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Elements of Electromagnetics 7E Sadiku Solution Manual
Electromagnetic Fields and Waves (Islamic University of Technology)
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CHAPTER 1
P. E. 1.1
A B ý ø1,0,3ù ø5,2,6 ù ý ø6,2,3ù
(a)
A B ý 36 4 9 ý 7
(b)
5 A B ý ø5,0,15ù ø5,2,6 ù ý ø0,2,21ù
(c)
The component of A along ay is
(d)
Ay = 0
3 A B ý ø3,0,9 ù ø5,2,6 ù ý ø8,2,3ù
A unit vector parallel to this vector is
ø8,2,3ù
a11 ý
64 4 9
ý ø0.9117a x 0.2279a y 0.3419a z ù
P. E. 1.2 (a) rp ý a x 3a y 5a z
rR ý 3a y 8a z
(b)
The distance vector is
rQR ý rR rQ ý (0,3,8) (2, 4, 6) ý 2a x a y 2a z
(c)
The distance between Q and R is
| rQR |ý 4 1 4 ý 3
Consider the figure shown on the next page:
40
uZ ý uP uW ý 350a x
ø a x a y ù
2
ý 378.28a x 28.28a y km/hr
P. E. 1.3
or
uz ý 379.3ð175.72 km/hr
Where up = velocity of the airplane in the absence of wind
uw = wind velocity
uz = observed velocity
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N
y
up
x
uW
uz
W
E
S
P. E. 1.4
Using the dot product,
cosñ AB ý
A÷ B
ý
AB
13
13
ý
50
10 65
ñ AB ý 120.66
P. E. 1.5
(a) E F ý ø E a F ùa F ý
øE F ùF
F
2
ý
10ø4,10,5ù
141
ý 0.2837a x 0.7092a y 0.3546a z
ax
ay
(b) E F ý 0
3
10
4
az
4 ý ø55,16,12 ù
5
a E F ý ø0.9398,0.2734,0.205ù
P. E. 1.6
a + b + c = 0 showing that a, b, and c form the sides of a triangle.
a b ý 0,
hence it is a right angle triangle.
1
1
1
ab ý bc ý ca
2
2
2
1
1 4 0 1 1
ab ý
ý ø3,17,12ù
2
21 3 4
2
Area ý
Area ý
1
9 289 144 ý 10.51
2
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P. E. 1.7
ø x2 x1 ù ø y2 y1 ù ø z2 z1 ù
2
(a) P1 P2 ý
2
2
ý 25 4 64 ý 9.644
ø
(b) rP ý rP1 ü rP2 rP1
ù
ý ø1,2,3ù ü ø 5,2,8ù
ý ø1 5ü ,2 2ü ,3 8ü ù.
(c) The shortest distance is
d ý P1 P3 sin ñ ý P1 P3 a P1P2
ý
1
6
3 5
93 5 2 8
1
ø 14,73,27 ù ý 8.2
ý
93
Prob.1.1
rOP ý 4a x 5a y a z
arOP ý
rOP
(4, 5,1)
ý
ý 0.6172a x 0.7715a y 0.1543a z
| rOP |
(16 25 1)
Prob. 1.2
Method 1:
rAB ý rB rA ,
rBC ý rC rB ,
rCA ý rA rC
rAB rBC rCA ý rB rA rC rB rA rC ý 0
Method 2
rAB ý rB rA ý (2, 0,3) (4, 6, 2) ý (6, 6,1)
rBC ý rC rB ý (10,1, 7) (2, 0,3) ý (12,1, 10)
rCA ý rA rC ý (4, 6, 2) (10,1, 7) ý (6, 7,9)
rAB rBC rCA ý (0, 0, 0) ý 0
Prob. 1.3
(a)
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A 3B ý (4, 2, 6) 3(12,18, 8) ý (4, 2, 6) (36,54, 24)
ý (32, 56, 30)
(b)
2 A 5B ý 2(4, 2, 6) 5(12,18, 8) ý (68,86, 28)
| B |ý 122 182 82 ý 532 ý 23.065
(2 A 5B )/ | B |ý (68,86, 28) / 23.065 ý 2.948a x 3.728a y 1.214a z
(c )
ax A ý
1 0 0
ý 6a y 2a z
4 2 6
(d)
B ax ý
12 18 8
1
0
0
ý 8a y 18a z
( B a x ) ÷ a y ý 8
Prob. 1.4
(a) AðB ý (10, 6,8)ð(1, 0, 2) ý 10 16 ý 26
A B ý
(b)
10 6 8
1
0
2
ý (12 0)a x (8 20)a y (0 6)a z
ý -12a x 12a y 6a z
(c) 2 A 3B ý (20, 12,16) (3, 0, 6) ý 17a x 12a y 10a z
Prob. 1.5
(a) A B C ý (2,5,1) (1, 0, 3) (4, 6,10) ý (1, 1,8)
(b)
BC ý
1
0
3
4 6 10
ý (18, 2, 6)
A ÷ ( B C ) ý (2,5,1) ÷ (18, 2, 6) ý 36 10 6 ý 32
(c) cos ñ AB ý
A÷ B
2 0 3
ý
ý 0.05773
AB
4 25 1 1 9
Prob. 1.6
(a)
BC ý
1 1 1
ý a x 2a y a z
0 1 2
Að( B C ) ý (1, 0, 1)ð(1, 2,1) ý 1 0 1 ý 0
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ñ AB ý 86.69o
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(b)
A B ý
1 0 1
1 1
1
ý a x 2a y + a z
( A B )ðC ý (1, 2,1)ð(0,1, 2) ý 0 2 2 ý 0
(c ) A ( B C ) ý
1 0 1
ý 2a x 2a y 2a z
1 2 1
(d) ( A B ) C ý
1 2 1
ý 5a x 2a y a z
0 1 2
Prob.1.7
(a)
T = (4, 6, -1) and S = (10, 12, 8)
(b) rTS ý rS rT ý (10,12,8) (4, 6, 1) ý 6a x 6a y 9a z
(c ) TS ý| rTS |ý 36 36 81 ý 12.37
Prob. 1.8
(a) If A and B are parallel, B=kA, where k is a constant.
Bx ý kAx ,
By ý kAy ,
Bz ý kAz
3 ý k (1)
For Bz ,
k ý 3
Bx ý ñ ý kAx ý (3)(4) ý 12
By ý ò ý kAy ý (3)(2) ý 6
ñ ý 12, ò ý 6
Hence,
(b) If A and B are perpendicular to each other,
A÷ B ý 0
4ñ 2ò 3 ý 0
Prob. 1.9
(b)
10 5 2
ý 2a x 10a z
0 1 0
A ÷ a z ý 2
(c)
cos ñ z ý
(a)
A ay ý
A ÷ az
2
ý
100 25 4 11.358
ñ z ý 100.14o
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Prob. 1.10
(a) A B ý ABcos ñ AB
A B ý ABsin ñ AB an
ø A Bù
2
A B ý ø AB ù ø cos 2 ñ AB sin 2 ñ AB ù ý ø AB ù
2
2
2
(b) a x øa y a z ù ý a x a x ý 1. Hence,
a y az
a x a y az
ý
ax
ý ax
1
ay
az a x
ý
ý ay
a x a y az
1
ax ay
a x a y az
ý
az
ý az
1
Prob. 1.11
(a) P Q ý ø 6, 2, 0 ù , P Q R ý ø 7,1, 2 ù
P Q R ý 49 1 4 ý 54 ý 7.3485
2 1 2
(b) P.Q R ý 4 3 2 ý 2 ø 6 2 ù ø 8 2 ù 2 ø 4 3ù ý 8 10 14 ý 4
1 1 2
Q R ý
4 3 2
ý ø 4, 10, 7 ù
1 1 2
P.Q R ý ø 2, 1, 2 ù ø 4, 10, 7 ù ý 8 10 14 ý 4
(c) Q P ý
4
3
2
2 1 2
ý ø 4,12, 10 ù
Q P ðR ý ø 4,12, 10 ù ø 1,1, 2 ù ý 4 12 20 ý 4
1
or
Q P R ý R Q P ý 4
2
1
2
3
2 ý ø 6 2 ù ø 8 4 ù 2 ø 4 6 ù ý 4
1 2
(d) ø P Q ù øQ × R ù ý ø 4, 12,10 ù ø 4, 10, 7 ù ý 16 120 70 ý 206
(e) ø P × Q ù × øQ × R ù ý
4 12 10
ý 16ax 12a y 8az
4 10 7
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(f) cos ñ PR ý
ø 2 1 4 ù ý 7 ý 0.9526
PðR
ý
P R
4 1 4 11 4 3 6
ñ PR ý 162.3
(g) sin ñ PQ ý
PQ
16 144 100
260
ý
ý
ý 0.998
P Q
3 16 9 4
3 29
ñ PQ ý 86.45
Prob. 1.12
AðB ý (4, 6,1)ð(2, 0,5) ý 8 0 5 ý 13
(a) | B |2 ý 22 52 ý 29
AðB + 2 | B |2 ý 13 2 29 ý 71
(b)
aþ ý
Let
A B
| A B |
C ý A B =
aþ ý
4 6 1
2
0
5
ý (30, 18,12)
(30, 18,12)
C
ý
ý (0.8111a x 0.4867a y 0.3244a z )
|C |
302 182 122
Prob. 1.13
P ðQ ý (2, 6,5)ð(0,3,1) ý 0 18 5 ý 13
P Q ý
cos ñ PQ
2 6 5
ý 21a x - 2a y 6a z
0 3 1
P Q
13
ý
ý
ý 0.51
ñ PQ ý 120.66o
PQ
10 65
Prob. 1.14
P and Q are orthogonal if the angle between them is 90o. Hence
P ÷ Q ý PQ cos ñ ý 0
P ÷ Q ý (2, 4, 6) ÷ (5, 2,3) ý 10 8 18 ý 0
showing that they are perpendicular or orthogonal.
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Prob. 1.15
(a) Using the fact that
ø A B ù C ý ø A C ùB øB C ùA,
we get
A ø A B ù ý ø A B ù A ý øB AùA ø A AùB
ù
ø
ù
B
A
A
A
B
ùûø ÷
÷
ùû
B
A
A
A
A
A
A B
A A
=
(b) A × ø A × ø A × B ù ù =
B
A
A
=
2
ÿ ÷ Āÿ Āÿ ÷ Āÿ Ā
ÿ Ā
since AxA = 0
P2
Prob. 1.16
a
b
P1
c
P3
a ý rp 2 rp1 ý (1, 2, 4) (5, 3,1) ý (4,1,3)
(a) b ý rp 3 rp 2 ý (3,3,5) (1, 2, 4) ý (2,5,1)
c ý rp1 rp 3 ý (5, 3,1) (3,3,5) ý (2, 6, 4)
Note that a + b + c = 0
perpendicular
a ðb ý 8 5 3 ý 0
b c ý 4 30 4 ù 0
c a ý 8 6 12 ù 0
Hence P2 is a right angle.
1
1 4 1 3 1
| a b |ý
ý | (1 15)a x (6 4)a y (20 2)a z |
2
2 2 5 1 2
Area =
(b)
ý
1
1
| (14,10, 22) |ý
196 100 484 ý 13.96
2
2
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Prob. 1.17
Given rP ý ( 1, 4,8),
rQ ý (2, 1,3),
rR ý (1, 2,3)
(a) | PQ |ý 9 25 25 ý 7.6811
(b) PR ý 2a y 5a z
(c )
QP ý (1, 4,8) (2, 1,3) ý (3,5,5)
QR ý (1, 2,3) (2, 1,3) ý (3,3, 0)
9 15 0
QP ðQR
ý
ý 0.7365
| QP || QR |
59 18
ðPQR ý cos 1 ø 0.7365 ù ý 42.64o
(d) Area ý
1
1 3 5 5
QP QR ý
ý 0.5 (15, 15,8) ý 0.5 225 225 36 ý 10.677
2
2 3 3 0
(e) Perimeter ý PQ QR RP ý 59 18 29 ý 17.31
Prob.1.18
Let R be the midpoint of PQ.
1
rR ý {(2, 4, 1) (12,16,9)} ý (7,10, 4)
2
OR ý 49 100 16 ý 165 ý 12.845
OR 12.845
tý
ý
ý 42.82 ms
300
v
Prob. 1.19
Area = Twice the area of a triangue
= | D E |ý
4
1 5
1 2 3
ý| (3 10)a x (5 12)a y (8 1)a z |
ý| (7, 19,9) |ý 49 361 81 ý 22.16
Prob. 1.20
(a) Let P and Q be as shown below:
y
Q
ñ2
P
ñ1
x
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P ý cos 2 ñ 1 sin 2 ñ 1 ý 1, Q ý cos 2 ñ 2 sin 2 ñ 2 ý 1,
Hence P and Q are unit vectors.
(b) P Q ý (1)(1)cos(ñ 2 -ñ1 )
But P Q ý cos ñ1 cos ñ 2 sin ñ1 sin ñ 2 . Thus,
cos(ñ 2 ñ1 ) ý cos ñ1 cos ñ 2 sin ñ1 sin ñ 2
Let P1 ý P ý cosñ 1a x sin ñ 1a y and
Q1 ý cosñ 2 a x sin ñ 2 a y .
P1 and Q1 are unit vectors as shown below:
y
P1
ñ1
ñ1+ñ2
ñ2
x
Q1
P1 Q1 ý (1)(1) cos(ñ 1 ñ 2 )
But P1 Q1 ý cosñ 1 cosñ 2 sin ñ 1 sin ñ 2 ,
cos(ñ 2 ñ 1 ) ý cosñ 1 cosñ 2 sin ñ 1 sin ñ 2
Alternatively, we can obtain this formula from the previous one by replacing
ñ2 by -ñ2 in Q.
(c )
1
1
| P Q |ý | (cos ñ1 cos ñ 2 )ax (sin ñ1 sin ñ 2 )a y |
2
2
ý
1
cos2 ñ 1 sin 2 ñ 1 cos2 ñ 2 sin 2 ñ 2 2 cos ñ 1 cos ñ 2 2 sin ñ 1 sin ñ 2
2
1
1
2 2 cos(ñ 2 ñ 1 )
2 2(cos ñ 1 cos ñ 2 sin ñ 1 sin ñ 2 ) ý
2
2
Let ñ 2 ñ 1 ý ñ , the angle between P and Q.
ý
1
1
| P Q |ý
2 2 cos ñ
2
2
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But cos 2A = 1 – 2 sin 2A.
1
1
| P Q |ý
2 2 4sin 2 ñ / 2 ý sin ñ / 2
2
2
Thus,
ñ ñ
1
| P Q |ý| sin 2 1 |
2
2
Prob. 1.21
ý
(1, 2, 2)
3
u ý r ý
ý (1, 2, 2), r ý rp ro ý (1,3, 4) (2, 3,1) ý (1, 6,3)
1 2 2
ý (18, 5, 4)
1 6 3
u ý 18ax 5a y 4az
Prob. 1.22
r1 ý (1,1,1),
r2 ý (1, 0,1) (0,1, 0) ý (1, 1,1)
r1 r2 (1 1 1) 1
ý
ý
ñ ý 70.53o
3
r1r2
3 3
Prob. 1.23
T S ø 2, 6,3ù ø1, 2,1ù 7
(a) Ts ý T as ý
ý
ý
ý 2.8577
S
6
6
øS T ùT ý 7ø2,6,3ù
(b) S T ý ø S a T ùa T ý
72
T2
ý 0.2857a x 0.8571a y 0.4286a z
cos ñ ý
(c) sin ñ TS ý
TS
T S
ý
2 6 3 ø 12,1,10ù
245
ý
ý
ý 0.9129
1 2 1
7 6
7 6
ñ TS ý 65.91
Prob. 1.24
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Let
A ý ABð AB þ
ABð ý ( A a B )a B ý
A B
B
BB
Hence,
ABþ ý A ABð ý A
A B
B
BB
Prob. 1.25
(a)
A ÷ B ý 20 0 10 ý 10
(b)
A B ý
(c)
20 15 10
ý 15a x 30a y 15a z
1 0
1
( A ÷ B ) B 10(a x a z )
AB ý ( A ÷ a B )a B ý
ý
ý 5a x 5a z
B2
2
Prob. 1.26
A ÷ a x ý Ax ý A cos ñ
cos ñ ý
Ax
2
ý
ý 0.2673
A
4 16 36
4
ý 0.5345
ò ý 122.31o
A
56
A
6
cos ÷ ý z ý
ý 0.8018 ÷ ý 36.7o
A
56
cos ò ý
Ay
ý
Prob.1.27
(a) H (1,3, 2) ý 6a x a y 4a z
aH ý
(6,1, 4)
ý 0.8242a x 0.1374a y 0.5494a z
36 1 16
(b) | H |ý 10 ý 4 x 2 y 2 ( x z ) 2 z 4
or
100 ý 4 x 2 y 2 x 2 2 xz z 2 z 4
Prob. 1.28
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ñ ý 74.5o
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R ý Ra R ,
aR ý
P Q
| P Q |
P Q ý
aR ý
Rý4
2 4 1
1
2
0
ý 2a x + a y 8az
2a x + a y 8az
ý 0.2408a x 0.1204a y 0.9631a z
4 1 64
R ý Ra R ý 4(0.2408a x 0.1204a y 0.9631a z ) ý 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 ý a x 2a y 6a z ,
H ý 6a x 3a y 3a z
G ý 1 4 36 ý 6.403
H ý 36 9 9 ý 7.348
(b) G ðH ý 6 6 18 ý 18
(c )
18
G ðH
ý
ý 0.3826
6.403 7.348
GH
ý 112.5o
cos ñGH ý
ñGH
Prob. 1.30
(a)
H ý 10(2)(16)a x 8(8)a y 12(4)a z ý 320a x 64a y 48a z
(b)
F
Let
ý ax a y
H F ý ( H ÷ a F )a F ý
( H ÷ F ) F (320 64)(1, 1, 0)
ý
ý 128a x 128a y
11
F2
Prob. 1.31
(a) At (1,2,3), E = (2,1,6)
E ý 4 1 36 ý 41 ý 6.403
(b) At (1,2,3), F = (2,-4,6)
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E F ý ( E aF ) aF ý
( E F )F
F
2
ý
36
( 2,4,6)
56
ý 1.286a x 2.571a y 3.857az
(c) At (0,1,-3), E = (0,1,-3), F = (0,-1,0)
EF ý
a E F ý
0
3
1
0 1
0
ý (3,0,0)
EF
ý ax
EF
Prob. 1.32
(a) At P, x = -1, y = 2, z = 4
D ý 8a x 4a y - 2a z ,
E ý 10a x 24a y 128a z
C ý D E ý 2a x 20a y 126a z
(b)
C ða x ý C cos ñ x
cos ñ x ý
C ða x
2
ý
ý 0.01575
2
C
2 202 1262
ñ x ý 90.9o
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CHAPTER 2
P. E. 2.1
(a) At P(1,3,5),
òý
x = 1,
x y
2
2
=
y = 3,
10 ,
z =5,
z = 5,
ö ý tan 1 y / x ý tan 1 3 ý 71.6o
. o ,5)
P( ò , ö , z) ý P( 10 , tan 1 3,5) ý P(3.162,716
Spherical system:
rý
x2 y2 z2 ý
35 ý 5.916
ñ ý tan 1 x 2 y 2 z ý tan 1 10 5 ý tan 1 0.6325 ý 32.31ð
P (r ,ñ , ) ý P (5.916,32.31ð, 71.57ð)
At T(0,-4,3),
x=0
y =-4,
z =3;
ò ý x y ý 4, z ý 3, ý tan y / x ý tan 4 / 0 ý 270ð
T ( ò, , z) ý T (4,270ð ,3).
2
1
2
1
Spherical system:
rý
x 2 y 2 z 2 ý 5,ñ ý tan 1 ò / z ý tan 1 4 / 3 ý 5313
. ð.
T (r ,ñ , ) ý T (5,5313
. ð ,270ð ).
At S(-3-4-10),
x =-3, y =-4, z =-10;
ö 4 ö
ò ý x 2 y 2 ý 5, ö ý tan 1 ÷ ÷ ý 233.1ð
ø 3 ø
S ( ò , ö , z ) ý S (5, 233.1, 10).
Spherical system:
r ý x 2 y 2 z 2 ý 5 5 ý 11.18.
ñ ý tan 1 ò z ý tan 1
5
ý 153.43ð;
10
S (r , ñ , ö ) ý S (11.18,153.43ð, 233.1ð).
(b)
In Cylindrical system,
Qx ý
ò
ò z2
2
;
ò ý x2 y 2 ;
yz ý z ò sin ö ,
z ò sin ö
Qy ý 0;
Qz ý
ò 2 z2
;
15
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ù Qò ù ù cos ö
ú ú ú
ú Qö ú ý ú sin ö
úû Qz úû úû 0
Qò ý Qx cos ö ý
ò cos ö
ò z
2
0ù ù Qx ù
úú ú
0ú ú 0 ú ;
1úû úû Qz úû
sin ö
cos ö
0
2
,
Qö ý Qx sin ö ý
ò sin ö
ò 2 z2
Hence,
Qý
ò
ò 2 z2
(cos ö a ò sin ö a ö z sin ö a z ).
In Spherical coordinates:
r sin ñ
Qx ý
ý sin ñ ;
r
1
Qz ý r sin ö sin ñ r cosñ ý r sin ñ cosñ sin ö .
r
ùQr ù ù sin ñ cos ö sin ñ sin ö cosñ ù ùQx ù
ú ú ú
úú ú
úQñ ú ý úcosñ cos ö cosñ sin ö sin ñ ú ú 0 ú ;
ú ú
cos ö
0 úû úûQz úû
ûQö û úû sin ö
Qr ý Qx sin ñ cos ö Qz cosñ ý sin 2 ñ cos ö r sin ñ cos2 ñ sin ö .
Qñ ý Qx cosñ cos ö Qz sin ñ ý sin ñ cosñ cos ö r sin 2 ñ cosñ sin ö .
Qö ý Qx sin ö ý sin ñ sin ö.
ü Q ý sin ñ ø sin ñ cos ö r cos2 ñ sin ö ù a
r
sin ñ cosñ (cos ö r sin ñ sin ö )añ sin ñ sin ö aö .
At T :
4
12
a x a z ý 0.8a x 2.4a z ;
5
5
4
Q ( ò , ö , z ) ý (cos 270ð a ò sin 270ð aö 3sin 270ða z
5
ý 0.8aö 2.4 a z ;
Q ( x, y , z ) ý
4
45
4 3
20
4
Q (r , ñ , ö ) ý (0 (1))ar ( )(0 (1))añ (1)aö
5
25
5 5
5
5
36
48
4
ý a r añ aö ý 1.44ar 1.92añ 0.8 aö ;
25
25
5
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Note, that the magnitude of vector Q = 2.53 in all 3 cases above.
P.E. 2.2 (a)
ù Ax ù ùcosö sinö 0ù ù òz sinö ù
ú A ú ý úsinö cosö 0ú ú 3ò cosö ú
ú yú ú
úú
ú
úû Az úû úû 0
0
1úû úûò cosö sinöúû
A ý (òz cosö sinö 3ò cosö sinö) ax (òz sin2 ö 3ò cos2 ö) ay ò cosö sinö az .
y
x
y
But ò ý x2 y2 , tanö ý , cosö ý
, sinö ý
;
2
2
2
2
x
x y
x y
Substituting all this yields:
1
Aý
[(xyz 3xy)ax (zy2 3x2 ) ay xy az ].
2
2
x y
ù Bx ù ùsin ñ cos ö
ú B ú ý ú sin ñ sin ö
ú yú ú
úû Bz úû úû cos ñ
cos ñ cos ö
cos ñ sin ö
sin ñ
Since r ý x 2 y 2 z 2 , tan ñ ý
and sin ñ ý
and sin ö ý
x2 y 2
x y z
2
2
y
x y
2
2
,
2
sin ö ù
cos ö úú
0 úû
ù r2 ù
ú
ú
ú 0 ú
úsin ñ ú
û
û
x2 y2
y
, tan ö ý ;
z
z
z
, cos ñ ý
cos ö ý
x y2 z2
2
x
x y2
2
;
;
y
1
ý ( r 2 x y ).
r
r
x
1
B y ý r 2 sin ñ sin ö sin ñ cos ö ý ry ý (r 2 y x ).
r
r
1
Bz ý r 2 cos ñ ý r z ý ( r 2 z ).
r
Bx ý r 2 sin ñ cos ö sin ñ sin ö ý rx
Hence,
Bý
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 ].
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P.E.2.3 (a) At:
(1, ð / 3, 0),
H ý (0, 0.06767,1)
1
a x ý cos ö a ò sin ö aö ý (a ò 3 aö )
2
H ÷ a x ý 0.0586.
(b)
At:
(1, ð / 3, 0),
añ ý cos ñ aò sin ñ a z ý a z .
H
aò
az ý 0
0
(c)
(d)
aö
az
0.06767 1 ý 0.06767 a ò .
0
1
( H ÷ a ò ) a ò ý 0 aò .
H az ý
aò
aö
az
0
0.06767
1
0
0
1
ý 0.06767 a ò .
H a z ý 0.06767
A
P.E. 2.4
(a)
ðB ý (3, 2, 6) ÷ ø 4, 0,3ù ý 6.
(b)
A B ý
3 2 6
ý 6 ar 33añ 8aö .
4 0 3
B
A
Thus the magnitude of
ý 34.48.
(c )
At (1, ð / 3, 5ð / 4), ñ ý ð / 3,
a z ý cos ñ a r sin ñ añ ý
1
3
ar
añ .
2
2
3 ö
ö3
öö 1
( Aða z )a z ý ÷ 3 ÷ ÷÷ ar
añ ÷÷ ý 0.116ar 0.201añ
2
ø2
øø 2
ø
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Prob. 2.1
(a)
ò ý x 2 y 2 ý 4 25 ý 5.3852,
ö ý tan 1
y
ý tan 1 2.5 ý 68.2o
x
x2 y 2
5.3852
ý tan 1
ý 79.48o
1
z
o
o
P (r , ñ , ö ) ý P(5.477, 79.48 , 68.2 )
r ý x 2 y 2 z 2 ý 4 25 1 ý 5.477,
P ( ò , ö , z ) ý P(5.3852, 68.2o ,1),
ñ ý tan 1
(b)
ò ý x 2 y 2 ý 9 16 ý 5,
r ý x 2 y 2 z 2 ý 5,
ö ý tan 1
y
4
ý tan 1
ý 360o 53.123o ý 306.88o
3
x
x2 y 2
ý tan 1 ý 90o
z
P(r , ñ , ö ) ý P(5,90o ,306.88o )
ñ ý tan 1
Q( ò , ö , z ) ý Q(5,306.88o , 0),
(c )
ò ý x 2 y 2 ý 36 4 ý 6.325,
ö ý tan 1
y
2
ý tan 1 ý 18.43o
x
6
r ý x 2 y 2 z 2 ý 36 4 16 ý 7.483,
x2 y 2
6.325
ý tan 1
ý 180o 57.69o ý 122.31o
4
z
o
R( ò , ö , z ) ý R(6.325,18.43 , 4),
R(r , ñ , ö ) ý R(7.483,122.31o ,18.43o )
ñ ý tan 1
Prob. 2.2
(a)
x ý ò cos ö ý 2 cos 30ð ý 1.732;
y ý ò sin ö ý 2sin 30ð ý 1;
z ý 5;
P1 ( x, y, z ) ý P1 (1.732,1, 5).
(b)
x ý 1cos 90ð ý 0;
y ý 1sin 90ð ý1;
z ý 3.
P2 ( x, y, z ) ý P2 (0, 1, 3).
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(c)
(d)
x ý r sin ñ cos ö ý 10sin(ð / 4) cos(ð / 3) ý 3.535;
y ý r sin ñ sin ö ý 10sin(ð / 4) sin(ð / 3) ý 6.124;
z ý r cos ñ
ý 10 cos(ð / 4)
ý 7.0711
P3 ( x, y, z ) ý P3 (3.535, 6.124, 7.0711).
x ý 4sin 30ð cos 60ð ý 1
y ý 4sin 30ð sin 60ð ý1.7321
z ý r cos ñ ý 4 cos 30ð ý 3.464
P4 ( x, y, z ) ý P4 (1,1.7321,3.464).
Prob. 2.3
ò ý x 2 y 2 ý 4 36 ý 6.324
6
y
ý tan 1 ý 71.56o
x
2
o
P is (6.324, 71.56 , 4)
(a) ö ý tan 1
r ý x 2 y 2 z 2 ý 4 36 16 ý 7.485
x2 y2
6.324
4
ý tan 1
ý 90o tan 1
ý 122.3o
(b) ñ ý tan
6.324
4
z
o
o
P is (7.483,122.3 , 71.56 )
1
Prob. 2.4
(a)
x ý ò cos ö ý 5cos120o ý 2.5
y ý ò sin ö ý 5sin120o ý 4.33
z ý1
Hence Q ý (2.5, 4.33,1)
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(b)
r ý x 2 y 2 z 2 ý ò 2 z 2 ý 25 1 ý 5.099
ñ ý tan 1
x2 y2
ò
5
ý tan 1 ý tan 1 ý 78.69o
z
z
1
ö ý 120o
Hence Q ý (5.099, 78.69o ,120o )
Prob. 2.5
T (r ,ñ , ö )
r ý 10, ñ ý 60o , ö ý 30o
x ý r sin ñ cos ö ý 10sin 60o cos 30o ý 7.5
y ý r sin ñ sin ö ý 10sin 60o sin 30o ý 4.33
z ý r cos ñ ý 10 cos 60o ý 5
T ( x, y, z ) ý (7.5, 4.33,5)
ò ý r sin ñ ý 10sin 60o ý 8.66
T ( ò , ö , z ) ý (8.66,30o ,5)
Prob. 2.6
(a)
x ý ò cos ö ,
y ý ò sin ö ,
V ý ò z cos ö ò 2 sin ö cos ö ò z sin ö
(b)
U ý x2 y 2 z 2 y 2 2 z 2
ý r 2 r 2 sin 2 ñ sin 2 ö 2r 2 cos 2 ñ
ý r 2 [1 sin 2 ñ sin 2 ö 2 cos 2 ñ ]
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Prob. 2.7
(a)
ù Fò ù ù cos ö
ú ú ú
ú Fö ú ý ú sin ö
úû Fz úû úû 0
Fò ý
Fö ý
Fz ý
_
F ý
1
ò z
2
[ ò cos 2 ö ò sin 2 ö ] ý
2
1
ò 2 z2
4
ò z2
2
1
ò z2
2
sin ö 0 ù
cos ö 0 úú
0
1 úû
ù
ù
x
ú 2
ú
ú ò z2 ú
ú
ú
y
ú
ú
ú ò 2 z2 ú
ú
ú
4
ú
ú
ú 2
2 ú
û ò z û
ò
ò z2
2
;
[ ò cos ö sin ö ò cos ö sin ö ] ý 0;
;
( ò aò 4 az )
In Spherical:
ù Fr ù
ùsinñ cosö sinñ sinö cosñ ù
ú ú
ú
ú
úFñ ú ý úcosñ cosö cosñ sinö sinñ ú
úFö ú
úû sinö
cosö
0 úû
û û
ù xù
úr ú
ú ú
ú yú
úr ú
ú4ú
ú ú
ûú r ûú
r
r
4
4
Fr ý sin2 ñ cos2 ö sin2 ñ sin2 ö cosñ
ý sin2 ñ cosñ;
r
r
r
r
4
4
Fñ ýsinñ cosñ cos2 ö sinñ cosñ sin2 ö sinñ ý sinñ cosñ sinñ ;
r
r
Fö ý sinñ cosö sinö sinñ sinö cosö ý 0;
_
4
4
ü F ý (sin2 ñ cosñ ) ar sinñ (cosñ )añ
r
r
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(b)
ùGò ù ù cos ö
ú G ú ý ú sin ö
ú öú ú
úû Gz úû úû 0
ò2
Gò ý
sin ö
0ù
cos ö 0 úú
0
1 úû
ù xò 2 ù
ú 2
ú
ú ò z2 ú
ú
ú
2
ú yò
ú
ú ò 2 z2 ú
ú
ú
ú zò 2 ú
ú 2
ú
2
úû ò z úû
[ ò cos 2 ö ò sin 2 ö ] ý
ò 2 z2
ò3
ò 2 z2
;
Gö ý 0;
zò 2
Gz ý
ò 2 z2
ò2
Gý
ò 2 z2
;
( ò aò z az )
Spherical :
Gý
ò2
( xa x ya y za z ) ý
r
r 2 sin 2 ñ
rar ý r 2 sin 2 ñ ar
r
Prob. 2.8
B ý ò ax
y
ò
a y za z
ù Bò ù ù cos ö
ú B ú ý ú sin ö
ú öú ú
úû Bz úû úû 0
Bò ý ò cos ö
sin ö
cos ö
0
y
ò
Bö ý ò sin ö
0ù ù ò ù
0 úú úú y / ò úú
1 úû úû z úû
sin ö
y
ò
cos ö
Bz ý z
But y ý ò sin ö
Bò ý ò cos ö sin 2 ö , Bö ý ò sin ö sin ö cos ö
Hence,
B ý ( ò cos ö sin 2 ö )a ò sin ö (cos ö ò )aö za z
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Prob. 2.9
ù Ax ù ùcos ö
ú A ú ý ú sin ö
ú yú ú
úû Az úû úû 0
sin ö
cos ö
0
At P, ò ý 2,
ö ý ð / 2,
0ù ù2ù
0 úú úú 3 úú
1 úû úû 4 úû
z ý 1
Ax ý 2 cos ö 3sin ö ý 2 cos 90o 3sin 90o ý 3
Ay ý 2sin ö 3cos ö ý 2sin 90o 3cos 90o ý 2
Az ý 4
Hence, A ý 3a x 2a y 4a z
Prob. 2.10
(a)
ù Ax ù ùcos ö
ú ú ú
ú Ay ú ý ú sin ö
ú Ay ú úû 0
û û
sin ö
cos ö
0
0 ù ù ò sin ö ù
0 úú úú ò cos ö úú
1 úû úû 2 z úû
Ax ý ò sin ö cos ö ò cos ö sin ö ý 0
Ay ý ò sin 2 ö ò cos 2 ö ý ò ý x 2 y 2
Az ý 2 z
Hence,
A ý x 2 y 2 a y 2 za z
(b)
ù Bx ù
úB ú ý
ú yú
úû Bz úû
ùsin ñ c o sö c o sñ c osö
ú sin ñ sin ö c osñ sin ö
ú
úû c osñ
sin ñ
sin ö ù
ú
c o sö ú
0 úû
ù4 r c o sö ù
ú
ú
r
ú
ú
úû 0 úû
Bx ý 4 r sin ñ c os2 ö r c osñ c osö
By ý 4 r sin ñ sin ö c osö r c osñ sin ö
Bz ý 4 r c osñ c osö r sin ñ
But
sin ñ ý
r ý x y z ,
2
sin ö ý
2
2
y
x y2
2
,
c osö ý
x2 y 2
,
r
x
c osñ ý
z
r
x2 y 2
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Bx ý 4 x2 y 2
x2
x2 y 2
By ý 4 x2 y 2
xy
x y2
x y2
zy
2
x
Bz ý 4 z
Bý
zx
2
x y
2
1
x y
2
2
2
x2 y 2
x2 y 2
ùû x(4 x z)a x y(4 x z)a y (4 xz x2 y 2 )a z ùû
Prob. 2.11
Method 1:
ù Fx ù ùsin ñ cos ö
ú ú ú
ú Fy ú ý ú sin ñ sin ö
úû Fz úû úû cos ñ
Fx ý
4
sin ñ cos ö ,
r2
r 2 ý x2 y2 z 2 ,
sin ö ý
Fx ý
cos ñ cos ö
cos ñ sin ö
sin ñ
y
x2 y2
4
2
x y2 z2
4
Fy ý 2
x y2 z2
,
Fy ý
sin ö ù ù 4 / r 2 ù
ú
ú
cos ö úú ú 0 ú
0 úû úû 0 úû
4
sin ñ sin ö ,
r2
sin ñ ý
cos ö ý
x2 y 2
x y z
2
2
x
x2 y2
x2 y 2
x y z
cos ñ ý
,
z
x y2 z2
2
x2 y 2
x2 y2 z 2
2
2
4
cos ñ
r2
x
x2 y 2
2
Fz ý
y
2
x y
2
2
ý
4x
( x y 2 z 2 )3/2
ý
4y
( x y 2 z 2 )3/ 2
2
2
z
4
4z
ý 2
2
2
2
2
2
x y z ( x y z ) ( x y 2 z 2 )3/2
Thus,
4
ù xa x ya y za z ùû
Fý 2
2
( x y z 2 )3/2 û
Fx ý
2
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Method 2:
4a r 4ra
F ý 2r . ý 3 r
r r
r
4
ù xa x ya y za z ùû
Fý 2
( x y 2 z 2 )3/2 û
Prob. 2.12
(a)
(b)
ñ ý ð / 2, ö ý 3ð / 2
B ý 2sin(ð / 2)ar 4 cos(3ð / 2)aö ý 2ar
r ý 2,
ù Bx ù ù sin ñ cos ö cos ñ cos ö sin ö ù ù r sin ñ ù
ú B ú ý ú cos ñ cos ö cos ñ sin ö cos ö ú ú
ú
0
ú yú ú
úú
ú
úû Bz úû úû cos ñ
0 úû úû r 2 cos ö úû
sin ñ
Bx ý r sin 2 ñ cos ö r 2 sin ö cos ö ,
By ý r sin ñ cos ñ cos ö r 2 cos 2 ö
Bz ý r sin ñ cos ñ
But
z
ò
r ý x y z , cos ñ ý ,sin ñ ý ý
r
r
2
cos ö ý
x
ò
2
x
ý
x2 y2
Bx ý x 2 y 2 z 2
ý
,
sin ö ý
x2 y 2
x2 y 2 z 2
y
ò
ý
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 ý x 2 y 2 z 2
ý
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 ý x y z 2
ý
x y2 z2
2
x
2
z x2 y 2
x2 y2 z 2
B ý Bx a x By a y Bz a z
Prob. 2.13
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x ý ò cos ö
(a)
B ý ò cos ö a z
x ý r sin ñ cos ö
(b) B ý r sin ñ cos ö a z ,
Bx ý 0 ý By , Bz ý r sin ñ cos ö
ù Br ù ù sin ñ cos ö sin ñ sin ö cos ñ ù ù
0
ù
ú ú ú
ú
ú
ú
0
ú Bñ ú ý úcos ñ cos ö cos ñ sin ö sin ñ ú ú
ú
ú Bö ú úû sin ö
cos ö
0 úû úû r sin ñ cos ö úû
û û
Br ý r sin ñ cos ñ cos ö ý 0.5r sin(2ñ ) cos ö
Bñ ý r sin 2 ñ cos ö ,
Bö ý 0
B ý 0.5r sin(2ñ ) cos ö ar r sin 2 ñ cos ö añ
Prob. 2.14
(a)
a x a ò ý (cos ö a ò sin ö aö ) a ò ý cos ö
a x aö ý (cos ö a ò sin ö aö ) aö ý sin ö
a y a ò ý (sin ö a ò cos ö aö ) a ò ý sin ö
_
_
a y aö ý (sin ö a ò sin ö aö ) aö ý cos ö
(b) and (c)
In spherical system :
a x ý sin ñ cos ö ar cos ñ cos ö añ sin ö aö .
a y ý sin ñ sin ö a r cos ñ sin ö añ cos ö aö .
a z ý cos ñ a x sin ñ añ .
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Hence,
a x a r ý sin ñ cos ö ;
a x añ ý cos ñ cos ö ;
a y a r ý sin ñ sin ö ;
a y añ ý cos ñ sin ö ;
_
_
a z a r ý cos ñ ;
_
_
a z añ ý sin ñ ;
Prob. 2.15
(a)
a ò ý cos ö a x sin ö a y ,
a ò aö ý
cos ö
sin ö
az aò ý
0
0
cos ö sin ö
aö a z ý
sin ö
0
aö ý sin ö a x cos ö a y
sin ö
cos ö
0
ý (cos 2 ö sin 2 ö )a z ý a z
0
1
ý sin ö a x cos ö a y ý aö
0
cos ö
0
0
ý cos ö a x sin ö a y ý a ò
1
(b)
ar ý sin ñ cos ö a x sin ñ sin ö a y cos ñ a z
añ ý cos ñ cos ö a x cos ñ sin ö a y sin ñ a z
aö ý sin ö a x cos ö a y
ar añ ý
sin ñ cos ö
cos ñ cos ö
sin ñ sin ö
cos ñ sin ö
cos ñ
sin ñ
ý ( sin 2 ñ sin ö cos 2 ñ sin ö )a x (cos 2 ñ cos ö sin 2 ñ cos ö )a y
(sin ñ cos ñ sin ö cos ö sin ñ cos ñ sin ö cos ö )a z
ý sin ö a x cos ö a y ý aö
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aö ar ý
sin ö
cos ö
sin ñ cos ö sin ñ sin ö
0
cos ñ
ý cos ñ cos ö a x cos ñ sin ö a y ( sin ñ sin 2 ö sin ñ cos 2 ö )a z
ý cos ñ cos ö a x cos ñ sin ö a y sin ñ a z ý añ
añ aö ý
cos ñ cos ö
sin ö
cos ñ sin ö
cos ö
sin ñ
0
ý sin ñ cos ö a x sin ñ sin ö a y (cos ñ cos 2 ö cos ñ sin 2 ö )a z
ý sin ñ cos ö a x sin ñ sin ö a y cos ñ a z ý ar
Prob. 2.16
(a)
rý
ò 2 z2 .
x2 y2 z2 ý
ò
ñ ý tan 1 ;
z
ö ý ö.
or
òý
x 2 y 2 ý r 2 sin 2 ñ cos2 ö r 2 sin 2 ñ sin 2 ö .
ý r sin ñ ;
z ý r cosñ ;
ö ý ö.
(b) From the figures below,
cosñ aò
z
z
aò
aò
az
ar
-az
ñ
sin ñ a ò
ñ
sin ñ ( az )
c o sñ a z
ò
ò
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a r ý sin ñ a ò c o sñ a z ;
añ ý c o sñ a ò sin ñ a z ;
aö ý aö .
Henc e ,
ù ù
ù ù
ú a r ú ù sin ñ 0 c osñ ù úa ò ú
ú ú ú
úú ú
ú a ñ ú ý úc osñ 0 sin ñ ú ú aö ú
ú ú ú 0
1
0 úû ú ú
úaö ú û
úa z ú
úû úû
úû úû
From the fig ures b elow,
a ò ý c osñ añ sin ñ a r ; a z ý c osñ a r sin ñ añ ; aö ý aö .
z
z
sin ñ ar
az
sin ñ ( añ )
cosñ añ
añ
aò
cosñ ar
ar
ñ
ñ
añ
ar
ò
ù ù
úa ò ú ù sin ñ
ú ú ú
ú aö ú ý ú 0
ú ú ú cos ñ
ú az ú û
û û
cos ñ
0
sin ñ
ò
0ù
1 úú
0 úû
ù ù
úar ú
ú ú
úañ ú
ú ú
ú az ú
û û
Prob. 2.17
At P(2, 0, 1),
ö ý 0,
ö
ö
ö 1 ö
÷ ý cos 1 ÷
ý 116.56o
÷
2
2
2
÷ x y z ÷
ø 5ø
ø
ø
ñ ý cos 1 ÷
z
(a) a ò ÷ a x ý cos ö ý 1
(b) aö ÷ a y ý cos ö ý 1
(c) ar ÷ a z ý cos ñ ý 0.4472
30
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Prob. 2.18
If A and B are perpendicular to each other, AðB = 0
AðB ý ò 2 sin 2 ö ò 2 cos 2 ö -ò 2
=ò 2 (sin 2 ö cos 2 ö )-ò 2
=ò 2 ò 2
=0
As expected.
Prob. 2.19
(a) A B ý 8a ò 2aö 7az
(b) AðB = 15 + 0 - 8 = 7
(c ) A B =
3 2
1
5 0 8
=-16a ò (5 24)aö 10a z
=-16a ò 29aö 10a z
A B
7
7
ý
ý
AB
9 4 1 25 64
14 89
=0.19831
(d ) cosñ AB ý
ñ AB =78.56o
Prob. 2.20
ù Gx ù ùcos ö sin ö
úG ú ý ú sin ö cos ö
ú yú ú
úû Gz úû úû 0
0
Gx ý Gò cos ö Gö sin ö
0 ù ùGò ù
0 úú úú Gö úú
1 úû úû Gz úû
ý 3ò cos ö ò cos ö sin ö
ý 3x x sin ö ý 3(3) (3) sin(306.87o ) ý 11.4
Gx ý Gx a x ý 11.4a x
31
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Prob. 2.21
ùGò ù ù cos ö sin ö 0 ù ù yz ù
ú G ú ý ú sin ö cos ö 0 ú ú xz ú
ú öú ú
úú ú
úû Gz úû úû 0
0
1 úû úû xy úû
Gò ý yz cos ö xz sin ö
x ý ò cos ö , y ý ò sin ö ,
yz ý ò z sin ö , xz ý ò z cos ö
Gò ý ò z sin ö cos ö ò z cos ö sin ö ý 2 ò z sin ö cos ö ý ò z sin 2ö
Gö ý yz sin ö xz cos ö ý ò z (cos 2 ö sin 2 ö ) ý ò z cos 2ö
Gz ý xy ý ò 2 cos ö sin ö ý 0.5ò 2 sin 2ö
G ý ò z sin 2ö a y ò z cos 2ö aö 0.5ò 2 sin 2ö a z
Prob. 2.22
ù cos ö sin ö 0 ù ù Aò ù
ù Ax ù
ú ú ú
ú
ú ú
ú Ay ú ý ú sin ö cos ö 0 ú ú Aö ú
úû 0
úû Az úû
0
1úû úû Az úû
x
y
ù
ù
0ú
ú 2
2
2
2
x y
ú
ú x y
y
x
ú
ú
ý ú 2
0ú
2
2
2
x y
ú
ú x y
0
0
1ú
ú
úû
úû
ù Ax ù
ùsin ñ c o sö c osñ c o sö
ú A ú ý ú sin ñ sin ö
c o sñ sin ö
ú yú
ú
úû A z úû
úû c o sñ
sin ñ
ù
x
ú
2
ú x y 2 z2
ú
ú
y
ú
2
2
2
ý
ú x y z
ú
z
ú
ú x2 y 2 z2
úû
sin ö ù
c o sö úú
0 úû
ù Ar ù
ú ú
ú Añ ú
ú Aö ú
û û
xz
x2 y 2 x2 y 2 z2
yz
x2 y 2 x2 y 2 z2
x2 y 2
x2 y 2 z2
ù Aò ù
ú ú
ú Aö ú
úû Az úû
ù
ú
2
2 ú
x y
ú
ú
x
2
2 ú
x y ú
ú
ú
0
ú
úû
y
ù Ar ù
ú ú
ú Aö ú
ú Aö ú
û û
32
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Prob. 2.23 (a) Using the results in Prob.2.14,
Aò ý òz sin ö ý r 2 sin ñ cosñ sin ö
Aö ý 3ò cos ö ý 3r sin ñ cos ö
Az ý ò cos ö sin ö ý r sin ñ cos ö sin ö
Hence,
ù Ar ù
ú ú
ú Añ ú ý
úA ú
û öû
ù sin ñ
ú
ú cos ñ
úû 0
0 cos ñ ù ù r 2 sin ñ cos ñ sin ö ù
úú
ú
0 sin ñ ú ú 3r sin ñ cos ö ú
1
0 úû úû r sin ñ cos ö sin ö úû
ø
ù
A(r,ñ ,ö ) ý r sin ñ ùsin ö c osñ ø r sin ñ c o sö ù a r sin ö r c os2 ñ sin ñ c osö añ 3c o sö aö ù
û
û
At (10 , ð / 2,3ð / 4),
r ý 10 , ñ ý ð / 2, ö ý 3 ð / 4
A ý 10(0ar 0.5añ
(b)
Br ý r 2 ý (ò 2 z 2 ),
3
aö ) ý 5añ 21.21aö
2
Bñ ý 0 ,
ù Bò ù ù sin ñ
ú ú ú
ú Bö ú ý ú 0
úû Bz úû úû cos ñ
Bö ý sin ñ ý
cos ñ
0
sin ñ
ò
ò 2 z2
0 ù ù Br ù
úú ú
1ú ú Bñ ú
0 úû úû Bö úû
ö
ö
ò
z
B( ò , ö , z ) ý ò 2 z 2 ÷ ò aò 2
a
a
÷
z
ö
ò z2
ø
ø
At (2, ð / 6 ,1),
ò ý 2, ö ý ð / 6 , z ý 1
B ý 5(2a ò 0.4aö a z ) ý 4.472a ò 0.8944aö 2.236a z
Prob. 2.24
(a) d ý
(b)
(6 2) 2 ( 1 1) 2 (2 5) 2 ý
29 ý 5.385
d 2 ý 32 52 2(3)(5) cos ð ( 1 5) 2 ý 100
d ý 100 ý 10
33
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(c)
ð
ð
ð
ð
ð 3ð
d 2 ý 102 52 2(10)(5)c os c o s 2(10)(5)sin sin c os(7
)
4
6
4
6
4 4
ý 125 100(0.7071)(0.866) 100(0.7071)(0.5)(0.2334)
ý 125 61.23 35.33 ý 99.118
d ý 99.118 ý 9.956.
Prob. 2.25
Using eq. (2.33),
d 2 ý r12 r22 2r1r2 cos ñ1 cos ñ 2 2r1r2 sin ñ1 sin ñ 2 cos(ö2 ö1 )
ý 16 36 2(4)(6) cos 30o cos 90o 2(4)(6) sin 30o sin 90o cos(180o )
ý 16 36 0 48(0.5)(1)(1) ý 52 24 ý 76
d ý 8.718
Prob. 2.26
ù aò ù ù cos ö
ú a ú ý ú sin ö
ú öú ú
úû a z úû úû 0
sin ö 0 ù ù a x ù
cos ö 0 úú úúa y úú
0
1 úû úû a z úû
a ò ý cos ö a x sin ö a y , aö ý sin ö a x cos ö a y
At (0, 4, -1), ö ý 90o
a ò ý sin 90o a y ý a y
aö ý sin 90o a x ý a x
Prob. 2.27
At (1, 60o , 1),
ò ý 1, ö ý 60o , z ý 1,
(a) A ý (2 sin 60o )aò (4 2 cos 60o )aö 3(1)(1)a z
ý 2.866 a ò 5aö 3a z
B ý 1cos 60o a ò sin 60o aö a z ý 0.5aö 0.866aö a z
AðB ý 1.433 4.33 3 ý 5.897
AB ý 2.8662 26 9 0.25 1 0.8662 ý 9.1885
cos ñ AB ý
AðB 5.897
ý
ý 0.6419
AB 9.1885
ñ AB ý 50.07o
34
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Let D = A B. At (1,90o , 0),
ò ý 1, ö ý 90o , z ý 0
(b) A ý sin 90o a ò 4aö ý a ò 4aö
B ý 1cos 90o a ò sin 90o aö a z ý aö a z
aò aö
D ý A B ý 1 4
0 1
aD ý
az
0 ý 4a ò aö a z
1
D
(4,1, 1)
ý
ý 0.9428a ò 0.2357aö 0.2357a z
D
16 1 1
Prob.2.28
ö ý 90ð;
ù Bx ù ù cos ö sin ö 0 ù
ú B ú ý ú sin ö cos ö 0 ú
ú yú ú
ú
At P(0, 2, 5),
úû Bz úû
úû 0
0
ù 0 1 0 ù
ý úú1 0 0 úú
ûú0 0 1 ûú
B ý a x 5a y 3a z
ù Bò ù
úB ú
ú öú
1 úû úû Bz úû
ù 5ù
ú1ú
ú ú
ûú 3úû
(a) A B ý (2, 4,10) (1, 5, 3)
ý ax a y 7 az .
(b ) c osñ AB ý
A ÷B
ý
AB
ñ AB ý c os1(
(c) AB ý A ÷ a B ý
52
4200
52
4200
) ý 143.36ð.
A÷ B
52
ý
ý 8.789.
B
35
Prob. 2.29
35
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B ÷ a x ý Bx
ù Bx ù ù cos ö sin ö 0 ù ù Bò ù
ú B ú ý ú sin ö cos ö 0 ú ú B ú
ú yú ú
ú ú öú
úû Bz úû úû 0
0
1 úû úû Bz úû
Bx ý Bò cos ö Bö sin ö ý ò 2 sin ö cos ö ( z 1) cos ö sin ö
ý 16(0.5) (2)(0.5) ý 8 1 ý 9
Prob. 2.30
_
G ý c os2 ö a x
_
2 r c osñ sin ö _
a y (1 c o s2 ö ) a z
r sin ñ
_
_
_
ý c os2 ö a x 2c ot ñ sin ö a y sin 2 ö a z
ö Gr ö
ù sin ñ c o sö sin ñ sin ö
÷ ÷
ú
÷ Gñ ÷ ý úc osñ c osö c osñ sin ö
÷G ÷
úû sin ö
c osö
ø öø
c osñ ù
sin ñ úú
0 úû
ù c os2 ö ù
ú
ú
ú2c o t ñ sin ö ú
ú sin 2 ö ú
û
û
G r ý sin ñ c o s3 ö 2c o sñ sin 2 ö c osñ sin 2 ö
ý sin ñ c o s3 ö 3c o sñ sin 2 ö
Gñ ý c osñ c os3 ö 2c o t ñ c osñ sin 2 ö sin ñ sin 2 ö
Gö ý sin ö c os2 ö 2c ot ñ sin ö c osö
_
G ý [sin ñ c os3 ö 3c osñ sin 2 ö ] a r
[c osñ c o s3 ö 2c o t ñ c osñ sin 2 ö sin ñ sin 2 ö ]añ
sin ö c o sö(2 c ot ñ c o sö ) a ö
Prob. 2.31
(a)
An infinite line parallel to the z-axis.
(b)
Point (2,-1,10).
(c)
A circle of radius r sin ñ ý 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
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Prob. 2.32
(a) J z ý ( J ÷ a z )a z .
At (2, ð / 2, 3ð / 2), a z ý cos ñ ar sin ñ añ ý añ .
J z ý cos 2ñ sin ö añ ý cos ð sin(3ð / 2) añ ý añ .
ñ
ð
(b) Jö ý tan ln r aö ý tan ln 2 aö ý ln 2aö ý 0.6931aö .
2
4
(c) J t ý J J n ý J J r ý añ ln 2 aö ý añ 0.6931aö
(d )
J P ý ( J ÷ a x )a x
a x ý sin ñ cos ö ar cos ñ cos ö añ sin ö aö ý aö .
At (2, ð / 2, 3ð / 2),
J P ý ln 2aö .
Prob. 2.33
H ða x ý H x
ù H x ù ù cos ö
ú H ú ý ú sin ö
ú yú ú
úû H z úû úû 0
sin ö
cos ö
0
0 ù ù ò 2 cos ö ù
ú
ú
0 úú ú ò sin ö ú
1 úû úû 0 úû
H x ý ò 2 cos 2 ö ò sin 2 ö
At P, ò ý 2, ö ý 60o , z ý 1
H x ý 4(1/ 4) 2(3 / 4) ý 1 1.5 ý 2.5
Prob. 2.34
(a) 5 ý r a x r a y ý x y
(b)
10 ý rxa z ý
a p la ne
x y z
ý| y a x xa y | ý x2 y 2 ý ò
0 0 1
a c ylind er of infinite leng th
37
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CHAPTER 3
P. E. 3.1
ö ý 60ð
(a) DH ý
ö
r sin ñ dö
ý 45ð
ñ ý 90 ð
(b) FG ý
rdñ
rý5
ñ ý 60 ð
ð
ð
ð
ý 3(1)[ ] ý
ý 0.7854.
r ý 3,90
3 4
4
o
ð ð
5ð
ý 5( ) ý
ý 2.618.
2 3
6
(c)
ñ ý 90 ð ö ý 60 ð
AEHD ý
r
2
sin ñ dñ dö
ñ ý 60 ð ö ý 45 ð
rý3
ö ý 60 ð
90 ð
ý 9 ( cos ñ )|ññ ýý 60
ð ö |ö ý 45 ð
1 ð
3ð
ý 9 ( )( ) ý
ý 1178
. .
2 12
8
(d)
ABCD ý
r ý 5 ñ ý90
r ý3
r2 r ý 5 ð ð
4ð
rdñ dr ý
( )ý
ý 4.189.
2 r ý3 2 3
3
ñ ý 60
(e)
Volume ý
r ý5
r ý3
ý
ö ý 60ð
ö
ý 45ð
ñ ý90
ñ
r 2 sin ñ dr dñ dö ý
ý 60
r3
3
r ý5
r ý3
( cos ñ )
ñ ý90ð
ñ ý 60ð
ö
ö ý 60ð
ö ý 45ð
49ð
ý 4.276 .
36
P.E. 3.2
y
3
2
60o
1
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x
ý
1
1 ð
(98)( )
3
2 12
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ð A÷ dl ý (
L
1
2
) A÷ dl ý C1 C2 C3
3
2
Along (1), C1 ý A÷ dl ý ò cos ö d ò |ö ý0 ý
0
ò2
20
ý 2.
Along (2), dl ý ò dö aö , A÷ dl ý 0, C2 ý 0
0
Along (3), C3 ý ò cos ö d òö ý60ð ý
ò2
2
ð A÷ dl ý C
1
2
0
2
1
( ) ý 1
2
C2 C3 ý 2 0 1 ý 1
l
P.E. 3.3
ñU ý
(a)
öU
öU
öU
ax
ay
az
öx
ö y
öz
ý y (2 x z ) a x x( x z ) a y xy a z
ñV ý
(b)
öV
öV
1 öV
aò
aö
az
öò
òöö
öz
ý ( z sin ö 2 ò ) a ò ( z cos ö
z2
ò
sin 2ö ) aö ( ò s inö 2 z cos 2 ö ) a z
(c)
ñf ý
1ö f
1 öf
öf
ar
añ
aö
r öñ
r sin ñ öö
ör
cos ñ sin ö
sin ñ sin ö ln r
(cos ñ cos ö ln r r 2 )
2rö )ar
añ
aö
r
r
r sin ñ
sin ñ sin ö ln r
ö cos ñ sin ö
ö
ö cot ñ cos ö ln r
ö
añ ÷
ý÷
2rö ÷ a r
r cos ecñ ÷ aö
r
r
r
ø
ø
ø
ø
ý(
P.E. 3.4
ñö ý ( y z ) a x ( x z ) a y ( x y ) a z
At (1, 2,3), ñö ý (5, 4,3)
(2, 2,1) 21
ý
ý 7,
3
3
where (2, 2,1) ý (3, 4, 4) (1, 2,3)
ñö ÷ a1 ý (5, 4,3) ÷
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P.E. 3.5
g ý x log z y 2 4,
Let f ý x y z 3,
2
ñ f ý 2 xy a x x 2 a y a z ,
ñ g ý log z a x 2 y a y
x
az
z
At P ( 1 , 2,1),
(4 a x a y a z )
(4 a y a z )
ñf
ñg
,
ý
ý
ng ý
| ñf |
| ñg |
18
17
( 5)
cos ñ ý n f . n g ý
18 17
T ake positive value to get acute angle.
5
ñ ý cos 1
ý 73.39 ð
17.49 3
nf ý
P.E. 3.6
(a) ñ ÷ A ý
At
ö Ax ö Ay ö Az
ý 0 4 x 0 ý 4 x.
öx
öy
öz
(1, 2,3), ñ ÷ A ý 4.
(b)
ñ÷B ý
ý
1 ö
ò öò
1
ò
( ò Bò )
2 ò z sin ö
1
ò
1 ö Bö
ò öö
ö Bz
öò
3ò z 2 sin ö ý 2 z sin ö 3 z 2 sin ö
ý (2 3 z ) z sin ö .
At (5,
(c )
ð
2
,1) ,
ñ ÷ B ý (2 3)(1) ý 1.
1 ö 2
1 ö
1 ö Cö
(
)
(
sin
)
ñ
r
C
C
r
ñ
r2 ö r
r sin ñ öñ
r sin ñ öö
1
ý 2 6r 2 cos ñ cos ö
r
ý 6 cos ñ cos ö
ñ÷C ý
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ð ð
At (1,
ñ ÷ C ý 6 cos
, ),
6 3
ð
cos
6
ð
3
ý 2.598.
P.E. 3.7 This is similar to Example 3.7.
ý ð
D ÷ dS ý t b c
S
t ý 0 ý b since D has no z-component
c ý ò 2 cos 2 öò dö dz ý ò 3
ö ý 2ð
ö
cos 2 ö dö
ý0
z ý1
dz ò ý 4
z ý0
ý (4) ð (1) ý 64ð
ý 0 0 64ð ý 64ð
3
By the divergence theorem,
ð D ÷ dS ý ð ñ ÷ Ddv
S
V
ñ÷ D ý
1 ö
ò öò
ý 3ò cos 2 ö
z
ò
ý ñ ÷ Ddv ý
V
4
ý 3 ò d ò
2
0
ý 3(
1 ö
( ò 3 cos 2 ö )
ö Az
dz
cos ö .
(3ò cos 2 ö
V
2ð
cos
0
ò öö
z sin ö
1
2
ö dö dz
0
z
ò
cos ö ) ò dö dzd ò
4
2ð
1
0
0
0
d ò cos ö dö zdz
3
4
) ð (1) ý 64ð .
3
P.E. 3.8
(a)
ñ A ý a x (1 0) a y ( y 0) a z (4 y z )
ý a x y a y (4 y z ) a z
_
At (1, 2,3) , ñ A ý a x 2 a y 11a z
(b)
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ñ B ý a ò (0 6 ò z cos ö ) aö ( ò sin ö 0) a z
1
ò
(6 ò z 2 cos ö ò z cos ö )
ý 6 ò z cos ö a ò ò sin ö aö (6 z 1) z cos ö a z
At (5,
ð
2
, 1) , ñ B ý 5 aö
(c)
_
1
2r cos ñ sin ö 3 1/ 2 a ö
añ
r ) (0 2r sin ñ cos ö )
(r 1/ 2 cos ñ 0)
(
sin ñ
2
r sin ñ
r
r
3
ý r 1/ 2 cot ñ a r (2 cot ñ sin ö r 1/ 2 ) añ 2sin ñ cos ö aö
2
ñ C ý ar
ð ð
, ), ñ C ý 1.732 a r 4.5 añ 0.5 aö
6 3
At (1,
P.E. 3.9
ð A ÷ dl ý (ñ A) ÷ dS
L
S
But (ñ A) ý sin ö a z
z cos ö
ò
a ò and
d S ý ò dö d ò a z
(ñ A) ÷ dS ý ò sin ö dö d ò
S
ý
ò2
|
2
2
0
60ð
( cos ö ) |
0
1
ý 2( 1) ý 1.
2
P.E. 3.10
ñ ñV ý
ax
ay
az
ö
öx
öV
öx
ö
öy
öV
öy
ö
ý
öz
öV
öz
ö 2V
ö 2V
ö 2V
ö 2V
ö 2V
ö 2V
) ax (
)ay (
)az ý0
ý(
ö yö z ö yö z
ö xö z ö zö x
ö xö y ö yö x
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P.E. 3.11
(a)
ñ 2U ý
ö
ö 2
ö
(2 xy yz )
( x xz )
( xy )
öx
öy
öz
ý 2 y.
(b)
ñ 2V ý
ý
1 ö
ò öò
1
ò
ò( z sin ö 2 ò)
( z sin ö 4 ò)
ý 4 2 cos2 ö
(c)
2z 2
ò2
1
ò2
1
ò
2
( òz sin ö 2 z 2
ö
ö
sin ö cos ö ) ( ò sin ö 2 z cos2 ö )
öò
öz
( zò sin ö 2 z 2 cos 2ö ) 2 cos2 ö.
cos 2ö.
1 ö 21
1
ö
[r cos ñ sin ö 2r 3ö ] 2
[ sin 2 ñ sin ö ln r ]
2
r ör
r
r sin ñ ö ñ
1
2 2 [ cos ñ sin ö ln r ]
r sin ñ
1
ý 2 cos ñ sin ö (1 2 ln r csc 2 ñ ln r ) 6ö
r
ñ2 f ý
P.E. 3.12
If B is conservative , ñ B ý 0 must be satisfied.
ax
ö
ñ B ý
öx
y z cos xz
ay
ö
öy
x
az
ö
öz
x cos xz
ý 0 a x (cos xz xz sin xz cos xz xz sin xz ) a y (1 1) a z ý 0
Hence B is a conservative field.
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Prob. 3.1
(a)
dl ý ò dö ;
òý3
ð
2
ð ð
3ð
L ý dl ý 3 dö ý 3( ) ý
ý 2.356
2 4
4
ð
4
(b)
dl ý r sin ñ dö;
r ý 1, ñ ý 30ð ;
L ý dl ý r sin ñ
(c)
ð
3
0
ð
dö ý (1) sin 30ð [( ) 0] ý
3
0.5236.
dl ý rdñ
ð
Lý
2
4ð
ð ð
ý 4.189
d l ý r dñ ý 4( ) ý
2 6
3
ð
6
Prob. 3.2
(a)
d S ý ò dö dz
5
S ý d S ý ò dö dz ý 2 dz
0
ð
2
ð dö
ð
ð
ý 2(5)[ ] ý
2 3
10ð
ý 5.236
6
3
(b)
In cylindrical, dS ý ò d ò dö
ð
S ý d S ý
3
4
1
0
ò d ò dö ý
( ) ý 3.142
4 2 1
d S ý r 2 sin ñ dö dñ
(c) In spherical,
S ý d S ý 100
ð ò2 3
2ð
3
2ð
ð
0
2ð
3
sin ñ dñ dö ý 100 (2ð )( cos ñ ) ð|
4
ý 200ð (0.5 0.7071) ý 758.4
4
(d)
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d S ý r dr dñ
ð
2
r2 4 ð ð
8ð
S ý dS ý rdr dñ ý |0 ( ) ý
ý 4.189
2 2 3
6
ð
0
4
3
Prob.3.3
(a ) dV ý dxdydz
1
2
3
0
1
3
V ý dxdydz ý dx dy dz ý (1) (2 1)(3 3) ý 6
(b) dV ý ò dö d ò dz
5
ð
4
V ý ò d ò dz dö ý
5
|
2
ð
1
2
ò2
2
ð
1
2ð
(4 1)(ð ) ý (25 4)(5)( ) ý 35ð ý 110
3
2
3
3
(c) dV ý r 2 sin ñ drdñ dö
ð
3
ð
2
2
6
2
ð /3 ð
r3 3
ð
V ý r dr sin ñ dö ý |1 ( cos ñ ) |ð / 2 ( )
3
2 6
ð
ð
1
3
2
ý
1 ð
26 ð
1
(27 1)( )( ) ý
ý 4.538
2 3
18
3
Prob.3.4
L ý dl ý
L
ð /6
ò dö
ö ý0
ò ý4
ý 4(ð / 6) ý 2.094
Prob. 3.5
dS ý r 2 sin ñ dñ dö
S ý r2
ð /2
dö
0
ý
ð /4
sin ñ dñ
0
r ý5
ý 25(ð / 2)( cos ñ )
ð /4
0
25ð
( cos(ð / 4) 1) ý 11.502
2
Prob. 3.6
dv ý ò d ò dö dz
Vý
10
z ý0
30
dz
ö ý0
dö
5
ò ý2
ö ò 2 5 ö 5ð
(25 4) ý 54.98
÷ý
ø 2 2ø 6
ò d ò ý 10(ð / 6) ÷
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Prob. 3.7
dl ý dxa x dya y
I ý H ÷dl = ( xy 2 dx x 2 ydy )
L
But on L, x ý y 2
dx ý 2 ydy
ö 2 y6 y4 ö 4
÷
I ý y (2 ydy ) y dy ý (2 y dy y dy ) ý ÷
4 ø1
ø 6
y ý1
1
2 1
ý (4096) 64 ý 1428.75
3
6 4
4
4
3
5
3
Prob. 3.8
The line joining P and Q is y ý x 2, dy ý dx
I ý (2 x 2 4 xy )dx (3xy 2 x 2 y )dy ý
L
3
ùû2 x
x ý1
2
4 x( x 2) ùûdx ùû3 x( x 2) 2 x 2 ( x 2) ùûdx
ö 3
x4 ö 3
2
ý ùû3x 2 x 2 x ùûdx ý ÷ x x ÷ ý 27 9 81/ 2 ý 4.5
4 ø1
ø
x ý1
3
2
3
Prob. 3.9
(a)
1
F ÷ dl ý ( x
y ý0
2
z )dy|
2
x ý 0, z ý 0
xý2
2 xydx |
x ý0
y ý1, z ý 0
z ý3
(3xz
z ý0
x2 2
z3 3
3(2)
|
|
2 0
3 0
ý 0 4 54 ý 50
ý 0 2(1)
(b)
Let x ý 2t. y ý t , z ý 3t
dx ý 2dt , dy ý dt , dz ý 3dt ;
F ÷ dl ý
1
(8t
2
5t 2 162 t 3 ) dt
0
1
ý (t 3 40.5t 4 ) ý 39.5
0
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2
)dz|
x ý 2, y ý1
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Prob.3.10
W ý F ÷ dl ý
L
ð /4
z ò dö
z ý 0, ò ý 2
ö ý0
3
cos ö dz ö ð
ý /4
z ý0
ý 0 cos(ð / 4)(3) ý 3cos 45o ý 2.121 J
Prob. 3.11
H ÷ dl ý
0
( x y )dx
x ý1
y ý 0, z ý 0
( x 2 zy )dy 5yzdz
0
2
1
0
ý xdx ( y
1
5 yzdz x ý 0, y ý 0
z ý0
x ý 0, z ý 1 y / 2
0
y2
)dy (10 z 10 z 2 )dz
2
1
ý 1.5
Prob. 3.12
Method 1:
1
1
ð B dl ý
yzdy
But z ý y
dz ý dy on the last segment (or integral).
y ý0
L
ð B dl
ý 0
L
ý
zý0
xzdz
z ý0
x ý1
( yzdy xzdz )
x ý1
0
z2 1
y3 y 2 0
1
( y 2 y )dy ý ( )
2 0 y ý1
2
3
2 1
1 1 1 1
ý ý 0.333
2 3 2 3
Method 2:
ð B dl ý ñ BðdS
L
S
ö
ñ B = öx
xy
ñ BðdS ý
S
ö
öy
-yz
1
ö
öz ý ya x za y xa z ,
xz
y
y ý0 z ý0
1
ydzdy ý y 2 dy ý
0
dS ý dydza x
y3 1 1
ý ý 0.333
3 0 3
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Prob. 3.13
1
2
z2 2
ý AðdS ý zdxdz ý dx zdz ý (1)
ý2
2 0
S
0
0
Prob. 3.14
ýð
D ÷ dS ý ñ ÷ Ddv
S
v
ñ÷D ý
öDx öDy öDz
ý 2 xz 3 y 2 2 yz
öx
öy
öz
ý ñ ÷ Ddv ý (2 xz 3 y 2 2 yz )dxdydz
v
1
4
3
1
4
3
1
4
3
1
0
1
ý 2 xdx dy zdz 3 dx y dy dz 2 dx ydy zdz
2
1
0
1
1
0
1
ö y 4ö
ö y 4öö z 4ö
ý 0 3(2) ÷
÷ (3) 2(2) ÷
÷÷
÷ ý 6(64) 16(9 1) ý 384 128
ø 3 0ø
ø 2 0øø 2 0ø
ý 512
3
2
2
Prob. 3.15
ýð
A ÷ dS ý ñ ÷ Adv
S
v
ö
1 ö 2
1
1
3cos ñ
5
(r r )
(3sin ñ )
(5) ý 3
2
r ör
r sin ñ öñ
r sin
r sin ñ r sin ñ
2
dv ý r sin ñ dñ drdö
ñ÷ A ý
ý 3 r 2 sin ñ d ñ d ö dr 3 r cos ñ d ñ d ö dr 5d ñ d ö dr
4
ý 3 r 2 dr
0
ð /2
sin ñ dñ
0
3ð /2
0
4
ð /2
0
0
dö 3 rdr
cos ñ dñ
3ð /2
0
4
ð /2
0
0
dö 5 dr
dñ
3ð /2
dö
0
ö r 4ö
ö r 4ö
ð /2
ð /2
ý 3÷
(3ð / 2) 3 ÷
(3ð / 2) 5(4)(ð / 2)(3ð / 2)
÷ ( cos ñ )
÷ (sin ñ )
0
0
ø 3 0ø
ø 2 0ø
ý 96ð 36 ð 15ð 2 ý 336.54
3
2
Prob. 3.16
(a) dv = dxdydz
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2
1
1
xydv ý
z ý0 y ý0 z ý0
v
ý
2
1
1
z
0
0
0
xydxdydz ý xdx ydy dz
2
x 1 y 1 2
z ý (1/ 2)(1/ 2)(2) ý 0.5
2 0 2 0 0
(b)
dv ý ò d ò dö dz
ò zdv ý
v
ý
ð
2
3
ö ý 0 z ý 0 ò ý1
3
2
ð
1
0
0
ò z ò d ò dödz ý ò 2 d ò zdz dö
ò 3 3 z2 2
1
(ð ) ý (9 )(2ð ) ý 54.45
3 1 2 0
3
Prob. 3.17
öV1
öV
öV
a x 1 a y 1 az
öx
öy
öz
ý (6 y 2 z )a x 6 xa y (1 2 x)a z
(a ) ñV1 ý
ñV3 ý
(c )
öV2
öV
1 öV2
aò
aö 2 a z
öò
öz
ò öö
ý (10 cos ö z )a ò 10sin ö aö ò a z
(b) ñV2 ý
öV3
1 öV3
1 öV3
ar
añ
aö
ör
r öñ
r sin ñ öö
2
1 ö 2
ö
cos ö ar 0
÷ sin ö ÷ aö
2
r
r sin ñ ø r
ø
2
2sin ö
ý 2 cos ö ar 2
aö
r
r sin ñ
ý
Prob. 3.18
(a)
ñV ý (10 yz 4 xz )a x 10 xza y (10 xy 2 x 2 )a z
x ý 1, y ý 4, z ý 3. Hence,
At P,
ñV ý (120 12)a x 30a y (40 2)a z ý 132a x 30a y 42a z
(b)
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ñU ý (2sin ö z )aò 2 cos ö aö ò a z
At Q, ò ý 2, ö ý 90o , z ý 1
ñU ý (2 1)aò 0aö 2a z ý a ò 2a z
(c )
8
1 4 cos ñ cos ö
1
añ
sin ñ cos ö ar
3
2
r
r
r
r sin ñ
At R, r ý 1, ñ ý ð / 6, ö ý ð / 2
4
ñW ý 3 aö ý 4aö
(1)
ñW ý
ö 4sin ñ sin ö ö
÷
÷ aö
r2
ø
ø
Prob. 3.19
r ý x2 y 2 z 2 ,
r n ý ( x 2 y 2 z 2 )n / 2
Method 1:
ñr n ý
n
ör n
ör n
ör n
ax
ay
a z ý ( x 2 y 2 z 2 ) n / 21 (2 x)a x
öx
öy
öz
2
ý n( x 2 y 2 z 2 )
n2
2
( xa x ya y za z ) ý nr n 2 r
Method 2:
ör n
r
ñr n ý
ar ý nr n 1 ý nr n 2 r
ör
r
Prob. 3.20
At
a
ñ T ý 2x a x 2 y a y
z
ø1,1, 2 ù , ñ T ý (2, 2, 1).
The mosquito should move in the direction of
2ax 2a y az
Prob. 3.21
ñ F ý a x 2a y a z
an ý
a x 2a y a z
ñF
ý
ý 0.4082a x 0.8165a y 0.4082a z
| ñF |
1 4 1
Prob. 3.22
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Method 1:
öT
1 öT
1 öT
ar
añ
aö
ör
r öñ
r sin ñ öö
ý sin ñ cos ö ar cos ñ cos ö añ sin ö aö
ñT ý
At P, r ý 2,ñ ý 60o , ö ý 30o
ñT ý sin 60o cos 30o ar cos 60o cos 30o añ sin 30o aö
ý 0.75ar - 0.433añ 0.5aö
| ñT |ý 0.752 0.4332 0.52 ý 1
The magnitude of T is 1 and its direction is along ñT.
Method 2:
T ý r sin ñ cos ö ý x
ñT ý a x
| ñT |ý 1
Prob. 3.23
öf
öf
öf
ñf ý a x a y a z ý (2 xy 2 y 2 )a x ( x 2 4 xy )a y 3 z 2 a z
öx
öy
öz
At point (2,4,-3), x ý 2, y ý 4, z ý 3
ñf ý (16 32)a x (4 32)a y 27a z ý 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
ñf ða ý (16, 28, 27)ð (1, 2, 1) ý
ý 40.42
6
6
aý
ý
Prob. 3.24
(a) Let f = ax + by + cz – d = 0
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ñf ý aa x ba y ca z
aa ba y ca z
ñf
ý x
| ñf |
a 2 b2 c2
Let g ý ñ x ò y ÷ z ô
ñg ñ a x ò a y ÷ a z
ý
an 2 ý
| ñg |
ñ2 ò2 ÷ 2
an1 ý
cos ñ ý an1 ðan 2 ý
ñ ý cos 1
añ bò c÷
a 2 b2 c 2 ñ 2 ò 2 ÷ 2
añ bò c÷
(a b 2 c 2 )(ñ 2 ò 2 ÷ 2 )
2
a ý 1, b ý 2, c ý 3
(b) ñ ý 1, ò ý 1, ÷ ý 0
ñ ý cos 1
1 2 0
(12 22 32 )(12 12 02 )
ý cos 1
3
ý cos 1 0.5669 ý 55.46o
28
Prob. 3.25
öV
öV
öV
ax
ay
a z ý 4 ye z a x 4 xe z a y 4 xye z a z
öx
öy
öz
At (3,1,-2), x ý 3, y ý 1, z ý - 2
ñV ý
ñV ý 4e 2 a x 12e 2 a y 12e 2 a z ý 0.5413a x 1.624a y 1.624a z
This is the direction. The maximum rate of change is
| ñV |ý e 2 42 122 122 ý 0.1353 17.44 ý 2.36
Prob. 3.26
(a)
ö
ö
ö
ñUV ý (UV )a x (UV )a y (UV )a z
öx
öy
öz
öU
ö öV
ý ÷U
V
öx
ø öx
ý U ñV V ñU
ö öV
öU
ö
V
÷ ax ÷U
öy
ø
ø öy
ö
öU
ö öV
V
÷ a y ÷U
öz
ø öz
ø
(b)
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ö
÷ az
ø
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ñV ý 10 xya x (5 x 2 2 z )a y 2 ya z
U ñV ý 30 x 2 y 2 za x (15 x 3 yz 6 xyz 2 )a y 6 xy 2 za z
ñU ý 3 yza x 3 xza y 3xya z
2
z
2
y
6
+
V ñU ý (15 x 2 y 2 z
a x (15 x3 yz 6 xyz 2 )a y (15 x3 y 6 xy 2 z )a z
︶
U ñV V ñU ý (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 ý 15 x 3 y 2 z 6 xy 2 z 2
ñ(UV ) ý (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
öAx öAy öA z
ý 3y x
öx
öy
öz
1
1
ñ B ý 2 ò z2 ò 2 sin ö c osö 2 ò sin 2 ö
(a) ñ A ý
ò
(b)
ò
ý 2 z sin 2ö 2 ò sin 2 ö
2
(c ) ñ C ý
1 2
3r 0 ý 3
r2
Prob. 3.28
öAx öAy öAz
ý 2 xy 0 2 y ý 2 y (1 x)
öx
öy
öz
(a) At (3, 4, 2), x ý 3, y ý 4
ñðA ý 2(4)(1 3) ý 16
ñðA ý
ñðB ý
1 ö
1 öBö öBz 1 ö
3ò 2 sin ö ù 0 8 z cos 2 ö
ý
ò Bò ù
ø
ø
ò öò
ò öö öz ò öò
ý 6sin ö 8 z cos 2 ö
(b)
At
(5,30o ,1), z ý 1, ö ý 30o
ñðB ý 6sin 30o 8(1) cos 2 30o ý 3 6 ý 9
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1 ö 2
1 öCö 1 ö 4
( r Cr ) 0
(r cos ñ ) 0 ý 4r cos ñ
ý
2
r ör
r sin ñ öö r 2 ör
(c ) At (2, ð / 3, ð / 2), r ý 2, ñ ý ð / 3
ñðC = 4(2)cos(ð /3) = 4
ñðC ý
Prob. 3.29
2
ñ÷ H ý k ñ÷ ñ T ý k ñ T
ö 2T ö 2T
ðx
ð y ð2 ð2
ñ Tý
ý 50sin
)ý 0
cos h
(
ö x2 ö y
2
2
4
4
2
2
Hence, ñ÷ H ý 0
Prob. 3.30
(a)
ö
ö
ö
(V Ax )
(V Ay )
(V Az )
öx
ö y
öz
ö Ay
ö Ax
ö Az
öV
öV
öV
ý ( Ax
V
) ( Ay
V
) ( Az
V
)
öx
öx
ö y
ö y
öz
öz
ö Ax ö Ay ö Az
öV
öV
öV
ý V(
) Ax
Ay
Az
öx ö y öz
öx
ö y
öz
ñ ÷ (V A) ý
ý V ñ ÷ A A÷ ñ V
(b)
ñ ÷ A ý 2 3 4 ý 1;
ñ V ý yz a x xz a y xy a z
A
ñ ÷ (V A) ý V ñ ÷ A÷ ñV
ý xyz 2 xyz 3 xyz 4 xyz ý 2 x y z
Prob. 3.31
(a)
(ñ÷ r )T ý 3T ý 6 yz a y 3 xy 2 a y 3 x 2 yz a z
(b)
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x
öT
öT
öT
y
z
ý x (y 2 a y 2 xyz a z ) y(2 z a x 2 xy a y x2 z a z )
öx
öy
öz
z(2y a x x2 y a z )
ý 4yz a x 3 xy 2 a y 4 x2 yz a z
(c)
ñ ÷ r (r ÷ T ) ý 3 (2 xyz xy 3 x 2 yz 2 )
ý 6 xyz 3xy 3 3 x 2 yz 2
(d)
( r ÷ ñ) r 2 ý ( x
ö
ö
ö
y
z )( x2 y 2 z2 )
öx
öy
öz
ý x(2 x) y(2y ) z(2 z)
ý 2( x2 y 2 z2 ) ý 2 r 2
Prob. 3.32
We convert A to cylindrical coordinates; only the ò-component is needed.
Aò ý Ax cos ö Ay sin ö ý 2 x cos ö z 2 sin ö
But x ý ò cos ö ,
Aò ý 2 ò cos 2 ö z 2 sin ö
ý A dS ý Aò ò dö dz ý ùû 2 ò 2 cos 2 ö ò z 2 sin ö ùûdö dz
S
ý 2(2)
ð /2
2
0
1
1
1
(1 cos 2ö )dö dz 2 z 2 dz
2
0
0
ð /2
sin ö dö
0
ð /2
ð /2
1
z 1
( cos ö )
ý 4(ö sin 2ö )
2
ý 2ð 2 / 3 ý 5.6165
0
0
2
3 0
3
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Prob. 3.33
z
Z=1
y
Z=1
x
(a)
ð D ÷ d Sý [
zý1
zý1
ò
]D ÷ d S
ý5
ý ò 2 c os2 ö d ö d ò ò 2 c os2 ö d ö d ò 2 ò 2 z2 d ö d z|
ý 2(5)2
2ð
1
0
1
2
dö z d z ý 50(2ð )(
3
z 1
| 1 )
3
200 ð
ý 209.44
3
1 ö
(2 ò 2 z 2 ) ý 4 z 2
(b) ñ ÷ D ý
ý
ò öò
1
5
2ð
0
0
ñ ÷ Dd v ý 4 z ò d ò dö d z ý 4 z d z ò d ò dö
2
2
1
ý 4x
3
z
3
1
1
ò
2
2
5
0
(2ð ) ý
200ð
ý 209.44
3
Prob. 3.34
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ò ý5
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ð / 2 2ð
H ðdS ý ñ ö 10 cosñ r
2
sin ñ dñ dö
ý0 ý0
S
ý 10(1)
2ð
2
r ý1
ð /2
ð /2
0
0
dö sin ñ cos ñ dñ ý 10(2ð ) sin ñ d (sin ñ )
0
ö sin ñ ö ð / 2
ý 20ð ÷
ý 10ð ý 31.416
÷
ø 2 ø 0
2
Prob. 3.35
ð H d S ý ñ Hd v
S
v
ð H d S ý 2 xyd yd z x ý 0 2 xyd yd z x ý 1 (x
2
z2 )d xd z
S
( x2 z2 )d xd z
y ý2
2yzd xd y
y ý1
2yzd xd y
z ý 1
zý3
2
3
1
2
1
2
1
1
0
1
0
1
ý 0 2 yd y d z 2 d x yd y 6 d x yd y
ý 12 3 9 ý 24
ñH ý
öHx öHy öHz
ý 2y 0 2y ý 4y
öx
öy
öz
1
2
3
0
1
1
ñ Hd v ý 4yd xd yd z ý 4 d x yd y d z
V
ý 4(1)
2
y 2
(3 1) ý 24
2 1
Prob. 36
ýð
H ÷ dS ý ñ ÷ Hdv
S
To find
v
ð H ÷ dS ,
let
S
= t + b s
wher t , b , and s are the fluxes from the top, bottom, and side of the cylinder.
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dS ý ò d ò dö a z ,
For t ,
t ý 2 z ò d ò dö
2ð
0
0
ý 6 ò d ò dö ý 12ð
ò 2 10
2 0
ý 600ð
dS ý ò d ò dö (a z ) ,
For t ,
b ý 2 z ò d ò dö
zý0
ý0
dS ý ò dö dzaò ,
For s ,
t ý 4ò
z ý3
10
3
3
2ð
0
0
dö dz ò ý 10 ý 4000 dz dö ý 4000(3)(2ð ) ý 24000ð
ý 600ð 0 24000ð ý 23400ð ý 73,513.27
ñ ÷ Hdv,
To get
dv ý ò dö d ò dz
v
1 ö
(4 ò 3 ) 2 ý 12 ò 2
ò öò
ñ÷ H ý
10
2ð
3
10
2ð
3
0
0
0
0
0
0
ý (12 ò 2)ò dö d ò dz ý 12 ò 2 d ò dö dz 2 ò d ò dö dz
(1000)
(100)
(2ð )(3) 2
(2ð )(3) ý 24000ð 600ð ý 73,513.27
3
2
ý 12
Prob. 3.37
Let
=ð
AðdS= ñðAdv
v
S
öAx öAy öAz
ý 2( x y z )
öx
öy
öz
ñðA = 2( ò cos ö ò sin ö z ),
dv ý ò d ò dö dz
ñðA =
ý 2( ò cos ö ò sin ö z ) ò d ò dö dz
1
4
2ð
0
2
0
ý 0 0 2 ò d ò zdz dö ý
2ò 2 1 z 2 4
1
(2ð ) ý (16 4)(2ð )
2 0 2 2
2
ý 12ð ý 37.7
Prob. 3.38
1 ö
1 ö
(r 4 )
(r sin 2 ñ cos ö )
2
r ör
r sin ñ öñ
ý 4 r 2 cos ñ cos ö
ñ÷ A ý
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ñ ÷ Adv ý 4r
3
sin ñ dñ dö dr 2r 2 sin ñ cos ñ cos ö dñ dö dr
ð
ð /2 ð
r4 3
2r 3 3 cos 2 ñ ð / 2
2
(
cos
)
(
)
(
)
sin
ñ
ö
|
|
|
|
|
0
0
0
0
0
4
2
3
2
1
ð
ý 81(1)( ) 18(0 )(1 0)
2
2
81ð
ý
9 ý 136.23
2
ý4
z
y
x
A ÷ dS ý [ö ö ð ñ ð
ý0
ý /2
r ý3
] A ÷ dS
ý /2
Since A has no ö component, the first two integrals on the right hand side vanish.
A ÷ dS ý
ð /2
ð /2
ö
ñ
ý0
ð
ý0
r 4 sin ñ dñ dö|
ð /2
r ý3
3
r ý0
ð /2
ö
ý0
r 2 sin 2 ñ cos ö drdö|
ð /2
ý 81 ( ) ( cos ñ )| 9(1) sin ö|
0
0
2
81ð
ý
9 ý 136.23
2
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ñ ýð / 2
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Prob. 3.39
Let ý
ð F ÷ dS ý
t
b o i
where t , b , o , i are the fluxes through the top surface, bottom surface,
outer surface ( ò ý 3), and inner surface respectively.
For the top surface, dS ý ò dö d ò a z ,
z ý 5;
F ÷ dS ý ò 2 z dö dz. Hence:
t ý
2ð
3
ö
ò
ý2
ý0
ò 2 z dö dz|z ý5 ý
z ý 0, dS ý ò dö d ò (
a
For the bottom surface,
190 ð
ý 198.97
3
z
)
F ÷ dS ý ò z dö d ò ý 0. Hence, b ý 0.
2
a
For the outer curved surface, ò ý 3, dS ý ò dö dz
ò
F ÷ dS = ò sin ö dö dz. Hence,
2
5
a ý
dz ò 3
z ý0
2ð
ö sin ö dö|ò
ý0
ý3
ý0
For the inner curved surface, ò ý 2, d S ý ò dö dz ( a ò )
F ÷ dS ý ò 3 sin ö dö dz. Hence,
a ý
2ð
5
dz ò ö sin ö dö|ò
3
z ý0
ý
ý
ý0
ý2
ý0
190 ð
190 ð
00 0ý
ý 198.97
3
3
ð F ÷ dS ý ñ ÷ FdV
ñ÷F ý
1 ö
òöò
ý 3 ò sin ö
z
ò
( ò 3 sin ö )
1 ö
ò öö
( z cos ö ) ò
sin ö ò
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V
z
(3ò sin ö ò sin ö ò ) ò dö d ò dz
ñ ÷ Fdv ý
5
2ð
3
0
0
2
ý 0 0 dz dö ò 2 d ò
ý
190 ð
ý 198.97
3
Prob. 3.40
ax
ö
(a) ñxA ý
öx
xy
ay
ö
öy
y2
az
ö
ý za y xa z
öz
xz
ö1
ö
1
ñxB ý ÷ 2 ò z2 sin ö c o sö 0 ÷ a ò (2 ò z 2 z sin 2 ö )aö
2 ò sin 2 ö 0 a
ò
øò
ø
ý 4 z sin ö c o sö a ò 2( ò z z sin 2 ö )aö 2 sin 2 ö a z
(b)
ø
ý 2 z sin 2ö a ò 2 z( ò sin 2 ö )aö 2 sin 2 ö a z
ñxC ý
1ù ö 2
ù ö
ù
2
2 ù
ú öñ (r c o s ñ sin ñ ú a r r ú ör (r c o s ñ ú añ
û
û
û
û
c os2 ñ
r
ùû(2c osñ )( sin ñ )sin ñ c o sñ (c o s2 ñ )ûù a r
(2 r)añ
r sin ñ
r
(c o s3 ñ 2 sin 2 ñ c osñ )
ý
a r 2c os2 ñ añ
sin ñ
ý
(c )
Prob. 3.41
(a)
ax
ö
öx
x2 y
az
ö
ö
ý
öy
öz
y 2 z 2 xz
y 2
x
2 z a y x2
a
ý
ay
a
A
ñ
1
r sin ñ
z
ñ÷ñ A ý 0
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(b)
A
ñ
ý(
1 ö Az
ò öö
ö Aö
ö A ö Az
1 ö ( ò Aò ) ö Aò
) aò ( ò
) aö (
)az
öz
öz
öò
ò
öò
öö
ý (0 0) a ò ( ò 2 3 z 2 ) aö
1
ò
(4 ò 3 0) a z
ý ( ò 2 3z 2 ) aö 4 ò 2 a z
A
ñ÷ñ
ý0
1 ù sin ö ù
1 ù 1 cos ö
1 ù ö cos ö ) ù
ù
0 2 ú ar ú
0 ú añ ú (
0 ú aö
2
ú
r sin ñ û
r û
r û sin ñ r
r û ör
r
û
û
sin ö
cos ö
cos ö
ý 3
ar 3
añ 3 aö
r sin ñ
r sin ñ
r
ñ A ý
(c )
ñ÷ñ A ý
sin ö
sin ö
0 4
ý0
4
r sin ñ
r sin ñ
ñ÷ñ Aý 0
Prob. 3.42
ñ H ý 0a ò 1aö
1
ò
(2 ò cos ö ò cos ö )a z ý aö cos ö a z
ö 1
ö
1
1
1
ñ ñ H ý ÷ sin ö 0 ÷ a ò 0aö (1 0)a z ý - sin ö a ò a z
ò
ò
ò
ø ò
ø
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Prob. 3.43
Method 1: We can express A in spherical coordinates.
a
r
A ý 3 ar ý 2r ,
r
r
2
öa ö
ö1ö
ñ A ý ñ ÷ 2r ÷ ý ñ ÷ 2 ÷ ar ý 3 ar ar ý 0
r
ør ø
ør ø
Method 2:
x
y
z
a 3 a y 3 az
3 x
r
r
r
ö
ö
ö
3
öx öy öz ü 3
ü
ñ A ý
ý ý z ( x 2 y 2 z 2 ) 5/ 2 (2 y ) y ( x 2 y 2 z 2 ) 5 / 2 (2 z ) ý a x ...
2
x
y
z
þ 2
þ
3
3
3
r
r
r
ý0
Aý
Prob. 3.44
y
1
1
2
3
0
1
2
(a)
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x
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ð F ÷ d l ý (
) F ÷ d l
1
L
2
3
_
_
_
For 1, y ý x dy ý dx, dl ý dx a x dy a y ,
1
3
F ÷ d l ý x dx xdx ý
1
0
1
4
_
_
_
For 2, y ý x 2, dy ý dx, dl ý dx a x dy a y ,
2
3
2
F ÷ d l ý ( x 2 x x 2)dx ý
2
1
17
12
For 3,
0
2
F ÷ d l ý x ydx|
3
2
1
ð F ÷ d l ý 4
L
y ý0
ý0
17
7
0 ý
12
6
(b)
ñ F ý x2 a z ;
1 x
ý ( x2 )d xd y ý
2
x d yd x
S
(ñ F) ÷ d
d S ý d xd y( a z )
1
ý
x
2
x y| d x
0
0
0 0
2
2
x y|
1
x2
0
dxý
2
1
x2
x2 d yd x
y ý0
x1
|
4 0
2
x ( x 2)d x ý
2
1
7
6
(c) Yes
Prob. 3.45
1
ð A ÷ d l ý ò
ý2
ý
ò sin ö d ò
ð/2
ö ý0
ö
ò ò dö
2
ý0
2
0
ò sin ö d ò
ò 3dö
o
ò ý 1 ò ý1
ö ý 90 ö ýð / 2
ò ý2
ð
1
ð
(4 1) 8( ) ý 9.4956
2 2
2
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Prob. 3.46
y
45o
0
2
ð/4
2
ð F ÷ d l ý 2 ò zd ò
0
ý ò2
2
0
(6c osö )
ñxF ý
1
ð/4
0
z ý1
ò2
0
2
x
3 z sin ö ò d ö
0
0
ò ý 2, z ý 1
2 ò zd ò z ý 1
2
ý (4 0) 6( c osð / 4 1) (0 4) ý 1.757
ù3 z sin ö 0 ùû a z ...
òû
2
ð/4
(ñxF) d S ý ò ö
ý0
ý0
3z
ò
sin ö ò d ö d ò
zý
ý 3(2)( c osö )
ð/4
0
ý 6( c osð 1) ý 1.757
Prob. 3.47
ñ A ý 8 xe y 8 xe y ý 16 xe y
ñ(ñ A) ý 16e y a x 16 xe y a y
ax
ay
ö
ö
ñxñ(ñ A) ý
öx
öy
y
16 xe y
16e
Should be expected since ñxñV ý 0 .
Prob. 3.48
(a) ñV ý
az
ö
ý (16e y 16e y )a z ý 0
öz
0
sin ñ c o sö
c o sñ c osö
sin ö
ar
añ 2 a ö
2
2
r
r
r
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(b) ñxñV ý 0
(c)
1 ö
1
c o sñ c osö
1
sin ñ c o sö
ö
( sin ñ c osö ) 2
(sin ñ
) 2 2 (
)
2
r
r
r ör
r sin ñ öñ
r sin ñ
c o sö
c osö
(1 2 sin 2 ñ ) 3
ý0 3
r sin ñ
r sin ñ
2 sin ñ c osö
ý
r3
ñ ñ V ý ñ 2V ý
Prob.3.49
r
Qý
r sin ñ [(cos ö sin ö ) a x (cos ö sin ö ) a y ]
r sin ñ
ý r (cos ö sin ö ) a x r (cos ö sin ö ) a y
ù Qr ù ù sin ñ cos ö
ú ú ú
úQñ ú ý ú cos ñ cos ö
úQö ú úû sin ö
û û
sin ñ sin ö
cos ñ sin ö
cos ö
cos ñ ù
sin ñ úú
0 úû
ùQx ù
ú ú
úQ y ú
úû Qz úû
Q ý r sin ñ a r r cos ñ añ r a ö
(a)
dl ý ò dö aö ,
1
2
ò ý r sin 30ð ý 2( ) ý 1
z ý r cos 30ð ý 3
Qö ý r ý
ð Q ÷ dl ý
2ð
ò 2 z2
ò 2 z 2 ò dö ý 2(1)(2ð ) ý 4ð
0
(b)
ñ Q ý cot ñ a r 2 añ cos ñ aö
For S1 , dS ý r 2 sin ñ dñ dö a r
(ñ Q) ÷ dS ý r
2
sin ñ cot ñ dñ dö|
S1
2ð
30ð
0
0
ý 4 dö
r ý2
cos ñ dñ ý 4ð
(c)
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dS ý r sin ñ dñ dr añ
For S2 ,
(ñ Q) ÷ dS ý 2 r sin ñ dö dr|ñ
S2
2
2ð
0
0
ý 2sin 30 rdr
ý 30ð
dö
ý 4ð
(d)
Fo r S1, d S ý r 2 sin ñ d ö dñ a r
S1
Q ÷ d S ý r 3 sin 2 ñ dñ d ö|
2ð
ý 8 dö
30
0
ý 4ð [
ð
3
r ý2
ð
sin
2
ñ dñ
0
3
] ý 2.2767
2
(e)
_
Fo r S2 , d S ý r sin ñ d ö d r añ
S2
Q ÷ d S ý r 2 sin ñ c osñ d ö d |r
ý
(f)
ñ ý 30ð
4ð 3
ý 7.2552
3
ö
1 ö
r
(r 3 sin ñ )
(sin ñ cos ñ ) 0
2
r ör
r sin ñ ö ñ
ý 2sin ñ cos ñ cot ñ
ñ÷Q ý
ñ ÷ Q d v ý (2 sin ñ c o sñ c ot ñ )r
2
sin ñ dñ d ö d r
30
r3 2
ý
(2ð ) (1 sin 2 ñ )dñ
3 0
0
ý
4ð
3
(ð
) ý 9.532
3
2
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ñ ÷ Q d v ý (
Che c k :
S1
ý 4ð [
ý
ð
3
Q ÷d S
S2
3
3
]
2
3
4ð
3
[ð
]
3
2
(It c hec ks!)
Prob. 3.50
Sinc e u ý r,
ñ u ý ñ ( r). Fro m Ap p end ix A.10,
ñ ( A B) ý A(ñ ÷ B) B(ñ ÷ A) (B ÷ ñ) A ( A ÷ ñ) B
ñ u ý ñ ( r )
_
ñ ( r ) ý (ñ ÷ r ) r (ñ ÷ ) (r ÷ ñ) ( ÷ ñ) r
ý (3) ý 2
or ý
1
ñ u.
2
Alternatively, let
uý
x ý r cos t ,
y ý r sin t
öx
öy
ax
ay
öt
öt
ý r sin t a x r cos t a y
ý y ax x a y
ö
ö
ñu ý ö x ö y
y x
i.e., ý
ö
ö z ý 2 a z ý 2
0
1
ñu
2
Note that we have used the fact that ñ ÷ ý 0,
( r ÷ ñ) ý 0,
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( ÷ ñ)r ý
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Prob. 3.51
(a) ñ ÷ H ý
öH x öH y öH z
ý 2 z 5x 8
öx
öy
öz
ö
ö
ö
öz
(b) ñ H ý öx öy
ý 8a x 2 xa y 5 ya z
2 xy 5 xy 8( y z )
Prob. 3.52
1 ö 2
1
1 öBö
ö
(r Br )
( Bñ sin ñ )
2
r ör
r sin ñ öñ
r sin ñ öö
ö
1 ö
1
ý 2 (r 4 )
(4 rsin ñ cos 2ñ ) 0
r ör
r sin ñ öñ
1
ý 4r
4r cos ñ cos 2ñ 4r sin ñ (2sin 2ñ )ý
sin ñ
ñ ÷ B ý 4r 4r cot ñ cos 2ñ 8r sin 2ñ
ñ÷B ý
öBö ù
ù ö
ù
1 ù 1 öBr ö
1ù ö
öBr ù
ú öñ ( Bö sin ñ ) öö ú ar r ú sin ñ öö ör (rBö ) ú añ r ú ör (rBñ ) öñ ú aö
û
û
û
û
û
û
All terms are zero except one.
ñ B =
1
r sin ñ
1ù ö
1 ö
ù
(4r 2 cos 2ñ )aö ý 8cos 2ñ aö
ñ B = 0 ar 0añ ú (rBñ ) 0 ú aö ý
r û ör
r ör
û
Prob. 3.53
(a)
ö öV
öV
öV ö
ax V
ay V
az ÷
ñð(VñV)=ñð÷ V
öy
öz ø
ø öx
ý
ö ö öV
÷V
öx ø öx
ö ö ö öV
÷ ÷V
ø öy ø öy
ö ö ö öV ö
÷ ÷V
÷
ø öz ø öz ø
2
2
ö 2V
ö 2V
ö 2V ö öV ö ö öV ö ö öV ö
ý V 2 V 2 V 2 ÷
÷ ÷
÷ ÷
÷
öx
öy
öz
ø öx ø ø öy ø ø öz ø
ý V ñ 2V | ñV |2
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(b)
ö
ö
ö
ñ VA ý öx öy öz
VAx VAy VAz
ùö
ù
ùö
ù
ö
ö
ö
ùö
ù
ý ú (VAz ) (VAy ) ú a x ú (VAx ) (VAz ) ú a y ú (VAy ) (VAx ) ú a z
öz
öx
öy
û öz
û
û öy
û
û öx
û
öAy ù
ù öV
öA
öV
ý ú Az
V z Ay
V
ú ax
ö
ö
ö
ö
y
y
z
z
û
û
öA
öV
öA ù
ù öV
ú Ax
V x Az
V z ú ay
öz
öx
öx û
û öz
öA
ù öV
öA ù
öV
ú Ay
V y Ax
V x ú az
öx
öy
öy û
û öx
ùö öA öA ö
ö öA öA ö ù
ö öA öA ö
ñ VA ý V ú÷ z y ÷ a x ÷ x z ÷ a y ÷ y x ÷ a z ú
öz ø
öx ø
öy ø û
ø öz
ø öx
ûø öy
ö öV
ö öV
öV ö
öV ö
öV ö
ö öV
÷ Az
Ay
Az
Ax
÷ a x ÷ Ax
÷ az
÷ a y ÷ Ay
öz ø
öz
öx ø
öx
öy ø
ø
ø öy
ø
ý V ñ A ñV A
Prob. 3.54
(a)
ñðB ý
öBx öBy öBz
ý 2 xy 1 1 ý 2 2 xy
öx
öy
öz
(b)
ö
ö
ö
öy
öz ý (1 0)a x (0 0)a y (4 x x 2 )a z
ñ B ý öx
x 2 y (2 x 2 y ) ( z y )
ý a x x(4 x)a z
(c )
ñ(ñðB) ý ñ(2 2 xy ) ý 2 ya x 2 xa y
(d)
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ö
ö
ñ ñ B ý öx öy
1 0
ö
öz
ý 0a x - (4 - 2x)a y 0a z
2
(4 x x )
ý 2( x 2)a y
Prob. 3.55
(a)
V1 ý x 3 y 3 z 3
ö 2V1 ö 2V1 ö 2V1
ñ V1 ý
öx 2 öy 2 öz 2
ö
ö
ö
( 3x 2 )
3 y 2 ù (3z 2 )
ý
ø
öx
öy
öx
2
ý 6 x 6 y 6z ý 6( x y z)
(b)
V2 ý ò z2 sin 2ö
1 ö
ñ 2V2 ý
ý
ò öò
2
z
ý(
ò
( ò z2 sin 2ö )
sin 2ö
3 z2
ò
2
4z
ò
4 z2
ò
sin 2ö
ö
(2 ò z sin 2ö )
öz
sin 2ö 2 ò sin 2ö
2 ò )sin 2ö
(c)
V3 ý r 2(1 c o sñ sin ö )
ñ 2V3 ý
1 ö
[2 r 3(1 c o sñ sin ö )]
r2 ö r
1
ö
1
( sin 2 ñ sin ö )r 2 2 2 r 2 ( cosñ sin ö )
r sinñ öñ
r sin ñ
2 sinñ
cosñ sin ö
ý 6(1 cosñ sin ö )
cosñ sin ö
sinñ
sin 2 ñ
cosñ sin ö
ý 6 4 cosñ sin ö
sin 2 ñ
2
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Prob. 3.56
(a)
U ý x3 y 2 e xz
ö
ö
ö 4 2 xz
(3 x2 y 2 e xz x3 y 2 ze xz )
(2 x3 y e xz )
(x y e )
öx
öy
öz
ñ2U ý
ý 6 xy 2 e xz 3 x2 yze xz 3 x2 y 2 ze xz x3 y 2 z2 e xz 2 x3 e xz x5 y 2 e xz
ý e xz(6 xy 2 3 x2 y 2 z 3 x2 y 2 z x3 y 2 z2 2 x3 x5 y 2 )
At (1, 1,1),
ñ 2U ý e1 (6 3 3 1 2 1) ý 16e ý 43.493
(b)
V ý ò 2 z (cos ö sin ö )
1 ö
ñ 2V ý
[2 ò 2 z (cos ö sin ö )] z (cos ö sin ö ) 0
ò öò
ý 4 z (cos ö sin ö ) z (cos ö sin ö )
ý 3z (cos ö sin ö )
ð
At (5,
6
, 2), ñ 2V ý 6(0.866 0.5)ý 8.196
(c )
W ý e r sin ñ cos ö
ñ 2W ý
ö
1 ö
e r
2 r
ñ
ö
(
sin
cos
)
cos ö
(sin ñ cos ñ )
r
e
2
2
öñ
r ör
r sin ñ
e r sin ñ cos ö
r 2 sin 2 ñ
1
(2re r sin ñ cos ö ) e r sin ñ cos ö
r2
e r cos ö
e r cos ö
2
(1 2sin 2 ñ ) 2
r sin ñ
r sin ñ
2 2
ñ 2W ý e r sin ñ cos ö (1 2 )
r r
ý
At
(1, 60ð,30ð),
ñ W ý e 1 sin 60 cos 30(1 2 2) ý 2.25e 1 ý 0.8277
2
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Prob. 3.57
(a ) Let V ý 1nr ý 1n x2 y 2 z2
öV 1 1
x
ý
2 x ù ( x2 y 2 z2 )1/ 2 ý 2
ø
öx r 2
r
ñV ý
öV
öV
ax
ox
oy
ay
öV
oz
az ý
x a x y ay z a z
r
2
ý
r
r2
r 1
ý ar in sp heric a l c o o rd ina tes.
r2 r
1 ö 2
1 ö
( r A r ) ý 2 ( r)
ñ 2(1nr) ý ñ ðñ (1nr) ý ñ ðA ý 2
ör
r
r ör
1
ý 2
r
(b ) Le t ñV ý A ý
Prob. 3.58
(a)
(b)
ñV ý
ñU ý
öU
öU
öU
öU
ax
ay
a z ý y 2 z 3a x 2 xyz 3a y 3 xy 2 z 2
az
öx
öy
öz
öz
ñ 2U ý
ö ö öU
÷
öx ø öx
ö ö ö öU
÷ ÷
ø öy ø öy
ö ö ö öU
÷ ÷
ø öz ø öz
ö
3
2
3
2
÷ ý 0 2 xz 6 xy z ý 2 xz 6 xy z
ø
1 öV
sin ö
cos ö
1
öV
öV
aò
aö
a z ý 2 a ò 2 aö 0 ý 2 ùû sin ö a ò cos ö aö ùû
öò
öz
ò öö
ò
ò
ò
ñ 2V ý
ý
1 ö ö öV
ò
ò öò ÷ø öò
sin ö
ò
3
1
ò3
ö 1 ö 2V ö 2V 1 ö
1 ö sin ö ö
2 ý
ò 1 sin ñ ù 2 ÷
0
ø
÷ 2
2
öz
ò öò
ò ø ò ÷ø
ø ò öö
sin ö ý 0
öW
1 öW
1 öW
ar
añ
aö
ör
r öñ
r sin ñ öö
ý 2r sin ñ cos ö ar r cos ñ cos ö añ r sin ö aö
ñW ý
( c)
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1 ö ö 2 öW ö
1
1
ö ö
öW ö
ö 2W
ñ W ý 2 ÷r
÷
÷ sin ñ
÷
r ör ø ör ø r 2 sin ñ öñ ø
öñ ø r 2 sin 2 ñ öö 2
1 ö
1
1
ö 2
r sin ñ cos ñ cos ö ù 2 2 (r 2 sin ñ cos ö )
ý 2 ø 2r 3 sin ñ cos ö ù 2
ø
r ör
r sin ñ öñ
r sin ñ
1
cos ö
(cos 2 ñ cos ö sin 2 ñ cos ö )
ý 6sin ñ cos ö
sin ñ
sin ñ
cos ö
cos ö
ý 6sin ñ cos ö
2sin ñ cos ö
ý 4sin ñ cos ö
sin ñ
sin ñ
2
Prob. 3.59
ñV ý
1 öV
öV
öV
aò
aö
a z ý 2 ò z cos ö a ò - ò zsinö aö ò 2 cos ö a z
öò
öz
ò öö
ö 1 ö 2V ö 2V 1 ö
1
2 ò 2 z cos ö ù 2 ò 2 z cos ö 0
2 ý
ø
÷ 2
2
öz
ò öò
ò
ø ò öö
ý ø 4 1ù z cos ö ý 3z cos ö
ñ 2V ý
1 ö ö öV
ò
ò öò ÷ø öò
Prob.3.60
1 öV
1 öV
öV
ar
añ
aö
r öñ
r sin ñ öö
ör
10
5sin ö
ý 3 cos ö ar 3
aö
r
r sin ñ
ñV ý
(a)
ö
öV
ö 2V
1 ö 2 öV
1
1
ñðñV ý ñ V ý 2
(r
) 2
(sin ñ
) 2 2
r ör
r sin ñ öñ
r sin ñ öö 2
ör
öñ
1 ö 2 10 cos ö
1
5cos ö
(b)
r (
ý 2
) 0 2 2 (
)
3
r ör
r
r sin ñ
r2
10 cos ö 5cos ö
ñðñV ý
4 2
r4
r sin ñ
2
(c )
ñ ñV ý 0, see Example 3.10.
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Prob. 3.61
öU
öU
öU
ñU ý
ax
ay
a z ý 4 yz 2 a x (4 xz 2 10 z )a y (8 xyz 10 y )a z
öx
öy
öz
ö
ö
ö
ñðñU ý (ñU x ) (ñU y ) (ñU z ) ý 0 0 8 xy ý 8 xy
öx
öy
öz
ñ 2U ý
ö 2U ö 2U ö 2U
ý 0 0 8 xy ý 8 xy
öx 2 öy 2 öz 2
Hence, ñ 2U ý ñðñU
Prob. 3.62
Method 1
ñ 2G
ò
ý ñ 2Gò
2 öGö Gò
ò 2 öö ò 2
1 ö
2 ò sin ö
8 ò sin ö 2 ò sin ö
0
ø 2 ò sin ö ù
2
ò öò
ò
ò2
ò2
2sin ö 2sin ö 8sin ö 2sin ö
6sin ö
ý
ý
ý
ò
ñ 2G
ö
ò
ý ñ 2Gö
ò
ò
ò
2 öGò Gö
ò 2 öö ò 2
1 ö
1
4 ò cos ö 4 ò cos ö
(4 ò cos ö ) 4 ò cos ö 0
ò öò
ò
ò2
ò2
4 cos ö 4 cos ö 4 cos ö 4 cos ö
ý
ý0
ý
ò
ñ 2G
z
ò
ý ñ 2 Gz ý
ý
1
ò
ò
ò
1 ö
ö
ùû ò ( z 2 1) ùû 0 (2 z ò )
ò öò
öz
( z 2 1) 2 ò
Adding the components together gives
ñ 2G ý
6sin ö
ò
ù
ù
1
a ò ú 2 ò ( z 2 1) ú a z
ò
û
û
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Method 2:
ñ 2G ý ñ(ñðG ) ñ (ñ G )
1 ö
1
Let V ý ñðG =
(2 ò 2 sin ö ) (4 ò sin ö ) 2 z ò ý 2 z ò
ò öò
ò
ñ(ñðG ) ý ñV ý 2 za ò 2 ò a z
ù1
ù
ù
1ùö
ý ñ G ý ú 0 0 ú a ò ûù0 ( z 2 1) ûù aö ú (4 ò 2 cos ö ) 2 ò cos ö ú a z
ò û öò
ûò
û
û
2
ý ( z 1)aö 6 cos ö a z
A
Let
ù 6
ù
ù
1ùö
ñ ñ G ý ñ A ý ú sin ö 2 z ú a ò (0 0)aö ú ( ò ( z 2 1)) 0 ú a z
ò û öò
û ò
û
û
ù
ù
6
1
ý ú 2 z sin ö ú a ò ( z 2 1)a z
ò
ò
û
û
ñ 2G ý ñ V ñ A
ù
ù
6
1
ý 2 za ò 2 ò a z ú 2 z sin ö ú a ò ( z 2 1)a z
ò
ò
û
û
ý
ù
ù
1
sin ö a ò ú 2 ò ( z 2 1) ú a z
ò
ò
û
û
6
Prob. 3.63
ö
ö
ö
( xz ) ( z 2 ) ( yz ) ý z y
öx
öy
öz
ñ(ñðA) = a y a z
ñðA ý
ñ 2 A ý ñ 2 Ax a x ñ 2 Ay a y ñ 2 Az a z ý 0 2a y 0 ý 2a y
ñ(ñðA) - ñ 2 A ý a y a z
ö
ñ A ý öx
xz
ö
öy
z2
ö
ñ ñ A ý öx
z
(1)
ö
öz ý za x xa y
yz
ö
öy
x
ö
öz ý a y a z
0
(2)
From (1) and (2),
ñ ñ A = ñ( ñðA) - ñ 2 A
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Prob. 3.64
ñðA ý
öAx öAy öAz
ý 111 ý 3 ù 0
öx
öy
öz
ö
ñ A = öx
x
ñðB ý
ö
öy
y
ö
öz ý 0
z
1 ö
1 öBö öBz
ý 4 cos ö 4 cos ö ý 0
( ò Bò )
ò öò
ò öö öz
öB ù
ù 1 öBz öBö ù
ù öB öB ù
1ùö
ñ B = ú
a ò ú ò z ú aö ú ( ò Bò ) ò ú a z
ú
öz û
öò û
öö û
ò û öò
û ò öö
û öz
1
ý 0aò 0aö 8 ò sin ö 2 ò sin ö ý a z ý 6sin ö a z ù 0
ò
1 ö 2
2sin ñ
ù0
(r sin ñ ) 0 0 ý
2
r ör
r
ö 2
1 ù ö
1ù
ù
ù
2
ñC =
ñ
(
sin
)
0
0
(r sin ñ ) ú añ
a
r
r
ú
ú
ú
r sin ñ û öñ
r û ör
û
û
1
0 cos ñ ý aö
r
cosñ
ý 2 cos ñ ar - 2sinñ añ aö ù 0
r
(a) B is solenoidal.
(b) A is irrotational.
ñðC ý
Prob. 3.65 (a)
ax
ay
ö
ö
ñG ý
öx
öy
16 xy z 8 x 2
az
ö
öz
x
ý 0 a x (1 1) a y (16 x 16 x) a z ý 0
Thus, G is irrotational.
(b) Assume that represents the net flux.
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ýð
G ÷ dS ý ñ ÷ Gdv
ñ ÷ G ý 16 y 0 0 ý 16 y
1
1
1
0
0
0
ý 16 ydxdydz ý 16 dx dz ydy ý 16(1)(1)(
(c)
y2 1
)ý8
2 0
y
1
0
ð G ÷ dl
ý
x ý0
L
y ý1
x ý1
(16 xy z )dx|
1
y ý0
z ý0
ý 0 8(1) y| 16(1)
0
1
8x dy|
2
y ý0
x ý1
z ý0
x
(16 xy z)dx|
x ý1
x2 0
| 0
2 1
ý 88 ý 0
This is expected since G is irrotational, i.e.
ð G ÷ dl ý (ñ G ) ÷ dS ý
0
ñ ÷ T ý 6 0 ý 6
Prob. 3.66
1 ö
1 ö ö ü
( ò Eò ) ý
ò öò
ò öò ÷ø 2ðõ
Hence, E is solenoidal.
öE
öE
ñ E ý ò aö ò a z ý 0
öz
öö
showing that E is conservative.
ñ÷ E ý
y ý0
x ý0
ö
÷ý0
ø
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y ý1
z ý0
8 x dy|
2
y ý1
x ý0
z ý0
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Prob. 3.67
ñ H ý 0
ð H ðdl ý (ñ H)ðdS ý 0
L
S
Prob. 3.68
ö
öx
ñ B ý
2
3x z y 2
ö
öy
2 xy
ö
öz ý (0 0)a x (3x 2 3x 2 )a y (2 y 2 y )a z ý 0
x3
showing that B is conservative.
Prob. 3.69
ñ÷ D ý
1
( ò 0)
1
(0) 0 ý 0
ò
ò
We conclude that D is solenoidal.
Prob. 3.70
1
1ö 1
1ù ö
2k ( sin ñ ) ù
ö
(0 0)ar ÷
0 0 ÷ añ ú (kr 2 sin ñ )
úû aö
r sin ñ
r ø sin ñ
r û ör
r3
ø
1 ù 2k sin ñ 2k sin ñ ù
ý 0 ú
aö ý 0
rû
r3
r 3 úû
showing that E is conservative.
ñ E ý
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CHAPTER 4
P. E. 4.1
ù 5 109 [(1, 3, 7) (2, 0, 4)]
ù
ú
ú
3
[(1, 3, 7) (2, 0, 4)]
9
ú
ú
110
9
ú
ú
(a) F ý
(
2
10
)[(1,
3,
7)
(
3,
0,5)]
ö 109 ö ú
ú
3
4ð ÷
[(1, 3, 7) (3, 0,5)] ú
÷ú
ø 36ð ø ú
úû
û
ý[
45(1, 3,3) 18(4, 3, 2)
] nN
193/ 2
293/ 2
ý 1.004a x 1.284a y 1.4 a z nN
(b)
Eý
F
ý 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
ñ
T
A
F1
d/2
F2
mg
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At point A,
T sin ñ cos 30ð ý F1 F2 cos 60ð
q2
q2
1
ý
2 ( )
2
4ð õ 0 d
4ð õ 0 d 2
3 q2
ý
8ð õ 0d 2
T cos ñ ý mg
Hence,
sin ñ ý
But
Thus,
or
tan ñ cos 30ð ý
q2 ý
h
d
tan ñ ý
ý
l
3l
d
3
l2
d2
3
d
3
( )
3q 2
3 2 ý
8ðõ 0 d 2 mg
d2
2
l
3
4ð õ0 d3 mg
3 l2
d2
3
Q
3
12 ð õ 0 d 3 m g
but q ý
Q2 ý
3q 2
8ðõ 0 d 2 mg
q2 ý
Q
. Hence,
9
d2
l
3
2
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P.E. 4.3
_
d2 l
eE ý m 2
dt
_
_
d 2 x _ d 2 y _ d 2z _
ax 2 a y 2 az )
dt 2
dt
dt
E0 ý 200 kV / m
_
eE0 ( 2a x a y ) ý m(
where
d 2z
ý0
z ý ct c2
dt 2
2eE0 t 2
d2x
c3 t c4
m 2 ý 2eE0
xý
2m
dt
d2y
eE0 t 2
c5 t c6
m 2 ý eE0
yý
2m
dt
At t ý 0, ( x, y, z ) ý (0, 0, 0) c1 ý 0 ý c4 ý c6
dx dy dz
, , ) ý (0, 0, 0)
dt dt dt
c1 ý 0 ý c3 ý c5
At t ý 0
Also, (
Hence,
( x, y ) ý
eE0t 2
(2,1)
2m
i.e. 2 | y | ý | x |
Thus the largest value of is
80 cm ý 0.8 m
P.E. 4.4
(a)
Consider an element of area dS of the disk.
The contribution due to dS ý ò dö d ò is
dE ý
ò s dS
ò s dS
ý
2
4ðõ 0 r
4ðõ 0 ( ò 2 h 2 )
The sum of the contribution along ò gives zero.
òs
Ez ý
4ð õ 0
a
ò
ý0
2ð
hò
h ò d ò dö
ö ý0 ( ò 2 h2 )3/ 2 ý 2 õ 0s
a
ò
0
ò dò
( ò h 2 )3/ 2
2
a
h òs
hòs
2
2 3/ 2
2
2
2 1/ 2
(
)
(
)
(
2(
)
|
ý
ý
h
d
h
ò
ò
ò
4 õ 0 0
4õ 0
0
a
ý
òs
h
[1 2
]
2õ 0
(h a 2 )1/ 2
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(b)
As a
Eý
,
òs
az
2õ 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.
ù
ù
ú
ú
òs ú
1
ú ý òs
1
Ez ý
2õ o ú
a 2 ú 2õ o
ú
1 2 ú
h û
û
ò ù ða 2 ù
Q
=
ý s ú
2 ú
2õ o û 2ðh û 4ðõ o h 2
ù
2
ú1 ö÷1 a
ú ÷ø h 2
úû
1
ù
ö 2ú
òs ù a2 ù
÷÷
a
ú
ú
0 , but ò sða 2 ý Q
2õ o û 2h 2 û
ø úú
û
This is in keeping with original Coulomb’s law.
P. E. 4.5
QS ý
ò
2
S
dS ý
2
12 | y| dx dy
2 2
2
ý 12(4) 2 y dy ý 192 mC
0
Eý
ò s dS
ò dS | r r' |
a ý s
2 r
4ðõ r
4ðõ o | r r' |3
where r r' ý (0, 0,10) ( x, y, z ) ý ( x, y,10).
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Eý
2
x ý2
12 | y |103 ( x, y,10)
109 2 2
y ý2
4ð (
)( x y 100)3/ 2
36ð
2
2
ý 108(106 )[ | y |
2
10 a z
2
xdx dy a x
( x 2 y 2 100)3/ 2 2
2
2
2
2
(x
2 2
E ý 108(107 ) a z
y | y | dy dx a y
2
y 2 100)3/ 2
(x
2
| y | dx dy
]
y 2 100)3/ 2
1
d ( y2 )
[2 2 2 2
]dx
( x y 100)3/ 2
0
2
2
2
ý 216(107 ) a z
2
2
2
[
2
1
1
2
]dx
1/ 2
( x 104)
( x 100)1/ 2
ý 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
ý 216 (107 )a z (7.6202 (103 ) )
ý 216 (107 )a z (ln(
E ý 16.46 a z MV/m
P.E. 4.6
y = -3 plane
z
line
charge
O
y
P
x
x=2 plane
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E1 and E2 remain the same as in Example 4.6.
E3 ý
òL
aò
2ðõ 0 ò
This expression, which represents the field due to a line charge,
is modified as follows. To get a ò , consider the z ý 1 plane. ò ý 2
a ò ý a x cos 45ð a y sin 45ð
1
(a x a y )
2
10(109 ) 1
E3 ý
(a x a y )
109 2
2ð (
)
36ð
ý
ý 90 ð (a x a y ).
Hence,
E ý E 1 E 2 E3
ý 180 ð a x 270 ð a y 90 ð a x 90 ð a y
ý 282.7 a x 565.5a y V/m
P.E. 4.7
ò
Q
a s an
2 r
4ð r
2
[(0, 4,3) (0, 0, 0)] 10 109
ay
5
2
D ý DQ Dò ý
30 109
ý
4ð (5) 2
30
(0, 4,3) 5 a y nC / m 2
ý
500ð
ý 5.076a y 0.0573a z nC / m 2
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P.E. 4.8
(a) òv ý ñ ÷ D ý 4 x
òv (1, 0,3) ý 4 C/m3
(b) = ð
D ÷ dS ý D ÷ dS
xý0
D ÷ dS
x ý1
D ÷ dS
yý0
D ÷ dS
1 1
1 1
1 1
1 1
1 1
0 0
0 0
0 0
0 0
0 0
y ý1
D ÷ dS
ý (2 y 2 z )dydz (2 y 2 z )dydz 4 x(1)dxdz (x)dxdz (x)dxdz
1 1
ý 4 x dxdz ý 4(1/ 2)(1) ý 2C
0 0
1 1 1
(c) ý Q ý ò v dv ý 4 xdxdydz
0 0 0
ý 4(1)(1)(1/ 2) ý 2 C
Qýý2C
P.E. 4.9
Q ý ò vdv ý ý
ð D ÷ dS
For 0 r 10,
Dr ( 4ð r 2 ) ý
2r
(r 2 ) sin ñ dñ dr dö
2r 4 r
| ) ý 2ð r 4
4 0
r2
Eý
ar nV/m
2õ 0
Dr (4ð r 2 ) ý 4ð (
Dr ý
r2
2
E (r ý 2) ý
For r ó 10,
4(109 )
ar ý 72ð ar ý 226 ar V/m
9
2( 1036ð )
Dr (4ð r 2 ) ý 2ð r0 4 ,
4
r
Dr ý 0 2
2r
E (r ý 12) ý
r0 ý 10m
r0 4
Eý
ar nV/m
2õ 0 r 2
104 (109 )
ar ý 1250ð ar
109
2(
)(144)
36ð
ý 3.927ar kV/m
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P. E. 4.10
V (r ) ý õ k ý1
3
Qk
C
4ðõ 0 | r rk |
At V () ý 0,
C ý0
| r r1 | ý| (1,5, 2) (2, 1,3) |ý 46
| r r2 |ý| (1,5, 2) (0, 4, 2) |ý 18
| r r3 |ý| (1,5, 2) (0, 0, 0) |ý 30
4
106
5
3
[
]
V (1,5, 2) ý
9
10
46
18
30
4ð (
)
36ð
ý 10.23 kV
P.E. 4.11
Vý
Q
4ðõ 0 r
C
If V (0, 6, 8) ý V (r ý 10) ý 2;
5(109 )
C
109
4ð (
)(10)
36ð
2ý
C ý 2.5
(a)
5(109 )
VA ý
109
4ð (
)|( 3,2,6) (0,0,0)|
36ð
ý 3.929 V
2.5
(b)
VB ý
(c )
45
7 12 52
2
2.5 ý 2.696 V
V AB ý VB VA ý 2.696 3.929 ý 1233
.
V
P.E. 4.12
(a)
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W
ý E ÷ dl ý (3x 2 y )dx xdy
Q
ý
2
(3x
2
y )dx |
y ý5
0
1
x dy
5
|
xý2
ý 18 12 ý 6 kV
W ý 6 Q ý 12 mJ
(b)
dy ý 3 dx
2
W
ý E ÷ dl ý (3x 2 5 3 x)dx x(3)dx
Q
0
2
= (3x 2 6 x 5)dx ý 8 12 10 ý 6
0
W ý 12 mJ
P.E. 4.13
(a)
(0, 0,10)
Vý
100 cos 0
(1012 ) ý
2
4ðõ 0 (10 )
(r ý 10, ñ ý 0, ö ý 0)
1012
ý 9 mV
109
4ð (
)
36ð
100(1012 )
[2 cos 0 ar sin 0 añ ]
109 3
4ð (
)10
36ð
ý1.8 ar mV/m
Eý
(b)
At (1,
ð ð
, ),
3 2
100 cos
ð
(1012 )
3
ý 0.45 V
109
2
4ð (
)(1)
36ð
ð
ð
100 (1012 )
Eý
(2 cos ar sin añ )
9
10
3
3
4ð (
)(1) 2
36ð
Vý
ý 0.9 ar 0.7794añ V/m
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P.E. 4.14
After Q1 , W1 ý 0
After Q2 , W2 ý Q2V21 ý
ý
Q2 Q1
4 ð õ 0 |(1,0 ,0 ) (0 ,0 ,0 )|
1( 2) (10 18 )
ý 18 nJ
1
9
4 ð (10 )
36 ð
After Q3 ,
W3 ý Q3 (V31 V32 ) Q2V21
ü
2
1
ý 3(9 )(10 9 ) ý
þ |(0 ,0 , 1) (0 ,0 ,0 )| |(0 ,0 , 1) (1,0 ,0 )|
2
ý 27 (1
) 18
2
ý 29.18 nJ
18 nJ
After Q4 ,
W4 ý Q4 (V41 V42 V43 ) Q3 (V31 V32 ) Q2V21
ü
ü
1
3
2
ý 4(9 )(10 9 ) ý
ý W3
þ |(0 ,0 ,1) (0 ,0 ,0 )| |(0 ,0 ,1) (1,0 ,0 ) |(0 ,0 ,1) (0 ,0 , 1) þ
2 3
ý 36 (1
) W3
2 2
ý 39.09 29.18 nJ ý 68.27 nJ
P.E. 4.15
E ý ñV ý ( y 1)a x (1 x)a y 2a z
At (1,2,3), E ý 3a x 2a z V/m
1 1 1
1
1
W ý õ o E ÷ Edv ý õ o ( x 2 y 2 2 x 2 y 6)dxdydz
2
2 1 1 1
1
1
1
1
ù
1 ù
ý õ o ú x 2 dx dydz y 2 dy dxdz 2 xdx dydz 2 ydy dxdz 6(2)(2)(2) ú
2 û 1
1
1
1
û
ù 80õ o
1 ù x3 1
(2)(2) 0 0 6(8) ú ý
ý õ o ú2
2 û 3 1
3
û
ý 0.2358 nJ
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Prob. 4.1
FQ1 ý
Q1Q2 (rQ1 r Q2 )
4ð õ rQ1 rQ2
3
(7, 2, 5)
20(1012 )[(3, 2,1) (4, 0, 6)]
ý
ý 180
103
9
10
3
688.88
4ð
(3, 2,1) (4, 0, 6)
36ð
ý 1.8291a x 0.5226 a y 1.3065 a z mN
Prob. 4.2 (a)
E ( 5, 0, 6) ý
Q2 [(5, 0, 6) (2, 0,1)]
[(5, 0, 6) (4, 0, 3)]
3
4 ð õ 0 | (5, 0, 6) (4, 0, 3) |
4 ð õ 0 | (5, 0, 6) (2, 0,1) |3
Q1
Q1 (1, 0,9)
Q2 (3, 0,5)
4 ð õ 0 ( 82)3 4 ð õ 0 (34)3/ 2
If Ez ý 0, then
ý
9 Q1
5 Q2
1
1
ý 0
3/ 2
4 ð õ 0 (82)
4 ð õ 0 (34)3/ 2
5
82
5 82
Q1 ý Q2 ( )3/ 2 ý 4 ( )3/ 2 nC
9
34
9 34
ý 8.3232 nC
(b)
F (5, 0, 6) ý qE (5, 0, 6)
If Fx ý 0, then
qQ1
3qQ2
ý0
3/ 2
4ðõ 0 (82)
4ðõ 0 (34)3/ 2
82 3/ 2
82
) ý 12( )3/ 2 nC
34
34
Q1 ý 44.945 nC
Q1 ý 3Q2 (
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Prob. 4.3
Q (0, 0, 0) (a, 0, 0) ý
Q (0, 0, 0) ( a, 0, 0) ý
Q(r rk' )
ý
' 3
3
4ðõ o | (0, 0, 0) (a, 0, 0) | 4ðõ o | (0, 0, 0) ( a, 0, 0) |3
k ý1 4ðõ o | r rk |
2
E ýõ
(a)
ý
(b) E ý
ý
Q( a, 0, 0) Q(a, 0, 0)
Q
ax
ý
3
3
4ðõ o a
4ðõ o a
2ðõ o a 2
Q (0, a, 0) (a, 0, 0) ý
Q (0, a, 0) ( a, 0, 0) ý
3
4ðõ o | (0, a, 0) (a, 0, 0) | 4ðõ o | (0, a, 0) ( a, 0, 0) |3
Q
Q( a, a, 0)
Q(a, a, 0)
ý
ax
2 3/ 2
2 3/ 2
4ðõ o (2a )
4ðõ o (2a )
4 2ðõ o a 2
(c )
Eý
Q (a, 0, a ) (a, 0, 0) ý
4ðõ o | (a, 0, a) (a, 0, 0) |
ý
3
Q (a, 0, a ) (a, 0, 0) ý
4ðõ o | (a, 0, a) ( a, 0, 0) |3
Q
Q(0, 0, a ) Q(2a, 0, a )
Q
a
ý
x
3
2 3/ 2
4ðõ o a
4ðõ o (5a )
4ðõ o a 2
10 5ðõ o a 2
Prob. 4.4
F ý qE ý mg
Eý
mg 2 9.8
ý
ý 4.9 kV/m
4 103
q
Prob. 4.5
5
(a)
5
Q ý ò L dl ý 12 x 2 dx ý 4 x3 | mC ý 0.5 C
0
0
(b)
Qý
ò S dS ý
4
2ð
ò z 2ò dö dz | ý 9(2ð )
zý0 ö ý0
òý 3
z3 4
| nC
30
ý 1.206 ý C
(c)
10
r 2 sin ñ dñ dö dr
r sin ñ
2ð
4
ð
42
ý 10 dö dñ rdr ý 10(2ð ) (ð )
2
0
0
0
Q ý òV dV ý
ý 1579.1 C
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1 ù
ù
ú1
ú az
û 5 5û
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Prob. 4.6
ö x 2 a ö a3 òo
dxdydz ý (a)(a) ò o ÷
÷ý
0
2
2
a
a
ø
ø
òo x
a a a
Q ý ò v dv ý
v
0 0 9
Prob. 4.7
Q ý ò v dv ý
v
2
1
ö ð
ò
5 ò 2 z òdö d ò dz mC
ý0 z ý0 ý / 6
ò 4 2 z2 1
=5
ð /2
4 0 2 0
ö
ð /2 5
10ð
ý (16)(1)(ð / 2 ð / 6) ý
ð /6 8
3
Q ý 10.472 mC
Prob. 4.8
Q ý ò s dS ý 6 xydxdy
ý
2
x
x ý0
y ý0
6 xydxdy
4
x4
xý2
y ý2
6 xydxdy
4
y2 x
y2 x 4
ý6 x
dx 6x
dx
0
2 0
2
x ý0
x ý2
2
2
ý6
4
x(
x ý0
x2
0)dx 3 x ùû(4 x) 2 0 ùû dx
2
xý2
2
4
x3
ý 6 dx 3 (16 x 8 x 2 x 3 )dx
2
0
2
x4 2
8 x3 x 4 4
2
ý3
6(8 x
)
4 0
3
4 2
ý 12 3(128 32
ý 12 3(96
512 64
64 4)
3
3
448
60)
3
Q ý 32 C
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Prob. 4.9
y
y=3x/2, z=4 plane
3
0
2
Q ý ò s dS ý 10 x 2 yzdxdy
S
2
x
zý4
mC ý 10(4)
2 3x/2
x 2 ydydx
x ý0 y ý0
2
y 3x / 2
x5 2
ö9 ö
ý 9(32) ý 288 mC
dx ý 20 x 2 ÷ x 2 ÷dx ý 20(9 / 4)
0
2 0
4
5
ø
ø
x ý0
2
ý 40 x 2
x ý0
Prob. 4.10
Q ý ò v dv ý 4 ò 2 z cos öò d ò dö dz
nC
v
2
1
ð /4
0
0
0
ý 4 ò 3 d ò zdz
cos ödö ý ò 4
2 z2 1
0 2 0
(sin ö )
ð /4
0
ý (16)(0.5)(sin ð / 4) ý 5.657 nC
Prob. 4.11
1
12 x 3 1
(a) Q ý ò L dl ý 12 x dx ý
ý 4 nC
3 0
L
0
(b) Method 1:
2
Eý
ò L dl
r
4ðõ r 3
r = (0, 0, h) - (x, 0, 0) = (-x, 0, h), r ý| r |ý h 2 x 2
12 x 2 ( x, 0, h)dx
4ðõ (h 2 x 2 )3/ 2
Since x<<h, we can ignore the x-component.
Eý
1
Eý
h
4ðõ
az
12 x 2 dx
Q
az
ý
3
4ðõ h 2
h
Method 2:
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The line charge can be regarded as a point charge because it is very far from the
observation point.
Eý
Q
4ðõ r 2
ar ý
4 109
4(9)
a z ý 6 a z ý 3.6 105 a z V / m
9
10
10
4ð
(1000) 2
36ð
Prob. 4.12
ò dl
E ý L 2 aR
4ðõ o R
L
R ý aa ò ha z , R ý| R |ý a 2 h 2 , dl ý adö
Eý
ò L ( aa ò ha z )
adö
4ðõ o (a 2 h 2 )3/ 2
Due to symmetry, the ò-component cancels
ò L 2ð haa z dö
ò L haa z
ý
(2ð )
Eý
2
2 3/ 2
4ðõ o 0 (a h )
4ðõ o (a 2 h 2 )3/ 2
out.
4 103 12 106 4 3a z
(2ð ) ý 260.58a z N
F ý QE ý
109 3
5
4ð
36ð
Prob. 4.13
ò s dS
R, R = ò (-aò ) ha z , dS ý ò dö d ò
3
R
ðõ
4
o
S
òs
ò dö d ò
Eý
(- ò a ò ha z )
4ðõ o S ( ò 2 h 2 )3/ 2
Due to symmetry, the ò -component vanishes.
Eý
Eý
2ð
ò s ha z b
2
2 3/ 2
h
d
(
)
ò
ò
ò
0 dö
4ðõ o a
Let u=ò 2 h 2 , du ý 2 ò d ò
bö
ò s ha z
ò s ha z 1/ 2u 1/ 2 ò s ha z ö
1 3/ 2
1
÷
÷
Eý
ý
(2ð ) u du ý
2õ o 1/ 2
2õ o ÷ø
4ðõ o
2
ò 2 h 2 a ÷ø
Eý
ù
òs h ù
1
1
ú 2
ú az
4ðõ o û a h 2
b2 h2 û
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Prob. 4.14
(a) From eq. (4.26),
ü 678.58a z V/m, z>0
ò
12 109
E ý s an ý
an ý 18ð (12)an ý ý
9
10
2õ o
þ678.58a z V/m, z<0
2
36ð
(b) Similarly,
Eý
ü678.58a z V/m, z>4
òs
12 109
an ý
an ý 18ð (12)an ý ý
9
10
2õ o
þ 678.58a z V/m, z<4
2
36ð
For z > 4 and z < 0, the fields cancel out. For 0 < z < 4, they add up. Thus
E ý 2 678.58a z V/m= 1.357a z kV/m
Prob. 4.15
y
an
x
Let f ( x, y ) ý x 2 y 5;
ñf ý a x 2 a y
(a x 2 a y )
ñf
ý
| ñf |
5
Since point ( 1, 0,1) is below the plane,
an ý
_
an ý
Eý
(a x 2 a y )
.
5
òs
6(109 )
(a x 2 a y )
(
)
an ý
9
2õ 0
2(10 / 36ð )
5
ý 151.7 a x 303.5 a y V/m
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Prob. 4. 16
òs
òs
x
x=0
x=a
(a) For x < 0,
E ý E1 E2 ý
òs
( ò s )
( a x )
( a x ) ý 0
2õ o
2õ o
(b) For 0 < x < a,.
ò
( ò s )
ò
E ý s ax
( a x ) ý s a x
õo
2õ o
2õ o
(c ) For x > a,
ò
( ò s )
E ý s ax
(a x ) ý 0
2õ o
2õ o
Prob. 4.17
(a) At P(5,-1,4),
ò sk
20 106
10 106
30 106
ank ý
a
a
(
)
(
)
( a z )
x
y
109
109
109
k ý1 2õ o
2
2
2
36ð
36ð
36ð
3
ý 36ð (5, 10, 15) 10 ý 565.5a x 1131a y 1696.5a z kV/m
3
E ýõ
(b) At R(0,-2,1)
E ý 36ð ùû5(a x ) 10(a y ) 15(a z ) ùû 103 ý 565.5a x 1131a y 1696.5a z kV/m
(c ) At Q(3,-4,10),
E ý 36ð ùû5a x 10(a y ) 15a z ùû 103 ý 565.5a x 1131a y 1696.5a z kV/m
Prob. 4.18
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Fe ý
e2
4ðõ o r 2
ar
Fe
1
e2
ý
ý
Fg 4ðõ o Gm 2 4ð 109
ý 4.17 10
ö 1.6 1019 ö
÷
31 ÷
6.67 1011 ø 9.1 10 ø
1
36ð
2
42
Prob. 4.19
Let Q1 be located at the origin. At the spherical surface of radius r,
2
Q1 ý ð
DðdS ý õ Er (4ð r )
Or
Q1
ar by Gauss's law
4ðõ r 2
If a second charge Q2 is placed on the spherical surface, Q2 experiences a force
Eý
Q1Q2
ar
4ðõ r 2
which is Coulomb’s law.
F ý Q2 E ý
Prob. 4.20
Q
R
4ð R 3
For the given three point charges,
For a point charge,
Dý
1 ù QR1 QR2 2QR3 ù
3
ú
ú
4ð û R13
R2
R33 û
R1 ý (0, 0) (1, 0) ý (1, 0), R1 ý 1
Dý
R2 ý (0, 0) (1, 0) ý (1, 0), R2 ý 1
R3 ý (0, 0) (0,1) ý (0, 1), R3 ý 1
Dý
Q
Q
Q
(1, 0) (1, 0) 2(0, 1)ý ý (0, 2) ý a y
4ð
4ð
2ð
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Prob. 4.21
(a) Assume for now that the ring is placed on the z=0 plane.
z
(0,0,h)
R
y
a
ò L dl
x
ò L dl R
Dý
4ð R
Dý
òL
4ð
ö ý 2ð
ö
ý0
3
, R ý a aò h az
adö ( a a ò h a z )
(a 2 h 2 )3/ 2
Due to symmetry, the ò component vanishes.
ò L a (2 ð h) a z
òL a h a z
ý
2
2 3/ 2
4 ð (a h )
2(a 2 h 2 )3/ 2
a ý 2, h ý 3, ò ý 5 ý C/m
Dý
L
_
Since the ring is actually placed in x ý 0, a z becomes a x .
Dý
(6)(5) a x
ý 0.32 a x ý C/m 2
2(4 9)3/ 2
(b)
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Q [(3, 0, 0) (0, 3, 0)]
Q [(3, 0, 0) (0,3, 0)]
3
4 ð | (3, 0, 0) (0, 3, 0) |
4 ð | (3, 0, 0) (0,3, 0) |3
6 Q(1, 0, 0)
Q(3,3, 0) Q(3, 3, 0)
ý
ý
3/ 2
3/ 2
4ð (18)
4ð (18)
4 ð (18)3/ 2
D ý DR DQ ý 0
DQ ý
0.32(106 )
6Q
ý0
4 ð (18)3/ 2
ü Q ý 0.32(4ð )(183/ 2 )106
Prob. 4.22
(a) ò v ý ñ ÷ D ý
1
ý 51.182ý C
6
öDx öDy öDz
ý 0 2 x 4 ý 2 x 4 nC/m3
öx
öy
öz
(b)
ý D ÷ dS ý y 2 dydz
S
xý3
5
6
0
0
ý dz y 2 dy ý (5)
y3 6 5 3
ý (6) ý 360 nC
3 0 3
Prob. 4.23
ý D ÷ dS ý õ o E ÷ dS ,
S
õo
dS ý d ò dzaö
S
ý E ÷ dS ý 6 ò z sin ö d ò dz
S
ý 6sin 90
o
2
5
0
0
ö ý 90o
ö ò 2 2 ö ö z2 5 ö
1
÷÷
÷ ý 3(4 0) (25 0) ý 150
2
ø 2 0øø 2 0ø
ò d ò zdz ý 6 ÷
ý 150õ o ý 150
109
ý 1.326 nC
36ð
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Prob. 4.24
(a)
1 ö 2
1
1
ö
(r sin ñ sin ö )
(sin ñ cos ñ sin ö )
( sin ö )
2
r ör
r sin ñ öñ
r sin ñ
2
sin ö
sin ö
(cos 2 ñ sin 2 ñ )
ý sin ñ sin ö
sin ñ
r
r sin ñ
2
2
2
But cos ñ sin ñ ý 1 2sin ñ
òv ý ñ ÷ D ý
2
r
òv ý sin ñ sin ö
At
sin ö 2sin ñ sin ö sin ö
ý0
sin ñ
sin ñ
r
r ý 2, ñ ý 30o , ö ý 60
(2,30o , 60o ),
o
òv ý 0
ý D ÷ dS ,
dS ý r 2 sin ñ dñ dö ar
S
ý
60o 30o
r
ö ñ
2
sin 2 ñ sin ö dñ dö
ý0 ý0
60o
ý 4 sin ö dö
(b)
0
30o
sin
2
rý2
ñ dñ
0
1
sin 2ñ ý (1 cos 2ñ ).
2
o
ö
60o ö 1 30
1
ö
öð /6
ý 4 ÷ cos ö
(1 cos 2ñ )dñ ý 2( cos 60o 1) ÷ ñ sin 2ñ ÷
÷
÷
2
0 ÷ø 2 0
ø
ø 0
ø
1
ö
ö
ý 2(1/ 2 1) ÷ ð / 6 sin(ð / 3) 0 ÷ ý 0.5236 0.433 ý 0.0906 nC = 90.6 pC
2
ø
ø
But
Prob. 4.25
öDx öDy öDz
ý 8 y C/m 2
öx
öy
öz
1 ö
1 öDö öDz
(b) ò v ý ñðD =
( ò Dò )
ò öò
ò öö
öz
ý 8sin ö 2sin ö 4 z
(a)
òv ý ñðD =
ý 6sin ö 4 z C/m3
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ö
1 ö 2
1
(r D r )
(Dñ sinñ ) 0
2
r ör
r sin ñ öñ
2
2cosñ
= 4 cosñ
r
r4
= 0
ò v =ñðD =
(c)
Prob. 4.26
ö x2 2 ö ö y2 2 ö ö z 2 2 ö
(a) Q ý ò v dv ý 12 xyzdxdydz ý 12 ÷
÷÷
÷÷
÷ ý 12(2)(2)(2) ý 96 mC
ø 2 0øø 2 0øø 2 0ø
0 0 0
v
(b) =Q= 96 mC
2 2 2
Prob. 4.27
Gaussian surface
r
a
b
Apply Gauss’s law,
ð DðdS ý Q
enc
Dr 4ð r 2 ý ò s1 4ð a 2 ò s 2 4ð b 2 ý 8 109 4ð (1) 2 (6 103 ) 4ð (2) 2 ý 0.3016
Dr ý
0.3016 ý 0.3016
ý
ý 0.0027
4ð r 2
4ð (3) 2
D ý -2.7ar mC/m 2
Prob. 4.28
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For r <a.
ð DðdS ý Q
enc
S
ý òv dv
v
ð
2ð
r
0
0
0
Dr (4ð r 2 ) ý 5r1/ 2 r 2 sin ñ dñ dö dr ý 5 sin ñ dñ dö r 5/ 2 dr
ý 5(2)(2ð )
r 40ð r
r
ý
7/2 0
7
7/2
7/2
40ð r 7 / 2
10
7
ý r 3/ 2
Dr ý õ o Er ý
2
4ð r
7
10 3/ 2
Er ý
r ,0 ü r ü a
7õ o
For r > a,
40 7 / 2
ða
7
40 7 / 2
ða
Dr ý õ o Er ý 7
4ð r 2
10a 7 / 2
,r þ a
Er ý
7õ o r 2
Thus,
ü 10 3/ 2
r ar , 0 ü r ü a
ÿ
ÿ 7õ o
Eýý
7/2
ÿ 10a a , r þ a
ÿ 7õ o r 2 r
þ
Dr (4ð r 2 ) ý
Prob. 4.29
(a) ò v ý ñ ÷ D ý
öDx öDy öDz
ý 2 y C/m3
öx
öy
öz
(b) ý D dS ý x 2 dxdz
1
y ý1
1
ý x 2 dx dz ý
0
0
1
1
1
0
0
0
1
C
3
(c ) Q ý ò v dv ý 2 ydxdydz ý 2 dx ydy dz ý 1 C
v
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Prob. 4.30
(a)
òV ý ñ ÷ D ý
1 ö
( ò Dò )
1 ö Dö
ò öò
ò öö
òV ý 4 ( z 1) cos ö ( z 1) cos ö 0
ö Dz
öz
òV ý 3( z 1) cos ö ý C/m3
(b)
Qenc ý òV dv ý 3( z 1) cos ö ò dö d ò dz
ð /2
ö1
cos ö dö ý 3 ÷ ò 2
0
0
0
ø2
ý 3(2)(8 4)(1 0) ý 72ý C
2
4
ý 3 ò d ò ( z 1) dz
4
ð /2
2 ö z2
÷ ( z )|0(sin ö|0 )
0ø 2
(c)
Let ý 1 2 3 4 5 ý ð
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
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For S1 ò ý 2, dS ý ò dö dza ò
4
ð /2
0
0
1 ý 2 ò ( z 1) cos ö dS|ò ý 2 ý 2(2) ( z 1)dz cos ö dö
2
ý 8(12)(1) ý 96
For S2 , z ý 0, dS ý ò dö d ò ( a z )
2
2 ý ò cos ö ò dö d ò ý ò d ò
2
3
0
ý
ò
4
ð /2
cos ö dö
0
2
| (1) ý 4
4
0
For S3, z ý 4, dS ý ò dö d ò a z , 3 ý 4
For S4 , ö ý ð / 2, dS ý d ò dzaö
ö
ø
4 ý ò ( z 1) sin ö d ò dz|ö ýð /2 ý ÷ sin
ý
ò
2
2
|
2
0
ð ö2
4
÷ ò d ò ø z 1ùdz
2ø0
0
(12) ý (2)(12) ý 24
For S5 , ö ý 0, dS ý d ò dz (aö ) , 5 ý ò ( z 1) sin ö d ò dz|
ö ý0
ý 96 4 4 24 0 ý 72ý C
This is exactly the answer obtained in part (b).
Prob. 4.31
e
R
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F ý eE
ò0 ý
e
ý
3
R
3
4ð
3e
4 ð R3
ü ò 0,
0ürüR
þ0,
elsewhere
òV ý ý
ð D ÷ dS ý Qenc ý òV dV ý
Er ý
3 e 4ð r 3
ý Dr (4ð r 2 )
4ð R 3 3
3e r
12ðõ 0 R 3
F ý eE ý
e2 r
4ð õ 0 R 3
Prob. 4.32
Using Gauss’ law, when ò ü 0,
Eý0
Qenc ý ð
D ÷ dS ýõ o ð E ÷ dS ýõ o Eò (2ðò L)
For a ü ò ü b,
S
S
where L is the length of the cable. But
Q enc ý ò s dS ý
S
2ð
òo
ò dö dz ý ò o 2ð L
ò
ý0 z ý0
L
ö
ò o 2ð L ý õ o Eò (2ðò L)
Eò ý
òo
õoò
For ò >b, Q enc ý 0. Thus,
ü òo
aò ,
aü ò üb
ÿ
E ý ýõ o ò
ÿ 0,
otherwise
þ
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Prob. 4.33
(a)
ý Qenc
at r ý 2
10
Qenc ý òV dV ý
r
2ð
2
2
r 2 sin ñ dñ dr dö
ð
sin ñ dñ drdö
ý 10
r ý1 ö ý 0 ñ ý 0
ý 10 (1) (2ð ) (2) ý (40 ð ) mC
Thus, ý 125.7 mC
At r ý 6;
2ð
4
Qenc. ý 10
ð
dr ö dö ñ sin ñ dñ
r ý1
ý0
ý0
ý 10 (3)(2ð ) (2) ý 120 ð mC
ý 377 mC
(b)
ý Qenc
But ý
At
ð D ÷ dS ý
r ý 1,
Qenc ý 0
At r ý 5,
Dr ý
2
Dr ð
dS ý Dr (4ð r )
Dý0
Qenc ý 120 ð
Qenc
120 ð
ý
ý 1.2
2
4ð r
4 ð (5) 2
D ý 1.2 ar mC/m 2
Prob. 4.34
òv ý
Q
Q
3Q
ý
ý
3
volume 4ð a / 3 4ð a 3
For r < a,
ð DðdS ý Q
Dr 4ð r 2 ý
enc
ý òv dv
3Q 4ð r 3 Qr 3
ý 3
4ð a 3 3
a
Dr ý
Qr
4ð a 3
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ð DðdS ý Q
For r > a,
Dr 4ð r 2 ý Q
Dr ý
Q
4ð r 2
Hence,
ü Qr
ÿÿ 4ð a 3 ar ,
Dýý
ÿ Q a,
ÿþ 4ð r 2 r
rüa
rþa
Prob. 4.35
ù
109 ù
2
4
ý
VP ý
ú
9 ú
10 û | (1, 2,3) (1, 0,3) | | (1, 2,3) (2,1,5) | û
4ðõ o r1 4ðõ o r2
4ð
36ð
4
ù2
ù
ý 9ú
ú ý 1.325 V
994û
û2
Prob. 4.36
Q
,
V ý4
r ý a 2 a 2 h 2 ý 22 22 32 ý 17 cm
4ðõ o r
Q1
Vý
Q2
4 8 109
ý 6.985 kV
109
2
4ð
17 10
36ð
Prob. 4.37 (a)
Q/2
Vý
ý
2
Q
2
4ð õ0 r
ý
Q/2
Q
4ð õ0 r
6
60(10 )
ý 135 kV
109
4ð
x4
36ð
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(b)
Q
3( )
3 ý 135 kV
Vý
4ð õ0 r
(c)
Q
2ð (4)
Q
ò L dl
8
ð
Vý
ý
ý 135 kV
4ð õ0 r
4ð õ 0 r 4ð õ0 r
Prob. 4.38
(a)
VP ý
õ
Qk
4ð | r p r k |
103
3(103 )
2 (103 )
| (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 ð õ 0 (103 ) V p ý
ý
| (1,1, 2) | | (1, 4,1) | (4,5, 4) |
6
18
57
4 ð õ oV p ý
4ð
109
(103 ) V p ý 0.3542
36 ð
üV p ý 3.008 106 V
(b)
VQ ý
õ
Qk
4 ðõ o | r p r k |
2 (103 )
103
3(103 )
| (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 ð õ 0 (103 ) V p ý
ý
| (1, 2, 1) | | (3, 3, 2) | | (2, 6, 3) |
6
22 7
4 ð õ oVQ ý
109
4ð
(103 ) V p ý 0.410
36ð
VQ ý 3.694 (106 )V
üVPQ ý VQ VP ý 0.686 (106 ) ý 686 kV
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Prob. 4.39
öV
1 öV
öV
aò
aö
az
öò
öz
ò öö
E ý ñV ý
ý 2 ò e z sin ö a ò ò e z cos ö a ö ò 2 e z sin ö a z
At (4,ð /4,-1), ò =4,ö =ð /4,z ý 1
E ý 2(4)e1 sin(ð / 4)a ò 4e1 cos(ð / 4)a ö 16e1 sin(ð / 4)a z
ý 2(4)(2.7183)(0.7071)a ò 4(2.7183)(0, 7071)a ö 16(2.7183)(0, 7071)a z
E ý 15.38a ò 7.688a ö 30.75a z V/m
Prob. 4.40
(a)
öV
öV
öV
E ý (
ax
ay
az )
öx
ö y
öz
ý 2 xy ( z 3)a x x 2 ( z 3) a y x 2 y a z
x ý 3, y ý 4, z ý 6,
At (3, 4, 6),
E ý 2(3)(4)(3) a x 9 (3)a y 9(4) a z
ý 72 a x 27 a y 36a z V/m
(b)
òV ý ñ ÷ D ý õ 0ñ ÷ E ý õ 0 (2 y ) ( z 3)
ý Qenc ý
ò
V
1
ý 2õ 0 dx
0
dV ý 2õ 0 y ( z 3)dx dy dz
1
1
0
0
y dy ( z 3)dz ý 2õ 0 (1)(1/ 2)(
1
z2
3z ) |
0
2
1
7 109
(
)
ý õ 0 ( 3) ý
2
2 36 ð
Qenc ý 30.95 pC
Prob. 4.41
(a)
ö r2 ö
Q ý ò v dv ý ò o ÷1 2 ÷r 2 sin ñ dñ dö dr
ø a ø
v
2ð
ð
a
ö
ö a 3 a 3 ö 8ð 3
r4 ö
a òo
ý òo dö sin ñ dñ ÷ r 2 2 ÷dr ý ò o (2ð )(2) ÷ ÷ ý
a ø
ø 3 5 ø 15
0
0
0ø
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(b) Outside the nucleus, r >a,
ð DðdS ý Q
enc
Eý
S
Qenc
ar
4ðõ o r 2
8ð a 3 ò o
2a 3 ò o
15
Eý
ar ý
ar
4ðõ o r 2
15õ o r 2
V ý E ðdl ý Er dr ý
V () ý 0, C1 ý 0.
Since
Vý
2a 3 ò o
C1
15õ o r
2a ò o
15õ o r
3
(c ) Inside the nucleus, r <a
ö r3 r5 ö
Qenc ý 4ðòo ÷ 2 ÷
ø 3 5a ø
Qenc
òo ö r r 3 ö
Eý
a
ý
÷
÷a
r
õ o ø 3 5a 2 ø r
4ðõ o r 2
òo ö r 2
r4 ö
V ý Er dr ý ÷
÷C
õ o ø 6 20a 2 ø 2
V (r ý a ) ý
2a 2 ò o ò o ö a 2 a 2 ö
ý
÷ ÷C
15õ o
õ o ø 20 6 ø 2
2a 2 ò o 7 a 2 ò o a 2 ò o
C2 ý
ý
15õ o
60õ o
4õ o
Vý
òo ö r 4
r 2 ö a 2 òo
÷
÷
õ o ø 20a 2 6 ø 4õ o
(d) E is maximum when
ò ö 1 3r 2 ö
dE
ý0ý o ÷ 2 ÷
õ o ø 3 5a ø
dr
rý
9r 2 ý 5a
2
5
a ý 0.7454a
3
We are able to say maximum because
6r
d 2E
ý 2 ü 0.
2
dr
5a
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Prob. 4.42
z
h
R
b
ò
a
x
Dý
òs ý
Dý
ò s dSR
,
4ð R 3
R ý ò aò ha z ,
R ý| R | ý ò 2 h 2 , dS ý ò dö d ò
Q
Q
ý
2
S ð (b a 2 )
ò s ò dö d ò ( ò a ò ha z )
4ð
( ò 2 h 2 )3/ 2
Due to symmetry, the component along aò vanishes.
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b
ò s h b 2ð ò d ö d ò
òs h
ý
(2ð ) ( ò 2 h 2 ) 3/ 2 ò d ò
Dz ý
4ð òý a öý0 ( ò 2 h 2 )3/ 2 4ð
a
ù
1 ù b ò s h ù
1
1
ý
ú 2
ú
ú
ú
2 úû ò h 2 úû a
2 û a2 h2
b2 h2 û
ù
ù
1
1
Qh
Dý
ú az
2
2 ú
2ð (b a ) û a 2 h 2
b2 h2 û
ý
òs h ù
Prob. 4.43
öDx öDy öDz
ý 4 20 y 2 z
öx
öy
öz
At P(1,2,3), x=1, y= 2, z=3
òv ý ñðD ý
òv ý 4 20(2) 2(3) ý 30 C/m3
Prob. 4.44
r=3cm
r=5cm
For r < 3cm, Qenc ý 0
For 3 < r < 5cm,
ð DðdS ý Q
enc
Dý0
ý 10 nC
Dr 4ð r 2 ý 10 nC
Dr ý
10
nC/m 2
4ð r 2
For r > 5 cm,
ð DðdS ý Q
enc
ý 10 -5 = 5nC
Dr ý
5
nC/m 2
2
4ð r
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Thus,
ü
ÿ
0,
r ü 3 cm
ÿ
ÿ 10
Dýý
a nC/m 2 , 3 ü r ü 5 cm
2 r
ÿ 4ð r
5
ÿ
2
r þ 5 cm
ÿ 4ð r 2 ar nC/m ,
þ
Prob. 4.45
òv ý ñðD ý ñðõ o E ý
1 ö ö õ o Eo ò 2 ö
÷
÷
ò öò ø a ø
ý
2õ o Eo
,0 ü ò ü a
a
Prob. 4.46
Let us choose the following path of two segments.
(2,1, 1) (5,1, 1) (5,1, 2)
W ý q E ÷ dl
5
W
ý E ÷ dl ý 2 xyzdx
q
xý2
z ý 1, y ý 1
2
x 2 ydz
z ý1
x ý 5, y ý 1
2
x2 5
ý 2(1)(1)
(5) 2 (1) z
ý 21 75 ý 54
1
2 2
W ý 54q ý 108 ý J
Prob. 4.47
(a)
òv ý ñ ÷ D ý ñ ÷ õ E
òv
1 ö
1
ý ñ÷ E ý
(12 ò 2 z cos ö ) (6 ò z cos ö ) 0
õ
ò öò
ò
ý 24 z cos ö 6 z cos ö ý 18 z cos ö
At A(2,180o , 1),
ò ý 2, ö ý 180o , z ý 1
òv ý 18õ z cos ö ý 18(1) cos(180o )
109 109
ý
ý 0.1592 nC/m3
36ð
2ð
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(b)
W ý Q E ÷ dl ,
dl ý ò dö aö
L
ö
0o ö
W ý Q 6 ò z sin ö ò dö
ý Q6(2) (1) ÷ cos ö
÷ ý 24Q(1 1)
÷
ò ý 2, z ý 1
180o ÷ø
ö ý180o
ø
0o
2
ý 48Q ý 48 10 106 ý 480 ý J
Prob. 4.48
(a)
From A to B, dl ý rdñ añ ,
90ð
WAB ý Q
10 r cos ñ r dñ
|
ñ ý30ð
ý Q (10)(5) 2 (sin ñ )
r ý5
90o
1250 nJ
30o
(b)
From A to C , dl ý dr a r ,
WAC ý Q
10
ö 1 ö 10
ý Q(20)(sin 30o ) ÷ r 2 ÷
ý 3750 nJ
ø2 ø 5
ý 30ð
20 r sin ñ dr ñ |
r ý5
(c )
From A to D, dl ý r sin ñ dö aö ,
WAD ý Q 0(r sin ñ ) dö ý 0 J
(d)
WAE ý WAD WDF WFE
where F is (10,30o , 60o ). Hence,
90
üÿ 10
ü
WAE ý Q ý 20 r sin ñ dr | 10 r cos ñ r dñ | ý
r ý10 þ
ñ ý30ð
ÿþ r ý5
ñ ý30o
75 100
] nJ ý 8750 nJ
ý 100[
2
2
o
Prob. 4.49
B
5
10
dr
2
r
1
VAB ý E ðdl ý
A
10
ý
r
5
1
1
ý 10( 1) ý 8V
5
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Prob. 4.50
Method 1:
W ý Q E ðdl ,
dl ý ò dö aö
L
ù Eò ù ù cos ö sin ö 0 ù ù Ex ù
ú E ú ý ú sin ö cos ö 0 ú ú E ú
ú öú ú
úú yú
úû Ez úû úû 0
0
1 úû úû Ez úû
Eö ý Ex sin ö E y cos ö ý 20 x sin ö 40 y cos ö
x ý ò cos ö , y ý ò sin ö
Eö ý 20 ò cos ö sin ö 40 ò sin ö cos ö ý 20 ò cos ö sin ö
W ý Q E ðdl = -2 10-3 20 ò cos ö sin öò dö
L
ý 2(20)(2)2
ð /2
sin ö d (sin ö ) mJ ý 160
0
ò ý2
sin 2 ö ð / 2
ý 80 mJ
0
2
Method 2:
W
ý E ðdl = 20xdx + 40ydy
Q L
y ý 2 x, dy ý dx
0
W
ý 20xdx + 40(2 - x)(-dx) = (60x - 80)dx
Q
x=2
ý
0
60 x 2
80 x ý 40
2
2
W ý 40Q ý 80 mJ
Method 3:
ö
ö
ö
öy
öz ý 0
ñ E ý öx
20 x 40 y 10 z
V ý E ðdl = -10x 2 20 y 2 5 z 2 C
L
W ý Q(V2 V1 ) ý Q(20 4 10 4) ý 40Q
W ý 40Q ý 80 mJ
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Prob. 4.51
W ý Q E ðdl
L
ò
ò
E ý s an ý s a x ,
2õ o
2õ o
dl ý dxa x
Qòs
Qòs
Qòs
W ý
dx ý
(2) ý
2õ o 3
2õ o
õo
1
ý
10 106 40 109
ý 400 36ð 106 ý 45.24 mJ
109
36ð
Prob. 4.52
(a) E ý ñV ý
öV
öV
öV
ax
ay
a y ý 4 xa x 8 ya y
öx
öy
öz
òv ý ñðD ý õ oñðE ý õ o (4 8) ý 12õ o ý 106.25 pC/m3
öV
öV
1 öV
aò
aö
az
öò
öz
ò öö
ý (20 ò sin ö 6 z )aò 10 ò cos ö aö 6 ò a z
(b) E ý ñV ý
ö 1
òv ý ñðD ý õ oñðE ý õ o ÷
ø ò
ö
40 ò sin ö 6 z ý 10sin ö ÷
ø
ö
6z ö
ý ÷ 30sin ö ÷ õ o C/m3
ò ø
ø
1 öV
1 öV
öV
ar
añ
aö
ör
r öñ
r sin ñ öö
ý 10r cos ñ sin ö ar 5r sin ñ sin ö añ 5r cot ñ cos ö aö
(c) E ý ñV ý
òv ý ñðD ý õ oñðE
òv 1
5r sin ö
5r cot ñ sin ö
2sin ñ cos ñ
ý 2 (30r 2 cos ñ sin ö )
õo r
r sin ñ
r sin ñ
òv ý õ o (5sin ö csc2 ñ cos ñ 20 cos ñ sin ö ) C/m3
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Prob. 4.53
(a)
E ý ñV ý
1 öV
öV
öV
aò
aö
a z ý e z sin ö a ò e z cos ö aö ò e z sin ö a z
öò
öz
ò öö
E ý e z sin ö a ò e z cos ö aö ò e z sin ö a z
(b)
.
0
öE ù
ù 1 öEz öEö ù
ù öE
öE ù
1ùö
ñ E ý ú
a ò ú ò z ú aö ú ( ò Eö ) ò ú a z
ú
öz û
öò û
öö û
ò û öò
û ò öö
û öz
ù1
ù
1
ý ú ò e z cos ö e z cos ö ú a ò ûù e z sin ö e z sin ö ûù aö ûù e z cos ö e z cos ö ûù a z ý 0
ò
ûò
û
since each component is zero. Alternatively,
ñ E ý ñ ñV ý
Prob. 4.54
V ý r 3 sin ñ cos ö
E ý ñV ý
öV
1 öV
1 öV
ar
añ
aö
ör
r öñ
r sin ñ öö
r 4 sin ñ
( sin ö )aö
sin ñ
3
1
sin ö
E
ý 4 sin ñ cos ö ar 4 cos ñ cos ö añ 4 aö
r
r
r
o
o
o
o
At (1,30 , 60 ), r ý 1,ñ ý 30 , ö ý 60
ý 3r 4 sin ñ cos ö ar r 4 cos ñ cos ö añ
E
ý 3sin 30o cos 60o ar cos 30o cos 60o añ sin 60o aö
ý 0.75ar 0.433añ 0.866aö
109
(0.75ar 0.433añ 0.866aö )
36ð
ý 6.635ar 3.829añ 7.657aö pC/m 2
D ý õo E ý
Prob. 4.55
For a < r < b, we apply Gauss’s law.
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ð DðdS ý Q
enc
ýQ
S
Dr (4ð r 2 ) ý Q
Er ý
b
Vab ý E ðdl ý
a
Dr
õo
ý
Q
4ðõ o r 2
b
1
Q 1b
Q ö1 1ö
dr
ý
ý
÷ ÷
2
4ðõ o a r
4ðõ o r a
4ðõ o ø a b ø
Q
Prob. 4.56
ð /2
ò s dS
ò s 2ð ð / 2 r 2 sin ñ dñ dö
òs
ý
ý
(2ð ) sinñ dñ
V ý
4ðõ o r 4ðõ o öý0 ñ ý0
4ðõ o r
r
S
0
ð / 2ö
òs ö
÷ cos ñ
÷
0 ørýa
2õ o r ø
ò
ý s
2õ o a
ý
V
Prob. 4.57
ñ E ý 0
ñ D ý 0
öD ù
ù 1 öDz öDö ù
ù öD öD ù
1ùö
aò ú ò z ú aö ú ( ò Dö ) ò ú a z
ñ D ý ú
ú
ò û öò
öz û
öò û
öö û
û ò öö
û öz
1
ý 0 a ò 0aö 2 ò cos ö a z ù 0
ò
Hence D is not a genuine EM field.
ý DðdS ý
S
ý 2 cos ö
ð /4 1
ð /4
1
2ò sin öò dö dz ý 2 sin ö dö dz ò
ö
ý0 z ý0
ð /4
0
0
2
0
ò ý1
(1)(1) 2 ý 2(cos ð / 4 1) ý 0.5858 C
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Prob. 4.58
(a)
d2y
m 2 ý eE ; divide by m , and integrate once, one obtains :
dt
dy
eEt
uý
ý
c0
dt
m
e E t2
c0t c1
2m
"From rest" implies c1 ý 0 ý c0
y ý
(1)
V
or V ý E d .
d
Substituting this in (1) yields :
2m d
t2 ý
eE
Hence :
At t ý t0 ,
y ý d, E ý
2md
eE
eE
m
uý
that is, u ñ
ý
2 e Ed
2eV
ý
m
m
V
uý k V
or
(b)
k ý
2e
ý
m
2 (1.603) 1019
9.1066 (1031 )
ý 5.933 105
(c )
1
u m
100 ý 2.557 k V
Vý
ý
2e
2 (1.76) (1011 )
2
9(1016 )
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Prob. 4.59
(a)
This is similar to Example 4.3.
eEt
uy ý
, u x ý u0
m
e E t2
, x ý u0 t
y ý
2m
x 10 (102 )
ý 10 ns
tý ý
107
u0
Since x ý10 cm when y ý1cm,
Eý
2m y
2 (102 )
ý
ý 1.136 kV/m
1.76 (1011 ) (1016 )
et2
E ý 1.136 a y kV/m
(b)
u x ý u0 ý 107 ,
uy ý
2000
(1.76)1011 (108 ) ý 2(106 )
1.76
u ý (a x 0.2a y ) (107 ) m/s
Prob. 4.60
p cos ñ
k cos ñ
ý
Vý
2
4ð õ0 r
r2
At (0, l nm),
ñ ý 0,
r ý 1 nm, V ý 9;
k (1)
, ü k ý 9(1018 )
1(1018 )
cos ñ
V ý 9(1018 ) 2
r
ñ ý 45ð,
At (1,1) nm, r ý 2 nm,
that is,
Vý
9ý
9(1018 ) cos 45ð
9
ý
ý 3.182 V
18
2
10 ( 2)
2 2
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Prob. 4.61
The dipole is oriented along y axis.
Vý
p÷r
; p ÷ r ý Q d a y ÷ a r ý Qd sin ñ sin ö
4ðõ 0 r 2
Vý
Qd sin ñ sin ö
4ð õ0 r 2
E ý ñV ý
ý
Qd
4ð õ0
Eý
Qd
1öV
1 öV
öV
ar
añ
aö
r öñ
r sin ñ ö ö
ör
ü 2sin ñ sin ö
ü
cos ñ sin ö
cos ö
ÿ
ar
añ 3 a ö ý
ý
3
3
r
r
r
þ
ÿþ
_
4ð õ 0r 3
(2sin ñ sin ö a r cos ñ sin ö añ cos ö aö )
Prob. 4.62
Using eq. (4.81),
p ÷ (r r ')
Vý
4ðõ o | r r ' |3
r r ' ý (4, 0,1) (2,3, 1) ý (2, 3, 2)
| r r ' |ý 4 9 4 ý 17
p ÷ (r r ') ý (2, 6, 4) ÷ (2, 3, 2) ý 4 18 8 ý 22
Vý
22 106
198
ý
kV= 2.825 kV
9
10
70.093
3/ 2
17
4ð
36ð
Prob. 4.63
E ý k (2 cos ñ ar sin ñ añ )
ù Eò ù ù sin ñ cos ñ
úE ú ý ú 0
0
ú öú ú
úû Ez úû úûcos ñ sin ñ
Ez ý 2k cos 2 ñ k sin 2 ñ
0 ù ù 2k cos ñ ù
1 úú úú k sin ñ úú
0 úû úû 0 úû
ý 2k cos 2 ñ k (1 cos 2 ñ ) ý 3k cos 2 ñ k
Setting this to zero gives
3cos 2ñ ý 1
cos ñ ý
1
ý 0.5773
3
ñ ý 54.74o , 125.26o
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Prob. 4.64
W ý W1 W2 ý 0 Q2V21 ý Q2
Q1
4ðõ o | (2, 0, 0) (0, 0,1) |
40 109 (50) 109 40 9 (50) 109
ý
ý
109
4 1
4ð
| (2, 0, 1) |
36ð
ý 8.05 ý J
Prob. 4.65
E ý ñV ý
öV
öV
ax
a y ý 4 xa x 12 ya y V/m
öx
öy
1
1
1
1
1
W ý õ o | E |2 dv ý õ o (16 x 2 144 y 2 )dxdydz
2 v
2 z ý1 y ý1 x ý1
x3 1
y 3 1 ù 1 109
1 ù
1
ý õ o ú16(4)
144(4)
(160)(4) (1 1)
úý
2 û
3 1
3 1û 2 36ð
3
ý 1.886 nJ
Prob. 4.66
Given that E ý 2r sin ñ cos ö ar r cos ñ cos ö añ r sin ö aö
E 2 ý 4r 2 sin 2 ñ cos 2 ö r 2 cos 2 ñ cos 2 ö r 2 sin 2 ö
ý r 2 cos 2 ö ø 4sin 2 ñ cos 2 ñ ù r 2 sin 2 ö
r 2 cos 2 ö 3r 2 cos 2 ö sin 2 ñ r 2 sin 2 ö
ý
ý r 2 (1 3cos 2 ö sin 2 ñ )
Wý
õ
ý
E r
2
õ
2 2
ð ð
2
2 0
sin ñ drdñ dö
4
r dr
(1 3cos
ñ ö
2
ö sin 2 ñ ) sin ñ dñ dö
ý0
ð
ý
16õ
3ð
(ð sin ñ
sin 2ñ )dñ
5 0
2
ý
16 109
16
x
(4ð ) ý
nJ= 0.36 nJ
5 36ð
45
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Prob. 4.67
Method 1:
Wý
1
1
V
ò sVdS ý ò s dS ý QV
2S
2S
2
Vý
But
Wý
Q
Q
4ðõ o a
2
8ðõ o a
Method 2:
1
1
W ý DðEdv ý õ o E 2 dv
2v
2 v
2
ö Q ö 2
1
r sin ñ dñ drdö
ý õ o ÷
2 ÷
2
ø 4ðõ o r ø
2ð
ð
õo
Q2
Q2
1
ð
dr
(2
)(2)
ý õ o dö sin ñ dñ
ý
2 0
16ð 2õ o2 r 2
2
16ð 2õ o2 a
r ýa
0
Wý
Q2
8ðõ o a
Prob. 68
Wý
2W
õo
1
1
1
D ÷ Edv ý õ o | E |2 dv ý õ o ø y 4 4 x 2 y 2 16 z 2 ù dxdydz
2v
2 v
2 v
2
4
1
2
4
1
2
4
1
ý 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
ö y 1ö
ö x 2 öö y 1 ö
ö z3 4 ö
ý 2(4) ÷ 2
4(4)
2
16(2)(2)
÷
÷
÷÷
÷
÷
÷
ø 5 0ø
ø 3 0 øø 3 0 ø
ø 3 0ø
16 256 64 64
ý
ý 1396.98
5
9
3
1 109
1396.98 ý 6.176 nJ
Wý
2 36ð
5
3
3
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Prob. 4.69
(a)
ö öV
öV ö
1 öV
E ý ñV ý ÷
aò
aö
az ÷
ò öö
öz ø
ø öò
ý e z sin ö a ò e z cos ö aö ò e z sin ö a z
E ðE ý| E |2 ý e2 z sin 2 ö e2 z cos 2 ö ò 2 e 2 z sin 2 ö ý e2 z ò 2 e2 z sin 2 ö
1
W ý õ o | E |2 dv
2 v
2W
õo
ý | E |2 dv ý ø e2 z ò 2 e2 z sin 2 ö ùò d ò dö dz
v
1
(b)
2ð
2
0
0
1
2ð
2
0
0
ý ò d ò dö e2 z dz ò 3 d ò sin 2 ö dö e 2 z dz
0
0
ö e 2 ö ò 1 ö ö sin 2ö ö 2ð ö e2 z 2 ö
(2ð ) ÷
÷
÷
÷
÷
÷
2 0
4 ø 0 ø 2 0 ø
ø 2 0 ø 4 0 ø 2
1
5ð
ý ð (1/ 2)(e4 1) (ð 0)(1/ 2)(e 4 1) ý
(1 e 4 ) ý 1.9275
4
8
9
1
1 10
(1.9275) ý 8.512 pJ
W ý õ o (1.9275) ý
2
2 36ð
ý
ò 1
2
2 z
4
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CHAPTER 5
dS = ò dö dzaò
P. E. 5.1
Iý
2ð
J ÷ dS ý
10(2)
5
10 z sin
2
öòdzdö|ò ý 2
=
ö ý0 zý1
z2
2
5 2ð
1
1
2 (1 cos 2ö )dö ý 10(5
0
2
1ö
1
ö 2ð
12 ) ÷ ö sin 2ö ÷
ý 10(24)ð ý 240ð
2ø
2
ø 0
I = 754 A
P. E. 5.2
I ý ò s wu ý 0.5 106 0.110 ý 0.5ý A
V ý IR ý 0.5 106 1014 ý 50 MV
P. E. 5.3 ó ý 5.8 107 S/m
8 106
J
ý 0.138 V/m
J ýóE
Eý ý
ó 5.8 107
8 106
uý
ý
ý 4.42 104 m/s
J ý òvu
10
òv 1.8110
P. E. 5.4 The composite bar can be modeled as a parallel combination of resistors as
shown below.
J
RL
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Rc
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l
,
ó L SL
RL ý
For the lead,
RL ý
l ý 4m, S L ý d 2 ð r 2 ý 9
4
ö ðö
(5 10 ) ÷ 9 ÷ 104
4ø
ø
ð
4
cm 2 , ó L ý 5 106 S/m
ý 973.8ý
6
Rc ý
For copper,
Rc ý
4
5.8 10
ð
l
,
ó c Sc
4
Sc ý ð r 2 ý
ð
4
, ó c ý 5.8 107 S/m cm2
ý 878.5 ý
10
4
RR
973.8 878.5
Rý L c ý
ý 461.8ý
RL Rc 973.8 878.5
7
òPs ý P ÷ a s ý a x2 b
P. E. 5.5
ò p s xý0 ý P ÷ ( a x ) xý0 ý (a x2 b )
ò p s xý L ý P ÷ a x
xýL
ý (a x2 b )
xý0
xýL
ýb
ý a L2 b
Qs ý ò ps dS ý bA (aL2 b) A ý AaL2
ò p v ý ñ ÷ P ý
ò pv
xý0
ý 0,
d
(a x2 b ) ý 2a x
dx
ò pv
xý L
ý 2aL
L
Qv ý ò pv dv ý ( 2ax ) Adx ý AaL2
0
Hence,
QT ý Qv Qs ý AaL2 AaL2 ý 0
P. E. 5.6
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V
103
ax ý
a x ý 500a x kV/m
d
2 x10 3
109
x0.5 x106 a x ý 6.853a x ý C / m 2
P ý ó e õ o E ý (2.55 1)x
36ð
Eý
ò p s ý P ÷ a x ý 6.853 ýC/ m 2
P. E. 5.7 (a) Since P ý õ o ó e E ,
óe ý
Px ý õ o ó e E x
3 x109 1
Px
ý
x36ð x109 ý 2.16
10ð 5
õ o Ex
36ð 109 1
ý
(3, 1, 4)109 ý 5ax 1.67 a y 6.67 az V/m
(b) E ý
2.16 10ð
ó eõ o
(c )
õ P 3.16 ö 1 ö
nC / m2 ý 139.7 a x 46 .6 a y 186 .3az
D ý õ oõ r E ý r ý
÷ (3, 1,4)
÷
óe
2.16 ø 10 ð ø
P
P. E. 5.8 From Example 5.8,
Fý
But
òs2 S
2õ o
òs ý õ o E ý õ o
òs2 ý
òs2 ý
Vd
.
d
2õ o F õ o 2Vd 2
ý
S
d2
2õ o F
S
Hence
Vd 2 ý
2 Fd 2
õoS
i.e.
Vd ý V1 V2 ý
2 Fd 2
õoS
as required.
P. E. 5.9 (a) Since
D1n ý 12a x ,
E 2 t ý E1t
an ý a x ,
D1t ý 10 a x 4a z ,
D2 t ý
D2 n ý D1n ý 12a x
õ 2 D1t
1
ý
( 10 a y 4a z ) ý 4a y 16
. az
õ1
2.5
D2 ý D2 n D2 t ý 12a x 4a y 1.6a z nC/m2.
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pC / m2
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D
tan ñ 2 ý 2 t ý
D2 n
( 4) 2 (16
. )2
ý 0.359
12
ñ 2 ý 19.75 o
(b ) E1t ý E 2t ý E2 sin ñ 2 ý 12 sin 60 o ý 10.392
E2
E 2t
ñ2
E2n
E1n ý
õr2
1
12 cos 60 o ý 2.4
E2n ý
2.5
õ r1
E1 ý
tan ñ1 ý
x
E1t 2 E1n 2 ý 10.67
E1t
E2t
õ E
õ
2.5
tan 60o ý 4.33
ý
ý r1 2t ý r1 tan ñ 2 ý
E1n (õ r1 / õ r 2 ) E2 n õ r 2 E2 n õ r 2
1
Note that ñ1 þ ñ 2 .
P. E. 5.10
10 9
D ý õ oE ý
(60 ,20 , 30 ) x10 3 ý 0.531a x 0.177 a y 0.265az pC/m2
36 ð
10 9
ò s ý Dn ý | D| ý
(10 ) 36 4 9 (10 3 ) ý 0.619 pC/m2
36 ð
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Prob. 5.1
I ý J ÷ dS ,
dS = dydza x
I ý e x cos(4 y )dydz
xý2
ý e 2
ð /3
0
4
cos(4 y )dy dz
0
ö sin 4 y ð / 3 ö 2 ö
4ð
ö
ý 4e 2 ÷
÷ ý e ÷ sin( ) 0 ÷ ý 0.1172 A
0 ø
3
ø
ø
ø 4
Prob. 5.2
Method 1:
I ý J ÷ dS =
10 103 t 2
e r sin dñ dö
t ý 2ms, r ý 4m
r
ý 10(4)e 10 210
3
3
ð
ñ
sin ñ dñ
ý0
2ð
dö ý 40e
ö
2
(2)(2ð ) ý 160ð e 2
ý0
ý 68.03 A
Method 2:
3
10 103 t
10
e dS ý e 10 t (4ð r 2 )
r
r
since r is constant on the surface.
I ý J ÷ dS =
I=40ð re 2 ý 160ð e 2
ý 68.03 A
Prob. 5.3
I ý J dS ý
10
ò
5
ð
0
0
sin öò dö dz ý 10 dz sin ö dö
ý 10(5)( cos ö )
ð
0
ý 100 A
Prob. 5.4
I ý J ÷ dS
ý5
a
2ð
ò ö
ý0 ý0
2ð
e 10 ò ò dö d ò ý 5 dö
0
a
ò
ò e10 ò d ò
ý0
10 ò
öe
ö a 10ð 10 a
ý 5(2ð ) ÷
(10 ò 1) ÷ ý
ûù e (10a 1) 1(0 1) ûù ,
ø 100
ø 0 100
ý
ð
ð
ùûe 0.04 (0.04 1) 1ùû ý (0.00078) ý 244.7 ý A
10
10
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Prob. 5.5
I ý J ÷ dS ,
dS ý r 2 sin ñ dñ dö ar
S
2ð
ð /2
20 cos ñ 2
20(9)
ö ý0 ñ ýð / 4 r 3 r sin ñ dñ dö r ý 3 ý 6
Iý
ý 30(2ð )
ð /2
cos ñ d ( cos ñ ) ý 60ð
ð /4
2ð
ð /2
0
ð /4
dö
cos ñ sin ñ dñ
cos 2 ñ ð / 2
ý 60ð (0 cos 2 (ð / 4)) ý 60ð (1/ 2) ý 30ð
2 ð /4
I ý 94.2 A
Prob. 5.6
2 102
l
l
óý
ý 6
ý 3.978 104 S/m
Rý
3 2
óS
RS 10 (ð )(4 10 )
8 102
8
l
Prob. 5.7 (a) R ý
ý
ý
ý 33.95m
4
6
ó S 3 10 ð (25)10
75ð
(b) I ý V / R ý 9
75ð
ý 265.1 A
8
(c ) P = IV = 2.386 kW
Prob. 5.8
(a) E ý
V
9
ý
ý 90 mV/m
l 100
(b) R ý
9
V
ý
ý 30
I 0.3
l
óS
óý
Rý
100
l
ý
ý 2.653 105 S/m
6
2
RS 30(ð 2 )10
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Prob. 5.9
If R and S are the same,
ó
1 ý 2 1
R1 ý 1 ý R2 ý 2
ó 1S
ó 2S
ó2
If 1 corresponds to copper and 2 to silver,
ó 1 ý 5.8 107 S/m, ó 2 ý 6.1107 S/m
5.8
ý 0.951 2
6.1
That is, the copper wire is shorter than silver wire or the silver wire is longer.
1 ý 2
Prob. 5.10
V
I
,
S ý ð r2
ý
óý
SV
óS I
7
2(5)
10
óý
ý
ý 6.635 104 S/m
2
6
ð (2 10 )(12) 48ð
Rý
Prob. 5.11 (a) Si ý ð ri ý ð (1.5)2 x10 4 ý 7.068 x10 4
2
So ý ð (ro 2 ri 2 ) ý ð (4 2.25) 104 ý 5.498 104
RI ý
Ro ý
òIl
SI
ý
òo l
So
11.8 10 8 10
ý 16.69 104
4
7.068 X10
ý
1.77 108 10
ý 3.219 104
4
5.498 10
Ri Ro
16.69 3.219 104
R ý Ri // Ro ý
ý
ý 0.27m
Ri Ro
16.69 3.219
(b)
V ý I i Ri ý I o Ro
I i I o ý 11929
.
I o ý 60
I o ý 50.3 A
Ii
R
0.3219
ý o ý
ý 0.1929
1669
.
Io
Ri
A
(copper),
I i ý 9.7 A
(steel)
Alternatively, using the principle of current division,
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I o ý 60
Ri
ý 50.3 A
Ri Ro
I i ý 60
Ro
ý 9.7 A
Ri Ro
(c)
10 1.77 108
ý 0.141m
Rý
ð (22 ) 104
Prob. 5.12
From eq. (5.16),
ò ò
R1 ý 1 ý 1 ,
S1
ab
R2 ý
ò2
S2
ý
ò2
ac
ò1 ò 2
ò1 ò 2 2
R1 R2
R ý R1 R2 ý
ý ab ac ý
R1 R2 ò1 ò 2 ac ò1 abò 2
ab ac
ò1 ò 2
Rý
a (c ò1 bò 2 )
Prob. 5.13
V
ý 2 ý
óS ðr ó I
ð r 2ó 12ð (0.84 103 ) 2 6.1107
ý
ý 130.86 A
I ýV
12.4
Rý
Prob. 5.14
| P |ý n | p |ý nQd ý 2ned ý ó eõ o E
(Q ý 2e)
2ned 2 5 1025 1.602 1019 1018
óe ý
ý
ý 0.000182
109
õoE
4
10
36ð
õ r ý 1 ó e ý 1000182
.
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Prob. 5.15
N
Pý
õ qi di
i ý1
v
N
ý
õp
i
i ý1
v
N
| p |ý 2 1019 1.8 1027 ý 3.6 108
v
| P |ý
P ý| P| a x ý 3.6 108 a x C/ m 2
But
P ý ó eõ o E
or
3.6 108
P
óe ý
ý
ý 0.0407
õ o E (109 / 36ð )105
õ r ý 1 ó e ý 10407
.
Prob. 5.16
Q
Eý
ar
4ðõ oõ r r 2
P ý ó eõ o E ý
ó eQ
3(10)103
ý
a
ar ý 596.8ar ý C/m 2
r
4ðõ r r 2
4ð (4)12
Prob. 5.17
P ý ó eõ o E
Eý
P
ó eõ o
ý
100 109
aò ý 2.261a ò kV/m
109
2.5
(2)
36ð
109
D ý õ oõ r E = 3.5
2.261103 a ò ý 70a ò nC/m 2
36ð
Prob. 5.18
1 ö
( p ò 2 ) ý 2 po
ò öò o
The surface polarization charge is
ò pv ý ñ P ý
ò ps ý P a ò
ò ýa
ý po a
Prob. 5.19
(a)
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Qs1 ý P dS , dS ý r 2 sin dñ dö (ar )
S
ý 4r r 2 sin ñ dñ dö
2ð
r ý 1.2cm
ð
ý 4(1.2) (106 ) dö sin ñ dñ (1012 )
3
0
ý 6.912(2ð )(2) 10
0
18
ý 86.86 1018 C
(b)
Qs 2 ý P dS , dS ý r 2 sin dñ dö (ar )
S
ý 4r r 2 sin ñ dñ dö
r ý 2.6cm
2ð
ý 4(2.6) (10 ) dö sin ñ dñ (1012 )
6
3
0
ý 4(2.6) (2ð )(2) 1018 ý 883.5 1018 C
3
(c )
ò pv ý ñ P ý
1 ö
(4r 3 ) pC/m3 ý 12pC/m3
2
r ör
ð
2ð
2.6
0
0
1.2
Qv ý ò pv dv ý 12 dv ý 12 sin ñ dñ
v
ý 12(2)(2ð )
18
2
dö r dr (10 )
3
r 2.6 18
(10 ) ý 16ð (2.63 1.23 )(1018 )
3 1.2
ý 796.611018 C
Prob. 5.20
109
(6,12, 20) ý 0.1114a x 0.2228a y 0.3714a z nC/m 2
36ð
109
(6,12, 20) ý 0.0584a x 0.1167a y 0.1945a z nC/m 2
P ý ó eõ o E ý 1.1x
36ð
D ý õ oõ r E ý 2.1x
Prob. 5.21
At P ø 2,5,3ù ,
x ý 2, y ý 5, z ý 3
V ý 4 ø 4 ùø 5 ùø 27 ù ý 2.16 kV
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ö öV
öV
öV ö
E ý ñV ý ÷
ax
ay
a z ÷ ý ø 8 xyz 3a x 4 x 2 z 3a y 12 x 2 yz 2 a z ù
öy
öz ø
ø öx
At P,
E ý (16)(5)(27)a x 4(4)(27)a y 12(4)(5)(9)a z ý 2.16a x 0.432a y 2.16a z kV/m
P ý ó eõ o E ý
7 109
(2160, 432, 2160) ý 133.69a x 26.74a y 133.69a z nC/m 2
36ð
Prob. 5.22
ö öV
öV
öV ö
(a) E ý ñV ý ÷
ax
ay
a z ÷ ý 20 xyza x 10 x 2 za y 10( x 2 y z )a z V/m
x
y
z
ö
ö
ö
ø
ø
(b) D ý õ E ý 5õ o E ý 0.8842 xyza x 0.4421x 2 za y 0.4421( x 2 y z )a z nC/m 2
(c ) P ý ó eõ o E ý 4õ o E ý 0.7073xyza x 0.3537 x 2 za y 0.3537( x 2 y z )a z nC/m 2
(d) ò v ý õñ 2V
ö
ö
ö
ñ 2V ý (20 xyz ) (10 x 2 z ) (10 x 2 y 10 z ) ý 20 yz 10
öx
öy
öz
ò v ý 5õ o10(2 yz 1) ý 0.8854 yz 0.4427 nC/m3
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Prob. 5.23
Using Gauss' law,
Qenc ý
D ÷ dS
For r < a, Qenc ý 0
E ý0ý DýP
For a < r <b, Qenc ý 4C
4 ý Dr (4ð r 2 )
Eý
D
õ
ý
Dý
a
4
a ý r2
2 r
4ð r
ðr
ar
D
ý
2õ o 2õ oð r 2
ar
2ð r 2
ý 4 6 ý 2C
P ý ó eõ o E ý (1)õ o E =
For b < r < c, Qenc
2 ý Dr (4ð r 2 )
Eý
D
õ
ý
Dý
a
2
a ý r 2
2 r
4ð r
2ð r
ar
D
ý
5õ o
10õ oð r 2
2ar
4 ar
ý2
10ð r
5ð r 2
ý 4 6 10 ý 8C
P ý ó eõ o E =(4)õ o E = For r >c, Qenc
8 ý Dr (4ð r 2 )
Eý
D
ý
Dý
8
2a
a ý r2
2 r
ðr
4ð r
2a r
õ oð r 2
õo
P ý ó eõ o E = (0)õ o E = 0
Prob. 5.24 (a) Applying Coulomb’s law, we obtain the electric field intensity due to a
point charge as
Q
ü Dr
ÿÿ õ ý 4ðõ r 2 , r þ b
o
Er ý ý o
ÿ Dr ý Q , a ü r ü b
ÿþ õ
4ðõ r 2
Pý
õ r 1
D
õr
(ý D õ O E )
Hence
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Pr ý
õr 1 Q
.
,
õ r 4ð r 2
aü rüb
ò p v ý ñ ÷ P ý
(b)
(c )
ò p s ý P ÷ ( a r ) ý
ò p s ý P ÷ (a r ) ý
1 d 2
(r Pr ) ý 0
r2 d r
Q õr 1
(
),
4ð a 2 õ r
Q õr 1
(
),
4ð b 2 õ r
rýa
rýb
Prob. 5.25
F1 ý
Q1Q2
Q1Q2
ý 2.6 nN,
ý 1.5 nN
F2 ý
2
4ðõ o d
4ðõ oõ r d 2
F1 2.6
ý
ý õ r ý 1.733
F2 1.5
Prob. 5.26
(a) By Gauss’s law,
D dS ý Q
Dr ý
enc
S
Er ý
Dr
õ
ý
Q
4ð r 2
Q
4ðõ r 2
1
W ý õ | E |2 dv,
2
v
dv ý r 2 sin ñ drdö dñ
2ð ð
1
Q2
Q2
2
ý
sin
ñ
ö
ñ
Wý õ
r
drd
d
2 ö ý0 ñ r ý a 16ð 2õ 2 r 4
8ðõ a
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(b) Dr remains the same but
D
Q
Q
ý
Er ý r ý
2
2
õ
4ðõ o ø r a ù
ö aö
4ð r 2õ o ÷1 ÷
ø rø
2ð
ð
1
1
Q 2 r 2 sin ñ drdñ dö ö r a ö
2
õo ÷
W ý õ | E | dv ý
÷
2
2 ö ý0 ñ ý0 r ý a 16ð 2õ 2 ø r a ù4
ø r ø
v
2
1 ö Q2 1
Q2
dr
Q2 ö
ð
(4
)
ý
ý
÷
a ø r a ù2 8ðõ o ø r a a ÷ø ý 8ðõ o 2a
32ð 2õ o
Q2
Wý
16aðõ o
Prob. 5.27
(a)
üò o ,
òv ý ý
þ 0,
0ü rü a
rþa
For r < a, õEr (4ðr 2 ) ý òo
4 ðr 3
3
V ý E ÷ dl ý
For r > a, õ o E r (4ðr 2 ) ý òo
At r = a,
òo r
3õ
òo a 3
3õ o r
Er ý
òo a 3
3õ o r 2
c2
V ý 0 and c2 ý 0
V(a+) = V(a-)
òo a 2
ò a2
c1 ý o
6õ o õ r
3õ o
V(r=0) = c 1 ý
(b)
Er ý
òo r 2
c1
6õ
4ða 3
3
V ý E ÷ dl ý
As r
,
c1 ý
òo a 2
(2õ r 1)
6õ o õ r
òo a 2(2õ r 1)
6õ o õ r
ò oa 2
V (r ý a ) ý
3õ o
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Prob. 5.28
ù Dx ù
ù4 1 1ù ù 1 ù
ú D ú ý õ E ú1 3 1 ú ú 1 ú
o o ú
ú yú
úú ú
ûú Dz ûú
ûú1 1 2 ûú úû 1ûú
Dx ý õ o Eo (4 1 1) ý 4õ o Eo
Dy ý õ o Eo (1 3 1) ý 3õ o Eo
Dz ý õ o Eo (1 1 2) ý 0
D ý õ o Eo (4a x 3a y ) C/m 2
Prob. 5.29
Since
öòv
ý 0,
öt
ñ ÷ J ý 0 must hold.
(a)
ñ ÷ J ý 6 x2 y 0 6 x2 y ý 0
(b)
ñ ÷ J ý y ( z 1) ù 0
(c)
ñ÷J ý
(d)
ñ÷J ý
1 ö
ò öò
( z 2 ) cos ö ù 0
1 ö
(sin ñ ) ý 0
r2 ö r
This is possible.
This is not possible.
This is not possible.
This is possible.
Prob. 5.30
öJ x öJ y öJ z
öò
ý 2e 2 y cos 2 x 2e 2 y cos 2 x 1 ý 1 ý v
öx
öy
öz
öt
öò v
ý 1 C/m3 s
Hence,
öt
ñJý
Prob. 5.31
(a) ñ ÷ J ý
1 ö 100
100
(
)ý 3
ò öò
ò
öòv
100
ý ñ÷J ý 3
öt
ò
ò
öò v 100
ý 3 C/m3 .s
öt
ò
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(b)
I ý J ÷ d S ý
100
ò2
ò d ö d z ò ý2 ý
100
2
2ð
1
0
0
dö d z ý 100ð ý 314.16 A
Prob. 5.32
ò ý òvo e t /T
r
where òvo is the initial value (t=0). When t ý 80 ý s,
6
1
òvo ý òvo e (8010 )/Tr
2
80 106
Tr ý
ý 115.42 ý s
ln 2
Tr ý
õ
ó
óý
õ
Tr
ý
(80 106 ) / Tr ý ln 2
109
ln 2
36ð
ý 5.746 107 S/m
6
80 10
7.5
Prob. 5.33
From the continuity equation,
öò
ñ÷J ý v
(1)
öt
But
J ý òv u
Applying the vector identity
ñ • (VA) ý V ñ ÷ A A ÷ ñV
ñ ÷ J ý ò v (ñ ÷ u) u ÷ ñò v
(2)
Substituting (2) into (1) gives
öò
(u ÷ ñ) òv ò v (ñ ÷ u) v ý 0
öt
as required.
Prob. 5.34
1
2
ò ý òvo . Then,
öòv
öJ
ý ñ ÷ J ý x ý 0.5ð cos ð x
öt
öx
At P(2,4,-3), x= 2
öòv
ý 0.5ð cos(2ð ) ý 0.5ð ý 1.571 C/m 2 s
öt
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Prob. 5.35
(a)
õ
ý
ó
3.1
10
109
36ð ý 2.741 104 s
15
109
õ
36ð ý 5.305 104 s
ý
ó
1015
6
(b)
õ
ý
ó
(c)
109
36ð ý 7.07 ý s
104
80
Prob. 5.36
Tr ý
2.5 109
õ
ý
ý 4.42 ý s
ó 5 106 36ð
òvo ý
1
Q
ý
ý 29.84 kC / m3
V 4ð 106 8
3
òv ý òvo e t / T ý 29.84e2 / 4.42 ý 18.98 kC/m3
r
Prob. 5.37
The normal component of a solenoidal vector field is continuous across an interface.
Hence, since ñ ÷ J =0,
J 1n ý J 2 n
Similarly, the tangential component of a curl-free vector field is continuous.
Hence, since ñ (J /ó )=0,
J1t ó 1
ý
J 2t ó 2
Alternatively,
E1n ý E2 n
J1t
ó1
ý
J 2t
ó2
or
J1t ó 1
ý
J 2t ó 2
as required.
Prob. 5.38
(a)
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P1 ý ó e1õ o E1 ý 3
109
(60, 100, 40) ý 1.591a x 2.6526a y 1.061a z nC/m 2
36ð
(b)
E2t ý E1t ý 60a x 100a y
D2 n ý D1n
E2 n ý
õ 2 E2 n ý õ1 E1n
õ1
4
(40a z ) ý 21.33a z
E1n ý
õ2
7.5
E2 ý 60a x 100a y 21.33a z
109
(60, 100, 21.33) ý 3.979a x 6.631a y 1.414a z nC/m 2
D2 ý õ oõ r 2 E2 ý 7.5
36ð
Prob. 5.39
(a)
E2t ý E1t ý 10a y 8a z
D2 n ý D1n
E1n ý
õ 2 E2 n ý õ1 E1n
2õ
õ2
E2 n ý o (6a x ) ý 3a x
4õ o
õ1
E1 ý 3a x 10a y 8a z
109
P1 ý ó e1õ o E1 ý 3
(3, 10,8) ý 79.6a x 265.3a y 212.2a z nC/m 2
36ð
109
(6, 10,8) ý 53.05a x 88.42a y 70.74a z nC/m 2
P2 ý ó e 2õ o E2 ý 1
36ð
(b)
1
1
w1 ý õ1 E12 ý (4õ o )(9 100 64) ý 346õ o ý 3.0593 nJ/m 2
2
2
1
1
w2 ý õ 2 E22 ý (2õ o )(36 100 64) ý 200õ o ý 1.7684 nJ/m 2
2
2
Prob. 5.40
f(x,y)= 4x +3y –10=0
(4a x 3a y )
ñf
ý
ý 0.8a x 0.6a y
| ñf |
5
The minus sign is chosen for an because it is directed toward the origin.
ñf ý 4a x 3a y
an ý
D1n ý ( D1 an )an ý (1.6 2.4)an ý 0.64a x 0.48a y
D1t ý D1 D1n ý 2.64a x 3.52a y 6.5a z
D2 n ý D1n ý 0.64a x 0.48a y
E2t ý E1t
D2t
õ2
ý
D1t
õ2
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D2t ý
õ2
2.5
(2.64, 3.52, 6.5) ý (6.6, 8.8,16.25)
D1t ý
1
õ1
D2 ý D2 n D2t ý 5.96a x 9.28a y 16.25a z nC/m 2
ñ 2 ý cos 1
D2 an
ý 87.66o
| D2 |
Prob. 5.41
(a)
Let f ( x, y, z ) ý x 2 y z 1 ý 0
1
ñf
ý
( a x 2a y a z )
| ñf |
6
1
1
E1n ý ( E1 ÷ an )an ý
(20 20 40)
(a x 2a y a z ) ý 6.667a x 13.33a y 6.667a z V/m
6
6
E1t ý E1 E1n ý 13.3a x 23.3a y 33.3a z V/m
ñ f ý a x 2a y a z
an ý
(b)
E2t ý E1t ý 13.3a x 23.3a y 33.3a z
D2 n ý D1n
E2 n ý
õ 2 E2 n ý õ1 E1n
õ1
2
E1n ý (6.667,13.33, 6.667) ý 2.7a x 5.3a y 33.3a z
õ2
5
E2 ý E2t + E 2 n ý 16a x 18a y 36a z V/m
Prob. 5.42
109
(10, 6,12) ý 0.1768a x 0.1061a y 0.2122a z nC/m 2
(a) P1 ý õ o ó e1 E1 ý 2
36ð
(b)
E1n ý 6a y ,
D2 n ý D1n
or
E2 n ý
E2t ý E1t ý 10a x 12a z
õ 2 E2 n ý õ1 E1n
3õ o
õ1
(6a z ) ý 4a y
E1n ý
4.5õ o
õ2
E2 ý 10a x 4a y 12a z V/m
E
tan ñ 2 ý 2 t ý
E2 n
10 2 12 2
ý 3.905
4
ñ 2 ý 75.64 o
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(c ) wE ý
w E1 ý
w E2
1
1
D÷E ý õ | E |
2
2
2
1
1
109
(102 62 122 ) ý 3.7136 nJ/ m 3
õ1| E1| 2 ý x3 x
2
2
36ð
1
1
109
2
(102 42 122 ) ý 5.1725 nJ/ m 3
ý õ 2 | E2 | ý x4.5 x
2
2
36ð
Prob. 5.43 (a) D2 n ý 12a ò ý D1n ,
E 2t ý E 2t
D1t ý
D1t
õ1
ý
D2t ý 6aö 9a z
D2t
õ2
3.5õ o
õ1
(6aö 9a z ) ý 14aö 21a z
D2t ý
1.5õ o
õ2
D1 ý 12a ò 14aö 21a z nC/m 2
(12, 14, 21) 109
ý 387.8a ò 452.4aö 678.6a z
109
3.5
36ð
0.5õ o
D
(12, 6,9) ý 4aò 2aö 3a z nC/m 2
(b) P2 ý õ o ó e 2 E2 ý 0.5õ o 2 ý
õ 2 1.5õ o
E1 ý D1 / õ1 ý
òv 2 ý ñ ÷ P2 ý 0
(c)
w E1 ý
w E2 ý
1
1 D1 ÷ D1 1 (122 142 212 )x10 18
ý
ý 12.62 ý J/ m 2
D1 ÷ E1 ý
9
2
2 õ o õ r1
2
10
3.5 x
36ð
1 D2 ÷ D2 1 (122 62 92 )x10 18
ý
ý 9.839 ý J/ m 2
9
2 õ oõ r 2
2
10
1.5 x
36ð
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Prob. 5.44
.
Gaussian surface
.
.
Q ý D.dS ý õ1 Er
4ð r 2
4ð r 2
õ 2 Er
ý 2ð r 2 (õ1 õ 2 ) Er
2
2
Q
ü
,
rþa
ÿ
E r ý ý 2ð (õ1 õ 2 )r 2
ÿ
0,
rüa
þ
Prob. 5.45
E1t ý E2t ý 5cos ñ añ
D1n ý D2 n
õ1 E1n ý õ 2 E2 n
6õ
õ2
E2 n ý o (10sin ñ )a r ý 30sin ñ ar
2õ o
õ1
E1t ý E1t + E1n ý 30sin ñ ar 5cos ñ añ
E1n ý
D1 ý õ1 E1 ý 2õ o E1 ý õ o (60sin ñ ar 10 cos ñ añ )
Prob. 5.46
(a) The two interfaces are shown below
oil
1
glass
2
glass
air
2
oil-glass
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glass-air
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E1n ý 2000 ,
E1t ý 0 ý E2 t ý E 3 t
D1n ý D2 n ý D3 n
õ 1 E1n ý õ 2 E 2 n ý õ 3 E 3 n
E2 n ý
õ1
3.0
E1 n ý
(2000 ) ý 705.9 V / m, ñ 2 ý 0 o
õ2
8.5
E3 n ý
õ1
3.0
E1 n ý
(2000 ) ý 6000 V / m, ñ 3 ý 0 o
.
õ3
10
(b)
oil
glass
glass
ñ2
air
ñ 1 ý 75 o
1
2
2
E1n ý 2000 cos75 o ý 517 .63,
E1t ý 2000 sin75 o ý E 2 t ý E 3 t ý 1931.85
E2n ý
õ1
3
E1n ý
(517 .63) ý 1827
. ,
õ2
8.5
E3n ý
õ1
3
E1n ý (517 .63) ý 1552.89
õ3
1
E2 ý
E2 n 2 E2 t 2 ý 1940.5,
ñ 2 ý tan 1
E2 t
ý 84.6 o ,
E2 n
E3 ý
E3 n 2 E3 t 2 ý 2478.6 ,
ñ 3 ý tan 1
E3 t
ý 51.2 o
E3 n
Prob. 5.47
109
2900
302 402 202 103 ý
pC/m 2
36ð
36ð
2
ý 0.476 pC/m
ò s ý Dn ý õ o E ý
Prob. 5.48 (a) ò s ý Dn ý õ o En ý
(b)
10 9
15 2 8 2 ý 0.1503 nC / m2
36 ð
Dn ý ò s ý 20 nC/m 2
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D ý Dn an ý ( 20 nC)(-a y ) ý 20 a y nC / m2
Prob. 5.49
At the interface between õ o and 2õ o ,
E1n ý Eo cos 30o ,
E1t ý Eo sin 30o
E2t ý E1t ý 0.5 Eo
D2 n ý D1n
E2 n ý
õ1 E1n ý õ 2 E2 n
õ
õ1
E1n ý o (0.866 Eo ) ý 0.433Eo
õ2
2õ o
The angle E makes with the z-axis is
E
0.5
ñ1 ý tan 1 2t ý tan 1
ý 49.11o
E2 n
0.433
At the interface between 2õ o and 3õ o ,
E3t ý E2t ý 0.5 Eo
D3n ý D2 n
E3n ý
2õ
õ2
E2 n ý o (0.433Eo ) ý 0.2887 Eo
3õ o
õ3
The angle E makes with the z-axis is
E
0.5
ñ 2 ý tan 1 3t ý tan 1
ý 60o
E3n
0.2887
At the interface between 3õ o and õ o ,
E4t ý E3t ý 0.5 Eo
D4 n ý D3n
E4 n ý
õ3
3õ
E3n ý o (0.2887 Eo ) ý 0.866 Eo
õ4
õo
The angle E makes with the z-axis is
E
0.5
ñ3 ý tan 1 4t ý tan 1
ý 30o
E4 n
0.866
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CHAPTER 6
P. E. 6.1
ñ 2V ý
ò
õ
ò x
d 2V
ý o
2
õa
dx
ò o x3
Vý
Ax B
6õa
ö ò x2
ö
dV
Eý
ax ý ÷ o A ÷ ax
dx
ø 2õ a
ø
If E = 0 at x =0, then
0 ý0 A
Aý0
If V= 0 at x =a, then
0ý
òo a 3
B
6 õa
Bý
òo a 2
6õ
Thus
òo
(a 3 x 3 ),
6õ a
Vý
V1 ý A1 x B1 ,
P. E. 6.2
V1 ( x ý d ) ý Vo ý A1d B1
V1 ( x ý 0 ) ý 0 ý 0 B2
A1a Vo A1d ý
õ1
aA1
õ2
B1 ý Vo A1d
B2 ý 0
õ 1 A1 ý õ 2 A2
òo x 2
ax
2õ a
V2 ý A2 x B2
V1( x ý a ) ý V2( x ý a )
D1n ý D2 n
Eý
a A1 B1 ý A2 a
A2 ý
õ1
A1
õ2
ö
õ ö
Vo ý A1 ÷ a d 1 a÷
õ2 ø
ø
or
A1 ý
Vo
,
d a õ 1a / õ 2
A2 ý
õ1
õ 1Vo
A1
õ2
õ 2 d õ 2 a õ 1a
Hence
E1 ý A1a x ý
Vo a x
,
d a õ1a / õ 2
E2 ý A2 a x ý
Vo a x
a õ 2 d / õ1 õ 2 a / õ1
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P. E. 6.3 From Example 6.3,
V
E ý o aö , D ý õ o E
òöo
ò s ý Dn (ö ý 0 ) ý
Vo õ
òö o
The charge on the plate ö ý 0 is
Q ý ò s dS ý
Cý
Vo õ
öo
L
b
zý0 òý a
Võ
1
dzdò ý o L ln(b / a )
öo
ò
| Q| õL b
ý
ln
Vo ö o a
4mm
a
45o
a
a sin
Cý
45 o
ý2
2
aý
2
sin 22.5 o
ý 5.226 mm
109
36ð 5ln 1000 ý 444 pF
ð
5.226
4
1.5
Q ý CVo ý 444 10 12 50 C ý 22.2 nC
P. E. 6.4 From Example 6.4,
Vo ý 50,
ñ2 ý 45o ,
ñ1 ý 90o ,
r ý 32 42 22 ý 29 ,
5
ñ ý 68.2o ;
ta n 45o ý 1
2
50ln(ta n 34.1o )
ý 22.125 V,
Vý
ln(ta n 22.5o )
50añ
Eý
ý 11.35añ V/ m
29 sin68.2o ln(ta n 22.5o )
ta n 1
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ñ ý ta n 1
ò
z
ý
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P. E. 6.5
From Example 6.5,
V( x, y ) ý
sin(nð x / b )sinh[ nð y / b ]
n sinh(nð a / b )
ð nýod d
4Vo
õ
E ý ñV ý
ý
4Vo
b
öV
öV
ax
ay
öx
öy
1
õ sinh nð a/b ùûcos(nð x/b)sinh (nð y/b) a
n ý odd
x
sin(nð x/b)cosh(nð y/b)a y ùû
(a) At (x,y) = (a, a/2),
Vý
400
ð
(0.3775 0.0313 0.00394 0.000585 ...) ý 44.51 V
E ý 0a x (115.12 19.127 3.9411 0.8192 0.1703 0.035 0.0074 ...)a y
ý 99.25a y V/ m
(b) At (x,y) = (3a/2, a/4),
Vý
400
ð
(0.1238 0.006226 0.00383 0.0000264 ...) ý 16.50 V
E ý (24.757 3.7358 0.3834 0.0369 0.00351 0.00033 ...)a x
(66.25 4.518 0.3988 0.03722 0.00352 0.000333 ...)a y
ý 20.68a x 70.34a y V/ m
P. E. 6.6
V ( y ý a ) ý Vo sin(7 ðx / b) ý
õc
n
sin(nðx / b) sinh(nða / b)
ný1
By equating coefficients, we notice that cn = 0 for n ù 7 . For n=7,
Vo sin(7 ðx / b) ý c7 sin(7 ðx / b) sinh(7 ða / b)
c7 ý
Vo
sinh(7 ða / b)
Hence
V ( x, y) ý
Vo
sin(7 ð x / b) sinh(7 ð y / b)
sinh(7 ð a / b)
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P. E. 6.7 Let V (r , ñ , ö ) ý R(r )F (ñ )ö (ö ).
Substituting this in Laplace’s equation in spherical coordinates gives
öF d ö 2 dR ö
Rö d ö
dF ö
RF d 2 ö
sin
ñ
r
ý0
÷
÷
÷
÷
dñ ø r 2 sin 2 ñ dö 2
r 2 dr ø dr ø r 2 sin ñ dñ ø
Dividing by RFö / r 2 sin 2 ñ gives
sin 2 ñ d 2
sin ñ d
1 d 2ö
ø sin ñ F 'ù ý
ý ü2
r R'ù
ø
R dr
F dñ
ö dö 2
ö '' ü 2ö ý 0
d
1 d 2
1
ø sin ñ F 'ù ý ü 2 / sin 2 ñ
r R'ù
ø
R dr
F sin ñ dñ
1 d 2
ü2
1
d
ø sin ñ F 'ù ý ý 2
r R'ù ý
ø
2
sin ñ F sin ñ dñ
R dr
2 rR' r 2 R'' ý ý 2 R
or
R''
2
ý2
R ' 2 R ý 0
r
r
sin ñ d
ø sin ñ F 'ù ü 2 ý 2 sin 2 ñ ý 0
F dñ
or
F '' cos ñ F ' ( ý 2 sin ñ ü 2 cs cñ ) F ý 0
P. E. 6.8 (a) This is similar to Example 6.8(a) except that here 0 ü ö ü 2 ð instead of
0 ü ö ü ð / 2 . Hence
b
Vo ln a
2ð tVo ó
Iý
and
Rý
ý
I 2ð tó
ln(b / a )
(b) This to similar to Example 6.8(b) except that here 0 ü ö ü 2 ð . Hence
Vó
Iý o
t
b 2ð
a
Vo ó ð (b 2 a 2 )
0 ò dò dö ý
t
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and R ý
Vo
t
ý
2
I ó ð (b a 2 )
From Example 6.9,
P. E. 6.9
J1 ý
Iý
ó 1Vo
,
b
ò ln
a
J2 ý
ù ð
ú öý0 J1ò dö
zý0 ú
û
L
J ÷ dS ý
b
ln
Vo
a
Rý
ý
I
ðl ó 1 ó 2
P. E. 6.10 (a)
Cý
ù
2ð
J ò dö úúdz ý
2
öýð
û
Vo l
ðó 1 ðó 2
b
ln
a
ý
ý
4ð õ
,
1 1
a b
ö
109 ÷
2.5
C1 ý 4ð x
÷ 3
36ð ÷ 10 103
÷
3
ø 2
Cý
ó 2Vo
b
ò ln
a
C1 and C2 are in series.
ö
÷
÷ ý 5/ 3 p F,
÷
÷
ø
ö
109 ÷
3.5
C 2 ý 4ð x
÷ 3
36ð ÷ 10 103
÷
2
ø 1
C1C2
(5 / 3)(7 / 9 )
ý
ý 0.53 pF
C1 C2 (5 / 3) (7 / 9 )
2ð õ
, C1 and C2 are in parallel.
1 1
a b
ö
ö
ö
109 ÷ 2.5 ÷
109 ÷ 3.5
C1 ý 2ð
÷
÷ ý 5 / 24 pF, C2 ý 2ð
÷
36ð ÷ 103 103 ÷
36ð ÷ 103 103
÷
÷
÷
3 ø
3
ø 1
ø 1
(b)
ö
÷
÷ ý 7/ 9 p F
÷
÷
ø
Cý
ö
÷
÷ ý 7 / 24 pF
÷
÷
ø
C ý C1 C2 ý 0.5 pF
P. E. 6.11 As in Example 6.8, the solution of Laplace’s equation yields
V ( ò ) ý A ln ò B
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Using the boundary conditions V (ò ý a ) ý 0 ,
0 = Alna + B
Vo = Aln b + B
Solving this yields
V ý Vo
ln ò / a
,
ln b / a
E ý ñV ý
Võ
Q ý õE ÷ dS ý o
ln b / a
Cý
L 2ð
zý0 ö ý0
V (ò ý b) ý Vo ,
Vo
a
ò ln b / a ò
V 2 ðõL
1
dzòdö ý o
ò
ln b / a
Q 2ð õ L
ý
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 ý
2ðõ r1õ o 2ð x2.5 10 9
ý
ý 342.54 p F/ m ,
ln b / c
ln 3 / 2 36ð
C2 ý
2ðõ r 2õ o 2ð x3.5 10 9
ý
ý 280.52 p F/ m
ln c / a
ln 2 36ð
342.54 x280.52
C1C 2
ý
ý 154.22 p F
C1 C 2 342.54 280.52
C T ý Cl ý 1.54 nF
Cý
(b) C1 and C2 are in parallel.
C ý C1 C 2 ý
ðõ r1õ o
ln b / a
CT ý Cl ý 1.52 nF
ðõ r 2õ o
ln b / a
ý
ð (õ r1 õ r 2 )õ o
ln b / a
6ð 109
ý
ý 151.7 p F/ m
ln 3 36ð
P. E. 6.13 Instead of Eq. (6.31), we now have
a
V ý
Qdr
4ðõr
b
2
a
ý
Qdr
4ð 10õ
b
r
o
ý
r2
Q
40 ðõ o
ln b / a
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40 ð 10 9
Q
Cý
ý
ý 113
. nF
|V | ln 4 / 1.5 36 ð
P. E. 6.14 Let
F ý F1 F2 F3 F4 F5
i ý 1,2,...,5
where Fi ,
Fý
Q2
4ðõ o r 2
ay
are shown on in the figure below.
Q 2(a x sin 30o a y c o s30o )
4ðõ o (2 r c os30o )2
Q 2(a x c os30o a y sin 30o )
4ðõ o (2 r)2
Q 2(a x c os30o a y sin 30o )
Q 2a x
4ðõ o (2 r c o s30o )2
4ðõ o r 2
ý
ù
3a y
1 ö ax
ú
÷
a
y
2
3÷ 2
2
4ðõ o r ú
ø
û
Q2
ö 1 ö 3a
a ö 1
3a x a y ù
x
÷ ÷
y ÷ ax
ú
÷ 4÷ 2
2 ÷ø 3
2
2ú
ø
ø
û
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ù ö1 5 3ö
ö 5
3 öù
ý 9x105 ú a x ÷
ay ÷
÷
÷÷ ú ý 52.4279a x 30.27a y ýN
÷
÷
÷
2
8
8
6
ø
ø
ø úû
ûú ø
| F| ý 60.54 ýN
Note that the force tends to pull Q toward the origin.
Prob. 6.1
(a)
öV
öV ö
ö öV
E ý ñV ý ÷
ax
ay
az ÷
öz
öz ø
ø öx
ý (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 ý 15(9)(1)(2)a x 10(27)(1)(2)a y 5(27)(1)a z ý 270a x 540a y 135a z V/m
òv ý ñ ÷ D
(b)
ñ 2V ý
or
òv ý õñ 2V
ö 2V ö 2V ö 2V ö
ö
ö
2 2 ý (15 x 2 y 2 z ) (10 x3 yz ) (5 x3 y 2 )
2
öx
öy
öy
öx
öy
öz
ý 30 xy 2 z 10 x3 z
At P,
òv ý õñ 2V ý 2.25
109
30(3)(1)(2) 10(27)(2)ý ý 14.324 nC/m3
36ð
Prob. 6.2
(a)
20
10
1 10
cos ñ sin ö ar 3 sin ñ sin ö añ +
cos ñ cos ö aö
3
r
r
rsinñ r 2
At P(1,60o ,30o ), r ý 1, ñ ý 60o , ö ý 30o
E ý ñV ý
20
10
1 10 cos 60o cos 30o
o
o
o
o
cos
60
sin
30
sin
60
sin
30
a
a
aö
r
ñ
13
13
sin60o
13
ý 5ar 4.33añ 5aö V/m
Eý
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(b)
ö ö 10sin 2 ñ sin ö ö
1 ö ö 20 cos ñ sin ö ö
1
÷
÷
÷
÷ 2
r 2 ör ø
r
r2
ø r sin ñ öñ ø
ø
1
10 cos ñ sin ö
2 2
r sin ñ
r2
20 cos ñ sin ö 20sin ñ cos ñ sin ö 10 cos ñ sin ö
ý
r4
r 4 sin ñ
r 4 sin 2 ñ
10 cos ñ sin ö
ý
r 4 sin 2 ñ
10õ o cos ñ sin ö
ò
ñ 2V ý v
ò v ý õ ñ 2V ý
r 4 sin 2 ñ
õ
At P, r =1, ñ =60o , ö =30o
ñ 2V ý
ò v =10
10-9 cos60o sin 30o
=29.47 pC/m3
4
2
o
36ð 1 sin 60
Prob. 6.3
ñ 2V ý
òv
õ
d 2V
y 109
y 109
ý
ý
ý 2.25 y
dy 2
4ð 4õ o
4ð 109
4
36ð
2
dV
y
ý 2.25 B
dy
2
3
V ý 0.375 y By C
V (1) ý 0 ý 0.375 B C
V (3) ý 50 ý 10.125 3B C
From (1) and (2), B=29.875 and C=-29.5
V ý 0.375 y 3 29.875 y 29.5
V (2) ý 27.25 V
(1)
(2)
Prob. 6.4
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E ý ñV ý
ñ 2V ý
dV
a ò ý (0.8)(10) ò 0.2 a ò ý 8(0.6) 0.2 a ò ý 8.861a ò
dò
òv 1 d
1 d ö dV ö
1
8ò 0.8 ù ý (8)0.8ò 0.2 ý 6.4 ò 1.2
ý
ø
÷ò
÷ý
ò dò ø dò ø
õo ò d ò
ò
òv ý õ oñ 2V ý 6.4
109 1.2
6.4(0.6) 1.2
nC/m3 ý 0.1044 nC/m3
ò ý
36ð
36ð
Prob. 6.5
ñ 2V ý
where
ký
òv
d 2V
50(1 y 2 )x10 6
ý
ý
ý k(1 y 2 )
2
õ
õ
dx
50 106
ý 600ð 103
109
3
36ð
dV
ý k ( y y 3 / 3) A
dy
ö y2 y4 ö
V ý k ÷
÷ Ay B ý 50 ð.10 3 y 4 300 ð.10 3 y 2 Ay B
2
12 ø
ø
When y=2cm,
V=30X103,
30 103 ý 50ð 103 16 106 300ð 103 4 104 Ay B
or
30,376.77 ý 0.02A B
When y=-2cm,
(1)
V=30x103,
30,376.77 ý 0.02A B
From (1) and (2), A=0, B=30,376.77.
(2)
Thus,
V ý 157.08 y 4 942.5 y 2 30.377 kV
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Prob. 6.6
òv
ò z
d 2V
ý o
2
õ
õd
dz
2
ò z
dV
ý o A
dz
2õ d
ò z3
V ý o Az B
6õ d
ñ 2V ý
z ý 0,V ý 0
z ý d ,V ý Vo
0 ý 0 B,
Vo ý
Vo ò o d
d
6õ
Hence,
i.e. B ý 0
òo d
Ad
6õ
2
Aý
òo z 3 ö Vo òo d ö
V ý
z
6õ d ÷ø d
6õ ÷ø
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Prob. 6.7
ñ 2V ý
òv
ý
õ
10
ò
3.6
Let ñ ý 0.1ð .
ñ 2V ý
ñ ý
1012
9
10
36ð
ý
0.1ð
ò
ñ 1 d ö dV ö
ý
ò
ò ò d ò ÷ø d ò ÷ø
d ö dV ö
÷ò
÷
dò ø dò ø
dV
ý ñò A
dò
dV
A
ý ñ
dò
ò
V ý ñò A ln ò B
ò
At ò =2, V=0
0 ý 2ñ A ln 2 B
At ò =5, V=60
60 ý 5ñ A ln 5 B
Subtracting (1) from (2),
60 3ñ
60 ý 3ñ A ln 5 / 2
Aý
ý 66.51
ln 2.5
From (1),
B ý 2ñ A ln 2 ý 45.473
dV
A
66.51
)a ò
Eý
a ò ý (ñ )a ò ý (0.3142
dò
ò
ò
(1)
(2)
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Prob. 6.8
òo
1 d ö dV ö
÷ò
÷ý
ò dò ø dò ø
õ
òo ò
d ö dV ö
÷ò
÷ý
õ
dò ø dò ø
Integrating gives
ò ò2
dV
ý o
A
dò
õ 2
Integrating again,
ò
V ý
ò
dV
A
ý o ò
dò
2õ
ò
òo ò 2
A ln ò B
4õ
where A and B are integration constants.
Prob. 6.9
òv
10 109
60ð
ö
ý
ý
ý
÷
9
10
r
õ
ø
6r
36ð
d ö 2 dV ö
dV
r2
ý 30ð r 2 A
÷r
÷ ý 60ð r
dr ø dr ø
dr
dV
A
A
ý 30ð 2
V ý 30ð r B
dr
r
r
(1)
V (r ý 1) ý 0 0 ý 30ð A B
ñ 2V ý
1 d ö 2 dV
÷r
r 2 dr ø dr
V (r ý 4) ý 50 50 ý 120ð A / 4 B
Solving (1) and (2) yields A ý 443.66, B ý 537.91
Thus,
443.66
537.91
V ý 30ð r
r
443.66
537.91 ý 127.58 V
V (r ý 2) ý 60ð
2
(2)
Prob. 6.10
(a)
ö 2V1 ö 2V1 ö 2V1
ý 002 ù 0
öx 2 öy 2 öz 2
It does not satisfy Laplace's equation.
ñ 2V1 ý
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(b)
ñ 2V2 ý
1 ö
1 ö 10sin ö ö 10 2
10sin ö
ù ò ( ò 210sin ö ) ûù 2 ÷
ý0
÷ ý ( ò ) sin ö
û
ò öò
ò ø
ò ø ò
ò3
It does satisfy Laplace's equation.
( c)
1 ö 2
1
(5cos ñ ) ù
ö ù
ùû r ( r 2 5sin ñ ) ùû 2
sin ñ
2
ú
r ör
r sin ñ öñ û
r 2 úû
5
ý 0 3
(1 2sin 2 ñ ) ù 0
r sin ñ
It does not satisfy Laplace's equation.
ñ 2V3 ý
Prob. 6.11
ö 2U ö 2U ö 2U
ñ U ý 2 2 2 ý 6 xy 0 2c ý 0
öx
öy
öz
c ý 3xy
2
Prob. 6.12
(a)
öV
ý 4 xyz,
öx
ö 2V
öx 2
öV
ý 2 x 2 z 3 y 2 z,
öy
öV
ý 2x2 y y3 ,
öz
ý 4 yz
ö 2V
öy 2
ö 2V
öz 2
ý 6 yz
ý0
ñ 2V ý 4 yz 6 yz 0 ý 2 yz
ñ 2V ù 0 ,
V does not satisfy Laplace’s equation.
(b)
ñ 2V ý
òv
ý 2 yz
õ
1 1 1
òv ý 2 yzõ
Q ý ò v d v ý (2yzõ )d xd yd z ý 2õ (1)
0 0 0
y 2 1 z2 1
ý õ / 2 ý 2õ o / 2 ý õ o
2 0 2 0
Q = 8.854 pC
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Prob. 6.13
ñ 2V ý
1 d ö dV ö
ò
ý0
ò d ò ÷ø d ò ÷ø
V ý A ln ò B
Let a = 1 cm, b = 1.5 cm, Vo ý 50V
V ( ò ý b) ý 0
V ( ò ý a) ý Vo
0 ý A ln b B or B ý A ln b
Vo ý A lna Alnb ý Aln
V ý A ln ò A lnb ý Aln
ò
V
a
or A ý o
a
b
ln
b
b
A
V
dV
aò ý aò ý o aò
a
ò
dò
ò ln
b
9
10
õ oõ rVo 50(4) 36ð
400(50)
ý 2
ý
nC/m 2 ý 436.14 nC/m 2
ò s ý Dn ý õ En ý
b 10 ln1.5 36ð ln1.5
a ln
a
E ý ñV ý
Prob. 6.14
ñ 2V ý
d 2V
dz 2
ý0
V ý Az B
When z=0, V = 0
B=0
When z=d, V = Vo
Vo=Ad or A = Vo/d
Hence,
Vý
Vo z
d
V
dV
az ý o az
dz
d
V
D ý õ E ý õ oõ r o a z
d
Since Vo = 50 V and d = 2mm,
E ý ñV ý
V = 25z kV, E = - 25az kV/m
Dý
109
(1.5)25 103 a z ý 332a z nC/m 2
36ð
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ò s ý Dn ý 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 ñ 2V ý 0 when V ý V (ò) leads to
Vo ln ò / a
ln(a / ò )
ý Vo
ln b / a
ln(a / b )
Vo
Vo
E ý ñV ý
aò ý
aò ,
ò ln b / a
ò ln a / b
Vý
ò s ý Dn ý
õ o õ rVo
ò ln b / a
D ý õEý
õ o õ rVo
aò
ò ln b / a
ò ý a ,b
In this case, Vo=100 V, b=5mm, a=15mm, õ r ý 2. Hence at ò = 10mm,
Vý
100ln(10 / 15)
ý 36.91 V
ln(5/ 15)
100
a ò ý 9.102a ò kV/ m
10 x103 ln 3
109
2a ò ý 161a ò nC/ m 2
D ý 9.102 x103 x
36ð
Eý
ò s (ò ý 5 mm) ý
10 5
10 9
( 2)
ý 322 nC / m2
5 ln 3
36 ð
ò s (ò ý 15 mm) ý
10 5
10 9
( 2)
ý 107 .3 nC / m2
15 ln 3
36 ð
Prob. 6.16
1 d 2V
ý0
ò dö 2
V ý Aö B
0 ý 0 B
d 2V
ý0
dö 2
dV
ýA
dö
Bý0
50 ý Að / 2
E ý ñV ý
Aý
100
ð
A
1 dV
100
aö ý aö ý
a
ò dö
ò
ðò ö
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Prob. 6.17
(a)
öV
a2
ý Vo (1 2 ) sin ö
ò
öò
ò
a2
öV
ý Vo ( ò ) sin ö
ò
öò
ö ö öV
÷ò
öò ø öò
ö
a2
(1
) sin ö
V
ý
÷ o
ò2
ø
1 ö ö öV ö
1 a2
ò
ý V ( ) sin ö
ò öò ÷ø öò ÷ø o ò ò 3
a2
ö 2V
ò
V
(
) sin ö
ý
o
ò
öö 2
1 ö 2V
1 a2
V
(
ý
) sin ö
o
ò 2 öö 2
ò ò3
ñ 2V ý
1 ö ö öV
ò
ò öò ÷ø öò
ö ö 2V
÷ 2 ý 0
ø öö
(b)
2
a 2 , then a
If ò 2
E ý ñV ý
ò2
1 and V
Vo ò sin ö
öV
1 öV
aò
a ý Vo sin ö aò Vo cos ö aö
ò öö ö
öò
Prob. 6.18
d 2V
V ý Ax B
ñ 2V ý 2 ý 0
dx
At x ý 20 mm ý 0.02 m, V ý 0
0 ý 0.02 A B
(1)
E = -100 a x
From (1)
Then
At x ý 0
dV
ax
dx
A ý 110
(2)
B ý 0.02 A ý 2.2
V ý 110 x 2.2
V ý 2.2V
At x ý 50 mm ý 0.05 m,
V ý 110 0.05 2.2 ý 3.3V
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Prob. 6.19
ñ 2V ý 0
V ý 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 ý
100
150
r
E ý ñV ý
100
A
a ý 2 ar V/m
2 r
r
r
Prob. 6.20 From Example 6.4,
Vý
Vo ý 100 ,
ö tan ñ / 2 ö
Vo ln÷
÷
ø tan ñ 1 / 2 ø
ö tan ñ 2 / 2 ö
ln÷
÷
ø tan ñ 1 / 2 ø
ñ 1 ý 30 o ,
ñ 2 ý 120 o ,
r ý 3 2 0 2 4 2 ý 5,
ñ ý tan 1 ò / z ý tan 1 3 / 4 ý 36 .87 o
ö tan 18.435 o ö
ln÷
÷
ø tan 15 o ø
. V
V ý 100
ý 117
ö tan 60 o ö
ln÷
÷
ø tan 15 o ø
sec 2 ñ / 2
Vo añ
Vo añ
1 öV
tan ñ1 / 2
Eý
añ ý
ý
r öñ
ö tan ñ 2 ö tan ñ / 2
ö tan ñ 2 ö
r ln ÷
÷ tan ñ / 2 r ln ÷
÷ 2sin(ñ / 2) cos(ñ / 2)
1
ø tan ñ1 ø
ø tan ñ1 ø
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Eý
Vo añ
100añ
ý
ý 17.86añ V/m
ö tan ñ 2 / 2 ö 5sin 36.87 o ln 6.464
r sin ñ ln ÷
÷
ø tan ñ1 / 2 ø
Prob. 6.21
(a)
ñ 2V ý
1 ö ö öV ö
÷ ý0
÷ò
ò öò ø öò ø
V (ò ý b ) ý 0
V ( ò ý a) ý Vo
V ý
V ý A ln ò B
0 ý A ln b B
Vo ý A ln a / b
B ý A ln b
Aý
V ln b / ò
Vo
ln ò / b ý o
ln b / a
ln b / a
V (ò ý 15 mm) = 70
Vo
ln b / a
ln2
ý 12.4 V
ln50
(b) As the electron decelerates, potential energy gained = K.E. loss
e[70 12.4] ý
1
m[(107 ) 2 u 2 ]
2
u 2 ý 1014
1014 u 2 ý
2e
57.6
m
2 1.6 1019
57.6 ý 1012 (100 20.25)
31
9.1 10
u ý 8.93 106 m/s
Prob. 6.22 This is similar to case 1 of Example 6.5.
X ý c1 x c2 ,
But
X (0 ) ý 0
Y ý c3 y c4
0 ý c2 ,
Y (0 ) ý 0
0 ý c4
Hence,
V ( x , y ) ý XY ý ao xy ,
Also,
Thus,
V ( xy ý 4) ý 20
ao ý c1c3
20 ý 4ao
ao ý 5
V ( x, y ) ý 5 xy and E ý ñV ý 5 ya x 5 xa y
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At (x,y) = (1,2),
V ý 10 V, E ý 10a x 5a y V/m
Prob. 6.23 (a) As in Example 6.5, X ( x ) ý A sin(nðx / b)
For Y,
Y ( y ) ý c1 cosh(nðy / b) c2 sinh(nðy / b)
Y (a ) ý 0
Vý
õa
n
0 ý c1 cosh(nða / b) c2 sinh(nða / b)
c1 ý c2 tanh(nða / b)
sin(nðx / b) sinh(nðy / b) tanh(nða / b) cosh(nðy / b)ý
ný1
V( x, y ý 0) ý Vo ý õ a n ta nh(nð a / b )sin(nð x / b )
n ý1
ü 4Vo
b
, n ý od d
2
ÿ
a n ta nh(nð a / b ) ý Vo sin(nð x / b )d x ý ý nð
b0
ÿþ 0, n ý e ven
Hence,
V ý
ý
ù sinh(nð y / b )
c o sh(nð y / b ) ù
sin(nð x / b ) ú
ú
n
ð nýod d
û n ta nh(nð a / b )
û
4Vo
õ
sin(nð x / b )
ù sinh(nð y / b )c osh(nð a / b ) c osh(nð y / b )sinh(nð a / b )ùû
ð nýo d d n sinh(nð a / b ) û
4Vo
õ
ý
sin(nð x / b )sinh[ nð (a y )/ b ]
ð n ýo d d
n sinh(nð a / b )
4Vo
õ
Alternatively, for Y
Y ( y ) ý c1 sinh nð( y c2 ) / b
Y (a ) ý 0
0 ý c1 sinh[nð(a c2 ) / b]
c2 ý a
Vý
õb
n
sin(nðx / b) sinh[nð( y a ) / b]
ný1
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where
4Vo
ü
ÿ
, n ý odd
bn ý ý nð sinh(nð a / b)
ÿþ
0,
n ý even
(b) This is the same as Example 6.5 except that we exchange y and x. Hence
V( x, y ) ý
sin(nð y / a )sinh(nð x / a )]
ð nýod d
n sinh(nð b / a )
4Vo
õ
(c) This is the same as part (a) except that we must exchange x and y. Hence
V( x, y ) ý
sin(nð y / a )sinh[ nð (b x)/ a ]
ð nýod d
n sinh(nð b / a )
4Vo
õ
Prob. 6.24 (a) X(x) is the same as in Example 6.5. Hence
V ( x, y) ý
õ sin(nðx / b)a
n
sinh(nðy / b) bn cosh(nðy / b)ý
ný1
At y=0, V = V1
V1 ý
õb
n
sin(nðx / b)
ný1
ü 4V1
ÿ nð , n ý odd
ÿ
bn ý ý
ÿ 0, n = even
ÿ
þ
At y=a, V = V2
V2 ý
õ sin(nðx / b)a
n
sinh(nða / b) bn cosh(nða / b)ý
ný1
ü 4V2
ÿ nð , n ý odd
ÿ
an sinh(nða / b) bn cosh(nða / b) ý ý
ÿ 0, n = even
ÿ
þ
or
4V2
ü
ÿ nð sinh(nða / b) øV2 V1 cosh(nða / b)ù ,
ÿ
an ý ý
ÿ
0, n = even
ÿ
þ
n ý odd
Alternatively, we may apply superposition principle.
y
0
V2
0
V2
0 0
0 0
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V
VA
V1
VB
0
x
V1
i.e. V ý V A VB
VA is exactly the same as Example 6.5 with Vo ý V2 , while VB is exactly the same
as Prob. 6.19(a). Hence
4 sin(nð x / b)
Vý
V sinh[nð (a y ) / b] V2 sinh(nð y / b)
õ
ð n ý odd n sinh(nð a / b) 1
ý
(b)
V ( x , y ) ý (a1e ñ x a2 e ñ x )(a3 sin ñy a4 cos ñy )
lim V ( x , y ) ý 0
x
a2 ý 0
V ( x, y ý 0) ý 0
a4 ý 0
V ( x, y ý a) ý 0
ñ ý nð / a ,
n ý 1,2,3,...
Hence,
V ( x, y) ý
õae
n
nð x / a
sin(nðy / a )
ný1
V ( x ý 0 , y ) ý Vo ý
V ( x, y) ý
4Vo
ð
õ
ný1
an sin(nð y / a )
ü 4Vo
ÿ
, n ý odd
an ý ý nð
ÿþ 0 ,
n ý even
sin(nð y / a )
exp( nð x / a )
n
n ý odd
õ
(c ) The problem is easily solved using superposition theorem, as illustrated below.
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y
V3
a
V2
V
V4
x
0
0
b
V1
V2
0
0
VI
VII
0
0
V1
0
V3
0
VIV
0
VIII
V4
0
0
Therefore,
V ý VI VII VIII VIV
ü sin(nð x / b)
V1 sinh(nð (a y ) / b) V3 sinh(nð y / b)ý üÿ
ÿ
4
1 ÿ sinh(nð a / b)
ÿ
ý õ ý
ý
ð n ýodd n ÿ sin(nð x / a)
sinh(
/
)
sinh(
(
)
/
)
ð
ð
V
n
y
a
V
n
b
x
a
ýÿÿ
4
ÿþ sinh(nð b / a) 2
þ
where
VI ý
4V1
ð
VII ý
4V2
ð
sin(nðx / b) sinh[nð(a y ) / b]
n sinh(nða / b)
n ý odd
õ
sin(nðx / a ) sinh(nðy / a )
n sinh(nðb / a )
n ý odd
õ
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VIII ý
VIV ý
sin(nðx / b) sinh(nðy / b)
n sinh(nða / b)
n ý odd
õ
4V3
ð
4V4
ð
sin(nðy / a ) sinh[nð(b x) / a ]
n sinh(nðb / a )
n ý odd
õ
Prob. 6.25
E ý ñV ý
öV
öV
ax
ay
öx
öy
nð sin(nð y / a)
exp( nð x / a)
n
ð n ýodd a
4V nð cos(nð y / a)
Ey ý o õ
exp(nð x / a )
n
ð n ýodd a
4V
E ý o õ exp( nð x / a ) ùûsin(nð y / a )a x cos(nð y / a)a y ùû
a n ýodd
Ex ý
4Vo
õ
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
ö nð x ö ö nð y ö
V ( x, y ) ý õ cn sinh ÷
÷ sin ÷
÷
ø b ø ø b ø
Applying the condition at x=a, we get
öð y ö
ö nð a ö ö nð y ö
Vo sin ÷
÷ ý õ cn sinh ÷
÷ sin ÷
÷
ø b ø
ø b ø ø b ø
This yields
ö nð a ö üVo ,
cn sinh ÷
÷ýý
ø b ø þ 0,
Hence,
n ý1
n ù1
öðx ö öð y ö
sinh ÷
÷
÷ sin ÷
b ø ø b ø
ø
V ( x, y ) ý Vo
öða ö
sinh ÷
÷
ø b ø
Prob. 6.27
ñ 2V ý
1 ö ö öV ö 1 ö 2V
ý0
÷ò
÷
ò öò ø öò ø ò 2 öö 2
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If we let V (ò, ö) ý R(ò)ö(ö),
ö ö
1
(òR') 2 Rö'' ý 0
ò öò
ò
or
ö''
ò ö
(òR') ý
ýü
R öò
ö
Hence
ö'' üö ý 0
and
ö
üR
(òR')
ý0
öò
ò
or
R''
R' üR
ý0
ò ò2
Prob. 6.28
ñ 2V ý
1 ö ö 2 öV ö
1
öV
ö
(sin ñ ) ý 0
÷r
÷ 2
2
öñ
r ör ø ör ø r sin ñ öñ
If V (r , ñ) ý R(r ) F (ñ),
F
r ù 0,
d 2
R d
(r R')
(sin ñF ') ý 0
sin ñ dñ
dr
Dividing through by RF gives
1 d 2
1
d
(r R') ý
(sin ñF ') ý ü
R dr
F sin ñ dñ
Hence,
sin ñF '' cos ñF ' üF sin ñ ý 0
or
F '' cot ñF ' üF ý 0
Also,
d 2
(r R') üR ý 0
dr
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or
R''
2 R' ü
2 Rý0
r
r
Prob. 6.29 If the centers at ö ý 0 and ö ý ð / 2 are maintained at a potential difference
of Vo, from Example 6.3,
Eö ý
2Vo
,
ðò
J ý óE
Hence,
Iý
2V ó
J ÷ dS ý o
ð
b
t
òý a zý0
2V ót
1
dòdz ý o ln(b / a )
ò
ð
and
Rý
Vo
ð
ý
2ót ln(b / a )
I
Prob. 6.30 If V (r ý a ) ý 0 ,
Eý
Vo
r 2 (1 / a 1 / b)
V (r ý b) ý Vo
, from Example 6.9,
J ý óE
,
Hence,
Iý
Vo ó
J ÷ dS ý
1/ a 1/b
ñ
2ð
ñ ý0 ö ý0
1
r
2
r 2 sin ñdñdö ý
2ðVo ó
( cos ñ)|0 ñ
1/ a 1/b
1 1
Vo
a b
Rý
ý
I
2 ðó (1 cos ñ )
Prob. 6.31
This is the same as Problem 6.30 except that ñ = ð . Hence,
Rý
1
1 ö1 1ö
ö1 1ö
÷ ÷ý
÷ ÷
2ðó (1 cos ð ) ø a b ø 4ðó ø a b ø
Prob. 6.32 For a spherical capacitor, from Eq. (6.38),
1 1
Rý a b
4ðó
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For the hemisphere, R' ý 2 R since the sphere consists of two hemispheres in parallel. As
b
,
ù 1 1ù
2ú ú
ûa b û ý 1
R' ý lim
b
4ðó
2ð aó
G ý 1 / R' ý 2 ðaó
C ý 4ðõa.
Alternatively, for an isolated sphere,
õ
ó
RC ý
R' ý 2 R ý
Rý
1
2 ðaó
But
1
4ðaó
G ý 2 ðaó
or
Prob. 6.33
(a) For the parallel-plate capacitor,
V
E ý o ax
d
From Example 6.11,
Cý
1
Vo
2
õ| E |2 dv ý
1
Vo2
õ
Vo2
d
2
dv ý
õ
d
2
Sd ý
õS
d
(b) For the cylindrical capacitor,
Vo
aò
ò ln b / a
From Example 6.8,
Eý
Cý
1
Vo2
õVo2
øò ln b / aù
2
òdòdödz ý
b
2 ðõL
ø ln b / a ù
2
a
dò 2 ðõL
ý
ò ln b / a
(c )For the spherical capacitor,
Eý
Vo
ar
r (1/ a 1/ b)
2
From Example 6.10,
Cý
1
Vo2
r
õVo2
4
ø 1 / a 1 / bù
r sin ñdñdrdö ý
2
2
b
õ
ø 1 / a 1 / bù
2
4ð
dr
r
a
2
ý
4ðõ
1 1
a b
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Prob. 6.34
Assume V(ò =a) = 0 and V(ò =b)=Vo . Following Example 6.8,
Vo
ò
ln
b a
ln
a
J ý ó E ý óñV ,
dS ý ò dö dza ò
V ý A ln ò B ý
I ý J ÷ dS ý
S
2ð
L
ö
ý0 z ý0
ó Vo
ò ln
b
a
ò dö dz ý
ó Vo
ln
b
a
(2ð L)
b
ln
Vo
a
Rý ý
I 2ðó L
I 2ðó L
Gý ý
b
Vo
ln
a
The conductance per unit length is
G 2ðó
G' ý ý
L ln b
a
Prob. 6.35
From eq. (6.37) or from previous problem
b
a
Rý
2ðó L
V 2 2ðó LV 2
P ý VI ý
ý
b
R
ln
a
The power loss per unit length is
ln
P' ý
P 2ðó V 2
ý
b
L
ln
a
Prob. 6.36
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Cý
õS
d
Sý
Cd
õ oõ r
ý
2 109 106 2
m ý 0.5655 cm 2
9
4 10 / 36ð
Prob. 6.37
This can be regarded as three capacitors in parallel.
C1
C2
C ý C1 C2 C3 ý õ
C3
õ oõ rk S k
dk
õo
ù3 15 102 20 102 5 15 102 20 102 8 15 102 20 102 ùû
2 103 û
109 15 102 20 102
[3 5 8] ý 2.122 nF
ý
36ð
2 103
ý
Prob. 6.38
The structure may be treated as consisting of three capacitors in series.
C1 ý
õo A
C2 ý
,
õ rõ o A
C3 ý
,
õo A
a
a
a
a
a
a
1 1
1
1
ý
ý
C C1 C2 C3 õ o A õ oõ r A õ o A
A
2
1
2õ 1
ý
ý r
aC õ o õ oõ r
õ oõ r
Cý
õ oõ r A
a(1 2õ r )
Prob. 6.39
A
C1
C1
d
C3
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C3
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C2
C2
From the figure above,
Cý
C1C2
C3
C1 C2
here
C1 ý
õo A / 2 õo A
,
ý
d /2
d
C2 ý
õoõr A
,
d
C3 ý
õo A
2d
õ o 2õ r A 2 / d 2 õ o A õ o A ö 1
õ r ö 109 10 104 ö 1 6 ö
ý
Cý
÷
÷ý
÷ ÷ 6 pF
õ o (õ r 1) A / d 2d
d ø 2 õ r 1 ø 36ð 2 103 ø 2 7 ø
Prob. 6.40
Cý
õoS
d
S ý
Cd
õo
11103
ý 36ð 106
109 / 36ð
S ý 1.131 108 m 2
Sý
Prob. 6.41
Fdx ý dWE
WE ý
1
2õ| E |
2
dv ý
Fý
dWE
dx
1
1
õ o õ r E 2 xad õ o E 2 da (1 x )
2
2
where E ý Vo / d .
dWE 1 Vo 2
ý õ o 2 (õ r 1)da
dx
2
d
Alternatively,
WE ý
C ý C1 C2 ý
1
CVo 2 ,
2
Fý
õ o (õ r 1)Vo 2 a
2d
where
õ oõ r ax õ oõ r ( L x)
d
d
2
dWE 1 Vo a
ý õo
(õ r 1)
dx
d
2
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Fý
õ o (õ r 1)Vo 2 a
2d
Prob. 6.42
(a)
õoS
Cý
d
ý
109 200 104
ý 59 pF
36ð 3 103
(b) ò s ý Dn ý 10 6 nC / m2 . But
Dn ý õE n ý
or
Vo ý
õ oVo
ý òs
d
òs d
ý 106 3 103 36ð 109 ý 339.3 V
õo
(c )
ò S 1012 200 104 36ð 109
Q2
ý s ý
ý 1.131 mN
Fý
2Sõ o
2õ o
2
2
Prob. 6.43
C1 ý
õ oõ r S
d
C2 ý
,
C1
ý õr
C2
õoS
d
õr ý
56ý F
ý 1.75
32ý F
Prob. 6.44
(a) C ý
õS
d
ý
109
0.5
36ð
ý 7.515 nF
4 103
6.8
Q
Q
, Cý
Q ý CV
S
V
(b)
CV
7.515 109 9
ý
ý 135.27 nC/m 2
òs ý
S
0.5
òs ý
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Prob. 6.45
Co ý
õS
õS
ý Co / 3
d
3d
Q
Co ý o
Qo ý CoV
V
Q ý CV ý (Co / 3)V ý Qo / 3
Eo ý
Cý
,
V
,
d
Eý
E
V
ý o
3d
3
1
1
W ý CV 2 ý (Co / 3)V 2 ý 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 ý
õ oõ r 1 S
d
,
C2 ý
õ oõ r 1 S õ o õ r 2 S
õ oõ r 2 S
d
õ oõ r1õ r 2 S
C1C2
109 40 104 4 6 4.8
d
d
d
ý
ý
ý
ý
Cý
nF=42.44 pF
C 1 C2 õ o õ r 1 S õ o õ r 2 S
õ r1 õ r 2 36ð 2 103 4 6 36ð
d
d
Q
Q ý CV ý 509.3 pC
(b) C ý
V
C1
+
V1
C2
-
+ V2
-
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(c) C1 and C2 are in series as shown above
Q ý C1V1 C2V2 ,
V1 V2 ý V ý 12
Solving these gives
C2
6
V ý (12) ý 7.2
V1 ý
C1 C2
10
V2 ý
C1
4
V ý (12) ý 4.8
C1 C2
10
V1
ý 3.6 kV/m
d
V
E 2 ý 2 ý 2.4 kV/m
d
D ý D1 ý D2 ý õ1 E1 ý 1.2732 107 C/m 2
E1 ý
P1 ý ó e1 õ o E1 ý 9.549 108 C/m 2
P2 ý ó e 2õ o E2 ý 1.061 108 C/m 2
Prob.6.47
(a)
109
4ðõ
36ð ý 25 pF
ý
Cý
1 1
1
1
2
a b 5 x10
10 x102
4ð 2.25
(b)
òs ý
Q = C Vo= 25x80 pC
Q
25 80
p C/ m 2 ý 63.66 nC/ m 2
ý
2
4ð r
4ð 25 10 4
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Prob. 6.48
C1
b
c
C2
d
C3
a
1
1
1
1
ý
C C1 C2 C3
4ðõ 3
where C1 ý
,
1 1
b a
C2 ý
4ðõ 2
,
1 1
c b
C3 ý
4ðõ1
,
1 1
d c
4ð 1/ b 1/ a 1/ c 1/ b 1/ d 1/ c
ý
C
õ3
õ2
õ1
4ð
Cý
õ1
õ
õ
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,
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Eý
Q
4ðõ o r 2
ar
The potential at r=a is
a
Va ý E dl ý
ý
Cý
Q
4ðõ o
Q
ý
Va
c
dr
r
2
c
b
a
Er dr Er dr Er dr
c
0
Q
4ðõ o
b
a
dr
r
b
2
ý
Q
4ðõ o c
Q ö1 1ö
÷ ÷
4ðõ o ø a b ø
1
1
1 ö1 1ö
÷ ÷
4ðõ o c 4ðõ o ø a b ø
Prob. 6.50
We can regard this as having two cylindrical capacitors in series.
2ðõ oõ r1 L
2ðõ oõ r 2 L
,
C1 ý
C2 ý
c
b
ln
ln
a
c
2ðõ oõ r1 L 2ðõ oõ r 2 L
c
b
ln
ln
2ðõ oõ r1õ r 2 L
CC
a
c
ý
Cý 1 2 ý
C1 C2 2ðõ oõ r1 L 2ðõ oõ r 2 L õ ln b õ ln c
r1
r2
c
b
c
a
ln
ln
a
c
Prob. 6.51
109
2ð 2.5
3 103
2ðõ L
36ð
Cý
ý
ý 0.8665 ý F
ln(b / a)
ln(8 / 5)
Prob. 6.52
Let the plate at ö=0 be 0, i.e. V(0)=0 and let the plate at ö=ð/4 be Vo , i.e. V(ð/4)=Vo.
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ñ 2V ý
1 d 2V
ý0
ò 2 dö 2
V (0) ý 0
0 ý 0 B
V (ð / 4) ý Vo
dV
ýA
dö
V ý Aö B
Bý0
Vo ý Að / 4
Aý
4Vo
ð
4V
A
1 dV
aö ý aö ý o aö
ò dö
ò
ðò
4õVo
D ýõE ý
aö
E ý ñV ý
ðò
4õ Vo
ò s ý Dn ý
ðò
Q ý ò s dS ý
Cý
b
4õ Vo
L
ò
ý a z ý0
ðò
d ò dz ý
4õ Vo
ð
L ln(b / a)
| Q | 4õ L
ý
ln(b / a)
ð
Vo
Prob. 6.53
Since V ý V (ö ),
ñ 2V ý 0
1 d 2V
ý0
ò 2 dö 2
d 2V
dV
ý0
ý A V ý Aö B
2
dö
dö
Bý0
For ö =0, V=0 0=0+B
V
Aý o
For ö =ñ , V=Vo
Vo =Añ
ò ù 0,
But
ñ
A
1 dV
aò ý aò
ò dö
ò
A
dS ý d ò dz
ò s ý Dn ý õ Eö ý õ ,
E ý ñV ý
ò
Q ý ò s dS ý
S
Cý
ò2
L
ò ò
ý
1
z ý0
A
ò
õ d ò dz ý Aõ L ln
ò 2 Vo
ò
ý õ L ln 2
ò1 ñ
ò1
Q õ L ò2
ln
ý
Vo ñ
ò1
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Prob. 6.54
Cý
2ðõ o L
ý
ln(b / a)
V ý Q/C ý
2ð
109
100 106
36ð
ý 1.633 1015 F
ln(600 / 20)
50 1015
ý 30.62 V
1.633 1015
Prob. 6.55
2ðõ1
2ðõ 2
C1 ý
, C2 ý
ln(b / a )
ln(c / b)
Since the capacitance are in series, the total capacitance per unit length is
2ðõ1õ 2
CC
Cý 1 2 ý
C1 C2 õ 2 ln(b / a ) õ1 ln(c / b)
Prob. 6.56
(a) This is similar to Example 6.10.
ö1 1ö
Vo ÷ ÷
r bø
ñ 2V ý 0 V ý ø
1 1
a b
Vo
dV
E ý ñV ý
ar ý
a
1ö r
dr
2ö1
r ÷ ÷
øa bø
õ oõ rVo
At r=a, ò s ý Dn ý õ Er ý
ö1 1ö
a2 ÷ ÷
øa bø
ö1 1ö
ö1 1ö 1
ò s a 2 ÷ ÷ 400 109 (4 104 ) ÷ ÷ 2
øa bø ý
ø 2 4 ø 10 ý 16(36ð )102 (1/ 4) ý 4.524
õr ý
9
10
õ oV
(100)
36ð
(b)
109
4ð
4.524
4ðõ oõ r
452.4
36ð
Cý
ý
ý
ý 201.1 nF
1 1
1 1ö 1
9(1/
4)
ö
÷ ÷ 2
a b
ø 2 4 ø 10
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Prob. 6.57
Each half has capacitance given by
2ðõ
2ðõ ab
ý
Cý
1 1
ba
a b
The two halves may be regarded as capacitors in parallel. Hence,
2ðõ1ab 2ðõ 2 ab 2ð (õ1 õ 2 )ab
C ý C1 C2 ý
ý
ba
ba
ba
Prob. 6.58
Eý
Q
4ðõ r 2
ar
Q
Vo
0
Wý
1
Q2
2
õ
E
ý
õ r 2 sin ñ dñ d ö d r
|
|
d
v
2 2 4
2
32ð õ r
b
Q2
d r Q2
(2
)(2)
ý
ð
c r 2 ý 8ðõ
32ð 2õ
ö 1 1ö
÷c b ÷
ø
ø
Q 2 (b c)
Wý
8 ð õ bc
Prob. 6.59
òs
(a x ) , where ò s is to be determined.
õ
d
d
òs
ò
1 d
Vo ý E ÷ d l ý
d x ý òs
d x ý s d ln( x d )
õ
õ dx
õ
0
0 o
(a) Method 1:
Eý
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Vo ý ò s d ln
Eý
2d
d
òs ý
Vo õ o
d ln 2
Vo
òs
ax ý
ax
( x d )ln 2
õ
Method 2: We solve Laplace’s equation
ñ ÷ ( õñ V ) ý
d
dV
(õ
)ý 0
dx dx
õ
dV
ý A
dx
dV A
Ad
c
ý ý
ý 1
dx õ õ o ( x d ) x d
V ý c1 ln( x d ) c2
V ( x ý 0) ý 0
V ( x ý d ) ý Vo
c1 ý
0 ý c1 ln d c2
c2 ý c1 ln d
Vo ý c1 ln 2d c1 ln d ý c1 ln 2
Vo
ln 2
V ý c1 ln
Eý
(b)
x d Vo
x d
ý
ln
ln 2
d
d
dV
Vo
ax ý
ax
dx
( x d )ln 2
õ o xVo
ö x d ö õ o Vo
1÷
P ý (õ r 1)õ o E ý ÷
ax ý
ax
d ( x d )ln 2
ø d
ø ( x d )ln 2
(c )
x=d
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x=0
ò p s| xý0 ý P ÷ (a x )| xý0 ý 0
ò p s| x ý d ý P ÷ a x | x ý d ý
E
(d)
ý
õ o Vo
2d ln 2
òs
Q
Q
ax ý
ax ý
ax
x
õ
õS
õ o (1 ) S
d
Q
dx
Q
V ý E dl ý
d ln 2
ý
õ o S a (1 x ) õ o S
d
õ S
Q
Cý ý o
V d ln 2
d
Prob. 6.60
We solve Laplace’s equation for an inhomogeneous medium.
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d ö dV ö
÷õ
÷ý0
dx ø dx ø
2
dV A
A ù öxö ù
1
ý ý
ú ÷ ÷ ú
dx õ 2õ o úû ø d ø úû
A
x3
Vý
(x 2 ) B
2õ o
3d
ñ (õñV ) ý
õ
dV
ýA
dx
When x=d, V=Vo ,
A
d
(d ) B
2õ o
3
Vo ý
When x = -d, V=0,
A
d
0ý
(d ) B
2õ o
3
0ý
Vo ý
Vo ý 2 B
Adding (1) and (2),
2 Ad
B
3õ o
2 Ad
B
3õ o
(1)
(2)
B ý Vo / 2
From (2),
Bý
2 Ad Vo
ý
3õ o
2
Aý
3õ oVo
4d
ù ö x ö2 ù
ú1 ÷ ÷ ú
2
3Vo ù ö x ö ù
3õ oVo ûú ø d ø ûú
dV
A
E ý ñV ý
ax ý ax ý
ax ý
ú1 ÷ ÷ ú a x
õ
dx
4d
2õ o
8d ûú ø d ø ûú
ò s ý D an ý õ E a x
xýd
Q ý ò s dS ý ò s S ý
S
ý A ý
3õ oVo
4d
3Sõ oVo
4d
| Q | 3õ o S
ý
Cý
Vo
4d
Prob. 6.61
Method 1: Using Gauss’s law,
Q ý D ÷ d S ý 4ð r 2 Dr
E ý D/ õ ý
Q
4ðõ o k
Dý
Q
ar ,
4ð r 2
õý
õok
r2
ar
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V ý E÷ d l ý
Cý
Q
4ðõ o
a
Q
dr ý
k
4ðõ
b
o
k
(b a )
Q 4ð õ o k
ý
|V | b a
Method 2: Using the inhomogeneous Laplace’s equation,
ñ ÷ ( õñ V ) ý 0
dV
ý A'
dr
V (r ý a ) ý 0
õ ok
V (r ý b) ý Vo
Eý
Vo ý Ab B ý A(b a )
V
dV
a r ý Aa r ý o a r
dr
b a
ò s ý Dn ý
Vo õ o k
|
b a r 2 r ý a ,b
Q ý ò s dS ý
Cý
1 d ö õ o k 2 dV ö
r
÷
÷ ý0
r 2 dr ø r 2
dr ø
dV
ý A or V ý Ar B
dr
0 ý Aa B
B ý Aa
Vo õ o k
b a
1
r
2
r 2 sin ñ dñ dö ý
Vo õ o k
4ð
b a
| Q| 4ð õ o k
ý
Vo
b a
Prob. 6.62
C ý 4ðõ o a ý 4ð
109
6.37 106 ý 0.708 mF
36ð
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Aý
Vo
b a
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Prob.6.63
Q
Cý
V
Q
Dý
aò
2ðò L
D
Q
E= ý
õ 2ðò Lõ o (3)(1 ò )
V ý E dl ý
Q
6ð Lõ o
b
dò
ò (1 ò )
a
A
B
1
ý
ò (1 ò ) ò 1 ò
Using partial fractions
A=1, B= -1
Let
ùb dò b dò ù
ú
ú
6ðõ 0 L û a ò a 1 ò û
Q
ý
ln ò ln(1 ò )ý
6ðõ 0 L
Q
Vý
ý
b
a
b
a ù
ù
ln
ln
ú
6ðõ 0 L û 1 b
1 a úû
Q
If a=1 mm, and b=5 mm
Q
Cý
|V |
6ðõ o
ý
b
a
ln
ln
1 b
1 a
9
10
6ð
36ð
ý
5
1
ln ln
6
2
1
109
1
6
ý
ý 1.9591 nF
ln 0.8333 ln 0.5 6
C ý 0.326 nF
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Prob. 6.64
Da D
ý 1 ý 11
a
a
109
ð
(4)
4
36ð
C
nF/m = 46.34 pF/m
ý
ln11
36 ln11
Prob. 6.65
(a) From eq. (6.46),
Qh
10 109 (10)
107
ý
ý
ý 12.107 pF/m 2
2ð [ x 2 y 2 h 2 ]3/2 2ð [4 16 100]3/2 2ð (120)3/2
(b) Qin ý Q ý 10 nC
òs ý
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 ý F13 F14 F23 F24
| F |ý
1018
ù 9 2(12) 16 ù
2 2 ú ý 5.25 nN
9
4ð 10 / 36ð úû 22
3
4 û
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,
(2a x 2a y 2a z ) (2a y 2a z ) ù
ù 2
2
2
3 ax 3 ay 3 a z
ú
ú
Q
2
2
2
123 / 2
83 / 2
ú
ú
Fý
4ðõ o ú (2a x 2a y ) (2a x 2a z )
ú
ú
ú
3/ 2
3/ 2
8
8
û
û
2
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ö 1
1
1 ö
ý 0.9(a x a y a z )÷
÷ ý 0.1092(a x a y a z ) N
ø 4 12 3 4 2 ø
Prob. 6.68
ö 360 o ö
Ný÷
1÷ ý 7
ø 45 o
ø
Prob. 6.69
(a)
E ý E E ý
(2, 2,3) (3, 2, 4) ù
ò L ö a ò 1 a ò 2 ö 16 109 ù (2, 2,3) (3, 2, 4)
÷
÷ý
9 ú
2
10 û | (2, 2,3) (3, 2, 4) | | (2, 2,3) (3, 2, 4) |2 úû
2ðõ o ø ò1 ò 2 ø
2ð
36ð
ù(1,0, 1) (1,0,7) ù
ý 18 x16 ú
ý 138.2a x 184.3a y V/ m
2
50 úû
û
(b) ò s ý Dn
ò
D ý D D ý L
2ð
ý
ö a ò1 a ò 2 ö 16 x109
÷
÷ý
2ð
ò2 ø
ø ò1
ù (5, 6,0) (3, 6,4)
(5, 6,0) (3, 6, 4) ù
ú| (5, 6,0) (3, 6,4)| 2 | (5, 6,0) (3, 6, 4)| 2 ú
û
û
8 ù (2, 0, 4) (2, 0, 4) ù
nC/m 2 ý 1.018a z nC/m 2
ú
ú
ð û 20
20 û
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ò s ý 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
òs
109
Eý õ
an ý
(30a y 20a y 20a y 30a y ) ý 18ð (60)a y
2õ o
2 x109 / 36ð
ý 3.4a y kV/ m
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CHAPTER 7
P.E. 7.1
z
ñ1
ñ2
2
27
a x a y
ö a a y ö
aö ý al aò = ÷ x
÷ az ý
2 ø
2
ø
ò ý 5, cos ñ1 ý 0, cos ñ 2 ý
H3 ý
ö ö a a y
10 ö 2
0 ÷÷ ÷ x
÷÷
4ð (5) ø 27
2
øø
ö
÷ ý 30.63a x 30.63a y mA/m
ø
P.E. 7.2
2 ö
3 ö
1
÷
÷ az ý 0.1458az A/m
4ð (2) ø
13 ø
12
ò ý 32 42 ý 5, ñ 2 ý 0, cos ñ1 ý ,
(b)
13
(a) H ý
ö 3a 4a z ö 4a x 3a z
aö ý a y ÷ x
÷ý
5
5
ø
ø
2 ö 12 ö ö 4a x 3az ö
1
Hý
ø 4ax 3az ù
÷1 ÷÷
÷ý
4ð (5) ø 13 øø
5
ø 26ð
= 48.97a x 36.73a z mA/m
P.E. 7.3
(a) From Example 7.3,
Ia 2
Hý
az
2(a 2 z 2 )3/ 2
At (0,0,-1cm), z = 2cm,
50 103 25 104
Hý
a z ý 400.2a z mA/m
2(52 22 )3/ 2 106
(b) At (0,0,10cm), z = 9cm,
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Hý
50 103 25 104
a z ý 57.3a z mA/m
2(52 92 )3/ 2 106
P.E. 7.4
2 103 50 103 (cos ñ 2 cos ñ1 )a z
NI
ñ
ñ
cos
cos
a
ý
ø
2
1ù z
2L
2 0.75
100
ý
øcos ñ 2 cos ñ 1 ùa z
1.5
0.75
(a) At (0,0,0), ñ ý 90 o , cos ñ 2 ý
0.75 2 0.05 2
ñ1
= 0.9978
100
Hý
ø 0.9978 0 ù az
1.5
Hý
= 66.52 az A/m
(b) At (0,0,0.75), ñ 2 ý 90 o ,cos ñ 1 ý 0.9978
100
ø 0 0.9978 ù a z
Hý
1.5
= 66.52az A/m
0.5
(c) At (0,0,0.5), cos ñ 1 ý
ý 0.995
0.5 2 0.05 2
0.25
cos ñ 1 ý
ý 0.9806
0.25 2 0.05 2
100
ø 0.9806 0.995ù a z
Hý
1.5
= 131.7az A/m
P.E. 7.5
Hý
(a)
(b)
1
K an
2
1
H (0, 0, 0) ý 50a z (a y ) ý 25a x mA/m
2
1
H (1,5, 3) ý 50a z a y ý 25a x mA/m
2
P.E. 7.6
ü NI
, ò a ü ò ü ò a, 9<ò ü 11
ÿ
H ý ý 2ðò
ÿ
0,
otherwise
þ
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ñ2
ñ1
ñ2
ñ1
ñ2
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(a) At (3,-4,0), ò ý 3 2 4 2 =5cm ‹ 9cm
H ý0
(b) At (6,9,0), ò ý 6 2 9 2 = 117 ‹ 11
H ý
103 100 103
ý 147.1 A/m
2ð 117 102
P.E. 7.7
(a) B ý ñ A ý ( 4 xz 0 )a x (0 4 yz )a y ( y 2 x 2 )a z
B (1, 2,5) ý 20a x 40a y 3a z Wb/m2
(b) ý B.dS ý
4
1
( y 2 x 2 )dxdy ý
y ý1 x ý 0
4
1
1
y 2 dy 5 x 2 dx
0
1
5
ý (64 1) ý 20 Wb
3
3
Alternatively,
1
4
0
0
1
1
ý A.dl ý x 2 (1)dx y 2 (1)dy x 2 (4)dx 0
5 65
ý
ý 20 Wb
3 3
P.E. 7.8
z
R
h
y
k
dS
x
H ý
kdS R
,
4ð R 3
dS ý dxdy, k ý k y a y ,
R ý ( x, y, h),
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k R ý (ha x xa z )k y ,
H ý
ý
k y (ha x xa z )dxdy
3
4ð ( x 2 y 2 h 2 )
k y ha x
dxdy
4ð
(x
2
2
y 2 h2 )
3
k y az
4ð
2
(x
xdxdy
2
y 2 h2 )
3
2
The integrand in the last term is zero because it is an odd function of x.
Hý
ý
k y ha x
4ð
2ð
ö ý0 ò ý0
ò dö d ò
( ò 2 h2 )
ö
1
ax ÷
÷ ( ò 2 h2 ) 12
2
ø
kyh
3
ý
k y h2ð a x
2
4ð
2
2
(ò h )
0
3
2
d (ò 2 )
2
ö ky
÷ 0 ý ax
÷
2
ø
1
Similarly, for point (0,0,-h), H = k y a x
2
Hence,
ù1
zþ0
ú k yax ,
H ýú2
1
zü0
ú k a ,
û2 y x
Prob. 7.1
(a) See text
(b) Let H = Hy + Hz
For H z ý
a ö ý a z
Hz ý
I
2ðò
aö
ò ý ( 3) 2 4 2 ý 5
( 3a x 4a y )
5
ý
( 3a y 4a x )
5
20
(4a x 3a y ) ý 0.5093a x 0.382a y
2 ð(25)
For H y ý
I
a ö , ò ý ( 3) 2 5 2 ý 34
2ðò
aö ý a y
(3a x 5a z ) 3a z 5a x
ý
34
34
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Hy ý
10
(5a x 3a z ) ý 0.234a x 0.1404a z
2ð (34)
H = Hy + Hz
= 0.7433ax + 0.382ay + 0.1404az A/m
Prob. 7.2
Hý
I
2ðò
òý
I
2ð H
ý
2
100
ý
ý 31.83 m
3
2ð (10 10 ) ð
Prob. 7.3
Let
H ý H1 H 2
where H1 and H 2 are respectively due to the lines located at (0,0) and (0,5).
H1 ý
I
2ðò
aö ,
ò ý 5,
aö ý a aò ý a z a x ý a y
a
10
ay ý y
ð
2ð (5)
I
ò ý 5 2, aö ý a a ò , a ý a z
H2 ý
aö ,
2ðò
5a 5a y a x a y
ý
aò ý x
5 2
2
ö a a y ö -a x a y
aö ý a z ÷ x
÷ý
2 ø
2
ø
10 ö -a x a y ö 1
( -a x a y )
H2 ý
÷
÷ý
2ð 5 2 ø
2 ø 2ð
H1 ý
H ý H1 H 2 ý
ay
ð
1
(-a x a y ) ý 0.1592a x 0.1592a y
2ð
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Prob. 7.4
H ý dH ý
For H1 ,
Idl R
ý H1 H 2
4ð R 3
R ý (3,1, 2) (0, 0, 0) ý (3,1, 2),
Idl R ý 4 105
R ý 9 1 4 ý 14
1 0 0
ý 4 105 (0, 2,1)
3 1 2
4 105 (0, 2,1)
ý (0, 0.01215, 0.006076)105
H1 ý
3/2
4ð (14)
R ý (3,1, 2) (0, 0,1) ý (3,1, 3),
For H 2 ,
Idl R ý 6 105
H1 ý
R ý 9 1 9 ý 19
0 1 0
ý 6 105 (3, 0, 3)
3 1 3
6 105 (3, 0, 3)
ý (0.017, 0, 0.017)105
3/2
4ð (19)
H ý H1 H 2 ý (0.0173, 0.01215, 0.01122)105 ý (0.173a x 1.215a y 0.1122a z ) ý A/m
Prob. 7.5
Let H =H y H z
For H z ,
Hz ý
I
2ðò
aö
where I = 20, ò = 3a x 4a y , ò = 32 42 ý 5
0
0 1
ö 3a x 4a y ö
ý 0.8a x 0.6a y
aö ý a aò ý a z ÷
÷ý
5
ø
ø 0.6 0.8 0
20
(0.8a x 0.6a y ) ý 0.5093a x 0.382a y
Hz ý
2ð (5)
I
For H y ,
Hy ý
aö
2ðò
where I = 10, ò = 3a x 5a z , ò = 32 52 ý 34
aö ý a aò ý
1 0 1 0
1
(5a x 3a z )
ý
34 3 0 5
34
10
(5a x 3a z ) ý 0.234a x 0.1404a z
2ð (34)
H =H y H z ý 0.7432a x 0.382a y 0.1404a z A/m
Hy ý
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Prob. 7.6
y
1 A
ñ2
6A
P
ò
B
ñ1
O
H ý
I
4ðò
x
1
ø cos ñ 2 cos ñ1 ù aö
1
2
2 ý
2
2
ö a x a y ö
ö a x a y ö
1 -1 1 0
ý ÷
ý
ý az
÷
÷
÷
2 -1 -1 0
2
2
ø
ø
ø
ø
ø cos 45o cos135o ù az ý 3 az
ñ1 ý 135o , ñ 2 ý 45o , ò ý
aö ý al a ò
H
6
ý
ð
2
2
H ø 0, 0, 0 ù ý 0.954a z A/m
4ð
Prob. 7.7
(a) At (5,0,0), ò ý 5,
Hý
aö ý a y ,
cos ñ1 ý 0,
cos ñ 2 ý
10
125
2
10
(
)a y ý 28.471a y mA/m
4ð (5) 125
(b) At (5,5,0), ò ý 5 2,
cos ñ1 ý 0,
cos ñ 2 ý
ö a x a y ö a x a y
aö ý a z ÷
÷ý
2 ø
2
ø
2
10 ö a x a y
(
)÷
Hý
4ð (5 2) 150 ø
2
10
150
ö
÷ ý 13(a x a y ) mA/m
ø
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(c ) At (5,15,0), ò ý 250 ý 5 10,
cos ñ1 ý 0,
cos ñ 2 ý
10
350
ö 5a x + 15a y
aö ý a z ÷
ø 5 10
ö 5a y - 15a x
÷ý
5 10
ø
2
10 ö 15a x 5a y ö
Hý
(
)÷
÷ ý 5.1a x 1.7a y mA/m
4ð (5 10) 350 ø
5 10
ø
d) At (5,-15,0), by symmetry,
H ý 5.1a x 1.7a y mA/m
Prob. 7.8
z
ñ1
x
A (2, 0, 0)
C (0, 0, 5)
y
ñ2
B (1, 1, 0)
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(a)
Consider the figure above.
AB
ý
AC
ý
ø1, 1, 0 ù
ø 0, 0, 5ù
AB AC
ý
ø 2, 0, 0 ù
ý
ø 2, 0, 0 ù
ý
2, i.e AB and AC are not perpendicular.
cos ø180o ñ1 ù
ý
BC
BA ý
H2 ý
ý
(b)
BC BA
BC BA
ý
BC ý ò ý
aö ý
AB AC
AB AC
ý
ø 0, 0, 5 ù ø1, 1, 0 ù
ø1, 1, 0 ù
cos ñ 2
i.e.
al aò
ý
ý
2
ý
2 29
ø 1,
1 1
BC BA
0
ø 1, 1, 5ù , ò ý 27
ø 1, 1, 0 ù ø 1, 1, 5ù
ý
ý
2
cos ñ1 ý
2
29
1, 5 ù
ý
27
ø 5, 5, 2 ù
54
ö
ø 5, 5, 2 ù A/m
2 ö ø 5, 5, 2 ù
5
ý
÷÷ 0
÷÷
27
29 ø
2 27
2ð 29
ø
27.37 a x 27.37a y 10.95 a z mA/m
10
4ð 27
H ý H1 H 2 H 3 ý
ý
ø 1, 1, 0 ù
ø 2, 0, 5ù
ø 0,
59.1, 0 ù
ø 27.37,
27.37, 10.95 ù
ø 30.63, 30.63, 0 ù
3.26 a x 1.1 a y 10.95a z mA/m
Prob. 7.9
y
(a)
Let H ý H x H y ý 2H x
Hx ý
I
4ðò
ø cos ñ 2 cos ñ1 ù aö
ñ2
O
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5A
ñ1
x
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where aö ý a x a y ý a z , ñ1 ý 180o , ñ 2 ý 45o
ø cos 45
4ð ø 2 ù
5
Hx ý
H
(b)
o
cos 180o ù ø a z ù ý 0.3397a z
ý 2 H x ý 0.6792 a z A/m
H ý Hx H y
where H x ý
5
ø1 0 ù aö ,
4ð ø 2 ù
aö ý a x a y ý a z
ý 198.9a z mA/m
H y ý 0 since ñ1 ý ñ 2 ý 0
H ý 0.1989 a z A/m
(c )
H ý Hx H y
where H x ý
Hy ý
5
ø1 0 ù ø a x a z ù
5
ø1 0 ù ø a y a z ù
4ð ø 2 ù
4ð ø 2 ù
ý 198.9 a y mA/m
ý 198.9 a x mA/m
H ý 0.1989 a x 0.1989 a y A/m.
Prob. 7.10
For the side of the loop along y-axis,
I
H1 ý
ø cos ñ 2 cos ñ1 ù aö
4ðò
where aö ý a x , ò ý 2 tan 30o ý
2
, ñ 2 ý 30o , ñ1 ý 150o
3
5
3
15
ax
cos 30o cos 150o ù ø a x ù ý
ø
4ð 2
8ð
H ý 3H1 ý 1.79a x A/m
H1 ý
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Prob. 7.11
3
Let H ý H1 H 2 H 3 H 4
4
where H n is the contribution by side n.
H ý 2H1 H 2 H 4 since H1 ý H 3
(a)
H1 ý
H2 ý
I
4ðò
ø cos ñ 2 cos ñ1 ù aö
10 ö 6
1 ö 1
÷ az
4ð ø 2 ù ÷ø 40
2ø
ý
10 ö
2 ö
10 ö
1 ö
2
2
az , H 4 ý
÷
÷
÷
÷ az
4ð ø 6 ù ø
4ð ø 2 ù ø
40 ø
2ø
ù 5 ö 3
1 ö
5
ý ú ÷
÷
2ø
6ð 10
û 2ð ø 10
At ø 4, 2, 0 ù , H ý 2 ø H1 H 4 ù
H
(b)
H1 ý
ý
H
(c )
At
10
4ð ø 2 ù
8
10
az , H 4 ý
4ð ø 4 ù
20
2 5 ö 1ö
az ý
1
ð ÷ø 4 ÷ø
ø 4, 8, 0 ù ,
ù
ú a z = 1.964a z A/m
2ð 2 û
5
4
az
20
1.78a z A/m
H ý H1 2H 2 H 3
H1 ý
10 ö
4 ö
10 ö 8
1 ö
2
az , H 2 ý
÷
÷
÷
÷ az
4ð ø 8 ù ø 4 5 ø
4ð ø 4 ù ø 4 5
2ø
H3 ý
10 ö 2 ö
ø a z ù
4ð ø 4 ù ÷ø 2 ÷ø
ý
H
(d )
2
At
5
8ð
ø az ù
4
4 ö
ö 1
÷
÷
5
2ø
ø 5
ý
0.1178a z A/m
ø 0, 0, 2 ù ,
H1 ý
10 ö 8
ö
0 ÷ ø ax az ù
÷
4ð ø 2 ù ø 68
ø
H2 ý
10
ö 4
ö
ö 2a 8a x ö
0 ÷ ay ÷ z
÷
÷
4ð 68 ø 84
68 ø
ø
ø
ý
10
ay
ð 68
ý
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5 ø a x 4a z ù
17ð 84
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ö 2a 4a y ö
10
ö 8
ö
0 ÷ ax ÷ x
÷ ý
÷
4ð 20 ø
84
20 ø
ø
ø
5a x
10 ö
4 ö
ý
÷0
÷ ø a y a z ù ý
ð 20
4ð 2 ø
20 ø
H3 ý
H4
H
ð 21
5
5 ö
10 ö
2 ö
ö
ö 1
ö 20
ý ÷
÷ az
÷ ax ÷
÷ ay ÷
ø 34ð 21 ð 21 ø
ø 34ð 21 ð 20 ø
ø ð 21 ð 68 ø
ý 0.3457 ax 0.3165 a y 0.1798 az A/m
Prob. 7.12
H ý 4 H1 ý 4
I
4ðò
(cos ñ 2 cos ñ1 )aö
ò ý a ý 2cm, I ý 5mA, ñ 2 ý 45o , ñ1 ý 90o 45o ý 135o
aö ý a a ò ý a y (a x ) ý a z
Hý
a y 2a z
1
2I
2 5 103
I 1
(
)a z ý
az ý
a ý 0.1125a z
ða 2
ða
ð 2 102 z
2
Prob. 7.13
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Consider one side of the polygon as shown. The angle
(a )
subtended by the Side At the center of the circle
360o
2ð
ý
n
n
The field due to this side is
H1 ý
where ò ý r ,
I
øcos ñ 2 cos ñ1 ù
4ðò
ð
ð
cos ñ 2 ý cos(90 ) ý sin
n
n
cos ñ1 ý sin
H1 ý
I
ð
n
2 sin
4ð r
ð
nI
ð
sin
n
2ð r
3I
ð
For n ý 3, H ý
sin
2ð r
3
2
r cot 30o ý 2 r ý
3
n
H ý nH1 ý
(b)
3 5
2ð 2
H ý
For n ý 4, H ý
3
3
2
ý
45
8ð
ý 1.79 A/m.
4I
ð
45 1
ý
sin
2ðr
4
2ðø2 ù 2
ý 1.128 A/m.
(c)
As n ,
H ý lim
n
ð
nI
ý
sin
2ðr
n
I
nI ð
ý
2r
2ðr n
From Example 7.3, when h ý 0,
H ý
I
2r
which agrees.
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Prob. 7.14
4
ò
1
ñ2
2
Let H ý
H1 H 2 H 3 H 4
I
10
az ý
a z ý 62.5 a z
4a
4 4 102
I
4
cos ñ 2 cos 90o ù a z , ñ 2 ý tan 1
ý H4 ý
2 ø
4ð 4 10
100
ý 19.88 a z
H1 ý
H2
H3 ý
I
4ð ø1ù
2 cos ò a z , ò ý tan 1
100
4
ý 87.7 o
10
2 cos 87.7o a z ý 0.06361 a z
4ð
ý ø 62.5 2 19.88 0.06361ù a z
ý
H
ý 102.32 a z A/m.
Prob. 7.15
y
3
I
öo
0
ò1
4
2
1
ò2
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x
ý 2.29o
3
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Let
H ý 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 ý H1
Idl R
,
dl ý ò1dö aö , R ý ò1a ò
4ð R 3
I ò12 dö aö (a ò ) Idö (a z )
ý
dH 4 ý
4ðò13
4ðò1
dH 4 ý
For H 4 ,
ö
H4 ý
I ö (a )
I (a z ) o
dö ý o z
4ðò1 0
4ðò1
Similarly, for H 2 ,
H2 ý
Iöo (a z )
4ðò 2
H ý H1 H 2 H 3 H 4 ý
Iöo ö 1 1 ö
÷ ÷ az
4ð ø ò1 ò 2 ø
Prob. 7.16
From Example 7.3,
H1 ý
I
az ,
2a
H2 ý
Ia 2
az
2[a 2 d 2 ]3/2
ù
ù
10 ù 1
32
I ù1
a2
2
H = H1 H 2 ý ú 2
a ý ú
a
2 3/2 ú z
2 3/2
2
2 ú z
2 û a [a d ] û
2 û 3 10
[3 4 ] (10 ) û
ù1 9 ù
a z ý 202.67a z A/m
ý 500 ú
û 3 125 úû
Prob. 7.17
ñ2
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ñ2
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H
ý
nI
ø cos ñ 2 cos ñ1 ù
2
cos ñ 2 ý -cos ñ1 ý
H
ý
ø
nI
2
2 a
2
4
ø
ù
1
2
2
a2
ý
2
4ù
1
2
0.5 150 2 102
2 103 42 102
ý 69.63 A/m
(b)
ñ1
ñ2
a
4
ý
ý 0.2 ñ 2 ý 11.31o
b
20
150 0.5
ý
cos 11.31o ý 36.77 A/m
2
ñ1 ý 90o , tan ñ 2 ý
H
ý
nI
cos ñ 2
2
Prob. 7.18
e
y
÷ P (4, 3, 2)
x
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Let
H ý Hl H p
Hl ý
ò ý
aò ý
Hl
Hp
H
1
2ðò
aö
ø 4, 3, 2 ù
3a x 5a y
34
(1, -2, 2) ý (3, 5, 0),
ò ý
ò
ý
34
, al ý a z
3a y 5a x
ö 3a 5a y ö
aö ý a l a ò ý a z ÷ x
÷ ý
34
34 ø
ø
20ð ö 5a x 3a y ö
-3
ý
÷
÷ x10 ý ( 1.47a y 0.88a y ) mA/m
2ð ø
34
ø
1
1
K an ý
ý
ø100 103 ù az ø -a x ù ý 0.05a y A/m
2
2
ý H l H p ý 1.47a x 49.12 a y mA/m
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Prob. 7.19
(a) See text
(b)
I
a
b
For ò ü a,
H dl
For a ü ò ü b,
Hö 2ðò ý
Hö ý
For ò þ b,
ý Ienc ý 0 H ý 0
Ið ø ò 2 a2 ù
ð ø b2 a 2 ù
ö ò 2 a2 ö
÷
÷
2ðò ø b2 a2 ø
I
Hö 2ðò ý I
Hö ý
I
2ðò
Thus,
Hö
ù
ú 0,
ú
ú I
ý ú
ú 2ðò
ú I
ú
û 2ðò
ò üa
ö ò 2 a2 ö
÷ 2 2÷ ,
ø b a ø
,
a ü ò üb
ò þb
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Prob.7.20
x
y
-1
1
1
H ý õ K an
2
1
1
ý (20a x ) (a y ) (20a x ) a y
2
2
ý 10(a z ) 10(a z )
ý 20a z A/m
Prob. 7.21
HP ý
HL ý
I
2ðò
aö ý
1
1
k an ý 10a x a z ý 5a y
2
2
I
I
(a x a z ) ý
ay
2ð (3)
6ð
H P H L ý 5a y
I
ay ý 0
6ð
I ý 30ð ý 94.25 A
Prob. 7.22
(a)
From eq. (7.29),
ü Iò
ÿÿ 2ð a 2 aö , 0 ü ò ü a
H=ý
I
ÿ
ò þa
aö ,
ÿþ 2ðò
(b)
ü I
a,
òüa
1 d
ÿ
( ò H ö )a z ý ý ð a 2 z
J = ñ H=
ò dò
ÿþ 0a z ,
ò þa
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Prob. 7.23
For 0 < ò < a
L
H dl ý I enc ý J dS
H ö 2ðò ý
2ð
ö ý0
ò
ò
ý0
Jo
ò
ò dö d ò
ý J o 2ðò
Hö ý J o
For ò > a
H dl = J
2ð
dS =
a
ö ò
=0
Jo
=0
ò
ò dö d ò
H ò 2ðò ý J o 2ð a
Hö ý
Joa
ò
ü J o , 0<ò <a
ÿ
Hence H ö ý ý J o a
ÿ ò , ò >a
þ
Prob. 7.24
(b)
J
(a)
ö
ý ñ H ý öx
y2
ö
öy
x2
ö
öz ý 2( x y )a z
0
dS ý dxdya z ,
5
2
2
5
ö x2 2 ö
ö y2 5 ö
I ý J ÷dS = (2 x 2 y )dxdy ý 2 dy xdx 2 dx ydy ý 2(4) ÷
2(2)
÷
÷
÷
ø 2 0ø
ø 2 1ø
1
0
0
1
S
ý 4(4 0) 2(25 1) ý 16 48 ý 32 A
Prob. 7.25
(a) J ý ñ H ý
2k
1 d
1 d
ò2
( ò H ö )a z ý
( ko
)a z ý o a z
a
a
ò dò
ò dò
(b)For ò>a,
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H dl ý I enc ý J dS ý
2ð
2ko
2ko
ò2 a
d
d
ý
(2
)
ò
ò
ö
ð
a
2 0
ò ý0 ö ý0 a
a
H ö 2ðò ý 2ð ko a
Hö ý
öaö
H ý ko ÷ ÷ aö ,
øòø
ò þa
ko a
ò
Prob. 7.26
ö
J ý ñ H ý öx
y2
ö
öy
x2
ö
öz ý (2 x 2 y )a z
0
At (1,-4,7), x =1, y = -4, z=7,
J ý 2(1) 2(4) ý a z ý 10a z A/m 2
Prob. 7.27
(a)
J ý ñ H ý
1 ö
1 ö
( ò Hö )a z ý
(103 ò 3 )a z
ò öò
ò öò
ý 3ò 103 a z A/m
2
(b)
Method 1:
I ý J dS ý 3ò ò dö d ò10 ý 3 10
3
S
ý 3 103 (2ð )
Method 2:
Iý
H
dl ý10
L
ò 2
3
3 2
2ð
3
ò
2
2ð
2
3
ò
0
2
d ò dö
0
ý 16ð 103 A ý 50.265 kA
ò dö ý 103 (8)(2ð ) ý 50.265 kA
0
Prob. 7.28
J ý ñ H ý
1 d
1 d
( ò H ö )a z ý
(4 ò 2 )a z ý 8a z
ò dò
ò dò
I ý J ÷ dS ýJS ý 8(ð a 2 ) ý 8ð 104 ý 2.513 mA
S
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Prob. 7.29
(a) B ý
ýo I
aö
2ðò
At (-3,4,5), ò=5.
Bý
(b)
4ð 107 2
aö ý 80aö nWb/m 2
2ð (5)
ý B ÷ dS ý
ýo I
2ð
d ò dz
ò
ý
6 4
4ð 107 2
ln ò z
2 0
2ð
ý 16 10 ln 3 ý 1.756 ý Wb
7
Prob. 7.30
Let H ý H1 H 2
where H1 and H 2 are due to the wires centered at x ý 0 and x ý 10cm respectively.
(a)
For H1 , ò ý 50 cm, aö ý al a ò ý a z a x ý a y
H1 ý
5
50
ay ý
a
2
ð y
2ð ø 5 10 ù
For H 2 , ò ý 5 cm, aö ý a z a x ý a y , H 2 ý H1
H ý 2H1 ý
100
ð
ay
ý 31.83 a y A/m
(b)
2a y a x
ö 2a a y ö
For H1 , aö ý a z ÷ x
÷ ý
5 ø
5
ø
ö a x 2a y ö
5
H1 ý
÷ ý 3.183a x 6.366a y
2 ÷
2ð 5 5 10 ø
5
ø
For H 2 , a ò ý a z a y ý a x
H2 ý
5
2ð ø 5 ù
a x ý 15.915a x
H ý H1 H 2
ý 12.3 a x 6.366a y A/m
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Prob. 7.31
(a) I ý J dS
ý
2ð
a
ò2
J o (1
a
ö ý0 ò ý0
(b)
H
a
0
a
0
0
ý J o dö ( ò
) ò d ò dö
2
ö ò2 ò4 ö
ý 2ð J o ÷
2÷
ø 2 4a ø
1
ý ð a2 Jo
2
2ð
ý
ò3
a2
)d ò
2ð ö 2 a 2 ö
Jo ÷ a ÷
2
2 ø
ø
dl = I enc ý J dS
For ò < a,
H ö 2ðò ý J dS
ö ò2 ò4 ö
= 2ð J o ÷
2÷
ø 2 4a ø
H ò 2ðò ý 2ð J o
Hò ý
ò2 ö
ò2 ö
2
÷
÷
a2 ø
4 ø
Jo ò ö
ò2 ö
2
÷
÷
4 ø
a2 ø
For ò > a,
H dl =
J o dS = I
1
H ö 2ðò ý ð a 2 J o
2
2
a Jo
Hö ý
4ò
ü Jo ò ö
ò2 ö
2
ÿ
÷
÷, ò ü a
a2 ø
ÿ 4 ø
Hence Hö ý ý
aJ o
ÿ
, ò >a
ÿþ
4ò
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Prob. 7.32
B ý
ý0 I
aö
2ðò
ý
B dS ý
da
ý0 I
dò dz
z ý 0 2ðò
ò
ýd
b
a
+ d
d
n
I
b π
I
0
μ 2
ý
Prob.7.33
For a whole circular loop of radius a, Example 7.3 gives
Hý
Ia 2 a z
2 ùû a 2 h 2 ùû
3/ 2
0
Let h
I
az
2a
For a semicircular loop, H is halfed
H=
H=
I
az
4a
B ý ýo H ý
ýo I
4a
az
Prob. 7.34
öBx öBy öBz
ý0
öx
öy
öz
showing that B satisfies Maxwell’s equation.
(a) ñ ÷ B ý
(b)
dS ý dydza x
4
ý B ÷ dS ý
1
z ý1 y ý 0
y 2 dydz ý
4
y3 1
( z ) ý 1 Wb
1
3 0
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(c ) ñ H = J
B
J ý ñ
ýo
ö
ö
ö
ñ B ý öx öy öz ý 2 za x 2 xa y 2 ya z
y 2 z 2 x2
2
J ý ( za x xa y ya z ) A/m 2
ýo
(d)
Since ñ ÷ B =0,
= B ÷ dS ý ñ ÷ Bdv ý 0
S
v
Prob. 7.35
h
ø ò aù
6
where H1 and H 2 are due to the wires centered at x ý 0 and x ý 10cm respectively.
On the slant side of the ring, z ý
ý
B.dS
ýo I
2ðò
ý
dò dz
ý
ýo I
2ð
ý
ýo Ih ö
abö
÷ b a ln
÷ as required.
a ø
2ð b ø
a b
h
( ò a )
b
z ý0
ò
ýa
dz dò
ò
ý
ýo Ih
2ð b
ö aö
1 ÷ dò
ýa ÷
ø òø
a b
ò
If a ý 30 cm, b ý 10 cm, h ý 5 cm, I ý 10 A,
ý
4ð 107 10 0.05 ö
4ö
÷ 0.1 0.3 ln ÷
2
3ø
2ð ø10 10 ù
ø
ý 1.37 108 Wb
Prob. 7.36
ý
B dS
ý ýo
0.2
50o
z ý0 ö ý0
106
ò
sin 2ö ò dö dz
ö cos 2ö ö
ý 4ð 107 106 ø 0.2 ù ÷
÷
2 ø
ø
ý 0.04ð ø1 cos 100o ù
ý
50o
0
0.1475 Wb
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Prob. 7.37
ð /4 2
ý B dS ý
ò
20
ö ý 0 ý1
S
ý 20(1)
ð /4
0
ò
2
sin ö ò d ò dö ý 20 d ò
2
1
ð /4
sin
2
ö dö
0
ð /4
1
1
(1 cos 2ö )dö ý 10(ö sin 2ö )
0
2
2
ð 1
ý 10( ) ý 2.854 Wb
4 2
Prob. 7.38
ý B dS , dS ý r 2sinñ dñ dö ar
S
ý
2ð
2
cos ñ r 2sinñ dñ dö
ý 2 dö
r ý1
r3
0
ð /3
ý 2(2ð ) sin ñ d (sin ñ ) ý 4ð
0
ð /3
cos ñ sin ñ dñ
0
sin ñ ð / 3
ý 2ð sin 2 (ð / 3)
0
2
2
ý 4.7123 Wb
Prob. 7.39
B ý ýo H ý
ýo J R
dv
4ð 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.
Bý
ý o ù qu R ù
4ð úû R 3 úû
In our case, u and R are perpendicular. Hence,
ýo qu 4ð 107 1.6 1019 2.2 106
1.6 1020
Bý
ý
ý
4ð R 2
4ð
(5.3 1011 ) 2
(5.3) 2 1022
ý 12.53 Wb/m 2
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Prob. 7.40
ý ya sin ax ù 0
A
ñ
A
(a)
ö
öx
ñ
ý
y cos ax
ö
öy
0
ö
öz
y e-x
ý a x e x a y cos axa z ù 0
A is neither electrostatic nor magnetostatic field
1 ö
1 ö
ò Bò ù ý
ø 20 ù ý 0
ø
ò öò
ò öò
ñ B ý 0
B can be E-field in a charge-free region.
(b)
ñB ý
(c )
ñ C ý
ö 2
1
(r sinñ ) = 0
r sin ñ öö
ö
1
1ö 3
ñC ý
r 2 sin 2 ñ ù ar (r sinñ )añ ù 0
ø
r sin ñ öñ
r ör
C is possibly H field.
Prob. 7.41
(a)
ñD ý 0
ö
öx
ñ D ý
y2z
ö
öy
ö
öz
2(x 1)yz -(x 1)z 2
ý 2(x 1)ya x . . . ù 0
D is possibly a magnetostatic field.
(b)
ñE ý
ñ E ý
1 ö
ö ö sin ö ö
ø ( z 1) cos ö ù ÷
÷ý0
öz ø ò ø
ò öò
1
ò2
cos ñ a ò . . . ù 0
E could be a magnetostatic field.
(c )
1 ö
1 ö ö sinñ ö
ø 2cosñ ù +
÷
÷ ù 0
2
rsinñ öñ ø r 2 ø
r ör
1 ù ö
2 sin ñ ù
r 1 sin ñ ù
añ ù 0
ñ F ý
ø
ú
r û ör
r 2 úû
F can be neither electrostatic nor magnetostatic field.
ñF ý
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Prob. 7.42
Aý
ýo Idl ýo ILaz
ý
4ð r
4ð r
This requires no integration since L << r.
öA
1 öAz
B ý ñ A ý
a ò z aö
ò öö
öò
But r ý ò 2 z 2
ýo ILa z
4ð ( ò 2 z 2 )1/ 2
ý IL 1
öAz ýo IL ö
ý
( ò 2 z 2 )1/ 2 ý o ( )( ò 2 z 2 ) 3/ 2 (2 ò )
öò
4ð öò
4ð
2
ýo IL ò aö
ý IL ò a
Bý
ý o 3ö
2
2 3/ 2
4ð ( ò z )
4ð r
Aý
Prob. 7.43
ö
öx
B ý ýo H ý ñ A ý
10sin ð y
Hý
ö
öz
ý ð sin ð xa y 10ð cos ð ya z
4 cos ð x
ö
ð ö
sin ð xa y 10 cos ð ya z ÷
÷
ýo ø
ø
ö
ð
öx
J ý ñ H ý
ýo
0
Jý
ö
öy
0
ö
ö
ö
ð ö
öy
öz
10ð sin ð ya x ð cos ð xa z ÷
ý
÷
ýo ø
ø
sin ð x 10 cos ð y
ð2
ø10sin ð ya x cos ð xa z ù
ýo
Prob. 7.44
(a)
ñ÷ A ý
ö
1 ö 2
1
1 ö ö 2 cos ñ
(r Ar )
( Añ sin ñ ) 0 ý 2 ÷
2
r ör
r sin ñ öñ
r ör ø r
ý
1
1
2 cos ñ 4
2sin ñ cos ñ ý 0
4
r
r sin ñ
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ö ö sin 2 ñ ö
1
ö
÷ 3 ÷
÷
ø r sin ñ öñ ø r ø
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(b)
öA
1 ù ö
( Aö sin ñ ) ñ
ú
r sin ñ û öñ
öö
öA ù
1ù ö
ú (rAñ ) r ú aö
öñ û
r û ör
B ý ñ A ý
ù
ù
1 ù 1 öAr ö
ú ar r ú sin ñ öö ör (rAö ) ú añ
û
û
û
1 ù ö ö sin ñ ö 2sin ñ ù
1 ù 2sin ñ 2sin ñ ù
ý 0 ar 0añ ú ÷ 2 ÷
aö ý ú
aö ý 0
ú
3
r û ör ø r ø
r û
rû
r3
r 3 úû
Prob. 7.45
ö
öx
B ý ñ A ý
(a)
ö
öy
ö
öz
2x 2 y yz xy 2 xz 3
6 xy 2z 2 y 2
B ý ( 6xz 4 x 2 y 3xz 2 )a x ø y 6yz-4xy 2 ù a y ø y 2 z 3 2 x 2 z ù a z Wb/m 2
(b)
ø 6 xz 4 x
2
ý
2
z ý0 y ý0
ý
2
y 3 xz 2 ù dy dz
ø 6 xz ù dy dz
0
4
2
2
2
0
0
0
ý 6 dz dy 4 dz
öy
ý 6(2) ø 2 ù 4(2) ÷
÷ 2
ø
ý 8 Wb
0
2
0
x ý1
x 2 y dy dz 3
2
ñ A ý
öA x
öx
öA y
öy
xz 2 dy dz
2
0
0
ö
öz ö
÷ +3(2) ÷
÷ ý
÷
÷
÷
3
0ø
ø 0ø
0
y dy 3 dy z 2 dz
2 2
(c )
3 2
öA z
öz
ý
-24+16+16
4xy 2xy 6 xy ý 0
ñ B ý 6 z 8 xy 3 z 3 6 z 8 xy 1 3z 3 1 ý 0
As a matter of mathematical necessity,
ñ ÷ B ý ñ ÷ (ñ A) ý 0
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Prob. 7.46
öA ù
ù
1 ù ö
1 ù 1 öAr ö
(rAö ) ú añ
( Aö sin ñ ) ñ ú ar ú
ú
öö û
r sin ñ û öñ
r û sin ñ öö ör
û
1 k ö
1 ö
ý
(sin 2 ñ )ar
(kr 1 ) sin ñ añ
2
r sin ñ r öñ
r ör
k
k sin ñ
k sin ñ
2k cos ñ
2sinñ cosñ ar
ý 3
añ ý
ar
añ
3
3
r sin ñ
r
r
r3
B ý ñ A ý
Prob. 7.47
öA
1 öAz
a ò z aö
ò öö
öò
B ý ñ A ý
ý
15
ò
ð
e ò cos ö a ò 15 e ò sin ö aö
1
1
ö
ö
B ÷ 3, , -10 ÷ ý 5 e 3
a ò 15 e 3
aö
2
2
ø 4
ø
B
107 15 3 ö 1
ö
ý
e ÷ a ò aö ÷
H ý
ýo
4ð
2
ø3
ø
H ý
ø14 a
ý
B dS
ý 15 z
ò
10
0
42 aö ù 104 A/m
ý
15
ò
ð
ø sin ö ù 0 2
e ò cos ö ò dö dz
e5 ý 150 e 5
ý 1.011 Wb
Prob. 7.48
ù ö
öAñ ù
ù
1 ù 1 öAr ö
ú öñ ( Aö sin ñ ) öö ú ar r ú sin ñ öö ör (rAö ) ú añ
û
û
û
û
öA ù
1ù ö
ú (rAñ ) r ú aö
öñ û
r û ör
1 10
1 ö
ý
2sin ñ cos ñ ar
(10) sin ñ añ 0aö
r sin ñ r
r ör
20
B ý 2 cos ñ ar
r
At (4, 60o , 30o ), r = 4, ñ =60o
B ý ñ A ý
Hý
B
ýo
ý
1
r sin ñ
1
ù 20
ù
cos 60o ar ú ý 4.974 105 ar A/m
7 ú 2
4ð 10 û 4
û
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Prob. 7.49
Applying Ampere's law gives
H ö 2ðò ý J o ðò 2
Jo
ò
2
Hö ý
ò
Bö ý ýo H ö ý ýo
ý
But B
ñ A
öAZ
öò
ý
öAZ
a . . .
öò
1
ý Jo ò
2
ý
Jo ò
2
AZ ý
ýo
1
or A ý - ýo J o ò 2 a z
4
Prob. 7.50
ö
B ý ñ A ý öx
0
ý
ð
2
sin
ö
öy
0
ðx
2
sin
ö
öA
öA
öz
ý z ax z a y
öy
öx
Az ( x, y )
ðy
2
ax
ð
2
cos
ðx
2
cos
ðy
2
ay
Prob. 7.51
ö
1
1 ö
( Aö sin ñ )ar
(rAö )añ
r sin ñ öñ
r ör
A
1 Ao
(2sin ñ cos ñ )ar o sin ñ (r 2 )añ
ý
2
r sin ñ r
r
A
ý 3o (2 cos ñ ar sin ñ añ )
r
B ý ñ A ý
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Jo ò 2
4
a
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Prob. 7.52
ö
ö
ö
öz ý (2 yz x 2 )a x (2 xz 2 xy )a z
(a) J ý ñ H ý öx öy
xy 2 x 2 z y 2 z
At (2,-1,3), x=2, y=-1, z=3.
J ý 2a x 16a z A/m 2
(b)
öò v
ý ñ ÷ J ý 0 2x 2x ý 0
öt
At (2,-1,3),
öòv
ý 0 C/m3s
öt
Prob. 7.53
(a) B ý ñ A
öA ù
ù 1 öAz öAö ù
ù öA öA ù
1ù ö
a ò ú ò z ú aö ú ( ò Aö ) ò ú a z
ýú
ú
öz û
öò û
öö û
ò û öò
û ò öö
û öz
öA
ý z aö ý 20 ò aö ý Wb/m 2
öò
B 20 ò
ý
Hý
aö ý A/m
ýo
ýo
1 ö
( ò Aö )a z
ò öò
40
1
ý
a z ý A/m 2
(40 ò )a z ý
J ý ñ H ý
ýo ò
ýo
I ý J dS ý
(b)
ý
40
ýo
2
2ð
40
ýo
ò d ò dö ý
0
0
2
2ð
ò ö
ò dö d ò , dS = ò dö d ò a z
ý0 ý0
40 ò 2
ýo 2
2
0
(2ð )
80ð 2 106
ý
ý 400 A
4ð 107
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Prob. 7.54
H ý
ñVm
Ia 2
From Example 7.3, H ý
Ia 2
2
Vm ý
H dl ý mmf
Vm ý
2 ø z2 a 2 ù
øz
a2 ù
2
3
2
3
2
az
dz ý
Iz
2 ø z2 a 2 ù
1
c
2
As z , Vm ý 0 , i.e.
0 ý
I
c
2
c ý
I
2
Hence,
I
2
Vm ý
ù
ú1
û
ù
ú
z2 a 2 û
z
Prob. 7.55
H ý
I
aö
2ðò
ø J ý 0ù
H ý ñVm
But
I
2ðò
aö ý
1 öVm
a
ò öö
At ø10, 60o , 7 ù , ö ý 60o ý
or
C ý
ø 4, 30 , 2 ù ,
o
, Vm ý 0
0 ý
I
I
ö
2ð
6
ö ý 30o ý
ð
6
,
I
I ð
2ð 6
6
ý 1A
Vm ý
Vm
3
I
ö C
2ð
I
6
Vm ý
At
ð
Vm ý
ý
I
12
ý
12
12
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I ð
C
2ð 3
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Prob. 7.56
Fo r a n infinite c urre nt sheet,
1
1
H ý
K an ý
50a y a n
2
2
But
H ý
ý 25a x
ø J ý 0ù
ñVm
öVm
a n Vm ý 25x c
öx
At the orig in, x ý 0, Vm ý 0, c ý 0, i.e.
25 a x
ý
Vm
ý
25 x
(a )
At
(b )
At
ø 2, 0, 5 ù ,
ø10, 3, 1ù ,
Vm ý 50A.
Vm ý 250A.
Prob. 7.57
(a)
ö öV
1 öV
öV ö
ñ ñV ý ñ ÷
aò
aö
az ÷
öz ø
ò öö
ø öò
ý
ö 1 ö 2V
ö ö 2V
1 ö 2V ö
a
÷
÷
÷
ò dzöö ø ò
ø ò öööz
ø özöò
1 ö ö 2V
÷
ò ø öòöö
(b)
ö 2V ö
÷ aö
öòöz ø
ö 2V ö
÷ az ý 0
öö d ò ø
ùö 1 öAz
ñ ø ñ A ù ý ñ ú÷
ûø ò öö
ö öA
÷ ò
ø öz
öAö ö
÷ aò
öz ø
öAz ö
1ö ö
÷ aö
÷
öò ø
ò ø öò
øò A ù
ö
2
1 ö 2 Az
1 ö ö öAö ö
1 ö Aò
ò
÷
÷
ò öò ø öz ø
ò öööz
ò öòöö
ö
ö ö1 ö
ö ö 1 öAò ö
ò Aö ù ÷
ø
÷
÷
÷
öz ø ò öò
öz ø ò öö ø
ø
ö 2 Aö
ö 2 Aö
1 öAö
1 öAö
ý
öòöz
özöò
ò öz
ò öz
ý
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öAò ö ù
÷ az ú
öö ø û
1 ö 2 Az
ò öööò
ý 0
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Prob. 7.58
ñ'
1
R
1
ö ö
ö ù
ö
ö
2
2
2 2
ù
x
x'
y
'
z
z'
y
a
a
a
ø
ù
ø
ù
ø
ù
÷
x
y
z ÷
û
öy '
öz ' ø û
ø öx '
3
2
2
2 2
ö 1ö
ù
ù
2
x
x'
x
x'
y
'
z
z'
y
a
ù x ûø
ù ø
ù ø
ù û a y and a z terms
÷
÷ ø ù ø
ø 2ø
R
R3
ý
ý
ý
R ý
ñ
1
R
ý
ý
ý
1
2
2
2
2
ý ùúø x x' ù ø y y 'ù ø z z ' ù ùú
û
û
1
ö ö
ö
ö ö ù
2
2
2 2
ù
y
a
a
a
x
x'
y
'
z
z'
ø
ù ø
ù ø
ùû
x
y
z ÷
÷
öy
öz ø û
ø öx
r r'
3
1
2
2
2 2
ù
ù
2 ø x x' ù a x ø x x' ù ø y y' ù ø z z' ù
a y and az terms
û
û
2
ùûø x x'ù a z ø y y 'ù a y ø z z'ù a z ùû
R
ý 3
R3
R
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CHAPTER 8
P.E. 8.1
öu
ý QE ý 6a z N
öt
öu
ö
(b)
ý 6a z ý (u x , u y , u z )
öt
öt
öu x
ý 0 ux ý A
öt
öu y
ý 0 uy ý B
öt
öu z
ý 6 u z ý 6t C
öt
A=B=C=0
Since u ø t ý 0 ù ý 0 ,
(a) F ý m
ux = 0 = uy, uz = 6t
öx
ux ý
ý0 xý A
öt
öy
uy ý
ý0 yý B
öt
öz
ý 6t z ý 3t 2 C1
uz ý
öt
At t = 0, (x,y,z) = (0,0,0) A1 = 0 = B1 = C1
Hence , (x,y,z) = (0,0,3t2),
u ý 6ta z at any time. At P(0,0,12), z = 12 =3t2 t =2s
t =2s
(c) u ý 6ta z ý 12a z m/s .
öu
aý
ý 6a z m 2
s
öt
(d) K .E ý
1
1
2
m u ý ø1ùø144 ù ý 72 J
2
2
P.E. 8.2
(a)
ma ý eu B = (eBouy, -eBoux, 0)
d 2 x eBo dy
dy
ý
ý
2
dt
m dt
dt
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d2y
eBo dx
dx
ý
ý
2
dt
m dt
dt
(2)
d 2z
dz
ý 0;
ý C1
2
dt
dt
(3)
From (1) and (2),
d 3x
d2y
dx
ý
ý 2
3
2
dt
dt
dt
(D2 + w2 D)x = 0 Dx = (0, j)x
x = c2 + c3cost +c4sint
dy 1 d 2 x
ý
ý c3 cos t c4 sin t
dt dt 2
At t = 0, u ý (ñ , 0, ò ) . Hence,
c1 ý ò , c3 ý 0, c4 ý
ñ
dx
dy
dz
ý ñ cos t , ý ñ sin t , ý ò
dt
dt
dt
(b)
Solving these yields
a
ñ
x ý sin t , y ý cos t , z ý ò t
The starting point of the particle is (0,
(c)
x2 + y2 =
ñ
,0)
ñ2
, z=òt
2
showing that the particles move along a helix of radius ñ
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.
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P.E. 8.4
By Newton’s 3rd law, F12 ý F21 , the force on the infinitely long wire is:
ýIIb 1
1
)a
Fl ý F ý o 1 2 (
2ð
òo òo a ò
4ð 107 50 3 ö 1 1 ö
ý
÷ ÷ a ò ý 5a ò ý N
2ð
ø 2 3ø
P.E. 8.5
m ý ISan ý 10 104 50
(2, 6, 3)
7
= 7.143 x 10-3 (2, 6, -3)
ý (1.429a x 4.286a y 2.143a z ) 102 A-m 2
P.E. 8.6
T ý mB ý
(a)
10 104 50 2 6 3
6 4 5
7 10
ý 0.03a x 0.02a y 0.02a z N-m
(b)
T ý ISB sin ñ
| T |max =
P.E. 8.7
(a)
ýr ý
(b)
Hý
(c)
T
max
ý ISB
50 10 -3
| 6a x 4a y 5a z |ý 0.04387 Nm
10
ý
ý 4.6, ó m ý ý r 1 ý 3.6
ýo
10 103 e y
a A / m ý 1730e y a z A/m
ý 4ð 107 4.6 z
M ý ó m H ý 6228e y a z A/m
B
ý
P.E. 8.8
3a 4a y 6a x 8a y
an ý x
ý
5
10
(6 32)(6a x 8a y )
B1n ý ( B1 ÷ an )an ý
1000
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ý 0.228a x 0.304a y ý B 2 n
B1t ý B1 B1n ý 0.128a x 0.096a y 0.2a z
B2t ý
ý2
B ý 10 B1t ý 1.28a x 0.96a y 2a z
ý1 1t
B2 ý B2 n B2t ý 1.052a x 1.264a y 2a z Wb/m2
P.E. 8.9
B1n ý B2 n ý1 H1n ý z ý2 H 2 n
(a)
or ý1 H1 ÷ an 21 ý ý2 H 2 ÷ an 21
(6 H 2 x 10 12)
(60 2 36)
ýo
ý 2ýo
7
7
35 ý 6 H 2 x
H 2 x ý 5.833 A/m
(b)
K ý ( H1 H 2 ) an12 ý an 21 ( H1 H 2 )
= an 21 ù (10,1,12) (35 , 5, 4) ù
6
û
û
=
2 3
1 6
7 25 6 6 8
K ý 4.86a x 8.64a y 3.95a z A/m
(c)
Since B ý ý H , B1 and H1 are parallel, i.e. they make the same angle with the
normal to the interface.
H ÷a
26
ý 0.2373
cos ñ1 ý 1 n 21 ý
H1
7 100 1 144
ñ1 ý 76.27 o
cos ñ 2 ý
H 2 ÷ an 21
13
ý
ý 0.2144
H2
7 (5.833) 2 25 16
ñ 2 ý 77.62o
P.E. 8.10
(a)
L' ý ýo ý r n 2 S ý 4ð 107 1000 16 106 4 104
= 8.042 H/m
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(b)
Wm ' ý 1 L' I 2 ý 1 (8.042)(0.5 2 ) ý 1.005 J/m
2
2
From Example 8.11,
P.E. 8.11
Lin =
ýo l
8ð
Lext =
2 wm 1
ý 2
I2
I
=
l
1
4ð
2
2ð
ýI 2
4ð 2 ò 2 ò d ò dö dz
b
2 ýo
(1 ò ) ò dò
dz dö
0
0
a
ù1
2ýo
1 ù
dò
÷ 2ð l ú
2
4ð
ò (1 ò ) úû
a û
b
=
1 b ù
ý ol ù b
ln ln
ú
1 a úû
ð û a
ý l ý lù b
1 bù
L = Lin + Lext = o o úln ln
ð û a
8ð
1 a úû
=
P.E. 8.12
(a)
L’in =
4ð 107
ýo
=
= 0.05 ýH/m
8ð
8ð
L’ext = L’ – L’in = 1.2 – 0.05 = 1.15 ýH/m
(b)
L’ =
ln
d aù
ýo ù 1
ln
ú
a úû
2ð û 4
2ð 1.2 106
d a 2ð L '
ý
0.25 ý
0.25
ýo
4ð x107
a
ý 6 0.25 ý 5.75
d a
ý e 5.75 ý 314.19
a
d a ý 314.19a ý 314.19
2.588 10 3
ý 406.6mm
2
d ý 407.9mm ý 40.79cm
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P.E. 8.13
This is similar to Example 8.13. In this case, however, h=0 so that
ýo I1a 2b
A1 ý
aö
4b3
ý I a2
ý ðI a 2
ö12 ý o 12 ÷ 2ðb ý o 1
2b
4b
ýoð a
ö12
m12 ý
ý
2b
I1
= 2.632 ýH
2
4ð 107 ð 4
ý
23
P.E. 8.14
Lin =
4ð 107 10 102
ýo
ý 2ðò o
l= o
=
8ð
8ð
4
= 31.42 nH
P.E. 8.15
(a) From Example 7.6,
Bave ý
ýo NI
l
ö ý Bave ÷ S ý
or I ý
ý
ýo NI
2ðòo
ýo NI
÷ ða 2
2ðò o
2 ò oö 2 10 102 0.5 103
ý
ýa 2 N
4ð 10 7 10 4 103
= 795.77A
Alternatively, using circuit approach
l
2ðò o
2ðòo
Rý
ý
ý
ýS ýo S ýoða 2
öò 2 ò oö
ñ ý NI ý
ý
, as obtained before.
N
ýa 2 N
òý
2 òo
2 10 102
ý
ý 1.591 109
2
7
4
ýa
4ð 10 10
ñ ý öò = 0.5x10 x1.591x10 =7.9577x10
-3
Iý
(b)
9
5
ñ
ý 795.77 A as obtained before.
N
If ý=500ýo,
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Iý
795.77
ý 1.592 A
500
P.E. 8.16
ñý
B 2 a S (1.5) 2 10 104 22500
= 895.25N
ý
ý
2ýo
8ð
2 4ð 10 7
Prob. 8.1
At P, x = 2, y = 5, z = -3
E ý 2(2)(5)(3)a x (2)2 (3)a y (2)2 (5)a z ý 60a x 12a y 20a z
B ý (5) 2 a x (3) 2 a y 22 a z ý 25a x 9a y 4a z
F ý Q( E u B )
u B ý
1.4 3.2 1
ý 3.8a x 30.6a y + 92.6a z
25
9
4
E u B ý (60, 12, 20) (3.8, 30.6,92.6) ý (63.8, 42.6,112.6)
F ý Q( E u B ) ý 4( E u B ) mN
= 255.2a x 170.4a y 450.4a z mN
Prob. 8.2
F ý m 2 r ý 9.111031 (2 1016 ) 2 (0.4 1010 ) ý 14.576 nN
Prob. 8.3
(a)
F ý Q(u B ) ý 103
10 2
6
0
25
0
ý 103 (50a x 250a y )
= 0.05a x 0.25a y N
(b) Constant velocity implies that acceleration a = 0.
F = ma ý 0 ý Q( E u B )
E = -u B ý
50a x + 250 a y V/m
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Prob. 8.4
Fe ý qE,
Fm ý qu B
Fe
20 103
E
ý
ý
ý 80
Fm uB 0.5 108 5 106
Prob. 8.5
ma ý Qu B
103 a ý 2 103
ux
0
uy
6
uz
0
d
(u x , u y , u z ) ý (12u z ,0,12u x )
dt
i.e.
du x
ý 12u z
dt
du y
ý 0 u y ý A1
dt
du z
ý 12u x
dt
(1)
(2)
(3)
From (1) and (3),
ux ý 12u z ý 144u x
or
ux 144u x ý 0 u x ý c1 cos12t c 2 sin 12t
From (1), uz= - c1sin12t + c2cos12t
At t=0,
Hence,
ux=5, uy=0, uz=0 A1=0=c2, c1=5
u ý (5cos12t , 0, 5sin12t )
u(t ý 10s ) ý (5cos120, 0, 5sin120) = 4.071a x 2.903a z m/s
dx
ý 5 cos12t x ý 5 sin 12t B1
12
dt
dy
uy ý
ý 0 y ý B2
dt
dz
uz ý
ý 5 sin 12t z ý 5 cos12t B3
12
dt
ux ý
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At t=0, (x, y, z) = (0, 1, 2) B1=0, B2=1, B3=
19
12
5
19 ö
ö 5
( x , y , z) ý ÷ sin 12t ,1, cos 12t ÷
ø 12
12
12 ø
(4)
At t=10s,
5
19 ö
ö 5
( x , y , z) ý ÷ sin 120 ,1, cos 120 ÷ = (0.2419, 1, 1.923)
ø 12
12
12 ø
By eliminating t from (4),
x 2 ( z 19 ) 2 ý ( 5 ) 2 , y ý 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 ý e(u B )
u
m d
(u x , u y , u z ) ý x
0
e dt
uy
0
uz
ý u y Bo ax Bo u x a y
Bo
du z
ý 0 uz ý c ý 0
dt
du x
Be
Be
ý u y o ý u y w , where w = o
m
dt
m
du y
ý ux w
dt
Hence,
ux ý wu y ý w2u x
or ux w2u x ý 0 u x ý A cos wt B sin wt
uy ý
u x
ý A sin wt B cos wt
w
At t=0, ux = uo, uy = 0 A = uo, B=0
Hence,
u
dx
x ý o sin wt c1
dt
w
dy
u
u y ý uo sin wt ý
y ý o cos wt c2
dt
w
u x ý uo cos wt ý
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At t=0, x = 0 = y c1=0, c2=
xý
uo
. Hence,
w
uo
u
sin wt , y ý o (1 cos wt )
w
w
2
u 2o
u
öu ö
(cos2 wt sin 2 wt ) ý ÷ o ÷ ý x 2 ( y o ) 2
2
w
w
øwø
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 ý qE
qý
mg 0.4 103 9.81
ý
ý 26.67 nC
1.5 105
E
Prob. 8.8
F ý Idl B ý IL B
ax
ay
az
ý 4.5(0.2)a x (2.5)(a y a z )10 ý 4.5(0.5) 1
0
0
1
0 mN
1
3
F ý 2.25(a y a z ) mN
Prob. 8.9
qE ý quB
Bý
E 12 103 1200
ý
ý
ý 85.714 Wb/m 2
u
140
14
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Prob. 8.10
ý I I a a
F
ý I1al B2 ý o 1 2 l ö
L
2ðò
7
a (a y )4ð 10 (100)(200)
F21 ý z
ý 4a x mN/m (repulsive)
2ð
ñ ý IL B ñ ý
(a)
(b)
(c)
(d)
F12 ý F21 ý 4a x mN/m (repulsive)
4
3
3
4
al aö ý a z ( a x a y ) ý a x a y , ò ý 5
5
5
5
5
7
4
4ð 10 (3 10 ) ö 3
4 ö
F31 ý
÷ ax a y ÷
2ð (5)
5 ø
ø 5
ý 0.72a x 0.96a y mN/m (attractive)
F3 ý F31 F32
4ð 107 6 104 )
ø az a y ù ý 4ax mN/m(attractive)
2ð (3)
F3 ý 3.28a x 0.96a y mN/m
F32 ý
(attractive due to L2 and repulsive due to L1)
Prob. 8.11
Fý
ýo I1 I 2 4ð 107 (10)10
ý
ý 100 ý N
2ðò
2ð (20 102 )
Prob. 8.12
F ý Ldl B ý 3(2a z ) cos ö
W ý F ÷ dl ,
F ý 6 cos ö
2ð
3
= -1.8sin
aö
aö N
W ý 6 cos ö
0
3
3
òo dö ý 6 òo 3sin ö 3
2ð
0
J
2ð
= -1.559 J
3
Prob. 8.13
(a)
F1 ý
ýo I1 I 2
4ð 107
ò
d
(2)(5) ln 6 a z
a
a
ý
ò
ö
ò ý2 2ðò
2
2ð
6
ý 2 ln 3a z ý N = 2.197a z ý N
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(b)
F2 ý I 2 dl2 B1
ýo I1 I 2
2ð
ýII
ý o 1 2
2ð
ý
1
ò ùû d ò aò dza ùû aö
z
1
ò ùûd ò a
z
dzaò ùû
But ò = z+2, dz=dò
4ð 107
1
(5)(2) ùû d ò a z dza ò ùû
F2 ý
2ð
ò ý4 ò
2
2 ln 2 (a z aò ) ý N ý 1.386aò 1.386a z a z ý N
4
ýII 1
F3 ý o 1 2 ùû d ò a z dza ò ùû
ò
2ð
But z = -ò + 6, dz = -dò
4
4ð 107
1
(5)(2) ùû d ò a z dza ò ùû
F3 ý
2ð
ò ý6 ò
2 ln 4 (a z aò ) ý N ý 0.8109a ò 0.8109a z ý N
6
F ý F1 F2 F3
ý a ò (ln 4 ln 4 ln 9) a z (ln 9 ln 4 ln 4 ln 9)
ý 0.575a ò ý N
Prob. 8.14
A
fBC
From Prob. 8.7,
f ý
C
ýo I1 I 2
aò
2ðò
f ý f AC f BC
60o
B
4ð 107 75 150
ý 1.125 103
2ð 2
o
f ý 2 1.125cos 30 a x mN/m
| f AC |ý| f BC |ý
ý 1.949a x mN/m
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30o
fAC
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Prob. 8.15
The field due to the current sheet is
Bý
ý
2
K an ý
ýo
2
10a x (a z ) ý 5ý o a y
L
F ý I 2 dl2 B ý 2.5 dxa x (5ýo a y ) ý 2.5L 5ýo (a z )
0
F
ý 12.5 4ð 107 (a z ) ý 15.71a z ý N/m
L
Prob. 8.16
F ý Idl B ý IL B ý 5(2a z ) 40a x 103 ý 0.4a y N
Prob. 8.17
m ý ISan ý 10(2 6)(a x ) ý 120a x
T ý m B ý 120a x 4.5(a y a z ) ý 540
Prob. 8.18
f ( x, y, z ) ý x 2 y 5 z 12 ý 0
an ý
1 0
0
0 1 1
ý 540(a y +a z ) N.m
ñf ý a x 2a y 5a z
a x 2a y 5a z
ñf
ý
| ñf |
30
m ý NISan ý 2 60 8 104
(a x 2a y 5a z )
30
ý 17.53a x 35.05a y 87.64a z mAm
Prob. 8.19
m ý IS
Iý
Iý
m
m
ý 2
S ðr
8 1022
ý 6.275 108 ý 627.5 MA
ð (6370 103 ) 2
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Prob. 8.20
Let F ý F1 F2 F3
0
F1 ý Idl B ý 2dxa x 30a z mN
5
0
ý 300a y mN
5
=-60a y x
5
F2 ý 2dya y 30a z mN
0
=60a x y
5
0
ý 300a x mN
5
F3 ý 2(dxa x dza z ) 30a z mN
0
5
ý 300a y mN
0
F ý F1 F2 F3 ý 300a y +300a x -300a y mN=300a x mN
=60(-a y ) x
1
T ý m B ý ISan B ý 2( )(5)(5)a y 30a z 103 ý 0.75a x N.m
2
Prob. 8.21
For each turn, T = m B, m = ISan
For N turns,
T ý NISB ý 50 4 12 104 100 103 ý 24 mNm
Prob. 8.22
F ý Idl B
F ý IB ý 520 0.4 103 30 103
F ý 6.24 mN
Prob. 8.23
M ý óm H ý óm
B
ýo ýr
ý
óm B
ýo (1 ó m )
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Prob. 8.24
M ý óm H ý óm
(a )
B
ýo ý
4999
1.5
5000
4ð 107
M ý
ý
1.193 106 A/ m
N
M ý
(b )
õm
k ý1
k
v
If we a ssum e tha t a ll m k a lig n with the a p p lied B fie ld ,
Nmk
v
M ý
mk ý
mk ý
M
N
v
ý
1.404 1023 A m 2
Prob. 8.25
ýr ý ó m 1 ý 6.5 1 ý 7.5
24 y 2
az
Hý
ý
óm
6.5
M
M ý óm H
At y = 2cm,
24 4 104
a z ý 1.477a z mA/m
6.5
ö
ö
ö
öx öy
öz
48 y
J ý ñ H ý
ax
ý
2
6.5
24 y
0
0
6.5
At y=2cm,
Hý
Jý
48 2 102
a x ý 0.1477a x A/m 2
6.5
Prob. 8.26
(a)
(b)
(c )
ý ý 80 ýo
ó m ý ýr 1 ý 79
Hý
B
ý
ý
ýr ý 80
20 xa y 103
80(4ð 107 )
ý 198.9 xa y A/m
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1.193 106
8.5 1028
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(d )
( e)
M ý ó m H ý 15.713 xa y kA/m
ö
ö
öy
J b ý ñ M ý öx
0 15.713x
ö
öz ý 15.713a z kA/m
0
Prob. 8.27
When H = 250,
2H
2(250)
Bý
ý
ý 1.4286 mWb/m 2
100 H 100 250
But B=ýo ýr H
ýr ý
1.4286 103
B
ý
ý 4.54
ýo H 4ð 107 250
Prob. 8.28
H dl
ý Ienc
Hö 2ðò ý
ðò 2
I
ð a2
M ý óm H ý
Jb ý ñ M ý
Iò
2ð a 2
Hö ý
Iò
aö
2ð a 2
ø ýr 1ù
1 ö
ò öò
ø òM ù a
ö
z
ý
ø ýr 1ù
I
a
ð a2 z
Prob. 8.29
(a)
From H1t – H2t = K and M = ómH, we obtain:
M1t
ó m1
M2t
óm2
ýK
Also from B1n – B2n = 0 and B = ýH = (ý/óm)M, we get:
ý1 M 1n ý 2 M 2 n
ý
ó m1
ó m2
(b)
From B1cosñ1 = B1n = B2n = B2cosñ2
B sin ñ1
B sin ñ 2
ý H1t ý K H2 t ý K 2
and 1
ý1
ý2
Dividing (2) by (1) gives
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(1)
(2)
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tanñ1
ý1
i.e.
ý
k
ký 2 ö
tanñ 2 tanñ 2 ö
÷
÷÷1
ý
B2 cosñ 2
ý2
ý 2 ø B2 sin ñ 2 ÷ø
ký 2
tan ñ 1 ý1 ö
÷÷1
ý
tan ñ 2 ý 2 ø B2 sin ñ 2
ö
÷÷
ø
Prob. 8.30
B2 n ý B1n ý 1.8a z
H 2t ý H1t
B2t ý
B2t
ý2
ý
B1t
ý1
4 ýo
ý2
B1t ý
(6a x 4.2a y ) ý 9.6a x 6.72a y
ý1
2.5ýo
B2 ý B2 n B2t ý 9.6a x 6.72a y 1.8a z mWb/m 2
H2 ý
B2
ý2
ý
103 (9.6, 6.72,1.8)
4 4ð 107
ý 1,909.86 a x 1,336.9a y 358.1a z A/m
z
B2n
ñ2
B2t
B2 n
1.8
ý
ý 0.1536
B2t
9.62 6.722
ñ 2 ý 8.73o
tan ñ 2 ý
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Prob. 8.31
B2 n ý B1n ý 12a z
H 2t ý H1t
B2t ý
B2t
ý2
ý
B1t
ý1
2ý
ý2
B1t ý o (4a x 10a y ) ý 1.6a x 4a y
ý1
5ý o
B2 ý B2 n B2t ý 1.6a x 4a y 12a z mWb/m 2
B23 (1.62 42 122 ) 106
1
3
w2 ý ý2 H 2 ý
ý
ý 32.34 J/m3
7
2
2 ý2
2(2)(4ð 10 )
Prob. 8.32
ñ1
1
2
ñ2
tan ñ1 ý
But
H1t
,
H 1n
tan ñ 2 ý
H 2t
H 2n
H1t ý H 2t
B1n ý B2t
ý1 H1n ý ý2 H 2 n
H 2n ý
ý1
H
ý 2 1n
H 2t
H1t
6.5H1t
ý
ý
1
H 2n
H1n
H 1n
6.5
If ñ1 ý 42o ,
tan 42o ý
H1t
H 1n
tan ñ 2 ý 6.5(0.9004) ý 5.832
H1t
ý 0.9004
H1n
ñ 2 ý 80.3o
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Prob. 8.33
x <0
K
x>0
( H1 H 2 ) an12 ý K
H1 a x ý K H 2 a x
H1 a x ý (10a x 6a z ) a x ý 6a y
H 2 ax ý ( H 2 x , H 2 y , H 2 z ) ax ý H 2 z a y H 2 y az
6a y ý 12a y H 2 z a y H 2 y a z
Equating components,
6 ý 12 H 2 z
H 2 z ý 6,
H2 y ý 0
Also,
H 2n ý
B1n ý B2 n
ý1 H1n ý ý2 H 2 n
2ý
ý1
H1n ý o (10a x ) ý 5a x
4ýo
ý2
H 2 ý 5a x 6a z A/m
Prob. 8.34
H 2t ý H1t ý ñ a x ô a z
B2 n ý B1n
ý2 H 2 n ý ý1 H1n
ý1
ý
H1n ý r1 ò a y
ý2
ýr 2
ý
H ý ñ a x r1 ò a y ô a z
ýr 2
H 2n ý
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Prob. 8.35
B1n ý B 2n ý 15aö
(a)
B1t
H1t ý H 2t
ý1
ý
B2t
ý2
ý1
2
B2t ý
ø10aò 20az ù ý 4aò 8az
5
ý2
B1t ý
Hence,
B1 ý 4a ò 15aö 8a z mWb/m 2
(b)
w m1
B12
1
ý
B1 H1 ý
2
2ý1
w m1 ý 60.68 J / m3
B22
ý
2 ý2
w m2
Prob. 8.36
ý
ø10
2
ý
ø4
2
152 82 ù 106
2 2 4ð 107
152 202 ù 106
2 5 4ð 10
7
ý 57.7 J / m3
f ( x, y , z ) ý x y 2 z
ñ f ý a x a y 2a z
an ý
1
ñf
( a x a y 2a z )
ý
| ñf |
6
(a)
H1n ý ( H1 an )an ý (40 20 60)
(a x a y 2a z )
6
ý 6.667a x 6.667a y 13.333a z A/m
(b)
H 2 ý H 2 n H 2t
B 2 n ý B1n
ý2 H 2 n ý ý1 H1n
B2 ý ý2 H 2 ý ý2 H 2 n ý2 H 2t ý ý1 H1n ý2 H 2t ý ýo (2 H1n 5 H 2t )
But
ý 4ð 107 (13.333,13.333, 26.667) (233.333, 66.666, 83.333ý
ý 4ð 107 (220,80, 110)
ý 276.5a x + 100.5a y 138.2a z ý Wb/m 2
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Prob. 8.37
an ý a ò
B2 n ý B1n ý 22 ýo aò
H 2t ý H1t
B2t ý
B2t
ý2
ý
B1t
ý1
ýo
ý2
(45ýo aö ) ý 0.05625ýo aö
B1t ý
800 ýo
ý1
B2 ý ýo (22a ò 0.05625aö ) Wb/m 2
Prob. 8.38
H1n ý 3a z ,
H1t ý 10a x 15ay
H 2t ý H1t ý 10a x 15a y
H 2n ý
ý1
1
H1n ý
(3a z ) ý 0.015a z
200
ý2
H 2 ý 10a x 15a y 0.015a z
B2 ý ý2 H 2 ý 200 4ð 107 (10,15, 0.015)
B2 ý 2.51a x 3.77a y 0.0037a z mWb/m2
tan ñ ý
B2 n
B2t
or ñ ý tan 1
0.0037
2.51 3.77
2
2
= 0.047o
Prob. 8.39
(a)
H ý 1 K an ý 1 (30 40)a x (a z ) ý 5a y A/m
2
2
B ý ýo H ý 4ð 107 (5a y ) ý 6.28a y ý Wb/m2
(b)
H ý 1 (30 40)a y ý 35a y A/m
2
B ý ýo ýr H ý 4ð 107 (2.5)(35a y ) ý 110a y ý Wb/m 2
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(c)
H ý 1 (30 40)a y ý 5a y
2
B ý ýo H ý 6.283a y ý Wb/m2
Prob. 8.40
r = a is the interface between the two media.
Bo1 (1 1.6) cos ñ ar ý Bo 2 cos ñ ar
B2 n ý B1n
2.6 Bo1 ý Bo 2
(1)
H 2t ý H1t
B2t
ý2
ý
B1t
ý1
ý2 B1t ý ý1 B2t
ý2 Bo1 (0.2) sin ñ añ ý ýo Bo 2 ( sin ñ )añ
ýB
(2)
ý2 ý o o 2
0.2 Bo1
Substituting (1) into (2) gives
ý2 ý
ýo
0.2
(2.6) ý 13ýo
Prob. 8.41
(a)
The square cross-section of the toroid is shown below. Let (u,v) be the local
coordinates and ò o =mean radius. Using Ampere’s law around a circle passing
through P, we get
v
u
(0, ò o )
H (2ð )( ò o v) ý NI
Hý
NI
2ð ( òo v)
The flux per turn is
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ý
a/2
a/2
Bdudv ý
u ý a / 2 v ý a / 2
Lý
ýo NIa ö òo a / 2 ö
ln ÷
÷
2ð
ø òo a / 2 ø
N ýo N 2 a ö 2 òo a ö
ý
ln ÷
÷
I
2ð
ø 2 òo a ø
(b)
The circular cross-section of the toroid is shown below. Let (r,ñ) be the local
coordinates. Consider a point P( r cos ñ , ò o r sin ñ ) and apply Ampere’s law
around a circle that passes through P.
H (2ð )( ò o r sin ñ ) ý NI
Hý
r
NI
NI ö r sin ñ ö
÷1
÷
2ð ( ò o r sin ñ ) 2ðòo ø
òo ø
ñ
(0, ò o )
Flux per turn ý
a 2ð
r ý0 ñ
Lý
ý NI ö r sin ñ ö
ý NI a 2
(2ð )
÷1
÷rdrdñ ý
2ðò o ø
2ðò o 2
òo ø
N ý N 2a2
ý
I
2 òo
Or from Example 8.10,
L ý L' l ý
ýo N 2lS
l2
ý
ýo N 2ða 2 ýo N 2a 2
ý
2ðò o
2 òo
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Prob. 8.42
1
2
a ý 2 cm
òo ý (3 5) ý 4cm
Lý
ýo N 2 a ù 2 ò o a ù
ln ú
ú
2ð
û 2 òo a û
N2 ý
2ð L
2ð (45 106 )
ý
ý 22, 023.17
ù 2 òo a ù
ö8 2ö
7
2
ýo a ln ú
ú 4ð 10 (2 10 ) ln ø÷ 8 2 ø÷
û 2 òo a û
N ý 148.4 or 148
Prob. 8.43
Lý
ýo 4ð 107 (40)
ý
ý 20 107 ý 2 ý H
8ð
8ð
Prob. 8.44
Lin ý
ýo
,
8ð
ýo
ln(b / a )
2ð
ýo ýo
ln(b / a )
ý
ð
8ð
Lext ý
If Lin = 2Lext
1
b
ý e1/ 8 ý 1.1331
8
a
b ý 1.1331a ý 7.365 mm
ln(b / a ) ý
Prob. 8.45
ýo ù 2 ù 4ð 107 (10) ù 2 10
ù
1ú ý 2 106 (ln1000 1)
ln 1ú ý
ln
Lý
2
ú
ú
2ð û a
2ð
û
û 2 10
û
ý 2(5.908) ý H ý 11.82 ý H
Prob. 8.46
12 ý B1 ÷ dS ý
òo a
ýo I
ý Ib a òo
dzd ò ý o ln
2ðò
2ð
òo
z ý0
b
ò ò
ý
o
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M 12 ý
For N = 1,
M 12 ý
ý
N 12 N ýo b a ò o
ý
ln
I
2ð
òo
12
I1
ý
ýo b a ò o
ln
2ð
òo
4ð 107
(1) ln 2 = 0.1386ý H
2ð
Prob. 8.47
We may approximate the longer solenoid as infinite so that B1 ý
the second solenoid is:
2 ý N 2 B1S1 ý
M ý
2
I1
ý
ýo N1 I1
ýo N1N 2
l1
ýo N1I1
l1
÷ ð r1 N 2
2
l1
÷ ðr1
2
Here we assume air-core solenoids.
Prob. 8.48
For a straight infinitely long conductor,
Bö ý
ýo I
2ðò
h a b w
ù
ýo I ù h b w 1
1
dzd ò ú
ý B ÷ dS ý
ú dzd ò
ò
2ð ûú z ý0 ò ýb ò
S
z ý0 ò ý a b
ûú
ý
bw
a b wù ýo Ih ù b w
ýo Ih ù
a b wù
ln
ln ò
ln
úln ò
úý
ú
b
a b û 2ð û
2ð û
b
a b úû
ý
ýo Ih ù (a b)(b w) ù
ln ú
ú
2ð
û b(a b w) û
Prob. 8.49
Hý
I
aò
2ðò
1
1
I2
wm ý ý | H |2 ý ý 2 2
2
2 4ð ò
I2
1
1
W ý wm dv ý ý 2 2 ò dö d ò dz ý
ý I 2 L ln(b / a)
2 4ð ò
4ð
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. The flux linking
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ý
1
4 4ð 107 (625 106 )3ln(18 /12) ý 304.1 pJ
4ð
Alternatively,
Wý
1 2 1 ýL b 2 ýI 2L b
ln I ý
ln
LI ý
a
2
2 2ð a
4ð
Prob. 8.50
ýr ý ó m 1 ý 20
1
1
ýH H
B1 H1 ý
2
2
1
ý ø 25x 4 y 2 z 2 100x 2 y 4 z 2 225x 2 y 2 z 4 ù
ý
2
ý wm dv
wm ý
Wm
1
2
2
1
2
2
1 ù
ý ú 25 x 4 dx y 2 dy z 2 dz 100 x 2 dx y 4 dy z 2 dz
1
1
0
0
0
2 û 0
1
2
2
225 x 2 dx y 2 dy zdz ù
úû
0
0
1
1
2
2
1
2
2
25ý ù x 5
y3
z3
x3
y5 z 3
ý
4
ú
2 ú 5 0 3 0 3 1
3 0 5 0 3 1
û
1
2
2
x3
y3 z 5 ù
9
ú
3 0 3 0 5 1 ú
û
ý
ý
25ý
2
ý
3600
25
4ð 10 7 20
45
2
4 32 9
9 8 33 ö
ö1 8 9
÷ ÷
3 3 3
3 3 5ø
ø5 3 3
Wm ý 25.13 mJ
Prob. 8.51
1
| B |2
1
W ý B ÷ Hdv ý
dv ý
2v
2ý
2(15)4ð 107
v
ý
4
3
2
(4
2
122 )106 dxdydz
z ý0 y ý0 x ý0
6
10
320
ý 101.86 J
(16 144)(2)(3)(4) ý
7
ð
(30)4ð 10
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Prob. 8.52
NI ý Hl ý
Bl
ý
l
òo
1.5 0.6ð
ý
Ný
ýo ýr I 4ð 107 600 12
ý 313 turns
Bl
N
Prob. 8.53
F = NI = 400 x 0.5 = 200 A.t
Ra ý
100
MAt/Wb,
4ð
Fa ý
R 1 ý R2 ý
6
MAt/Wb,
4ð
R3 ý
1.8
MAt/Wb
4ð
Ra F
ý 190.8 A.t
Ra R3 R1 // R2
Ha ý
Fa
190.8
ý
ý 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
Rc ý
= 0.115 x 107 A.t/m
ý
7
4
ý o ý r S 4ð 10 1500 2 10
la
Ra ý
ý
0.6 102
= 2.387 x 107 A.t/m
7
4
4ð 10 (1) 2 10
ýo ý r S
R = Ra + Rc = 2.502 x 107 A.t/m
ý
ñ
20,000
ýa ýc ý
= 8 x 10-4 Wb/m2
R
2.502 107
Ra
2.387 20,000
ñý
ý 19,081 A.t
2.502
R a Rc
Rc
0.115 20,000
ñc ý
ñý
ý 919 A.t
R a Rc
2.502
ña ý
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Prob. 8.55
Rc
F
F = NI = 500 x 0.2 = 100 A.t
Rc ý
lc
42 102
42 106
ý
ý
ý S 4ð 107 103 4 104
16ð
Ra ý
la
103
108
ý
ý
ýo S 4ð 107 4 104 16ð
1.42 108
Ra Rc ý
16ð
ý
Ba ý
16ð 100 16ð
F
ý
ý
ý Wb
Ra Rc 1.42 108 1.42
S
ý
16ð 106
ý 88.5 mWb/m 2
4
1.42 4 10
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Prob. 8.56
The equivalent circuit is shown below.
R1
R2
R3
N1 I1
R4
N2 I2
R5
Prob. 8.57
4.4 103
0.2282 106
ý
ý
ý 7.2643 104 A.t/Wb
Rý
7
2
ýo S 4ð 10 (4.82 10 )
ð
Prob. 8.58
Fý
2
4 106
B2S
ý
ý
ý 53.05 kN
2 ýo 2ýo S 2 4ð 107 0.3 104
Prob. 8.59
(a)
F = NI = 200 x 10-3 x 750 = 150 A.t.
10
la
ý
ý 3.183 107
6
ýo S 25 10 ýo
lt
2ð 0.1
= 6.7 x 107
Rt ý
ý
6
ýo ý r S ýo 300 25 10
Ra ý
ý
3
Rt
ñ
150
ñ
ý 7
ý 15.23 10 7
Ra Rt 10 (3.183 20 / 3)
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Fý
2
2.32 1012
B2S
ý
ý
2 ýo 2ýo S 2 4ð 107 25 106
= 37 mN
(b)If ýt , Rt ý 0, ý
ñ
150
ý
Ra 3.183 107
F2 ý I 2 dl2 ÷ B1 ý I 2 dl2
1
S
ý
2 103 5 103 150
3.183 107 25 10 6
F2 = 1.885 ýN
Prob. 8.60
2
2
1
1
ñ
Ra
ñ
Ra
Ra
Ra
Ra/2
1 ý 2 2 , 1 ý
ñ
3 R
2 a
ý
ñ
2ñ
2 ý
3Ra
3Ra
2
ö 2 ö
ñ2
3 1
ñ ý 2÷ 2 ÷ 1 ý
ý
2
ø 2ýo S ø 2 ýo S 4ýo S 3Ra ýo S
ýo S ñ2 4ð 107 200 104 9 106
ý
ý
2
3 106
3la
ý 24ð 10 3 ý mg m ý
24ð 10 3
ý 7694 kg
9.8
Prob. 8.61
ñ
Rs
ñ = NI
Ra
Rs
Ra
Rs/2
Since ý for the core (see Figure) , Rc = 0.
a
R ö ( 2 x)
ö
ñ ý NI ý ÷ Ra s ÷ ý
ýo S
2 ø
ø
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ý
(2 x a)
2ýo S
N 2 I 2 4ýo2 S 2
B2S
1
1
2
ñý
ý
ý
÷
2 ýo
2 ýo S 2 ýo S
(a 2 x)2
ý
2 N 2 I 2 ýo S
(a 2 x)2
F ý Fa x since the force is attractive, i.e.
2 N 2 I 2 ý o Sa x
Fý
(a 2 x) 2
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CHAPTER 9
P.E. 9.1
(a) Vemf ý ø u B ù dl ý uBl ý 8 ø 0.5 ùø 0.1ù ý 0.4 V
(b) I ý
Vemf
R
ý
0.4
ý 20 mA
20
(c) Fm ý Il B ý 0.02 ø 0.1a y 0.5a z ù ý a x mN
(d) P ý FU ý I 2 R ý 8 mW
Pý
or
Vemf
R
ý
ø0.4ù2
20
ý 8 mW
P.E. 9.2
(a) Vemf ý ø u B ù dl
where B ý Bo a y ý Bo ø sin ö a ò cos ö aö ù , Bo ý 0.05 Wb/m2
ø u B ù dl ý ò Bo sin ö dz ý 0.2ð sin øt ð 2 ù dz
Vemf ý
0.03
ø u B ù dl ý 6ð cos ø100ð t ù mV
0
At t = 1ms,
Vemf ý 6ð cos 0.1ð ý 17.93 mV
Vemf
ý 60ð cosø100ð .t ù mA
R
At t = 3ms, i ý 60ð cos 0.3ð ý 110.8 mA
iý
(b) Method 1:
ò o zo
ý B dS ý Bot ø cos ö a ò sin ö aö ù d ò dzaö ý Bo t sin ö d ò dz ý Bo òo zo t sin ö
0 0
where Bo ý 0.02 , ò o ý 0.04 , zo ý 0.03
ö ý t ð 2
ý Bo òo zot cos t
Vemf ý
ö
ý Bo ò o zo cos t Bo òo zo t sin t
öt
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ý ø 0.02 ùø 0.04 ùø 0.03ù cos t t sin t ý
ý 24 cos t t sin t ý ýV
Method 2:
öB
÷ dS (u B ).dl
öt
B ý Bo ta x ý Bot (cos ö a ò sin ö aö ), ö ý t ð
Vemf ý
2
öB
ý Bo (cos ö aò sin ö aö )
öt
Note that only explicit dependence of B on time is accounted for, i.e. we make ö
= constant because it is transformer (stationary) emf. Thus,
òo zo
0
0 0
zo
Vemf ý Bo (cos ö a ò sin ö aö ) d ò dzaö òo Bot cos ö dz
ý Bo òo zo (sin ö t cos ö ), ö ý t ð
2
ý Bo òo zo (cos t t sin t ) as obtained earlier.
At t = 1ms,
Vemf ý 24[cos18o 100ð 103 sin 18o ]ýV
= 20.5ýV
At t = 3ms,
i ý 240[cos 54o .03ð sin 54o ]mA
= -41.93 mA
P.E. 9.3
d
d
, V2 ý N 2
dt
dt
V2 N 2
N
300 120
ý
V2 ý 2 V1 ý
ý 72V
V1 N 1
N1
500
V1 ý N 1
P.E. 9.4
(a)
Jd ý
öD
ý 20õ o sin(t 50 x)a y A / m 2
öt
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(b)
öH z
a y ý 20õ o sin(t 50 x)a y
öx
20õ o
or H ý
cos(t 50 x)a z
50
ñ H ý Jd
ý 0.4õ o cos(t 50 x)a z A/m
(c)
ñ E ý ýo
öE y
öH
a z ý 0.4 ýo 2õ o sin(t 50 x)a z
öt
öx
1000 ý 0.4ýoõ o 2 ý 0.4
2
or = 1.5 x 1010 rad/s
c2
P.E. 9.5
ø
2
2
ù
2ð45o ù
jö
÷÷ ý j ú
ý j 2 ð143.13o
j ÷÷
oú
5
úû 5ð 26.56 úû
ø2 jø
3ö 1
(a)
ù
= 0.24 + j0.32
(b)
o
6ð30o j 5 3 e j 45 ý 5.196 j 3 j 5 3 0.7071(1 j )
= 2.903 + j8.707
P.E. 9.6
ø
ù
P ý 2sin(10t x ð )a y ý 2 cos 10t x ð ð a y , w ý 10
4
4
2
ø
ý Re 2e
i.e. Ps ý 2e
j ( x 3ð )
4
j ( x 3ð )
4
ù
a y e jwt ý Re ø Ps e jwt ù
ay
Q ý Re ø Qs e jwt ù ý Re ø e j ( x wt ) (a x a z ) ù sin ð y
ý sin ð y cos( wt x)(a x a z )
P.E. 9.7
ý
öH
1
ö
1 ö
ý ñ E ý
( Eö sin ñ )ar
(rEö )añ
r sin ñ öñ
r ör
öt
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ò
2 cos ñ
cos(t ò r )ar sin ñ sin(t ò r )añ
2
r
r
ò
2 cos ñ
H ý
sin(t ò r )ar
sin ñ cos(t ò r )añ
ý r 2
ý r
=
òý
c
ý
6 107
ý 0.2 rad/m
3 108
1
1
H ý
cos ñ sin(6 107 0.2r )ar
sin ñ cos(6 107 0.2r )añ
2
12ð r
120ð r
P.E. 9.8
3
ý
Eý
ýõ
ý
3c
ýrõ r
1
ý
9 10 8
10
ý 2.846 10 8 rad/s
6
ñ Hdt ý
cos(t 3 y )a
õ
õ
x
6
cos(t 3 y )a x
9 10 109
÷
(5)
10 36ð
E ý 476.86 cos(2.846 108 t 3 y )a x V/m
=
Prob. 9.1
V ý
8
ö
ö
öB
ý B ÷ dS ý
÷S
öt
öt
öt
= 3770 sin377t x ð(0.2)2 x 10-3
= 0.4738 sin377t V
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Prob. 9.2
ö
, ý B ÷ dS ý BS
öt
öB
Vemf ý
S
öt
V
(4 20sin 20t )(2 104 ) 160
i (t ) ý emf ý
ý
sin 20t (104 )
R
20 30
50
i (t ) ý 0.32sin 20t mA
Vemf ý
Prob.9.3
ý B S ý (0.2) 2 ð 40 103 sin104 t
ö
V ý
ý 16ð cos104 t
öt
V 16ð
cos104 t
iý ý
4
R
ý 12.57 cos104 t A
Prob.9.4
Measuring the induced emf in the clockwise direction,
Vemf ý (u B ) dl
0
1.2
=
(5a
0
x
0.2a z ) dya y (15a x 0.2a z )dya x
1.2
1.2
0
0
1.2
= - (1) dy (3)dy
ý 1.2 1.2 3 ý 1.2 3.6
ý 2.4 V
Prob. 9.5
Vemf ý (u B ) ÷ dl ý
ý
1.6
(2a x 10 cos ò ya z ) ÷ dya y ý
y ý0
1.6
20 cos ò ydy
y ý0
20sin ò y 1.6
20sin1.6ò
ý
0
ò
ò
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Prob. 9.6
Bý
ýo I
( a x )
2ð y
ýo I
2ð
ý B ÷ dS ý
Vemf ý
a ò a
ò
z ý0 y ý
dzdy ýo Ia ò a
ý
ln
ò
y
2ð
ö öò
ö
ý Ia d
ý o uo
[ln( ò a ) ln ò ]
÷
ý
2ð
öò öt
öt
dò
ý o Ia ù 1
ý o a 2 Iu o
1ù
uo ú
ý
úý
2ð
û ò a ò û 2ðò ( ò a)
where ò ý òo u o t
Prob. 9.7
ò a
Vemf ý
ò 3a
z
ýo I
3ý I ò a
aö ÷ d ò a ò ý o ln
ò
2ðò
2ð
60
4ð 10 7
ý
15 3 ln
ý 9.888ýV
20
2ð
Thus the induced emf = 9.888ýV, point A at higher potential.
Prob. 9.8
V ý ( u B ) dl
u ý ò aö ,
Vý
B ý Bo a z
1
2
ò Bo d ò ý Bo ò 2
ò ý0
ý
0
ý
Bo 2
2
30
60 103 (8 102 ) 2 ý 5.76 mV
2
Prob. 9.9
Vemf ý N
ö
ö
dS
ý N B dS ý NB
öt
öt
dt
d
dö
ý NBò
( òö ) ý NBò
dt
dt
ý 50(0.2)(30 104 )(60) ý 1.8V
ý NB
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Prob. 9.10
Method 1:
We assume that the sliding rode is on ü y ü
ý x / 3 ý 5t / 3
ö 25t 2 ö
2
Vemf ý (u B ) dl ý 5a x 0.6a z ÷ dya y ý 3 x dy ý 3 x(2) ý 6 ÷
÷ ý 86.6025t
ø 3 ø
Method 2:
The flux linkage is given by
ý
5t
x. 3
0.6 xdxdy ý 0.6
x ý o y ý x / 3
Vemf ý
2
125t 3 / 3 ý 28,8675t 3
3
d
ý 86.602t 2
dt
Prob. 9.11
u
B
B
u
ñ
Vemf ý (u B ) dl ý uBl cos ñ
ö 120 103
ö
m / s ÷ ø 4.3 105 ù ø1.6 ù cos 65o
ý÷
ø 3600
ø
o
ý 2.293cos 65 ý 0.97 mV
Prob. 9.12
Vemf ý uB ý 410 0.4 106 36 ý 5.904 mV
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Prob. 9.13
d = 0.64 – 0.45 = 0.19, dt = 0.02
d
ö 0.19 ö
ý 10 ÷
÷ ý 95V
dt
ø 0.02 ø
Vemf ý N
Iý
ö 95 ö
ý ÷ ÷ ý 6.33 A
R
ø 15 ø
Vemf
Using Lenz’s law, the direction of the induced current is counterclockwise.
Prob. 9.14
V ý (u B ) ÷ dl , where u ý ò aö , B ý Bo a z
Vý
ò2
ò B d ò ý
ò
Bo
o
2
1
( ò 2 2 ò 21 )
60 15
Vý
÷ 103 (100 4) ÷ 104 ý 4.32 mV
2
Prob. 9.15
J ds ý jDs J ds
ý
max
ý õE s ý õ
Vs
d
109 2ð 20 106 50
36ð
0.2 10 3
ý 277.8 A/m2
I ds ý J ds ÷ S ý
1000
2.8 10 4 ý 77.78 mA
3.6
Prob. 9.16
Jc ý ó E,
öD
öE
ýõ
öt
öt
| J d |ý õ | E |
Jd ý
| J c |ý ó | E |,
If I c ý I d , then | J c |ý| J d |
ý 2ð f ý
f ý
ó
ý
2ðõ
ó ý õ
ó
õ
4
109
2ð 9
36ð
ý 8 GHz
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Jc
óE
ó
ý
ý
J d õE õ
Prob. 9.17
ó
(a)
ý
õ
(b)
ó
ý
õ
ó
(c)
ý
õ
2 103
2ð 109 81
9
10
36ð
25
109
2ð 10 81
36ð
ý 0.444 103
ý 5.555
9
2 104
2ð 109 5
ý 7.2 104
9
10
36ð
Prob. 9.18
Jc ý ó E
Jd ý õ
öE
öt
| J cS | ó
ý
ý
| J ds | õ
J cs ý ó Es
J ds ý jõ Es
4
2ð 107 81
9
10
36ð
ý
400(18)
ý 88.89
81
Prob. 9.19
J d õ E õ
ý
ý
ý1
J
óE
ó
2ð f ý 12ð 105
ý
104
ó
ý
ý 12ð 105
9
10
õ
3
36ð
f ý 600 kHz
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Prob. 9.20
J c ý ó E ý 0.4 cos(2ð 108 t )
Eý
0.4
ó
cos(2ð 108 t )
öE
0.4õ
ý
(2ð 108 ) sin(2ð 108 t )
öt
ó
109
0.4 4.5
36ð (2ð 108 ) sin(2ð 108 t )
ý
4
10
ý 100sin(2ð 108 t ) A/m 2
Jd ý õ
Prob. 9.21
öD
ý J Jd
öt
Since the regin is source-free, J = 0.
ñ H ý J
ö
J d ý ñ H ý öx
H x ( z)
ö
öy
0
ö
öH x
öz ý
a y ý H o ( ò ) sin(t ò z )a y
öz
0
ý ò H o sin(t ò z )a y
Prob. 9.22
Jd ý
öD
öE
ýõ
öt
öt
öE
109
25 103 cos103 t (2 5) ý 2.2105 108 cos103 t
I ý J d ÷ dS ý J d S ýõ
Sý
öt
36ð
S
I
ý 22.1cos103 t nA
Prob. 9.23
(a)
ñ ÷ Es ý
òs
õ ,ñ ÷ Hs ý 0
ñ E s ý jý H s ,
ñ H s ý (ó jõ ) E s
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(b)
öDx öDy öDz
ý òv
öx
öy
öz
öB öBy öBz
ñ÷B ý 0 x
ý0
öx
öy
öz
öB
öE öE y
öB
ñ E ý
z
ý x
öt
öy
öz
öt
ñ ÷ D ý òv
(1)
(2)
(3)
öBy
öEx öEz
ý
öx
öt
öz
öE y öE x
öBz
ý
öt
öy
öx
ñ H ý J
öH x
öz
öH y
öx
(4)
(5)
öD
öH z öH y
öD
ý Jx x
öt
öy
öz
öt
öD y
öH z
ý Jy
öx
öt
öH x
öD z
ý Jz
öy
öt
(6)
(7)
(8)
Prob. 9.24
If J ý 0 ý òv , then
ñ÷B ý 0
(1)
(2)
ñ ÷ D ý òv
öB
ñ E ý
öt
öD
ñ H ý J
öt
Since ñ ÷ ñ A ý 0 for any vector field
(3)
(4)
A,
ö
ñ÷B ý 0
öt
ö
ñ÷ñ H ý ñ÷ D ý 0
öt
showing that (1) and (2) are incorporated in (3) and (4). Thus Maxwell’s equations can be
ñ÷ñ E ý
reduced to (3) and (4), i.e.
ñ E ý
öB
öD
, ñ H ý
öt
öt
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Prob. 9.25
ñ E ý0
(1)
ñ H ý0
(2)
ñ E ý ý
öH
öt
ùö
ú öx
ú
ñ E ý ú
ú0
ú
úû
ý
öE y
öx
H ý
ö
öy
(3)
öù
öz ú
ú
ú
0 úú
úû
E y ( x, t )
a z ý Eo sin x cos ta z
Eo
1
ñ Edt ý
ý
ý
ñ H ý õ
öE
öt
ùö
ú öx
ú
ñ H = ú
ú0
ú
úû
E
sin sin ta z
o
ö
öy
0
(4)
ö
öz
ù
ú
ú
ú
H z ( x, t ) úú
ûú
ý
E
öH z
a y ý o cos x sin ta y
öx
ýo
1
Eo
=
ñ Hdt ý
cos x cos ta
õ
ýõ
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 ýoõ ý 1 which is not true.
Prob. 9.26
ñ E ý
ññ E ý
öB
öt
ö
ö
öJ
ö2 E
ñ B ý ý ñ H ý ý
ýõ 2
öt
öt
öt
öt
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But
ñ ñ E ý ñ(ñ ÷ E ) ñ 2 E
ñ(ñ ÷ E ) ñ 2 E ý ý
öJ
ö2 E
ýõ 2 ,
öt
öt
J ýó E
In a source-free region, ñ ÷ E ý ò v / õ ý 0 . Thus,
ñ 2 E ý ýó
öE
ö2 E
ýõ 2
öt
öt
Prob. 9.27
ñ ÷ J ý (0 0 3 z 2 ) sin 10 4 t ý
öòv
öt
òv ý ñ ÷ Jdt ý 3z 2 sin104 tdt ý
3z 2
cos104 t Co
4
10
If ò v |z ý0 ý 0, then Co ý 0 and
òv ý 0.3z 2 cos104 t mC/m3
Prob. 9.28
ñ ÷ D ý õñ ÷ E ý òv
ò v ý õñ ÷ E = õ o
öEz
ý õ o Eo sin z cos t
öz
Prob. 9.29
=108
Let
öD
öE
1
ý 0õ
E ý ñ Hdt
öt
öt
õ
ö
ö
ö
öy
öz ý 40 ò cos(t ò x)a z
ñ H ý öx
0 40sin(t ò x) 0
1
40ò
sin(t ò x)a z
E ý ñ Hdt ý
ñ H ý J
õ
õ o
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öB
öH
ý ý
öt
öt
ö
ö
ö
öx öy
öz
40ò 2
cos(t ò x)a y
ñ E ý
ý
40ò
õ o
0
0
sin(t ò x)
ñ E ý
(1)
õ o
öH
ý 40ýo cos(t ò x)a y
öt
Equating (1) and (2) gives
ýo
40ýo ý
40ò 2
ò 2 ý ý o õ o 2
õ o
(2)
ò ý õ o ýo ý 10
m
/
d
a
r
3
3
3
.
0
109 1
4ð 10
ý ý
36ð 3
7
8
Prob. 9.30
öD
öE 50õ o
4.421 102
Jd ý
ý õo
ý
(108 ) sin(108 t kz )a ò ý
sin(108 t kz )a ò A/m
öt
öt
ò
ò
ñ E ý ýo
ñ E ý
H ý
Hý
öEò
öz
1
ýo
k2 ý
aö ý
50k
ò
sin(108 t kz )aö
1
ñ Edt ý 4ð 10
50k
cos(108 t kz )aö
108 ò
7
2.5k
cos(108 t kz )aö A/m
2ðò
ñ H ý
ñ H ý Jd
öH
öt
öHö
öz
2ð
4.421 102
2.5
aò ý
2.5k 2
sin(108 t kz )a ò
2ðò
4.421x102
ò
2.5k 2
sin(10 t kz )a ò ý
sin(108 t kz )a ò
2ðò
8
k ý 0.333
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Prob. 9.31
ñ E ý ýo
öH
öt
H ý
1
ýo
ñ Edt
1 ö
1 ö
(rEñ )aö ý
10sin ñ cos(t ò r )ý aö
r ör
r ör
10ò
ý
sin ñ cos(t ò r )aö
r
10ò
sin ñ sin(t ò r )dtaö
H ý
ýr
10ò
ý
sin ñ cos(t ò r )aö
ýo r
ñ E =
Prob. 9.32
öD
D ý J d dt
dt
60 103
cos(109 t ò z )a x ý 60 1012 cos(109 t ò z )a x C/m 2
D=
109
öH
öH
D
ñ E ý ý
ñ ý ý
öt
öt
õ
ö
ö
ö
öx öy öz
1
D 1
ñ ý
ý (60)(1) 1012 sin(109 t ò z )a x
õ õ
õ
Dx 0
0
(a) J d ý
ý
H ý
1
60ò
õ
1012 sin(109 t ò z )a y
ñ
D
dt ý
1
(1)
60 ò 1012
9 cos(109 t ò z )a y
õ
10
ý
õ
ý
60 ò
ý
1021 cos(109 t ò z )a y A/m
ýõ
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(b)
ñ H ý J Jö ý 0 J d
ö
öx
ö
öy
ö
öz
Jd ý ñ H ý
ý
Hy
0
( ò )(1)60 ò
0
ýõ
(1021 ) sin(109 t ò z )a x
Equating this with the given J d
3
60 10 ý
60 ò 2
ýõ
1021
109 2000
ò ý ýõ 10 ý 2 4ð 10 10
ý
36ð
9
ò ý 14.907 rad/m
2
7
18
Prob. 9.33
ñ ÷ D ý òv ,
ñ H ý J
ñ÷ñ H ý ñ÷ J
Or
ñ÷J = -
öD
öt
öñ ÷ D
ý0
öt
ñ÷J
öò v
öt
öò v
ý0
öt
Prob. 9.34
From Maxwell's equations,
ñ ÷ D ý ò v ý 0,
ñ ÷ B ý 0,
ö
ñ H
öt
ñ ÷ E =0,
J ý ó E,
ñ ñ E ý ý
But
ñ 2 E ý ý
ñ 2 E ýó
öD
öt
öö
öD ö
ñ(ñ ÷ E ) ñ 2 E ý ý ÷ J
÷
öt ø
öt ø
D ýõE
ñ E ý
öB
,
öt
ñ H ý J
öö
öõ E ö
öE
ö2 E
ý
ó
ýó
ýõ
E
÷
÷
öt ø
öt ø
öt
öt 2
öE
ö2 E
ýõ 2 ý 0
öt
öt
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Prob. 9.35
(a) ñ ÷ A ý 0
ö
öx
ö
öy
ö
öz
ñ A ý
ý
0
E z ( x, t )
0
ö E z ( x, t )
ay ù 0
öx
Yes, A is a possible EM field.
(b)
ñ÷ B ý 0
ñ B ý
1 ö
ò öò
10 cos(t 2 ò )ý az ù 0
Yes, B is a possible EM field.
(c)
ñ÷C ý
ñC ý
1 ö
ò öò
1 ö
ø 3ò
ò öò
3
ù
cot ö sin t
ø cos ö sin t ù a z 3ò 2
sin ö sin t
ò2
ù0
ö
(cot ö sin t )a z ù 0
öö
No, C cannot be an EM field.
ö
1
sin(t 5r ) (sin 2 ñ ) ù 0
(d) ñ ÷ D ý 2
r sin ñ
öñ
ñ D ý
ö Dñ
1 ö
1
(rDñ )aö ý sin ñ (5) sin(t 5r )aö ù 0
ar
r ör
r
öö
No, D cannot be an EM field.
Prob. 9.36
From Maxwell’s equations,
öB
ñ E ý
öt
öD
ñ H ý J
öt
Dotting both sides of (2) with E gives:
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öD
(3)
ö
t
But for any arbitrary vectors A and B ,
ñ ÷ ( A B ) ý B ÷ (ñ A) A ÷ (ñ B )
Applying this on the left-hand side of (3) by letting A ú H and B ú E , we get
ö
H ÷ (ñ E ) ñ ÷ ( H E ) ý E ÷ J 1
( D ÷ E ) (4)
2 öt
From (1),
ö öB ö 1 ö
H ÷ (ñ E ) ý H ÷ ÷
÷ ý 2 (B ÷ H )
öt
ø öt ø
Substituting this in (4) gives:
ö
ö
1
(B ÷ H ) ñ ÷ (E H ) ý J ÷ E 1
(D ÷ E)
2 öt
2 öt
Rearranging terms and then taking the volume integral of both sides:
E ÷ (ñ H ) ý E ÷ J E ÷
ö
ñ ÷ ( E H )dv ý öt 1 2 ( E ÷ D H ÷ B)dv J ÷ Edv
v
v
Using the divergence theorem, we get
( E H ) ÷ dS ý
s
or
öW
J ÷ Edv
öt v
öW
ý ( E H ) ÷ dS E ÷ Jdv as required.
öt
s
v
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Prob. 9.37
ñ E ý ýo
ö
ñ E ý öx
0
öH
öt
ö
öy
0
H ý
1
ýo
ñ Edt
ö
öE
öE
öz
ý z ax z a y
öy
öx
E z ( x, y )
ý ñ cos(12ð x) cos(1011 t ñ y )a x 12ð sin(12ð x) sin(1011 t ñ y )a y
H ý
ý
1
ýo
ñ Edt
1 ù
ñ cos(12ð x) cos(1011 t ñ y )dta x 12ð sin(12ð x) sin(1011 t ñ y )dta y ùû
û
ý
o
ý
ñ
ýo10
But
11
cos(12ð x) sin(1011 t ñ y )a x
ñ H ý
12ð
sin(12ð x) cos(1011 t ñ y )a y
10
ýo10
öD
öE
ý õo
öt
öt
öE
ý õ o (1011 ) cos(12ð x) cos(1011 t ñ t )a z
öt
ö
ö
ö
ö öH y öH x
öx
öy
öz ý ÷
ñ H ý
öx
öy
ø
H x ( x , y ) H y ( x, y ) 0
õo
(1)
ö
÷ az
ø
ù (12ð ) 2
ù
ñ2
11
x
cos(12
)
co
ð
ñ
s(10
)
cos(12ð x) cos(1011 t ñ y ) ú a z
t
y
ýú
11
11
ýo10
û ýo 10
û
Equating (1) and (2),
õ o (1011 ) ý
ñ2
(12ð ) 2
ýo1011 ýo1011
ýoõ o (1022 ) ý (12ð ) 2 ñ 2
109
106
ý 144ð 2 ñ 2
ñ ý 331.2
(1022 ) ý
36ð
9
12ð
12ð 1011
ñ
331.2 1011
3
ý
ý
ý
ý 3 103
10
,
2.636
11
7
11
7
ýo10
4ð 10
ýo10
4ð 10
4ð 107
Thus,
H ý 2.636 cos(12ð x) sin(1011 t ñ y )a x 3sin(12ð x) cos(1011 t ñ y )a y mA/m
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Prob. 9.38
ñ E ý ýo
öH
öt
ö
ñ E ý öx
Ex ( z )
ö
öy
0
H ý
1
ýo
ñ Edt
ö
öE
öz ý x a y ý ò Eo cos(1200ð t ò z )a y
öz
0
ýE
öH
ý o o (1200ð ) cos(1200ð t ò z )a y
ø
öt
Setting (1) and (2) equal,
ýE
1200ðýo
ò Eo ý o o (1200ð ) ò ý
ýo
ø
(2)
(3)
ø
öE
öt
ö
öy
H y ( z)
(1)
But ñ H ý õ o
ö
ñ H ý öx
0
ö
öH y
ò Eo
öz ý
cos(1200ð t ò z )a x
ax ý
ø
öz
0
öE
ý õ o (1200ð ) Eo cos(1200ð t ò z )a x
öt
Setting (4) and (5) equal,
ò =õ o (1200ð )ø
(6)
õo
From (3) and (6),
1200ðýo
ý õ o (1200ð )ø
ø
øý
ø2 ý
(5)
ýo
õo
ýo
4ð 107
ý
ý 120ð ý 377
109
õo
36ð
From (3),
òý
1200ðýo
ø
1200ð 4ð 107
ý
ý 40ð 107 ý 1.257 105 rad/m
120ð
Prob. 9.39 Using Maxwell’s equations,
ñ H ý ó E õ
öE
öt
(ó ý 0)
Eý
1
õ
ñ Hdt
But
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ñ H ý
Eý
12sin ñ
ý
õo
1 ö Hñ
1 ö
12sin ñ
ò sin(2ð 108 t ò r )aö
ar
(rHñ )aö ý
r sin ñ öö
r ör
r
ò sin(2ð 108 t ò r )dtaö
12sin ñ
ò cos(t ò r )aö ,
õ o r
ý 2ð 108
Prob. 9.40
With the given A, we need to prove that
ö2 A
2
ñ A ý ýõ 2
öt
2
ñ A ý ýõ ( j )( j ) A ý 2 ýõ A
Let ò 2 ý 2 ýõ , then ñ 2 A ý ò 2 A is to be proved. We recognize that
Aý
ýo jt jò r
e e az
4ð r
ý
e jò r
,
A ý o e jt a z
r
4ð
ö ù 1 ù ö 2 ö j ò 1 ö jò r ù
1 ùö 2
ñ 2 ý 2
2 ÷e
(r sin ñ
) ý
(r ) ÷
ú
ú
r sin ñ û ör
dr úû r 2 úû ör
r ø
ø r
û
jò r
e
1
ý 2 ø ò 2 r j ò j ò ù e jò r ý ò 2
ý ò 2
r
r
ñ2 A ý ò 2 A
Therefore,
We can find V using Lorentz gauge.
1
1
ñ ÷ Adt ý
ñ÷ A
Vý
jýoõ o
ý oõ o
Assume
ý
ý
ö ö ýo j ò r jt ö
1
ö j ò 1 ö jò r jt
2 ÷ e e cos ñ
e e ÷ý
÷
÷
jýoõ o ör ø 4ð r
r ø
ø jõ o (4ð ) ø r
Vý
1
cos ñ ö
1 ö j (t ò r )
÷ jò ÷ e
j 4ðõ o r ø
rø
Prob. 9.41
Take the curl of both sides of the equation.
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ö
ñ A
öt
But ñ ñV ý 0 and B =ñ A. Hence,
ñ E ý ñ ñV
öB
öt
which is Faraday's law.
ñ E =
Prob. 9.42
ñ A ý
(a)
Hence,
(b)
öAz x
ý ,
öz c
öV
ý xc,
öt
öV
ñ A = ýoõ o
öt
E ý ñV
ý oõ o
öV
x
x
ý 2cý
öt c
c
öA
öV ö
ö öV
ý ÷
ax
a z ÷ xa z ý ( za x xa z ) xa z
öt
öz ø
ø öx
ý za x
E
Prob. 9.43
ñ A ý 0 ý ýõ
öV
öt
V ý constant
öA
ý 0 Ao cos(t ò z )a x
öt
ý Ao cos(t ò z )a x
(a) E ý ñV
(b) Using Maxwell’s equations, we can show that
ò ý ý oõ o
Prob. 9.44
(a)
z ý 4ð30o 10ð50o ý 3.464 2 j 6.427 j 7.66 ý 2.963 j 5.66
ý 6.389ð 117.64o
z1/ 2 ý 2.5277ð 58.82o
(b)
1 j2
2.236 ð 63.43 o
2.236 ð 63.43 o
ý
ý
6 j 8 7 ð 15 o 6 j 8 7 .761 j1.812 9.841ð 265.57 o
ý 0.2272ð 202.1o
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(c)
(5ð 53.13 o ) 2
25ð 106 .26 o
ý
zý
12 j7 6 j10 18.028 ð 70.56 o
ý 1.387 ð 176 .8 o
(d)
1.897 ð 100 o
o
o
o ý 0.0349 ð 68
(576
. ð 90 )(9.434ð 122 )
Prob. 9.45
(a) H ý Re ùû H s e jt ùû ,
H ý Re ùû 10e j (t ð /3) a x ùû
ý 106
H s ý 10e jð /3a x
(b) E s ý 4 cos(4 y )e j 2 x a z
(c) sinA ý cos(A 90o )
D ý 5cos(t ð / 2 ð / 3)a x 8cos(t ð / 4)a y
Ds ý 5e jð /6 a x 8e jð / 4 a y
Prob. 9.46
(a)
As ý 10e jð /2 a x 20e jð /2 a y
A ý Re[ As e jt ] ý Re ùû10e j (t ð /2) a x 20e j (t ð /2) a y ùû
ý 10 cos(t ð / 2)a x 20 cos(t ð / 2)a y
A ý 10sin ta x 20sin ta y
(b)
B ý Re[ Bs e jt ] ý Re ùû 4e j (t 2 x ð /2) a x 6e j (t 2 x ) a y ùû ý 4 cos(t 2 x ð / 2)a x 6 cos(t 2 x)a y
ý 4sin(t 2 x)a x 6sin(t 2 x)a y
(c)Cs ý 2e jð /2 e 10 z e jð /4 a z
C ý Re[C s e jt ] ý 2 Re[e j (t ð / 2ð /4) e 20 z a z ] ý 2 e 20 z cos(t ð / 4 ð / 2)a z
ý 2 e 20 z sin(t ð / 4)a z
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Prob. 9.47
Hs ý
(a)
1
ò
e j 3 z aö
öE
öt
ñ H ý õo
(b)
öH ö
1
õo
ñ Hdt
1 ö
3
( ò H ö )a z ý sin(t 3 z )a ò 0
ò öò
ò
öz
3
3
sin(t 3z )a ò dt ý
cos(t 3 z )a ò V/m
But
ñ H ý
Eý
1
aò
õo ò
òõ o
ñ E ý
(c )
Eý
öEò
öz
aö ý
9
òõ o
sin(t 3 z )aö
öB
öH ýo
ý ýo
ý
sin(t 3 z )aö
ò
öt
öt
ý
öB
9
ñ E ý
ý o
öt
òõ o
ò
2 ý
9
ý oõ o
ý
3
ý oõ o
ý 3 3 108 ý 9 108 rad/s
Prob. 9.48
We can use Maxwell’s equations or borrow ideas from chapter 10.
1 120ð
ý
ý øo
ý
øý
9
õ
õr
Ho ý
Eo
ø
ý
10 9
ý 0.2387
120ð
ò ý ýõ ý
c
õr ý
2ð 109
81 ý 60ð ý 188.5 rad/m
3 108
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Prob. 9.49
ñ H s ý jõ o E s
ö
ñ H s ý öx
0
ö
ö
öy
öz ý j12ñ e jñ x a y
0 12e jñ x
Es ý
ñ Hs
12ñ jñ x
ý
e ay
õ o
jõ o
But
ñ E s ý jýo H s
ö
ñ E s ý öx
0
Hs ý
ö
öy
Esy ( x)
ö
öE
12ñ 2 jñ x
öz ý sy a z ý j
e az
õ o
öx
0
ñ Es
12ñ 2 jñ x
ý 2
e az
jýo ýoõ o
Equating this with the given H,
12=
12ñ 2
2 ýoõ o
ñ 2 ý 2 ý oõ o
ñ 2 ý (109 ) 2 4ð 107
109
1016
ý (109 ) 2
36ð
9
10
ý 3.333
3
12ñ jñ x
12(10 / 3) j10 x /3
e ay ý
e
Es ý
a y ý 40(36ð )e j10 x /3a y ý 4.533e j 3.33 x a y kV/m
9
10
õ o
109
36ð
ñý
Prob. 9.50
( j ) 2 Y 4 jY Y ý 2ð0o ,
ý3
Y ( 2 4 j 1) ý 2
Yý
2
2
2
ý
ý
ý 0.0769 j 0.1154
4 j 1 9 j12 1 8 j12
2
ý 0.1387ð 123.7 o
y (t ) ý Re(Ye jt ) ý 0.1387 cos(3t 123.7 o )
Prob. 9.51 We begin with Maxwell’s equations:
ñ ÷ D ý òv / õ ý 0,
öB
ñ E ý
,
öt
ñ÷B ý 0
ñ H ý J
öD
öt
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We write these in phasor form and in terms of Es and Hs only.
ñ ÷ Es ý 0
(1)
ñ ÷ Hs ý 0
(2)
ñ E s ý jý H s
(3)
ñ H s ý (ó jõ ) E s
(4)
Taking the curl of (3),
ñ ñ E s ý jýñ H s
ñ(ñ ÷ E s ) ñ 2 E s ý jý (ó jõ ) E s
ñ 2 E s ( 2 ýõ jýó ) E s ý 0
ñ 2 Es ÷ 2 Es ý 0
Similarly, by taking the curl of (4),
ñ ñ H s ý (ó jõ )ñ E s
ñ(ñ ÷ H s ) ñ 2 H s ý jý (ó jõ ) H s
ñ 2 H s ( 2 ýõ jýó ) H s ý 0
ñ2 H s ÷ 2 H s ý 0
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CHAPTER 10
P. E. 10.1 (a)
Tý
2ð
ý
2ð
ý 31.42 ns,
2 108
ü ý uT ý 3 108 31.42 109 ý 9.425 m
k ý ò ý 2 ð / ü ý 0.6667 rad/m
(b) t1 = T/8 = 3.927 ns
(c )
H (t ý t1 ) ý 0.1cos(2 108
ð
8 108
2 x / 3)a y ý 0.1cos(2 x / 3 ð / 4)a y
as sketched below.
P. E. 10.2 Let xo ý
ñý
or
xo 2 ý
1 (ó / õ ) 2 , then
ý oõ o
ý r õ r ( xo 1) ý
2
c
xo 1 ý
16
xo 1
2
1
ñ c 1/ 3 3 108
ý
ý
8
8
10 8
8
81
ý 1 (ó / õ ) 2
64
xo ý 9 / 8
ó
ý 0.5154
õ
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tan 2ñ ø ý 0.5154
xo 1
ý
xo 1
ò
ý
ñ
ñ ø ý 13.63 o
17
ò ý ñ 17 ý
(a)
17
ý 1.374 rad/m
3
ó
ý 0.5154
õ
ý / õ 120 ð 2 / 8
(c ) | ø | ý
ý
ý 177 .72
xo
9/8
(b)
ø ý 177 .72ð 13.63 o
(d)
uý
108
ý
ý 7.278 107 m/s
ò 1.374
(e) a H ý ak a E
Hý
az ax ý aH
aH ý a y
0.5 z / 3
e sin(108 t ò z 13.63o )a y ý 2.817e z / 3 sin(108 t ò z 13.63o )a y mA/m
177.5
P. E. 10.3 (a) Along -z direction
(b) ü ý
2ð
ý 2 ð / 2 ý 3.142 m
ò
10 8
f ý
ý
ý 15.92 MHz
2ð
2ð
ò ý ý õ ý ý oõ o ý r õ r ý
or õ r ý ò c / ý
(c ) ñ ø ý 0 ,| ø | ý
ak ý a E a H
c
(1) õ r
3 108 2
ý6
108
ý /õ ý
ý o / õ o 1/ õ r ý
a z ý a y a H
õ r ý 36
120 ð
ý 20 ð
6
aH ý ax
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Hý
50
sin(t ò z )a x ý 795.8sin(108 t 2 z )a x mA/m
20ð
P. E. 10.4 (a)
ó
ý
õ
ñ
ò
102
109
10 ð 4
36ð
ý 0.09
9
ýõ ù
2
ù
ó
1ö ó ö
109 ð
ý
ýr õ r
(2)(0.09) ý 0.9425 Np/m
ú1 ÷
÷ 1ú ý
õ 2 3 108
2 ûú 2 ø õ ø
úû 2c
ýõ ù
2
ù 109 ð
1ö ó ö
1
1
2[2 0.5(0.09) 2 ] ý 20.965 rad/m
ú
úý
÷
÷
8
2 úû 2 ø õ ø
úû 3 10
E ý 30e 0.9425 y cos(109 ð t 20.96 y ð / 4)a z
At t = 2ns, y = 1m,
E ý 30e 0.9425 cos(2ð 20.96 ð / 4)a z ý 2.844a z V/m
(b) ò y ý 10 o ý
10 ð
rad
180
or
yý
ð 1
ð
ý
ý 8.325 mm
18 ò 18 20.965
(c ) 30(0.6) = 30 e ñy
yý
1
1
1
ln(1 / 0.6 ) ý
ln
ý 542 mm
0.9425 0.6
ñ
(d)
| ø |
ý /õ
60ð
ý
ý 188.11
1
1.002
2
[1 (0.09) ]
4
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2ñ ø ý tan 1 0.09
ñ ø ý 2.571o
a H ý ak a E ý a y a z ý a x
Hý
30 0.9425 y
e
cos(109 ð t 20.96 y ð / 4 2.571o )a x
188.11
At y = 2m, t = 2ns,
H ý (0.1595)(0.1518) cos(34.8963rad )a x ý 22.83a x mA/m
P. E. 10.5
w
w
0 0
0
0
I s ý J xs dydz ý J xs (0) dy e z (1 j ) / ô dz ý
| Is|ý
J xs (0) wô
1 j
J xs (0 ) wô
2
P. E. 10.6 (a)
Rac
a a
1.3 103
ð f ýó ý
ð 107 4ð 107 3.5 107 = 24.16
ý
ý
Rdc 2ô 2
2
(b)
Rac 1.3 103
ð 2 109 4ð 107 3.5 107 ý 341.7
ý
Rdc
2
P. E. 10.7
E ý Re[ E s e jt ] ý Re ùû Eo e jt e j ò z a x Eo e jð / 2 e jt e jò z a y ùû
ý Eo cos(t ò z )a x Eo cos(t ò z ð / 2)a y
ý Eo cos(t ò z )a x Eo sin(t ò z )a y
At z = 0, Ex ý Eo cos t , E y ý Eo sin t
2
2
ö E ö ö Ey ö
÷ x ÷ ÷ ÷ ý1
cos t sin t ý 1
ø Eo ø ø Eo ø
which describes a circle. Hence the polarization is circular.
2
2
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P. E. 10.8
1
2
Pave ý ø H o a x
2
(a) Let f(x,z) = x + y –1 = 0
an ý
ax a y
ñf
,
ý
| ñf |
2
Pt ý P.dS =P.San =
1
=
2 2
dS = dSan
a a
1
ø H o 2ax . x y
2
2
(120 ð )(0.2) 2 (0.1) 2 ý 53.31 mW
(b) dS = dydzax , Pt ý P.dS =
Pt ý
1
(120 ð )(0.2) 2 ð (0.05) 2 ý 59.22 mW
2
P. E. 10.9
ôý
1
ø Ho 2 S
2
ø1 ý øo ý 120ð ,ø 2 ý
ý øo
ý
õ
2
2ø 2
ø ø1
ý 2 / 3, ÷ ý 2
ý 1/ 3
ø2 ø1
ø2 ø1
Ero ý ÷ Eio ý
Ers ý
10
3
10 j ò1z
e a x V/m
3
where ò 1 ý / c ý 100 ð / 3 .
20
Eto ý ô Eio ý
3
Ets ý
20 j ò2 z
a x V/m
e
3
where ò 2 ý õ r / c ý 2ò 1 ý 200 ð / 3 .
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P. E. 10.10
ñ1 ý 0, ò1 ý
ó2
ý
õ 2
c
ýr õ r ý
0.1
109
7.5 10 4
36ð
2
ý 5
ý 5c / 2 ý 7.5 108
c
ý 1.2ð
8
ñ2 ý
c
4
2
1 1.44 ð 2 1 ý 6 .021
ò2 ý
c
4
2
1 1.44 ð 2 1 ý 7 .826
|ø 2 |ý
ý
60 ð
4
1 1.44 ð 2
ý
ý 95.445, ø 1 ý 120 ð õ r 1 ý 754
tan 2ñ ø 2 ý 1.2 ð ñ ø 2 ý 37 .57 o
ø 2 ý 95.445ð 37 .57 o
(a)
÷ ý
ø 2 ø 1 95.445ð 37 .57 o 754
ý 0.8186 ð 17108
ý
. o
ø 2 ø 1 95.445ð 37 .57 o 754
ô ý 1 ÷ ý 0.2295ð 33.56 o
sý
(b)
1 | ÷ | 1 0.8186
ý 10.025
ý
1 | ÷ | 1 0.8186
Ei ý 50sin(t 5 x)a y ý Im( Eis e jt ) , where Eis ý 50 e j 5 x a y .
Ero ý ÷ Eio ý 0.8186 e j 171.08 (50 ) ý 40.93e j 171.08
o
o
Ers ý 40.93e j 5 x j171.08 a y
o
Er ý Im( Ers e jt ) ý 40.93sin(t 5 x 171.1o )a y V/m
a H ý a k a E ý a x a y ý a z
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Hr ý -
40.93
sin(t 5 x 171.1o )a z ý 0.0543sin(t 5 x 171.1o )a z A/m
754
(c )
o
o
Eto ý ô Eio ý 0.229 e j 33.56 (50 ) ý 11.475e j 33.56
Ets ý 11.475e j ò2 x j 33.56 e ñ 2 x a y
o
Et ý Im( Ets e jt ) ý 11.475e 6.021x sin(t 7.826 x 33.56o )a y V/m
a H ý ak a E ý a x a y ý a z
Ht ý
11.495 6.021x
sin(t 7.826 x 33.56o 37.57o )a z
e
95.445
ý 0.1202e 6.021x sin(t 7.826 x 4.01o )a z A/m
(d)
2
2
P1ave
E
E
1
[502 a x 40.932 a x ] ý 0.5469a x W/m2
ý io a x ro (a x ) ý
2ø1
2ø1
2(240ð )
P2ave
E
(11.475) 2
cos 37.57 o e 2(6.021) x a x ý 0.5469e 12.04 x a x W/m2
ý to e 2ñ 2 x cos ñø2 a x ý
2 | ø2 |
2(95.445)
2
P. E. 10.11 (a)
k ý 2a y 4a z
k ý 22 42 ý 20
ý kc ý 3 108 20 ý 1.342 109 rad/s ,
ü ý 2ð / k ý 1.405m
(b) H ý
ak E
øo
ý
(2a y 4a z )
20(120ð )
(10a y 5a z ) cos(t k.r )
ý 29.66 cos(1.342 x109 t 2 y 4 z )a x mA/m
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(c ) Pave ý
| Eo |2
125 (2a y 4a z )
ak ý
ý 74.15a y 148.9a z
2øo
2(120ð )
20
mW/m2
P. E. 10.12 (a)
y
ki
ñi
kt
ñr
z
kr
tan ñ i ý
sin ñ t ý
kiy
kiz
ý
2
ñ i ý 26 .56 ý ñ r
4
ý1 õ1
ý2 õ2
sin ñ i ý
(b) ø 1 ý ø o , ø 2 ý ø o / 2
1
sin 26 .56 o ñ t ý 12.92 o
2
E is parallel to the plane of incidence. Since ý 1 ý ý 2 ý ý o ,
we may use the result of Prob. 10.42, i.e.
÷ \\ ý
tan(ñ t ñ i ) tan( 13.64 o )
ý
ý 0.2946
tan(ñ t ñ i ) tan(39.48 o )
ô \\ ý
2 cos 26 .56 o sin 12.92 o
ý 0.6474
sin 39.48 o cos( 13.64 o )
(c) kr ý ò1 sin ñ r a y ò1 cos ñ r a z .
Once kr is known, Er is chosen such that
kr .Er ý 0 or ñ.Er ý 0. Let
Er ý Eor ( cos ñ r a y sin ñ r a z ) cos(t ò1 sin ñ r y ò1 cos ñ r z )
Only the positive sign will satisfy the boundary conditions. It is evident that
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Ei ý Eoi (cos ñi a y sin ñi a z ) cos(t 2 y 4 z )
Since ñ r ý ñ i ,
Eor cos ñ r ý ÷ / / Eoi cos ñ i ý 10 ÷ / / ý 2.946
Eor sin ñ r ý ÷ / / Eoi sin ñ i ý 5÷ / / ý 1.473
ò 1 sin ñ r ý 2, ò 1 cos ñ r ý 4
i.e.
Er ý (2.946 a y 1.473az ) cos( t 2 y 4 z)
E1 ý Ei Er ý (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 ý ò 2 sin ñt a y ò 2 cos ñt a z .
Since k r ÷ Er ý 0 , let
Et ý Eot (cos ñt a y sin ñt a z ) cos(t ò 2 y sin ñt ò 2 z cos ñt )
ò 2 ý ý 2 õ 2 ý ò 1 õ r 2 ý 2 20
1
1
sin ñ i ý
,
cos ñ t ý
2
2 5
19
ò 2 cos ñ t ý 2 20
ý 8.718
20
sin ñ t ý
9
20
Eot cos ñ t ý ô / / Eoi cos ñ t ý 0.6474 125
19
ý 7 .055
20
Eot sin ñ t ý ô / / Eoi sin ñ t ý 0.6474 125
1
ý 16185
.
20
Hence
E2 ý Et ý (7.055a y 1.6185a z ) cos(t 2 y 8.718 z ) V/m
(d) tan ñ B / / ý
õ2
ý 2 ñ B / / ý 63.43 o
õ1
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P.E. 10.13
Si =
1 0.4 14
.
ý
= 2.333
1 0.4 0.6
So =
1 0.2 12
.
ý
= 1.5
1 0.2 0.8
Prob. 10.1 (a) Wave propagates along +ax.
(b)
2ð
ý 1ý s
2ð 106
2ð 2ð
ý
ý 1.047m
üý
6
ò
Tý
uý
2ð
ý
2ð 106
ý
ý 1.047 106 m/s
6
ò
(c ) At t=0, E z ý 25 sin( 6 x ) ý 25 sin 6 x
At t=T/8, E z ý 25 sin(
ð
2ð T
6 x ) ý 25 sin( 6 x )
4
T 8
At t=T/4, E z ý 25 sin(
At t=T/2, E z ý 25 sin(
2ð T
6 x ) ý 25 sin( 6 x 90 o ) ý 25 cos 6 x
T 4
2ð T
6 x ) ý 25 sin( 6 x ð ) ý 25 sin 6 x
T 2
These are sketched below.
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Prob. 10.2
c 3 108
ý 5 106 m
(a) ü ý ý
f
60
3 108
ý 150 m
2 106
3 108
(c ) ü ý
ý 2.5 m
120 106
3 108
(d) ü ý
ý 0.125 m
2.4 109
(b) ü ý
Prob. 10.3
(a) ý 108 rad/s
(b) ò ý
(c ) ü ý
ý
c
2ð
ò
108
ý 0.333 rad/m
3 108
ý 6ð ý 18.85 m
(d) Along -ay
At y=1, t=10ms,
1
(e) H ý 0.5cos(108 t 10 109 3) ý 0.5cos(1 1)
3
ý 0.1665 A/m
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Prob. 10.4
(a)
öE
ý sin( x t ) sin( x t )
öx
ö2 E
ý cos( x t ) cos( x t ) ý E
öx 2
öE
ý sin( x t ) sin( x t )
öt
ö2 E
ý 2 cos( x t ) 2 cos( x t ) ý 2 E
öt 2
2
ö2 E
2 ö E
u
ý 2 E u 2 E ý 0
2
2
öt
öx
2
if ý u 2 and hence, eq. (10.1) is satisfied.
(b) u ý
Prob. 10.5 If
÷ 2 ý j ý (ó j õ ) ý 2 ý õ j ý ó
|÷ 2 |ý
i.e.
( ñ 2 ò 2 ) 4ñ 2 ò 2 ý
and ÷ ý ñ jò , then
(ñ 2 ò 2 ) 2 ý ñ 2 ò 2
ñ 2 ò 2 ý ý (ó 2 2 õ 2 )
(1)
Re( ÷ ) ý ñ ò ý ý õ
2
2
2
2
ò 2 ñ 2 ý 2ý õ
(2)
Subtracting and adding (1) and (2) lead respectively to
ñý
2
ù
ýõ ù
ö ó ö
ú 1 ÷ ÷ 1ú
ø õø
2 ú
úû
û
òý
2
ù
ýõ ù
ö ó ö
ú 1 ÷ ÷ 1ú
ø õø
2 ú
úû
û
(b) From eq. (10.25), E s ( z ) ý Eo e÷ z a x .
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ñ E ý jý H s
Hs ý
But H s ( z ) ý H o e ÷ z a y , hence Ho ý
øý
j
ý
ñ Es ý
j
ý
(÷ Eo e÷ z a y )
Eo
j÷
ý
E
ý o
ø
j ý
÷
(c) From (b),
j ý
ý
j ý (ó j õ )
øý
ý /õ
|ø|ý
4
ö ó ö
1 ÷ ÷
ø õø
j ý
ý
ó j õ
ö õö
, tan 2ñ ø ý ÷ ÷
ø ó ø
2
ý /õ
1 j
1
ý
ó
õ
ó
õ
Prob. 10.6 (a)
From eq. (10.18),
÷ = jý (ó jõ ) ý
Assuming that
ó
õ
jó
jó ö
ö
) ý j ýõ ÷1
jý jõ (1
÷
õ
ø õ ø
1/2
1, we include up to the second power in
neglect higher-order terms.
÷
ý
ó
2
ö
ý ö ó2 ö
ó2 ö
÷1
÷ j ýõ ÷1 2 2 ÷ ý ñ j ò
õ ø 2õ ø
ø 8 õ ø
Thus,
ö
ó2 ö
ò = ýõ ÷1 2 2 ÷
ø 8 õ ø
Prob. 10.7
(a)
ó
ý
õ
8 102
109
2ð 50 10 3.6
36ð
ý8
6
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and
õ
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ñ ý
ò ý
ýõ ùú
2
ù
2ð 50 106
öó ö
ú
=
1 ÷
1
÷
2 ú
3 108
ú
ø õ ø
û
û
2.1 3.6
[ 65 1] ý 5.41
2
ýõ ù
2
ù
ó ö
ö
ú 1 ÷
ú = 6.129
1
÷
2 ú
ú
ø õ ø
û
û
÷ ý ñ jò ý 5.41 j6 .129 /m
2ð
2ð
.
ý
ý 1025
m
ò
6 .129
(b)
üý
(c)
2ð 50 106
uý ý
ý 5.125 107 m/s
6.129
ò
ý
õ
(d) | ø | ý
4
ö ó ö
1 ÷ ÷
ø õø
tan 2ñ ø ý
ý
2
2.1
3.6
ý 101.4
65
120 ð
4
ó
ý 8 ñ ø ý 41.44 o
õ
ø ý 101.41ð 41.44 o
(e)
H s ý ak
Es
ø
o
6
6
ý a x e ÷ z a z ý e÷ z a y ý 59.16e j 41.44 e÷ z a y mA/m
ø
ø
Prob. 10.8
(a) tan ñ ý
ó
ý
õ
102
ý 1.5
109
2ð 12 10 10
36ð
4
10
(b) tan ñ ý
ý 3.75 102
9
10
2ð 12 106 4
36ð
6
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(c ) tan ñ ý
4
109
2ð 12 10 81
36ð
ý 74.07
6
Prob. 10.9
(a)
ýõ ù
2
ù 2ð 15 109 1 9.6
ó ö
ö
ù 1 9 108 1ù
ú 1 ÷
úý
1
÷
8
û
û
2 ú
3
10
2
õ
ú
ø
ø
û
û
ñ ý
ö1
ö
ý 100ð 4.8 ÷ 9 108 ÷ ý 0.146
ø2
ø
1
ô ý ý 6.85 m
ñ
(b) A ý ñ ý 0.146 5 103 ý 0.73 103 Np
Prob. 10.10
If ó =õ , the loss tangent is
ñ ý
But
ñý
ýõ ù
2
ù
ó ö
ýõ ù
ýõ ù
ú 1 ÷ö
ú
ý
1
1 12 1ù ý
2 1ûù
÷
û
2 ú
2 û
2 û
õ ø
ú
ø
û
û
2ð 2ð c
2ð c
ý
ý
üo ý
ò
2ð c
ýõ
üo
2
ò ý
ó
ý1
õ
2 1 ý
üo
2ð 3 108
12 102
ýõ ù
1
109
4ð 107 4
(0.6436) ý 47.66 Np/m
2
36ð
2
ù 2ð c ýõ
ó ö
ú 1 ö÷
÷ 1ú ý
2 ú
üo
2
ú
ø õ ø
û
û
2 1
2ð 3 108 1
109
7
ð
4
10
4
(1.5538)
12 102
2
36ð
ý 115.06 rad/m
ý
2ð c
2ð 3 108
ý
ý 1.3652 108 m/s
uý ý
2
ò üo ò 12 10 115.06
Prob. 10.11
For silver, the loss tangent is
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ó
ý
õ
6.1 107
ý 6.1 18 108
9
10
2ð 108
36ð
Hence, silver is a good conductor
For rubber,
ó
ý
õ
1015
ý
1
18
1014
3.1
1
10
2ð 10 3.1
36ð
Hence, rubber is a poor conductor or a good insulator.
9
8
Prob. 10.12
4
ó
ý
ý 9, 000 þþ 1
5
õ 2ð 10 80 109 / 36ð
ñ ýò ý
ýó
2
2ð 105
ý 5 105 m/s
0.4ð
(a)
u ý/ò ý
(b)
ü ý 2ð / ò ý
(c ) ô ý 1 / ñ ý
2ð 105
4ð 107 4 ý 0.4ð
2
ý
2ð
ý5m
0.4 ð
1
ý 0.796 m
0.4 ð
(d) ø ý | ø | ð ñ ø , ñ ø ý 45 o
ý
õ
| ø |ý
4
ø ý 0.4443ð45o
öó ö
1 ÷
÷
ø õ ø
2
ý õ
4ð 107 2ð 105
ý
ý 0.4443
õ ó
4
Prob. 10.13 (a)
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T ý 1 / f ý 2ð / ý
xý
(b) Let
ñ ö x 1ö
ý÷
÷
ò ø x 1ø
But
ö ó ö
1 ÷ ÷
ø õø
ñý
ö x 1ö
òý÷
÷
ø x 1ø
2
1/ 2
c
x 1 ý
2ð
ý 20 ns
ð x10 8
ý rõ r
2
x 1
ñc
0.1 3 108
ý
ý 0.06752
x ý 1.0046
ýr õ r
ð 108 2
2
1/ 2
ü ý 2ð / ò ý
ö 2.0046 ö
ñý÷
÷
ø 0.0046 ø
1/ 2
0.1 ý 2.088
2ð
ý 3m
2.088
ý /õ
|ø|ý
xý
ö ó ö
1 ÷ ÷ ý 10046
.
ø õø
x
ý
377
ý 188.1
2 10046
.
(c )
2
ó
ý 0.096 ý tan 2ñ ø ñ ø ý 274
. o
õ
ø ý 188.1ð 274
. o
Eo ý ø H o ý 12 188.1 ý 2257.2
a E a H ý ak
a E a x ý a y
aE ý az
E ý 2.257e 0.1 y sin(ð 108 t 2.088 y 2.74o )a z kV/m
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(d) The phase difference is 2.74o.
Prob. 10.14
This is a lossy medium in which ý=ýo.
öó ö
Let x ý ÷
÷
ø õ ø
o
jý j 2ð 109 4ð
øý
ý
ý 35.31ð26.57
÷
100 j 200
Eo ý 0.05 35.31 ý 1.765
2
a E ý a H a k ý a z
Thus, we obtain
E = -1.765cos (2ð 109 t 200 x 26.57o )a z V/m
õ r (1/ 3) ý
õr
ñ c 100 3 108 15
ý
ý
2ð 109
ð
1
ý 4.776
3
tan 2ñø ý
ó 4
ý
õ 3
õ r ý 14.32
ñø ý 26.57 o
377
ý 14.32 ý 77.175
| ø |ý 4
1 x
5/3
Eo ý| ø | H o ý 77.175 50 103 ý 3.858
ý /õ
a E ý ( a k a H ) ý (a x a y ) ý a z
E ý 3.858e 100 x cos(2ð 109 t 200 x 26.57 o )a z V/m
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Prob. 10.15
ó
=
õ
1
109
2ð 10 4
36ð
=4.5
9
ñ ý
ýõ ù
2
ù
ó ö
ú 1 ÷ö
ú
1
õ ÷ø
2 ú
ú
ø
û
û
ý 2ð 10
9
4ð
109 ù
7
1 4.52 1ù
10 4 9
û
2
36ð û
ý 20ð 2[ 21.25 1] ý 168.8 Np/m
ò ý
ýõ ù
2
ù
ó ö
ú 1 ö÷
ú ý 20ð 2[ 21.25 1] ý 210.5 rad/m
1
2 ú
õ ÷ø
ú
ø
û
û
tan 2ñø ý
| ø |ý
ó
ý 4.5
ñø ý 38.73o
õ
ý /õ
120ð 9 / 4
öó ö
4 1
÷
÷
ø õ ø
2
ý
4
ý 263.38
1 4.52
ø ý 263.38ð38.73o
uý
2ð 109
ý
ý 2.985 107 m/s
ò
210.5
Prob. 10.16
ò ý 6.5 ý ýoõ o ý
(a)
c
ý ò c ý 6.5 3 10 ý 1.95 109 rad/s
8
2ð
ý 0.9666 m
ò 6.5
(b) For z=0, Ez ý 0.2 cos t
üý
2ð
ý
2ð ü
) ý 0.2 cos t
ü 2
The two waves are sketched below.
For z=ü /2, Ez ý 0.2 cos(t
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300
z = 0
z = ü/2
Amplitude (mV / m)
200
100
0
-100
-200
-300
-3
-2
-1
0
1
Time t (ns)
2
(c) H ý H o cos(t 6.5 z )a H
Ho ý
Eo
øo
ý
0.2
ý 5.305 104
377
a E a H ý ak
ax aH ý az
aH ý a y
H ý 0.5305cos(t 6.5 z )a y mA/m
Prob. 10.17
üý
2ð
ò
ý
2ð
ýõ
ý o õ oõ r
üo
ý
ý õr
ü
ýoõ o
õr ý
üo 6.4
ý
ý 2.286
ü 2.8
õ r ý 5.224
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Prob. 10.18 (a) Along -x direction.
(b) ò ý 6,
ò ý ýõ ý
ý 2 108 ,
c
ý rõ r
6 3 108
ý9
õr ý òc / ý
2 108
õ ý õ oõ r ý
õ r ý 81
109
81 ý 7.162 1010 F/m
36ð
(c ) ø ý ý / õ ý ýo / õ o ýr / õ r ý
120ð
ý 41.89
9
Eo ý H oø ý 25 103 41.88 ý 1.047
a E a H ý ak
a E a y ý a x
aE ý az
E ý 1.047 sin(2 108 t 6 x)a z V/m
Prob. 10.19 (a)
ó
ý
õ
106
109
2ð 107 5
36ð
ý 3.6 104 üü 1
Thus, the material is lossless at this frequency.
(b) ò ý ýõ ý
üý
2ð 107
5 750 ý 12.83 rad/m
3 108
2ð
2ð
ý
ý 0.49 m
ò
12.83
(c ) Phase difference = ò l ý 25.66 rad
(d) ø ý ý / õ ý 120ð
ýr
750
ý 120ð
ý 4.62 k
õr
5
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Prob. 10.20
(a)
108 ð
ð
ý ý 1.0472 rad/m
ò ý ýõ ý ý
8
c 3 10
3
(b)
sin(108 ð to ò xo ) ý 0 ý sin(nð ), n ý 1, 2,3,...
E ý0
108 ð to ò xo ý ð
108 ð 5 103
ð
3
xo ý ð
xo
5 105 m
(c )
H ý H o sin(108 ð t ò x)a H
50 103
Ho ý
ý
ý 132.63 ý A/m
120ð
ø
a H ý a k a E ý a x a z ý a y
Eo
H ý 132.63sin(108 ð t 1.0472 x)a y ý A/m
Prob. 10.21
This is a lossless material.
ý
ý
ý 377 r ý 105
õ
õr
øý
uý
(1)
c
ý
ý
ý 7.6 107
ò ýõ
ýr õ r
(2)
From (1),
ýr 105
ý
ý 0.2785
õ r 377
(1)a
From (2),
1
ýr õ r
ý
7.6 107
ý 0.2533
3 108
(2)a
Multiplying (1)a by (2)a,
1
õr
ý 0.2785 0.2533 ý 0.07054
õ r ý 14.175
Dividing (1)a by (2)a,
ýr ý
0.2785
ý 1.0995
0.2533
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Prob. 10.22
ax
ay
ö
ö
ñ E ý
öx
öy
(a)
Ex ( z, t ) E y ( z, t )
az
öE
öE
ö
ý y ax x a y
öz
öz
öz
0
ý 6 ò cos(t ò z )a x 8ò sin(t ò z )a y
But ñ E ý ý
Hý
6ò
ý
öH
öt
H ý
sin(t ò z )a x
8ò
ý
1
ý
ñ Edt
cos(t ò z )a y
2ð f
2ð 40 106
4.5 ý
4.5 ý 1.777 rad/m
c
3 108
2ð
2ð
ý
ý 3.536 m
üý
ò 1.777
(b) ò ý ýõ ý
øý
ý 120ð
ý
ý 177.72
õ
4.5
uý
c
3 108
ý
ý
ý 1.4142 108 m/s
4.5
4.5
ýõ
1
Prob. 10.23
(a) E ý Re[ E s e jt ] ý (5a x 12a y )e 0.2 z cos(t 3.4 z )
At z = 4m, t = T/8, t ý
2ð T ð
ý
T 8 4
E ý (5a x 12a y )e 0.8 cos(ð / 4 13.6)
| E |ý 13e 0.8 | cos(ð / 4 13.6) |ý 5.662 V/m
(b) loss ý ñ z ý 0.2(3) ý 0.6 Np. Since 1 Np = 8.686 dB,
loss = 0.6 x 8.686 = 5.212 dB
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(c ) Let
ö ó ö
1 ÷ ÷
ø õø
xý
ñ ö x 1ö
ý÷
÷
ò ø x 1ø
1/ 2
ý 0.2 / 3.4 ý
x 1
ý 1 / 289
x1
ñ ý ýõ / 2 x 1 ý
õr
2
ý
ñc
x 1
| ø |ý
2
ý
ýo 1
.
õo õr
x
c
1
17
x ý 100694
.
õr /2 x 1
0.2 3 108
ý 7.2
108 0.00694
ý
õ r ý 103.68
120ð
ý 36.896
103.68 1.00694
ó
ý x 2 1 ý 0.118
õ
ø ý 36.896ð3.365o
tan 2ñ ø ý
ñ ø ý 3.365 o
Prob. 10.24
ñ ýò ý
ýó
2
ø ý| ø | ðñø ý
| ø |ý
2ñ 2
2 122
óý
ý
ý 36.48
ý 2ð 106 4ð 107
ý
ð45o
ó
2ð 106 4ð 107
ý
ý
ý 0.4652
36.48
ó
Eo ý| ø | H o ý 0.4652 20 103 ý 9.305 103
a E ý a H a k ý a y ( a z ) ý a x
E ý Eo e ñ z sin(t ò z )a E
ý 9.305e 12 z sin(2ð 106 t 12 z 45o )a x mV/m
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Prob. 10.25 For a good conductor,
(a)
ó
ý
õ
102
109
2ð 8 10 15
36ð
ó
þ þ 1,
õ
ý 1.5
say
ó
þ 100
õ
lossy
6
No, not conducting.
(b)
(c )
ó
ý
õ
0.025
ó
ý
õ
25
109
2ð 8 10 16
36ð
No, not conducting.
ý 3.515
lossy
ý 694.4
conducting
6
109
2ð 8 10 81
36ð
Yes, conducting.
6
Prob. 10.26
ñ ýò ý
But
1
ô
uý
ý ô ý 0.02 103 2ð 100 106 ý 4ð 103 ý 1.256 104 m/s
ò
Prob. 10.27 (a)
Rdc ý
l
l
600
ý
ý
ý 2.287
2
7
ó S óð a 5.8 10 ð (1.2) 2 106
l
.
ó 2 ð aô
(see Table 10.2).
(b) Rac ý
Rac ý
(c )
At 100 MHz, ô ý 6.6 103 mm =6.6 10-6 m mm for copper
600
ý 207.61
5.8 10 2ð (1.2 103 ) 6.6 106
Rac
a
ý
ý1
Rdc 2ô
7
66.1103
ô ý a/2ý
f
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f ý
66.1 2 103 66.1 2
ý
a
1.2
f ý 12.137 kHz
Prob. 10.28
(a) Copper is a good conductor.
ñ ý ð f ýó ý ð 1010 4ð 107 5.8 107 ý 2ð 105 5.8
ñ ý 1.513 106 Np/m
(b)
(c )
1
ý 6.609 107 m
6
ñ 1.513 10
1 j
1 j
ý
ý 26.09(1 j ) 103
øý
7
7
óô 5.8 10 6.609 10
ôý
Prob. 10.29
1
ôý
ð f ýó
f ý
1
ý
f ý
1
ô ðýó
2
1
ý 1.038 kHz
4 10 ð 4ð 107 6.1 107
6
Prob. 10.30
ôý
1
1
1
ý
ý
ý 8.531 mm
ð f ýó
ð 5.8 107 (60)4ð 107 2ð 5.8(60)
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Prob. 10.31
This is a good conductor.
ý
ð45o
ó
øý
2ð 12 106 4ð 107
24.6 ý
ó
ý
16ð 2 (0.6)
ó
16ð (0.6)
ý 0.1566 S/m
24.62
1
1
ò ýñ ý ý
ý 8.333 rad/m
ô 0.12
2ð
üý
ý 2ð (0.12) ý 0.754 m
2
óý
ò
u ý ý ô ý 2ð 12 106 0.12 ý 9.05 106 m/s
ò
Prob. 10.32
ó
ý
õ
0.12
109
2ð 2.42 10 5.5
36ð
This is a lossy material.
ý 0.1623
9
ýõ ù
2
ù
ó ö
ö
ú 1 ÷
ú
1
õ ÷ø
2 ú
ú
ø
û
û
ñ ý
ý 2ð 2.42 109
4ð
109 ù
1 0.16232 1ù
107 5.5
û
û
2
36ð
ý 15.21(109 )(108 ) 0.01308
ý 17.39
ôý
1
ñ
ý 57.5 mm
Prob. 10.33
t ý 5ô ý
5
5
ý
ý 2.94 106 m
9
7
7
ð f ýó
ð 12 10 4ð 10 6.110
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Prob. 10.34
ó
ý
õ
4
109
2ð 2 10 24
36ð
ý 1.5
9
ýõ ù
ù
ó ö
ú 1 ö÷
ú
1
ý 2ð 2 109
÷
õ
2 ú
ú
ø
ø
û
û
ý 130.01 Np/m
ñ ý
2
109
4ð 10 24
36ð ù 1 ø1.5 ù2 1ù
úû
ûú
2
7
105 Eo ý Eo e ñ d
Taking the log of both sides gives
-5ln10 ý ñ d
dý
5ln10
ñ
ý
5ln10
ý 0.0886 m
130.01
Prob. 10.35
(a) Linearly polarized along az
(b) ý 2ð f ý 2ð 107
ò ý ýõ ý ýoõ o õ r ý
(c )
Ho ý
c
ò c 3 3 108
ý
ý 14.32
2ð 107
H ý H o sin(t 3 y )a H
õr ý
Let
f ý 107 ý 10 MHz
Eo
ø
,
øý
õr
õ r ý 205.18
ý øo 120ð
ý
ý
ý 26.33
õ
õ r 14.32
12
ý 0.456
26.33
a H ý ak a E ý a y a z ý a x
(d) H o ý
H ý 0.456sin(2ð 107 t 3 y )a x A/m
Prob. 10.36
E ý (2a y 5a z ) sin(t ò x)
The ratio E y / Ez remains the same as t changes. Hence the wave is linearly polarized
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Prob. 10.37
(a)
Ex ý Eo cos(t ò y ),
E y ý Eo sin(t ò y )
Ex (0, t ) ý Eo cos t
cos t ý
Ex (0, t )
Eo
E y (0, t ) ý Eo sin t
sin t ý
E y (0, t )
Eo
2
2
ö E ö ö Ey ö
÷ x ÷ ÷ ÷ ý1
cos t sin t ý 1
ø Eo ø ø Eo ø
Hence, we have circular polarization.
2
2
(b)
Ex ý Eo cos(t ò y ),
E y ý 3Eo sin(t ò y )
In the y=0 plane,
Ex (0, t ) ý Eo cos t
cos t ý
E y (0, t ) ý Eo sin t
sin t ý
2
Ex (0, t )
Eo
E y (0, t )
3Eo
2
ö E ö 1ö E ö
cos t sin t ý 1
÷ x ÷ ÷ y ÷ ý1
ø Eo ø 9 ø Eo ø
Hence, we have elliptical polarization.
2
2
Prob. 10.38
(a) We can write
E ý Re( Es e jt ) ý (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.
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Prob. 10.39
(a) The wave is elliptically polarized.
(b)
Let E ý E1 E2 ,
where E1 ý 40 cos(t ò z )a x ,
E2 ý 60sin(t ò z )a y
H1 ý H o1 cos(t ò z )a H 1
H o1 ý
40
øo
ý
40
ý 0.106
120ð
a H 1 ý ak a E ý a z a x ý a y
H1 ý 0.106 cos(t ò z )a y
H 2 ý H o 2 sin(t ò z )a H 2
H o1 ý
60
øo
ý
60
ý 0.1592
120ð
a H 2 ý ak a E ý a z a y ý a x
H 2 ý 0.1592sin(t ò z )a x
H ý H1 H 2 ý 159.2sin(t ò z )a x 106 cos(t ò z )a y mA/m
Prob. 10.40
We can write Es as
E s ý E1 ( z ) E2 ( z )
where
1
Eo (a x ja y )e j ò z
2
1
E2 ( z ) ý Eo (a x ja y )e jò z
2
We recognize that E1 and E2 are circularly polarized waves. The problem is therefore proved.
E1 ( z ) ý
Prob. 10.41
(a)
When ö ý 0,
E ( y, t ) ý ( Eo1a x Eo 2 a z ) cos(t ò y )
The two components are in phase and the wave is linearly polarized.
(b)
When ö ý ð / 2,
Ez ý Eo 2 cos(t ò y ð / 2) ý Eo 2 sin(t ò y )
We can combine Ex and Ez to show that the wave is elliptically polarized.
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(c )
When ö ý ð ,
E ( y, t ) ý Eo1 cos(t ò y )a x Eo 2 cos(t ò y ð )a z
ý ( Eo1 a x Eo 2 a y ) cos(t ò y )
Thus, the wave is linearly polarized.
Prob. 10.42
Let E s ý Er jEi
H s ý H r jH i
and
E ý Re(E s e jt ) ý E r cos t Ei sin t
Similarly,
H ý H r cos t H i sin t
1
P ý E H ý Er H r cos 2 t Ei H i sin 2 t ( Er H i Ei H r ) sin 2t
2
T
T
T
T
1
1
1
1
Pave ý Pdt ý 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
ý
1
1
( Er H r Ei H i ) ý Re[( Er jEi ) ( H r jH i )]
2
2
Pave ý
1
*
Re(E s H s )
2
as required.
Prob. 10.43
(a)
ò ý ýõ ý ýoõ o õ r ý
õr ý
c
õr
ò c 8 3 108
ý
ý 2.4
109
õ r ý 5.76
(b) ø ý
(c ) u ý
ýo 1
ý
377
ý
ý
ý 157.1
õ
õ o õ r 2.4
109
ý
ý 1.25 108 m/s
8
ò
(d)
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H ý H o cos(109 t 8 x)aH
Let
Eo
150
ý 0.955
ø 157.1
a H ý a k a E ý a x a z ý a y
Ho ý
ý
H ý 0.955cos(109 t 8 x)a y A/m
(e)
P ý E H = -150(0.955)cos 2 (109 t 8 x)a x
ý -143.25cos 2 (109 t 8 x)a x W/m 2
Prob. 10.44
P ý EH ý
Pave ý
Eo2
ø
cos 2 (t 10 z )a z
T
Eo2
1
P
ý
az
dt
2ø
T 0
P ý Pave ÷ dS ý
S
Eo2 S (40) 2 ð (1.5) 2 (60) 2
ý
ý
ý 15 W
2ø
2 120ð
240
Prob. 10.45
(a)
H
Let H s ý o sin ñ e j 3r a H
r
E
10
1
Ho ý o ý
ý
øo 120ð 12ð
a H ý ak a E ý ar añ ý aö
Hs ý
1
sin ñ e j 3r aö A/m
12ð r
(b)
1
2
Pave ý Re( E s H s ) ý
Pave ý Pave dS ,
10
sin 2 ñ ar
2 12ð r 2
dS ý r 2 sin ñ dñ döar
S
10
Pave ý
24ð
ð ð /6
r
ö ñ
ý0 ý0
2
sin 3 ñ dñ
5 5 3
ý
ý 0.007145
r ý 2 8 32
ý 7.145 mW
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Prob. 10.46
(a) Pave ý
|E |
1
1
82 0.2 z
Re( Es H s* ) ý Re( s ) ý
e
2
2
|ø |
2 |ø |
ñ ý
ò ý
ýõ ù
2
ù
ó ö
ú 1 ÷ö
ú
1
õ ÷ø
2 ú
ú
ø
û
û
ýõ ù
2
ù
ó ö
ö
ú 1 ÷
ú
1
õ ÷ø
2 ú
ú
ø
û
û
öó ö
Let x = 1 ÷
÷
ø õ ø
2
x 1
ñ
ý
ý 0.1/ 0.3 ý 1/ 3
ò
x 1
x 1 1
ý
x 1 9
x ý 5/ 4
5
öó ö
ý 1 ÷
÷
4
ø õ ø
2
ý
õ
| ø |ý
ý
ó
ý 3/ 4
õ
120ð / 81
ý 37.4657
5
4
öó ö
1 ÷
÷
ø õ ø
64
Pave ý
e 0.2 z ý 0.8541e 0.2 z W/m 2
2(37.4657)
2
4
(b) 20dB ý 10 log
P1
P2
P1
ý 100
P2
P2
1
ý e 0.2 z ý
e0.2 z ý 100
100
P1
z ý 5log100 ý 23 m
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Prob. 10.47
(a) u ý / ò
øý
òc
4.5
ý
2 3 108
ý 2.828 108 rad/s
4.5
120 ð
ý 177 .7
4.5
H ý ak
E
ø
ý
az
ø
(b) P ý E H ý
(c )
ý uò ý
Pave =
4.5
ò2
40
ò
9
ò
2
sin(t 2 z )a ò ý
0.225
ò
sin(t 2 z )aö A/m
sin 2 (t 2 z )a z W/m 2
a z , dS = ò dö dò az
Pave = Pave ÷ dS = 4.5
3mm
2mm
dò
ò
2ð
dö = 4.5ln(3/2)(2ð ) = 11.46 W
0
Prob. 10.48
P= EH ý
Pave ý
Eo2
sin 2 ñ sin 2 (t r / c)ar
ør2
T
Eo2
1
sin 2 ñ ar
dt
ý
P
2
2ø r
T 0
Prob. 10.49
òý
Eý
c
1
õ
ý ò c ý 40(3 108 ) ý 12 109 rad/s
ñ Hdt
ö
ö
ö
öy
öz
ñ H = öx
0 10sin(t 40 x) 20sin(t 40 x)
ý 800 cos(t 40 x)a y 400 cos(t 40 x)a z
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Eý
1
800
ñ Hdt ý
sin(t 40 x)a
õ
õ
800
y
400
õ
sin(t 40 x)a z
400
sin(t 40 x)a z
9
10
9 10
12 10
12 10
36ð
36ð
ý 7.539sin(t 40 x)a y 3.77 sin(t 40 x)a z kV/m
ý
9
sin(t 40 x)a y
9
P ý EH ý
0 Ey
0 Hy
Ez
ý ( E y H z E z H y )a x
Hz
ý ùû 20(7.537) sin 2 (t 40 x) 37.7 sin 2 (t 40 x) ùû a x 103
1
20(7.537) 37.7ý a x 103 ý 94. 23 a x kW/m2
2
Pave ý
Prob. 10.50
Pý
Eo2
2øo
Eo2 ý 2øo P ý 2(120ð )10 103 ý 7.539
Eo ý 2.746 V/m
Prob. 10.51
Let T ý t ò z.
ö
öB
ý ñ E ý öx
öt
cos T
ý
ö
öy
sin T
ö
öz
0
öH
ý ò cos Ta x ò sin Ta y
öt
H ý
ò
ò
ò
ùûcos Ta x sin Ta y ùûdt ý
sin Ta x
cos Ta y
ý
ý
ý
cosT
P ý EH =
ý
ò
sin T
ý
sinT
0
ò
ý
(cos 2 T sin 2 T )a z
ò
cos T 0 ý
ý
ò
õ
az ý
a
ý
ý z
which is constant everywhere.
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Prob. 10.52
Eo2
Pý
2øo
Eo2 S (2.4 103 ) 2 450 104
ý
ý 343.8 W
P= PS ý
2øo
2 377
Prob. 10.53
P ý EH ý
Pave ý
(a)
Vo I o
sin 2 (t ò z )a z
2ðò ln(b / a)
2
T
T
Vo I o
Vo I o
1
1
1
az
dt
sin 2 (t ò z )dta z ý
P
ý
2
2
2ðò ln(b / a) T 0
2ðò ln(b / a ) 2
T 0
ý
Vo I o
az
4ðò ln(b / a)
2
(b)
Pave ý Pave dS ,
dS ý ò d ò dö a z
S
2ð b
Vo I o
Vo I o
1
ò d ò dö ý
ý
(2ð ) ln(b / a)
2
4ð ln(b / a ) ö ý0 ò ý a ò
4ð ln(b / a )
1
ý Vo I o
2
Prob. 10.54
E 2
(a) Pi ,ave ý io ,
2ø 1
Pr ,ave ý
Ero 2
,
2ø 1
Pt ,ave ý
Pr ,ave Ero 2
ö ø ø1ö
2
÷
Rý
ý
ý÷ 2
2 ý ÷
Pi ,ave Eio
ø ø2 ø1 ø
ö
÷
÷
Rý ÷
÷
ø
ýo
õ2
ýo
õ2
Eto 2
2ø 2
2
2
ýo ö
2
÷
ö ý oõ 1 ý oõ 2 ö
õ1 ÷
÷÷
ý ÷÷
ýo ÷
ø ý oõ 1 ý oõ 2 ø
÷
õ1 ø
Since n1 ý c ý 1õ 1 ý c ý o õ 1 ,
n2 ý c ý o õ 2 ,
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ö n n ö
Rý÷ 1 2 ÷
ø n1 n2 ø
Pt , ave
Tý
Pi ,ave
ý
2
2
ø1 Eto ø1 2 ø1
4n1n2
ý ô ý (1 ÷) 2 ý
2
(n1 n2 ) 2
ø 2 Eio ø2
ø2
(b) If Pr ,ave ý Pt ,ave RPi ,ave ý TPi .ave R ý T
i.e. (n1 n2 ) 2 ý 4n1n2
n12 6 n1n2 n2 2 ý 0
2
ön ö
ön ö
or ÷ 1 ÷ 6 ÷ 1 ÷ 1 ý 0, so
ø n2 ø
ø n2 ø
n1
ý 3 8 ý 5.828
or
0.1716
n2
(Note that these values are mutual reciprocals, reflecting the inherent symmetry of the
problem.)
Prob. 10.55
ø1 ý
÷ý
2 ýo øo
ý1
ý
ý ,
8õ o
2
õ1
ø2 ý
ø1 ø2 øo / 4 øo / 2
1
ý
ý ,
3
ø1 ø 2 øo / 4 øo / 2
ýo
ø
ý2
ý
ý o
16õ o
4
õ2
ô ý ÷ 1 ý
2
3
1
1
Er ý (60) sin(t 10 z )a x (30) sin(t 10 z ð / 6)a y
3
3
ý 20sin(t 10 z )a x 10sin(t 10 z ð / 6)a y V/m
2
2
Et ý (60) sin(t 10 z )a x (30) sin(t 10 z ð / 6)a y
3
3
ý 40sin(t 10 z )a x 20sin(t 10 z ð / 6)a y V/m
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Prob. 10.56
ø1 ý øo ý 120ð ,
÷ý
ø2 ý
ø 2 ø1 øo / 3 øo
1
ý
ý ,
ø2 ø1 øo / 3 øo
2
Er ý ÷Eo ý Eo / 2,
Piave ý
Ptave
ýo øo
ý
ý
ý
ý 40ð
õ
9õ o
3
2
ô ý 1 ÷ ý
1
2
Et ý ô Eo ý Eo / 2
2
o
| Eo |
E
ý
2ø1
2øo
1 2
Eo
| Et |2
E2 3
4
ý
ý
ý o
2ø2
2(øo / 3) 2øo 4
Ptave 3
ý ý 0.75
Piave 4
Prob. 10.57
ø1 ý øo
(a)
Ei ý Eio sin(t 5 x)a E
Eio ý H ioøo ý 120ð 4 ý 480ð
a E a H ý ak
a E a y ý a x
a E ý a z
Ei ý 480ð sin(t 5 x)a z
ø2 ý
÷ ý
ýo 120ð
ý
ý 60ð
4õ o
4
ø 2 ø 1 60 ð 120 ð
ý 1 / 3,
ý
ø 2 ø 1 60 ð 120 ð
ô ý 1 ÷ ý 2 / 3
Ero ý ÷Eio ý (1/ 3)(480ð ) ý 160ð
Er ý 160ð sin(t 5 x)a z
E1 ý Ei Er ý 1.508sin(t 5 x)a z 0.503sin(t 5 x)a z kV/m
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(b) Eto ý ô Eio ý (2 / 3)(480 ð ) ý 320 ð
E
(320ð ) 2
P ý to a x ý
a x ý 2.68a x kW/m 2
2ø 2
2(60ð )
1 | ÷ | 1 1 / 3
ý
(c) s ý
ý2
1 | ÷ | 1 1 / 3
2
Prob. 10.58 ø 1 ý
÷ ý
ø2 ø1
ý 1 / 3,
ø2 ø1
ý1
ý ø o / 2,
õ1
ô ý 1 ÷ ý 4 / 3
Eot ý ô Eio ý 20 / 3
Eor ý ÷Eio ý (1/ 3)(5) ý 5 / 3,
òý
(a)
108
ýr õ r ý
4 ý 2/3
3 108
c
5
Er ý cos(108 t 2 y / 3)a z
3
E1 ý Ei Er ý 5cos(108 t
2
(b)
ø2 ý øo
Pave1 ý
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 ) ý
(1 )(a y ) ý 0.0589a y W/m 2
2ø1
2ø1
2(60ð )
9
2
(c ) Pave2
E
400
(a y ) ý 0.0589a y W/m 2
ý to (a y ) ý
2ø 2
9(2)(120ð )
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Prob. 10.59
ò1 ý ý1õ1 ý ýoõ o ýr1õ r1 ý
ò 2 ý ýoõ o ý
ø1 ý
÷ý
16 ý
c
90 109 (4)
ý 300(4) ý 1200
3 108
90 10
ý 300
3 108
ýo
ý1
ý
õ1
õo
9
ý r1 ø o
ý
ý 30ð ,
õ r1 4
ø 2 ý øo ý 120ð
ø 2 ø1 øo øo / 4 3
ý
ý ,
ø2 ø1 øo øo / 4 5
ô ý 1 ÷ ý
8
5
P ý EH
P1 ý Pi Pr ý
602
ø1
3
(60 ) 2
5 cos 2 (t ò z )a
cos 2 (t ò1 z )a x
1
x
ø1
602
602
ý
ý 38.197,
ø1 30ð
3
(60 ) 2
5 ý 13.75
ø1
P1 ý 38.197 cos (t ò1 z )a x 13.75cos 2 (t ò1 z )a x W/m 2 , where ò1 ý 1200
2
8
(60 ) 2
5 cos 2 (t ò z )a
P2 ý Pt ý
1
x
øo
8
(60 ) 2
2
5 ý 96 ý 24.46
120ð
øo
P2 ý 24.46 cos 2 (t ò 2 z )a x W/m 2 , where ò 2 ý 300
Prob. 10.60
(a) In air, ò 1 ý 1, ü 1 ý 2 ð / ò 1 ý 2 ð ý 6 .283 m
ý ò1c ý 3 108 rad/s
In the dielectric medium, is the same .
ý 3 108 rad/s
ò2 ý
c
õ r2 ý ò 1 õ r2 ý
3
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2ð 2ð
ý
ý 3.6276 m
ò2
3
E
10
ý 0.0265
(b) Ho ý o ý
ø o 120 ð
a H ý a k a E ý a z a y ý a x
ü2 ý
H i ý 26.5cos(t z )a x mA/m
(c ) ø1 ý øo ,
÷ ý
ø 2 ý øo / 3
ø 2 ø 1 (1 / 3 ) 1
ý
ý 0.268 ,
ø 2 ø 1 (1 / 3 ) 1
(d) Eto ý ô Eio ý 7 .32,
ô ý 1 ÷ ý 0.732
Ero ý ÷ Eio ý 2.68
E1 ý Ei Er ý 10 cos(t z )a y 2.68cos(t z )a y V/m
E2 ý Et ý 7.32 cos(t z )a y V/m
Pave1 ý
1
1
2
2
(a z )[ Eio Ero ] ý
(a z )(102 2.682 ) ý 0.1231a z W/m 2
2ø1
2(120ð )
2
E
3
(7.32) 2 (a z ) ý 0.1231a z W/m 2
ý to (a z ) ý
2ø 2
2 120ð
Pave2
Prob. 10.61
ø1 ý øo ý 120ð
For seawater (lossy medium),
ø2 ý
÷ý
jýo
ý
ó jõ
j 2ð 108 4
109
4 j 2ð 10 81
36ð
ý 10.44 j 9.333
8
ø2 ø1
ý 0.9461ð177.16
ø2 ø1
| ÷ |2 ý 0.8952,
Pr
ý 89.51%,
Pi
1 | ÷ |ý 0.1084
Pt
ý 10.84%,
Pi
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÷ý
ø2 ø1 7.924ð43.975 377
ý
ý 0.9702ð178.2o
ø2 ø1 7.924ð43.975 377
The fraction of the incident power reflected is
Pr
ý| ÷ |2 ý 0.97022 ý 0.9413
Pi
The transmitted fraction is
Pt
ý 1 | ÷ |2 ý 1 0.97022 ý 0.0587
Pi
Prob. 10.62
(a)
ý 120ð
ø1 ý 1 ý
ý 188.5,
õ1
4
ý2 120ð
ý
ý 210.75
õ2
3.2
ø ø 210.75 188.5
2ø 2
2 210.75
÷ý 2 1 ý
ý 0.0557, ô ý
ý
ý 1.0557
ø2 ø1 210.75 188.5
ø2 ø1 210.75 188.5
Ero ý ÷Eio ý (0.0557)(12) ý 0.6684
Eto ý ô Eio ý 1.0557(12) ý 12.668
ò1 ý ý1õ1 ý
c
ø2 ý
ý
4
ò1 c
2
ý
40ð (3 108 )
ý 6ð 109 rad/s
2
(b)
6ð 109 3.2
ò 2 ý ý 2õ 2 ý
3.2 ý
ý 112.4
c
3 108
Er ý Ero cos(t 40ð x)a z ý 0.6684 cos(6ð 109 t 40ð x)a z V/m
Et ý Eto cos(t ò 2 x)a z ý 12.668cos(6ð 109 t 112.4 x)a z V/m
Prob. 10.63 (a) ý ò c ý 3 3 108 ý 9 108 rad/s
(b) ü ý 2ð / ò ý 2ð / 3 ý 2.094 m
(c )
4
ó
ý
ý 2ð ý 6.288
8
õ 9 10 80 109 / 36ð
tan 2ñ ø ý
ó
ý 6 .288
õ
ý2 /õ2
|ø 2 |ý
4
ö ó ö
1 ÷ 2 ÷
ø õ2ø
2
ý
377 / 80
4
1 4ð 2
ñ ø ý 40.47 o
ý 16 .71
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ø 2 ý 16 .71ð 40.47 o
(d)
ø 2 ø1 16.71ð40.47o 377
÷ý
ý
ý 0.935ð179.7 o
o
ø2 ø1 16.71ð40.47 377
Eor ý ÷Eoi ý 9.35ð179.7o
Er ý 9.35sin(t 3 z 179.7)a x V/m
ù
2
ù
ö ó2 ö
ý r 2õ r 2 ú
9 108 80 ù
ú
ñ2 ý
1 ÷
1 4ð 2 1ù ý 43.94 Np/m
÷ 1 ý
8
û
û
ú 3 10
2 ú
2
c
ø õ 2 ø
û
ò2 ý
û
9 108
3 108
80 ù
1 4ð 2 1ù ý 51.48 rad/m
û
û
2
2ø2
2 16.71ð40.47o
ý
ý 0.0857ð38.89o
ôý
o
ø2 ø1 16.71ð40.47 377
Eot ý ô Eo ý 0.857ð38.89o
Et ý 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
ó= 0 Standing waves of H
ó =
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.
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Prob. 10.65 Since ý o ý ý 1 ý ý 2 ,
sin ñ t 1 ý sin ñ i
õ o sin 45 o
ý
ý 0.3333
õ1
4.5
sin ñt 2 ý sin ñt1
õ1 1 4.5
ý
ý 0.4714
õ 2 3 2.25
ñ t 1 ý 19.47 o
ñt 2 ý 28.13o
Prob. 10.66
Es ý
e
2
jk y y
j ( kx x k y y )
e
20(e jkx x e jk x x ) (e
j2
ý j 5 ùûe
j ( kx x k y y )
e
jk y y
)
az
j ( kx x k y y )
e
j ( kx x k y y )
ù az
û
which consists of four plane waves.
ñ E s ý jýo H s
Hs ý
Hs ý
j
ýo
ñ Es ý
ö Ez ö
j ö ö Ez
ax
ay
÷
ýo ø ö y
ö x ÷ø
j 20
ù k y sin(k x x) sin(k y y )a x k x cos(k x x) cos(k y y )a y ùû
ýo û
Prob. 10.67
ø1 ý øo ý 377
For ø 2 ,
ó2
ý
õ 2
4
109
2ð 1.2 10 50
36ð
ý 1.2
9
tan 2ñø2 ý
ó2
ý 1.2
õ 2
ý /õ
| ø2 |ý
4
öó ö
1 ÷ 2 ÷
ø õ 2 ø
2
ñø2 ý 25.1o
ý
120ð 1/ 50
4
1 1.22
ý 42.658
ø 2 ý 42.658ð25.1o
ø ø 42.658ð25.1o 377
÷ý 2 1 ý
ý 0.8146ð174.4o
o
ø2 ø1 42.658ð25.1 377
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Prob. 10.68
(a)
Pt ý (1 | ÷ |2 ) Pi
sý
1 | ÷ |
1 | ÷ |
| ÷ |ý
s 1
s 1
2
Pt
4s
ö s 1 ö
ý 1 ÷
÷ ý
2
Pi
ø s 1 ø ( s 1)
(b) Pi ý Pr Pt
P ö s 1 ö
Pr
ý 1 t ý ÷
÷
Pi
Pi ø s 1 ø
2
Prob. 10.69
If A is a uniform vector and ö (r ) is a scalar,
ñ (öA) ý ñö A ö(ñ A) ý ñö A
since ñ A ý 0.
ñ E ý (
ö
ö
ö
j (k xk
ax
ay
a z ) Eo e
öx
öy
öz
x
y y k z z t )
ý j (k x a x k y a y k z a z )e j ( k ÷r t ) Eo
ý jk Eo e j ( k ÷r t ) ý jk E
Also,
öB
ý jý H .
öt
Hence ñ E ý
öB
becomes k E ý ý H
öt
ak a E ý a H
From this,
Prob. 10.70
k
k ý| |ý 1242 1242 2632 ý 316.1
2ð
üý
ý 19.88 mm
k
2ð f
kc 316.1 3 108
f ý
ý
ý 15.093 GHz
k ý ýõ ý
2ð
2ð
c
124
cos ñ x ý
ñ x ý 66.9o ý ñ y
k ÷ a x ý k cos ñ x
316.1
ñ z ý cos 1
263
ý 33.69o
361.1
Thus,
ñ x ý ñ y =66.9o , ñ z ý 33.69o
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Prob. 10.71
k ý 3.4a x 4.2a y
k E ý0
0 ý 3.4 Eo 4.2
4.2
ý 1.235
3.4
Eo ý
k ý| k |ý ò ý (3.4) 2 (4.2) 2 ý 5.403
2ð
üý
ò
f ý
c
ü
Hs ý
ý
ý
ý
2ð
ý 1.162
5.403
3 108
ý 258 MHz
1.162
1
k Es ý
ý
1
k Es
ý kc
3.4 4.2
0
1
A
8
1 3 j4 o
4ð 10 5.403 3 10 Eo
7
Ao ý e j 3.4 x 4.2 y
where
H s ý 4.91Ao 104 ùû 4.2(3 j 4)a x 3.4(3 j 4)a y (3.4 4.2 Eo )a z ùû
ý 0.491 ùû(12.6 j16.8)a x (10.2 j13.6)a y 8.59a z ùû e j 3.4 x 4.2 y mA/m
Prob.10.72
ñ÷ E ý (
ö
ö
ö
j (k xk
ax
ay
a z ) ÷ Eo e
öx
öy
öz
x
ý jk ÷ Eo e j ( k ÷r t ) ý jk ÷ E ý 0
y y k z z t )
ý j (k x a x k y a y k z a z )e j ( k ÷r t ) ÷ Eo
k÷E ý0
Similarly,
ñ ÷ H ý jk ÷ H ý 0
k÷H ý0
It has been shown in the previous problem that
ñ E ý
öB
öt
k E ý ý H
Similarly,
ñ H ý
öD
öt
kxH ý õ E
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From k E ý ý H ,
ak a E ý a H
From k H ý õ E ,
a k a H ý a E
Prob. 10.73
ý o ý ý 1 ý ý 2, ø1 ý
If
÷ \\ ý
1
cos ñ t
õ r2
1
cos ñ t
õ r2
õ r 1 sin ñ i ý
and
øo
øo
,ø2 ý
õ r1
õ r2
1
cos ñ i
õ r1
1
cos ñ i
õ r1
õ r 2 sin ñ t
õ r2
õ r1
ý
sin ñ i
sin ñ t
sin ñ i
cos ñ i
sin ñ t cos ñ t sin ñ i cos ñ i
sin ñ t
÷ \\ ý
ý
sin ñ i
sin ñ t cos ñ t sin ñ i cos ñ i
cos ñ t
cos ñ i
sin ñ t
sin 2ñt sin 2ñi sin(ñt ñi ) cos(ñt ñi ) tan(ñt ñi )
ý
ý
ý
sin 2ñt sin 2ñi cos(ñt ñi ) sin(ñt ñi ) tan(ñt ñi )
cos ñ t
Similarly,
ô \\ ý
2
cos ñ i
õ r2
2 cos ñ i
ý
1
1
sin ñ i
cos ñ t
cos ñ i cos ñ t
cos ñ i
sin ñ t
õ r2
õ r1
2 cos ñ i sin ñ t
sin ñ t cos ñ t (sin ñ i cos2 ñ i ) sin ñ i cos ñ i (sin 2 ñ t cos2 ñ t )
2 cos ñ i sin ñ t
ý
(sin ñ i cos ñ t sin ñ t cos ñ i )(cos ñ i cos ñ t sin ñ i sin ñ t )
ý
ý
2
2 cos ñ i sin ñ t
sin(ñ i ñ t ) cos(ñ i ñ t )
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1
sin ñ i
cos ñ t cos ñ i
cos ñ t
õ r1
sin(ñ t ñ i )
sin ñ t
ý
ý
1
sin ñ i
sin(ñ t ñ i )
cos ñ t cos ñ i
cos ñ t
sin ñ t
õ r1
÷þ ý
1
cos ñ i
õ r2
1
cos ñ i
õ r2
ôþ ý
2
cos ñ i
õ r2
2 cos ñ i
2 cos ñ i sin ñ i
ý
ý
1
1
sin ñ i
sin(ñ t ñ i )
cos ñ i
cos ñ t cos ñ i
cos ñ t
sin ñ t
õ r2
õ r1
Prob. 10.74
n2 ý c ý2õ 2 ý c 6.4õ o ýo ý 6.4 ý 2.5298
(a) n1 ý 1,
sin ñt ý
n1
1
sin ñi ý
sin12o ý 0.082185
n2
2.5298
ñt ý 4.714o
1
ý 47.43ð
6.4
ø cos ñt ø1 cos ñi 47.43ð cos 4.714o 120ð cos12o
Ero
ý÷ý 2
ý
ø 2 cos ñt ø1 cos ñi 47.43ð cos 4.714o 120ð cos12o
Eio
ø1 ý 120ð ,
ø2 ý 120ð
47.27 117.38
ý 0.4258
47.27 117.38
2ø 2 cos ñi
Eto
2 x 47.43cos12o 92.787
ýô ý
ý
ý
ý 0.5635
ø 2 cos ñt ø1 cos ñi 47.27 117.33 164.65
Eio
ý
Prob. 10.75
(a) ki ý 4a y 3az
ki ÷ an ý ki cos ñi
cos ñi ý 4 / 5
ñi ý 36.87o
(b)
2
E
1
( 82 62 ) 2 (4a y 3a z )
*
Re( E s H s ) ý o ak ý
ý 106.1a y 79.58a z mW/m 2
2
2ø
2 120ð
5
o
(c ) ñ r ý ñ i ý 36 .87 . Let
Pave ý
Er ý ( Ery a x Erz a z ) sin(t kr ÷ r )
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z
kt
Er
kr
Et
ñr
ki
ñt
ñi
y
Ei
From the figure, kr ý krz a z kry a y . But
k rz ý k r sin ñ r ý 5(3 / 5) ý 3,
Hence,
sin ñ t ý
k r ý ki ý 5
k ry ý k r cos ñ r ý 5(4 / 5) ý 4,
kr ý 4a y 3a z
c ý 1õ 1
n1
3/5
sin ñ i ý
sin ñ i ý
ý 0.3
n2
c ý 2õ 2
4
ñt ý 17.46, cos ñt ý 0.9539,
ø1 ýøo ý 120ð ,ø 2 ý øo / 2 ý 60ð
ø
o
(0.9539) øo (0.8)
Ero ø2 cos ñt ø1 cos ñi
÷ // ý
ý
ý 2
ý 0.253
Eio ø 2 cos ñt ø1 cos ñi øo (0.9539) ø (0.8)
o
2
Ero ý ÷ / / Eio ý 0.253(10 ) ý 2.53
But
3
4
( Ery a y Erz a z ) ý Ero (sin ñ r a y cos ñ r a z ) ý 2.53( a y a z )
5
5
Er ý (1.518a y 2.024a z ) sin(t 4 y 3 z ) V/m
Similarly, let
Et ý ( Ety a y Etz a z ) sin(t kt ÷ r )
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k t ý ò 2 ý ý 2 õ 2 ý 4ý o õ o
But
ki ý ò 1 ý ý o õ o
kt
ý2
ki
k t ý 2 ki ý 10
k ty ý k t cos ñ t ý 9.539 ,
k tz ý kt sin ñ t ý 3,
k t ý 9.539 a y 3az
Note that kiz ý k rz ý k tz ý 3
ô \\ ý
ø o (0.8 )
Eto
2 ø 2 cos ñ i
ý
ý
ý 0.6265
ø
Eio ø 2 cos ñ t ø 1 cos ñ i
o
(0.9539 ) ø o (0.8 )
2
Eto ý ô \\ Eio ý 6.265
But
( Ety a y Etz a z ) ý Eto (sin ñt a y cos ñt a z ) ý 6.256(0.3a y 0.9539a z )
Hence,
Et ý (1.879a y 5.968a z ) sin(t 9.539 y 3 z ) V/m
Prob. 10.76
(a)
k
1
tan ñ i ý ix ý
kiz
8
sin ñ t ý sin ñ i
(b) ò1 ý
(c )
c
õ r1 1
ý ( 3) ý 1
õ r2 3
õ r1 ý
ü ý 2ð / ò ,
ñ i ý ñ r ý 19.47 o
ñ t ý 90 o
109
3 ý 10 ý k 1 8 ý 3k
3 108
k ý 3.333
ü 1 ý 2 ð / ò 1 ý 2 ð / 10 ý 0.6283 m
ò 2 ý / c ý 10 / 3,
ü2 ý 2ð / ò 2 ý 2ð 3 /10 ý 1.885 m
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(d)
(a x 8a z )
3
ý (23.6954a x 8.3776a z ) cos(109 t kx k 8 z ) V/m
Ei ý ø1 H x ak ý 40ð (0.2) cos(t k ÷ r )a y
(e) ô / / ý
2 cos ñ i sin ñ t
2 cos 19.47 o sin 90 o
ý
ý6
sin(ñ i ñ t ) cos(ñ t ñ i ) sin 19.47 o cos 19.47 o
cot 19.47 o
÷// ý
ý 1
cot 19.47 o
Et ý Eio (cos ñt a x sin ñt a z ) cos(109 t ò 2 x sin ñt ò 2 z cos ñt )
Let
where
Et ý Eio (cos ñi a x sin ñi a z ) cos(109 t ò1 x sin ñi ò1 z cos ñi )
sin ñ t ý 1,
cos ñ t ý 0 ,
ò 2 sin ñ t ý 10 / 3
Eto sin ñ t ý ô \\ Eio ý 6 (24 ð )(3)(1) ý 1357 .2
Hence,
Et ý 1357 cos(109 t 3.333 x)a z V/m
Since ÷ ý 1,
ñr ý ñi
Er ý (213.3a x 75.4a z ) cos(109 t kx k 8 z ) V/m
(f) tan ñ B / / ý
õ2
ý
õ1
õo
ý 1/ 3
9õ o
ñ B / / ý 18.43 o
Prob. 10.77
(a) Ei þ ý 5cos(t 0.5ð x 0.866ð z )a y
Ei ý (4a x 3a z ) cos(t 0.5ð x 0.866ð z )
(b) Comparing Ei with eq. (10.115a),
4a x 3a z ý (cos ñi a x sin ñi a z ) Eio
tan ñi ý
sin ñi 3
ý
cos ñi 4
ñi ý 36.87o
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Prob. 10.78
tan ñ B ý
õõ
õ1
ý o r ý õr
õ2
õo
õ r ý tan 2 ñ B ý tan 2 68 ý 6.126
Prob. 10.79
c
(a) n ý ý ýr õ r ý 2.11 ý 1.45
u
(b) n ý ýr õ r ý 1 81 ý 9
(c ) n ý õ r ý 2.7 ý 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,
ù 2.5 0.5ù
T= ú
ú
û 0.5 0.3 û
Prob. 10.82
Since ZL = Zo , ÷ L = 0.
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÷ 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
ü = c/f =
3 108
8.4 109
= 35.71 mm
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CHAPTER 11
P.E. 11.1
Since Zo is real and ñ ù 0, this is a distortionless line.
R
G
Zo ý
or
(1)
L C
ý
R G
(2)
ñ ý RG
(3)
ò ý L
G L
ý
R Zo
(4)
(1) (3) R ý ñ Z o ý 0.04 80 ý 3.2 / m ,
(3) ø (1) G ý
ñ
ý
0.04
ý 5 104 S / m
80
Zo
1.5 80
ò Zo
ý 38.2 nH / m
Lý
ý
2ð 5 10 8
Cý
0.04
1
LG 12
ý 10 8
ý 5.97 pF/m
ð
80 0.04 80
R
P.E. 11.2
(a) Zo ý
R û j L
ý
G û j C
0.03 û j 2ð 0.1 10 3
0 û j 2 ð 0.02 10 6
ý 70.73 j1688
.
ý 70.75 1.367 o
øR û
(b) ÷ ý
j LùøG û j C ù ý
ø0.03 û
j0.2 ð ùø j0.4 10 4 ð ù
ý 2.121 10 4 û j 8.888 10 3 / m
(c) u ý
2ð 103
ý
ý 7.069 105 m/s
3
ò 8.888 10
P.E. 11.3
(a) Zo ý Zl Zin ý Zo ý 30 û j60
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(b) Vin ý Vo ý
I in ý I o ý
Vg
Zin
Vg ý
ý 7 .5 0 o Vrms
Zin û Zo
2
Vg
Z g û Z in
ý
Vg
2Z o
ý
150o
2 ø 30 û j 60 ù
ý 0.2236 63.43o A
assuming that Zg = 0.
(c ) Since Zo = Zr, ý 0 Vo ý 0 ,Vo û ý Vo
The load voltage is VL ý Vs ø z ý l ù ý Vo û e ÷l
e
÷l
Vo û
7 .5 0 o
1.5 48 o
ý
ý
VL 5 48 o
e ñ l e jòl ý 1.548 o
1
1
e ñ l ý 1.5 ñ ý lnø 1.5ù ý
lnø 1.5ù ý 0.0101
40
l
o
1 48 o
e jòl ý e j 48 ò ý
ðrad ý 0.02094
l 180 o
÷ ý 0.0101 û 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 ñ = 300o . At P,
ý
OP 2.1cm
ý
ý 0.228
OQ 9.2cm
O
ý 0.228 300 o
Also at P, zL =1.15-j0.48,
S
R
ZL =Zo zL =70(1.15-j0.48) = 80.5-j33.6
=0.6ü 0.6 720o =432o =360o +73o
From P, move 432o to R. At R, zin ý 0.68 j025
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Zin ý Zo Zin ý 70 (0.68 j0.25) ý 47 .6 j17 .5
(b) The minimum voltage (the only one) occurs at ñ ý 180 o ; its distance from the
180 60
ü
load is
ü ý ý 0.1667 ü
720
6
Values obtained using formulas are as follows:
s 1
ý
300o ý 0.2308300o
s û1
Z L ý 80.5755 j 34.018
Z in ý 48.655 j17.63
These are pretty close.
P.E. 11.5
Z Zo 60 û j60 60
j
ý
ý
ý 0.4472 63.43 o
(a) ý L
Z L û Zo 60 û j60 û 60 2 û j
sý
1û
1 û 0.4472
ý
ý 2.618
1
1 0.4472
Let x ý tanø òl ù ý tan
2 ðl
ü
ù Z û jZ o tanø òl ù ù
Zin ý Z o ú L
ú
úû Zo û jZ L tanø òl ù úû
ù 60 û j 60 û j 60 x ù
120 j 60 ý 60 ú
ú
û 60 û j ø 60 û j 60 ù x û
Or 2 j ý
1 û jø1 û xù
1 x û jx
1 x û j ø 2 x 2ù ý 0
Or x ý 1 ý tanø òl ù
ð
2ðl
û nð ý
ü
4
i.e
lý
ü
ø 1 û 4nù , n ý 0 ,1,2,3...
8
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(b) z L ý
Z L 60 û j 60
ý 1û j
60
Zo
Locate the load point P on the Smith chart.
ý
OP 4.1cm
ý
ý 0.4457 ,ñ ý 62 o
OQ 9.2cm
ý 0.4457 62 o
Locate the point S on the Smith chart. At S, r = s = 2.6
Zin ý
Zin 120 û j60
ý
ý 2 j,
60
Zo
which is located at R on the chart. The angle between OP
and OR is 64o-(-25o) = 90o which is equivalent to
Hence
ü
ü ü
l ý û n ý ø 1 û 4nù , n ý 0 ,1,2........
8
2 8
90 ü ü
ý
720 8
.
Q
64º
P
ø Zin ù max ý sZo ý 2.618ø60ù ý 157.08
S
O
( Zin ) min ý Zo / s ý 60 / 2.618 ý 22.92
lý
62 o
720 o
R
-26º
ü ý 0.0851ü
P.E. 11.6
ZL 57.6 º
Vmax
S = 1.8
ü
ý 37 .5 25 ý 12.5cm or ü ý 25cm
2
2ü
l ý 37 .5 35.5 ý 2cm ý
25
l ý 0.08 ü 57 .6 o
z L ý 1.184 û j 0.622
Z L ý Z o z L ý 50 ø1.184 j 0.622 ù
ý 59.22 û j 31.11
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See the Smith chart
P.E. 11.7
100 j80
ý 1.33 j1.067
75
132 o 65
lA ý
ü ý 0.093ü
72
132 o û 64 o
lB ý
ý 0.272ü
720 o
91º
zL ý
dA ý
A
lB
yL
zL
dB
A´ B
91
ü ý 0.126 ü
720
d B ý 0.5ü d A
Ys ý ñ
lA
B´
1 + j0.97
Y = 1 + jb
dA
-91º
ý 0.374ü
1 - j0.97
j0.95
ý ñ j12.67 mS
75
P.E. 11.8
(a) G ý
1
Z Zo
lim
ý 1
, L ý Z L 0 L
Z L û Zo
3
Vg
V
ZL
12
lim
Vg ý 0,
I ý zL
0
ý g ý
ý 120mA
ZL û Zg
Z g û Z L Z g 100
Thus the bounce diagrams for current and waves are as shown below.
lim
V ý zL
0
G =
L = –1
1
3
4V
V=4
–4V
2t1
– 43
V = –1.33
V = 0.444
6t1
t1
V=0
3t1
4
3
4t1
V=0
4
9
94
274
(Voltage)
V=0
5t1
V=0
L = 1
G= 13
80mA
I = 80
2t1
– 803
803
I = 133.3
4t1
80
9
6t1
80
27
(Current)
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I = 160
3t1
80
9
I = 115.6
I=0
t1
80
I = 106.27
5t1
I = 124.4
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V (l,t)
0V
t (ýs)
V(0,t)
4V
4V
4
3
4
9
t (ýs)
0
2
4
6
– 43
–4
I(l,t)
160mA
124.45
106.67
80
80
9
0
2
6
4
t (ýs)
803
I(0,t)
133.33
115.5
80
80mA
80
9
0
2
4
6
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(b) G ý
1
Z Zo
lim
ý1
, L ý Z L L
3
Z L û Zo
ZL
V ý Vg ý 12V ,
Z L û Zg g
V ý z L
lim
I ý zL
lim
Vg
ZL û Zg
ý0
The bounce diagrams for current and voltage waves are as shown below.
G =
L = 1
1
3
4
V=4
t1
4
3
4t1
3t1
6t1
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
I = 26.67
5t1
4
9
–80
2t1
V = 10.67
4
9
V = 11.11
I = 80
V=8
4
3
V = 9.33
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 (ýs)
6
I (l,t)
0A
2
4
6
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12V
11.11
9.333
V(0,t)
4V
4V
4
3
0
2
4
9
8
6
4
t (ýs)
I(0,t)
80
80mA
80
80
80
3
2
0
9
3
80
9
t (ýs)
6
4
80 3
–80
P.E. 11.9
1
1
G ý , L ý , t1 ý 2ý s
2
7
øV ù
ø I o ù max ý Z
g max
g
û Zo
ý
10
ý 100 mA
100
The bounce diagrams for maximum current are as shown below.
=
=
1
2
1
7
100mA
– 100
t1
7
2t1
50 7
50
3t1
49
4t1
25
49
25 343
6t1
5t1
12.5 343
7t1
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I (0,t)mA
100
1.521
0
2
6
4
8
10
8
10
t (ýs)
–21.43
I (l,t)mA
85.71
0
2
4
6
–6.122
P.E. 11.10
(a) For w / h ý 0.8 ,
(b) Zo ý
õ eff
4.8 2.8 ù
12 ù
ý
û
1û
ú
2
2 û 0.8 úû
1
2
ý 275
.
60
ö 8 0.8 ö
ln÷
û
÷ ý 36 .18 ln 10.2 ý 84.03
ø 0.8 4 ø
.
275
3 10 8
(c) ü ý 10
ý 18.09 mm
.
10 275
P.E. 11.11
Rs ý
ðfý o
ý
óc
ð 20 10 9 4 ð 10 7
5.8 107
ý 3.69 10 2
R
8.686 3.69 10 2
ñ c ý 8.685 s ý
wZ o
2.5 10 3 50
ý 2.564 dB / m
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Prob. 11.1
1
1
ý
ð f ýó c
ð 500 106 4ð 107 7 107
ôý
ô ý 2.6902 106
Rý
2
wôó c
ý
2
ý 0.0354 / m
0.3 2.6902 106 7 107
ý o d 4 ð 10 7 1.2 10 2
Lý
ý
ý 50.26 nH / m
0.3
w
Cý
õ o w 10 9
0.3
ý
ý 221 pF / m
d
36 ð 1.2 10 2
Since ó ý 0 for air,
Gý
ów
d
ý0
Prob. 11.2
ôý
1
1
ý
ý 7.744 106
6
7
7
ð f ýcó c
ð 80 10 4ð 10 5.28 10
1
1
ù
ù
û
3
3
3 ú
ú
1 ù1 1ù
0.8
10
2.6
10
û
û ý 10 (1.25 û 0.3836) ý 0.6359 /m
û
ý
Rý
2ðôó c úû a b úû 2ð 7.744 106 5.28 107
2569.09
ý b 4ð 107 2.6
Lý
ln ý
ln
ý 2.357 107 H/m
2ð a
2ð
0.8
Gý
2ðó 2ð 105
ý
ý 5.33 105 S/m
b
2.6
ln
ln
a
0.8
109
2ð 3.5 ú
2ðõ
36ð ý 1.65 1010 F/m
Cý
ý
b
2.6
ln
ln
a
0.8
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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
Eò ý
aüò üb
,
2ðõò
b
Q
ln(b / a)
2ðõ
V ý E dl ý
a
J ýóE ý
óQ
2ðõò
I ý J dS ý
S
2ð
óQ
óQ
(1) ò dö ý
2ðõò
õ
ý0
ö
óQ
õ
I
2ðó
ý
ý
Q
V
ln(b / a) ln(b / a)
2ðõ
Method 2:
Gý
Consider a section of unit length, Assume that a total current of I flows from inner
conductor to outer conductor. At any radius ò between a and b,
I
Jý
aò ,
aü ò üb
2ðò
J
I
E= ý
a
ó 2ðóò ò
b
V = - E dl ý
a
Gý
I
2ðó
ln(b / a )
2ðó
I
ý
V ln(b / a )
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Prob. 11.4
(a)
Rý
ôý
1
2
wôó c
ð f ó c ýc
ý
1
ð 200 106 5.8 107 4ð 107
ý
103
ý 4.67 106
2ð 200(5.8)
2
20
ý
ý 0.2461 /m
6
7
30 10 4.67 10 5.8 10
3(4.67)5.8
Rý
3
ýd
4ð 107 2 103 4ð 107
ý
ý 83.77 nH/m
w
30 103
15
ó w 103 30 103
Gý
ý
ý 15 mS/m
d
2 103
109
4
30 103
õw
60 109
ð
36
ý
ý
ý 0.5305 nF/m
Cý
2 103
36ð
d
Lý
ý
÷ ý ( R û j L)(G û jC )
(b)
÷ 2 ý (0.2461 û j 2ð 200 106 83.77 109 )(15 103 û j 2ð 200 106 0.5305 109 )
ý (0.2461 û j105.3)(15 103 û j 0.6667)
÷ ý 0.104 û j8.379 /m
Zý
( R û j L)
(0.2461 û j105.3)
ý
ý 12.565 û j 0.1266
(G û jC )
(15 103 û j 0.6667)
Prob. 11.5
Cý
ð õl
ðõl
cosh (d / 2a ) ln(d / a )
1
since (d/2a)2 = 11.11 >> 1.
Cý
ôý
ð
109
16 103
36ð
ý 0.2342 pF
ln(2 / 0.3)
1
1
ý
ý 2.09 105 m << a
7
7
7
ð f ýó c
ð 10 4ð 10 5.8 10
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Rac ý
l
ð aôó c
ý
16 103
ý 1.4 102
3
5
7
ð 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
ö d ö
ý
ý 12.5, ÷ ÷ ý 39.1 1
a
1.2
ø 2a ø
d
d
Hence, cosh 1
ln ý ln12.5 ý 2.526
a
2a
d
ý
ý d 4ð 107 2.526
L ý cosh 1
ý ln ý
ý 10.103 107 ý 1.01 ý H/m
ð
ð
2a ð a
109
ðõ
ðõ
ð
Cý
ý 44.105 pF/m
4
ý
ý
d 2.526
ð
36
1 d
cosh
ln
a
2a
L
1.01 106
Zo ý
ý
ý 151.16
C
44.105 1012
Prob. 11.7
V ( z û z , t ) ý V ( z , t ) Lz
öI ( z , t )
öt
or
öI ( z , t )
V ( z û z , t ) V ( z , t )
ý L
z
öt
öV ( z û z , t )
I( z û z , t ) ý I ( z , t ) C z
öt
or
öV ( z û z , t )
I( z û z , t ) I ( z , t )
ý C
z
öt
As z 0, we obtain
(1)
(2)
öV
öI
ý L
öz
öt
öI
öV
ý C
öz
öt
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Prob.11.8
(a)
÷ ý ( R û j L)(G û jC ) ý j LC (1 û
ý j LC 1
R
j L
)(1 û
G
)
jC
RG
R
G
û
û
2
LC j L jC
As R<< L and G<<C, dropping the 2 term gives
ù
R
G
R
G ù
÷ j LC 1 û
û
j LC ú1 û
û
ú
j L jC
û 2 j L 2 jC û
öR C G L ö
ý ÷÷
û
÷÷ û j LC
ø2 L 2 Cø
(b)
1û
R
R û j L
L
L
j L
Zo ý
ý
ý
G û jC
C 1û G
C
jC
1/ 2
ö
R ö
÷1 û
÷
j L ø
ø
ö
G ö
÷1 û
÷
jC ø
ø
1/ 2
öö
ö
Lö
R
G
Lö
R
G
ö
û ... ÷ ÷1
û ... ÷ ý
ûj
û ... ÷
÷1 û
÷1 j
C ø 2 j L
j 2C
Cø
2 L
2C
ø
øø
ø
L
C
ù
ú1 û
û
R öù
ö G
j÷
÷ú
ø 2C 2 L ø û
Prob. 11.9
(a) R û j L ý 0.2 û j 2ð 12 106 40 106 ý 0.2 û j 24ð (40) ý 0.2 û j 3015.93
G û jC ý 4 103 û j 2ð 12 106 25 106 ý 4 103 û j 50ð (12)
ý 4 103 û j1884.96
÷ ý ( R û j L)(G û jC ) ý (0.2 û j 3015.93)(4 103 û j1884.96)
ý 8.159 102 û j 2.384 103 /m
(b) Since R<< L and G<< L, we may use the result in Problem 11.8(a).
L 40 106
ý
ý 1.6,
LC ý 40(25) 1012 ý 109
C 25 106
öR C G L ö
ö 0.1 4 103
ö
÷ ý ÷÷
û
û
1.6 ÷ û j 2ð 12 106 40 106 25 106
÷÷ û j LC ý ÷
2
ø 1.6
ø
ø2 L 2 Cø
3
2
ý 8.159 10 û j 2.384 10 /m
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Prob. 11.10
I1 (t ò z ) travels along +z direction.
(a)
(b) From eq. (11.18),
Voû
Vo
Zo ý û ý
Io
Io
Hence,
V ( z , t ) ý Z o I1 (t ò z ) Z o I 2 (t û ò z )
Prob. 11.11
R û j L
G û jC
Zo ý
(1)
÷ ý ñ û j ò ý ( R û j L)(G û jC )
ñ ý 0.04 dB/m ý
(2)
0.04
Np/m = 0.00461 Np/m
8.686
Multiplying (1) and (2),
Z o (ñ û j ò ) ý R û j L
50(0.00461 û j 2.5) ý R û j L
R ý 50 0.00461 ý 0.2305 /m
Lý
50 2.5
ý 0.3316 ý H/m
2ð 60 106
Dividing (2) by (1),
ñ û jò
ý G û jC
Zo
Gý
Cý
ñ
Zo
ý
0.00461
ý 92.2 ýS/m
50
2.5
ò
ý
ý 0.1326 nF/m
Z o 2ð 60 106 50
Prob. 11.12
L
ýd d
d ý
ý
.
ý
c
w õw w õ
d
Z o ý øo ý 78
w
Zo ý
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Z o ' ý øo
d
ý 75
w'
78 w'
ý w' ý 1.04 w
75 w
i.e. the width must be increased by 4%.
Prob. 11.13
6.8 û j 2ð 103 3.4 103
R û j L
(a) Z o ý
ý
0.42 106 û j 2ð 103 8.4 109
G û jC
ý 103
6.8 û j 21.36
ý 644.3 j 97
0.42 û j 52.78
÷ ý ( R û j L)(G û jC ) ý 103 (6.8 û j 21.36)(0.42 j 52.78)
ý (5.415 û j 33.96) 103 /mi
2ð 103
ý
ý 1.85 105 mi/s
3
ò 33.96 10
2ð
2ð
(c) ü =
ý
ý 185.02 mi
ò 33.96 103
(b)
uý
Prob. 11.14
Using eq. (11.42a),
Z in ý jZ o cot ò
c
3 108
ý
ý 0.75 m
f 400 106
2ð
2ð 0.1
ý
ý 48o
ò ý
ü
0.75
Z in ý j (250) cot 48o ý j 225.1
Z o ý 250,
üý
Prob. 11.15
Assume that the line is lossless.
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Zo ý
L
C
Lý
From Table 11.1,
L ýö 1
bö
ln ÷
ý ÷
C õ ø 2ð a ø
Zo ý
ln
ý b
ln ,
2ð a
Cý
2ðõ
b
ln
a
2
øo
L
b
b
1
ý
ln
ln
ý
ý
C 2ð a
õ 2ð õ r a
Z
b
75
ý 2ð õ r o ý 2ð 2.25
ý 1.875
a
120ð
øo
b
ý e1.875
a
a ý be 1.875 ý 3e 1.875 mm = 0.46 mm
Prob. 11.16
(a) For a lossless line, R = 0 = G.
÷ ý j LC
uý
ý cý
ò
ò ý LC ý ý o co ý
c
1
LC
(b) For lossless line, R = 0 =G
Lý
d
ý
cosh 1
,C ý
ð
2a
Zo ý
ý
ðõ
cosh 1
d
2a
L
d 120ð
d
ý 1
. cosh 1
cosh 1
ý
ý
C
2a ð õ r
2a
ð ðõ
120
d
cosh 1
2a
õr
Yes, true for other lossless lines.
Prob. 11.17
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Lý
ý
d
0.32
cosh 1
ý 4 10 7 cosh 1
2a
ð
0.12
L ý 0.655 ý H / m
10 9
3.5
ðõ
ð
36
Cý
ý
Cosh 1 2.667
1 d
Cosh
2a
ð
C ý 59.4 pF / m
L
0.655 106
ý
ý 105
Zo ý
C
59.4 1012
or
120
cosh 1 2.667 ý 105
Zo ý
3.5
Prob. 11.18
For a distortionless cable,
R G
ý
RC ý LG
L C
L
ý 60
Zo ý
C
1
4
ý ý
uý ý
6
ò
LC to 80 10
ñ ý 0.24dB ý
(1)
(2)
(3)
0.24
Np ý 0.0276
8.686
ñ ý RG ý 0.00069
(4)
From (2) and (3),
1
60 4
ý
C 80 106
From (2),
Cý
8 105
ý 333.3 nF/m
240
L ý (60) 2 C ý 3600 333.3 109 ý 1.20 mH/m
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From (1) and (4),
LG
C
0.000692
2
G
ý
ý
G 0.000692
602
0.00069
Gý
ý 11.51 ýS/m
60
From (4),
R=
0.000692
ý 0.00069 60 ý 0.0414 /m
G
Prob. 11.19
R
R G
20 63 10 12
(a) ý G ý C ý
L
L C
0.3 10 6
G ý 4.2 10 3 S / m
ñ ý RG ý 20 4.2 10 3 ý 0.2898
ò ý LC ý 2ð 120 10 6 0.3 10 6 63 10 12
ý 3.278
÷ ý 0.2898 û j 3.278 / m
uý
2ð 120 106
ý
ý 2.3 108 m/s
3.278
ò
L
0.3 106
ý
ý 69
C
63 1012
Zo ý
(b) Let Vo be its original magnitude
Vo e ñ z ý 0.2Vo e ñ z ý 5
zý
1
ln 5 ý 5.554 m
ñ
(c) ò l ý 45o ý ð
4
l ý
ð
4
ý
ý 0.3051 m
4ò 4 3.278
Prob. 11.20
(a) ñ ý 0.0025 Np/m, ò ý 2 rad/m,
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10 8
uý ý
ý 5 107 m / s
ò
2
V
60 1
ý
(b) ý oû ý
Vo
120 2
Z L Zo
1 300 Zo
Zo ý 100
ý
Z L û Zo
2 300 û Zo
'
120 0.0025l '
60
cos ø108 û 2l ' ù e 0.0025l cos ø108 t 2l ' ù
I øl ' ù ý
e
Zo
Zo
But ý
ý 1.2e0.0025l cos ø108 û 2l ' ù 0.6e 0.0025l cos ø108 t 2l ' ù A
'
'
Prob. 11.21
ñ ý 103 , ò ý 0.01
÷ ý ñ û j ò ý 0.001 û j 0.01 ý (1 û j10) 103 /m
uý
2ð 104
ý
ý 6.283 106 m/s
ò
0.01
Prob. 11.22
L
0.6 106
ý
ý 85.54
C
82 1012
RC
RC ý LG
Gý
L
RC
R 10 103
ý
ý
ý 1.169 104 Np/m
ñ ý RG ý R
L
Zo
85.54
Zo ý
ò ý LC ý 2ð 80 106 0.6 106 82 1012
ý 3.5258 rad/m
÷ ý 1.169 104 û j 3.5258 /m
Prob. 11.23
R û j L ý 6 .5 û j 2 ð 2 10 6 3.4 10 6 ý 6 .5 û j 4273
.
G û jC ý 8.4 103 û j 2ð 2 106 21.5 1012 ý ø 8.4 û j 0.27 ù 103
Zo ý
R û j L
ý
G û j C
6 .5 û j 4273
.
ø8.4 û j0.27 ù 10 3
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Zo ý 7171
. 39.75 o ý 55.12 û j 45.85
øR û
÷ ý
j LùøG û j C ù
ý
ø43.1981.34 ùø8.4 10
o
3
= 0.45 + j0.4 /m
ñ
ò
,
ò
tý
l
u
tý
0.39 5.6
òl
ý
ý 0.1783ý s
2ð 2 106
, but u ý
Prob. 11.24
(a) For a lossy line,
ù Z û Z o tanh ÷ ù
Z in ý Z o ú L
ú
û Z 0 û Z L tanh ÷ û
For a short-circuit, Z L ý 0.
Z sc ý Z in
tanh ÷ ý
ZL ý 0
ý Z o tanh ÷
Z sc 30 j12
ý
ý 0.168 j 0.276
Z o 80 û j 60
÷ ý ñ û j ò ý tanh 1 (0.168 j 0.276) ý 0.1571 j 0.2762
0.1571
ý 0.0748 Np/m
2.1
0.2762
ý 0.1316 rad/m
òý
2.1
ñý
(b)
ù (40 û j 30) û (80 û j 60)(0.168 j 0.276) ù
Z in ý (80 û j 60) ú
ú
û (80 û j 60) û (40 û j 30)(0.168 j 0.276) û
ý 61.46 û j 24.43
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ù
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Prob. 11.25
ù Z û Z o tanh ÷ ù
Z in ý Z o ú L
ú
û Z 0 û Z L tanh ÷ û
÷ ý ñ û j ò ý 1.4 0.5 û j 2.6 0.5 ý 0.7 û j1.3
tanh ÷ ý 1.4716 û j 0.3984
ù 200 û (75 û j 60)(1.4716 û j 0.3984) ù
Z in ý (75 û j 60) ú
ú
û (75 û j 60) û 200(1.4716 û j 0.3984) û
ý 57.44 û j 48.82
Prob. 11.26
2Z L I L
VL
ZL IL
ý
ý
û
1 øV û Z I ù Z L I L û Z o I L
Vo
o L
2 L
2Z L
ý
Z L û Zo
(a) TL ý
1û L ý 1û
(b) (i) ô L ý
Z L Zo
2Z L
ý
Z L û Z o Z L û Zo
2nZ o
2n
ý
nZ o û Z o n û 1
lim
(ii) ô L ýYL
0 ý
lim
(iii) ô L ý Z L
0 ý
(iv) ô L ý
ý2
ZL
2Z L
ý0
Z L û Zo
2Z o
ý1
2Z o
Prob. 11.27
Z Zo
(a) ý L
Z L û Zo
Hence,
1û
2
Zo
ZL ý
Z L û Z o ý Z L Z o
Z L ( 1) ý ( û 1) Z o
1û
Zo
1
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1 û 0.645o
1.424 û j 0.424
(50) ý
(50) ý (1.252 û j1.6586)(50)
o
1 0.645
0.5751 j 0.424
ý 62.6 û j83.03
ZL ý
(b)
Prob. 11.28
L ý
sý
Z L Z o 200 j 240 120 80 j 240
ý
ý
ý 0.52 j 0.36 ý 0.6325 34.7o
Z L û Z o 200 j 240 û 120 320 j 240
1û | L | 1 û 0.6325
ý
ý 4.4415
1 | L | 1 0.6325
Prob. 11.29
÷ ý ( R û j L)(G û jC ) ý (3.5 û j 2ð 400 106 2 106 )(0 û j 2ð 400 106 120 1012 )
ý (3.5 û j 5026.55)( j 0.3016) ý 0.0136 û j 38.94
ñ ý 0.0136 Np/m,
uý
Zo ý
ò ý 38.94 rad/m
2ð 400 106
ý
ý 6.452 107 m/s
38.94
ò
3.5 û j 5026.55
R û j L
ý
ý 129.1 j 0.045
G û jC
j 0.3016
Prob. 11.30
From eq. (11.33)
Z sc ý Z in Z L ý0 ý Z o tanh ÷ l
Zoc ý Zin
ZL ý
ý
Zo
ý Zo cothø ÷ l ù
tanh ÷ l
For lossless line, ÷ ý jò , tanø ÷ l ù ý tanhø jò l ù ý j tanø ò l ù
Z sc ý jZo tanø ò l ù , Zoc ý jZo cotø ò l ù
Z sc ý jZ o tan ø ò l ù ý (65 û j 38) tanh[(0.7 û j 2.5)0.8] ý (65 û j 38) tanh(0.56 û j 2) ý 75.2530.3o
ý 7.86 60.3o ý 3.89 û j 6.83
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exp(0.56 û
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Prob. 11.31
ý
Z L Z o (1 û j 2) 1
190ð
j
ý
ý
ý
ý 0.7071 45o
ð
Z L û Z o (1 û j 2) û 1 1 û j
245
Prob. 11.32
Z Z o 120 50
(a) ý L
ý
ý 0.4118
170
Z L û Zo
For resistive load, s ý
(b) Z in ý Z o
òl ý
ZL
ý 2.4
Zo
Z L û jZ o tan øòl ù
Z o û jZ L tan øòl ù
2ð ü
. ý 60 o
ü 6
ø ù
ø ù
ù120 û j 50 tan 60 o ù
ý 34.63 40.65 o
Z in ý 50 ú
o ú
j
50
120
tan
60
û
û
û
Prob. 11.33
1
ø a ù ò l ý 100 ý 25 rad ý 1432.4o ý 352.4o
ù
Z in ý 60 ú
û
4
j 40 û j 60 tan 352.4o ù
ý j 29.375
60 40 tan 352.4o úû
V ( z ý 0) ý Vo ý
ý
Z in
j 29.375(100o
Vg ý
Z in û Z g
j 29.375 û 50 j 40
293.7590o
ý 5.75102o
51.116 12o
(b) Z in ý Z L ý j 40.
Voû ý
Vg
(e
jò l
û e j ò l )
(l is from the load)
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VL ý
Vg (1 û )
jò l
jò l
ý 12.620o V
( e û e )
1
ø c ù ò l ' ý 4 ý 1rad ý 57.3o
4
ù j 40 û j 60 tan 57.3o ù
Z in ý 60 ú
ý j 3471.88.
o ú
û 60 40 tan 57.3 û
Vý
Vg (e j û e j )
j 25
ý 22.740o V
( e û e )
(d) 3m from the source is the same as 97m from the load., i.e.
j 25
1
4
j 40 û j 60 tan 309.42o ù
ý j18.2
60 40 tan 309.42o úû
l ' ý 100 3 ý 97 m,
ò l ' ý 97 ý 24.25rad ý 309.42o
ù
Z in ý 60 ú
û
Vg (e j 97 / 4 û e j 97 / 4 )
ý 6.607180o V
Vý
(e j 25 û e j 25 )
Prob. 11.34
V1 ý Vs ( z ý 0) ý Voû û Vo
ûV e
û
o
o
V2 ý Vs ( z ý l ) ý V e
I1 ý I s ( z ý 0) ý
(1)
û ÷ l
o
÷l
o
V
V
Zo Zo
Voû ÷ l Vo ÷ l
I2 ý Is (z ý l) ý
e û
e
Zo
Zo
(2)
(3)
(4)
1
(1) û (3) Voû ý (V1 û Z o I1 )
2
1
(1) - (3) Vo ý (V1 Z o I1 )
2
û
Substituting Vo and Vo in (2) gives
1
1
V2 ý (V1 û Z o I1 )e ÷ l û (V1 Z o I1 )e÷ l
2
2
1
1
ý (e÷ l û e ÷ l )V1 û Z o (e ÷ l e÷ l ) I1
2
2
V2 ý cosh ÷ lV1 Z o sinh ÷ lI1
(5)
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Substituting Voû and Vo in (4),
I2 ý
ý
I2 ý
1
1
(V1 û Z o I1 )e ÷ l û
(V1 Z o I1 )e÷ l
2Z o
2Z o
1
1
(e÷ l e ÷ l )V1 û (e÷ l û e ÷ l ) I1
2Zo
2
1
sinh ÷ lV1 cosh ÷ lI1
Zo
(6)
From (5) and (6)
ù cosh ÷ l
ù V2 ù ú
ú I ú ý ú 1 sinh ÷ l
û 2û ú Z
û o
But
ù cosh ÷ l
ú 1
ú sinh ÷ l
úû Z o
Thus
Z o sinh ÷ l ù
ú ùV1 ù
cosh ÷ l ú úû I1 úû
úû
1
Z o sinh ÷ l ù
ù cosh ÷ l
ú ýú 1
ú sinh ÷ l
cosh ÷ l ú
úû
úû Z o
ù cosh ÷ l
ùV1 ù ú
ú I ú ý ú 1 sinh ÷ l
û 1û úZ
û o
Z o sinh ÷ l ù
ú
cosh ÷ l ú
úû
Z o sinh ÷ l ù
ú ù V2 ù
cosh ÷ l ú úû I 2 úû
úû
Prob. 11.35
(a)
za ý
Z a 80
ý
ý 1.6
Z o 50
(b)
zb ý
Z b 60 û j 40
ý
ý 1.2 û j 0.8
50
Zo
(c)
zc ý
Z c 30 j120
ý
ý 0.6 j 2.4
50
Zo
The three loads are located on the Smith chart, as A, B, and C as shown next.
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Prob. 11.36
zL ý
Z L 210
ý
ý 2.1 ý s
Z O 100
Or ý
sý
Z L Z O 110
,
ý
Z L û Z O 310
1û
1
But s ý
ý 2.1
Vmax
Vmax ý sVmin
Vmin
ü
ü
720 o
long,
ý 120 o
Since the line is
4
4
4
Hence the sending end will be Vmin , while the receiving end at Vmax
Vmin ý Vmax / s ý 80 / 2.1 ý 38.09
Vsending ý 38.0990o
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Prob. 11.37
ZL
ý 1û j2
Zo
We locate z L on the Smith chart.
Z L ý (1 û j 2) Z o
ü
zL ý
720o
ý 180o
4
4
o
We move 180 toward the generator and locate point Q at which
z ý 0.2 j 0.4
Z ý zZ o ý (0.2 j 0.4) Z o
Prob. 11.38
(a) Method 1:
At Y,
ù Z û jZ o tan ò ù
Z in ý Z o ú L
ú
û Z o û jZ L tan ò û
2ð ü
tan ð ý 0
ò ý
ýð,
ü 2
Z
Z in ý Z o L ý Z L ý 150
Zo
Z L ' ý 150
At X,
2ð ü
tan ð / 2 ý
ý ð / 2,
ü 4
ZL ' ù
ù
jZ
û
2
o
2
ú
tan ò ú Z o (75)
Z in ý lim Z o ú
ý
ý 37.5
úý
tan ò
ú jZ 'û Z o ú Z L ' 150
úû L tan ò úû
ò ý
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Method 2: Using the Smith chart,
Z ' 150
zL ý L ý
ý3
Zo
50
Since =ü /2, we must move 360o toward the generator. We arrive
at the same point. Hence,
Zin ý Z L ý 150
zL ' ý
ý
ü
150
ý2
75
180o
4
We move 180o toward the generator.
zin ý 0.5
Z in ý 75(0.6) ý 37.5
(b) From the Smith chart,
s = 3 for section XY
s = 2 for section YZ
Z Z o 150 50
ý
ý 0.5
(c) = L
ZL Z o 150 û 50
Prob. 11.39
zin ý
üý
Z in 100 j120
ý
ý 2 j 2.4
Zo
50
u 0.8 3 108
ý
ý 0.4 m,
f
6 108
If ü 720 ,
o
then
ý 0.1 m =
ü / 4 180
ü
4
o
We move 180o counterclockwise from zin to get z L on the Smith chart as shown below.
z L ý 0.2 û j 0.25
From the Smith chart,
Z L ý Z o z L ý 50(0.2 û j 0.25) ý 10 û j12.5
s ý 5.19
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zL
s
zin
Prob. 11.40
40 j 25
Z
ý 0.8 j 0.5
zL ý L ý
50
Zo
We locate this at point P on the Smith chart shown below
OP 2.4 cm
ý
ý 0.3, ñ ý 96o
| L |ý
OQ
8 cm
L ý 0.3 96o
At S, s = r = 1.81
ý 0.27ü
0.27 720o ý 194.4o
From P, we move 194.4o toward the generator to G. At G,
z in ý 1.0425 û j 0.6133
Z in ý Z o z in ý 50(1.0425 û j 0.6133) ý 52.13 j 30.66
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69.323o
G
O
S=1.81
P
-96.277o
Prob. 11.41
ü 720o implies that 0.2ü 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 ý 0.46 û j 0.26,
ý| | ñ ý
Z L ý 100(0.46 û j 0.26) ý 46 û j 26
OP
3.5cm
144o ý
144o ý 0.4167144o
OQ
8.4cm
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zL
s=2.4
Prob. 11.42
(a)
0.12ü
0.12 720o ý 86.4o
We draw the s=4 circle and locate Vmin . We move from that
location 86.4o toward the load.
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0.37ü
Min
O
Max
s=4
P
-93.6o ZL =22.294 -j41.719
At P, z L ý 0.45 j 0.83
Z ý 50(0.45 j 0.83) ý 22.3 j 41.72
(b) the load is capacitive.
(c) Vmin and Vmax are ü /4 apart. Hence the first maximum
occurs at
0.12ü +0.25ü ý 0.37ü
Prob. 11.43
(a)
zL ý
Z L 75 û j 60
ý
ý 1.5 û j1.2
50
Zo
OP 3.8cm
ý
ý 0.475,
8cm
OQ
ý 0.475420
| |ý
ñ ý 42o
(Exact value = 0.4688 41.76o )
(b) s=2.8
(Exact value = 2.765)
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(c )
0.2ü 0.2 x720o ý 144o
zin ý 0.55 j 0.65
Z in ý Z o zin ý 50(0.55 û j 0.65) ý 27.5 û j 32.5
(d) Since ñ ý 42o , Vmin occurs at
42
ü ý 0.05833ü
720
(e) same as in (d), i.e.. 0.05833ü
Prob. 11.44
If ü 720o , then
ü
6
120o
zin ý 0.35 û j 0.24
Prob. 11.45
Z 100 û j 60
ý
ý 2 û j1.2
zý
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 ý yYo ý
1
(0.36 j 0.22) ý 7.4 j 4.4 mS
50
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Prob. 11.46
u 0.5 3 108
ý
ý 0.9375 m
160 106
f
Z
50 û j 30
ý 1 û j 0.6
zL ý L ý
50
Zo
üý
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 ý 1.8
Z in ý Z o zin ý 50(1.8) ý 90
The angular distance between P and Q is ñ ý 73.3o.
If
ý
ü 720o ,
73.3o
ü
720o
73.3o
73.3o
0.9375 ý 0.0954 m
720o
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Prob. 11.47
60
ý 0.6
100
If ü 720o , then
zL ý
ü / 2 360o . This means that the input impedance is
located on the same spot as z L on the Smith chart.
Zin' ý Z o z L ý 60
ü / 4 180o ,
zin' ý 60 / 50 ý 1.2
zin ý 0.84
Z in ý 50(0.84) ý 42
Prob. 11.48
Z Z o 0.5 j 1
(a) ý L
ý
ý 0.0769 j 0.6154 ý 0.6202 82.87 o
Z L û Z o 0.5 j û 1
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(b)
ý
zL 1
zL û 1
zL ý
1 û 1 û 0.425o
ý
1 1 0.425o
z L ý 1.931 û j 0.7771
Z L ý (1.931 û j 0.7771) Z o
Prob. 11.49
(a )
Z in ý
Z in
1 0 0 j1 2 0
ý
ý 1 .2 5 j1 .5
80
Zo
0 .8 3 1 0 8
u
ý
ý 20m
12 106
f
22ü
ý 1 .1 ü 7 2 0 ð û 7 2 ð
l1 ý 2 2 m ý
20
28ü
ý 1 .4 ü 7 2 0 ð û 7 2 ð û 2 1 6 ð
l2 ý 2 8 m ý
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 ð to w a rd th e lo a d . A t P ,
OP
5 .1c m
L ý
ý
ý 0 .5 5 4 3
OQ
9 .2 c m
ñ ý 72ð 47ð ý 25ð
ü ý
L ý 0 .5 5 4 3 2 5 ð
( E x a c t v a lu e = 0 .5 6 2 4 2 5 .1 5 o )
Z in , m a x ý s Z o ý 3 .7 (8 0 ) ý 2 9 6
( E x a c t v a lu e = 2 8 5 .5 9 )
Zo
80
ý
ý 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 ý
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Z L ý 2.3 û j1.55
(b) Also, at P,
Z L ý 80(2.3 û j1.55) ý 184 û j124
(Exact value = 183.45+j128.25 )
s ý 3.7
At S,
To Locate Z , we move 216ð from Zin toward the geneator.
'
in
At Zin' ,
zin' ý 0.48 û j 0.76
Z in' ý 80(0.48 û j 0.76) ý 38.4 û j 60.8
(Exact=37.56+j61.304 )
(c) Between ZL and Zin , we move 2 revolutions and 72ð. 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
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(a)
ü
2
ý 120cm ü ý 2.4m
3 108
uý fü f ý ý
ý 125MHz
ü
2.4
40ü ü
720ð
ý
ý 120ð
(b) 40cm ý
240 6
6
Z L ý Z o z L ý 150(0.48 û j 0.48)
u
ý 72 û j 72
(Exact value = 73.308+j70.324 )
s 1 1.6
ý
ý 0.444,
(c) ý
s û 1 3.9
ý 0.444120ð
Prob. 11.51
(a)
Z
Z L j 60
j 40
ý
ý j 0.75,
zin ý in ý
ý j 0.5
Zo
Zo
80
80
The two loads fall on the r=0 circle, the outermost resistance circle. The shortest
distance between them is
zL ý
106.26o
126.87o
360o (126.87o 106.26o )ü
ý 0.4714ü
720o
(b) s ý ,
L ý 1106.26o
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Prob. 11.52
Z
40 û j 25
ý 0.8 û j 0.5
zL ý L ý
Zo
50
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.
ñ P ý 97 o. The angular distance between P and Q is
ñ =180 û ñ P ý 277 o.
720o ü
277 o ý
ü
720o
277o ý 0.3847ü
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Prob. 11.53
(a) From Eq. (11.43), Z in 2
Z in1 ý
Z o22
ý
ZL
Z2
Z2
Z o21
ý Z o , i.e. Z in 2 ý o1 ý o 2
Zo
ZL
Z in 2
Z o1 ý Z o 2
Zo
50
ý 30
ý 24.5.
75
ZL
(b) Also,
Z Z
Z o ö Z o2 ö
÷÷ Z o 2 ý o L
ý ÷÷
Z o1
Z o1 ø Z L ø
2
ö
÷÷ øZ o 2 ù3 ý Z o1 Z L2
ø
Z o3 Z L3
3
2
From (1) and (2), ( Z o 2 ) ý Z o1Z L ý 3
Z o1
öZ
Z
Also, o1 ý ÷÷ o 2
Z o2 ø Z L
(1)
(2)
(3)
or Z o1 ý 4 Z o3 Z L ý 4 (50)3 (75) ý 55.33
From (3), Z o 2 ý 3 Z o1Z L2 ý 3 (55.33)(75) 2 ý 67.74.
Prob. 11.54
ü
74
1
ý 1.48,
ý 0.6756
zL
4
50
This acts as the load to the left line. But there are two such loads in parallel due to
180ð, z L ý
the two lines on the right. Thus
ö 1 ö
÷ ÷
Z
'
Z L ý 50 ø L ø ý 25(0.6756) ý 16.892
2
16.892
1
z L' ý
ý 0.3378, z in ý ' ý 2.96
zL
50
Z in ý 50(2.96) ý 148.
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Prob. 11.55
From the previous problem, Z in ý 148
I in ý
Vg
Z g û Z in
ý
120
ý 0.5263 A
80 û 148
1
1
2
I in Rin ý (0.5263) 2 (148) ý 20.5W
2
2
Since the lines are lossless, the average power delivered to either antenna is 10.25W
Pave ý
Prob. 11.56
2ð ü ð
(a) ò l ý
. ý ,
4 4 2
tan ò l ý
ö ZL
ö
û jZ o ÷
÷
ö Z û jZ o tan ò l ö
ø tan ò l
ø
Z in ý Z o ÷ L
÷ ý Zo
ö Zo
ö
ø Z o û jZ L tan ò l ø
÷ tan ò l û jZ L ÷
ø
ø
As tan ò l ,
Z in ý
Z o2 (50) 2
ý
ý 12.5
ZL
200
(b) If Z L ý 0,
Z o2
ý
(open)
0
25
25
ý
ý 25
(c) Z L ý 25 // ý
25 û 1 û 25
2
(50)
Z in ý
ý 200
12.5
Z in ý
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Prob. 11.57
ü
Z
Zo2
l1 ý Z in1 ý
or yin1 ý L2
Zo
4
ZL
yin1 ý
l2 ý
ü
8
yin 2 ý
l3 ý
200 û j150
ý 20 û j15 mS
(100) 2
Z in 2 ý Z L
7ð
ö
÷ Zi û jZo tan
4
ý Zo ø
7ð
ö
÷ Zo û jZi tan
4
ø
ö
÷
ø ý
ö
÷
ø
Z o ø Z i jZ o ù
ø Z o jZ i ù
y i ý y in1 û y in2 ý 20 û j5 mS
zi ý
yin 3 ý
ö
÷
÷ ý jZ o
÷
ø
1
1
ý
ý j10 mS
jZ o j100
7ü
Z in 3
8
But
ð
ö
÷ Z L û jZ o tan 4
lim
0 Z o ÷
ð
÷ Z o û jZ L tan
4
ø
1
1000
ý
ý 47.06 j11.76
yi
20 û j5
100-j47.06-11.76
Z o jZ in
ý
Z o ø Z in jZ o ù 100 ø 47.06-j11.76-j100 ù
ý 6.408 û j 5.189 mS
If the shorted section were open,
yin1 ý 20 û j15 mS
yin 2 ý
j tan ð
1
4 ý j ý j10 mS
ý
Z in 2
Zo
100
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7ð ö
ö
÷
÷ Z i û jZ o tan
7ü
4 ø Z o øZ i j Z o ù
ø
Z in 3 ý Z o
l3 ý
ý
7ð ö
øZ o jZ i ù
8
ö
÷
÷ Z o û jZ i tan
4 ø
ø
y i ý y in1 û y in 2 ý 20 û j15 û j10 ý 20 û j 25 mS
Zi ý
1
1000
ý
ý 19.51 j 24.39
y i 20 û j 25
y in 3 ý
Z o jZ i
75.61 - j19.51
ý
Z o øZ i jZ o ù 100ø19.51 j124.39 ù
ý 2.461 û j 5.691 mS
Prob. 11.58
zL ý
Z L 75 û j100
ý
ý 1.5 û 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 ý j1.7 and at B, y s ý j1.7.
Locate points A' and B' where the stubs admittance is j1.7 and -j1.7
respectively. Calculate
dA ý
26.5
ü ý 0.0368ü
720
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B’, ys=-j1.7
0
yL
dA
A, ys=j1.7
Prob. 11.59
d A ý 0.12ü 0.12 720o ý 86.4o
l A ý 0.3ü
(a)
0.3 720o ý 216o
From the Smith Chart below,
z L ý 0.57 û j 0.69
Z L ý 60 ø 0.57 û j 0.69 ù
ý 34.2 û j 41.4
(b)
(c)
360o 86.4o
ü ý 0.38ü
720o
o
o
ü ø 62.4 82 ù
lB ý
ü ý 0.473ü
2
720o
s ý 2.65
dB ý
(Exact value = 2.7734 )
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0.4721 ü
ZL
O
YL
s = 2.773
A
0.38 ü
-421.97o
-82.033o
Prob.11.60
Z L 120 û j 220
ý
ý 2.4 û j 4.4
Zo
50
We follow Example 11.7. At A, ys=-j3 and at B, ys=+j3. The required stub
admittance is
ñ j3
Ys ý Yo ys ý
ý ñ j 0.06 S
50
The distance between the load and the stub is determined as follows. For A,value =
0.2308ü )
zL ý
For B,
180 û 10 û 17
ü ý 0.2875ü
720
The length of the stub line is determined as follows.
19
dA ý
ü ý 0.0264ü
720
(Exact value = 0.0515 ü)
360 19
ü ý 0.4736ü
dB ý
720
(Exact value = 0.4485ü )
lB ý
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0.2308 ü
A
33.863o
ZL
O
-159.96o
YL
0.05152 ü
-37.095o
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Prob. 11.61
ü
720 o
ý 180 o
4
4
65.274o
ZL
YL
O
A
0.4093 ü
At A,
0.07225 ü
-52.02o
yin ý 1 j1.561
ystub ý j1.5614
Position of the stub = 0.0723ü
Length of the stub = 0.4093ü
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Prob. 11.62
90o
V
4V
s ý max ý
ý4
Vmin 1V
ý
ü
2
s 1 3
ý ý 0.6
s û1 5
ý 25 cm 5 cm ý 20 cm
ñ180o
Vmin
0o
s=4
ü ý 40 cm
P
The load is l=5cm from Vmin, i.e.
lý
5ü ü
ý
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 )
ñ ý 270o or -90o
ý 0.6-90o
Prob. 11.63
sý
ü
2
Vmax 0.95
ý
ý 2.11
Vmin 0.45
ý 22.5 14 ý 8.5 ü ý 17 cm
l
3 10 8
ý 1.764 GHz
0.17
ü
3.2
l ý 3.2 cm ý
ü 135.5 o
17
f ý
c
ý
Vmin
÷
O
÷
S
P ÷ ZL
Q
–44.5o
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At P,
z L ý 1.4 j 0.8
ZL ý 50 ø1.4 j 0.8 ù ý 70 j 40
(Exact value = 70.606-j40.496 )
s 1 1.11
ý
ý
ý 0.357,
s û 1 3.11
ý 0.357 44.5o
ñ ý 44.5o
(Exact value = 0.3571-44.471o )
Prob. 11.64
s ý
Rg Ro
Rg û Ro
L ý
Prob. 11.65
L ý
g ý
t1 ý
ý
0 50
ý 1
0 û 50
RL Ro 80 50
ý
ý 0.231
RL û Ro 80 û 50
Z L Z o 0.5Z o Z o
1
ý
ý
Z L û Zo
1.5Z o
3
Z g Zo
Z g û Zo
ý
Zo 1
ý
3Z o 3
Z
l
ý 2ý s, Vo ý o ø 27 ù ý 9 V,
u
3Z o
Io ý
Vo
ý 180 mA
Zo
ZL
V
0.5
Vg ý
ø 27 ù ý 5.4 V, I ý ý 216 mA
Zg û ZL
ZL
2.5
The voltage and current bounce diagrams are shown below
V ý
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ry ý 1
3
zý0
ri ý 1 ry ý 1 3
3
zý0
z ý1
9V
180mA
t1
–3V
60mA
2t1
ri ý 1
3
z ý1
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
(Current bounce diagram)
(Voltage 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
9
3
12
8
-1V
-3V
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t(ýs)
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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 ý Rs ý 4 Z o ý 200
Vg ý I s Rs ý 10 200 103 ý 2V
g ý
L ý
Z g Zo
Z g û Zo
ý
4Zo Z o 3
ý
4Z o û Z o 5
Z L Z o 2Zo Z o 1
ý
ý
Z L û Z o 2Z o û Z o 3
10
ý
ý 50 ns
u 2 108
The bounce diagram is shown below.
t1 ý
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t(ýs)
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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
5 t1
V ( , t ) V ( , t )
ý
100
ZL
To get I(I,t), we just scale down V(l,t) by 100.
I ( , t ) ý
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Prob. 11.67
Z g Z o 32 75
g ý
ý
ý 0.4019
Z g û Z o 32 û 75
L ý
Z L Z o 2 106 75
1
ý
Z L û Z o 2 106 û 75
50 102
ý
ý 2.5ns
2 108
u
The bounce diagram is shown below.
t1 ý
At t =20 ns = 4t1 ,
V ý 8 û 8 3.2152 3.2152 ý 9.57 V
g =-0.4019
L = 1
8V
t1
8V
2t1
-3.2152
3t1
-3.2152
4t1
1,293
5 t1
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Prob. 11.68
l 40 102
t1 ý ý
ý 01.6ns,
u 2.5 108
Z Z o 100 50 1
L ý L
ý
ý ,
150
3
Z L û Zo
Vo ý
Z oVg
Zo û Z g
ý
50 ø12 ù
90
ý
g ý
Z g Zo
Z g û Zo
ý
40 50
1
ý ,
90
9
20
ý 6.668V
3
The bounce diagram is sketched below. From it, we obtain V(0,t) and V(L,t).
L ý
g ý
1
9
0
20/3
1
3
L
t1
20/9
2t1
-20/81
3t1
-20/243
4t1
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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 ý 8V ý
Zo
50
Vg ý
Vg
Zo û Z g
50 û 60
Vg ý
8 110
ý 17.6 V
50
u
8
6
ý ut1 ý 3 10 2 10 ý 600 m
2t1 ý 4 ý s
t1 ý 2 ý s ý
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Prob. 11.70
t1 ý
20
ý 107 ý 0.1ý s,
8
2 x10
g ý
Z g Zo
ý
Vo ý
Zo
50
Vg ý (12) ý 10V
Zo û Z g
60
10 50
ý 2 / 3
10 û 50
Z g û Zo
Z Z o 0 50
L ý L
ý
ý 1
Z L û Z o 0 û 50
The voltage bounce diagram is shown below
g ý 2 / 3
L ý 1
10V
t1
-10V
2t1
6.667V
-6.667 V
3t1
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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 ý
Zo
Vg
Zo û Z g
(1)
The reflection coefficient is
Zg Z o
(2)
=
Z g û 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 ý Vo û Vo û 2Vo û ...
ö 1 ö
ý Vo (1 û û 2 û 3 û ...) ý Vo ÷
÷
ø 1 ø
Substituting (1) and (2) into (3) gives
ö
÷
Zo
1
VT ý
Vg ÷
Z
÷
Zo û Z g
g Zo
÷÷ 1
ø Z g û Zo
(3)
ö
÷
Vg
1
÷ýZ V
ý
o g
÷
Z g û Zo Z g û Zo
2
÷÷
ø
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t(s)
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Prob.11.72
w ý 1.5cm, h ý 1cm,
(a)
w
ý 1.
h
õr - 1
0.6
ö 6 û1 ö
ý 1.6 û
÷ û
2 1 û 12h/w
1 û 12/1.5
ø 2 ø
õ eff ý ÷
377
1.8 (1.5 û 1.393 û 0.667 In (2.944))
Z0 ý
ñ c ý 8.686
(b)
Rs ý
1
281
3.613
ý 77.77
Rs
wZo
ý
óc ó
ý
ý 1.8
ýð f
óc
ý
19 2.5 109 4ð 10-3
1.1 107
ý 2.995 10-2
ñc ý
u ý
8.686 2.995 102
1.5 102 77.77
c
õ eff
u
c
ý
f
f õ eff
ý
3 108
2.5 109 1.8
0.8 ø 2.2 ù
2 102
1.2 1.8 8.944 102
ý
96.096
14.3996
ü ý
ñ d ý 27.3
ý 0.223dB/m
ñ d ý 6.6735 dB/m
(c)
ñ ý ñ c û ñ d ý 6.8965 dB/m
ñ ý 20dB ý
20
ñ
ý
20
ý 2.9 m
6.8965
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ý 8.944 102
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Prob. 11.73
(a ) Let x ý w h. If x ü l ,
50 ý
60
ö8
ln ÷
4.6 ø x
ö
û x÷
ø
ö8
ö
5 4.6 - 6ln ÷
û x÷ ý 0
øx
ø
we solve for x (e.g using Maple) and get x ý 2.027 or 3.945
which contradicts our assumptiom that x ü 1. If x þ 1,
120ð
4.6 x û 1.393 û 0.667 ln( x û 1.444)
50 ý
We solve this iteratively and obtain:
x ý 1.8628, w ý xh ý 14.9024 mm
For this w and h,
õ eff ý 3.4598
òý
(b)
õ eff
c
ò ý 450 ý
ý
ð
4
ý
õ eff
ðc
4 õ eff 2ð f
c
ý
8
3 108
3.4598 8 109
ý 0.00252 m
Prob. 11.74
For w = 0.4 mm,
w 0.4 mm
ý
ý 0.2 narrow strip
h
2m
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w
ý 0.2, õ eff ý 5.851, Z o ý 91.53
h
w
For
ý 0.4, õ eff ý 6.072, Z o ý 73.24
h
For
Hence,
73.24 ü Z o ü 91.53
Prob. 11.75
0.1 ö 5 1.2 ö
ln ÷
w ' ý 0.5 û
÷ ý 0.5 û 0.1279 ý 0.6279
3.2 ø 0.1 ø
Zo ý
ü
377
4 1.2 ù 8 1.2
ùü
û 3.354 ú ý
ln ý1 û
ú
2ð 2 þ ð 0.6177 û 4 0.6279
ûþ
ý 42.43ln 1 û 2.49(3.822 û 3.354 ý 42.43ln(20.255)
ý 127.64
Prob. 11.76
Suppose we guess that w/h ü 2
A ý
w
h
75 3.3
60 2
ý
û
1.3 ö
0.11 ö
÷ 0.23 û
÷ ý 1.117
3.3 ø
2.3 ø
8e A
24.44
ý
ý 3.331 w ý 3.331h ý 4mm
2A
e - 2
7.337
If we guess that w/h þ 2,
B=
60ð 2
Zo õ r
ý
60ð 2
ý 5.206
75 2.3
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w
2ù
1.3 ö
0.61 ö ù
4.266 ln 9.412 û
ý
÷ ln 4.206 û 0.39
÷
ú
h
4.6 ø
2.3 ø úû
ðû
ý 1.665 <2
Thus
w
h
ý 3.331 > 2
õ eff ý
u ý
3.3
û
2
3 108
1.953
1.3
12
2 1 û
3.331
ý 1.953
ý 2.1467 108 m/s
Prob. 11.77
Z L Z o 100 150
ý
ý 0.2
Z L û Zo
250
RL ý 20 log | |ý 13.98 dB
ý
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')
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2.05
2
1.95
ee
1.9
1.85
1.8
1.75
1.7
1.65
0
1
2
3
4
5
6
7
w/h
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8
9
10
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CHAPTER 12
P. E. 12.1 (a) For TE10, fc = 3 GHz,
1 ( f c / f )2 ý
òý
4ð f
c
0.96 ý
1 (3 / 15) 2 ý 0.96 , ò o ý / uo ý 4ð f / c
4ð 15 109
0.96 ý 615.6 rad/m
3 108
2ð 15 109
ý
ý 1.531 108 m/s
ò
615.6
ý
60 ð
'ý
ý 60ð , TE ý
ý 192.4
õ
0.96
uý
(b)For TM11, fc = 3 7.25 GHz,
òý
1 ( f c / f ) 2 = 0.8426
4ð f
4ð 15 109 (0.8426)
(0.8426) ý
ý 529.4 rad/m
c
3 108
2ð 15 109
uý ý
ý 1.78 108 m/s
ò
529.4
TM ý 60 ð (0.8426 ) ý 158.8
P. E. 12.2 (a) Since E z 0 , this is a TM mode
E zs ý E o sin(mð x / a ) sin(nð y / b)e jòz
mð
ý 40ð
a
i.e. TM21 mode.
Eo = 20,
m=2,
nð
ý 50ð
b
n=1
u'
3 108
( m / a ) 2 ( n / b) 2 ý
402 502 ý 1.5 41 GHz
2
2
2ð f
2ð 109
f 2 fc2 ý
225 92.25 = 241.3 rad/m.
ò ý ýõ 1 ( f c / f ) 2 ý
c
3 108
(c)
jò
Exs ý 2 (40ð )20 cos 40ð x sin 50ð ye j ò z
h
(b) f c ý
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E ys ý
Ey
Ex
jò
jò z
2 (50ð ) 20 sin 40ð x cos 50ð ye
h
ý 1.25 tan 40 ð x cot 50 ð y
P. E. 12.3 If TE13 mode is assumed, fc and ò remain the same.
fc = 28.57 GHz, ò = 1718.81 rad/m, ÷ ý jò
377 / 2
TE13 ý
= 229.69
1 (28.57 / 50) 2
For m=1, n=3, the field components are:
Ez= 0
H z ý H o cos(ð x / a ) cos(3ð y / b) cos( t ò z)
ý ö 3ð ö
E x ý 2 ÷ ÷ H o cos(ð x / a ) sin(3ð y / b) sin( t ò z)
h ø bø
ý ö ð ö
÷ ÷ H o sin(ð x / a) cos(3ð y / b) sin(t ò z )
h2 ø a ø
ò öðö
H x ý 2 ÷ ÷ H o sin(ð x / a ) cos(3ð y / b) sin( t ò z)
h ø aø
Ey ý
ö 3ð ö
÷ ÷ H o cos(ð x / a ) sin(3ð y / b) sin( t ò z)
ø aø
ò
Given that H ox ý 2 ý 2 (ð / a ) H o ,
h
ò
H oy ý 2 (3ð / b) H o ý 6a / b ý 6(15
. ) / 8 ý 1125
.
h
Hy ý
ò
h2
2 14.51ð 2 104 1.5 102
ý 7.96
1718.81ð
òð
ý ö ð ö
2 ý
ý 2 ÷ ÷ Ho ý
ý 2 TE ý 459.4
ø
ø
ò
h a
3a
ý E oy
ý 459.4(4.5 / 0.8) ý 2584.1
b
H oz ý H o ý
E oy
E ox
2h 2 a
ý
E x ý 2584.1cos(ð x / a ) sin(3ð y / b) sin( t ò z) V/m,
E y ý 459.4sin(ð x / a) cos(3ð y / b) sin(t ò z ) V/m,
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Ez= 0,
H y ý 1125
. cos(ð x / a ) sin(3ð y / b) sin( t ò z ) A/m,
H z ý 7.96 cos(ð x / a ) cos(3ð y / b) cos( t ò z) A/m
P. E. 12.4
f c11 ý
u ' 1 1 3 108 102
2 ý
1/ 8.6362 1/ 4.3182 ý 3.883 GHz
2
2 a b
2
3 108
up ý
ug ý
1 (3.883 / 4)
2
ý 12.5 108 m/s,
9 1016
ý 7.2 107 m/s
12.5 108
P. E. 12.5 The dominant mode becomes TE01 mode
c
ý 3.75 GHz, TE ý 406.7
2b
f c 01 ý
From Example 12.2,
E x ý E o sin(3ð y / b) sin( t ò z) ,
Pave =
ò
a
xý0
where E o ý
ýb
Ho .
ð
| E xs |2
E o 2 ab
òy ý 0 2 dxdy ý 4
b
Hence Eo = 63.77 V/m as in Example 12.5.
Ho ý
ð Eo
ð 63.77
ý
ý 63.34 mA/m
10
ýb 2ð 10 4ð 107 4 102
P. E. 12.6 (a) For m=1, n=0, fc = u’/(2a)
ó
1015
1015
ý
ý
üü 1
õ 2ð 9 109 2.6 109 /(36ð ) 1.3
Hence,
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1
ý c / 2.6,
ýõ
u'
ñd ý
fc ý
ó '
2 1 ( fc / f )
1 0 15 3 7 7 /
ý
2
3 108
ý 2.2149 GHz
2 2.4 102 2.6
2 .6
2 1 ( 2 .2 1 4 9 / 9 )
2
ý 1 .2 0 5 1 0 1 3 Np/m
For n = 0, m=1,
ñc ý
2 Rs
1 b
[
( f c / f )2 ]
2 2
a
b ' 1 ( f c / f )
=
2 2.6 ð 9 109 1.1 107 4ð 107
2
377 1.5 10 1.110
7
1 (2.2149 / 9)
2
[0.5 (2.4 /1.5)(2.2148 / 9) 2 ] ý 2 102 Np/m
(b)Since ñ c þþ ñ d , ñ ý ñ c ñ d ñ c ý 2 x102
loss = ñ l ý 2 102 0.4 ý 0.8 102 Np = 0.06945 dB
P. E. 12.7 For TE11 , m = 1 = n,
H zs ý H o cos(ð x / a ) cos(ð y / b)e ÷ z
j
(ð / b) H o cos(ð x / a) sin(ð y / b)e ÷ z
2
h
jý
E ys ý 2 (ð / a) H o sin(ð x / a ) cos(ð y / b)e÷ z
h
Exs ý
jò
(ð / a ) H o sin (ð x / a ) co s(ð y / b ) e ÷ z
h2
jò
ý 2 (ð / b ) H o co s(ð x / a ) sin (ð y / b ) e ÷ z
h
ý0
H xs ý
H
ys
E zs
For the electric field lines,
dy E y
ý
ý (a / b) tan(ð x / a ) cot(ð y / b)
dx E x
For the magnetic field lines
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dy H y
ý
ý (a / b) cot(ð x / a ) tan(ð y / b)
dx H x
Ey Hy
Notice that ( )( ) ý 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' ý
ý
ýõ
c
õr
1.5 1010
1/ 25 0 1/100 ý 1.936 GHz
3
1
QTE101 ý
, where
61ô
1
1
ôý
ý
ý 1.5 106
9
7
7
ð f101ýó c
ð 1.936 10 4ð 10 5.8 10
fTE101 ý
QTE101 ý
106
ý 10,929
611.5
P. E. 12.9
(a) By Snell’s law, n1 sin ñ 1 = n2 sin ñ 2 . Thus
ñ 2 = 90o
sin ñ 2 = 1
sin ñ 1 = n2/n1,
(b) NA =
ñ 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
ñ 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 %
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Prob. 12.1
(a) For TE10 mode, f c ý
u'
3 108
ý
ý 2.5 GHz
2a 2 6 102
(b) f ý 3 f c ý 7.5 GHz
f cmn
u1
ý
2
m2 n2
a 2 b2
2
2
3 108 102 ö m ö ö n ö
ý
÷
÷ ÷ ÷
2
ø 6 ø ø 4ø
2
2
öm ö ön ö
ý 15 ÷
÷ ÷ ÷ GHz
ø 6 ø ø 4ø
2
f c 20 ý 15 ý 5 GHz
6
f c 01 ý 3.75 GHz,
f c 02 ý 7.5 GHz
f c10 ý 2.5 GHz,
f c 20 ý 5.0 GHz
f c 21 ý 6.25 GHz,
f c 30 ý 7.5 GHz
2
2
2
2
ö1ö ö2ö
f12 ý 15 ÷ ÷ ÷ ÷ ý 7.91 GHz
ø6ø ø4ø
ö1ö ö1ö
f11 ý 15 ÷ ÷ ÷ ÷ ý 4.507 GHz
ø6ø ø4ø
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
1
u'
u' 1
u'
2 ý
, f c11 ý
2
f c10 ý
2
2a
2 a
2a
a
Since the guide can only propagate TE10 mode,
f c10 ü f ü f c11
u'
u'
ü 2a ü
2
f
f
ü
2
üaü
u'
u'
ü f ü
2
2a
2a
u ' ü 2af ü u ' 2
ü ü 2a ü ü 2
ü
2
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Prob. 12.3
2
2
u' ö mö ö nö
3 108
fc ý
ý
÷ ÷ ÷ ÷
2 ø a ø øbø
2 2.25 102
2
2
15 ùö m ö ö n ö ù
ý
ú÷
÷ ÷
÷ ú
2.25 ûúø 2.28 ø ø 1.01 ø ûú
ùö m ö 2 ö n ö 2 ù
ú÷
÷ ÷
÷ ú
úûø 2.28 ø ø 1.01 ø úû
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,
2
u' ö 1 ö
3 108
fc ý
ý 6.25 GHz
÷ ÷ ý
2 øaø
2 2.4 102
For TE 01 mode,
2
u' ö1ö
3 108
ý
ý 12.5 GHz
÷ ÷
2 øbø
2 1.2 102
For TE 20 mo de,
fc ý
2
u' ö 2ö
÷ ÷ ý 2 6.25 ý 12.5 GHz
2 øaø
For TE 02 mode,
fc ý
2
fc ý
u' ö 2ö
÷ ÷ ý 2 12.5 ý 25 GHz
2 øbø
(b) Since f = 12 GHz, only TE10 mode will propagate.
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Prob. 12.5 a/b = 3
f c10 ý
u'
2a
a = 3b
aý
u'
3 108
ý
m = 0.833cm
2 f c10 2 18 109
A design could be a = 9mm, b = 3mm.
Prob. 12.6
fc ý
For the dominant mode,
c 3 108
ý
ý 18.75 MHz
2a
28
(a) It will not pass the AM signal, (b) it will pass the FM signal.
Prob. 12.7 (a) For TE10 mode, f c ý
u'
2a
u'
3 108
Or a ý
ý
ý 3 cm
2 f c 2 5 109
u'
For TE01 mode, f c ý
2b
u'
3 108
ý
ý 1.25 cm
bý
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 ' ý
For TE11
1
c
ý 2 108 m/s
ýõ
2.25
modes,
ý
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fc ý
3 108
2 102 2.25
1
1
ý 8.67 GHz
2
3 1.252
Prob. 12.8
2
2
ö f ö
ö 25 ö
Let F12 ý 1 ÷ c12 ÷ ý 1 ÷ ÷ ý 0.7806
ø 40 ø
ø f ø
3 108
c
ý
ý 0.0075 m ý 7.5 103 m
f 40 109
ü' ý
ü'
ü12 ý
F12
ý
7.5 103 m
ý 9.608 103 m
0.7806
u ' 3 108
ý
ý 3.843 108 m/s
F12 0.7806
u12 ý
ò12 ý
2ð
ý
ü12
'
TE12 ý
F12
2ð
ý 653.95 rad/m
9.608 103
ý
120ð
ý 482.95
0.7806
Prob. 12.9
2
2
u' ö mö ö nö
fc ý
÷ ÷ ÷ ÷
2 ø a ø øbø
For TE10 mode, m=1, n=0,
fc ý
3 108
u'
c
ý
ý
ý 3GHz
2a 2a 2 5 102
2
2
ö fc ö
109
785.4
ö 3 ö
7
9
ò ý ýõ 1 ÷ ÷ ý 2ð 12.5 10 4ð 10
1 ÷
(0.9708)
÷ ý
36ð
3
ø 12.5 ø
ø f ø
ò ý 254.15 rad/m
uý
2ð 12.5 109
ý
ý 3.09 108 m/s
ò
254.15
'
120ð
TE ý
ö f ö
1 ÷ c ÷
ø f ø
2
ý
0.9708
ý 388.3
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Prob. 12.10
u'
c
3 108
fc ý
ý
ý
ý 3.75 GHz
2a 2a 2 4 102
2
ö fc ö
2ð 24 109
480ð
ö 3.75 ö
ò ý ýõ 1 ÷ ÷ ý
1 ÷
(0.9877)
÷ ý
8
3 10
3
ø 24 ø
ø f ø
ò ý 496.48 rad/m
üý
2ð
ò
ý
2
2ð
ý 0.0127 m
496.48
Prob. 12.11
uý
u'
3 108
ý
ý
ý 6.975 108 m/s
2
2
ò
1 ( fc / f )
1 (6.5 / 7.2)
ug ý
tý
9 1016
ý 1.2903 108 m/s
u
2l
300
ý
ý 2.325 ý s
u g 1.2903 108
Prob. 12.12
u'
c
3 108
ý
ý
ý 6.25 GHz
fc ý
2a 2a 2 2.4 102
f ý 1.25 f c ý 7.813 GHz
'
ý
ö f ö
1 ÷ c ÷
ø f ø
2
ý
377
ö 1 ö
1 ÷
÷
ø 1.25 ø
2
ý
377
ý 628.32
0.6
Prob. 12.13
u'
c
(a) f c10 ý
ý
2a 2a õ r
3 108
2 1.067 102 6.8
30
GHz
ý
2 1.067 6.8
ý 5.391 GHz
ý
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lOMoARcPSD|23609374
412
(b)
ö f ö
F ý 1 ÷ c ÷
ø f ø
2
ö 5.391 ö
ý 1 ÷
÷
ø 6 ø
ý 0.439
2
u'
c
3 108
ý
ý
ý 2.62 108 m/s
F F õ r 0.439 6.8
uò ý
(c)
ü'
üý
F
ý
u'
ý
f
F
ý
c
ý
fF õ r
3 108
0.439 6 109 6.8
1
10
ý 0.04368 m ý 4.368 cm
2 0.439 6.8
Prob.12.14
In evanescent mode,
ö mð ö ö nð ö
k ý ýõ ü ÷
÷ ÷
÷
ø a ø ø b ø
2
2
ö mð ö ö nð ö
2
2
2
2
÷ ýñ ý ÷
÷ ÷
÷ k ý 4ð ýõ f c ýõ
ø a ø ø b ø
2
ò ý 0,
2
2
2
ñ ý ýõ 4ð f 4ð f ý 2ð ýõ f c
2
2
c
2
2
ö f ö
1 ÷ ÷
ø fc ø
2
Prob. 12.15
E z 0 . This must be TM23 mode (m=2, n=3). Since a= 2b,
fc ý
10 12
c
3 108
4
36
15.81
f
ý
ý
ý 159.2 GHz
GHz,
ý
m 2 4n 2 ý
2ð
2ð
4b
4 3 102
TM ý 377 1 (15.81/159.2) 2 ý 375.1
P
ave ý
| Exs |2 | E ys |2
2TM
az
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ý
ò 2 Eo 2
ùû (2ð / a) 2 cos 2 (2ð x / a) sin 2 (3ð y / b) (3ð / b) 2 sin 2 (2ð x / a) cos 2 (3ð y / b) ùû a z
4
2h TM
Pave ý ò Pave .dS ý
a
b
ò òP
ave
dxdya z
x ý0 y ý0
ò 2 Eo 2 ab ù 4ð 2 9ð 2 ù ò 2 Eo 2 ab
ý 4
2 úý
2h TM 4 úû a 2
b û 8h 2TM
But
1012
1 ( fc / f ) ý
1 (15.81/159.2) 2 ý 3.317 103
òý
8
c
3 10
2
2
2
4ð
9ð
10ð
h 2 ý 2 2 ý 2 ý 1.097 105
a
b
b
2
(3.317) 2 106 52 18 104
ý 1.5 mW
Pave ý
8 (1.098 105 ) 375.1
Prob. 12.16 (a) Since m=2 and n=1, we have TE21 mode
(b) ò ý ò ' 1 ( f c / f ) 2 ý ý o õ o 1 ( c / ) 2
òc ý
fc ý
2 2c
c
ý
2ð
(c ) TE ý
f2
c2 ý
2 ò 2c2
144 9 1016
ò 2c2
18
36
10
ý
ý 5.973 GHz
4ð 2
4ð 2
1 ( fc / f )
2
ý
377
1 (5.973 / 6 ) 2
ý 3978
(d)For TE mode,
Ey ý
ý
(mð / a ) Ho sin(mð x / a ) cos(nð y / b) sin( t ò z)
h2
Hx ý
ò
(mð / a ) Ho sin(mð x / a ) cos(nð y / b) sin( t ò z)
h2
ò ý 12, m = 2, n =1
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ý
ò
(mð / a ) Ho
2 ( mð / a ) Ho , Hox ý
h
h2
Eoy ý 2ð 6 109 4ð 107
ý
ý
ý
ý 4ð 2 100
H ox
12
ò
Eoy ý
TE
H ox ý
Eoy
TE
5
ý 1.267 mA/m
4ð 100
ý
2
H x ý 1.267 sin(mð x / a ) cos(nð y / b) sin( t ò z) mA/m
Prob. 12.17 (a) Since m=2, n=3, the mode is TE23.
(b)
ò ý ò ' 1 ( f c / f )2 ý
2ð f
c
1 ( f c / f )2
But
fc ý
u'
3 108
( m / a ) 2 ( n / b) 2 ý
(2 / 2.86) 2 (3 /1.016) 2 ý 46.19 GHz, f = 50 GHz
2
2 102
òý
2ð 50 109
1 (46.19 / 50) 2 ý 400.68 rad/m
8
3 10
÷ ý jò ý j400.7 /m
(c ) ý
'
1 ( fc / f )
2
ý
377
1 (46 .19 / 50 ) 2
ý 985.3
Prob. 12.18 In free space,
1 ý
1 ý
2 ý
o
1 ( f c / f )2
377
1 (3 / 8) 2
,
fc ý
c
3 108
ý
ý 3 GHz
2a 2 5 102
ý 406.7
'1
1 ( fc / f )
2
,'ý
120 ð
u'
ý 80 ð , f c ý
, u' ý
2a
2.25
c
õr
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80ð
3 108
ý 2 GHz, 2 ý
ý 259.57
2
2 5 10 2.25
1 (2 / 8) 2
ý 2 1 ý 0.2208
2 1
fc ý
sý
1 | |
ý 1.5667
1 | |
Prob. 12.19 Substituting E z ý Rö Z into the wave equation,
RZ
öZ d
(ò R' ) 2 ö '' Rö Z '' k 2 Rö Z ý 0
ò dò
ò
Dividing by Rö Z ,
Z ''
1 d
ö ''
2
(ò R')
ý kz2
2 k ý
Z
Rò dò
öò
i.e.
Z '' k z Z ý 0
2
1 d
ö ''
(ò R')
(k 2 kz2 ) ý 0
Rò dò
öò2
ò d
ö ''
(ò R') ( k 2 k z 2 )ò 2 ý
ý kö 2
R dò
ö
or
ö '' k ö 2ö ý 0
ò
d
(ò R') ( kò 2ò 2 k ö 2 ) R ý 0 , where kò 2 ý k 2 k z 2 . Hence
dò
ò 2 R'' ò R' ( kò 2ò 2 k ö 2 ) R ý 0
Prob. 12.20
The MATLAB code and the plot of the phase and group velocities are presented below.
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%
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
70
frequency f
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80
90
100
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Prob. 12.21
(a)
For TE10 mode,
fc ý
3 108
u'
ý
ý 2.083 GHz
2a 2 7.2 102
2
ö f ö
ö 2.083 ö
Let F ý 1 ÷ c ÷ ý 1 ÷
÷ ý 0.942
ø 6.2 ø
ø f ø
2
F
2ð 6.2 109 0.942
ý
ý 122.32 rad/m
ò ý ýõ F ý
3 108
c
c 3 108
ý 3.185 108 m/s
up ý ý ý
ò F 0.942
u g ý u ' F ý 3 108 (0.942) ý 2.826 108 m/s
TE ý
'
F
ý
377
ý 400.21
0.942
(b)
3 108
ý
ý
ý 2 108
2.25
ýõ
õr
u'ý
fc ý
c
1
2 108
u'
ý
ý 1.389 GHz
2a 2 7.2 102
2
2
ö f ö
ö 1.389 ö
Let F ý 1 ÷ c ÷ ý 1 ÷
÷ ý 0.9746
ø 6.2 ø
ø f ø
ò ý ýõ F ý
F õr
ý
2ð 6.2 109 0.9746 1.5
ý 189.83 rad/m
3 108
c
2ð 6.2 109
ý 2.052 108 m/s
up ý ý
189.83
ò
u g ý u ' F ý 2 108 (0.9746) ý 1.949 108 m/s
TE ý
'
F
ý
377
ý 257.88
1.5 0.9746
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Prob. 12.22
1
f c10 ý
ýõ
u'
c
3 108
ý
ý
ý
ý 1.315 GHz
2a
2a
2a õ r 2 7.214 102 2.5
2
2
ö f ö
ö 1.315 ö
Let F ý 1 ÷ c ÷ ý 1 ÷
÷ ý 0.9444
ø 4 ø
ø f ø
ò ý ýõ F ý
up ý
õr F
c
c
3 108
ý
ý
ý 2.009 108 m/s
ò F õ r 0.9444 2.5
ug ý u ' F ý
cF
õr
ý
3 108 0.9444
ý 1.792 108 m/s
2.5
Prob. 12.23
ü
up ý ý
ò 2ð
üo2 cüo2 ò 3
,
ý òup ý òc 2 ý
4ð 2
ü
(ü ý 2ð / ò )
ö cü 2 ö
d
öü ö
ý 3 ÷ o2 ÷ ò 2 ý 3c ÷ o ÷ ý 3u p
ug ý
dò
øü ø
ø 4ð ø
2
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Prob. 12.24
ö f ö
1
ug ý c ý u ' 1 ÷ c ÷
4
ø f1 ø
ö f ö
1
ug ý c ý u ' 1 ÷ c ÷
3
ø f2 ø
Dividing (1) by (2),
1/ 4
ý
1/ 3
ö f ö
1 ÷ c ÷
ø f1 ø
2
ö f ö
1 ÷ c ÷
ø f2 ø
2
2
(1)
2
(2)
ö
1 ÷
2
ö3ö
ø
÷ ÷ ý 0.5625 ý
4
ø ø
ö
1 ÷
ø
fc ö
÷
f1 ø
2
fc ö
÷
f2 ø
2
2
ù ö f ö2 ù
ö fc ö
1 ÷ ÷ ý 0.5625 ú1 ÷ c ÷ ú
úû ø f 2 ø úû
ø f1 ø
Assuming f c is in GHz,
0.5625 f c2
f c2
ý 0.5625
f c2 ý 98.44
144
225
From (1),
0.25c
0.25c
ý
ý 0.4444c
u'ý
2
2
ö fc ö
ö 9.9216 ö
1 ÷
1 ÷ ÷
÷
ø 12 ø
f
ø 1ø
1
But
c
õr
u'ý
1
ýõ
ý
f c ý 9.9216 GHz
c
õr
2
ý 0.4444c
ö 1 ö
õr ý ÷
÷ ý 5.0625
ø 0.4444 ø
Prob. 12.25
fc ý
u'
2a
2
ö f ö
ug ý u ' 1 ÷ c ÷
ø f ø
2
ö
ö
2
2
8 ÷
÷
u
ö gö
ö fc ö
1.8 10
÷ ý 0.208
÷ ÷ ý 1 ÷ ÷ ý 1 ÷
8
3
10
f
u
'
÷
÷
ø ø
ø ø
÷
÷
ø 2.2 ø
f c ý 0.208 f ý 2.0523 GHz
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aý
3 108
u'
ý
ý 4.927 cm
2 f c 2 2.2 2.053 109
Prob. 12.26
Let F ý
u'ý
1
ýõ
1 ( f c / f )2 ý
ý
1 (16 / 24) 2 ý 0.7453
3 108
ý 2 108 ,
2.25
up ý
u'
,
F
u g ý u ' F ý 2 108 0.7453 ý 1.491108
m/s
377
ý 337 .2
1.5 x0.7453
TE ý '/ F ý
Prob. 12.27
For the TE10 mode,
ö ð x ö jò z
H zs ý H o cos ÷
÷e
ø a ø
jò a
ö ð x ö jò z
H xs ý
H o sin ÷
÷e
ð
ø a ø
jý a
ö ð x ö jò z
E ys ý
H o sin ÷
÷e
ð
ø a ø
Exs ý 0 ý Ezs ý H ys
E s H s* ý
0
H
*
xs
0
E ys
0
H zs*
jý a
ý E ys H zs* a x E ys H xs* a z
ýò a 2 2 ö ð x ö
öðx ö öðx ö
ý
H cos ÷
H o sin ÷
÷ az
÷ sin ÷
÷ ax
ð
ð2
ø a ø
ø a ø ø a ø
ýò a 2 2 2 ö ð x ö
1
*
ù
ù
H o sin ÷
Pave ý Re û E s H s û ý
÷ az
2
2ð 2
ø a ø
2
2
o
Prob. 12.28
Pave ý
| Exs |2 | E ys |2
2
2 ý 2ð 2 2 2
H o sin ð y / ba z
az ý
2 b 2 h 4
where ý TE10 .
Pave ý ò Pave .dS ý
2 ý 2ð 2 2 a b
H o ò ò sin 2 ð y / bdxdy
2 4
2 b h
x ý0 y ý0
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Pave
2ý 2 ð 2 2
H ab / 2
ý
2 b2h4 o
h 2 ý (mð / a ) 2 (nð / b) 2 ý
But
Pave ý
ð2
,
b2
2 ý 2 ab 3 Ho 2
4ð 2
Prob. 12.29
Rs ý
ðý f
ð 12 109 4ð 107
ý
ý 2.858 102
7
óc
5.8 10
f c10 ý
u'
3 108
ý
ý 4.651 GHz
2a 2 2.6 2 102
1/ 2
1ù
u' ù 1
f c11 ý ú 2 2 ú ý 10.4 GHz
2 ûa
b û
377
ý
'ý
ý
ý 233.81
õ
2.6
(a) For TE10 mode, eq.(12.57) gives
ñ d j ò d ý 2 ýõ k x 2 k y 2 jýó d
ý
2 / u2
ð2
j ý ó d
a2
2
ö 2ð 12 109 ö
ð2
(2.6)
ý ÷
j 2ð 12 109 4ð 107 104
÷
2 2
8
3
10
(2
10
)
ø
ø
= 0.012682 + j373.57
ñ d ý 0.012682 Np/m
ñc ý
ù 1 b fc 2 ù
ú ( ) ú
b ' 1 ( f c / f ) 2 û 2 a f û
2 Rs
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ý
2 x 2.858 x102
ù 1 1 4.651 2 ù
) ú ý 0.0153 Np/m
ú (
û
10 (233.81) 1 (4.651 /12) û 2 2 12
2
2
(b) For TE11 mode,
ñ d jò d ý
2 / u 2 1 / a 2 1 / b 2 j ý ó d
ð2
139556 .21
j 9.4748 ý 0.02344 j 202.14
(10 2 ) 2
ý
ñ d ý 0.02344 Np/m
ñc ý
ù (b / a ) 3 1 ù
ù (1/ 8) 1 ù
2 2.858 102
=
ú
ú
2
ú
ú
b ' 1 ( f c / f ) 2 û (b / a ) 1 û 102 (233.81) 1 (10.4 /12) 2 û (1/ 4) 1 û
2 Rs
ñ c ý 0.0441 Np/m
Prob. 12.30
õ c ý õ ' jõ '' ý õ j
ó
Comparing this with
ó
ý 16õ o x10 4
õ ý 16õ o ,
u ' ù m2 n2 ù
fc ý ú 2 2 ú
b û
2 ûa
õ c ý 16õ o (1 j104 ) ý 16õ o j16õ o 104
For TM21 mode,
1/ 2
ý 2.0963 GHz,
f ý 1.1 f c ý 2.3059 GHz
ó ý 16õ o 104 ý 16 2ð 2.3059 109
'ý
ñd ý
109
104 ý 2.0525 104
36ð
ý
ý 30ð
õ
ó '
2 1 ( f c / f )2
Eo e ñ d z ý 0.8 Eo
ý
4.1104 30ð
ý 0.0231 Np/m
2 1 1/1.12
zý
1
ñd
ln(1/ 0.8) ý 9.66 m
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Prob. 12.31
For TM21 mode,
ñc ý
Rs ý
ñc ý
2 Rs
b ' 1 ( f c / f ) 2
1
ó cô
ý
ð fý
ð 2.3059 109 4ð 107
ý
ý 0.0246
1.5 107
óc
2 0.0246
ý 0.0314 Np/m
4ð 102 30ð 0.4166
Eo e (ñc ñ d ) z ý 0.7 Eo
zý
1
ln(1/ 0.7) ý 6.5445 m
ñc ñd
Prob. 12.32
u'
3 108
ý
ý 2.5 GHz
2a 2 6 102
377
'
ý
ý
ý 483
2
2
ö f ö
ö 2.5 ö
1 ÷
1 ÷ c ÷
÷
ø 4 ø
ø f ø
f c10 ý
TE
From Example 12.5,
Eo2 ab (2.2) 2 106 6 3 104
Pave ý
ý
ý 9.0196 mW
2
2 483
Prob. 12.33
For TE10 mode,
fc ý
u'
3 108
ý
ý 2.151 GHz
2a 2 2.11 4.8 102
(a) loss tangent ý
ó
ýd
õ
ó ý dõ ý 3 104 2ð 4 109 2.11
'ý
120ð
ý 259.53
2.11
109
ý 1.4086 104
36ð
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ñd ý
ó '
2 1 ( fc / f )
(b) Rs ý
2
ý
1.4067 104 259.53
2 1 (2.151/ 4)
2
ý 2.165 102 Np/m
ý fð
ð 4 109 4ð 107
ý
ý 1.9625 102
7
4.110
óc
ñc ý
3.925 102 (0.5 0.5 0.2892)
ù1 b
2ù
f
f
(
/
)
ý
c
ú
úû
2.4 102 259.53 0.8431
b ' 1 ( f c / f ) 2 û 2 a
2 Rs
ý 4.818 103 Np/m
Prob.12.34
2
ð fý
óc
ù1 b ö f ö ù
ù 1 1 ö f ö2 ù
c
ñc ý
ú ÷ ÷ úý
ú ÷ c÷ ú
2
2
2
a
f
ø ø úû
ö f ö úû
ö f ö úû 2 2 ø f ø úû
b ' 1 ÷ c ÷
b ' 1 ÷ c ÷
ø f ø
ø f ø
1
2 4ð 107 ð f
ù ö f ö2 ù
2
ý
ú1 ÷ c ÷ ú
2
ö f ö úû ø f ø úû
0.5 102 (120ð / 2.25) 5.8 107 1 ÷ c ÷
ø f ø
2
2 Rs
ù ö f ö2 ù
ý
ú1 ÷ c ÷ ú
2
ö f ö úû ø f ø úû
30 (5.8 / 2.25) 1 ÷ c ÷
ø f ø
105 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
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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
70
frequency (GHz)
Prob. 12.35
The cutoff frequency of the dominant mode is
3 108
u
f c10 ý
ý
ý 6.56 GHz
2a 4.576 102
The surface resistance is
Rs ý
ð fý
ð 8.4 109 4ð 107
ý
ý 23.91 103
óc
5.8 107
For TE10 mode,
2
ù
b ö fc ö ù
ñc ý
ú 0.5 ÷ ÷ ú
2
a ø f ø úû
ö f ö úû
b ' 1 ÷ c ÷
ø f ø
f c 6.56
ý
ý 0.781, ' ý o ý 377
8.4
f
2 Rs
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80
90
100
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2 23.91103
1.016
2ù
ù
úû0.5 2.286 ø 0.781ù úû
1.016 10 377 1 0.781
47.82 103 (0.5 0.2711)
ý
ý 15.42 103 Np/m
3.83 0.6245
ý15.42 103 8.686 dB/m = 0.1339 dB/m
ñc ý
2
2
Prob. 12.36
f c10 ý
(a)
3 108
u'
ý
ý 3.947 GHz
2a 2 3.8 102
2
ö f ö
u g ý u ' 1 ÷ c ÷ ý 3 108 1 (0.3947) 2 ý 2.756 108 m/s
ø f ø
(b) ñ ý ñ d ñ c
ñ d ý 0 since the guide is air-filled.
ð fý
ð 1010 4ð 107
ý
ý 2.609 102
5.8 107
óc
Rs ý
2
ù
b ö fc ö ù
ñc ý
ú0.5 ÷ ÷ ú
2
a ø f ø úû
ö f ö úû
b ' 1 ÷ c ÷
ø f ø
2 Rs
ý
2 2.609 102
1.6 10 (377) 1 ø 0.3947 ù
2
2
1.6
5.218 0.5656
2ù
ù
úû 0.5 3.8 ø 0.3947 ù úû ý
554.23
ý 5.325 103 Np/m
ñ c (dB) ý 8.686 5.325 103 ý 0.04626 dB/m
Prob.12.37
f c10 ý
ò' ý
u'
ý
u'
c
3 108
ý
ý
ý 3.991 GHz
2a 2a õ r ýr 2 2.5 102 2.26
2ð f õ r
c
2
2
ö f ö
ö 3.991 ö
F ý 1 ÷ c ÷ ý 1 ÷
÷ ý 0.8467
ø 7.5 ø
ø f ø
ò ý ò 'F ý
2ð 7.5 109 2.26
0.8467 ý 199.94 rad/m
3 108
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ñd ý
ó '
2F
ý
óo
104 (377)
ý
ý 1.481 102 Np/m
2 F õ r 2 0.8467 2.26
2
ù
b ö fc ö ù
ñc ý
ú 0.5 ÷ ÷ ú
2
a ø f ø úû
ö f ö úû
b ' 1 ÷ c ÷
ø f ø
2 Rs
ð fý
ð 7.5 109 4ð 107
Rs ý
ý
ý 0.0519
1.1 107
óc
2
ù
1.5 ö 3.991 ö ù
2 0.0519 ú 0.5
÷
÷ ú
2.5 ø 7.5 ø ûú 0.1038 0.6698
ú
û
ñc ý
ý
377
3.1848
1.5 102
0.8467
2.66
ý 0.02183 Np/m
up ý
u'
c
3 108
ý
ý
ý 2.357 108 m/s
F F õ r 0.8467 2.26
3 108 0.8467
ý 1.689 108 m/s
2.26
u'
c
3 108
ý
ý 0.05 m = 5 cm(ý 2a, as expected)
üc ý ý
f c f c õ r 3.991 109 2.26
ug ý u ' F ý
Prob. 12.38 (a) For TE10 mode,
fc ý
fc ý
(b)
u'
,
2a
u' ý
c
2.11
3 108
ý 4.589 GHz
2.11(2 2.25 102 )
ñ cTE 10 ý
Rs ý
' ý
ù1 b
2ù
úû 2 a ( f c / f ) úû
b ' 1 ( f c / f )
2 Rs
2
ð fý
ð 5 109 4ð 107
ý
ý 3.796 102
7
1.37 10
óc
377
ý 259.54
2.11
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ñc ý
1.5
(4.589 / 5) 2 ]
2.25
ý 0.05217 Np/m
4
1.5 10 (259.54) 1 (4.589 / 5) 2
2 3.796 102 [0.5
Prob. 12.39 For TE10 mode,
ñc ý
ù 1 b fc 2 ù
ú ( ) ú
b ' 1 ( f c / f ) û 2 a f û
2 Rs
2
But a = b, Rs ý
1
ý
ó cô
ð fý
óc
ð fý
óc
ù1
f ù
k f ú ( c )2 ú
f û
ù1
fc 2 ù
û2
ñc ý
(
)
ý
ú
ú
f û
a ' 1 ( f c / f ) 2 û 2
1 ( f c / f )2
2
where k is a constant.
dñ c
ý
df
k [1 (
f c 2 1/ 2 1
) ] [ f
4
f
For minimum value,
Prob. 12.40
1/ 2
3 2
f f
2 c
5 / 2
k 1 1/ 2
[ f f c2 f
2 2
1 ( f c / f )2
]
3 / 2
](2 f c 2 f
3
)[1 (
f c 2 1/ 2
) ]
f
dñ c
ý 0 . This leads to f = 2.962 fc.
df
For the TE mode to z,
E zs ý 0 , Hzs ý Ho cos(mð x / a ) cos(nð y / b) sin( pð z / c)
E ys ý
j ý
÷ E zs j ý Hzs
2
ý 2 (mð / a ) Ho sin(mð x / a ) cos(nð y / b) sin( pð z / c)
2
h
h y
h
x
as required.
E xs ý
j ý
÷ E zs j ý Hzs
2
ý 2 (nð / b) Ho cos(mð x / a ) sin(nð y / b) sin( pð z / c)
2
h
h x
h
y
From Maxwell’s equation,
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jý H s ý ñ E s ý x
y
z
Exs
E ys
0
H xs ý
1
1 E ys
ý 2 (mð / a )( pð / c) Ho sin(mð x / a ) cos(nð y / b) cos( pð z / c)
h
j ý z
Prob. 12.41 Maxwell’s equation can be written as
j õ E zs ÷ Hzs
h2 y h2 x
For a rectangular cavity,
H xs ý
h 2 ý k x 2 k y 2 ý (mð / a ) 2 (nð / b) 2
For TM mode, Hzs = 0 and
E zs ý Eo sin(mð x / a ) sin(nð y / b) cos( pð z / c)
Thus
jõ
jõ Ezs
ý 2 (nð / b) Eo sin(mð x / a) cos(nð y / b) cos( pð z / c)
2
h
h y
as required.
H xs ý
H xs ý
j õ E zs ÷ Hzs
h2 x h2 y
j õ
(mð / a ) Eo cos(mð x / a ) sin(nð y / b) cos( pð z / c)
h2
From Maxwell’s equation,
ý
jõ E s ý ñ H s ý x
y
z
H xs
H ys
0
E ys ý
1 H xs
1
ý 2 (nð / b)( pð / c ) Eo sin(mð x / a ) cos(nð y / b) cos( pð z / c)
j õ z
h
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Prob. 12.42
fr ý
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 ý
The lowest TE mode is TE011 with f r ý
1
u' 1
2
2 a
b2
u' 1 1 u' 1 1
ü
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 ý
The lowest TE mode is TE101 with f r ý
u' 1
1
2
b2
2 a
u' 1 1 u' 1 1
þ
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 ý
The lowest TE mode is TE101 with f r ý
1
u' 1
2
2 a
b2
u' 1 1 u' 1
1
2 2 ü
2
c
b2
2 a
2 a
Hence the dominant mode is TE101.
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Prob. 12.43
(a)
1
u'ý
ýõ
2
c
ý
õr
ý
3 108
4.6
2
2
u' ö mö ö nö ö p ö
fc ý
÷ ÷ ÷ ÷ ÷ ÷
2 ø a ø øbø ø c ø
For the dominant mode, m = 1, n=0, p=1
2
2
u' ö 1 ö ö1ö
3 108
fc ý
÷ ÷ ÷ ÷ ý
2 øaø øcø
2 4.6
(b)
Qý
ôý
1
Qý
1
1
3 1010
(0.37267) ý 2.606 GHz
ý
9 104 36 104 2(2.1447)
(a 2 c 2 )abc
ô ùû 2b(a 3 c3 ) ac(a 2 c 2 ) ùû
ð f r101óýo
ý
1
ð 2.606 10 1.57 10 4ð 10
9
7
7
ý 2.49 106 m
(9 36)(72) 102
32.42
ý
ý 4727.7
ô 8(27 216) 18(9 36) 2.49 106 (2754)
Prob. 12.44
(a)
2
2
u' ö mö ö nö ö p ö
fr ý
÷ ÷ ÷ ÷ ÷ ÷
2 ø a ø øbø ø c ø
2
2
2
ö1ö ö1ö
f rTE101 ý 1.5 1010 ÷ ÷ ÷ ÷ ý 1.5 1010 0.1736 ý 6.25 GHz
ø3ø ø 4ø
f rTE 011 ý 1.5 1010
1
1
ý 1.5 1010 0.2225 ý 7.075 GHz
6.25 16
f rTE110 ý 1.5 1010
1
1
ý 1.5 1010 0.2711 ý 7.81 GHz
9 6.25
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Prob. 12.45
u'ý
c
3 108
ý
ý
ý 1.897 108
2.5
2.5
ýõ
1
2
2
2
2
2
u ' ö m ö ö n ö ö p ö 1.897 108 102 ö m ö ö n ö ö p ö
fc ý
÷ ÷ ÷ ÷ ÷ ÷ ý
÷ ÷ ÷ ÷ ÷ ÷
2 ø a ø øbø ø c ø
2
ø 1 ø ø2ø ø 3 ø
ý 9.485 m 2 0.25n 2 0.111 p 2 GHz
f r101
ý 9.485 1 0 0.111 = 10 GHz
f r 011
ý 9.485 0 0.25 0.111 = 5.701 GHz
f r 012
ý 9.485 0 0.25 0.444 = 7.906 GHz
f r 013
ý 9.485 0 0.25 0.999 = 10.61 GHz
f r 021
ý 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
Qý
ô ùû 2b(a 3 c3 ) ac(a 2 c 2 ) ùû
When a = b = c,
Qý
a
2a 2 a 3
2a 5
ý
ý
4
3
2
2
ô ùû 2a 2a a 2a ùû 6ô a 3ô
Prob. 12.47
(a) Since a > b < c, the dominant mode is TE101
u' 1
1 3 108 102
f r101 ý
0 2 ý
c
2 a2
2
1 1
ý 16.77 GHz
22 12
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(a 2 c 2 )abc
(b) QTE101 ý
ô ûù 2b(a 3 c3 ) ac(a 2 c 2 ) ùû
(400 100)20 8 10 103
3.279 103
ý
ý
ô [16(8000 1000) 200(400 100)]
ô
104
ý
ý
But ô ý
m
ð f r101ýoó
ð 16.77 109 4ð 107 6.1107 200.961
200.961
QTE101 ý 3.279 103
ý 6589.51
104
1
1
Prob. 12.48
c
m2 n 2 p 2
2a
The lowest possible modes are TE101, TE011, and TM110. Hence
fr ý
c
fr ý
2
2a
aý
c
fr
3 108
ý
ý 7.071 cm
2
2 3 109
a = b = c = 7.071 cm
Prob. 12.49
(a) a = b = c
u'
fr ý
m2 n2 p 2
2a
For the dominant mode TE101 ,
fr ý
aý
u'
c
11 ý
2
2a
2a
3 108 2
c 2
ý
ý 0.03788 m
2 f r 2 5.6 109
a ý b ý c ý 3.788 cm
(b)
For õ r ý 2.05,
aý
u'ý
c
õr
0.03788
c 2
ý
ý 0.02646
2 fr õ r
2.05
a ý b ý c ý 2.646 cm
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Prob. 12.50
(a)
This is a TM mode to z. From Maxwell’s equations,
ñ E s ý jý H s
Hs ý
1
jý
ñ Es ý
j
ý
x
y
z
0
0
Ezs ( x, y )
ý
Ezs ö
j ö Ezs
ax
a
÷
ý ø y
x y ÷ø
But
Ezs ý 200sin 30ð x sin 30ð y,
1
ý
ý
1
102
ý
6 109 4ð 107 24ð
j102
Hs ý
200 30ð sin 30ð x cos 30ð ya x cos 30ð x sin 30ð ya y
24ð
H = Re (Hs e jt )
H ý 2.5 sin 30ð x cos 30ð ya x cos 30ð x sin 30ð ya y sin 6 109 ð t A/m
(b)
E ý Ez a z ,
H ý H xax H y a y
E H ý0
Prob. 12.51
(a) a ý b ý c
f r101 ý
3 108
ý 12 109
a 2
3 108
aý
ý 1.77 cm
2 12 109
(b) QTE101 ý
ý
a a ð f r101ýó
ý
3ô
3
1.77 102 ð 12 109 4ð 107 5.8 107
ý 9767.61
3
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Prob. 12.52
2
2
u' ö mö ö nö ö p ö
fr ý
÷ ÷ ÷ ÷ ÷ ÷
2 ø a ø øbø ø c ø
f r101 ý
3 108
2
2
1
1
ý 44.186 MHz
2
(10.2) (3.6) 2
f r 011 ý 150
1
1
MHz ý 45.093 MHz
2
(8.7) (3.6) 2
f r111 ý 150
1
1
1
MHz ý 47.43 MHz
2
2
(10.2) (8.7) (3.6) 2
f r110 ý 150
1
1
MHz ý 22.66 MHz
2
(10.2) (8.7) 2
f r102 ý 150
1
4
MHz ý 84.62 MHz
2
(10.2) (3.6) 2
f r 201 ý 150
4
1
MHz ý 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 ý n12 n22 ý 1.512 1.452 ý 0.1776 ý 0.421
Prob. 12.55
(a) NA =
n12 n2 2 =
2
162
. 2 1604
.
= 0.2271
(b) NA = sin ñ a = 0.2271 or ñ
(c) V =
a
= sin –1 0.2271 = 13.13o
ðd
ð 50 106 0.2271
= 27.441
NA =
ü
1300 109
N = V2/2 6 modes
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Prob. 12.56
ð d 2 3 ð 2 5 106
Vý
n1 n2 ý
1.482 1.462 ý 5.86
1300 109
ü
V2
ý 17.17 or 17 modes
Ný
2
Prob. 12.57
(a) NA = sin ñ a = n12 n2 2 =
ñ a = sin –1 0.4883 = 29.23o
153
. 2 145
. 2 = 0.4883
(b) P(l)/P(0) = 10- ñ l / 10 = 10-0.4X5/10 = 0.631
i.e. 63.1 %
Prob. 12.58
P () ý P(0)10ñ /10 ý 10 100.50.85 /10 ý 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,
ñ12 ý 8686ñ10
Prob. 12.60
ñ ý 10 log10
Pin
1.2 103
ý 10 log10
ý 30.792
1 106
Pout
0.4
Np/km
8.686
30.792 30.392 dB
ý
ý
ý 76.98 km
0.4 dB/km
ñ
ñ ý 0.4 dB/km =
Prob. 12.61
P(0) = P(l) 10 ñ
l/10
= 0.2 x 10 0.4 x 30/10 mW = 3.1698 mW
Prob. 12.62 See text.
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CHAPTER 13
P. E. 13.1
(a) For this case, r is at near field.
I dl sin ñ ö j ò 1 ö j ò r
2ð ü
, òr ý . ý 72 o
Hö s ý o
2 ÷e
÷
ü 5
4ð
ø r r ø
üý
2ð c
Hö s ý
ý
2ð 3 108
ý 6ð ,
108
(0.25)
2ð
òý
ý
ü
1
3
6ð
sin 30o ö
ö j 72o
1
j1/ 3
100
e
ý 0.2119ð 20.511o mA/m
÷
2 ÷
4ð
ø 6ð / 5 (6ð / 5) ø
H ý Im ø Hö s e jt aö ù
Im is used since I = Io sin t
ý 0.2119sin(108 20.5o )aö mA/m
(b) For this case, r is at far field. ò ý
Hö s ý
j (0.25)(
2ð
ü
)
Sin60o e j 0
ü 100
4ð (6ð 200)
e jt )
a
H ý Im ( H ö s
ö
2ð
ü
200ü ý 0o
o
ý 0.2871e j 90 ý A/m
o
ý 0.2871sin(108 90o )aö ý A/m
P. E. 13.2
ü
(a) l ý ý 1.5m ,
4
(b) Io = 83.3mA
(c) Rrad = 36.56 , Prad ý 1 (0.0833) 2 36 .56
2
= 126.8 mW.
(d) ZL = 36.5 + j21.25,
ý
36 .5 j 21.25 75
ý 0.3874ð 140.3 o
36 .5 j 21.25 75
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sý
1 0.3874
ý 2.265
1 0.3874
P. E. 13.3
Dý
4ðU max
Prad
(a) For the Hertzian monopole
U (ñ , ö ) ý sin 2 ñ , 0 ü ñ ü ð / 2,
ð
2 2ð
ò ò
Prad ý
sin 2 sin ñ dñ dö ý
ñ ý0 ö ý0
Dý
(b) For the
ü
4
0 ü ö ü 2ð , Umax = 1
4ð
3
4ð (1)
ý3
4ð
3
monopole,
ð
cos2 ( cos ñ )
2
U (ñ , ö ) ý
, Umax = 1
sin 2 ñ
ð
cos2 ( cos ñ )
2
ý ò ò
sin ñ dñ dö ý 2 ð (0.609 )
2
ñ
sin
ñ ý0 ö ý0
ð
Prad
Dý
2 2ð
4 ð (1)
ý 3.28
2 ð (0.609 )
P. E. 13.4
Prad = r Pin = 0.95(0.4)
4 ð U max
4 ð (0.5)
Dý
ý
ý 16 .53
Prad
0.4 0.95
4 ð (0.5)
ý 20.94
(b) D ý
0.3
(a)
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P. E. 13. 5
ð
Prad ý
2 2ð
ò ò
sin ñ sin ñ dñ dö ý
ñ ý0 ö ý0
Dý
ð2
, Umax = 1
2
4 ð (1)
ý 2.546
ð2
2
P. E. 13. 6
ù1
ù
(a) f (ñ ) ý cos ñ cosú ø ò d cos ñ ñ ù ú
û2
û
where ñ ý ð , òd ý
2ð ü
. ýð
ü 2
ù1
ù
f (ñ ) ý cos ñ cosú ø ð cos ñ ð ù ú
û2
û
unit pattern
group pattern
For the group pattern, we have nulls at
ð (cos ñ 1) ý ð
2
2
ñ ý ð 2
and maxima at
ð (cos ñ 1) ý 0, ð
2
cos ñ ý 1,1
ñ ý 0, ð
Thus the group pattern and the resultant patterns are as shown in Fig.13.15(a)
ù1
ù
(b) f (ñ ) ý cos ñ cosú ø ò d cos ñ ñ ù ú
û2
û
where ñ ý ð , ò d ý ð / 2
2
ù1 ö ð
öù
f (ñ ) ý cos ñ cos ú ÷ cos ñ ð ÷ ú
2
øû
û2 ø 2
unit pattern group pattern
For the group pattern, the nulls are at
ð (cos ñ 1) ý ð
4
2
ñ ý 180 o
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and maxima at
cosñ 1 ý 0
ñý0
Thus the group pattern and the resultant patterns are as shown in Fig.13.15(b)
P. E. 13.7
(a)
●
●
●
ü
●
●
●
●
2
ü
ü
●
●
●
●
2
ü
2
12
: :1
12
: :1
x
x
ü
●●
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 ý
e
e ............ e i ( N 1)
2!
3!
õ
N 1
ö Nö
õ ý õ ÷ø i ÷ø ý 1 N
where
i1
N 1 N(N 1)(N 2)
...........
2!
3!
ý (1 1) N 1 ý 2 N 1
( AF) n ý
ý
1
2 N 1
1 e j
1
2 N 1
N 1
2 cos
2
ý
1
2 N 1
N 1
j
j
e
2
ý cos
2
e
2
e
N 1
j
2
N 1
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P. E. 13.8
ü2
c 3 108
Ae ý
Gd , ü ý ý
ý3m
4ð
f
108
For the Hertzian dipole,
Gd ý 1.5 sin 2 ñ
ü2
(1.5 sin 2 ñ )
4ð
1.5ü 2 1.5 9
Ae,max ý
ý
ý 1074
m2
.
4ð
4ð
Ae ý
By definition,
Pr ý Ae Pave
Pr 3 10 6
ý
Ae
.
1074
Pave ý
ý 2793
.
ý W / m2
P. E. 13.9
(a) Gd ý
4ðr Pave
ý
Prad
ý
2
4 ðr 2
1 E2
2
ý
Prad
2 ðr 2 E 2
Prad
2ð 400 106 144 106
ý 0.0096
120ð 100 103
G ý 10 log10 Gd ý -20.18 dB
(b)
G ý r G d ý 0.98 0.0096 ý 9.408 103
P. E. 13.10
ù ü2 Gd 2 ó Prad
rýú
3
Pr
úû (4ð)
where ü ý
ù
ú
úû
1
4
c 3 10 8
ý
ý 0.05m
f 6 10 9
Ae ý 0.7 ð a 2 ý 0.7 ð (1.8 ) 2 ý 7 .125m2
Gd ý
4ðAe
ü
2
ý
4ð(7 .125)
25 10 4
ý 3.581 10 4
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ù 25 104 (3.581) 2 108 5 60 103 ù
rýú
ú
(4ð )3 0.26 103
û
û
1
4
ý 1168.4m ý 0.631 nm
At r ý
Pý
rmax
2
ý 584.2m,
Gd Prad 3.581 104 60 103
ý
ý 501 W/m 2
4ð r 2
4ð (584.2) 2
Prob. 13.1
Using vector transformation,
Ars ý Axs sin ñ cos ö , Añs ý Axs cos ñ cos ö , Aö s ý Axs sin ö
As ý
50e j ò r
(sin ñ cos ö ar cos ñ cos ö añ sin ö aö )
r
ñ As
ý
ý Hs ý
100 cos ñ sin ö j ò r
50
e ar 2 (1 j ò r ) sin ö e j ò r añ
2
ý r sin ñ
ýr
50
cos ñ cos ö (1 j ò r )e j ò r aö
ýr 2
At far field, only
Hs ý
1
r
term remains. Hence
j 50 jò r
ò e (sin ö añ cos ñ cos ö aö )
ýr
j 50 ò e j ò r
(sin ö aö cos ñ cos ö añ )
ýr
50ò
sin(t ò r )(sin ö aö cos ñ cos ö añ ) V/m
E ý Re ùû E s e jt ùû ý
ýr
E s ý ar H s ý
H ý Re ùû H s e jt ùû ý
50
ò sin(t ò r )(sin ö añ cos ñ cos ö aö ) A/m
ýr
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Prob. 13.2
(a)
rmin ý
üý
2d 2
ü
3 108
c
ý
ý 0.75 m
f 400 106
ý
2(0.02ü ) 2
ü
r
i.e. r is in the far field.
jI ò dl
Hö s ý o
sin ñ e jò r
4ð r
2ð
0.02ü sin 90o
3
I o ò dl
ü
ý 5 104 ý 0.5 mA/m
| H ö s |ý
sin ñ ý
4ð r
4ð (60)
|Eñ s |ý o | H ö s |ý 0.1885 V/m
(b)
| H ö s |ý 0.5 mA/m|
2
2
ö dl ö
(c) R rad ý 80ð 2 ÷ ÷ ý 80ð 2 ø 0.02 ù ý 0.3158
øüø
1
1
(d) Prad ý | I o |2 R rad ý (9)(0.3158) ý 1.421 W
2
2
Prob. 13.3
3 108
c
üý ý
ý6m
f 50 106
jI ò dl
sin ñ e jò r ,
Hö s ý o
Eñ s ý H ö s
4ð r
dl ý 10cm üü ü
2ð 2ð
ý
ýð /3
òý
6
ü
ð
20 0.1
I o ò dl
3
ý
ý 0.1667
4ð
4ð
j 0.1667
Hs ý
sin ñ e jð r /3aö A/m
r
Es ý
j 0.1667 377
j 62.83
sin ñ e jð r /3añ ý
sin ñ e jð r /3añ V/m
r
r
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Prob. 13.4
ü ýc/ f,
Prad ý
R rad
1 2
I o Rrad
2
ö dl ö
ý 80ð ÷ ÷
øüø
2
2
I o2 ý
2 Prad
Rrad
2
2
ö
ö 1
ö c ö 1
3 108
öüö 1
ý
ý
ý 348.93
2
2
12
I ý 2 Prad ÷ ÷
P
÷
rad ÷
÷
2 ÷
2
2
6
2
ø dl ø 80ð
ø fdl ø 80ð
ø 140 10 2 10 ø 80ð
I o ý 18.68 A
2
2
o
Prob. 13.5
e jòr
(a) Azs ý
4ðr
l
2
ò
l
I o (1
2z
l
)e jòz cosñ z
2
l
ù
2
2z
2z
e j ò r ùú 2
) cos( ò z cos ñ )dz j ò (1
) sin( ò z cos ñ )dz ú
I o ò (1
ý
4ð r ú l
l
l
ú
l
2
û 2
û
l
l
2
2z
e jòr
2 I o ò (1 ) cos(ò z cos ñ )dz
ý
4ð r
l
0
ý
2
I o e jòr
òl
. ùú 1 cos( 2 cos ñ ) ùú
2
2
û
2 ð rò cos ñ l û
E s ý jý As
Eñ s ý jý sin ñ Azs ý j ò sin ñ Azs
òl
ù
ù
j I o e jòr sin ñ úû 1 cos( 2 cos ñ ) úû
Eñ s ý
ð rl
ò cos2 ñ
òl
( cos ñ ) 2
òl
.
If òl 2 l , cos( 2 cos ñ ) ý 1 2
2!
Hence
Eñ s ý
Pave ý
j I o
ò le jòr sin ñ , Hö s ý Eñ s /
8ð r
Eñs
2
2ð ð
Prad ý
òò
0 0
2
,
Prad ý
òP
ave
dS
2
n ö I oò l ö 1
÷
÷ 2 sin 2 ñ r 2 sin ñ dñ dö
2 ø 8ð ø r
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2
ölö
ý 10 ð 2 I o 2 ÷ ÷ ý 1 2 I o 2 Rrad
ø üø
l
or Rrad ý 20 ð 2 ö÷ ö÷
2
ø üø
l
(b) 0.5 ý 20 ð 2 ö÷ ö÷
2
l ý 0.05ü
ø üø
Prob. 13.6
2
ö dl ö
R rad ý 80ð 2 ÷ ÷
øüø
3 108
ý 250m
ü ýc/ f ý
1.2 106
2
Rrad
0.5
ö dl ö
ý
ý 6.33 104
÷ ÷ ý
2
2
80ð
ø ü ø 80ð
dl
ý 2.516 102
dl ý 2.516 102 250 ý 6.29 m
ü
Prob. 13.7
40
I
24 V
Iý
+
-
Zin = 73+j42
24
V
ý
ý 0.1866 j 0.0694
Rs Z in 40 73 j 42
1 2
Rrad ý 73
| I | Rrad ,
2
1
ý (0.1991)2 73 ý 1.447 W
2
Prad ý
Prad
Prob. 13.8
Let us model this as a short Hertzian dipole.
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2
ö dl ö
Rrad ý 80ð ÷ ÷ ý 80ð 2 (1/ 8) 2 ý 12.34
øüø
1
Prad ý I o2 Rrad ý 4
I o ý 0.8052 A
2
2
Prob. 13.9
Change the limits in Eq. (13.16) to l 2 i.e.
As ý
ý
ý I o e jòzCosñ ø jò cos ñ cos ò t ò sin ò t ù
ò 2 cos2 ñ ò 2
4ð r
1
ý I o e jòr
2ð r ò sin 2 ñ
l
2
l
2
òl ö òl
ù òl
öù
ö
ö òl
ú sin 2 cos÷ø 2 cos ñ ÷ø cos ñ cos 2 sin÷ø 2 cos ñ ÷ø ú
û
û
But B ý ý H ý ñ A
Hös ý
1
ýr
Ar ù
ù
ú r rAñ ñ ú ,
û
û
ø ù
where Ao ý Az sin ñ , Ar ý Az cos ñ
òl ö òl
I o e jòr ö jò ö ù ò l
I
öù
ö
ö òl
Hö s ý
÷ ú sin cos÷ cos ñ ÷ cos ñ cos sin÷ cos ñ ÷ ú o 2 e jòr ø ......ù
÷
ø û 2ð r
ø 2
ø
ø 2
2 ð r ò ø sin ñ ø û
2
2
For far field, only the
Hö s ý
jI o jòr
e
2ð r
1
-term remains.
r
Hence
òl ö òl
ù òl
ö òl
ö
öù
ú sin 2 cos÷ø 2 cos ñ ÷ø cos ñ cos 2 sin÷ø 2 cos ñ ÷ø ú
û
û
sin ñ
òl
ö òl
ö
cos÷ cos ñ ÷ cos
ø 2
ø
2
(b) f (ñ ) ý
sin ñ
For l ý ü , f (ñ ) ý
cosø ð cos ñ ù 1
sin ñ
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ö 3ð
ö
cos÷
cos ñ ÷
ø 2
ø
3ü
, f (ñ ) ý
For l ý
2
sin ñ
For l ý 2ü , f (ñ ) ý
cos ñ sin ø 2ð cos ñ ù
sin ñ
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Prob. 13.10
(a)
c
3 108
ý 0.6667 m
üý ý
f 450 106
ý
ü
2
ý 0.333 m
(b)
ó
ý
õ
4
109
2ð 450 10 81
36ð
ý 1.975
6
ýõ ù
2
ù 2ð 460 106
ó ö
ú 1 ÷ö
ú
ý
1
õ ø÷
2 ú
c
ú
ø
û
û
ò ý
81 ù
2
1 ø1.975 ù 1ù
úû
2 úû
2ð 460 106
11.4086 ý 109.91
3 108
2ð
üý
ý 0.0572
ý
ý
ò
ü
2
ý 28.58 mm
Prob. 13.11
(a)
c
3 108
üý ý
ý 260.8 m
f 1.150 106
ý
ü
4
ý 65.22 m
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(b)
c
3 108
ý
ý 3.333 m
f 90 106
üý
ý
ü
ý 0.8333 m
4
(c )
c
3 108
ý
ý 3.75 m
f 80 106
üý
ý
ü
4
ý 0.9375 m
(d)
c
3 108
üý ý
ý 0.5 m
f 600 106
ý
ü
4
ý 0.125 m
Prob. 13.12
lü ,hence
it is a Hertzian monopole.
2
Rrad
2
ö dl ö
ö1ö
ý 80ð ÷ ÷ ý 80ð 2 ÷ ÷ ý 12.34
øüø
ø8ø
2
Prob. 13.13
(a) ü ý
c
3 108
ý
ý 30 m
f 10 106
E max
ð Io S
ý
rü 2
E max rü 2
Io ý
ðS
50 103 3 302
Io ý
ý 9.071 mA
120ð 2ð (0.2) 2100
(b)
Rrad ý
Prad ý
320ð 4 S 2
ü4
ý
(S=Nð r 2 )
320ð 4ð 2 (0.2) 4 104
ý 6.077
304
1 2
1
I o Rrad ý (9.071) 2 106 6.077
2
2
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ý 0.25 mW
Prob. 13.14
3 108
c
ý
ý 3.75 m
f 80 106
S ý N ðò o2
üý
Rrad ý
N2 ý
320ð 4 S 2
ü4
ý
320ð 4 N 2ð 2 òo4
ü4
(3.75) 4 8
ý 248006
320ð 6 (1.2 102 ) 4
N ý
ü 4 Rrad
2
320ð 6 (1.2 102 ) 4
N
498
320ð 4 (0.5027) 2
ý 1.26 m
Rrad ý
(50) 4
1
1
(b) Prad ý I o 2 Rrad ý (50)2 1.26 103 ý 1.575 W
2
2
2ð R
a
a
a
ð fýó =
(c) R ý
R dc =
=
2
2ô
2ô ó S 2óð a
2ð a
ý fð
ó
Prob. 13.15
(a)
Rrad ý
320ð 4 S 2
ü4
S ý ðòo2 ý ð (0.4) 2 ý 0.5027 m 2
c 3 108
ý 50 m
üý ý
f 6 106
0.4
4ð 107 6 106 ð
R ý fð
ý
ý 63.91 m
ó
4 103
5.8 107
a
R rad
1.26
=
ý
100% ý 1.933%
R rad R 1.26 63.91
R ý
Prob. 13.16
ö
öð
cos÷ cos ñ ÷
ø
ø2
(a) f (ñ ) ý
sin ñ
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(b) The same as for
ü
2
dipole except that the fields are zero for ññ
ð
2
as shown.
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 ý 20% ý
Prad 1
1
ý
Prad 1 Pohm1 5
But
4 Prad 1 ý Pohm1
(1)
1 2
I Rs z
2
1
ý I 2 Rs 2z ý 2Pohm1
2
Pohm1 ý
Pohm2
1
1
ö z ö
Prad 1 ý I o2 Rrad ý I o2 80ð 2 ÷ ÷
2
2
ø ü ø
(2)
2
1
1
ö 2z ö
Prad 2 ý I o2 Rrad ý I o2 80ð 2 ÷
÷ ý 4 Prad 1
2
2
ø ü ø
From (1) to (3),
2
r2 ý
Prad 2
P
4 Prad 1
ý
ý ohm1 ý 33.3%
Prad 2 Pohm 2 4 Prad 1 2 Pohm1 3Pohm1
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Prob. 13.18
r ý
Prad
Rrad
ý
Pin
Rrad R
Rrad ý 73,
ü
R ý
ý
ó c S ó cð a 2
3 108
ý 50m,
ý 25m
2
6 106
25
25
R ý
ý
ý 0.09528
6
2
6
58 10 ð (1.2) 10
262.4
73
r ý
ý 0.9987 ý 99.87%
73 0.09528
ý
ü ýc/ f ý
,
Prob. 13.19
(a) Let H s ý
a E a H ý ak
Hs ý
cos 2ñ j ò r
e aH
o r
añ a H ý ar
a H ý aö
cos 2ñ j ò r
e aö
120ð r
(b) Pave ý
| Es |2
cos 2 (2ñ )
ar ý
ar
2
2 r 2
Prad
1
ý
2
But cos 2ñ ý cos 2 ñ sin 2 ñ ý 2 cos 2 ñ 1
ð
1
(2 cos 2 ñ 1) 2 d (cos ñ )
Prad ý
120 ò0
ð
1
(4 cos 4 ñ 4 cos 2 ñ 1)d (cos ñ )
ý
ò
120 0
ý
ð
cos 2 2ñ 2
1
2
òò r 2 r sin ñ dñ dö ý 240ð (2ð )ò0 cos 2ñ sin ñ dñ
öð
1 ö 4 cos5 ñ 4 cos3 ñ
cos ñ ÷
÷
120 ø 5
3
ø0
1
4 4
4 4
1 14
[ 1 1] ý
( )
120 5 3
5 3
120 15
ý 7.778 mW
ý
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(c )
1
ý
120
Prad
ý
120o
ò
(2 cos 2 ñ 1) 2 d (cos ñ )
60o
ö 120o
1 ö 4 cos5 ñ 4 cos3 ñ
ñ
cos
÷
÷ o
120 ø 5
3
ø 60
1 4
1
4 1 1 4 1
4 1 1
1 1 1 1
[ ( ) ( ) ( ) ( ) ] ý [ ]
120 5 32 3 8 2 5 32 3 8 2 60 40 2 6
ý 5.972 mW
5.972
ý 0.7678 or 76.78%
7.778
ý
which is
Prob. 13.20
1
2
1
2
Pave ý Re( E s H s* ) ý | Hö s |2 ar
ò 2 I o2
1
sin 2 ñ cos 2 ö r 2 sin ñ dñ dö
Prad ý ò Pave dS ý òò
2 2
2
16ð r
2ð
ò 2 I o2 ð 3
ò 2 I o2 ö 4 ö 1 2ð
2
sin ñ dñ ò cos ö dö ý
(1 cos 2ö )dö
ý
÷ ÷
32ð 2 ò0
32ð 2 ø 3 ø 2 ò0
0
ò 2 I o2 ö 4 ö
ò 2 I o2
ð
ý
÷ ÷
32ð 2 ø 3 ø
24ð
2P
ò 2
Rrad ý rad
ý
I o2
12ð 2
Assuming free space, =120ð ,
ý
Rrad ý
10ò 2
ð
Prob. 13.21
(a) Prad ý ò Prad ÷ dS ý Pave .2ð r 2 (hemisphere)
Pave ý
Prad
200 103
ý
ý 12.73ýW / m 2
2
6
2ð r
2ð (2500 10 )
Pave ý 12.73ar ý W/m 2 .
(b)
Pave ý
( E max )
2
2
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E max ý 2 Pave ý 240 ð 1273
. 10 6
ý 0.098 V / m
Prob. 13.22
U (ñ , ö ) ý r 2 Pave ý k sin 2 ñ sin 3 ö
G d (ñ , ö ) ý
U ave
k
ý
4ð
U (ñ , ö )
U ave
ð
ð
ð
ù
k
k ù
3
3
2
ý
ý
U
d
d
d
d
(
,
)
sin
sin
sin
ñ
ö
ñ
ñ
ö
ö
ö
ñ
ñ
ú ò (1 cos ö )d ( cos ö ) ú
òò
ò
ò
4ð 0
4ð û 0
0
û
2
4k
ù4ù
ý
úû 3 úû
9ð
9ð
k sin 2 ñ sin 3 ö ý 7.069sin 2 ñ sin 3 ö
G d (ñ , ö ) ý
4k
D ý Gd ,max ý 7.069
ý
k
4ð
Prob. 13.23
3ü
2ð 3ü
ò ý
,
ý 3ð
ü 2
2
ù ö 2ð 1 3ü
ö
ö 2ð 1 3ü ö ù
cos ÷
cos ñ ÷ cos ÷
÷ú
ò r ú
jI e û ø ü 2 2
ø
ø ü 2 2 øû
Hö s ý o
2ð r
sin ñ
ù ö 3ð
ö
ö 3ð ö ù
cos ÷ cos ñ ÷ cos ÷ ÷ ú
òr ú
òr
jI e û ø 2
ø
ø 2 ø û jI o e cos ø1.5ð cos ñ ù
ý o
ý
2ð r
sin ñ
2ð r
sin ñ
Hence, the normalized radiated field pattern is
From Prob. 13.11, set =
cos ø1.5ð cos ñ ù
sin ñ
which is plotted below.
f (ñ ) ý
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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 oò dl
sin ñ e jòr
4ð r
2
ö dl ö
Rrad ý 80 ð 2 ÷ ÷
øüø
1
4ð r 2 .
Eñ s
2
4ð r Pave
2
Gd ý
ý
1 2
Prad
I o Rrad
2
(a) Eñs ý
4 ð r 2 1 ö ü ö 1 2 I o 2ò 2 ø dl ù sin 2 ñ
.
÷ ÷ .
I o 2 80 ð 2 ø dl ø
16 ð 2 r 2
2
ý
2
2
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Gd ý 1.5sin 2 ñ
(b) D ý Gd ,max ý 1.5
(c) Ae ý
ü2
1.5ü 2 sin 2 ñ
Gd ý
4ð
4ð
(d) Rrad
ö1ö
ý 80ð ÷ ÷ ý 3.084
ø 16 ø
2
2
Prob. 13.26
Gd (ñ , ö ) ý
4ð U (ñ , ö )
4ð f 2 (ñ )
ý
2
Prad
òò f (ñ )d
f (ñ ) ý sin ñ
Gd (ñ , ö ) ýý
4ð sin 2 ñ
2ð
ð
ò ò sin
ö ñ
3
ñ dñ
ý
4ð sin 2 ñ
ý 1.5sin 2 ñ
2ð (4 / 3)
ý0 ý0
D ý Gd ,max ý 1.5
Prob. 13.27
(a)
3 108
c
ý 250
üý ý
f 1.2 106
ý
ü
ý 62.5 m
4
(b) From eq. (13.30), Rrad ý 36.5
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(c )
For ü /4-monopole,
ð
cos( cos ñ )
2
f (ñ ) ý
,
sin ñ
0 üñ üð / 2
ð
4ð f 2 (ñ )
Gd (ñ , ö ) ý
ò
f 2 (ñ )d
4ð cos 2 ( cos ñ )
2
sin 2 ñ
ý
2ð ð /2
òò
0
0
ð
cos 2 ( cos ñ )
2
dñ d ö
sin ñ
ð
ð
4ð cos 2 ( cos ñ )
3.282 cos 2 ( cos ñ )
1
2
2
ý
ý
sin 2 ñ
2ð (0.6094)
sin 2 ñ
D ý Gd ,max ý 3.282
Prob. 13.28
(a) Umax = 1
Prad ò Ud
ý
4ð
4ð
U ave ý
ý
1
4ð
ò ò sin
2
2ñ sin ñ dñ dö
ð
ý
1
2
(2ð ) ò ø 2sin ñ cos ñ ù d ø cos ñ ù
4ð
0
ð
ý 2 ò øcos4 ñ cos2 ñ ùd ø cos ñ ù
0
ù cos5 ñ cos3 ñ ù
ý 2ú
3 úû
û 5
ð
0
ù 2 2ù 8
ý 2ú ú ý
û 5 3 û 15
U ave ý 0.5333
Dý
U max
ý 1.875
U ave
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(b)
Umax = 4
1
4
U ave ý
Ud ý
ò
4ð
4ð
ý
1
ð
ð
ð
sin ñ
ò ò sin
2
ò dö ò cos ecñ dñ ý
0
ð
2
ñ
dñ dö
ð
ln 3
ð
3
Uave = 0.5493
Dý
U max
16
ý
ý 9.7092
U ave 3ln 3
(c ) Umax = 2
1
1
2sin 2 ñ sin 2 ö sin ñ dñ dö
Ud ý
ò
4ð
4ð ò ò
ð
ð
1
2
ý
sin ö dö ò ø1 cos2 ñ ùd ø cos ñ ù
ò
2ð 0
0
U ave ý
ö
1 ð ö cos3 ñ
ý
. ÷
cos ñ ÷
2ð 2 ø 3
ø
ð
ý
0
1ù 2
ù 1
2
úû ý 3
4 úû 3
Uave = 0.333
Dý
U max
ý6
U ave
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Prob. 13.29
Gd (ñ , ö ) ý
(a)
U ave ý
1
4ð
10
ý
4ð
2ð
U (ñ , ö )
U ave
ð
ò ò 10sin ñ sin
ö ñ
2
ö sin ñ dñ dö
ý0 ý0
2ð
ò sin
0
ð
2
ö dö ò sin 2ñ dñ
0
10 1 ö
sin 2ñ ö 2ð 1 ö
sin 2ö ö ð
֖
÷
÷ö
÷
4ð 2 ø
2 ø 0 2ø
2 ø0
10
5ð
ý
(2ð 0)(ð 0) ý
16ð
4
2
40sin ñ sin ö
Gd (ñ , ö ) ý
ý 2.546sin ñ sin 2 ö
5ð
D ý Gd . max ý 2.546
ý
(b)
U ave ý
1
4ð
ð
ð
ò ò 2sin
ö ñ
2
ñ sin 3 ö sin ñ dñ dö
ý0 ý0
ð
ð
ð
ù
2
2 ù
3
3
2
ý
sin ö dö ò sin ñ dñ ý
ú ò (1 cos ö )d ( cos ö ) ú
ò
4ð 0
4ð û 0
0
û
2
2
öðù 1 ö4ö
1 ùö cos3 ö
16
ý
cos ö ÷ ú ý
ú÷
÷ ÷ ý
2ð ûø 3
ø 0 û 2ð ø 3 ø 18ð
18ð
Gd (ñ , ö ) ý
2sin 2 ñ sin 3 ö ý 2.25ð sin 2 ñ sin 3 ö
16
D ý Gd .max ý 7.069
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U ave
(c)
ý
1
ý
4ð
2ð
ð
ò ò 5(1 sin
ö ñ
2
ñ sin 2 ö ) sin ñ dñ dö
ý0 ý0
ð
ð
2ð
2ð
ù
5 ù
3
d
d
d
sin
sin
sin 2 dö ú
ñ
ñ
ö
ñ
ñ
úò
ò
ò
ò
4ð û 0
0
0
0
û
ù
ð 4 ö sin 2ö 2ð ù
) ú
ú 2ð ( cos ñ ) (
0 3 2
0û
4
û
5 ù
4 ù 20
4
ý
ð
ð ý
4ð úû
3 úû 3
3
Gd (ñ , ö ) ý 5(1 sin 2 ñ sin 2 ö ) ý 0.75(1 sin 2 ñ sin 2 ö )
20
D ý Gd .max ý 1.5
ý
5
4ð
Prob. 13.30
U max ý 4
1
1
ö
Ud ý
4sin 2 ñ sin sin ñ dñ dö
ò
òò
4ð
4ð
2
ð
ð
ð
ö
ö ð
1
1
ý ò sin 3 ñ dñ ò sin dö ý ò (1 cos 2 ñ )d ( cos ñ )(2 cos )
ð 0
ð 0
2
2 0
0
U ave ý
1 4
8
( )(2) ý
ð 3
3ð
U
3ð
D ý max ý 4
ý 4.712
U ave
8
ý
Prob. 13.31
P
ave ý
Prad
I 2 sin 2 ñ
| Er |2
ar ý o
ar
2
2 r 2
I2
ý o
2
ý
2ð
I o2
sin 2 ñ 2
sin
(2ð ) ò (1 cos 2 ñ )d ( cos ñ )
r
d
d
ñ
ñ
ö
ý
òò r 2
240ð
0
ð I2
I o2 cos3 ñ
I2
(
cos ñ ) ý o (1/ 3 1 1/ 3 1) ý o
0 120
120
3
90
I o2 ý 90 Pave ý 90 50 103
I o ý 2.121 A
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Prob. 13.32
U (ñ , ö ) ý r 2 Pave ý r 2
U (ñ , ö ) ý
| Eö |2
2
ý
r 2 1400ð 4 I o2 S 2 sin 2 ñ
,
r 2ü 4
2
where =120ð
120ð 3 I o2 S 2 sin 2 ñ 60ð 3 I o2 S 2 sin 2 ñ
ý
2ü 4
ü4
Prob. 13.33
Pave ý
E2 ñ 2I 2
ý
2 2 r 2
Rrad ý
2 Prad 2 Pave
8ð r 2 ñ 2 I 2 4ðñ 2 4ðñ 2 ñ 2
2
4
ý
ý
ý
ý
ý
ð
r
120ð 30
I2
I2
I 2 2 r 2
Prob. 13.34
According to eq. (13.10),
ð
Prad ý k ò sin 3 ñ dñ ý
0
4k
,
3
where k is a constant.
ù cos3 ñ
ùð / 3
1 1 ù 5
ù 1
3
ý
ñ
ñ
sin
d
k
ú 3 cos ñ ú 0 ý k ú 3 8 2 3 1ú ý 24 k
ò0
û
û
û
û
5
k
5
24
ý
ý 0.1562
Fraction =
4k
32
3
ð /3
'
ýk
Prad
Prob. 13.35
This is similar to Fig. 13.10 except that the elements are z-directed.
E s ý E s1 E s 2 ý
where r1 r
Es ý
ù
jò I o dl ù
e j ò r1
e j ò r2
a
añ 2 ú
sin
sin
ñ
ñ
1
2
ñ1
ú
r1
r2
4ð û
û
d
cos ñ ,
2
r2 r
d
cos ñ ,
2
ñ1 ñ 2 ñ ,
jò I o dl
sin ñ añ ùûe j ò d cosñ / 2 e j ò d cosñ / 2 ùû
4ð
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añ 1 añ 2 ý añ
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Es ý
jò I o dl
1
sin ñ cos( ò d cos ñ )añ
2ð
2
Prob. 13.36
ù1
ù
(a) AF = 2 cosú ø ò d cos ñ ñ ù ú ,
û2
û
ñ ý 0,
òd ý
2ð
ü ý 2ð
ü
AF = 2 cos( ð cos ñ )
(b)Nulls occur when
cos( ð cos ñ ) ý 0
or
ð cos ñ ý ð / 2, 3 ð / 2,...
ñ ý 60 o ,120 o
(c) Maxima and minima occur when
df
ý0
dñ
sin( ð cos ñ ) ð sin ñ ý 0
i.e. sin ñ ý 0
ñ ý 0 o ,180 o
cosñ ý 0
ñ ý 90 o
or
ñ ý 0 o ,90 o ,180 o
(d ) The group pattern is sketched below.
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Prob. 13.37
ù1
ù
f ø ñ ù ý cosú ø ò d cos ñ ñ ù ú
û2
û
(a) ñ ý ð 2 , òd ý
ø
2ð
. ü ý 2ð
ü
f ø ñ ù ý cos ð cos ñ ð 4
ù
ð 3ð
,... or ñ ý 75.5 o ,138.6 o
Nulls occur at ð cos ñ ð 4 ý ,
2
2
Maxima occur at
f
ý0
ñ
sin ñ ý 0
ñ ý 0 o ,180 o
ðö
ö
Or sin ÷ ð cos ñ ÷ ý 0
ñ ý 41.4 o ,104.5 o
4ø
ø
With f max ý 0.71,1 .
Hence the group pattern is sketched below.
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(b)
ñý
3ð
2ð ü ð
,ò d ý
. ý
ü 4 2
4
3ð ö
öð
f ø ñ ù ý cos÷ cos ñ
÷
ø4
8ø
Nulls occur at
3ð
ð
ð 3ð
cos ñ
ý ,
,...
4
8
2
2
ñ ý 60 o
3ð
öð
Minima and maxima occur at sin ñ cos ÷ cos ñ
8
ø4
ö
÷ý0
ø
i.e. ñ ý 0 o ,180 o f ø ñ ù ý 0.383,0.924
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2 ð 3ü 3 ð
.
ý
ü 4
2
ö
ö 3ð
f ø ñ ù ý cos÷
cos ñ ÷
ø
ø 4
3ð
ð 3ð
cos ñ ý ,
,... ñ ý 48.2 o ,131.8 o
It has nulls at
4
2
2
df
ö
ö 3ð
It has maxima and minima at
ý 0 sin ñ sin÷
cos ñ ÷ ý 0
ø
ø 4
dñ
o
o
o
i.e. ñ ý 0 ,180 f ø ñ ù ý 0.71,1 ,
ñ ý 90 , f øñ ù ý 1
(c) ñ ý 0 , ò d ý
ñ = 0 o,
ñýo,
ñ = 180o,
Prob. 13.38
ù1
ù
(a) For N = 2, f ø ñ ù ý cosú ø ò d cos ñ ñ ù ú
û2
û
ü
ñ ý 0 ,d ý
4
ù 1 ö 2ð ü
ö
öð
öù
f ø ñ ù ý cosú ÷
. cos ñ 0÷ ú ý cos÷ cos ñ ÷
ø
ø
ø
ø
4
û
û2 ü 4
Maxima and minima occur at
d ù öð
öù
cos÷ cos ñ ÷ ú ý 0
ú
øû
dñ û ø 4
öð
ö
sin ñ sin÷ cos ñ ÷ ý 0
ø4
ø
sin ñ ý 0 ñ ý ð ,0 and f ø ñ ù ý 0.707
öð
ö
sin÷ cos ñ ÷ cos ý 0 ñ ý 90 o , f ø ñ ù ý 1
ø4
ø
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abs(f) = 1/ö2=0.707
abs(f) = 1,
abs(f) = 1/ö2=0.707
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Nulls occur as
ð
4
cosñ ý
ð
2
,
3ð
,... (No Solution)
2
The group pattern is sketched below.
(b) For N = 4,
AF ý
Now,
sin 2ø ò d cos ñ 0 ù
1
sin ø ò d cos ñ 0 ù
2
sin 4ñ 2 sin 2ñ cos 2ñ
ý
ý 4 cos 2ñ cos ñ
sin ñ
sin ñ
ö
ö1
AF ý 4 cosø ò d cos ñ ù cos÷ ò d cos ñ ÷
ø
ø2
ö 1 2ð ü ö
ö
ö 2ð ü
f ø ñ ù ý cos÷
. cos ñ ÷ cos÷
÷ cos ñ
ø 2 ü 4ø
ø
ø ü 4
ö
öð
ö
öð
ý cos÷ cos ñ ÷ cos÷ cos ñ ÷
ø
ø4
ø
ø2
The plot is shown below.
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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.
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Prob. 13.40
(a) The resultant pattern is obtained as follows.
I ð 0o
I ð 0o
I ð 0o
I ð 0o
ü/2
ü/2
ü/2
ü
x
=
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(b) The array is replaced by by
+
ü
4
ð0 o
where + stands for
+
ð ð /2
ðð
ð0
Thus the resultant pattern is obtained as shown.
o
I ð 0o
I ð 90o
I ð 270o
I ð180o
x
ü/4
=
ü/4
ü/4
ü/2, ð
=
x
Prob. 13.41
Gd (dB) ý 20dB ý 10 log10 Gd
Gd ý 102 ý 100
3 108
c
ý 3 102
üý ý
9
f 10 10
9 104
ü2
100 ý 7.162 102 m 2
Ae ý
Gd ý
4ð
4ð
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Prob. 13.42
Ae ý
ý
Pr
Pr
ý
Pave | Er |2
ý
2
2 Pr
| Er |2
2 120ð 2 106 48ð
ý
ý 0.6031 m 2
6
2
25 10 10
250
Prob. 13.43
Friis equation states that
Pr
ö ü ö
ý Gr Gt ÷
÷
Pt
ø 4ð r ø
üý
2
c
3 108
ý
ý 1.5 m,
f 200 106
Gt (dB) ý 15dB ý 10 log10 Gt
r ý 238,857 1.609 103 ý 3.843 108
Gt ý 1015/10 ý 31.623
2
8
9
ö 4ð r ö Pr ö 4ð 3.843 10 ö 4 10
Gr ý ÷
ý
ý 34.55 1010
÷
÷
÷
3
P
1.5
120
10
ü
ø
ø t ø
ø
Gr (dB ) ý 10 log10 Gr ý 10 log10 34.55 1010 ý 115.384 dB
2
Prob. 13.44
Using Frii’s equation,
ù ü ù
Pr ý Gr Gt ú
Pt
û 4ð r úû
2
ù 4ð r ù Pr
Pt ý ú
û ü úû Gr Gt
c 3 108
ý 0.1,
üý ý
f 3 109
2
r ý 42 km
Gt (dB) ý 10 log10 Gt ý 25
Gt ý 102.5 ý 316.23
Gr (dB ) ý 10 log10 Gr ý 20
Gt ý 102 ý 100
2
ö 4ð 42 103 ö 3 106
ý 2.642 kW
Pt ý ÷
÷
0.1
ø
ø 31623
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Prob. 13.45
Gdt ý 10 4 , Gdr ý 10 3.2 ý 1585
üý
1
c
3 10 8
ý 0.02m ý
ý
50
f 15 10 9
2
2
0.02
ö
ö
ö ü ö
4
Pr ý Gdr Gdt ÷
÷ P ý 10 ø 1585ù ÷
÷ 320
ø 4ðr ø t
ø 4 ð 2.456741 107 ø
ý 2.129 1011 W ý 21.29 pW
Prob. 13.46
ö ü ö
Pr ý Gdt Gdr ÷
÷ Pt
ø 4ð r ø
3 108
c
üý ý
ý 15 m,
f 20 106
2
Gdt ý Gdr ý 1.64
2
3
Pr ö 4ð r ö
6 ö 4ð 80 10 ö
ý
0.5
10
Pt ý
÷
÷ ý 835.025 W
÷
÷
Gdt Gdr ø ü ø
ø 1.64 15 ø
2P
2P
2 835.025
ý 73
ý 22.8774
But Rrad ý rad
I o2 ý rad ý
2
73
73
Io
2
I o ý 4.783 A
Prob. 13.47
30 dB ý log
Pt
P
t ý 10 3 ý 1000
Pr
Pr
2
3
ö Gd ö
ö
÷ Pt ý Pt ÷
÷
ø 50 4 ð 12 ø
ø 800 ð ø
But Pr ý ø Gd ù 2 ö÷
2
ö 1 ö
Pr
1
ö Gd ö
ý
ý÷
÷
÷ ý
÷
ø 800 ð ø
Pt 1000 ø 10 10 ø
or Gd ý
800 ð
10 10
2
2
ý 79.476
Gd ý 10 log79.476 ý 19 dB
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Prob. 13.48
(a) Pi ý
E
2
2 o
Ei ý
ý
Prad Gd
4 ðr
2
Ei ý
240 ðPrad Gd
4 ðr 2
1
1
60 Prad Gd ý
60 200 103 3500
120 103
r
ý 1708
.
V/m
2
Ei ó
ý
4ð r 2
1708
. 28
ý 11.36 ý V / m
4 ð 14400 10 6
(b)
Es ý
(c)
1708
. 2
ø 8 ù ý 30.95 mW
Pc ý Pi ó ý
240 ð
(d)
2
ø 11.36 ù 2 10 12
E
Pi ý
ý
ý 1712
.
10 13 W / m2
2 o
240 ð
üý
3 10 8
15 10 8
ý 0.2m, A2 r ý
ü2 G 0.04 3500
ý
4ð
4ð
Pr ý Pa Aer ý 1712
.
10 13 1114
. ý 1.907 10 12
or Pr ý
ø üGd ù 2 óPrad
ø 4ðù 3 r 4
ý
ø 0.2 3500 ù 2 8 2 10 5
ø 4ðù 3 12 4 10 16
ý 1.91 10 12 W
Prob. 13.49
(ü Gd ) 2 ó Prad
Pr ý
(4ð )3 r 4
Gd (dB) ý 30dB ý 10 log10 Gd
Gd ý 103
üý
c 3 108
ý
ý 0.075 m
f 4 109
Pr ý
(0.075 103 ) 2 12 80 103
ý 272.1 pW
(4ð )3 (10 103 ) 4
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Prob. 13.50
2d 2 160 103 32
ý
ý 104 ý 1.067 ms
u
3 108
3
8
AóG P
3 10
ý 0.075m, Pr ý e d2 rad
(b) ü ý c / f ý
9
4 10
(4ð r ) 2
(a)
tr ý
But
Ae ý
ü 2Gd
4ð
Gd ý
4ð Ae
ü2
2
2
2
ö Ae ö
ö
ö
Pr ý 4ðó ÷
Prad ý 4ð 5 ÷
(60 103 ) ý 2.59 1014 W
2 ÷
2
6 ÷
ø 0.075 4ð 160 10 ø
ø 4ð r ü ø
ù ü 2Gd2ó Prad ù
rýú
ú
3
û (4ð ) Pr û
(c )
1/ 4
1/4
ý
ù ü 2ó (4ð ) 2 Ae2 Prad ù
ú
ú
3
ü4
Pr û
û (4ð )
ý
1/4
ù 5 4
60 103 ù
ýú
12 ú
2
û 4ð (0.075) 8 10 û
ý 38.167 km
Prob. 13.51
r4
Prad ý Pr
k
4
(2r )
r4
'
If R ý 2r ,
Prad
Pr ý 16 Pr ý 16 Prad
ý
k
k
i.e. the transmitted power must be increased 16 times.
kP
Pr ý rad
r4
Prob. 13.52
2
Prad ý
4 ð ö 4 ðr1r2 ö Pr
÷
÷
Gdt Gdr ø ü ø ó
But Gdt ý 36 dB ý 10 3.6 ý 39811
.
Gdr ý 20dB ý 102 ý 100
üý
c 3 10 8
ý
ý 0.06
f
5 10 9
r1 ý 3 km , r2 ý 5 km
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ù Ae2ó Prad ù
ú
ú
2
û 4ðü Pr û
1/ 4
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2
Prad
ö 4ð 15 10 6 ö 8 10 12
4ð
ý
÷
÷
39811
2.4
. 100 ø 6 10 2 ø
ý 1038
.
kW
Prob. 13.53
(a)
ð fL ð 300 106 50 109
ý
ý 2.356
Fý
R
20
IL ý 10 log10 (1 F 2 ) ý 10 log10 (1 2.3562 ) ý 8.164 dB
(b)
F ý ð fRC ý ð 300 106 10 103 60 1012 ý 180ð ý 565.5
IL ý 10 log10 (1 F 2 ) ý 10 log10 (1 565.52 ) ý 55.05 dB
Prob. 13.54
Zg
I1
I2
+
Vg
+
ù A Bù
úC
ú
û ù A DBû ù
ú C Dú
û
û
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 ý I 2 Z L ý
ý
I1Z L
CZ L D
Vg Z L
ö
V ö
÷÷ Z g 1 ÷÷øCZ L D ù
I1 ø
ø
ý
Vg Z L
ö
AV2 BI 2
÷÷ Z g
CV2 DI 2
ø
ö
÷÷øCZ L D ù
ø
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ý
ý
Vg Z L
ö
AZ L B ö
÷øCZ L D ù
÷÷ Z g
CZ L D ÷ø
ø
øZ øCZ
g
V2 ý
Vg Z L
L
D ù AZ L B ù
Vg Z L
øZ
g
ZL ù
Ratio and modulus give
øZ g øCZ L D ù AZ L B ù
V2
ý
V2
Zg ZL
Insertion loss =
IL =
20 log10
øZ g øCZ L D ù AZ L B ù
V2
ý 20 log10
V2
Zg ZL
which is the required result
Prob. 13.55
l
103
=
= 17.1 m /km
(a) Rdc =
óS
0.96 104 6.1 107
(b) Rac =
ô ý
Rac =
l
,
ô wó
1
=
ð fýó
ð a2 = 0.8 x 1.2 = 0.96 or a = 0.5528
1
ð 6 106 4ð 107 6.1107
ý
1
12.1 ð 103
1000 12.1 ð 103
= 51.93
1.2 102 6.1107
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Prob. 13.56
SE ý 20 log10
Ei
6
ý 20 log10
ý 20 log10 (3 105 )
6
Eo
20 10
ý 109.54 dB
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lOMoARcPSD|23609374
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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
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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
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
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
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(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)
ù 0.6357 01643
.
0.8ù
ú
ú
ú 01643
.
0.4357 0.6ú
úû 0.8
0.6
14
. úû
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)
ù 0.3607
0.05 0.4107ù
ú
ú
ú 0.05
0.7
0.75 ú
úû
úû 0.4107 0.75 11607
.
The global coefficient matrix is given by
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ù C (1) 11 C11 ( 2 )
ú
C21( 2 )
ú
C
ú C21(1) C31( 2 )
ú
C31(1)
úû
C12 ( 2 )
C12 (1) C13 ( 2 )
C22 ( 2 )
C32 ( 2 )
C23 ( 2 )
C22 (1) C33 ( 2 )
0
C32 ( 2 )
C13 (1) ù
ú
0 ú
C23 (1) ú
ú
C33 (1) úû
ù 0.9964 0.05 0.2464 0.8 ù
ú 0.05
0.7
0 úú
0.75
ú
ú 0.2464 0.75 1.596
0.6 ú
ú
ú
0
1.4 û
0.75
û 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;
ù 1.1333 - 0.0778 - 1.0556 ù
= úú - 0.0778 0.2259 - 0.1481úú
úû - 1.0556 - 0.1481 1.2037 úû
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 ;
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C
(2)
ù 0.5933 -0.9800 0.3867 ù
ú -0.9800 2.0400 -1.0600 ú
ú
ú
úû 0.3867 -1.0600 0.6733 úû
The global coefficient matrix is
ù C (1) 11
ú (1)
C21
C ú
ú 0
ú (1)
úû C31
C
C12 (1)
C22
(1)
C11
(2)
0
( 2)
C12
(1)
C32 C31( 2 )
ù 1.1333
ú - 0.0778
ú
ú 0
ú
û - 1.0556
C12
(2)
C22 ( 2 )
C32 ( 2 )
ù
ú
C23 C13 ú
ú
C23 ( 2 )
(1)
(2) ú
C33 C33 úû
- 0.0778
0
0.8193 - 0.9800
- 0.9800 2.0400
0.2385 - 1.0600
C13 (1)
(1)
(2)
- 1.0556ù
0.2385úú
- 1.0600 ú
ú
1.8770 û
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
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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
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FD
11.25
15.02
4.705
9.545
18.84
9.659
15.85
20.87
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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.
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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) ùûV ( x ) V ( x ) 2 ( x 1) ùû
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
2x
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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
ò 2 ò ò z 2
The equivalent finite difference expression is
V ( ò o ò , zo ) 2V ( ò o , zo ) V ( òo ò , zo ) 1 V ( ò o ò , zo ) V ( òo ò , zo )
(ò ) 2
2ò
òo
V ( ò o , zo z) 2V ( ò o , zo ) V ( ò o , zo z)
0
( z) 2
If z ò h, rearranging terms gives
V ( ò o , zo )
(1
1
1
h
V ( ò o , zo h) V ( ò o , zo h) (1
)V ( ò h, zo )
4
4
2ò o
h
)V ( ò h, zo )
2ò o
as expected.
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Prob. 14.5
2V
2V 1 V
1 2V
0,
ò 2 ò ò ò 2 ö 2
(1)
Vm1n 2Vm n Vm1n
2V
,
ò 2
( ò ) 2
(2)
Vm n 1 2Vm n Vm n 1
2V
,
( ö ) 2
ö 2
(3)
V
ò
m ,n
V n m 1 V n m 1
2 ò
.
(4)
Substituting (2) to (4) into (1) gives
V
2
=
1
( ò ) 2
V n m 1 V n m 1
m ò (2 ò )
+
Vm1n 2Vm n Vm1n
( ò ) 2
+
Vm n 1 2Vm n Vm n 1
ù
1
1
1
n
n
n
n 1
2Vm n Vm n 1
ú (1 2m )Vm1 2Vm (1 2m )Vm1
2 (Vm
m
(
ö
)
û
as required.
Prob. 14.6
Vo
(m ò ö ) 2
V1 V2 V3 V4 10 40 50 80
25V
4
4
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ù
)ú
û
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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 öV
ö
Va ÷ a 200 ÷
4ø 2
ø
Va 400 / 7 51.143 Vc
Va
28.57
2
Prob. 14.9
(a) We follow Example 6.5 with a=b.
ö nð x ö
ö nð y ö
sin ÷
÷ sinh ÷
÷
4Vo
ø a ø
ø a ø 4Vo
V V1 V2
õ
n sinh(nð )
ð n odd
ð
ö nð y ö
ö nð x ö
sin ÷
÷ sinh ÷
÷
ø a ø
ø a ø
õ
n sinh(nð )
n odd
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(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
50 109
h2 òs
104
0.18ð 0.5655
k
109
õ
36ð
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,
ù 4 1 1 0 ù ùV1 ù ù0.5655ù
ú
ú úV ú ú
ú
ú 1 4 0 1ú ú 2 ú ú0.5655ú
ú 2 0 4 1ú úV3 ú ú0.5655ú
ú
úú ú ú
ú
û 0 2 1 4 û úûV4 úû û0.5655û
Using a calculator or MATLAB, we obtain
V1 V2 0.3231 V, V3 V4 0.4039 V
Prob. 14.11
(a)
1
0
0ù
ù4 1 0
ú 1 4 1
0
1
0ú
ú
ú
ú0
1 4 0
0
1ú
ú
ú
ú 1 0 0 4 1 0 ú
ú0
1
0
1 4 1 ú
ú
ú
0
1
0
1 4û
û0
[A]
(b)
ú Va ú ù 200ù
úV ú ú
ú
ú b ú ú 100ú
ú Vc ú ú 100ú
ú ú ú
ú
ú Vd ú ú 100ú
ú Ve ú ú 0 ú
ú ú ú
ú
úûV f úû û 0 û
[B]
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(1)
(2)
(3)
(4)
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ù 4 1 0 1 0 0 0 0 ù ùV1 ù ù 30 ù
ú
ú úV ú ú
ú
ú 1 4 1 0 1 0 0 0 ú ú 2 ú ú 15 ú
ú 0 1 4 0 0 1 0 0 ú úV3 ú ú 30 ú
ú
úú ú ú
ú
ú 1 0 0 4 1 0 1 0 ú úV4 ú ú 7.5ú
ú 0 1 0 1 4 1 0 1 ú úV5 ú ú 0 ú
ú
úú ú ú
ú
ú 0 0 1 0 1 4 0 0 ú úV6 ú ú 7.5ú
ú 0 0 0 1 0 0 4 1 ú úV ú ú 0 ú
ú
úú 7ú ú
ú
úû 0 0 0 0 1 0 1 4 úû ûúV8 ûú úû 0 úû
[A]
[B]
Prob. 14.12 (a) Matrix [A] remains the same. To each term of matrix [B], we add
h2 òv / õ .
(b) Let x y h 0.25 so that nx = 5= ny.
òv x( y 1)109
36ð x( y 1)
109 / 36ð
õ
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:
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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
2
0
12.5
40.625
12.5
3.12
13.281
40.82
3
6.25
24.22
43.55
14.84
7.03
24.46
43.62
4
9.77
25.83
43.96
16.70
10.29
25.98
43.99
10.63
26.15
44.04
17.73
10.93
26.23
44.06
Prob. 14.14
1
c2
j
j 1
m ,n
m ,n 1
j 1 m ,n 2
( t ) 2
j m ,n 1 2
( z) 2
j
j
m ,n
j
m 1,n
j m1,n 2
( x) 2
j
m ,n
m ,n
If h x z , then after rearranging we obtain
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5
10.97
26.25
44.06
17.97
11.05
26.28
44.07
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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) 2
x
t
V ( x, t t ) 2V ( x, t ) V ( x, t t )
(t ) 2
2
ö t ö
V ( x, t t ) ÷ ÷ V ( x x, t ) 2V ( x, t ) V ( x x, t 2V ( x, t ) V ( x, t t )
ø x ø
or
V (i, j 1) V (i 1, j ) v(i 1, j ) 2(1 )V (i, j ) V (i, j 1)
2
ö t ö
where ÷ ÷ . Applying the finite difference formula derived above, the following
ø x ø
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.
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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')
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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
ðõ
cosh 1
d
2a
ð 109 / 36ð
cosh 1 5
12.12 pF/m
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Prob. 14.18
y
ñ
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 ñ , k=N+1, N+2, …,2 N,
xk = [h + (k-1/2)] cos ñ , k = N+1,N+2,… 2N
where h is determined from the gap g as
g
h
2 sin ñ / 2
(3)Calculate the matrices [V] and [A] with the following elements
ü Vo , k 1,..., N
Vk ý
þ Vo , k N 1,...2 N
ü
,i j
ÿ
Aij ý 4ðõ Rij
ÿþ 2 ln / a , i j
where Rij
( xi x j ) 2 ( yi y j ) 2
(4) Invert matrix [A] and find [ ò ] = [A]-1 [V].
(5) Find the charge Q on one conductor
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Q
õò
N
k
õ òk
k 1
(6) Find C = |Q|/2Vo
Taking N= 10, Vo = 1.0, a program was developed to obtain the following result.
ñ
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 õ Aij òi
j 1
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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.
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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
òin 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 2ðõ / ln( / a ) 8.02607 1011 F/m and Z o 41559
where a = 1cm and = 2cm. The numerical solution is shown below.
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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] = [A]-1 [B], [ ò v ] = [q]/(ht ), C
õq
i 1
i
10
The computed values of [ ò v ] and C are shown below.
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I
òvi (106 )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
ù 0.5909 0.1364 0.4545 ù
C úú 0.1364 0.4545 0.3182 úú
úû 0.4545 0.3182 0.7727 úû
A
(b)
P1 4, P2 4, P3 0, Q1 0, Q2 3, Q3 3
1
A ( P2Q3 P3Q2 ) 6
2
0 ù
ù 0.6667 0.6667
ú
C ú 0.6667 1.042
0.375úú
úû 0
0.375 0.375 úû
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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
5
4
4
[(1 / 4 3 / 2) 0 x y ]
( 125
. 2.5 y )
2
15
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
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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,
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A = (0 + 12 x 8)/2 = 48
Cij
C
(1)
1
[ P P QjQj ]
4 x 48 j i
ù 0.7969 0.125 0.6719 ù
úú 0.125 0.3333 0.2083úú
úû 0.6719 0.2083 0.8802 úû
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 ù
ù 1.2
ú
0.208 0.208úú
ú 0
úû 1.2 0.208 1.408 úû
ù C (1) 33
ú
C23 (1)
ú
C
ú 0
ú
(1)
úû C13
C23 (1)
C22 (1) C11( 2 )
C31( 2 )
C21(1) C21( 2 )
0
C13 ( 2 )
C33 ( 2 )
C23 ( 2 )
ù
C31(1)
(1)
(2) ú
C21 C12 ú
ú
C32 ( 2 )
(2)
(1) ú
C22 C11 úû
0
0.6719 ù
ù 0.8802 0.2083
ú 0.2083 1.533
1.2
0.125 úú
ú
ú 0
1.4083 0.2083ú
1.2
ú
ú
û 0.6719 0.125 0.2083 1.0052 û
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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
0.5ù
0
ù 4 0 4 ù ù 0.5
1ú
ú
ú
C ú 0 4 4 ú ú 0
0.5 0.5úú
8
úû 4 4 8 úû úû 0.5 0.5
1 úû
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
0.5ù
0
ù 8 0 8ù ù 0.5
1 ú
ú
ú
(2)
C ú 0 8 8ú ú 0
0.5 0.5úú
16
úû 8 8 16 úû úû 0.5 0.5
1 úû
The global coefficient matrix is
ù C11 C12
úC
C22
C ú 21
úC31 C32
ú
ûúC41 C42
C13
C23
C33
C43
C14 ù ùC11(1) C11(2)
ú
(2)
C24 úú ú C21
(1)
C34 ú úC21 C31(2)
ú ú
C44 ûú ûú C31(1)
C12(2)
C12(1) C13(2)
C22(2)
C32(2)
C23(2)
(1)
C33(2)
C22
0
C32(1)
C13(1) ù
ú
0 ú
C23(1) ú
ú
C33(1) ûú
0.5 0.5ù
0
ù 1
ú 0
0.5 0.5
0 úú
ú
ú 0.5 0.5 1.5 0.5ú
ú
ú
0
1 û
0.5
û 0.5
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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
C
(1)
1
[ P P QjQj ]
4A j i
ù 5 4 1ù
1ú
ú
ú 4 5 1ú
6
úû 1 1 2 úû
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)
ù 4 2 2ù
1ú
ú
ú 2 5 3ú
8
úû 2 3 5 úû
The global coefficient matrix is
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ù C (1) 11
ú
C12 (1)
ú
C
ú 0
ú
(1)
úû C13
C12 (1)
C22 (1) C22 ( 2 )
C23 ( 2 )
C23 (1) C21( 2 )
0
C23 ( 2 )
C33 ( 2 )
C31( 2 )
ù
C13 (1)
(1)
(2) ú
C23 C21 ú
ú
C31 ( 2 )
ú
C33 (1) C11( 2 ) úû
0
01667
.
ù 0.8333 0.667
ù
ú 0.6667 14583
0.375 0.4167ú
.
ú
ú
ú 0
0.375 0.625
0.25 ú
ú
ú
0.4167 0.25
0.833 û
.
û 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]
ù 10ù
ú 18ú
ú ú
The result is V ú 30ú
ú 20ú
úû úû
From this,
V2 = 18 V, V4 = 20 V
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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
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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
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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
lOMoARcPSD|23609374
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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
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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
lOMoARcPSD|23609374
509
Prob. 14.34
For element 1, the local numbering 1-2-3 corresponds with nodes with V1 , V2 , and
V3.
Vo
1
Coo
4
4
õVC
i 1
Coo õ Coj
j 1
(e)
i
io
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.
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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
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lOMoARcPSD|23609374
511
On the interface,
õ1
1
õ2
3
,
2(õ1 õ 2 ) 8
2(õ1 õ 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
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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
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lOMoARcPSD|23609374
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
V1
V2
V3
V4
V5
0
1
2
3
4
5…
100
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
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