1 
200  j150
 20  j15 mS
(100) 2


Z L  jZ o tan 


4  jZ
l2   Z in 2  Z L lim
 0 Z o 

o

8
 Z o  jZ L tan 
4

1
1
yin 2 

  j10 mS
jZ o j100
7 

Zi  jZo tan


7
4 

l3 
 Z in 3  Zo

7 
8

 Zo  jZi tan

4 

But
y i  y in1  y in2  20  j5 mS
zi 
yin 3 
1
1000

 47.06  j11.76
yi
20  j5
Z o  jZ in
100-j47.06-11.76

Z o  Z in  jZ o  100  47.06-j11.76-j100 
 6.408  j 5.189 mS
If the shorted section were open,
yin1  20  j15 mS
yin 2 
Z o  Z i  jZ o 
 Z o  jZ i 

j tan
1
4  j  j10 mS

Z in 2
Zo
100
382
7 

 Z i  jZ o tan

7
4  Z o Z i  j Z o 

 Z in 3  Z o

l3 
7 
Z o  jZ i 
8


 Z o  jZ i tan
4 

y i  y in1  y in 2  20  j15  j10  20  j 25 mS
Zi 
1
1000

 19.51  j 24.39
y i 20  j 25
y in 3 
Z o  jZ i
75.61 - j19.51

Z o Z i  jZ o  10019.51  j124.39 
 2.461  j 5.691 mS
Prob. 11.58
zL 
Z L 75  j100

 1.5  j 2
Zo
50
Locate z L on the Smith chart. Draw the s-circle passing through z L .
Extend the diameter through O to y L . Locate points A and B where the
s-circle intersects the g=1 circle. At A, y s   j1.7 and at B, y s  j1.7.
Locate points A' and B' where the stubs admittance is j1.7 and -j1.7
respectively. Calculate
dA 
26.5
  0.0368
720
383
B’, ys=-j1.7
0
yL
dA
A, ys=j1.7
Prob. 11.59
d A  0.12  0.12  720o  86.4o
l A  0.3
(a)
 0.3  720o  216o
From the Smith Chart below,
z L  0.57  j 0.69
Z L  60  0.57  j 0.69 
 34.2  j 41.4
(b)
(c)
360o  86.4o
  0.38
720o
o
o
  62.4  82 
lB  
  0.473
2
720o
s  2.65
dB 
(Exact value = 2.7734 )
384
0.4721 
ZL
O
YL
s = 2.773
A
0.38 
-421.97o
-82.033o
Prob.11.60
Z L 120  j 220

 2.4  j 4.4
Zo
50
We follow Example 11.7. At A, ys=-j3 and at B, ys=+j3. The required stub
admittance is
 j3
Ys  Yo ys 
  j 0.06 S
50
The distance between the load and the stub is determined as follows. For A,value =
0.2308 )
zL 
For B,
180  10  17
  0.2875
720
The length of the stub line is determined as follows.
19
dA 
  0.0264
720
(Exact value = 0.0515 )
360  19
dB 
  0.4736
720
(Exact value = 0.4485 )
lB 
385
0.2308 
A
33.863o
ZL
O
-159.96o
YL
0.05152 
-37.095o
386
Prob. 11.61

720 o

 180 o
4
4
65.274o
ZL
YL
O
A
0.4093 
At A,
yin  1  j1.561
ystub  j1.5614
Position of the stub = 0.0723
Length of the stub = 0.4093
0.07225 
-52.02o
387
Prob. 11.62
90o
V
4V
s  max 
4
Vmin 1V
 

2

s 1 3
  0.6
s 1 5
 25 cm  5 cm  20 cm
180o
Vmin
0o
s=4
  40 cm
P
The load is l=5cm from Vmin, i.e.
l
5 

40 8
 90 o
–90o
On the s = 4 circle, move 90o from Vmin towards the load and obtain ZL = 0.46 – j0.88 at
P.
ZL = Zo zL = 60(0.46 – j0.88) = 27.6 – j52.8 
(Exact value = 28.2353 –j52.9412 )
   270o or -90o
  0.6-90o
Prob. 11.63
s

2
Vmax 0.95

 2.11
Vmin 0.45
 22.5  14  8.5    17 cm
l
3  10 8
 1.764 GHz

0.17
3.2
l  3.2 cm 
  135.5 o
17
f 
c

Vmin

O

S
P  ZL
Q
–44.5o
388
At P,
z L  1.4  j 0.8
ZL  50 1.4  j 0.8   70  j 40
(Exact value = 70.606-j40.496 )
s  1 1.11
 

 0.357,
s  1 3.11
  0.357  44.5o
   44.5o
(Exact value = 0.3571-44.471o )
Prob. 11.64
s 
Rg  Ro
Rg  Ro
L 
Prob. 11.65
L 
g 
t1 

0  50
 1
0  50
RL  Ro 80  50

 0.231
RL  Ro 80  50
Z L  Z o 0.5Z o  Z o
1


Z L  Zo
1.5Z o
3
Z g  Zo
Z g  Zo

Zo 1

3Z o 3
Z
l
 2 s, Vo  o  27   9 V,
u
3Z o
Io 
Vo
 180 mA
Zo
ZL
0.5
V
Vg 
 27   5.4 V, I    216 mA
Zg  ZL
2.5
ZL
The voltage and current bounce diagrams are shown below
V 
389
ry  1
3
z0
ri   1 ry   1 3
3
z0
z 1
9V
180mA
t1
–3V
60mA
2t1
ri  1
3
z 1
t1
2t1
–1V
–20mA
3t1
1 V
3
4t1
–6.667mA
4t1
1 V
9
1
27
V
3t1
2.222mA
5t1
0.741mA
6t1
5t1
6t1
–0.25mA
(Voltage bounce diagram)
(Current bounce diagram)
From the bounce diagrams, we obtain V(0,t) and I(0,t) as shown below:
V(0,t)
9V
5.444V
5.395V
5V
1
1
4
3
12
8
-1V
-3V
9
t(s)
390
220mA
I(0,t)
215.5mA
180mA
216.05mA
60mA
4
12
8
-20mA
Prob. 11.66
Using Thevenin equivalent at z = 0 gives
R g  Rs  4 Z o  200
Vg  I s Rs  10  200  103  2V
g 
L 
Z g  Zo
Z g  Zo

4Zo  Z o 3

4Z o  Z o 5
Z L  Z o 2Zo  Z o 1


Z L  Z o 2Z o  Z o 3
10


 50 ns
u 2  108
The bounce diagram is shown below.
t1 
t(s)
391
g=0.6
L=1/3
2V
t1
2/3
0.4
3t1
0.133
0.08
5t1
The load voltage is sketched below.
V(l,t)
3.4136
3.2
2..667
2
2/3
0
t1
2 t1
3 t1
4 t1
V ( , t ) V ( , t )

ZL
100
To get I(I,t), we just scale down V(l,t) by 100.
I ( , t ) 
5 t1
392
Prob. 11.67
Z g  Z o 32  75
g 

 0.4019
Z g  Z o 32  75
L 
Z L  Z o 2  106  75
1

Z L  Z o 2  106  75
 50 102

 2.5ns
2 108
u
The bounce diagram is shown below.
t1 
At t =20 ns = 4t1 ,
V  8  8  3.2152  3.2152  9.57 V
g =-0.4019
L = 1
8V
t1
8V
2t1
-3.2152
3t1
-3.2152
4t1
1,293
5 t1
393
Prob. 11.68
l 40  102
t1  
 01.6ns,
u 2.5 108
Z  Z o 100  50 1
L  L

 ,
Z L  Zo
150
3
Vo 
Z oVg
Zo  Z g

50 12 
90

g 
Z g  Zo
Z g  Zo

40  50
1
 ,
90
9
20
 6.668V
3
The bounce diagram is sketched below. From it, we obtain V(0,t) and V(L,t).
L 
g  
1
9
0
20/3
1
3
L
t1
20/9
2t1
-20/81
3t1
-20/243
4t1
394
V(0,t)
8.889
8.642
6.667
t (ns)
0
2t1
4t1
6t1
V ( , t )
8.889
8.56
t (ns)
0
2t1
4t1
6t1
Prob. 11.69
Vo  8V 
Zo
50
Vg 
Vg
Zo  Z g
50  60

u
8
6
  ut1  3  10  2 10  600 m
2t1  4  s

 t1  2  s 

 Vg 
8 110
 17.6 V
50
395
Prob. 11.70
t1 
20
 107  0.1 s,
8
2 x10
g 
Z g  Zo

Vo 
Zo
50
Vg  (12)  10V
Zo  Z g
60
10  50
 2 / 3
10  50
Z g  Zo
Z  Z o 0  50
L  L

 1
Z L  Z o 0  50
The voltage bounce diagram is shown below
 g  2 / 3
 L  1
10V
t1
-10V
2t1
6.667V
-6.667 V
3t1
396
From the bounce diagram, we obtain V(0,t) as shown. V(l,t) =0 due to the short circuit.
V(0,t)
10
6.667
4.44
2.963
0
0.1
0.2
0.3
0.4
0.5
0.6
0.8
-6.667
-10
Prob. 11.71
The initial pulse on the line is
Vo 
Zo
Vg
Zo  Z g
(1)
The reflection coefficient is
Zg  Z o
(2)
=
Z g  Zo
The first reflected wave has amplitude Vo , while the second reflected wave
has amplitude  2Vo , etc. For a very long time, the total voltage is
VT  Vo  Vo   2Vo  ...
 1 
 Vo (1     2  3  ...)  Vo 

 1  
Substituting (1) and (2) into (3) gives
(3)




Vg
Zo
1
1
Z V
Vg 
VT 

o g
Zg  Z o 
Zo  Z g 
Z g  Zo  Z g  Zo
2
 1 

 Z g  Zo 
t(s)
397
Prob.11.72
w  1.5cm, h  1cm,
(a)
w
 1.
h
r - 1
0.6
 6 1 
 1.6 
 
2 1  12h/w
1  12/1.5
 2 
 eff  
377
1.8 (1.5  1.393  0.667 In (2.944))
Z0 
 c  8.686
(b)
Rs 
1
281
3.613
 77.77
Rs
wZo

c 

 1.8
 f
c

19  2.5  109  4  10-3
1.1 107
 2.995  10-2
c 
8.686  2.995 102
1.5 102  77.77
u 
c
 eff
u
c

f
f  eff

3  108
2.5  109 1.8
0.8  2.2 
2 102
1.2 1.8 8.944  102

96.096
14.3996
  
 d  27.3 
 0.223dB/m
 d  6.6735 dB/m
(c)
   c   d  6.8965 dB/m
   20dB   
20


20
 2.9 m
6.8965
 8.944 102
398
Prob. 11.73
(a ) Let x  w h. If x  l ,
50 
60
8
ln 
4.6  x

 x

8

5 4.6 - 6ln 
 x  0
x

we solve for x (e.g using Maple) and get x  2.027 or 3.945
which contradicts our assumptiom that x  1. If x  1,
120
4.6  x  1.393  0.667 ln( x  1.444) 
50 
We solve this iteratively and obtain:
x  1.8628, w  xh  14.9024 mm
For this w and h,
 eff  3.4598

(b)
  eff
c
   450 
 

4

   eff
c
4  eff 2 f
c

8 
3  108
3.4598  8  109
  0.00252 m
Prob. 11.74
For w = 0.4 mm,
w 0.4 mm

 0.2  narrow strip
h
2m
399
w
 0.2,  eff  5.851, Z o  91.53
h
w
For
 0.4,  eff  6.072, Z o  73.24
h
For
Hence,
73.24  Z o  91.53
Prob. 11.75
0.1  5  1.2 
ln 
w '  0.5 
  0.5  0.1279  0.6279
3.2  0.1 
Zo 

377
4  1.2  8  1.2

 3.354  
ln 1 

2 2    0.6177  4  0.6279

 42.43ln 1  2.49(3.822  3.354  42.43ln(20.255)
 127.64 
Prob. 11.76
Suppose we guess that w/h  2
A 
w
h
75 3.3
60 2


1.3 
0.11 
 0.23 
  1.117
3.3 
2.3 
8e A
24.44

 3.331  w  3.331h  4mm
2A
e - 2
7.337
If we guess that w/h  2,
B=
60 2
Zo  r

60 2
 5.206
75 2.3
400
w
2
1.3 
0.61  
4.266  ln 9.412 

 ln 4.206  0.39 


h
4.6 
2.3  

 1.665 <2
Thus
w
h
 3.331 > 2
 eff 
u 
3.3

2
3  108
1.953
1.3
12
2 1 
3.331
 1.953
 2.1467  108 m/s
Prob. 11.77
Z L  Z o 100  150

 0.2
250
Z L  Zo
RL  20 log |  | 13.98 dB

Prob. 11.78
The MATLAB code and the plot of the effective relative permittivity are presented
below.
%
Plot effective permitivity versus x = h/w
er = 2.2
e1 = (er + 1)/2; e2 = (er -1)/2;
% x = 0.1*0.1*100
for k=1:100
x(k)=0.1*k
fac=x(k);
ee(k) = e1 + e2/sqrt(1 + 12*fac);
end
plot(x,ee)
xlabel('w/h')
ylabel('ee')
401
2.05
2
1.95
ee
1.9
1.85
1.8
1.75
1.7
1.65
0
1
2
3
4
5
w/h
6
7
8
9
10
402
CHAPTER 12
P. E. 12.1 (a) For TE10, fc = 3 GHz,
1  ( f c / f )2 

4 f
c
0.96 
1  (3 / 15) 2  0.96 ,  o   / uo  4 f / c
4  15  109
0.96  615.6 rad/m
3 108
 2 15 109

 1.531 108 m/s

615.6
60 

'
 60 ,  TE 
 192.4

0.96
u
(b)For TM11, fc = 3 7.25 GHz,

1  ( f c / f ) 2 = 0.8426
4 f
4  15  109 (0.8426)
(0.8426) 
 529.4 rad/m
c
3  108
 2 15 109
u 
 1.78 108 m/s

529.4
 TM  60  (0.8426 )  158.8 
P. E. 12.2 (a) Since E z  0 , this is a TM mode
E zs  E o sin(m x / a ) sin(n y / b)e  jz
m
 40
a
i.e. TM21 mode.
Eo = 20,
m=2,
n
 50
b
n=1
u'
3  108
( m / a ) 2  ( n / b) 2 
402  502  1.5 41 GHz
2
2
2 f
2  109
    1  ( f c / f ) 2 
f 2  fc2 
225  92.25 = 241.3 rad/m.
c
3 108
(c)
 j
Exs  2 (40 )20 cos 40 x sin 50 ye  j  z
h
(b) f c 
403
E ys 
Ey
Ex
 j
 j z
2 (50 ) 20 sin 40 x cos 50 ye
h
 1.25 tan 40  x cot 50  y
P. E. 12.3 If TE13 mode is assumed, fc and  remain the same.
fc = 28.57 GHz,  = 1718.81 rad/m,   j
 TE13 
377 / 2
= 229.69 
1  (28.57 / 50) 2
For m=1, n=3, the field components are:
Ez= 0
H z  H o cos( x / a ) cos(3 y / b) cos( t   z)
   3 
E x   2   H o cos( x / a ) sin(3 y / b) sin( t   z)
h  b
   
  H o sin( x / a) cos(3 y / b) sin(t   z )
h2  a 
 
H x   2   H o sin( x / a ) cos(3 y / b) sin( t   z)
h  a
Ey 
  3 
  H cos( x / a ) sin(3 y / b) sin( t   z)
h2  a  o

Given that H ox  2   2 ( / a ) H o ,
h

H oy   2 (3 / b) H o  6a / b  6(15
. ) / 8  1125
.
h
Hy  
2 14.51 2  104  1.5  102
 7.96

1718.81
   
2 
E oy  2   H o  
 2 TE   459.4



h a
3a
E ox   E oy
 459.4(4.5 / 0.8)  2584.1
b
H oz  H o  
2h 2 a

E x  2584.1cos( x / a ) sin(3 y / b) sin( t   z) V/m,
E y  459.4sin( x / a) cos(3 y / b) sin(t   z ) V/m,
404
Ez= 0,
H y  1125
. cos( x / a ) sin(3 y / b) sin( t   z ) A/m,
H z   7.96 cos( x / a ) cos(3 y / b) cos( t   z) A/m
P. E. 12.4
f c11 
u ' 1 1 3  108 102
 2 
1/ 8.6362  1/ 4.3182  3.883 GHz
2
2 a b
2
3 108
up 
ug 
1  (3.883 / 4)
2
 12.5 108 m/s,
9 1016
 7.2 107 m/s
12.5 108
P. E. 12.5 The dominant mode becomes TE01 mode
c
 3.75 GHz,  TE  406.7
2b
f c 01 
From Example 12.2,
E x   E o sin(3 y / b) sin( t   z) ,
where E o 
b
Ho .

| E xs |2
E o 2 ab
dxdy 
Pave = 
x  0  y  0 2
4
a
b
Hence Eo = 63.77 V/m as in Example 12.5.
Ho 
 Eo
  63.77

 63.34 mA/m
10
b 2 10  4 107  4 102
P. E. 12.6 (a) For m=1, n=0, fc = u’/(2a)

1015
1015


 1
 2  9 109  2.6 109 /(36 ) 1.3
Hence,
405
u' 
d 
1
 c / 2.6,

fc 
 '
2 1  ( fc / f )
1 0 15  3 7 7 /

2
3  108
 2.2149 GHz
2  2.4 102 2.6
2 .6
2 1  ( 2 .2 1 4 9 / 9 )
2
 1 .2 0 5  1 0  1 3 Np/m
For n = 0, m=1,
c 
2 Rs
1 b
[
 ( f c / f )2 ]
2 2
a
b ' 1  ( f c / f )
=
2 2.6   9 109  1.1 107  4  107
2
377  1.5 10  1.110
7
1  (2.2149 / 9)
2
[0.5  (2.4 /1.5)(2.2148 / 9) 2 ]  2  102 Np/m
(b)Since  c   d ,    c   d   c  2 x102
loss =  l  2 102  0.4  0.8 102 Np = 0.06945 dB
P. E. 12.7 For TE11 , m = 1 = n,
H zs  H o cos( x / a ) cos( y / b)e  z
j
( / b) H o cos( x / a) sin( y / b)e  z
2
h
j
E ys   2 ( / a) H o sin( x / a ) cos( y / b)e z
h
Exs 
j
( / a ) H o sin ( x / a ) co s( y / b ) e   z
h2
j
H ys  2 ( / b ) H o co s( x / a ) sin ( y / b ) e   z
h
E zs  0
H xs 
For the electric field lines,
dy E y

 (a / b) tan( x / a ) cot( y / b)
dx E x
For the magnetic field lines
406
dy H y

  (a / b) cot( x / a ) tan( y / b)
dx H x
Ey Hy
Notice that ( )( )   1
Ex Hx
showing that the electric and magnetic field lines are mutually orthogonal. The field
lines are as shown in Fig. 12.14.
P. E. 12.8
1
u' 


c
r
1.5  1010
1/ 25  0  1/100  1.936 GHz
3
1
QTE101 
, where
61
1
1


 1.5 106
9
7
7
 f101 c
 1.936 10  4 10  5.8 10
fTE101 
QTE101 
106
 10,929
611.5
P. E. 12.9
(a) By Snell’s law, n1 sin  1 = n2 sin  2 . Thus
 2 = 90o
sin  2 = 1
sin  1 = n2/n1,
(b) NA =
 1 = sin –1 n2/n1 = sin –1 1.465/1.48 = 81.83o
n12  n2 2 = 148
. 2  1465
. 2 = 0.21
P. E. 12.10
 l = 10 log P(0)/P(l) = 0.2 X 10 = 2
P(0)/P(l) = 100.2, i.e. P(l) = P(0) 10-0.2 = 0.631 P(0)
i.e. 63.1 %
407
Prob. 12.1
(a) For TE10 mode, f c 
u'
3 108

 2.5 GHz
2a 2  6 102
(b) f  3 f c  7.5 GHz
u1
f cmn 
2
m2 n2

a 2 b2
2
2
3  108  102  m   n 


  
2
 6   4
2
2
m  n 
 15 
    GHz
 6   4
2
f c 20  15   5 GHz
6
f c 01  3.75 GHz,
f c 02  7.5 GHz
f c10  2.5 GHz,
f c 20  5.0 GHz
f c 21  6.25 GHz,
f c 30  7.5 GHz
2
2
2
2
1 2
f12  15       7.91 GHz
6 4
1 1
f11  15       4.507 GHz
6 4
The following modes are transmitted
TE01 , TE02 , TE10 , TE11 , TE20 , TE21 , TE30
TM 11 , TM 21
i.e. 7 TE modes and 2 TM modes
Prob.12.2
u'
u' 1
1
u'
 2 
f c10 
, f c11 
2
2
2a
2 a
a
2a
Since the guide can only propagate TE10 mode,
f c10  f  f c11
u'
u'
 2a 
2
f
f

2
a

2


u'
u'
 f 
2
2a
2a
  2a   2

u '  2af  u ' 2
408
Prob. 12.3
2
2
 m  2  n  2 
u'  m  n
3  108
fc 



   
 
 
2  a  b
2 2.25  102  2.28   1.01  
2
2
15  m   n  


 
 
2.25  2.28   1.01  
1/ 2
1/ 2
GHz
Using this formula, we obtain the cutoff frequencies for the given modes as shown below.
Mode
f c (GHz)
TE01
TE10
TE11
TE02
TE22
TM11
TM12
TM21
9.901
4.386
10.829
19.802
21.658
10.829
20.282
13.228
Prob. 12.4
(a)
For TE10 mode,
u'  1 
3  108
 6.25 GHz
fc 
  
2 a
2  2.4 102
For TE 01 mode,
2
u' 1
3  108

 12.5 GHz
 
2 b
2  1.2 102
For TE 20 mo de,
2
fc 
2
u'  2
   2  6.25  12.5 GHz
2 a
For TE 02 mode,
fc 
2
fc 
u'  2
   2 12.5  25 GHz
2 b
(b) Since f = 12 GHz, only TE10 mode will propagate.
409
Prob. 12.5 a/b = 3
f c10 
u'
2a
a = 3b
a
u'
3  108

m = 0.833cm
2 f c10 2  18  109
A design could be a = 9mm, b = 3mm.
Prob. 12.6
fc 
For the dominant mode,
c 3  108

 18.75 MHz
2a
28
(a) It will not pass the AM signal, (b) it will pass the FM signal.
Prob. 12.7 (a) For TE10 mode, f c 
u'
2a
u'
3 108
Or a 

 3 cm
2 f c 2  5 109
u'
For TE01 mode, f c 
2b
u'
3  108
b

 1.25 cm
Or
2 f c 2 12  109
(b) Since a > b, 1/a < 1/b, the next higher modes are calculated as shown below.
Mode
TE10
*TE20
TE30
TE40
*TE01
TE02
*TE11
TE21
fc (GHz)
5
10
15
20
12
24
13
15.62
The next three higher modes are starred ones, i.e. TE20, TE01, TE11
(c) u ' 
For TE11
1
c
 2  108 m/s

2.25
modes,

410
fc 
3  108
2  102 2.25
1
1

 8.67 GHz
2
3 1.252
Prob. 12.8
2
 f 
 25 
Let F12  1   c12   1     0.7806
 40 
 f 
2
3  108
c

 0.0075 m  7.5  103 m
f 40  109
' 
'
12 
F12

7.5 103 m
 9.608  103 m
0.7806
u ' 3 108

 3.843 108 m/s
F12 0.7806
u12 
12 
2

12
'
TE12 
F12
2
 653.95 rad/m
9.608 103

120
 482.95
0.7806
Prob. 12.9
2
2
u'  m  n
fc 
   
2  a  b
For TE10 mode, m=1, n=0,
fc 
3  108
u'
c


 3GHz
2a 2a 2  5  102
2
 fc 
109
785.4
 3 
7
9
    1     2 12.5 10 4 10 
1 
(0.9708)
 
36
3
 12.5 
 f 
  254.15 rad/m
u
2
 2 12.5 109

 3.09  108 m/s

254.15
120
'
TE 
 f 
1  c 
 f 
2

0.9708
 388.3 
411
Prob. 12.10
u'
c
3  108
fc 


 3.75 GHz
2a 2a 2  4 102
2
 fc 
2  24 109
480
 3.75 
1 
(0.9877)
    1    
 
8
3  10
3
 24 
 f 
  496.48 rad/m

2


2
2
 0.0127 m
496.48
Prob. 12.11
u

u'
3 108


 6.975 108 m/s
2
2

1  ( fc / f )
1  (6.5 / 7.2)
ug 
t
9  1016
 1.2903  108 m/s
u
2l
300

 2.325  s
u g 1.2903 108
Prob. 12.12
u'
c
3  108


 6.25 GHz
fc 
2a 2a 2  2.4 102
f  1.25 f c  7.813 GHz
'

 f 
1  c 
 f 
2

377
 1 
1 

 1.25 
Prob. 12.13
u'
c
(a) f c10 

2a 2a  r
3 108
2  1.067 102 6.8
30
GHz

2  1.067 6.8
 5.391 GHz

2

377
 628.32 
0.6
412
(b)
 f 
F  1  c 
 f 
2
 5.391 
 1 

 6 
 0.439
2
u'
c
3  108


 2.62  108 m/s
F F  r 0.439  6.8
u 
(c)
'

F

u'

f
F

c

fF  r
3 108
0.439  6 109  6.8
1
10
 0.04368 m  4.368 cm
2  0.439 6.8
Prob.12.14
In evanescent mode,
 m   n 
k     
 

 a   b 
2
2
 m   n 
2
2
2
2
   
 
  k  4  f c   
 a   b 
2
  0,
2
2
2
   4 f  4 f  2  f c
2
2
c
2
2
 f 
1  
 fc 
2
Prob. 12.15
E z  0 . This must be TM23 mode (m=2, n=3). Since a= 2b,
fc 

10 12
c
3  108
4
36
15.81
GHz,
f


 159.2 GHz
m 2  4n 2 


2
2
4b
4  3  102
TM  377 1  (15.81/159.2) 2  375.1 
P ave 
| Exs |2  | E ys |2
2TM
az
413

 2 Eo 2
 (2 / a) 2 cos 2 (2 x / a) sin 2 (3 y / b)  (3 / b) 2 sin 2 (2 x / a) cos 2 (3 y / b)  a z
4
2h TM
a
Pave   Pave .dS  
b
 P dxdya
ave
z
x 0 y 0
 2 Eo 2 ab  4 2 9 2   2 Eo 2 ab
 4
 2 
2h TM 4  a 2
b  8h 2TM
But

1012

1  ( fc / f ) 
1  (15.81/159.2) 2  3.317 103
8
c
3 10
2
2
2
4
9
10
h 2  2  2  2  1.097 105
a
b
b
2
(3.317) 2  106  52  18 104
 1.5 mW
Pave 
8  (1.098  105 )  375.1
Prob. 12.16 (a) Since m=2 and n=1, we have TE21 mode
(b)    ' 1  ( f c / f ) 2    o  o 1  ( c /  ) 2
c 
 2   2c
fc 
c

2
(c )  TE 
f2
 c2 
 2   2c2
144  9 1016
 2c2
18

36

10

 5.973 GHz
4 2
4 2

1  ( fc / f )
2

377
1  (5.973 / 6 ) 2
 3978 
(d)For TE mode,
Ey 

(m / a ) Ho sin(m x / a ) cos(n y / b) sin( t   z)
h2
Hx 

(m / a ) Ho sin(m x / a ) cos(n y / b) sin( t   z)
h2
  12, m = 2, n =1
414


(m / a ) Ho
2 ( m / a ) Ho , Hox 
h
h2
Eoy  2  6 109  4 107
TE 


 4 2 100
H ox

12
Eoy 
H ox 
Eoy
TE
5
 1.267 mA/m
4 100

2
H x   1.267 sin(m x / a ) cos(n y / b) sin( t   z) mA/m
Prob. 12.17 (a) Since m=2, n=3, the mode is TE23.
(b)
   ' 1  ( f c / f )2 
2 f
c
1  ( f c / f )2
But
fc 
u'
3 108
( m / a ) 2  ( n / b) 2 
(2 / 2.86) 2  (3 /1.016) 2  46.19 GHz, f = 50 GHz
2
2 102

2  50 109
1  (46.19 / 50) 2  400.68 rad/m
8
3 10
  j  j400.7 /m
(c )  
'
1  ( fc / f )
2

377
1  (46 .19 / 50 ) 2
 985.3
Prob. 12.18 In free space,
1 
1 
2 
o
1  ( fc / f )
377
1  (3 / 8) 2
,
2
fc 
c
3 108

 3 GHz
2a 2  5 102
 406.7
 '1
1  ( fc / f )
2
,'
120 
u'
 80  , f c 
, u' 
2a
2.25
c
r
415
3  108
80
 2 GHz,  2 
 259.57
2
2  5  10 2.25
1  (2 / 8) 2
 
  2 1  0.2208
2  1
fc 
s
1 |  |
 1.5667
1 |  |
Prob. 12.19 Substituting E z  R Z into the wave equation,
Z d
RZ
( R' )  2  '' R Z '' k 2 R Z  0
 d

Dividing by R Z ,
1 d
 ''
Z ''
2
( R') 
  kz2
2  k  
R d

Z
i.e.
Z '' k z Z  0
2
1 d
 ''
( R') 
 (k 2  kz2 )  0
R d
2
 d
 ''
( R')  ( k 2  k z 2 ) 2  
 k 2
R d

or
 '' k  2  0

d
( R')  ( k 2 2  k  2 ) R  0 , where k 2  k 2  k z 2 . Hence
d
 2 R''  R' ( k 2 2  k  2 ) R  0
Prob. 12.20
The MATLAB code and the plot of the phase and group velocities are presented below.
416
%
Plot U_p and U_g versus frequency f
(10<f<100) in GHz
c=3*10^8;
for k=1:91
f(k)=9+k
fac = sqrt( 1 - (8/f(k))^2 );
up(k) = c/fac;
ug(k) = c*fac;
end
plot(f,up, 'r', f, ug, 'k')
xlabel('frequency f')
ylabel('phase vel (red) & group vel (black)')
5.5
10 8
phase vel (red) & group vel (black)
5
4.5
4
3.5
3
2.5
2
1.5
10
20
30
40
50
60
frequency f
70
80
90
100
417
Prob. 12.21
(a)
For TE10 mode,
fc 
3 108
u'

 2.083 GHz
2a 2  7.2 102
2
 f 
 2.083 
Let F  1   c   1  
  0.942
 6.2 
 f 
2
F
2  6.2 109  0.942
    F 

 122.32 rad/m
3 108
c
 c 3 108
up   
 3.185 108 m/s
 F 0.942
u g  u ' F  3 108 (0.942)  2.826 108 m/s
TE 
'
F

377
 400.21 
0.942
(b)
3  108


 2  108
2.25

r
u'
1
c
fc 
2  108
u'

 1.389 GHz
2a 2  7.2 102
2
 f 
 1.389 
Let F  1   c   1  
  0.9746
 6.2 
 f 
    F 
F r
2

2  6.2 109  0.9746  1.5
 189.83 rad/m
3  108
c
 2  6.2 109
 2.052  108 m/s
up  
189.83

u g  u ' F  2  108 (0.9746)  1.949  108 m/s
TE 
'
F

377
 257.88 
1.5  0.9746
418
Prob. 12.22
1
f c10 

u'
c
3 108



 1.315 GHz
2a
2a
2a  r 2  7.214  102 2.5
2
 f 
 1.315 
Let F  1   c   1  
  0.9444
 4 
 f 
    F 
up 
2
 r F
c
c
3  108



 2.009  108 m/s
 F  r 0.9444  2.5
ug  u ' F 
cF
r

3  108  0.9444
 1.792  108 m/s
2.5
Prob. 12.23
 
up  
 2

o2 co2  3
  up  c 2 
,

4 2
 c 2 
d
 
 3  o2   2  3c  o   3u p
ug 
d
 
 4 
2
(  2 /  )
419
Prob. 12.24
 f 
1
ug  c  u ' 1   c 
4
 f1 
 f 
1
ug  c  u ' 1   c 
3
 f2 
2
(1)
2
(2)
Dividing (1) by (2),
1/ 4

1/ 3
 f 
1  c 
 f1 
2
 f 
1  c 
 f2 
2
2
 f 
1  c 
2
3
 f1 
   0.5625 
2
4
 fc 
1  
 f2 

2
  f 2 
 fc 
1     0.5625 1   c  
  f 2  
 f1 
Assuming f c is in GHz,
0.5625 f c2
f c2
 0.5625 

f c2  98.44
144
225
From (1),
0.25c
0.25c

 0.4444c
u'
2
2
 fc 
 9.9216 
1 
1  

 12 
f
 1
1
But
c
r
u'
1



f c  9.9216 GHz
c
r
2
 0.4444c

 1 
r  
  5.0625
 0.4444 
Prob. 12.25
fc 
u'
2a
2
 f 
ug  u ' 1   c 
 f 
2


f c  0.208 f  2.0523 GHz


2
2
8 

u
 g
 fc 
1.8 10
  0.208
   1    1 
8

3
10
f
u
'


 
 


 2.2 
420
a
u'
3  108

 4.927 cm
2 f c 2 2.2  2.053 109
Prob. 12.26
Let F 
u'
1

1  ( f c / f )2 

1  (16 / 24) 2  0.7453
3 108
 2  108 ,
2.25
up 
u'
,
F
u g  u ' F  2  108  0.7453  1.491108
m/s
377
 337 .2
1.5 x0.7453
 TE   '/ F 
Prob. 12.27
For the TE10 mode,
  x   j z
H zs  H o cos 
e
 a 
j a
  x   j z
H xs 
H o sin 
e

 a 
j a
  x   j z
E ys  
H o sin 
e

 a 
Exs  0  Ezs  H ys
E s  H s* 
0
H
*
xs
0
E ys
0
j a
H zs*
 E ys H zs* a x  E ys H xs* a z
 a 2 2 2   x 
x  x 

H cos 
H o sin 
 az
 sin 
 ax 

2
 a 
 a   a 
1
 a 2 2 2   x 
*


H o sin 
Pave  Re  E s  H s  
 az
2
2 2
 a 
2
o
Prob. 12.28
Pave 
| Exs |2  | E ys |2
2
 2  2 2 2 2
az 
H o sin  y / ba z
2 b 2 h 4
where   TE10 .
Pave   Pave .dS 
 2  2 2 2 a b
H o   sin 2  y / bdxdy
2 4
2 b h
x 0 y 0
421
 2 2  2 2
H ab / 2
Pave 
2 b2h4 o
h 2  (m / a ) 2  (n / b) 2 
But
Pave 
2
,
b2
 2  2 ab 3 Ho 2
4 2 
Prob. 12.29
Rs 
 f
 12 109  4 107

 2.858  102
7
c
5.8  10
f c10 
u'
3 108

 4.651 GHz
2a 2 2.6  2  102
1
u'  1
f c11   2  2 
2 a
b 
'
1/ 2
 10.4 GHz
377


 233.81

2.6
(a) For TE10 mode, eq.(12.57) gives
 d  j  d   2   k x 2  k y 2  j d

  2 / u2 
2
 j   d
a2
2
 2  12  109 
2
(2.6)
 

 j 2 12 109  4 107 104

2 2
8
3
10
(2
10
)




= 0.012682 + j373.57
 d  0.012682 Np/m
c 
 1 b fc 2 
  ( ) 
b ' 1  ( f c / f ) 2  2 a f 
2 Rs
422

2 x 2.858 x102
 1 1 4.651 2 
)   0.0153 Np/m
  (

10 (233.81) 1  (4.651 /12)  2 2 12
2
2
(b) For TE11 mode,
 d  j d 
  2 / u 2  1 / a 2  1 / b 2  j   d
2
 139556 .21 
 j 9.4748  0.02344  j 202.14
(10 2 ) 2

 d  0.02344 Np/m
c 
 (b / a ) 3  1 
 (1/ 8)  1 
2  2.858 102
=


2


b ' 1  ( f c / f ) 2  (b / a )  1  102 (233.81) 1  (10.4 /12) 2  (1/ 4)  1 
2 Rs
 c  0.0441 Np/m
Prob. 12.30
 c   ' j ''    j


Comparing this with
 c  16 o (1  j104 )  16 o  j16 o 104

 16 o x10 4

  16 o ,
u '  m2 n2 
fc   2  2 
2 a
b 
For TM21 mode,
1/ 2
 2.0963 GHz,
f  1.1 f c  2.3059 GHz
  16 o 104  16  2  2.3059 109 
'
d 
109
104  2.0525 104
36

 30 

 '
2 1  ( f c / f )2
Eo e   d z  0.8 Eo

4.1104  30
 0.0231 Np/m
2 1  1/1.12
z
1
d
ln(1/ 0.8)  9.66 m
423
Prob. 12.31
For TM21 mode,
c 
Rs 
c 
2 Rs
b ' 1  ( f c / f ) 2
1
 c

 f
  2.3059 109  4 107

 0.0246 
1.5  107
c
2  0.0246
 0.0314 Np/m
4  102  30  0.4166
Eo e  (c  d ) z  0.7 Eo
z
1
ln(1/ 0.7)  6.5445 m
c  d
Prob. 12.32
u'
3  108

 2.5 GHz
2a 2  6  102
377
'

 483 
TE 
2
2
 f 
 2.5 
1 
1  c 

 4 
 f 
f c10 
From Example 12.5,
Eo2 ab (2.2) 2  106  6  3  104
Pave 

 9.0196 mW
2
2  483
Prob. 12.33
For TE10 mode,
fc 
u'
3  108

 2.151 GHz
2a 2 2.11  4.8 102
(a) loss tangent 

d

  d  3 104  2  4 109  2.11
'
120
 259.53 
2.11
109
 1.4086 104
36
424
d 
 '
2 1  ( fc / f )
(b) Rs 
2

1.4067  104  259.53
2 1  (2.151/ 4)
2
 2.165  102 Np/m
 f
  4 109  4 107

 1.9625 102 
7
4.110
c
c 
3.925 102 (0.5  0.5  0.2892)
1 b
2
f
f


(
/
)
c


2.4  102  259.53  0.8431
b ' 1  ( f c / f ) 2  2 a
2 Rs
 4.818  103 Np/m
Prob.12.34
2
 f
c
1 b  f  
 1 1  f 2 
c
c 
    
   c 
2
2
2
a
f
  
 f  
 f   2 2  f  
b ' 1   c 
b ' 1   c 
 f 
 f 
2
2 Rs

2 4 107  
f 
1
2
0.5  102  (120 / 2.25) 5.8  107
  f 2 
1   c  
2
 f    f  
1  c 
 f 
  f 2 

1   c  
2
 f    f  
30 (5.8 / 2.25) 1   c 
 f 
105 f
The MATLAB code is shown below
k=10^(-5)/(30*sqrt(5.8/2.25));
fc=10^10;
for n=1:1000
f(n)=fc*(n/100+1);
fn=f(n);
num=sqrt(fn)*(1 +(fc/fn)^2 );
den=sqrt(1- (fc/fn)^2 );
alpha(n) =k*num/den;
end
plot(f/10^9,alpha)
xlabel('frequency (GHz)')
ylabel('attenuation')
grid
425
The plot of attenuation versus frequency is shown below.
attenuation (Np/m)
0.3
0.25
0.2
0.15
0.1
0.05
0
0
10
20
30
40
50
60
frequency (GHz)
Prob. 12.35
The cutoff frequency of the dominant mode is
u
3  108
f c10 

 6.56 GHz
2a 4.576  102
The surface resistance is
Rs 
 f
  8.4 109  4 107

 23.91 103 
c
5.8 107
For TE10 mode,
2

b  fc  
c 
 0.5    
2
a  f  
 f  
b ' 1   c 
 f 
2 Rs
f c 6.56

 0.781,
8.4
f
 '  o  377
70
80
90
100
426
2  23.91103
1.016
2

0.5  2.286  0.781 
1.016  10  377 1  0.781
47.82  103 (0.5  0.2711)

 15.42 103 Np/m
3.83  0.6245
15.42  103  8.686 dB/m = 0.1339 dB/m
c 
2
2
Prob. 12.36
f c10 
(a)
u'
3  108

 3.947 GHz
2a 2  3.8  102
2
 f 
u g  u ' 1   c   3 108 1  (0.3947) 2  2.756  108 m/s
 f 
(b)    d   c
 d  0 since the guide is air-filled.
 f
 1010  4 107

 2.609  102 
c
5.8 107
Rs 
2

b  fc  
c 
0.5    
2
a  f  
 f  
b ' 1   c 
 f 
2 Rs

2  2.609  102
1.6 10 (377) 1   0.3947 
2
2
1.6
5.218  0.5656
2

 0.5  3.8  0.3947   
554.23
 5.325 103 Np/m
 c (dB)  8.686  5.325 103  0.04626 dB/m
Prob.12.37
f c10 
' 

u'

u'
c
3 108


 3.991 GHz
2a 2a  r r 2  2.5 102 2.26
2 f  r
c
2
 f 
 3.991 
F  1  c   1 
  0.8467
 7.5 
 f 
   'F 
2
2  7.5  109 2.26
0.8467  199.94 rad/m
3  108
427
d 
 '
2F

o
104 (377)

 1.481 102 Np/m
2 F  r 2  0.8467 2.26
2

b  fc  
c 
 0.5    
2
a  f  
 f  
b ' 1   c 
 f 
2 Rs
 f
  7.5 109  4 107
Rs 

 0.0519 
c
1.1 107
2

1.5  3.991  
2  0.0519  0.5 

 
2.5  7.5   0.1038  0.6698


c 

377
3.1848
1.5  102 
 0.8467
2.66
 0.02183 Np/m
up 
u'
c
3 108


 2.357 108 m/s
F F  r 0.8467 2.26
3  108  0.8467
 1.689 108 m/s
2.26
u'
c
3 108

 0.05 m = 5 cm( 2a, as expected)
c  
f c f c  r 3.991 109 2.26
ug  u ' F 
Prob. 12.38 (a) For TE10 mode,
fc 
u'
,
2a
u' 
c
2.11
fc 
3 108
 4.589 GHz
2.11(2  2.25 102 )
(b)
 cTE 10 
1 b
2
 2  a ( f c / f ) 
b ' 1  ( f c / f )
Rs 
 f
  5 109  4 107

 3.796 102 
7
1.37  10
c
' 
377
 259.54 
2.11
2 Rs
2
428
c 
1.5
(4.589 / 5) 2 ]
2.25
 0.05217 Np/m
4
1.5 10 (259.54) 1  (4.589 / 5) 2
2  3.796  102 [0.5 
Prob. 12.39 For TE10 mode,
c 
 1 b fc 2 
  ( ) 
b ' 1  ( f c / f )  2 a f 
2 Rs
2
But a = b, Rs 
1

 c
 f
c
 f
c
1
f 
k f   ( c )2 
f 
1
fc 2 
2
c 

(
)



f 
1  ( f c / f )2
a ' 1  ( f c / f ) 2  2
2
where k is a constant.
1
3
f
k 1
f
k [1  ( c ) 2 ]1/ 2 [ f 1/ 2  f c 2 f 5 / 2 ]  [ f 1/ 2  f c 2 f 3 / 2 ](2 f c 2 f 3 )[1  ( c ) 2 ]1/ 2
d c
4
2
2 2
f
f

2
1  ( fc / f )
df
d c
For minimum value,
 0 . This leads to f = 2.962 fc.
df
Prob. 12.40
For the TE mode to z,
E zs  0 , Hzs  Ho cos(m x / a ) cos(n y / b) sin( p z / c)
E ys  
j 
  E zs j   Hzs
 2
  2 (m / a ) Ho sin(m x / a ) cos(n y / b) sin( p z / c)
2
h
h y
h
x
as required.
E xs  
j 
  E zs j   Hzs
 2
 2 (n / b) Ho cos(m x / a ) sin(n y / b) sin( p z / c)
2
h
h x
h
y
From Maxwell’s equation,
429

 j H s    E s   x

y

z
Exs
E ys
0
H xs 
1
1  E ys
  2 (m / a )( p / c) Ho sin(m x / a ) cos(n y / b) cos( p z / c)
h
j   z
Prob. 12.41 Maxwell’s equation can be written as
j   E zs   Hzs

h2  y h2  x
For a rectangular cavity,
H xs 
h 2  k x 2  k y 2  (m / a ) 2  (n / b) 2
For TM mode, Hzs = 0 and
E zs  Eo sin(m x / a ) sin(n y / b) cos( p z / c)
Thus
j
j  Ezs
 2 (n / b) Eo sin(m x / a) cos(n y / b) cos( p z / c)
2
h
h y
as required.
H xs 
H xs  
j   E zs   Hzs

h2  x h2  y
j 
(m / a ) Eo cos(m x / a ) sin(n y / b) cos( p z / c)
h2
From Maxwell’s equation,


j E s    H s   x

y

z
H xs
H ys
0
E ys 
1  H xs
1
 2 (n / b)( p / c ) Eo sin(m x / a ) cos(n y / b) cos( p z / c)
j   z
h
430
Prob. 12.42
fr 
u'
( m / a ) 2  ( n / b ) 2  ( p / c) 2
2
where for TM mode to z, m = 1, 2, 3,…, n=1, 2, 3, …., p = 0, 1, 2, ….
and for TE mode to z, m = 0,1, 2, 3,…, n=0,1, 2, 3, …., p = 1, 2, 3, … , (m  n)  0 .
(a) If a < b < c, 1/a > 1/b > 1/c,
The lowest TM mode is TM110 with f r 
The lowest TE mode is TE011 with f r 
u' 1
1
2 
b2
2 a
u' 1 1 u' 1 1



2 b2 c 2 2 a 2 b2
Hence the dominant mode is TE011.
(b) If a > b > c, 1/a < 1/b < 1/c,
The lowest TM mode is TM110 with f r 
The lowest TE mode is TE101 with f r 
1
u' 1
2 
2 a
b2
u' 1 1 u' 1 1



2 a 2 c2 2 a 2 b2
Hence the dominant mode is TM110.
(c) If a = c > b, 1/a = 1/c < 1/b,
The lowest TM mode is TM110 with f r 
The lowest TE mode is TE101 with f r 
Hence the dominant mode is TE101.
1
u' 1
2 
2 a
b2
1
u' 1 1 u' 1
2  2 
2 
2 a
2 a
c
b2
431
Prob. 12.43
(a)
1
u'

2
c

r

3 108
4.6
2
2
u'  m  n  p 
fc 
     
2  a  b  c 
For the dominant mode, m = 1, n=0, p=1
u'  1  1
3 108
fc 
    
2 a c
2 4.6
2
(b)
Q

1
Q
2
1
1
3 1010


(0.37267)  2.606 GHz
9  104 36 104 2(2.1447)
(a 2  c 2 )abc
  2b(a 3  c3 )  ac(a 2  c 2 ) 
 f r101o

1
  2.606 10 1.57 10  4 10
9
7
7
 2.49 106 m
(9  36)(72) 102
32.42

 4727.7
 8(27  216)  18(9  36) 2.49 106 (2754)
Prob. 12.44
(a)
2
2
u'  m  n  p 
fr 
     
2  a  b  c 
2
2
2
1 1
f rTE101  1.5 1010       1.5 1010 0.1736  6.25 GHz
3  4
f rTE 011  1.5 1010
1
1

 1.5 1010 0.2225  7.075 GHz
6.25 16
f rTE110  1.5 1010
1
1

 1.5 1010 0.2711  7.81 GHz
9 6.25
432
Prob. 12.45
u'
c
3  108


 1.897  108
2.5
2.5

1
u '  m   n   p  1.897  108 102  m   n   p 
fc 
      
     
2  a  b  c 
2
 1  2  3 
2
2
2
2
 9.485 m 2  0.25n 2  0.111 p 2 GHz
f r101
 9.485 1  0  0.111 = 10 GHz
f r 011
 9.485 0  0.25  0.111 = 5.701 GHz
f r 012
 9.485 0  0.25  0.444 = 7.906 GHz
f r 013
 9.485 0  0.25  0.999 = 10.61 GHz
f r 021
 9.485 0  1  0.111 = 10 GHz
Thus, the first five resonant frequencies are:
5.701 GHz(TE 011 )
7.906 GHz (TE 012 )
10 GHz (TE101 and TE 021 )
10.61 GHz (TE 013 or TM110 )
11.07 GHz (TE111 or TM111 )
Prob. 12.46
(a 2  c 2 )abc
Q
  2b(a 3  c3 )  ac(a 2  c 2 ) 
When a = b = c,
Q
a
2a 2 a 3
2a 5


4
3
2
2
  2a  2a  a  2a  6 a 3
Prob. 12.47
(a) Since a > b < c, the dominant mode is TE101
u' 1
1 3  108 102
f r101 
0 2 
c
2 a2
2
1 1
  16.77 GHz
22 12
2
2
433
(a 2  c 2 )abc
(b) QTE101 
  2b(a 3  c3 )  ac(a 2  c 2 ) 
(400  100)20  8  10 103
3.279 103


 [16(8000  1000)  200(400  100)]

104


But  
m
 f r101o
 16.77 109  4 107  6.1107 200.961
200.961
 6589.51
QTE101  3.279 103
104
1
1
Prob. 12.48
c
m2  n 2  p 2
2a
The lowest possible modes are TE101, TE011, and TM110. Hence
fr 
c
fr 
2
2a
a
c
fr
a = b = c = 7.071 cm
Prob. 12.49
(a) a = b = c
u'
fr 
m2  n2  p 2
2a
For the dominant mode TE101 ,
fr 
a
u'
c
11 
2
2a
2a
c 2
3  108 2

 0.03788 m
2 f r 2  5.6  109
a  b  c  3.788 cm
(b)
For  r  2.05,
a
u'
c
r
0.03788
c 2

 0.02646
2 fr  r
2.05
a  b  c  2.646 cm
3  108

 7.071 cm
2
2  3  109
434
Prob. 12.50
(a)
This is a TM mode to z. From Maxwell’s equations,
  E s   j H s
Hs  
1
j
  Es 
j


x

y

z
0
0
Ezs ( x, y )

1
102

6  109  4  107 24

 Ezs 
j   Ezs
ax 
a

   y
 x y 
But
Ezs  200sin 30 x sin 30 y,
1

j102
Hs 
 200  30 sin 30 x cos 30 ya x  cos 30 x sin 30 ya y 
24
H = Re (Hs e jt )
H  2.5 sin 30 x cos 30 ya x  cos 30 x sin 30 ya y  sin 6  109  t A/m
(b)
E  Ez a z ,
H  H xax  H y a y
E H 0
Prob. 12.51
(a) a  b  c


f r101 
3 108
 12 109
a 2
3 108
a
 1.77 cm
2 12  109
(b) QTE101 

a a  f r101

3
3
1.77 102  12  109  4  107  5.8 107
 9767.61
3
435
Prob. 12.52
2
2
u'  m  n  p 
fr 
     
2  a  b  c 
f r101 
3  108
2
2
1
1

 44.186 MHz
2
(10.2) (3.6) 2
f r 011  150
1
1

MHz  45.093 MHz
2
(8.7) (3.6) 2
f r111  150
1
1
1


MHz  47.43 MHz
2
2
(10.2) (8.7) (3.6) 2
f r110  150
1
1

MHz  22.66 MHz
2
(10.2) (8.7) 2
f r102  150
1
4

MHz  84.62 MHz
2
(10.2) (3.6) 2
f r 201  150
4
1

MHz  51 MHz
2
(10.2) (3.6) 2
Thus, the resonant frequences below 50 MHz are
f r110 , f r101, f r 011 , and f r111
Prob. 12.53
3 108
= 1.4286
n = c/um =
2.1 108
Prob. 12.54
NA  n12  n22  1.512  1.452  0.1776  0.421
Prob. 12.55
(a) NA =
n12  n2 2 =
2
162
. 2  1604
.
= 0.2271
(b) NA = sin  a = 0.2271 or  a = sin –1 0.2271 = 13.13o
(c) V =
d
  50 106  0.2271
= 27.441
NA =

1300 109
N = V2/2 6 modes
436
Prob. 12.56
 d 2 3   2  5 106
V
n1  n2 
1.482  1.462  5.86
1300 109

V2
N
 17.17 or 17 modes
2
Prob. 12.57
(a) NA = sin  a = n12  n2 2 =
 a = sin –1 0.4883 = 29.23o
153
. 2  145
. 2 = 0.4883
(b) P(l)/P(0) = 10-  l / 10 = 10-0.4X5/10 = 0.631
i.e. 63.1 %
Prob. 12.58
P ()  P(0)10  /10  10  100.50.85 /10  9.0678 mW
Prob. 12.59
As shown in Eq. (10.35), log10 P1/P2 = 0.434 ln P1/P2 ,
1 Np = 20 log10 e = 8.686 dB or 1 Np/km = 8.686 dB/km,
or 1Np/m = 8686 dB/km. Thus,
12  868610
Prob. 12.60
   10 log10
Pin
1.2 103
 10 log10
 30.792
1 106
Pout
0.4
Np/km
8.686
30.792 30.392 dB


 76.98 km
0.4 dB/km

  0.4 dB/km =
Prob. 12.61
P(0) = P(l) 10  l/10 = 0.2 x 10 0.4 x 30/10 mW = 3.1698 mW
Prob. 12.62 See text.
437
CHAPTER 13
P. E. 13.1
(a) For this case, r is at near field.
I dl sin   j  1   j  r
2 
, r  .  72 o
H s  o
 2 e

 5
4
 r r 

2 c


2  3 108
 6 ,
108
2



1
3
6
sin 30o 
  j 72o
1
j1/ 3
100
H s 

e
 0.2119  20.511o mA/m

2 
4
 6 / 5 (6 / 5) 
H  Im  H s e jt a  Im is used since I = Io sin t
(0.25)
 0.2119sin(108  20.5o )a mA/m
(b) For this case, r is at far field.  
H s 
j (0.25)(
2

)
Sin60o e  j 0
 100
4 (6  200)
a
H  Im ( H  s  e jt )
2

 200  0o
o
 0.2871e j 90  A/m
 0.2871sin(108  90o )a  A/m
P. E. 13.2

(a) l   1.5m ,
4
(b) Io = 83.3mA
(c) Rrad = 36.56  , Prad  1 (0.0833) 2 36 .56
2
= 126.8 mW.
(d) ZL = 36.5 + j21.25,
 
o
36 .5  j 21.25  75
 0.3874 140.3 o
36 .5  j 21.25  75
438
s
1  0.3874
 2.265
1  0.3874
P. E. 13.3
D
4U max
Prad
(a) For the Hertzian monopole
U ( ,  )  sin 2  , 0     / 2,

2 2
Prad  
0    2 , Umax = 1
4
 sin sin  d d  3
2
 0  0
D
(b) For the
4 (1)
3
4
3

monopole,
4

cos2 ( cos  )
2
U ( ,  ) 
, Umax = 1
sin 2 

cos2 ( cos  )
2
Prad   
sin  d d  2  (0.609 )
2

sin
 0  0

D
2 2
4  (1)
 3.28
2  (0.609 )
P. E. 13.4
(a)
Prad = r Pin = 0.95(0.4)
4  U max
4  (0.5)
D

 16 .53
Prad
0.4  0.95
4  (0.5)
 20.94
(b) D 
0.3
439
P. E. 13. 5

2 2
Prad  
 sin  sin  d d 
 0  0
D
2
, Umax = 1
2
4  (1)
 2.546
2
2
P. E. 13. 6
1

(a) f ( )  cos  cos   d cos     
2

where    , d 
2 
. 
 2
1

f ( )  cos  cos   cos     
2

unit pattern
group pattern
For the group pattern, we have nulls at
 (cos   1)   
2
2
   2
and maxima at
 (cos   1)  0, 
2
cos   1,1

   0, 
Thus the group pattern and the resultant patterns are as shown in Fig.13.15(a)
1

(b) f ( )  cos  cos   d cos     
2

where    ,  d   / 2
2
1  

f ( )  cos  cos   cos     
2

2  2
unit pattern group pattern
For the group pattern, the nulls are at
 (cos   1)   
4
2
  180 o
440
and maxima at
cos  1  0
0
Thus the group pattern and the resultant patterns are as shown in Fig.13.15(b)
P. E. 13.7
(a)
●
●

● 
●
●
●
●

● 2 ●
●
●
2

2 ●●
12
: :1
12
: :1
x
x

2
Thus, we take a pair at a time and multiply the patterns as shown below.
●

●
x

x
(b) The group pattern is the normalized array factor, i.e.
1
N ( N  1) i 2 N ( N  1)( N  2) i 3
1  Ne i 
( AF ) n 
e 
e  ............ e i ( N 1)
2!
3!

N 1
 N
N 1
    i   1  N  2 ! 
where
i1
N(N  1)(N  2)
 ...........
3!
 (1  1) N 1  2 N 1
( AF) n 
1
2
j
N 1 1  e
N 1

 N  1 2 cos
2
2
1

1
2
N 1
N 1 e
 j
j
2

 cos
2
e
N 1
2  e
N 1
j
2
441
P. E. 13.8
2
c 3 108
Ae 
Gd ,   
3m
4
f
108
For the Hertzian dipole,
Gd  1.5 sin 2 
2
(1.5 sin 2  )
4
1.5 2 1.5  9
Ae,max 

 1074
m2
.
4
4
Ae 
By definition,
Pr  Ae Pave
Pr 3  10 6

Ae
.
1074
Pave 
 2793
.
 W / m2
P. E. 13.9
(a) Gd 
4r Pave

Prad

2
4 r 2
1 E2
2 

Prad
2 r 2 E 2
Prad
2  400 106  144 106
 0.0096
120 100  103
G  10 log10 Gd  -20.18 dB
(b)
G   r G d  0.98  0.0096  9.408  103
P. E. 13.10
1
 2 Gd 2  Prad  4
r

3
Pr 
 (4)
where  
c 3  10 8

 0.05m
f 6  10 9
Ae  0.7  a 2  0.7  (1.8 ) 2  7 .125m2
Gd 
4Ae

2

4(7 .125)
25  10  4
 3.581  10 4
442
1
 25 104  (3.581) 2  108  5  60  103  4
r

(4 )3  0.26 103


 1168.4m  0.631 nm
r
At r  max
P
2
 584.2m,
Gd Prad 3.581 104  60 103

 501 W/m 2
4 r 2
4 (584.2) 2
Prob. 13.1
Using vector transformation,
Ars  Axs sin  cos  , As  Axs cos  cos  , A s   Axs sin 
As 
50e  j  r
(sin  cos  ar  cos  cos  a  sin  a )
r
  As


 Hs 
100 cos  sin   j  r
50
e ar  2 (1  j  r ) sin  e  j  r a
2
 r sin 
r
50
cos  cos  (1  j  r )e  j  r a
r 2
At far field, only
Hs 
1
term remains. Hence
r
j 50  j r
 e (sin  a  cos  cos  a )
r
 j 50  e  j  r
(sin  a  cos  cos  a )
r
50
sin(t   r )(sin  a  cos  cos  a ) V/m
E  Re  E s e jt  
r
E s   ar  H s 
H  Re  H s e jt  
50
 sin(t   r )(sin  a  cos  cos  a ) A/m
r
443
Prob. 13.2

(a)
rmin 
2d 2

3  108
c

 0.75 m
f 400  106

2(0.02 ) 2

r
i.e. r is in the far field.
jI  dl
H s  o
sin  e  j r
4 r
2
 0.02  sin 90o
3
I o  dl

 5 104  0.5 mA/m
| H  s |
sin  
4 r
4 (60)
|E s | o | H  s | 0.1885 V/m
| H  s | 0.5 mA/m|
(b)
2
2
 dl 
(c) R rad  80 2    80 2  0.02   0.3158 

1
1
(d) Prad  | I o |2 R rad  (9)(0.3158)  1.421 W
2
2
Prob. 13.3
c
3  108
 
6m
f 50 106
jI  dl
sin  e  j r ,
H s  o
E s   H  s
4 r
dl  10cm  

2


2
 /3
6

20   0.1
I o  dl
3

 0.1667
4
4
j 0.1667
Hs 
sin  e  j r /3a A/m
r
Es 
j 0.1667  377
j 62.83
sin  e  j r /3a 
sin  e  j r /3a V/m
r
r
444
Prob. 13.4
 dl 
R rad  80  

 c/ f,
Prad 
2
2
1 2
I o Rrad
2

I o2 
2 Prad
Rrad
2
2
2

 1
 c  1
3  108
 1



 348.93
I  2 Prad  
P
2
2
12

rad 

2 
2
2
6
2
 dl  80
 fdl  80
 140  10  2 10  80
I o  18.68 A
2
o
Prob. 13.5
l
2 z jz cos
e  jr
z
(a) Azs 
I o (1 
)e
4r  l
l
2

2
l

2
2z
2z
e  j  r  2
) cos(  z cos  )dz  j  (1 
) sin(  z cos  )dz 
I o  (1 

4 r   l
l
l

l
2
 2

l
l
2
2z
e  jr

2 I o  (1  ) cos( z cos  )dz
4 r
l
0

2
I o e  jr
l
.  1  cos( 2 cos  ) 
2
2

2  r cos  l 
E s   j As
E s  j sin  Azs  j  sin  Azs
l


j I o e  jr sin   1  cos( 2 cos  ) 
E s 
 rl
 cos2 
l
( cos  ) 2
l
.
If l 2  l , cos( 2 cos  )  1  2
2!
Hence
E s 
Pave 
j I o
 le  jr sin  , H s  E s / 
8 r
Es
2
2 
2
,
Prad   Pave dS
2
n  I l 1
Prad     o  2 sin 2  r 2 sin  d d
2  8  r
0 0
445
2
l
 10  2 I o 2    1 2 I o 2 Rrad
 
l
or Rrad  20  2  
2
 
l
(b) 0.5  20  2  
2
l  0.05
 
Prob. 13.6
2
 dl 
R rad  80 2  

3  108
 c/ f 
 250m
1.2  106
2
Rrad
0.5
 dl 

 6.33 104
  
2
2
80
   80
dl
 2.516  102

dl  2.516  102  250  6.29 m

Prob. 13.7
40
I
24 V
I
+
-
V
24

 0.1866  j 0.0694
Rs  Z in 40  73  j 42
1 2
Rrad  73
| I | Rrad ,
2
1
Prad  (0.1991)2  73  1.447 W
2
Prad 
Prob. 13.8
Let us model this as a short Hertzian dipole.
Zin = 73+j42
446
2
 dl 
Rrad  80    80 2 (1/ 8) 2  12.34

1


Prad  I o2 Rrad  4
I o  0.8052 A
2
2
Prob. 13.9
Change the limits in Eq. (13.16) to  l 2 i.e.
As 

 I o e jzCos  j cos  cos  t   sin  t  l 2
l
  2 cos2    2
2
4 r
1  l
 I o e jr
l  l
 l


sin cos cos    cos  cos sin cos   
2

 2

 2

2 r  sin  
2
2
But B   H    A
Hs 
 Ar 
1  
rA 
,
 r   r
  
 
where Ao   Az sin  , Ar  Az cos 
I o e  jr  j    l
I
l  l
 l


H s 

  sin cos cos    cos  cos sin cos     o 2 e  jr  ......
 2

 2
  2 r
2  r   sin   
2
2
For far field, only the
1
-term remains. Hence
r
l  l
 l
 l


sin cos cos    cos  cos sin cos   

 2

 2

2
2
jI

H s  o e  j  r
2 r
sin 
l
 l

cos cos    cos
 2

2
(b) f ( ) 
sin 
For l   , f ( ) 
cos  cos    1
sin 
447
 3

cos
cos  
 2

3
, f ( ) 
For l 
2
sin 
For l  2 , f ( ) 
cos  sin  2 cos  
sin 
448
Prob. 13.10
(a)
c
3 108
 0.6667 m
 
f 450 106


2
 0.333 m
(b)



4
109
2  450  10  81
36
 1.975
6
 
2
 2  460  106
 
 1  


1
 
2 
c




 
2  460  106
11.4086  109.91
3 108
2

 0.0572




2
 28.58 mm
Prob. 13.11
(a)
c
3 108
 
 260.8 m
f 1.150  106


4
 65.22 m
81 
2
1  1.975   1

2 
449
(b)
c
3 108

 3.333 m
f 90  106



 0.8333 m
4
(c )
c
3  108

 3.75 m
f 80  106



4
 0.9375 m
(d)
c
3  108
 
 0.5 m
f 600  106


4
 0.125 m
Prob. 13.12
l ,hence it is a Hertzian monopole.
2
2
 dl 
1
Rrad  80    80 2    12.34 

8
2
Prob. 13.13
(a)  
c
3 108

 30 m
f 10  106
  Io S
E max 
r 2
E max r 2
Io 
S
50 103  3  302
Io 
 9.071 mA
120 2 (0.2) 2100
(b)
Rrad 
Prad 
320 4 S 2
4

(S=N r 2 )
320 4 2 (0.2) 4  104
 6.077
304
1 2
1
I o Rrad  (9.071) 2 106  6.077
2
2
450
 0.25 mW
Prob. 13.14
c
3  108

 3.75 m
f 80  106
S  N  o2

Rrad 
N2 
320 4 S 2
4

320 4 N 2 2 o4
4
(3.75) 4  8
 248006
320 6 (1.2  102 ) 4

 N 
 4 Rrad
2
320 6 (1.2  102 ) 4

 N
498
320 4 (0.5027) 2
 1.26 m
Rrad 
(50) 4
1
1
(b) Prad  I o 2 Rrad  (50)2  1.26 103  1.575 W
2
2
2 R
a
a 
a
(c) R  
R dc =
=
 f =
2
2
2  S 2 a
2 a
 f

Prob. 13.15
(a)
Rrad 
320 4 S 2
4
S  o2   (0.4) 2  0.5027 m 2
c 3 108
 50 m
 
f 6 106
0.4
4  107  6 106  
R  f

 63.91 m

4 103
5.8 107
a
R rad
1.26
=

 100%  1.933%
R rad  R 1.26  63.91
R 
Prob. 13.16


cos cos  
2

(a) f ( ) 
sin 
451
(b) The same as for


dipole except that the fields are zero for  as shown.
2
2
Prob. 13.17
Let Prad1 and Prad2 be the old and new radiated powers respectively.
Let Pohm1 and Pohm2 be the old and new ohmic powers respectively.
 r1  20% 
Prad 1
1

Prad 1  Pohm1 5
But


4 Prad 1  Pohm1
(1)
1 2
I Rs z
2
1
Pohm2  I 2 Rs 2z  2Pohm1
2
Pohm1 
1
1
 z 
Prad 1  I o2 Rrad  I o2  80 2  
2
2
  
(2)
2
2
1
1
 2z 
Prad 2  I o2 Rrad  I o2  80 2 
  4 Prad 1
2
2
  
From (1) to (3),
 r2 
4 Prad 1
Prad 2
P

 ohm1  33.3%
Prad 2  Pohm 2 4 Prad 1  2 Pohm1 3Pohm1
(3)
452
Prob. 13.18
r 
Prad
Rrad

Pin
Rrad  R
Rrad  73,

R 



 c S  c a 2
3 108
 50m,
  25m
2
6 106
25
25

 0.09528
R 
6
6
2
58 10   (1.2)  10
262.4
73
r 
 0.9987  99.87%
73  0.09528

 c/ f 
,
Prob. 13.19
(a) Let H s 
cos 2  j  r
e aH
o r
a E  a H  ak


Hs 
a  a H  ar

 a H  a
cos 2  j  r
e a
120 r
(b) Pave 
| Es |2
cos 2 (2 )
ar 
ar
2
2 r 2

1
cos 2 2 2
1
(2 )  cos 2 2 sin  d
Prad 
r sin  d d 
2

2
240
r
0
But cos 2  cos 2   sin 2   2 cos 2   1

1
(2 cos 2   1) 2 d (cos  )
Prad  
120 0

1
(4 cos 4   4 cos 2   1)d (cos  )


120 0


1  4 cos5  4 cos3 

 cos  

120  5
3
0
1
4 4
4 4
1 14
[   1    1] 
( )
120 5 3
5 3
120 15
 7.778 mW

453
(c )
120o
1
(2 cos 2   1) 2 d (cos  )
Prad  

120 60o

 120o
1  4 cos5  4 cos3 

cos



 o
120  5
3
 60
1 4
1
4 1 1 4 1
4 1 1
1 1 1 1
[ ( )  ( )   ( )  ( )  ]  [   ]
120 5 32 3 8 2 5 32 3 8 2 60 40 2 6
 5.972 mW
5.972
 0.7678 or 76.78%
7.778

which is
Prob. 13.20
1
2
1
2
Pave  Re( E s  H s* )   | H s |2 ar
 2 I o2
1
sin 2  cos 2  r 2 sin  d d
Prad   Pave dS   
2 2
2
16 r
2
 2 I o2  3
 2 I o2  4  1 2
2
sin  d  cos  d 
(1  cos 2 )d

 
32 2 0
32 2  3  2 0
0
 2 I o2  4 
 2 I o2


 
32 2  3 
24
2P
 2
Rrad  rad

I o2
12 2
Assuming free space,  =120 ,

Rrad 
10 2

Prob. 13.21
(a) Prad   Prad  dS  Pave .2 r 2 (hemisphere)
Pave 
Prad
200  103

 12.73W / m 2
2
6
2 r
2 (2500  10 )
Pave  12.73ar  W/m 2 .
(b)
Pave 
( E max )
2
2
454
E max  2 Pave  240   1273
.  10 6
 0.098 V / m
Prob. 13.22
U ( ,  )  r 2 Pave  k sin 2  sin 3 
G d ( ,  ) 
U ( ,  )
U ave




k
k
k 
3
3
2


U ave 
U
d
d
d
d
(

,

)
sin



sin


sin


  (1  cos  )d ( cos  ) 



4
4 0
4  0
0

2
4k
k 4

4  3 
9
9
G d ( ,  ) 
k sin 2  sin 3   7.069sin 2  sin 3 
4k
D  Gd ,max  7.069

Prob. 13.23
3
2 3
,
 
 3
2
 2
  2 1 3

 2 1 3  
cos 
cos    cos 

 r 
jI e    2 2

  2 2 
H s  o
2 r
sin 
  3

 3  
cos  cos    cos   
r 
r
jI e   2

 2   jI o e cos 1.5 cos  
 o

2 r
sin 
2 r
sin 
Hence, the normalized radiated field pattern is
From Prob. 13.11, set  =
cos 1.5 cos  
sin 
which is plotted below.
f ( ) 
2
455
Prob. 13.24
The MATLAB code is shown below
N=20;
del= 2*pi/N;
sum=0;
for k=1:N
theta = del*k;
term = (1 – cos(theta))/theta;
sum = sum + term;
end
int = del*sum
When the program is run, it gives the value of 2.4335. The accuracy may be increased by
increasing N.
Prob. 13.25
j I o dl
sin  e  jr
4 r
2
 dl 
Rrad  80  2  

1
2
4 r 2 .
E s
2
4 r Pave
2
Gd 

1
2
Prad
I o Rrad
2
(a) Es 
4  r 2 1    1  2 I o 2 2  dl  sin 2 
.
  .
I o 2 80  2  dl  
16  2 r 2
2

2
456
Gd  1.5sin 2 
(b) D  Gd ,max  1.5
(c) Ae 
2
1.5 2 sin 2 
Gd 
4
4
2
1
(d) Rrad  80    3.084 
 16 
2
Prob. 13.26
Gd ( ,  ) 
4 U ( ,  )
4 f 2 ( )

2
Prad
 f ( )d 
f ( )  sin 
4 sin 2 
4 sin 2 
Gd ( ,  )  2 

 1.5sin 2 
2 (4 / 3)
3
  sin  d
 0  0
D  Gd ,max  1.5
Prob. 13.27
(a)
3 108
c
 250
 
f 1.2  106


 62.5 m
4
(b) From eq. (13.30), Rrad  36.5 
457
(c )
For  /4-monopole,

cos( cos  )
2
,
f ( ) 
sin 
0   / 2

4 f 2 ( )
Gd ( ,  ) 
2
 f ( )d 

4 cos 2 ( cos  )
2
sin 2 

cos 2 ( cos  )
2
d d 
0 0
sin 
2  /2


4 cos 2 ( cos  )
3.282 cos 2 ( cos  )
1
2
2


sin 2 
2 (0.6094)
sin 2 
D  Gd ,max  3.282
Prob. 13.28
(a) Umax = 1
Prad  Ud 

4
4
U ave 

1
sin 2 2 sin  d d
4  


1
2
(2 )   2sin  cos   d   cos  
4
0

 2  cos4   cos2  d  cos  
0
 cos5  cos3   
 2

3  0
 5
 2 2 8
 2    
 5 3  15
U ave  0.5333
D
U max
 1.875
U ave
458
(b)
Umax = 4
1
4
sin 
U ave 
Ud  
d d



4
4
sin 2 

1


2
 d  cos ec d 
 0

3

ln 3

Uave = 0.5493
D
U max
16

 9.7092
U ave 3ln 3
(c ) Umax = 2
1
1
2sin 2  sin 2  sin  d d
Ud  

4
4  


1
2

sin  d  1  cos2  d   cos  

2 0
0
U ave 
  1 2
1   cos3 
 1

. 
 cos       2 
2 2  3
 0 4 3
 3
Uave = 0.333
D
U max
6
U ave
459
Prob. 13.29
U ( ,  )
U ave
(a)
Gd ( ,  ) 
U ave 
1
10sin  sin 2   sin  d d
4 0  0
2


2

10

sin 2  d  sin 2 d

4 0
0

10 1 
sin 2  2 1 
sin 2  
 

 

4 2 
2  0 2
2 0
10
5
(2  0)(  0) 
16
4
2
40sin  sin 
Gd ( ,  ) 
 2.546sin  sin 2 
5
D  Gd . max  2.546


(b)
U ave 

1
2sin 2  sin 3   sin  d d


4  0  0




2
2 
3
3
2

sin  d  sin  d 
  (1  cos  )d ( cos  ) 

4 0
4  0
0

2
 1 4
1  cos3 
16

 cos    

  
2  3
 0  2  3  18
18
2sin 2  sin 3   2.25 sin 2  sin 3 
16
D  Gd .max  7.069
Gd ( ,  ) 
2
460
2

1
U ave 
5(1  sin 2  sin 2  )  sin  d d
4 0  0
(c)


2

2

5 
3

d

d



d

sin
sin
sin 2 d 




4  0
0
0
0


 4  sin 2 2 
5 
) 
 2 ( cos  )  ( 
0 3 2
0
4 
4
5 
4  20


 
4
4 
3  3
3
Gd ( ,  )  5(1  sin 2  sin 2  )  0.75(1  sin 2  sin 2  )
20
D  Gd .max  1.5

Prob. 13.30
U max  4
1
1

Ud  
4sin 2  sin sin  d d


4
4
2




 
1
1
  sin 3  d  sin d   (1  cos 2  )d ( cos  )(2 cos )
 0
 0
2
2 0
0
U ave 
1 4
8
( )(2) 
 3
3
U
3
D  max  4 
 4.712
8
U ave

Prob. 13.31
P ave 
I 2 sin 2 
| Er |2
ar  o
ar
2
2 r 2
2
I 2 sin 2 
I2
Prad  o  2 r 2 sin  d d  o (2 )  (1  cos 2  )d ( cos  )
2
240
r
0

 I2
I o2 cos3 
I2
(
 cos  )  o (1/ 3  1  1/ 3  1)  o
0 120
120
3
90
I o2  90 Pave  90  50 103

 I o  2.121 A
461
Prob. 13.32
U ( ,  )  r 2 Pave  r 2
U ( ,  ) 
| E |2
2

r 2 1400 4 I o2 S 2 sin 2 
,
2
r 2 4
where  =120
120 3 I o2 S 2 sin 2  60 3 I o2 S 2 sin 2 

2 4
4
Prob. 13.33
Pave 
E2  2I 2

2 2 r 2
Rrad 
2 Prad 2 Pave
8 r 2  2 I 2 4 2 4 2  2
2
4






r
120 30

I2
I2
I 2 2 r 2
Prob. 13.34
According to eq. (13.10),

Prad  k  sin 3  d 
0
4k
,
3
where k is a constant.
 /3
 cos3 
 / 3
1 1  5
 1
'
 k  sin 3  d  k 
 cos  
k
   1 
Prad
k
0

3
3
8
2
3
24




0
5
k
5
24

 0.1562
Fraction =
4k
32
3
Prob. 13.35
This is similar to Fig. 13.10 except that the elements are z-directed.
E s  E s1  E s 2 
where r1  r 
Es 

j I o dl 
e  j  r1
e  j  r2
sin
a
sin
a 2 



1
2
1

r1
r2
4 

d
cos  ,
2
r2  r 
d
cos  ,
2
j I o dl
sin  a e j  d cos / 2  e  j  d cos / 2 
4
1   2   ,
a 1  a 2  a
462
Es 
j I o dl
1
sin  cos(  d cos  )a
2
2
Prob. 13.36
1

(a) AF = 2 cos   d cos      ,
2

  0,
d 
2
  2

AF = 2 cos(  cos  )
(b)Nulls occur when
cos(  cos  )  0
or
 
 cos     / 2, 3  / 2,...
  60 o ,120 o
(c) Maxima and minima occur when
df
0
d
 
sin(  cos  )  sin   0
i.e. sin   0
 
  0 o ,180 o
cos  0
 
  90 o
or
  0 o ,90 o ,180 o
(d ) The group pattern is sketched below.
463
Prob. 13.37
1

f     cos   d cos     
2

(a)    2 , d 

2
.   2

f     cos  cos    4

Nulls occur at  cos    4  
Maxima occur at
f
0

 3
,
,... or   75.5 o ,138.6 o
2
2
sin   0


Or sin   cos     0
  41.4 o ,104.5 o
4

With f max  0.71,1 .
Hence the group pattern is sketched below.
  0 o ,180 o
464
(b)

3
2  
, d 
. 
 4 2
4
3 

f     cos cos  

4
8
Nulls occur at
3

 3
cos  
  ,
,...
4
8
2
2
 
3 

Minima and maxima occur at sin  cos  cos  
0
8 
4
i.e.   0 o ,180 o  f     0.383,0.924
  60 o
465
2  3 3 
.

 4
2
 3

f     cos
cos  
 4

3
 3
cos    ,
,...    48.2 o ,131.8 o
It has nulls at
4
2
2
df
 3

It has maxima and minima at
 0  sin  sin
cos    0
 4

d
(c)   0 ,  d 
i.e.   0 o ,180 o  f     0.71,1 ,
  90o ,  f    1
 = 0 o,
o,
 = 180o,
Prob. 13.38
1

(a) For N = 2, f     cos   d cos     
2


  0 ,d 
4
 1  2 



f     cos 
. cos   0   cos cos  




4
2  4

Maxima and minima occur at
d  

cos cos     0


d   4


sin  sin cos    0
4

sin   0     ,0 and f     0.707


sin cos    cos  0    90 o , f     1
4

abs(f) = 1/2=0.707
abs(f) = 1,
abs(f) = 1/2=0.707
466
Nulls occur as

4
cos  

2
,
3
,... (No Solution)
2
The group pattern is sketched below.
(b) For N = 4,
AF 
Now,
sin 2  d cos   0 
1
sin   d cos   0 
2
sin 4 2 sin 2 cos 2

 4 cos 2 cos 
sin 
sin 
1

AF  4 cos  d cos   cos  d cos  
2

 2 

 1 2  
f     cos
. cos   cos
 cos 
  4

 2  4




 cos cos   cos cos  
2

4

The plot is shown below.
467
Prob. 13.39
The MATLAB code is shown below.
for n=1:180
phi=n*pi/180;
p(n)=n;
sn=sin(2*pi*cos(phi));
cn=cos(0.5*pi*cos(phi));
sd=sin(0.5*pi*cos(phi));
fun=sn*cn*cn/sd;
f(n)= abs(fun);
end
polar(p,f)
The polar plot and the xy plot are shown below.
468
Prob. 13.40
(a) The resultant pattern is obtained as follows.
I  0o
I  0o
I  0o
I  0o
/2
/2
/2

x
=
469
(b) The array is replaced by by
+

4
0 o
where + stands for

+
  /2


0
Thus the resultant pattern is obtained as shown.
o
I  0o
I  90o
I  270o
I 180o
x
/4
=
/4
/4
/2, 
=
x
Prob. 13.41
Gd (dB)  20dB  10 log10 Gd


Gd  102  100
c
3 108
 3  102
 
9
f 10 10
9  104
2
Ae 
Gd 
100  7.162  102 m 2
4
4
470
Prob. 13.42
Ae 

Pr
Pr

Pave | Er |2

2
2 Pr
| Er |2
2 120  2 106 48

 0.6031 m 2
6
2
25 10  10
250
Prob. 13.43
Friis equation states that
Pr
  
 Gr Gt 

Pt
 4 r 

2
c
3 108

 1.5 m,
f 200  106
Gt (dB)  15dB  10 log10 Gt
r  238,857 1.609 103  3.843 108

 Gt  1015/10  31.623
2
2
8
9
 4 r  Pr  4  3.843 10  4  10

 34.55  1010
Gr  



3

P
1.5
120
10


 t 

Gr (dB )  10 log10 Gr  10 log10 34.55  1010  115.384 dB
Prob. 13.44
Using Frii’s equation,
2
  
Pr  Gr Gt 
Pt
 4 r 
2
 4 r  Pr
Pt  
   Gr Gt
c 3  108
 0.1,
 
f 3 109
r  42 km
Gt (dB)  10 log10 Gt  25

 Gt  102.5  316.23
Gr (dB )  10 log10 Gr  20

 Gt  102  100
2
 4  42  103  3 106
Pt  
 2.642 kW

0.1

 31623
471
Prob. 13.45
Gdt  10 4 , Gdr  10 3.2  1585

1
c
3  10 8
 0.02m 

50
f 15  10 9
2
2
0.02


  
4
Pr  Gdr Gdt 
 P  10  1585 
 320
 4r  t
 4   2.456741  107 
 2.129  1011 W  21.29 pW
Prob. 13.46
2
  
Pr  Gdt Gdr 
 Pt
 4 r 
c
3 108
 
 15 m,
f 20  106
Gdt  Gdr  1.64
2
2
3
Pr  4 r 
6  4  80  10 


Pt 
0.5
10

  835.025 W


Gdt Gdr   
 1.64 15 
But
Rrad 
2 Prad
 73
I o2

I o2 
2 Prad 2  835.025

 22.8774
73
73
I o  4.783 A
Prob. 13.47
30 dB  log
Pt
P
 t  10 3  1000
Pr
Pr
2
3
 Gd 

 Pt  Pt 

 50  4   12 
 800  
But Pr   Gd  2 
2
 1 
Pr
1
 Gd 




 
 800  
Pt 1000  10 10 
or Gd 
800 
10 10
2
 79.476
Gd  10 log79.476  19 dB
2
472
Prob. 13.48
(a) Pi 
E
2
2 o
Ei 

Prad Gd
4 r
2
 Ei 
240 Prad Gd
4 r 2
1
1
60 Prad Gd 
60  200 103  3500
120 103
r
 1708
.
V/m
2
Ei 

4 r 2
1708
. 28
 11.36  V / m
4   14400  10 6
(b)
Es 
(c)
1708
. 2
 8   30.95 mW
Pc  Pi  
240 
(d)
 11.36   10 12
E
.
Pi 

 1712
 10 13 W / m2
2 o
240 
2

2
3  10 8
15  10
 0.2m, A2 r 
8
2 G 0.04  3500

4
4
Pr  Pa Aer  1712
.
 10 13  1114
.  1.907  10  12
or Pr 
 Gd  2 Prad  0.2  3500  2  8  2  10 5
 4 3  12 4  10 16
 4 3 r 4
 1.91  10 12 W
Prob. 13.49
( Gd ) 2  Prad
Pr 
(4 )3 r 4
Gd (dB)  30dB  10 log10 Gd

 Gd  103

c 3 108

 0.075 m
f 4  109
Pr 
(0.075  103 ) 2 12  80 103
 272.1 pW
(4 )3 (10 103 ) 4
473
Prob. 13.50
2d 2  160 103 32

  104  1.067 ms
u
3 108
3
8
AG P
3 10
(b)   c / f 
 0.075m, Pr  e d2 rad
9
4  10
(4 r ) 2
(a)
tr 
But
Ae 
 2Gd
4

Gd 
4 Ae
2
2
2
2
 Ae 


Pr  4 
Prad  4  5 
(60  103 )  2.59  1014 W
2 
2
6 
 0.075  4  160  10 
 4 r  
  2Gd2 Prad 
r

3
 (4 ) Pr 
(c )
1/ 4
1/4

  2 (4 ) 2 Ae2 Prad 


3
4
Pr 
 (4 )
1/4
 5 4
60  103 

12 
2
 4 (0.075) 8 10 
 38.167 km
Prob. 13.51
r4
Prad  Pr

k
4
(2r )
r4
'
If R  2r ,
Prad

Pr  16 Pr  16 Prad
k
k
i.e. the transmitted power must be increased 16 times.
kP
Pr  rad
r4
Prob. 13.52
2
Prad 
4   4 r1r2  Pr


Gdt Gdr    
.
But Gdt  36 dB  10 3.6  39811
Gdr  20dB  102  100

c 3  10 8

 0.06
f
5  10 9
r1  3 km , r2  5 km

 Ae2 Prad 


2
 4 Pr 
1/ 4
474
2
 4  15  10 6  8  10 12
4
Prad 


39811
2.4
.  100  6  10 2 
 1038
.
kW
Prob. 13.53
(a)
 fL   300 106  50 109

 2.356
F
R
20
IL  10 log10 (1  F 2 )  10 log10 (1  2.3562 )  8.164 dB
(b)
F   fRC    300  106  10 103  60 1012  180  565.5
IL  10 log10 (1  F 2 )  10 log10 (1  565.52 )  55.05 dB
Prob. 13.54
Zg
I1
I2
+
Vg
+
 A B
C

  A DB 
 C D


V1
-
V2
ZL
-
By definition,
V1 = AV2 – BI2
I1 = CV2 – DI2
(1)
(2)
Let V2 and V2 be respectively the load voltages when the filter circuit is
present and when it is absent.
V2   I 2 Z L 

I1Z L
CZ L  D
Vg Z L

V 
 Z g  1 CZ L  D 
I1 


Vg Z L

AV2  BI 2 
 Z g 
CZ L  D 
CV2  DI 2 

475


Vg Z L

AZ L  B 
 Z g 
CZ L  D 
CZ L  D 

Vg Z L
Z CZ  D   AZ  B 
g
V2 
L
L
Vg Z L
Z  Z 
g
L
Ratio and modulus give
Z g CZ L  D   AZ L  B 
V2

V2
Zg  ZL
Insertion loss =
IL = 20 log10 V2  20 log10
Z CZ  D   AZ  B 
g
L
L
Zg  ZL
V2
which is the required result
Prob. 13.55
l
103
(a) Rdc =
=
= 17.1 m  /km
S
0.96  104  6.1 107
(b) Rac =
 
Rac =
l
,
 w
1
=
 f
 a2 = 0.8 x 1.2 = 0.96 or a = 0.5528
1
  6 106  4 107  6.1107
1000 12.1  103
= 51.93 
1.2  102  6.1107

1
12.1   103
476
Prob. 13.56
SE  20 log10
Ei
6
 20 log10
 20 log10 (3 105 )
6
20  10
Eo
 109.54 dB
477
CHAPTER 14
P. E. 14.1
The program in Fig. 14.3 was used to obtain the plot in Fig. 14.5.
P. E. 14.2
For the exact solution,
(D2 + 1) y = 0
y (0) = 0
y(1) = 1
y = A cos x + B sin x
A =0
1 = B sin 1 or B = 1/sin 1
Thus, y = sin x/sin 1
For the finite difference solution,
y( x   )  2 y( x)  y( x   )
 y 0
2
y’’ + y = 0
or
y( x   )  y( x   )
, y (0)  0, y (1)  1,   1 / 4
2  2
With the MATLAB program shown below, we obtain the exact result ye and FD result
y.
y( x) 
y(1)=0.0;
y(5)=1.0;
del=0.25;
for n=1:20
for k=2:4
y(k)=( y(k+1) +y(k-1) )/(2-del*del)
x=(k-1)*del;
ye=sin(x)/sin(1.0)
end
end
The results are listed below.
y(x)
N=5
y(0.25) 0.2498
y(0.5) 0.5242
y(0.75) 0.7867
N=10
N=15
N=20
Exact
ye(x)
0.2924
0.5682
0.8094
0.2942
0.5701
0.8104
0.2943
0.5702
0.8104
0.2940
0.5697
0.8101
478
P. E. 14.3 By applying eq. (14.16) to each node as shown below, we obtain the
following results after 5 iterations.
0
0
25
10.01
9.82
9.35
8.19
5.56
4.69
0
0
28.3
28.17
27.06
25
19.92
18.95
0
12.05
11.87
11.44
10.30
7.76
2.34
0
28.3
28.17
27.85
27.06
25.06
19.92
0
44.57
44.46
44.26
43.76
42.48
37.5
0
50
0
50
10.01
9.82
9.35
8.19
5.56
4.69
0
28.3
28.17
27.85
27.06
25
19.92
0
0
50
0
0
25
P. E. 14.4 (a) Using the program in Fig. 14.16 with nx = 4+1=5 and ny = 8+1=9, we
obtain the potential at center as
V(3,5) = 23.796 V
479
(b) Using the same program with nx = 12+1=13 and ny = 24+1=25, the potential at
the center is
V(7,13) = 23.883 V
P. E. 14.5 By combining the ideas in Figs. 14.20 and 14.24, and dividing each wire into
N segments, the results listed in Table 14.2 is obtained.
P. E. 14.6
(a)
3
4
2
2
1
1
For element 1, local 1-2-3 corresponds with global 1-3-4 so that A1 = 0.35,
P1 = 0.8, P2 = 0.6, P3 = -1.4, Q1 = -0.5, Q2 = 0.5, Q3 = 0
C
(1)
 0.6357 01643
.
 0.8


  01643
.
0.4357  0.6
  0.8
 0.6
14
. 
For element 2, local 1-2-3 corresponds with global 1-2-3 so that A2 = 0.7,
P1 = 0.1, P2 = 1.4, P3 = -1.5, Q1 = -1, Q2 = 0, Q3 = 1
C
(2)
 0.3607
0.05  0.4107


  0.05
0.7
 0.75 
  0.4107  0.75 11607

.
The global coefficient matrix is given by
480
 C (1) 11  C11 ( 2 )

C21( 2 )

C
 C21(1)  C31( 2 )

C31(1)

C12 ( 2 )
C12 (1)  C13 ( 2 )
C22 ( 2 )
C32 ( 2 )
C23 ( 2 )
C22 (1)  C33 ( 2 )
0
C32 ( 2 )
C13 (1) 

0 
C23 (1) 

C33 (1) 
 0.9964 0.05 0.2464 0.8 
 0.05
0.7
0.75
0 


 0.2464 0.75 1.596
0.6 


0
0.75
1.4 
 0.8
(b)
3
2
4
2
1
1
For element 1, local 1-2-3 corresponds with global 1-2-4 .
P1 = 0.9000 ; P2 = 0.6000 ; P3 = -1.5000
Q1 = -1.5000 ; Q2 = 0.5000; Q3 = 1;
A1 =
C
(1)
0.6750;
 1.1333 - 0.0778 - 1.0556 
=  - 0.0778 0.2259 - 0.1481
 - 1.0556 - 0.1481 1.2037 
For element 2, local numbering 1-2-3 corresponds with global numbering 2-3-4.
P1 = 0.8000; P2 = -0.9000 ; P3 = 0.1000 ;
Q1 = -0.5000 ; Q2 = 1.5000 ; Q3 = -1 ;
A2 =
0.3750 ;
481
C
(2)

 0.5933 -0.9800 0.3867 
 -0.9800 2.0400 -1.0600 


 0.3867 -1.0600 0.6733 
The global coefficient matrix is
 C (1) 11
 (1)
C21
C 
 0
 (1)
 C31
C

C12 (1)
C22
(1)
 C11
(2)
0
( 2)
C12
(1)
C32  C31( 2 )
C12
(2)
C22 ( 2 )
C32 ( 2 )


C23  C13 

C23 ( 2 )
(1)
(2) 
C33  C33 
C13 (1)
(1)
(2)
0
- 1.0556
 1.1333 - 0.0778
 - 0.0778 0.8193 - 0.9800 0.2385


 0
- 0.9800 2.0400 - 1.0600 


 - 1.0556 0.2385 - 1.0600 1.8770 
P. E. 14.7 We use the MATLAB program in Fig. 14.33. The input data for the
region in Fig. 14.34 is as follows:
NE = 32; ND = 26; NP = 18;
NL = [ 1 2 4
2 5 4
2 3 5
3 6 5
4 5 9
5 10 9
5 6 10
6 11 10
7 8 12
8 13 12
8 9 13
9 14 13
9 10 14
10 15 14
10 11 15
11 16 15
12 13 17
13 18 17
13 14 18
482
14 19 18
14 15 19
15 20 19
15 16 20
16 21 20
17 18 22
18 23 22
18 19 23
19 24 23
19 20 24
20 25 24
20 21 25
21 26 25];
X = [ 1.0 1.5 2.0 1.0 1.5 2.0 0.0 0.5 1.0 1.5 2.0 0.0 0.5 1.0 1.5 2.0 0.0 0.5 1.0 1.5
2 0.0 0.5 1.0 1.5 2.0];
Y = [ 0.0 0.0 0.0 0.5 0.5 0.5 1.0 1.0 1.0 1.0 1.0 1.5 1.5 1.5 1.5 1.5 2.0 2.0 2.0
2.0 2.0 2.5 2.5 2.5 2.5 2.5 ];
NDP = [ 1 2 3 6 11 16 21 26 25 24 23 22 17 12 7 8 9 4];
VAL = [0.0 0.0 15.0 30.0 30.0 30.0 30.0 25.0 20.0 20.0 20.0 10.0 0.0 0.0 0.0
0.0 0.0 0.0];
With this data, the finite element (FEM) solution is compared with the finite
difference (FD) solution as shown in the table below.
Node #
5
10
13
14
15
18
19
20
X
1.5
1.5
0.5
1.0
1.0
0.5
1.0
1.5
Y
0.5
1.0
1.5
1.5
1.5
2.0
2.0
2.0
FEM
11.265
15.06
4.958
9.788
18.97
10.04
15.32
21.05
FD
11.25
15.02
4.705
9.545
18.84
9.659
15.85
20.87
483
Prob. 14.1 (a) Using the Matlab code in Fig. 14.3, we input the data as:
>> plotit( [-1 2 1], [-1 0; 0 2; 1 0], 1, 1, 0.01, 0.01, 8, 2, 5 )
and the plot is shown below.
.
(b) Using the MATLAB code in Fig. 14.3, we input the required data as:
>> plotit( [1 1 1 1 1], [-1 -1; -1 1; 1 –1; 1 1; 0 0], 1, 1, 0.02, 0.01, 6, 2, 5 )
and obtain the plot shown below.
484
Prob.14.2
The exact solution is
1
1
1
V ( x)  x 3  x 2  x
6
2
3
so that V(0.5) = 0.3125. For the finite difference solution,
V( x  )  2V ( x)  V ( x   )
 x 1
2
which leads to
1
V( x)  V ( x  )  V ( x   )   2 ( x  1) 
2
We apply this at x = 0.25, 0.5, 0.75 for 5 iterations as tabulated below.
No. of iterations
0
1
2
3
4
5
V(0)
V(0.25)
V(0.5)
V(0.75)
V(1.0)
0
0
0
0
0
0
0
-0.03916
-0.07226
0.02254
0.06984
0.09349
0
-0.0664
0.1232
0.2178
0.2651
0.2888
0
0.4121
0.5069
0.5542
0.5779
0.5897
1
1
1
1
1
1
From the table, V(0.5) = 0.2888 which is smaller than the exact value due to the fact that
the number of iterations is not sufficiently large and also = 0.25 is large.
Prob. 14.3 (a)
dV V ( x  x)  V ( x  x)

dx
2x
485
For  x  0.05 and at x = 0.15,
dV 2.0134  100
.

 10.117
dx
0.05 X 2
d 2V V ( x  x)  2V ( x)  V ( x  x) 2.0134  1.0017  2 x1.5056


 1.56
dx 2
(x) 2
(0.05) 2
(b) V = 10 sinh x, dV/dx = 10 cosh x. At x = 0.15, dV/dx = 10.113
which is close to the numerical estimate.
d2V/dx2 = 10 sinh x. At x = 0.15, d2V/dx2 = 1.5056
which is slightly lower than the numerical value.
Prob. 14.4
 2V 
 2V 1  V  2V


0
 2    z 2
The equivalent finite difference expression is
V (  o   , zo )  2V (  o , zo )  V ( o   , zo ) 1 V (  o   , zo )  V ( o   , zo )

( ) 2
o
2

V (  o , zo   z)  2V (  o , zo )  V (  o , zo   z)
0
(  z) 2
If  z     h, rearranging terms gives
V (  o , zo ) 
 (1 
1
1
h
V (  o , zo  h)  V (  o , zo  h)  (1 
)V (   h, zo )
4
4
2 o
h
)V (   h, zo )
2 o
as expected.
486
Prob. 14.5
 2V 
 2V 1  V
1  2V


 0,
 2    2  2
(1)
Vm1n  2Vm n  Vm1n
 2V

,
 2
(  ) 2
(2)
Vm n 1  2Vm n  Vm n 1
 2V

,
(  ) 2
 2
(3)
V n m 1  V n m 1
V
.

2 
 m,n
(4)
Substituting (2) to (4) into (1) gives
 V 
2
=
V n m 1  V n m 1
m  (2   )
+
Vm1n  2Vm n  Vm1n
(  ) 2
(m    ) 2

1
1
1 
1
n 1
)Vm1 n  2Vm n  (1 
)Vm1n 
 2Vm n  Vm n 1 ) 
2  (1 
2 (Vm
2m
2m
(  ) 
(m  )

as required.
Prob. 14.6
Vo 
+
Vm n 1  2Vm n  Vm n 1
V1  V2  V3  V4 10  40  50  80

 25V
4
4
487
Prob. 14.7
Iteration

V1
V2
V3
V4
0
1
2
3
4
5
0.0000
25.0000
35.6250
38.9063
39.7266
39.9316
0.0000
26.2500
32.8125
34.4531
34.8633
34.9658
0.0000
16.2500
22.8125
24.4531
24.8633
24.9658
0.0000
15.6250
18.9063
19.7266
19.9316
19.9829
Prob. 14.8
1
1
(1)
Va   0  100  100  Vb   (Vb  200)
4
4
1
1
(2)
Vb   0  0  Va  Vc   (Va  Vc )
4
4
1
1
(3)
Vc  Vb  100  100  0  (200  Vb )
4
4
Using these relationships, we obtain the data in the table below.
Iteration
Va
Vb
Vc
1st
50
12.5
53.125
2nd
53.115
26.56
56.64
3rd
56.641
28.32
57.08
4th
57.08
28.54
57.135
5th
57.135
28.57
57.142
Alternatively, we can solve (1) to (3) simultaneously.
From (1) and (3), Va  Vc
From (2),
Vb 
Va
2
Thus (1) becomes
Vb 
1 V

Va   a  200 
4 2


Va  400 / 7  51.143  Vc
Va
 28.57
2
Prob. 14.9
(a) We follow Example 6.5 with a=b.
 n x 
 n y 
 n y 
 n x 
sin 
 sinh 
 4V  sin 
 sinh 

4Vo 
 a 
 a  o
 a 
 a 
V  V1  V2 


 n odd
n sinh(n )
 n odd
n sinh(n )
488
(b) At the center of the region, finite difference gives
V
1
V (a / 2, a / 2)  (0  0  Vo  Vo )  o  25 V
4
2
Prob. 14.10
h2 s
50 109
k
 104 
 0.18  0.5655
109

36
At node 1,
1
V1  [0  V2  V3  k ]


4V1  V2  V3  k
4
At node 2,
1
V2  [0  V1  V4  k ]

 4V2  V1  V4  k
4
At node 3,
1
V3  [0  2V1  V4  k ]


4V3  2V1  V4  k
4
At nde 4,
1
V4  [0  2V2  V3  k ]

 4V4  2V2  V3  k
4
Putting (1) to (4) in matrix form,
 4 1 1 0  V1  0.5655

 V  

 1 4 0 1  2   0.5655
 2 0 4 1 V3  0.5655

  

 0 2 1 4  V4  0.5655
Using a calculator or MATLAB, we obtain
V1  V2  0.3231 V, V3  V4  0.4039 V
Prob. 14.11
(a)
1
0
0   Va    200
4 1 0
 
 1 4 1
0
1
0   Vb    100




0
1 4 0
0
1   Vc    100
   


 1 0 0  4 1 0   Vd    100
0
1
0
1  4 1   Ve   0 
   


0
1
0
1  4 V f   0 
0
[A]
[B]
(b)
(1)
(2)
(3)
(4)
489
 4 1 0 1 0 0 0 0  V1   30 
 1 4 1 0 1 0 0 0  V   15 

 2 

 0 1 4 0 0 1 0 0  V3   30 

  

 1 0 0 4 1 0 1 0  V4    7.5
 0 1 0 1 4 1 0 1  V5   0 

  

 0 0 1 0 1 4 0 0  V6   7.5
 0 0 0 1 0 0 4 1  V   0 

 7 

 0 0 0 0 1 0 1 4  V8   0 
[A]
[B]
Prob. 14.12 (a) Matrix [A] remains the same. To each term of matrix [B], we add
h2 v /  .
(b) Let x  y  h  0.25 so that nx = 5= ny.
v x( y  1)109

 36 x( y  1)
109 / 36

Modify the program in Fig. 14.16 as follows.
H=0.25;
for I=1:nx –1
for J=1: ny-1
X = H*I;
Y=H*J;
RO = 36.0*pi*X*(Y-1);
V(I,J) = 0.25*( V(I+1,J) + V(I-1,J) + V(I,J+1) + V(I,J-1) + H*H*RO );
end
end
This is the major change. However, in addition to this, we must set
v1 = 0.0;
v2 = 10.0;
v3 = 20.0;
v4 = -10.0;
nx = 5;
ny = 5;
The results are:
490
Va = 4.6095 Vb= 9.9440 Vc= 11.6577
Vd = -1.5061 Ve =3.5090 Vf= 6.6867
Vg= -3.2592 Vh = 0.2366 Vi = 3.3472
Prob. 14.13
1
1
V1  (0  0  V2  V4 )  (V2  V4 )
4
4
1
1
V2  (0  50  V1  V3 )  (50  V1  V3 )
4
4
1
1
V3  (0  100  50  V2 )  (150  V2 )
4
4
1
1
V4  (0  50  V1  V5 )  (50  V1  V5 )
4
4
1
1
V5  (0  0  V4  V6 )  (V4  V6 )
4
4
1
1
V6  (0  50  V5  V7 )  (50  V5  V7 )
4
4
1
1
V7  (0  100  V6  50)  (150  V6 )
4
4
Initially set all free potentials equal to zero. Apply the seven formulas above iteratively
and obtain the results shown below.
n
V1
V2
V3
V4
V5
V6
V7
1
0
12.5
40.625
12.5
3.12
13.281
40.82
2
6.25
24.22
43.55
14.84
7.03
24.46
43.62
3
9.77
25.83
43.96
16.70
10.29
25.98
43.99
4
10.63
26.15
44.04
17.73
10.93
26.23
44.06
Prob. 14.14
1  j  1 m , n   j  1 m ,n  2  j m , n
 j m1,n   j m1,n  2 j m,n

c2
( t ) 2
(  x) 2

 j m , n  1   j m , n  1  2  j m ,n
(  z) 2
If h   x   z , then after rearranging we obtain
5
10.97
26.25
44.06
17.97
11.05
26.28
44.07
491
 j 1m ,n  2 j m, n   j 1m ,n   ( j m 1,n   j m 1,n  2 j m ,n )
 ( j m ,n 1   j m,n 1  2 j m, n )
where   (c t / h) 2 .
Prob. 14.15
 2V  2V
V ( x  x, t )  2V ( x, t )  V ( x  x, t )
 2



2
x
t
(x) 2
V ( x, t  t )  2V ( x, t )  V ( x, t  t )
(t ) 2
 t 
V ( x, t  t )    V ( x  x, t )  2V ( x, t )  V ( x  x, t   2V ( x, t )  V ( x, t  t )
 x 
2
or
V (i, j  1)   V (i  1, j )  v(i  1, j )   2(1   )V (i, j )  V (i, j  1)
 t 
where     . Applying the finite difference formula derived above, the following
 x 
2
programs was developed.
xd=0:.1:1;td=0:.1:4;
[t,x]=meshgrid(td,xd);
Va=sin(pi*x).*cos(pi*t);%Analytical result
subplot(211) ;mesh(td,xd,Va);colormap([0 0 0])
% Numerical result
N=length(xd);M=length(td);
v(:,1)=sin(pi*xd');
v(2:N-1,2)=(v(1:N-2,1)+v(3:N,1))/2;
for k=2:M-1
v(2:N-1,k+1)=-v(2:N-1,k-1)+v(1:N-2,k)+v(3:N,k);
end
subplot(212);mesh(td,xd,v);colormap([0 0 0])
The results of the finite difference algorithm agree perfectly with the exact solution as
shown below.
492
Prob. 14.16
The MATLAB code and the plot of F(u) are presented below.
% Integration using MATLAB
N=40;
del=pi/N;
for k=1:21
u(k)=0.1*(k-1);
sum=0.0;
for n=1:N
theta=del*n;
num= u(k)- cos(theta)
den=( 1 +u(k)^2 - 2*u(k)*cos(theta) )^1.5;
term = num/den;
sum=sum + term;
end
f(k) = sum*del;
end
plot(u,f)
title('f as a function of u')
493
f as a function of u
8
6
4
2
0
-2
-4
-6
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
2
Prob. 14.17 Combining the ideas in the programs in Figs. 14.20 and 14.24, we develop a
MATLAB code which gives
N = 20
C = 19.4 pF/m
N = 40
C = 13.55 pF/m
N = 100
C = 12.77 pF/m
For the exact value, d/2a = 50/10 = 5
C

cosh 1
d
2a

 109 / 36
cosh 1 5
 12.12 pF/m
494
Prob. 14.18
y

x
h
To find C, take the following steps:
(1)Divide each line into N equal segments. Number the segments in the lower conductor
as 1, 2, …, N and segments in the upper conductor as N+1, N+2, …, 2N,
(2) Determine the coordinate (xk, yk) for the center of each segment.
For the lower conductor, yk = 0, k=1, …, N, xk = h +  (k-1/2), k = 1,2,… N
For the upper conductor, yk = [h +  (k-1/2)] sin  , k=N+1, N+2, …,2 N,
xk = [h +  (k-1/2)] cos  , k = N+1,N+2,… 2N
where h is determined from the gap g as
g
h
2 sin  / 2
(3)Calculate the matrices [V] and [A] with the following elements
 Vo , k  1,..., N
Vk  
  Vo , k  N  1,...2 N
 
,i  j

Aij   4 Rij
 2 ln  / a , i  j
where Rij 
( xi  x j ) 2  ( yi  y j ) 2
(4) Invert matrix [A] and find [  ] = [A]-1 [V].
(5) Find the charge Q on one conductor
495
N
Q   k     k
k 1
(6) Find C = |Q|/2Vo
Taking N= 10, Vo = 1.0, a program was developed to obtain the following result.

10
20
30
40
50
60
70
80
90
100
110
120
130
140
150
160
170
180
C (in pF)
8.5483
9.0677
8.893
8.606
13.004
8.5505
9.3711
8.7762
8.665
8.665
10.179
8.544
9.892
8.7449
9.5106
8.5488
11.32
8.6278
Prob. 14.19 We may modify the program in Fig. 14.24 and obtain the result in the table
below. Z o  100  .
N
10
20
30
40
50
Zo, in 
97.2351
97.8277
98.0515
98.1739
98.2524
Prob. 14.20
We make use of the formulas in Problem 14.19.
2N
Vi   Aij i
j 1
496
where N is the number of divisions on each arm of the conductor.
The MATLAB code is as follows:
aa=0.001;
vo=10;
eo = (10^(-9))/(36*pi);
L=2.0;
N=10; %no.of divisions on each arm
NT=N*2;
delta=L/(NT);
x=zeros(NT,1);
y=zeros(NT,1);
%Second calculate the elements of the coefficient matrix
for i=1:N-1
y(i)=0;
x(i)=delta*(i-0.5)
end
for i=N+1:NT
x(i)=0;
y(i)=delta*(i-N-0.5);
end
for i=1:NT
for j=1:NT
if (i ~=j)
R=sqrt( (x(i)-x(j))^2 + (y(i)-y(j))^2 )
A(i,j)=-delta*R;
else
A(i,j)=-delta*(log(delta)-1.5);
end
end
end
%Determine the matrix of constant vector B and find rho
B=2*pi*eo*vo*ones(NT,1);
rho=inv(A)*B;
The result is presented below.
497
Segment
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
x
0.9500
0.8500
0.7500
0.6500
0.5500
0.4500
0.3500
0.2500
0.1500
0.0500
0
0
0
0
0
0
0
0
0
0
y
0
0
0
0
0
0
0
0
0
0
0.0500
0.1500
0.2500
0.3500
0.4500
0.5500
0.6500
0.7500
0.8500
in pC/m
89.6711
80.7171
77.3794
75.4209
74.0605
73.0192
72.1641
71.4150
70.6816
69.6949
69.6949
70.6816
71.4150
72.1641
73.0192
74.0605
75.4209
77.3794
80.7171
89.6711
Prob. 14.21(a) Exact solution yields
C  2 / ln( / a )  8.02607  1011 F/m and Z o  41559
. 
where a = 1cm and  = 2cm. The numerical solution is shown below.
498
N
10
20
40
100
Z o ( )
40.486
41.197
41.467
41.562
C (pF/m)
82.386
80.966
80.438
80.025
(b)For this case, the numerical solution is shown below.
N
10
20
40
100
Z o ( )
30.458
30.681
30.807
30.905
C (pF/m)
109.51
108.71
108.27
107.93
Prob. 14.22 We modify the MATLAB code in Fig. 14.24 (for Example 14.5) by
changing the input data and matrices [A] and [B]. We let
xi = h +  (i-1/2), i = 1,2,… N,
 = L/N
yi = h /2, j = 1,2,… N, zk = t/2, k = 1,2,… N
and calculate
Rij 
( x i  x j ) 2  ( y i  y j ) 2  ( zi  z j ) 2
We obtain matrices [A] and [B]. Inverting [A] gives
N
q
i
i 1
[q] = [A]-1 [B], [  v ] = [q]/(ht  ), C 
10
The computed values of [  v ] and C are shown below.
499
I
vi (106 )C / m3
1, 20
2, 19
3, 18
4, 17
5, 16
6, 15
7, 14
8, 13
9, 12
10,11
0.5104
0.4524
0.4324
0.4215
0.4144
0.4096
0.4063
0.4041
0.4027
0.4020
C = 17.02 pF
Prob. 14.23
From the given figure, we obtain
1 x
A1
1
1 

1 x2
A 2A
1 x3
y
1
y2 
[( x y  x 3 y 2 )  ( y 2  y 3 ) x  ( x3  x2 ) y]
2A 2 3
y3
as expected. The same applies for  2 and  3 .
Prob. 14.24
(a)
P1  1.5, P2  0.5, P3  2, Q1  1, Q2  1.5, Q3  0.5
1
( P2Q3  P3Q2 )  1.375
2
1
Cij 
[ PP
i j  Qi Q j ]
4A
 0.5909 0.1364 0.4545 
C   0.1364 0.4545 0.3182 
 0.4545 0.3182 0.7727 
A
(b)
P1  4, P2  4, P3  0, Q1  0, Q2  3, Q3  3
1
A  ( P2Q3  P3Q2 )  6
2
0 
 0.6667 0.6667

C   0.6667 1.042
0.375
 0
0.375 0.375 
500
Prob. 14.25 (a)
1 1/ 2 1/ 2
2A = 1 3 1 / 2 = 15/4
1 2
2
1 
4
1
4
[(6  1)  (  1 ) x  (  1) y ] 
(5  15
. x  y)
15
2
15
2 
4
3
3
4
[(1  1)  x  y ] 
(15
. x  15
. y)
15
2
2
15
3 
4
5
4
[(1 / 4  3 / 2)  0 x  y ] 
(  125
.  2.5 y )
15
2
15
V   1V1   2V2   3V3
Substituting V=80, V1 = 100, V2 = 50, V3 = 30,  1 ,  2 , and  3 leads to
20 = 7.5x + 10y + 3.75
Along side 12, y=1/2 so that
20 = 15x/2 + 5 + 15/4
x=3/2, i.e (1.5, 0.5)
Along side 13, x =y
20 = 15x/2 + 10x + 15/4
Along side 23, y = -3x/2 + 5
x=13/4, i.e. (13/14, 13/14)
20 = 15x/2 – 15 + 50 + 15/4
x=-5/2 (not possible)
Hence intersection occurs at
(1.5, 0.5) along 12 and (0.9286, 0.9286) along 13
(b) At (2,1),
1 
4
6
5
, 2 
, 3 
15
15
15
V (2,1)  1V1   2V2   3V3 = (400 + 300 + 150)/15 = 56.67 V
501
Prob. 14.26
1 0 0
2A = 1 2  1  9
1 1 4
1 
1
1
[(0  0)  (4  0) x  (0  1) y ]  (4 x  y )
9
9
2 
1
1
[(0  0)  (0  1) x  (2  0) y ]  ( x  2 y )
9
9
3 
1
1
[(8  1)  (  1  4) x  (1  2) y ]  (9  5x  y )
9
9
Ve   1Ve1   2Ve 2   31Ve 3
V(1,2) = 8(4-2)/9 + 12(1+4)/9 + 10(9-5-1)/9 = 96/9 = 10.667 V
At the center  1 =  2 =  3 = 1/3 so that
V(center) = (8 + 12 + 10)/3 = 10
Or at the center, (x, y) = (0 + 1 + 2, 0 + 4 –1)/3 = (1,1)
V(1,1) = 8(3)/9 + 12(3)/9 + 10(3)/9 = 10 V
Prob. 14.27
(3,12)
(3,12)
(8,12)
2
1
(8,0)
(8,0)
(0,0)
For element 1, local numbering 1-2-3 corresponds to global numbering 4-2-1.
P1 = 12, P2 = 0, P3 = -12, Q1 = -3, Q2 = 8, Q3 = -5,
502
A = (0 + 12 x 8)/2 = 48
Cij 
C
(1)
1
[ P P  QjQj ]
4 x 48 j i
 0.7969 0.125 0.6719 
  0.125 0.3333 0.2083
 0.6719 0.2083 0.8802 
For element 2, local numbering 1-2-3 corresponds to global numbering 2-4-3.
P1 = -12, P2 = 0, P3 = 12, Q1 = 0, Q2 = -5, Q3 = 5,
A = (0 + 60)/2 = 30
Cij 
C
(2)
1
[ P P  QjQj ]
4 x 48 j i
0
1.2 
 1.2

 0
0.208 0.208
 1.2 0.208 1.408 
 C (1) 33

C23 (1)

C
 0

(1)
 C13
C23 (1)
C22 (1)  C11( 2 )
C31( 2 )
C21(1)  C21( 2 )
0
C13 ( 2 )
C33 ( 2 )
C23 ( 2 )

C31(1)
(1)
(2) 
C21  C12 

C32 ( 2 )
(2)
(1) 
C22  C11 
0
0.6719 
 0.8802 0.2083
 0.2083 1.533
1.2
0.125 


 0
1.4083 0.2083
1.2


 0.6719 0.125 0.2083 1.0052 
503
Prob. 14.28
4
(0,2) 3
3

2
(2,2)
1
1
1
(0,0)
1
3
2

2
2
(4,0)
For element 1,
P1  0, P2  2, P3  2, Q1  2, Q2  0, Q3  2
1
A  (4  0)  2,
4A  8
2
0.5
0
 4 0 4   0.5
1


C   0 4 4    0
0.5 0.5
8
 4 4 8   0.5 0.5
1 
For element 2,
(1)
P1  2, P2  2, P3  0, Q1  2, Q2  2, Q3  4
1
A  (8  0)  4,
4 A  16
2
0.5
0
 8 0 8  0.5
1 


(2)
C   0 8 8   0
0.5 0.5
16
 8 8 16   0.5 0.5
1 
The global coefficient matrix is
 C11 C12
C
C22
C   21
C31 C32

C41 C42
C13
C23
C33
C43
C14  C11(1)  C11(2)

(2)
C24   C21
 (1)
C34  C21  C31(2)
 
C44   C31(1)
0.5 0.5
0
 1
 0
0.5 0.5
0 

 0.5 0.5 1.5 0.5


0
0.5
1 
 0.5
C12(2)
C12(1)  C13(2)
C22(2)
C32(2)
C23(2)
(1)
 C33(2)
C22
0
C32(1)
C13(1) 

0 
C23(1) 

C33(1) 
504
Prob. 14.29
4
(2,2)
3
(0,1)
1
1
4
(2,2)
1
1
2
2 2
(1,0)
2 2
(1,0)
3
3
(3,0)
For element 1, local numbering 1-2-3 corresponds to global numbering 1-2-4.
P1 = -2, P2 = 1, P3 = 1, Q1 = 1, Q2 = -2, Q3 = 1,
A = (P2 Q3 - P3 Q2 )/2 = 3/2, i.e. 4A = 6
Cij 
1
[ P P  QjQj ]
4A j i
(1)
 5  4  1
1

   4 5  1
6
  1  1 2 
C
For element 2, local numbering 1-2-3 corresponds to global numbering 4-2-3.
P1 = 0, P2 = -2, P3 = 2, Q1 = 2, Q2 = -1, Q3 = -1,
A = 2, 4A = 8
C
(2)
 4  2  2
1

   2 5  3
8
  2  3 5 
The global coefficient matrix is
505
 C (1) 11

C12 (1)

C
 0

(1)
 C13
C12 (1)
C22 (1)  C22 ( 2 )
C23 ( 2 )
C23 (1)  C21( 2 )
0
C23 ( 2 )
C33 ( 2 )
C31( 2 )

C13 (1)
(1)
(2) 
C23  C21 

C31 ( 2 )

C33 (1)  C11( 2 ) 
0
 01667
.
 0.8333  0.667

  0.6667 14583
 0.375  0.4167
.



 0
 0.375 0.625
 0.25 


0.833 
 0.4167  0.25
.
  01667
Prob. 14.30
We can do it by hand as in Example 14.6. However, it is easier to prepare
an input file and use the program in Fig. 14.54. The MATLAB input data is
NE = 2;
ND = 4;
NP = 2;
NL = [1 2 4
4 2 3];
X = [ 0.0 1.0 3.0 2.0];
Y = [ 1.0 0.0 0.0 2.0];
NDP= [ 1 3 ];
VAL = [ 10.0 30.0]
 10
 18
 
The result is V   30
 20
 
From this,
V2 = 18 V, V4 = 20 V
506
Prob. 14.31
Compare your finite element solution to the exact or finite difference solution:
V5 = 25 V
Prob. 14.32 As in P. E. 14.7, we use the program in Fig. 14.33. The input data based
on Fig. 14.64 is as follows.
NE =50; ND= 36; NP= 20;
NL = [1
8
7
1
2
8
2
9
8
2
3
9
3
10
9
3
4
10
4
11
10
4
5
11
5
12
11
5
6
12
7
14
13
7
8
14
8
15
14
8
9
15
9
16
15
9
10
16
10
17
16
10
11
17
11
18
17
11
12
18
13
20
19
13
14
20
14
21
20
14
15
21
15
22
21
15
16
22
16
23
22
16
17
23
17
24
23
17
18
24
19
26
25
19
20
26
20
27
26
20
21
27
507
21
28
27
21
22
28
22
29
28
22
23
29
23
30
29
23
24
30
25
32
31
25
26
32
26
33
32
26
27
33
27
34
33
27
28
34
28
35
34
28
29
35
29
36
35
29
30
36];
X = [0.0 0.2 0.4 0.6 0.8 1.0 0.0 0.2 0.4 0.6 0.8 1.0 0.0 0.2 0.4 0.6 0.8 1.0
0.0 0.2 0.4 0.6 0.8 1.0 0.0 0.2 0.4 0.6 0.8 1.0 0.2 0.4 0.6 0.8 1.0];
Y = [0.0 0.0 0.0 0.0 0.0 0.0 0.2 0.2 0.2 0.2 0.2 0.2 0.4 0.4 0.4 0.4 0.4
0.4 0.6 0.6 0.6 0.6 0.6 0.6 0.8 0.8 0.8 0.8 0.8 0.8 1.0 1.0 1.0 1.0 1.0 1.0];
NDP = [ 1 2 3 4 5 6 12 18 24 30 36 35 34 33 32 31 25 19 13 7];
VAL = [ 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 50.0 100.0 100.0 100.0
100.0 50.0 0.0 0.0 0.0 0.0];
With this data, the potentials at the free nodes are compared with the exact values as
shown below.
Node no.
8
9
10
11
14
15
16
17
20
21
22
23
26
27
28
29
FEM Solution
4.546
7.197
7.197
4.546
10.98
17.05
17.05
10.98
22.35
32.95
32.95
22.35
45.45
59.49
59.49
45.45
Exact Solution
4.366
7.017
7.017
4.366
10.60
16.84
16.84
10.60
21.78
33.16
33.16
21.78
45.63
60.60
60.60
45.63
508
Prob. 14.33 We use exactly the same input data as in the previous problem except that
the last few lines are replaced by the following lines.
VAL = [ 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 29.4 58.8 95.1 95.1
58.8 29.4 0.0 0.0 0.0 0.0];
The potential at the free nodes obtained with the input data are compared with the exact
solution as shown below.
Node no.
8
9
10
11
14
15
16
17
20
21
22
23
26
27
28
29
FEM Solution
3.635
5.882
5.882
3.635
8.659
14.01
14.01
8.659
16.99
27.49
27.49
16.99
31.81
51.47
51.47
31.81
Exact Solution
3.412
5.521
5.521
3.412
8.217
13.30
13.30
8.217
16.37
26.49
26.49
16.37
31.21
50.5
50.5
31.21
509
Prob. 14.34
For element 1, the local numbering 1-2-3 corresponds with nodes with V1 , V2 , and
V3.
Vo  
1 4
VC
Coo i 1 i io
4
Coo   Coj
j 1
(e)

1
1
(hh  hh)  2  2 (hh  0)  4  4
2
4h / 2
4h / 2
Co1 
2 1
2
[ P3 P1  Q3Q1 ]  2 [hh  0]  1
2
2h
2h
Co 2 
2 1
2
[ P1 P2  Q1Q2 ]  2 [ h  0  h  (h)]  1
2
2h
2h
Similarly, C03 = -1 = C04. Thus
Vo = ( V1 + V2 + V3 + V4 )/4
which is the same result obtained using FDM.
510
Prob. 4.35
V
1
V1  (0  0  100  V2 )  25  2
4
4
V V
1
V2  (0  100  V1  V3 )  25  1 3
4
4
1
V2
V3  (0  0  100  V2 )  25 
4
4
V5
1
V4  (0  0  100  V5 )  25 
4
4
(V  V )
1
V5  (0  0  100  V4  V6 )  25  4 6
4
4
V
1
V6  (0  0  100  V5 )  25  5
4
4
We initially set V1 = V2 = V3 = V4 =V5 = V6 = 0 and then apply above formulas
iteratively. The solutions are presented in the table below.
iteration
V1
V2
V3
V4
V5
V6
1st
25
31.25
32.81
25
31.25
32.81
V1  V4  35.71 V,
2nd
32.81
41.41
35.35
23.81
41.41
35.35
V2  V5  42.85V,
3rd
35.35
42.68
35.67
35.35
42.68
35.67
4th
35.67
42.83
35.71
35.67
42.83
35.71
5th
35.71
42.85
35.71
35.71
42.85
35.71
V3  V6  35.71 V
Alternatively, if we take advantage of the symmetry, V1  V3  V4  V6 and V2  V5 . We
need to find solve two equations, namely,
V1  25  V2 / 4
V2  25  V1 / 2
Solving these gives
V1  35.714
V2  42.857
Other node voltages follow.
Prob. 14.36
511
On the interface,
1
1
2
3
 ,

2(1   2 ) 8
2(1   2 ) 8
V 3V
V1  2  3  12.5
4
8
3V V
V2  12.5  4  1
8
4
1
V3  (V1  V4 )
4
1
V4  (V2  V3 )
4
Applying this iteratively, we obtain the results shown in the table below.
No. of iterations
V1
V2
V3
V4
0
1
2
3
4
5…
100
0
0
0
0
12.5
15.62
3.125
4.688
17.57
18.65
5.566
6.055
19.25
19.58
6.33
6.477
19.77
19.87
6.56
6.608
19.93
19.96
6.6634
6.649
20
20
6.667
6.667
512
Prob. 14.37
The MATLAB code is similar to the one in Fig.14.40. When the program is run, it gives
Z o  40.587  .
Prob. 14.38
The finite difference solution is obtained by following the same steps as in Example 14.8.
We obtain Z o  43 
Prob.14.39
1
1
V1  (V2  100  100  100)  V2  75
4
4
1
V2  (V1  V4  2V3 )
4
1
1
V3  (V2  V5  200)  (V2  V5 )  50
4
4
1
V4  (V2  V7  2V5 )
4
1
V5  (V3  V4  V6  V8 )
4
1
1
V6  (V5  V9  200)  (V5  V9 )  50
4
4
1
1
V7  (V4  2V8  0)  (V4  2V8 )
4
4
1
V8  (V5  V7  V9 )
4
1
1
V9  (V6  V8  100  0)  (V6  V8 )  25
4
4
Using these equations, we apply iterative method and obtain the results shown below.
1st
V1
V2
V3
V4
V5
V6
75
18.75
54.69
4.687
14.687
53.71
2nd
3rd
4th
5th
79.687
48.437
65.82
19.824
35.14
68.82
87.11
59.64
73.87
34.57
49.45
74.2
89.91
68.06
79.38
46.47
57.24
77.01
92.01
74.31
82.89
53.72
61.78
78.6
513
V7
V8
V9
1.172
4.003
39.43
6.958
20.557
47.34
18.92
28.93
50.78
26.08
33.53
52.63
30.194
36.153
53.69
Prob. 14.40
Applying the difference method,
V V
V1  3  2  25
4 2
1
V2  (V1  V4 )  50
4
1
V3  (V1  2V4 )
4
1
V4  (V2  V3  V5 )
4
V
V5  4  50
4
Applying these equations iteratively, we obtain the results below.
Iterations
0
1
2
3
4
5…
100
V1
V2
V3
V4
V5
0
0
0
0
0
25.0
56.25
6.25
15.63
53.91
54.68
67.58
17.58
35.74
58.74
64.16
74.96
33.91
41.96
60.49
70.97
78.23
38.72
44.86
61.09
73.79
79.54
40.63
45.31
61.37
74.68
80.41
41.89
45.95
51.49
2
N
y
up
uW
uz
W
x
E
S
P. E. 1.4
Using the dot product,
cos AB 
A B

AB
13
 13

50
10 65
 AB  120.66 
P. E. 1.5
(a) E F   E  a F a F 
E  F F   104,10,5
F
2
141
  0.2837a x  0.7092a y  0.3546a z
ax
ay
az
(b) E  F  0
3
4  55,16,12 
4
 10
5
a E F   0.9398,0.2734,0.205
P. E. 1.6
a + b + c = 0 showing that a, b, and c form the sides of a triangle.
a  b  0,
hence it is a right angle triangle.
1
1
1
ab  bc  ca
2
2
2
1
1 4 0 1 1
ab 
 3,17,12
2
21 3 4
2
Area 
Area 
1
9  289  144  10.51
2
3
P. E. 1.7
 x2  x1    y2  y1    z2  z1 
(a) P1 P2 
2
2
2
 25  4  64  9.644

(b) rP  rP1   rP2  rP1

 1,2,3    5,2,8
 1  5 ,2  2 ,3  8 .
(c) The shortest distance is
d  P1 P3 sin   P1 P3  a P1P2

1
6
3 5
93  5  2 8
1
 14,73,27   8.2

93
Prob.1.1
rOP  4a x  5a y  a z
arOP 
rOP
(4, 5,1)

 0.6172a x  0.7715a y  0.1543a z
| rOP |
(16  25  1)
Prob. 1.2
Method 1:
rAB  rB  rA ,
rBC  rC  rB ,
rCA  rA  rC
rAB  rBC  rCA  rB  rA  rC  rB  rA  rC  0
Method 2
rAB  rB  rA  (2, 0,3)  (4, 6, 2)  (6, 6,1)
rBC  rC  rB  (10,1, 7)  (2, 0,3)  (12,1, 10)
rCA  rA  rC  (4, 6, 2)  (10,1, 7)  (6, 7,9)
rAB  rBC  rCA  (0, 0, 0)  0
Prob. 1.3
(a)
4
A  3B  (4, 2, 6)  3(12,18, 8)  (4, 2, 6)  (36,54, 24)
 (32, 56, 30)
(b)
2 A  5B  2(4, 2, 6)  5(12,18, 8)  (68,86, 28)
| B | 122  182  82  532  23.065
(2 A  5B )/ | B | (68,86, 28) / 23.065  2.948a x  3.728a y  1.214a z
(c )
ax  A 
1 0 0
 6a y  2a z
4 2 6
(d)
B  ax 
12 18 8
1
0
0
 8a y  18a z
( B  a x )  a y  8
Prob. 1.4
(a) A B  (10, 6,8) (1, 0, 2)  10  16  26
A B 
(b)
10 6 8
1
0
2
 (12  0)a x  (8  20)a y  (0  6)a z
 -12a x 12a y  6a z
(c) 2 A  3B  (20, 12,16)  (3, 0, 6)  17a x  12a y  10a z
Prob. 1.5
(a) A  B  C  (2,5,1)  (1, 0, 3)  (4, 6,10)  (1, 1,8)
(b)
BC 
1
0
3
4 6 10
 (18, 2, 6)
A  ( B  C )  (2,5,1)  (18, 2, 6)  36  10  6  32
(c) cos  AB 
A B
2  0  3

 0.05773
AB
4  25  1 1  9
Prob. 1.6
(a)
BC 
1 1 1
 a x  2a y  a z
0 1 2
A ( B  C )  (1, 0, 1) (1, 2,1)  1  0  1  0

 AB  86.69o
5
(b)
A B 
1 0 1
1 1
1
 a x  2a y + a z
( A  B ) C  (1, 2,1) (0,1, 2)  0  2  2  0
(c ) A  ( B  C ) 
1 0 1
 2a x  2a y  2a z
1 2 1
(d) ( A  B )  C 
1 2 1
 5a x  2a y  a z
0 1 2
Prob.1.7
(a)
T = (4, 6, -1) and S = (10, 12, 8)
(b) rTS  rS  rT  (10,12,8)  (4, 6, 1)  6a x  6a y  9a z
(c ) TS | rTS | 36  36  81  12.37
Prob. 1.8
(a) If A and B are parallel, B=kA, where k is a constant.
Bx  kAx ,
By  kAy ,
Bz  kAz
3  k (1)
For Bz ,

k  3
Bx    kAx  (3)(4)  12
By    kAy  (3)(2)  6
  12,   6
Hence,
(b) If A and B are perpendicular to each other,
A B  0


4  2  3  0
Prob. 1.9
(b)
10 5 2
 2a x  10a z
0 1 0
A  a z  2
(c)
cos  z 
(a)
A ay 
A  az
2

100  25  4 11.358

 z  100.14o
6
Prob. 1.10
(a) A  B  ABcos  AB
A  B  ABsin  AB an
 A  B   A  B   AB   cos 2  AB  sin 2  AB    AB
2
2
2
2
(b) a x  a y  a z   a x  a x  1. Hence,
a y  az
a x  a y  az

ax
 ax
1
ay
az  a x

 ay
a x  a y  az
1
ax  ay
a x  a y  az

az
 az
1
Prob. 1.11
(a) P  Q   6, 2, 0  , P  Q  R   7,1, 2 
P  Q  R  49  1  4  54  7.3485
2 1 2
(b) P.Q  R  4 3 2  2  6  2    8  2   2  4  3  8  10  14  4
1 1 2
Q R 
4 3 2
  4, 10, 7 
1 1 2
P.Q  R   2, 1, 2    4, 10, 7   8  10  14  4
(c) Q  P 
4
3
2
2 1 2
  4,12, 10 
Q  P R   4,12, 10    1,1, 2   4  12  20  4
or
1
1
2
Q  P  R  R Q  P  4
3
2    6  2    8  4   2  4  6   4
2
1 2
(d)  P  Q   Q × R    4, 12,10    4, 10, 7   16  120  70  206
(e)  P × Q  × Q × R  
4 12 10
 16ax  12a y  8az
4 10 7
7
(f) cos  PR 
 2  1  4   7  0.9526
PR

P R
4 1 4 11 4 3 6
 PR  162.3
(g) sin  PQ 
PQ
16  144  100
260


 0.998
P Q
3 16  9  4
3 29
 PQ  86.45
Prob. 1.12
A B  (4, 6,1) (2, 0,5)  8  0  5  13
(a) | B |2  22  52  29
A B + 2 | B |2  13  2  29  71
(b)
a  
Let
A B
| A B |
C  A B =
a  
4 6 1
2
0
5
 (30, 18,12)
(30, 18,12)
C

 (0.8111a x  0.4867a y  0.3244a z )
|C |
302  182  122
Prob. 1.13
P Q  (2, 6,5) (0,3,1)  0  18  5  13
2 6 5
 21a x - 2a y  6a z
0 3 1
P Q
13
cos  PQ 

 0.51 
  PQ  120.66o
PQ
10 65
P Q 
Prob. 1.14
P and Q are orthogonal if the angle between them is 90o. Hence
P  Q  PQ cos   0
P  Q  (2, 4, 6)  (5, 2,3)  10  8  18  0
showing that they are perpendicular or orthogonal.
8
Prob. 1.15
(a) Using the fact that
 A  B   C   A  C B  B  C A,
we get
︵
2
  
 ︶︵





 ︶︵
 ︶︵
B
A
(b) A ×  A ×  A × B   =
︵
B A
A A
A - A
A
A
B
B
A B A
A
A A = =
A   A  B    A  B   A  B  AA   A  AB
︶
︶
since AxA = 0
P2
Prob. 1.16
a
b
P1
c
P3
a  rp 2  rp1  (1, 2, 4)  (5, 3,1)  (4,1,3)
(a) b  rp 3  rp 2  (3,3,5)  (1, 2, 4)  (2,5,1)
c  rp1  rp 3  (5, 3,1)  (3,3,5)  (2, 6, 4)
Note that a + b + c = 0

 perpendicular
a b  8  5  3  0
b  c  4  30  4  0
c  a  8  6  12  0
Hence P2 is a right angle.
1
1 4 1 3 1
| a  b |
 | (1  15)a x  (6  4)a y  (20  2)a z |
2
2 2 5 1 2
Area =
(b)

1
1
| (14,10, 22) |
196  100  484  13.96
2
2
9
Prob. 1.17
Given rP  ( 1, 4,8),
rQ  (2, 1,3),
rR  (1, 2,3)
(a) | PQ | 9  25  25  7.6811
(b) PR  2a y  5a z
(c )
QP  (1, 4,8)  (2, 1,3)  (3,5,5)
QR  (1, 2,3)  (2, 1,3)  (3,3, 0)
QP QR
9  15  0

 0.7365
| QP || QR |
59 18
PQR  cos 1  0.7365   42.64o
(d) Area 
1
1 3 5 5
QP  QR 
 0.5 (15, 15,8)  0.5 225  225  36  10.677
2
2 3 3 0
(e) Perimeter  PQ  QR  RP  59  18  29  17.31
Prob.1.18
Let R be the midpoint of PQ.
1
rR  {(2, 4, 1)  (12,16,9)}  (7,10, 4)
2
OR  49  100  16  165  12.845
OR 12.845
t

 42.82 ms
v
300
Prob. 1.19
Area = Twice the area of a triangue
= | D  E |
4
1 5
1 2 3
| (3  10)a x  (5  12)a y  (8  1)a z |
| (7, 19,9) | 49  361  81  22.16
Prob. 1.20
(a) Let P and Q be as shown below:
y
Q
P
2
1
x
10
P  cos 2  1  sin 2  1  1, Q  cos 2  2  sin 2  2  1,
Hence P and Q are unit vectors.
(b) P  Q  (1)(1)cos( 2 -1 )
But P  Q  cos 1 cos  2  sin 1 sin  2 . Thus,
cos( 2  1 )  cos 1 cos  2  sin 1 sin  2
Let P1  P  cos 1a x  sin  1a y and
Q1  cos 2 a x  sin  2 a y .
P1 and Q1 are unit vectors as shown below:
y
P1
1
1+2
x
2
Q1
P1  Q1  (1)(1) cos( 1   2 )
But P1  Q1  cos 1 cos 2  sin  1 sin  2 ,
cos( 2   1 )  cos 1 cos 2  sin  1 sin  2
Alternatively, we can obtain this formula from the previous one by replacing
2 by -2 in Q.
(c )
1
1
| P  Q | | (cos 1  cos  2 )ax  (sin 1  sin  2 )a y |
2
2

1
cos2  1  sin 2  1  cos2  2  sin 2  2  2 cos  1 cos  2  2 sin  1 sin  2
2
1
1
2  2(cos  1 cos  2  sin  1 sin  2 ) 
2  2 cos( 2   1 )
2
2
Let  2   1   , the angle between P and Q.

1
1
| P  Q |
2  2 cos 
2
2
11
But cos 2A = 1 – 2 sin 2A.
1
1
| P  Q |
2  2  4sin 2  / 2  sin  / 2
2
2
Thus,
 
1
| P  Q || sin 2 1 |
2
2
Prob. 1.21

 (1, 2, 2)
3
u  r 
 (1, 2, 2), r  rp  ro  (1,3, 4)  (2, 3,1)  (1, 6,3)
1 2 2
 (18, 5, 4)
1 6 3
u  18ax  5a y  4az
Prob. 1.22
r1  (1,1,1),
r2  (1, 0,1)  (0,1, 0)  (1, 1,1)
r  r (1  1  1) 1
cos   1 2 


   70.53o
3
r1r2
3 3
Prob. 1.23
T  S  2, 6,3  1, 2,1 7
(a) Ts  T  as 


 2.8577
S
6
6
(b) S T   S  a T a T 
S  T T   72,6,3
72
T2
  0.2857a x  0.8571a y  0.4286a z
(c) sin  TS 
TS
T S

2  6 3  12,1,10
245


 0.9129
1 2 1
7 6
7 6
  TS  65.91
Prob. 1.24
12
Let
A  AB  AB 
AB  ( A  a B )a B 
A B
B
BB
Hence,
AB  A  AB  A 
A B
B
BB
Prob. 1.25
(a)
A  B  20  0  10  10
(b)
A B 
(c)
20 15 10
 15a x  30a y  15a z
1 0
1
( A  B ) B 10(a x  a z )
AB  ( A  a B )a B 

 5a x  5a z
B2
2
Prob. 1.26
A  a x  Ax  A cos 

cos  
Ax
2

 0.2673
A
4  16  36
4
 0.5345 
  122.31o
A
56
A
6
 0.8018    36.7o
cos   z 
A
56
cos  
Ay

Prob.1.27
(a) H (1,3, 2)  6a x  a y  4a z
aH 
(6,1, 4)
 0.8242a x  0.1374a y  0.5494a z
36  1  16
(b) | H | 10  4 x 2 y 2  ( x  z ) 2  z 4
or
100  4 x 2 y 2  x 2  2 xz  z 2  z 4
Prob. 1.28

  74.5o
13
R  Ra R ,
aR 
P Q
| P Q |
P Q 
aR 
R4
2 4 1
1
2
0
 2a x + a y  8az
2a x + a y  8az
 0.2408a x  0.1204a y  0.9631a z
4  1  64
R  Ra R  4(0.2408a x  0.1204a y  0.9631a z )  0.9631a x  0.4815a y  3.852a z
An alternate choice of R is  0.9631a x  0.4815a y  3.852a z
Prob. 1.29
(a) At (1, -2, 3), x = 1, y = -2, z = 3.
G  a x  2a y  6a z ,
H  6a x  3a y  3a z
G  1  4  36  6.403
H  36  9  9  7.348
(b) G H  6  6  18  18
G H
18

 0.3826
GH
6.403  7.348
(c )
GH  112.5o
cos GH 
Prob. 1.30
H  10(2)(16)a x  8(8)a y  12(4)a z  320a x  64a y  48a z
F
(a)
Let
(b)
 ax  a y
H F  ( H  a F )a F 
( H  F ) F (320  64)(1, 1, 0)

 128a x  128a y
F2
11
Prob. 1.31
(a) At (1,2,3), E = (2,1,6)
E  4  1  36  41  6.403
(b) At (1,2,3), F = (2,-4,6)
14
E F  ( E  aF ) aF 
( E  F )F
F
2

36
( 2,4,6)
56
 1.286a x  2.571a y  3.857az
(c) At (0,1,-3), E = (0,1,-3), F = (0,-1,0)
EF 
a E F  
1
3
0 1
0
0
 (3,0,0)
EF
  ax
EF
Prob. 1.32
(a) At P, x = -1, y = 2, z = 4
D  8a x  4a y - 2a z ,
E  10a x  24a y  128a z
C  D  E  2a x  20a y  126a z
C a x  C cos  x
(b)
 x  90.9o

 cos  x 
C ax
2

 0.01575
2
C
2  202  1262