Solucionario Clayton R.Paul (Eletromagnetismo para Engenheiros)

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Solucionario 
 
Eletromagnetismo para Engenheiros 
 
Clayton R. Paul 
1-1
Chapter 1 
Problem Solutions 
111. .
(a) 25 10 250 10 2504 3× = × =Ω Ω Ωk
(b) 0 035 10 35 10 3504 2. .× = × =Ω Ω Ω
(c) 0 00045F 450 10 4506. = × =− F Fµ
(d) 0 003 10 0 3 107 9. .× = × =− −F F 0.3nF
(e) 0 005 10 50 10 502 6. × = × =− −H H Hµ
112. .
(a) 30 12in 254cm 1 1 48 28miles 5280ft
1mile 1ft 1in
m
100cm
km
1000m
km× × × × × =. .
(b) 1 12in 1000mils 12 000ft
1ft 1in
mils× × = ,
(c) 100yds 12in 2 54cm 1 9144× × × × =3ft
1yd 1ft 1in
m
100cm
m. .
(d) 5mm 100cm
254cm
1000 19685mils× × × × =1m
1000mm 1m
1in mils
1in.
.
(e) 20 100cm
254cm
1000 0 7874µ
µ
m 1m
10 m 1m
1in mils
1in
mils6× × × × =.
.
(f) 880yds 12in 2 54cm 1 804 67× × × × =3ft
1yd 1ft 1in
m
100cm
m. .
121. .
(a) λ = × =3 10 33333
8 m s
90 Hz
km = 2,071.2 miles. 
(b) λ = × =3 10 300
8 m s
10 Hz
km = 186.41 miles3
(c) λ = ×
×
=
3 10 85714
8 m s
350 10 Hz
m = 0.533 miles3 .
(d) λ = ×
×
=
3 10
2
250
8 m s
1 10 Hz
m = 820.2 ft6.
(e) λ = ×
×
=
3 10
35
8 57
8 m s
10 Hz
m = 28.12 ft6 .
1-2 
(f) λ = ×
×
=
3 10
110
2 73
8 m s
10 Hz
m = 8.95 ft6 . 
(g) λ = ×
×
=
3 10
335
89 55
8 m s
10 Hz
cm = 2.94 ft6 . 
(h) λ = ×
×
=
3 10
6
5
8 m s
10 Hz
cm = 1.97 in9 
(i) λ = ×
×
=
3 10
45
6 67
8 m s
10 Hz
mm = 262.5 mils9 . 
12 2. . 
(a) 
( )
λ
λ λ
λ
=
×
= ×
× × × ∴ = =
3 10 5 10
5 10 0 0161 1
62
8
6
6
m s
60 Hz
m 50 miles = 80.46 km
80.46 10 m = ? m3
length
1 244 344 1 24 34
? .
 
(b) 
( )
λ
λ
λ
=
×
×
=
× ∴ =
3 10 600
152 4 600 0 254
8 m s
500 10 Hz
m 500feet = 152.4 m
m = ? m
3
length
. ? .124 34 124 34
 
(c) 
( )
λ
λ
λ
=
×
×
=
× ∴ =
3 10 2 73 5
137 2 73 05
8 m s
110 10 Hz
m 4 feet = 1.37 m
m = ? m
6
length
. .
. . ? .123 124 34
 
(d) 
( )
λ
λ
λ
=
×
×
=
× ∴ =
15 10 75
508 75 0 677
8. .
. . ? .
m s
2 10 Hz
cm 2 inches = 5.08 cm
cm = ? cm
9
length
1234 124 34
 
 
 
 
1-3 
12 3. . 
(a) 
β ωβ λ
µ φ β
= × = = = × = =
= = =
−2 2 10 1 2 856 10 285 6
10 5 66
2 8. , . , .
.
rad
m
MHz, m
s
m,
s, = rad = 3781.5o
f v v
f
T d
v
d
 
(b) 
β ωβ λ
φ β
= = = = × = =
= = =
75 4 3 2 5 10 833
4 0 41 7 66
8. , . , .
. .
rad
m
GHz, m
s
mm,
inches = 0.102 m, ns, = rad = 438.92o
f v v
f
T d
v
d
 
(c) 
β ωβ λ
φ β
= = = = × = =
= = =
315 150 299 10 1995
20 20 4 19 2
8. , . , .
. .
rad
m
MHz, m
s
m,
feet = 6.1m, ns, = rad = 1100.2o
f v v
f
T d
v
d
 
(d) 
β ωβ λ
φ β
= × = = = × = =
= = =
−0126 10 3 15 10 50
50 0 54 1014
3 8. , . ,
. .
rad
m
kHz, m
s
km,
miles = 80.5 km, ms, = rad = 580.9o
f v v
f
T d
v
d
 
 2-1 
Chapter 2 
Problem Solutions 
211. . 
F m d1
2
= ω θsin , F mg2 =
sin
cos
θ
θ
, F F1 2= , ∴ =cosθ
ω
g
d2
, ω π π= × =2 2rpm
60
, 
( )cos
.
.
.θ
π
=
×
=
9 78
2 05
052 , ∴ =θ 603.
o 
 
 
 
 
 
 
 
 
 
212. . 
W 45o× = ×cos sin200 θ , sin cosθ = W
200
45o , ∴ =θ 813. o , 
GS + Wsin45o = 200cosθ , GS = 169.71 mi
hr
 
 
 
 
 
 
 
 
�
�
�
d
m
F2F1
mg
�
W
GS
200 mi/hr
45°
2-2 
213. . 
4 3sinθ = , sinθ = 3
4
, ∴ =θ 48 6. o 
 
 
 
 
 
 
214. . 
F N mv
r
= =sinθ
2
, − = =W N mgcosθ , ∴ =tanθ v
rg
2
, 
v = × × =60 5280 1 88mi
hr
ft
1mi
hr
3600s
ft
s
, ∴ =tan .θ 016 , ∴ =θ 916. o 
 
 
 
 
 
2 31. . 
C A B= + , 
( ) ( )∴ = • = + • + = • + • + • = + +C A B AB2 2 22 2C C A B A B A A B B A B cosα 
But α θ= −180o AB and ∴ = −cos cosα θ AB 
∴ = + −C A B AB AB
2 2 2 2 cosθ which is the law of cosines. 
 
 
 
 
�
3 mi/hr
4 mi/hr
N
�
�AB
C
B
A
N
W
—W
F
N
�
2-3 
2 41. . 
A B C+ + = 0 , ( )A A B C A A A B A C× + + = = × + × + × =0 0
0
123 
( )A B a a× = − = − −AB ABC n o C nsin sinα θ180 where an is a unit normal into the 
page. Also A C a a× = =AC ACB n B nsin sinα θ . ∴ =AB ACC Bsin sinθ θ giving the 
law of sines: B C
B Csin sinθ θ
= . Similarly for ( )B A B C× + + = 0 and 
( )C A B C× + + = 0 . 
 
 
 
 
 
 
2 4 2. . 
(a) ( )A B• gives a scalar which cannot be “crossed with” a vector. (c) ( )A B• gives a 
scalar which cannot be “dotted with” a vector. (d) ( )B C• gives a scalar which cannot be 
added to a vector. 
2 51. . 
(a) ( ) ( ) ( )( )A a a a a a a= − + − − + − − = − +3 0 4 2 5 4 3 6 9x y z x y z 
(b) ( ) ( ) ( )A = + − + =3 6 9 11222 2 2 . m 
(c) a A a a aA x y zA
= = − +0 27 053 0 8. . . 
2 5 2. . 
(a) A B a a a+ = + −3 4 3x y z , (b) B C a a a− = − + −2 2 3x y z , (c) 
A B C a a a+ 3 - 2 = − + −x y z8 9 , (d) A = + + =2 3 1 374
2 2 2 . , (e) 
( )a B a a a a a aB x y z x y zB= = + − = + −16 2 41 0 41 0 82. . . , (f) A B• = 7 , (g) B A• = 7 , 
(h) B C a a a× = − − −x y z7 4 , (i) C B a a a× = + +x y z7 4 , (j) A B C• × = −19
C
C
B
A
B
A
A
�B
�B
�C �C
�B�C
2-4 
2 5 3. . 
(a) A B A B• = − − = − =2 2 3 3 cosθ . Therefore B A B
A
cos .θ = • = =3
14
0 8 , 
(b) cos .θ θ= • = − ⇒ =A B
A B
3
14 6
1091o , (c) 
unit vector = ×
×
=
− − −
+ +
= − − −
A B
A B
a a a
a a ax y z x y z
7 5
1 49 25
0115 0808 0 577. . . . 
2 5 4. . 
A B• = + − =α 2 3 0 ∴ =α 1 
2 5 5. . 
( ) ( ) ( )A B a a a× = − − + + + − =18 3 3 9 2 0β α β αx y z . Hence α = −3 and β = −6 . 
2 5 6. . 
A B a a a× = − − +14 9x y z gives a vector that is perpendicular to the planes containing 
both A and B and hence is perpendicular to both A and B. The length of this vector is 
( ) ( ) ( )− + − + =14 9 1 16 672 2 2 . . Hence 
( )C a a a a a a= − − + = − − +1016 67 14 9 8 4 54 0 6. . . .x y z x y z . Check that A C• = 0 and 
B C• = 0 . 
2 5 7. . 
( ) ( ) ( )B C a a a× = − + − + −B C B C B C B C B C B Cy z z y x z x x z y x y y x z so that 
( ) ( ) ( )( )
( ) ( )( )
( ) ( )( )
A B C a
a
a
× × = − − −
+ − − −
+ − − −
A B C B C A B C B C
A B C B C A B C B C
A B C B C A B C B C
y x y y x z z x x z x
z y z z y x x y y x y
x z x x z y y z z y z
 
( ) ( ) ( )
( )
B A C a a
a
• = + + + + +
+ + +
B A C A C A C B A C A C A C
B A C A C A C
x x x y y z z x y x x y y z z y
z x x y y z z z
 
( ) ( ) ( )
( )
C A B a a
a
• = + + + + +
+ + +
C A B A B A B C A B A B A B
C A B A B A B
x x x y y z z x y x x y y z z y
z x x y y z z z
 
Matching components we find that ( ) ( ) ( )A B C B A C C A B× × = • − • . 
2-5 
2 5 8. . 
( ) ( ) ( )A B a a a× = − + − + −A B A B A B A B A B A By z z y x z x x z y x y y x z so that 
( ) ( ) ( )( )
( ) ( )( )
( ) ( )( )
A B C a
a
a
× × = − − −
+ − − −
+ − − −
A B A B C A B A B C
A B A B C A B A B C
A B A B C A B A B C
z x x z z x y y x y x
x y y x x y z z y z y
y z z y y z x x z x z
 
Comparing terms to the result in the previous problem we see that 
( ) ( )A B C A B C× × ≠ × × . 
2 5 9. . 
The distance is ( )( ) ( ) ( )( )D = − − + − − + − − =3 1 1 2 5 4 10 2962 2 2 . . By integration we 
integrate D dlP
P
= ∫
1
2 . A straight line between the two points is governed by y x= − +3
4
5
4
 
and z x= −9
4
7
4
. Hence dldx dx= + −

 +



 =1
3
4
9
4
2 574
2 2
. and 
D dx
x
x
= ∫ =
=−
=
2 574 10 296
1
3
. . . 
2 510. . 
The surface is drawn below and lies in the yz plane at x=1. Hence the surface area is 
( )A dydz z dz
z
z
y
y z
z
z
= ∫ ∫ = −∫ =
=
=
=
= −
=
=
1
2
1
2 1
1
2
2 2 1 . Directly it is the area of a triangle of height 1 and 
base 2 or ( )( )A = =1
2
2 1 1. 
 
 
 
 
 
 
(1, 3, 2)y = 2z – 1
(1, 3, 1)(1, 1, 1)
y
z
x
2-6 
2 61. . 
Drawing the rectangular and cylindrical coordinate system axes as shown below we see 
that x r= cosφ , y r= sin φ , and z z= . From this we form 
( )x y r2 2 2 2 2
1
+ = +cos sinφ φ
1 244 344
 and hence r x y= +2 2 . Similarly we form 
y
x
r
r
= =
sin
cos
tanφφ φ . 
 
 
 
 
 
 
 
2 6 2. . 
Draw the coordinate system and use the right-hand rule. 
2 6 3. . 
Drawing the vector in the xy plane as shown below shows that A A Ax r= −cos sinφ φφ 
and A A Ay r= +sin cosφ φφ . 
 
 
 
 
 
 
 
 
P(r, �, z)
�
y
z
z
x
r
r
�
�
Ay
Ax
Ar
A�
y
x
2-7 
2 6 4. . 
At point P, φ π= =
3
60o . Hence Ax = +2 3cos sinφ φ =3.598 and 
Ay = −2 3sin cosφ φ =0.232 and Bx = −4 6cos sinφ φ =-3.196 and 
By = +4 6sin cosφ φ =6.464. Directly in cylindrical coordinates 
A B• = − − = −8 18 2 12 . In rectangular coordinates 
( ) ( )A B• = × − + × − = −3598 3196 0 232 6 464 2 12. . . . . 
2 6 5. . 
From problem 2.6.1 x r= cosφ , y r= sin φ , z=z. At P1 2 2 1, ,
π


 , x y z1 1 10 2 1= = =, , 
and at P2 3 3
2= −

, ,
π , x y z2 2 215 2 598 2= = = −. , . , . Hence the distance between the two 
points is ( ) ( ) ( )D x x y y z z= − + − + − =2 1 2 2 1 2 2 1 2 3407. . 
2 6 6. . 
The surface is 1/6 of the surface of a cylinder of length 4-1=3 and radius 2. Hence the 
surface area is ( )S = × × =2 2 3 6 2π π . By direct integration we have 
( )S r d dr
dsz r
= ∫ =∫ =
= =1
4
0
3
2 2φ π
φ
π
1 24 34 . 
2 6 7. . 
The volume is 1/6 of the volume of a cylinder of radius 2 minus the volume of a cylinder 
of radius 1 or ( ) ( )( )V = × − × =16 2 1 1 1 22 2π π π . By direct integration, 
V rdrd dz
z dvr
= ∫ ∫ ∫ =
= = =0
1
0
3
1
2
2φ
π
φ π124 34 . 
2 71. . 
Drawing the rectangular and spherical coordinate system axes as shown below we see that 
x r= sin cosθ φ , y r= sin sinθ φ , and z r= cosθ . From this we form 
( )x y z r r r2 2 2 2 2 2 2 2 2 2
1
+ + = + + =sin cos sin cosθ φ φ θ
1 244 344
 and hence 
2-8 
r x y z= + +2 2 2 . Similarly we form y
x
r
r
= =
sin sin
sin cos
tanθ φ
θ φ φ and 
x y
z
r
r
2 2+
= =
sin
cos
tanθ
θ
θ . 
 
 
 
 
 
 
 
2 7 2. . 
Draw the coordinate system and use the right-hand rule. 
2 7 3. . 
Drawing the vector in the zx or zy plane as shown below shows that 
A A Az r= −cos sinθ θθ . The components parallel to the xy plane are A Ar sin cosθ θθ+ 
and the φ component, Aφ . Hence the x and y components of these are 
( )A A A Ax r= + −sin cos cos sinθ θ φ φθ φ and 
( )A A A Ay r= + +sin cos sin cosθ θ φ φθ φ . 
 
 
 
 
 
 
 
 
�
�
y
z
z
x
r sin �
r
z y
x
Ar sin �
A� sin �
Ar sin � + A� cos �
Ax
Ay
A� cos �
A�
A�
Ar cos � Ar
x, y plane
�
�
�
2-9 
2 7 4. . 
At point P, θ π= =2
3
120o and φ π= =
3
60o . Hence 
Ax = + −2 3sin cos cos cos sinθ φ θ φ φ =-0.75 and 
Ay = + +2 3sin sin cos sin cosθ φ θ φ φ =0.701 and Az = −2 3cos sinθ θ =-3.598. 
Similarly, Bx = 383. , By = 0 634. , and Bz = −3732. . Directly in spherical coordinates 
A B• = + − =8 6 3 11. In rectangular coordinates 
( ) ( ) ( )A B• = − × + × + − × − =0 75 383 0 701 0 634 3598 3732 11. . . . . . . 
2 7 5. . 
From problem 2.7.1 x r= sin cosθ φ , y r= sin sinθ φ , and z r= cosθ . At P1 2 2
2
3
, ,π π

 , 
x y z1 1 11 1732 0= − = =, . , and at P2 3 3 6
= −



, ,
π π , x y z2 2 22 25 1299 15= = − =. , . , . . 
Hence the distance between the two points is 
( ) ( ) ( )D x x y y z z= − + − + − =2 1 2 2 1 2 2 1 2 4 69. . 
2 7 6. . 
The surface is 1/8 of the surface of a sphere of radius 4. Hence the surface area is 
( )( )
S =
×
=
4 4
8 8
2π
π . By direct integration we have 
( )S r d d
dsr
= ∫ =∫ =
= =θ π
π
φ π
π
θ θ φ π
2
2
2
4 8sin1 2444 3444 . 
2 7 7. . 
The volume is ( )S r d d
dsr
= ∫ =∫ =
= =θ π
π
φ
π
θ φ θ
4
3 2
0
2
2 521sin .1 2444 3444 . 
2-10 
2 81. . 
F l•∫ = ∫ + ∫ − ∫
=− = =−
d xdx dy ydz
x y z
2 4
1
2
3
4
2
1
. Third integral with respect to z contains y. So we 
need to determine the equation of the path as y z= +1
3
11
3
. Hence the line integral 
becomes F l•∫ = ∫ + ∫ − +

∫ = −
=− = =−
d xdx dy z dz
x y z
2 4 1
3
11
3
7
21
2
3
4
2
1
. 
2 8 2. . 
W d xdx zdy dz
P
P
x y z
= •∫ = ∫ + ∫ + ∫
= = =
F l
1
2
2 3 4
0
1
0
1
0
2
. But along the path z y= 2 which when 
substituted into the second integral gives W=1+3+8=12J. 
2 8 3. . 
(a) F l•∫ = ∫ + ∫ − ∫
= = =
d xdx xydy ydz
P
P
x y z1
2
0
3
0
2
2
0
2 . Along this path x y= 3
2
 and y z= − + 2 . 
substituting these gives ( )F l•∫ = ∫ + ∫ − − +∫ =
= = =
d xdx y dy z dz
P
P
x y z1
2
0
3 2
0
2
2
0
3 2 29
2
. (b) The 
integral is the sum of the integrals along the two paths: 
( )F l•∫ = ∫ + ∫ − =∫ + ∫ + =

∫ − ∫
= = = = = =
d xdx xydy y dz xdx x y ydy ydz
P
P
x y z x y z1
2
0
0
0
0
2
0
0
3
0
2
0
0
2 0 2 3
2
=0+0+0+
9/2+8+0=25/2. Along the first segment of this path neither x nor y change and y=0. 
Hence the integral along this first path is zero. Along the second segment of the path, 
there is no change in z and we substitute the equation for the path, x y= 3
2
, into the y 
integration. 
2 8 4. . 
F l•∫ = ∫ + ∫ + ∫d rdr zrd dz
P
P
1
2
2 4φ . The two paths are sketched below. 
(a) ( ) ( )F l•∫ = =∫ + =∫ + ∫ = + − = −
= = =
d r dr z r d dz
P
P
r z1
2
2 0 0 4 0 0 12 12
0
0
0
0
3
0
φ
φ
 
(b) F l F l F l F l•∫ = •∫ + •∫ + •∫d d d d
P
P
P
P
P
P
P
P
1
2
1
3
3
4
4
2
 
2-11 
( )F l•∫ = ∫ + =∫ + ∫ = + + =
=
=
=
d rdr z rd dz
P
P
r z1
3
2 3 4 8 0 0 8
0
8
4
4
3
3
φ
φ π
π
 
( ) ( )F l•∫ = =∫ + =∫ + ∫ = + − = −
=
=
=
d r dr z r d dz
P
P
r z3
4
2 8 8 4 0 0 12 12
8
8
4
4
3
0
φ
φ π
π
 
( )F l•∫ = ∫ + =∫ + ∫ = − + + = −
=
=
=
d rdr z rd dz
P
P
r z4
2
2 0 4 8 0 0 8
8
0
4
4
0
0
φ
φ π
π
 
F l F l F l F l•∫ = •∫ + •∫ + •∫ = − − = −d d d d
P
P
P
P
P
P
P
P
1
2
1
3
3
4
4
2
8 12 8 12 
 
 
 
 
 
 
 
 
2 8 5. . 
F l•∫ = ∫ + ∫ − ∫d rdr rd zdz
P
P
1
2
2 φ . The two paths are sketched below. 
(a) F l F l F l•∫ = •∫ + •∫d d d
P
P
P
P
P
P
1
2
1
0
0
2
 
( ) ( )F l•∫ = =∫ + =∫ − ∫ = + + =
= = =
d r dr r d zdz
P
P
r z1
0
0 2 0 0 0 2 2
0
0
0
0
2
0
φ
φ
 
( )F l•∫ = ∫ + ∫ − =∫ = + + =
= = =
d rdr rd z dz
P
P
r z0
2
0
3
0
0
0
0
2 0 9
2
0 0 9
2
φ
φ
 
y
z
x
P3(2, 2, 3)
P4(2, 2, 0)
P2(0, 0, 0)
P1(0, 0, 3)2-12 
F l F l F l•∫ = •∫ + •∫ = + =d d d
P
P
P
P
P
P
1
2
1
0
0
2
2 9
2
13
2
 
(b) F l F l F l F l•∫ = •∫ + •∫ + •∫d d d d
P
P
P
P
P
P
P
P
1
2
1
3
3
4
4
2
 
F l•∫ = ∫ + ∫ − ∫ = + + =
=
=
=
d rdr rd dz
P
P
r z1
3
0
3
2
2
2
2
2 4 9
2
0 0 9
2
φ
φ π
π
 
( ) ( )F l•∫ = =∫ + =∫ − ∫ = + + =
=
=
=
d r dr r d dz
P
P
r z3
4
3 2 3 4 0 0 2 2
3
3
2
2
2
0
φ
φ π
π
 
( ) ( )F l•∫ = =∫ + =∫ − ∫ = − + = −
=
=
=
d r dr r d dz
P
P
r z4
2
3 2 3 4 0 3 0 3
3
3
2
0
0
0
φ π π
φ π
 
F l F l F l F l•∫ = •∫ + •∫ + •∫ = + − = −d d d d
P
P
P
P
P
P
P
P
1
2
1
3
3
4
4
2 9
2
2 3 13
2
3π π 
 
 
 
 
 
 
 
 
 
2 8 6. . 
F l•∫ = ∫ + ∫ + ∫d rdr rd r d
P
P
1
2
2 3 2θ θ φsin . The two paths are sketched below. 
(a) F l F l F l•∫ = •∫ + •∫d d d
P
P
P
P
P
P
1
2
1
0
0
2
 
y
z
x
P3(0, 3, 2)
P4(0, 3, 0)
3
3
2
P2(3, 0, 0)
P1(0, 0, 2)
2-13 
F l•∫ = ∫ + ∫ + ∫ = − + + = −
= = =
d rdr rd r d
P
P
r1
0
2 3 2 9 0 0 9
3
0
0
0
0
0
θ θ φ
θ φ
sin 
F l•∫ = ∫ + ∫ + ∫ = − + + =
= = =
d rdr rd r d
P
P
r0
2
2 3 2 9 0 0 9
0
3
2
2
0
0
θ θ φ
θ π
π
φ
sin 
F l F l F l•∫ = •∫ + •∫ = − + =d d d
P
P
P
P
P
P
1
2
1
0
0
2
9 9 0 
(b) F l F l F l•∫ = •∫ + •∫d d d
P
P
P
P
P
P
1
2
1
3
3
2
 
( ) ( ) ( )F l•∫ = =∫ + =∫ + =∫ = + + =
= = =
d r dr r d r d
P
P
r1
3
2 3 3 3 2 3 0 9
2
0 9
23
3
0
2
2
2
θ θ φ π π
θ
π
φ π
π
sin 
( ) ( ) ( )F l•∫ = =∫ + =∫ + = 

∫ = + − = −
= = =
d r dr r d r d
P
P
r3
2
2 3 3 3 2 3
2
0 0 3 3
3
3
2
2
2
0
θ π φ π π
θ π
π
φ π
sin 
F l F l F l•∫ = •∫ + •∫ = − =d d d
P
P
P
P
P
P
1
2
1
3
3
2 9
2
3 3
2
π
π
π 
 
 
 
 
 
 
 
 
 
 
y
z
x
P3(0, 3, 0)
3
3
3
P2(3, 0, 0)
P(0, 0, 3)
P0(0, 0, 0)
2-14 
2 8 7. . 
F l•∫ = ∫ + ∫ + ∫d rdr r d r d
P
P
1
2
2 32 θ θ φsin . The two paths are sketched below. 
(a) F l F l F l•∫ = •∫ + •∫d d d
P
P
P
P
P
P
1
0
1
3
3
0
 
( ) ( ) ( )F l•∫ = =∫ + =∫ + =∫ = + + =
= = =
d r dr r d r d
P
P
r1
3
2 2 2 3 2 0 2 0 2
2
2 2
0
4
2
2
θ θ φ π π
θ
π
φ π
π
sin 
F l•∫ = ∫ + ∫ + ∫ = − + + = −
= = =
d rdr r d r d
P
P
r3
0
2
0 2
4
4
2
2
2 3 2 0 0 2θ θ φ
θ π
π
φ π
π
sin 
F l F l F l•∫ = •∫ + •∫ = −d d d
P
P
P
P
P
P
1
0
1
3
3
0
2 2π 
(b) F l•∫ = ∫ + ∫ + ∫ = − + + = −
= = =
d rdr r d r d
P
P
r1
0
2
0 2
0
0
0
0
2 3 2 0 0 2θ θ φ
θ φ
sin 
 
 
 
 
 
 
 
2 91. . 
A sketch of the surface is given below. F s•∫ = ∫∫ + ∫∫ + ∫∫d xdydz ydxdz zdxdy . 
Top: ( )
y x
z dxdy
=− =−
∫ =∫ =
1
1
1
1
1 4 . Bottom: ( )− ∫ = −∫ =
=− =−y x
z dxdy
1
1
1
1
1 4 . 
Right: ( )
z x
y dxdz
=− =−
∫ =∫ =
1
1
1
1
1 4 . Left: ( )− ∫ = −∫ =
=− =−z x
y dxdz
1
1
1
1
1 4 . 
Front: ( )
z y
x dydz
=− =−
∫ =∫ =
1
1
1
1
1 4 . Back: ( )− ∫ = −∫ =
=− =−z y
x dydz
1
1
1
1
1 4 . 
Total=4+4+4+4+4+4=24. 
y
z
x
P0(0, 0, 0)
P1(0, 0, 2) 2
45°
P3(0, 2 , )2
2-15 
 
 
 
 
 
 
 
 
2 9 2. . 
A sketch of the surface is given below. F s•∫ = ∫∫ + ∫∫ − ∫∫d xydydz yzdxdz xzdxdy . 
Top: ( )− ∫ =∫ = −
= =y x
x z dxdy
0
2
0
1
3 3 . Bottom: ( )
y x
x z dxdy
= =
∫ =∫ =
0
2
0
1
0 0 . 
Right: ( )
z x
y zdxdz
= =
∫ =∫ =
0
3
0
1
2 9 . Left: ( )− ∫ =∫ =
= =z x
y zdxdz
0
3
0
1
0 0 . 
Front: ( )
z y
x ydydz
= =
∫ =∫ =
0
3
0
2
1 6 . Back: ( )− ∫ =∫ =
= =z y
x ydydz
0
3
0
2
0 0 . 
Total=-3+0+9+0+6+0=12. 
 
 
 
 
 
 
 
 
 
y
z
x
(–1, 1, 1)
(–1, 1, –1)
y
z
x
2
1
3
2-16 
2 9 3. . 
A sketch of the surface is given below. F s•∫ = ∫∫ + ∫∫ − ∫∫d r d dz drdz rdrd2 3 22 φ φ φ . 
Top: − ∫ ∫ = −
= =φ
π
φ π
0
2
0
2
2 2rdrd
r
. Bottom: 
φ
π
φ π
= =
∫ ∫ =
0
2
0
2
2 2rdrd
r
. 
Right: 
z r
drdz
= =
∫ =

∫ =0
3
0
2
3
2
9φ π π . Left: ( )− ∫ =∫ =
= =z r
drdz
0
3
0
2
3 0 0φ . 
Front: ( )
z
r d dz
= =
∫ =∫ =
0
3 2
0
2
2 2 12φ π
φ
π
. 
Total=-2π+2π+9π+0+12π=21π. 
 
 
 
 
 
 
 
 
 
2 9 4. . 
A sketch of the surface is given below. F s•∫ = ∫∫ + ∫∫ − ∫∫d r d dz drdz zrdrd3 2 φ φ φ . 
Top: ( )− ∫ =∫ =
=− =φ π
π
φ π
2
2
0
2
4 8z rdrd
r
. Bottom: ( )
φ π
π
φ
=− =
∫ =∫ =
2
2
0
2
0 0z rdrd
r
. 
Right: 
z r
drdz
= =
∫ =

∫ =0
4
0
2
2
4φ π π . Left: − ∫ = −

∫ =
= =z r
drdz
0
4
0
2
2
4φ π π . 
Front: ( )
z
z r d dz
= =−
∫ =∫ =
0
4 2
2
2
2 48φ π
φ π
π
. 
Total=8π+0+4π+4π+48π=64π. 
z
3
2
y
x
2-17 
 
 
 
 
 
 
 
 
2 9 5. . 
A sketch of the surface is given below. 
F s•∫ = ∫∫ + ∫∫ − ∫∫d r d d r drd rdrd3 2sin sinθ φ θ θ φ φ θ . 
Front: ( )
θ
π
φ
π
θ φ θ π
= =
∫ =∫ =
0
2 3
0
2
2 4r d dsin . Bottom: 
φ
π
π φ π
= =
∫ ∫ =
0
2
0
2
2
2
2r drd
r
sin . 
Right: − ∫ =

∫ = −
= =θ
π
φ π θ π
0
2
0
2 2
2 2
rdrd
r
. Left: ( )
θ
π
φ θ
= =
∫ =∫ =
0
2
0
2
0 0rdrd
r
. 
Total=4 2 2
0 6
2
2 2
π π π π π+ − + = − .
z
4
2
y
x
2
y
x
z
2-18 
2 9 6. . 
A sketch of the surface is given below. 
F s•∫ = ∫∫ − ∫∫ + ∫∫d r d d r drd rdrd3 2 3sin sinθ φ θ θ φ φ θ . 
Front: ( )
θ
π
φ π
π
θ φ θ π
= =−
∫ =∫ =
0
2 3
2
2
3 27r d dsin . Bottom: − ∫ ∫ = −
=− =φ π
π
π φ π
2
2
0
3
2
2
9r drd
r
sin . 
Right: 
θ
π
φ π θ π
= =
∫ =

∫ =0
2
0
3 2
3
2
27
8
rdrd
r
. Left: − ∫ = −

∫ =
= =θ
π
φ π θ π
0
2
0
3 2
3
2
27
8
rdrd
r
. 
Total= 27 9 27
8
27
8
18 27
4
2 2 2
π π
π π
π
π
− + + = + . 
 
 
 
 
 
 
 
3
y
x
z
 3-1 
Chapter 3 
Problem Solutions 
311. . 
Q dv zrdzd drv
r z
= ∫∫∫ = ∫ ∫ ∫ =
=
=
=
ρ φ π
φ π
π
0
1
4
2
0
2
2
2
C 
312. . 
The surface is shown below. ( )Q ds xy z dxdys
x y
x
= ∫∫ = ∫ =∫ =
= =
− +
ρ
1
2
1
2 5
3 2 13C . 
 
 
 
 
 
 
313. . 
The problem is sketched below. A vector from the second charge to the first is 
( )( ) ( ) ( )( )R a a a a a a21 1 1 1 0 1 2 2 3= − − + − + − − = + +x y z x y z whose length is 
R21 14= . A unit vector pointing from the second charge to the first is a
R
21
21
21
=
R
. 
Coulomb’s law yields 
( )( )
F a a a a21
9
6 6
21
2 219 10
100 10 50 10
1718 0 859 2577= ×
× ×
= + +
− −
R
x y z. . . N . 
 
 
 
 
 
y
z
x
y = –2x + 5
(1, 1, 2) (1, 3, 2)
(2, 1, 2)
y
z
x
(–1, 0, 2)
Q2 = 50 �C
Q1 = 100 �C
(1,1, 1)
3-2 
314. . 
(a) F=0, (b) 
( )
( )F a= × ×
×
×
−
4 9 10
100 10
2
459
6 2
2 cos
o
z , 
(c) 
( )
( )F a a1
9
6 2
29 10
100 10
1
90= ×
×
=
−
x x , 
( )
( )F a a2
9
6 2
29 10
100 10
1
90= ×
×
=
−
y y , 
( )F a a3 905= +cos sinθ θx y , ( )F a a4 905= +sin cosθ θx y . But cosθ = 25 and 
sinθ = 1
5
. Hence F F F F F a a= + + + = +1 2 3 4 11415 11415. .x x . 
 
 
 
 
 
 
 
 
315. . 
The problem is sketched below. For the forces exerted on Q3 by the other two charges to 
be equal and oppositely directed, we must have 
( )( ) ( )( )
( )9 10
18 10 8 10
9 10
72 10 8 10
0 03
9
6 6
2
9
6 6
2×
× ×
= ×
× ×
−
− − − −
d d.
. Solving for d gives 
d=1cm. 
 
 
 
 
 
F2
F4
F3
F1
–1
�
�
�
�
1
y
x
5
5
1
2
4
3
d
3 cm
Q1 = 18 �C Q3 = –8 �C Q2 = 72 �C
–1 1
3-3 
316. . 
The charge of an electron is e = − × −16 10 19. C. Placing the positive charge on the left 
and the negative charge on the right, the force exerted on the electron by the positive 
charge is directed to the left and is 
( )( )
( )F1
9
6 19
2 2
129 10
35 10 16 10
10 10
504 10= ×
× ×
×
= ×
− −
−
−
.
. N . The force exerted on the 
electron by the negative charge is of the same magnitude and in the same direction so that 
the net force is 1 10 11× − N directed toward the positive charge. 
317. . 
The problem is sketched below. In order that the forces balance, the Coulomb force 
acting in the horizontal direction is ( )F
Q
lo
=
2
24 2π ε θsin
. The component of the 
restraining force along the string that is horizontally directed is T sinθ and the force of 
gravity acting downward on the charges is T mgcosθ = . Hence F mg= sin
cos
θ
θ
. 
Equating the two and solving gives ( )Q l mgo2 2 31 16= cos sinθ πε θ . 
 
 
 
 
 
3 21. . 
The problem is sketched below. The electric field due to the positive charge is 
( )E a a1
9 1
29 10 4
2 8125= × =Q y y, . . The electric field due to the negative charge is 
( )E a a2
9 2
29 10 2
22 500= − × = −Q z z, . Hence the total electric field is 
E E E a a= + = −1 2 2 8125 22 500, . ,y z
V
m
.
�
2l sin �
QF
mg
Q
T l
3-4 
 
 
 
 
 
 
 
3 2 2. . 
The problem is sketched below. The angle θ is θ = 60o . The distances from the triangle 
vertices to the center is, according to the law of cosines, ( )5 2 22 2 2 2= + −d d d cos θ 
giving d=2.887 m. The vector contributions are 
( )E Q
d
Q
do o
o
= + =
4
2
4
60 1082 2π ε π ε
cos kV
m
. 
 
 
 
 
 
 
 
3 2 3. . 
The problem is sketched below. (a) First we determine the electric field along the z axis. 
Superimposing the fields due to the two charges gives 
E a a a a=
−




−
+




=
−



 +




=
−






1
4
2
1
4
2
2
2 2
2
4
2 2 2 2
2
2 2π ε π ε π ε π εo
z
o
z
o
z
o
z
Q
z l
Q
z l
Ql z
z l z l
Ql z
z l
z
y
(0, 4, 2)
Q1 = 5 �C
Q2 = –10 �C
z = 2
y = 4
E1
E2 E
5 m 5 m
5 m
Q
—Q —Q
d d
d
� �
3-5 
(b) Now we determine the electric field along the y axis. Superposing the fields as shown 
gives E a a= −
+





 +




= −
+






2
4
1
4
2
4
4
4
2
2
2
2
1
2
2
2
3
2
Q
y l
l
y l
Ql
y l
o
z
o
zπ ε
π ε
θsin
1 244 344
. 
 
 
 
 
 
 
 
3.2.4 
The problem is sketched below. Divide the charge into chunks of charge, {dQ adl
dl
= ρ φ . 
At a distance d from the center and on a line perpendicular to the ring, the horizontal 
components cancel out leaving only the vertical components so that 
 ( )E a= ∫
=
1
4 20
2
π ε
ρ φ
α
φ
π
o
l
z
ad
R
cos 
where R d a= +2 2 and ( )cos α = d
R
. Substituting these gives 
 
( )
( )
E a
a
=
+
∫
=
+
>
=
1
4
1
2
0
2 2
3
20
2
2 2
3
2
π ε
ρ φ
ε
ρ
φ
π
o
l
z
o
l
z
ad
d a
d
ad
d a
z
 
At a large distance from the center, d a>> , this result reduces to 
 
E a
a
= >>
= >>
1
2
2
4
2
2
ε
ρ
π ρ
π ε
o
l
z
l
o
z
a
d
d a
a
d
d a
 
y
l
2 �
� �
E+
E+
E–
E–
E
z
z
y
l
2
3-6 
 
 
 
 
 
 
 
 
3.2.5 The problem is sketched below. The chunks of charge are dQ dr rds
ds
= ρ φ123 and 
again, by symmetry, the horizontal components cancel leaving only the vertical (z-
directed) components. Summing these contributions gives 
 
( )
E a
a
a
= ∫ ∫
+ +
= −
+






>
= >>
= =r
a
s
o
R
z
s
o
z
s
o
z
rd dr
d r
d
d r
d
d a
z
a
d
d a
0 0
2
2 2
1
2 2
2 2
2
2
4
1
2
1 0
4
2
ρ φ
π ε
ρ
ε
π ρ
π ε
φ
π
α
124 34 1 24 34
cos
 
 
 
 
 
 
 
 
 
 
E
R
z = d
a
y
z
x
�
�
�l ad�
E
R
z = d
�s rdrd�
a
r
y
z
x
�
�
3-7 
3.2.6 
The problem is sketched in the xy plane below. Using the results of Example 3.3 and 
superpositioning the fields gives (a) 
E a a a=
−




−
+




=
−






ρ
π ε
ρ
π ε
ρ
π ε
l
o
y
l
o
y
l
o
y
y l y l
l
y l
2
1
2
2
1
2
2
4
2
2
. 
Similarly the fields along the x axis become E a= −
+ +
2
2
4
2
4
2
2
2
2
ρ
π ε
θ
l
o
y
x l
l
x l
cos
1 24 34
. 
 
 
 
 
 
 
 
 
 
3.2.7 
The problem is sketched below. Place the strip in the xz plane centered on the origin. 
Divide the strip into infinite line charges with distribution ρ ρl sdz=
C
m
. Use the result 
in Example 3.3, equation (3.10). Accounting for symmetry, E a= ∫
=
2
20
2 ρ
πε
αs
oz
W
y
dz
R
cos
`
 
where R z d= +2 2 and cosα = d
R
. Hence, using the integral 
1 1
2 2
1
d z
dz
d
z
d+
∫ = 


−tan , ( )E a a= +∫ =




=
−
ρ
πε
ρ
πε
s
o z
W
y
s
o
y
d
d z
dz
d
W
d
1
22 20
2 1
`
tan . 
Er–
Er+
Er–
Er+l
2
l
2
R
R
y
y
x
x
—�
�
�
�
3-8 
 
 
 
 
 
 
 
3 31. . 
The electric field intensity vector is E V
d
= = 105 V
m
. The polarization vector is 
P D Eo= − ε . Substituting D Er o= ε ε gives 
( ) ( )P Eo r= − = × − × =−ε ε π
µ1 1
36
10 54 1 10 3899 5. . C
m2
. 
3.4.1 
The problem is sketched below. (a) the total charge enclosed is 
Q r drd d kr drd d ka
r
a
v
dv r
a
enc = ∫ ∫ =∫ ∫ ∫ =∫
= = = = = =0 0
2 2
0 0 0
2 3
0
4
φ
π
θ
π
φ
π
θ
π
ρ θ φ θ θ φ θ πsin sin1 244 344 . (b) Using 
Gauss’ law, ε ε π πo od E r Q kaE s•∫ = = =4 2 4enc giving E a= ka
ro
r
4
24ε
. (c) The charge 
enclosed is Q krenc = π
4 . Hence the electric field is E a= kr
o
r
2
4ε
. 
 
 
 
 
 
 
 
 
�
�
E–
E+
z
d
R
R
W
—z
z
y
W
2
—
W
2
�
v
 = kr
a
r dv
y
z
x
3-9 
3 4 2. . 
No. No closed surface can be found for which the electric field is perpendicularto all 
sides and hence no simplification of D s•∫ d can be obtained. 
3 4 3. . 
The problem is sketched below. Since D is directed in the z direction, there is no flux 
through the sides. Hence Gauss’ law gives 
( ) ( )Q d d z rrdrd z rrdrd a
r
a
ds r
a
ds
enc
top bottom
C= •∫ + •∫ = ∫ =∫ − ∫ =∫ = =
= = = =
D s D s123 123 123 123φ
π
φ
π
φ φ π π
0
2
0 0
2
0
3
4 0 8
3
64
3
. 
 
 
 
 
 
 
 
 
3 4 4. . 
For r b≥ ( )Q dv kr r drd d k b av r a
b
dv
enc = ∫ = ∫ ∫ ∫ = −
= = =
ρ θ φ θ π
φ
π
θ
π
0
2 2
0
2 22sin1 244 344 . By symmetry, 
the field is radially directed. Hence D s•∫ =d Qenc so that ( )ε π πo rE r k b a4 22 2 2= − . 
Thus 
( )
E
k b a
r
r
o
=
−
2 2
22ε
. For r a≤ , Er = 0 since no charge is enclosed. For a r b≤ ≤ , 
( )Q k r aenc = −2 2 2π . Hence ( )ε π πo rE r k r a4 22 2 2= − . Thus E k a
r
r
o
= −





2 1
2
2ε
. 
3.5.1 
The problem is sketched below. The work required to move q around the paths is 
W q d q xdx q ydy q zdz= − •∫ = − ∫ − ∫ − ∫E l . (a) 
z
4
Dz
a
y
x Dz
3-10 
W q zdz q ydy q zdz q ydy
z y z y
= − ∫ − ∫ − ∫ − ∫ =
= = = =0
1
0
2
1
0
2
0
0 . (b) 
W q zdz q ydy q xdx q xdx
z z x x
= − ∫ − ∫ − ∫ − ∫ =
= = = =0
1
1
0
0
1
1
0
0 . 
 
 
 
 
 
 
 
3.5.2 
The problem is sketched below. Applying superposition and equation (3.37) yields 
V Q
o
= −



 =4
1
2
1
3
15
π ε
kV . 
 
 
 
 
 
 
3.5.3 
The problem is sketched below. Applying superposition and equation (3.37) yields 
( ) ( ) ( ) ( )
V Q Q
o o
=
+ −
−





 + + −
−





 = − − = −
1
2 2
2
2 24
1
5 5 3
1
3 4
1
5 5 2
1
2
13 287 42 1478256 28
π ε π ε
, . . ,070V
y
z
x
(0, 2 m, 1 m)
(0, 2 m, 0)
(0, 0, 1 m)
x = 1
y = 2
z = 1
(0, 0, 0)
(0, 3, 2)
–
+
y = 3
Q = 10 �C
V
z
y
3-11 
 
 
 
 
 
 
 
 
3.5.4 
The problem is sketched below. Applying superposition and equation (3.37) yields 
V Q Q
o o
=
+
−





 +
+
−





 = − − = −1 2 2
2
2 24
1
5 2
1
2 4
1
5 3
1
3
28 287 42 7 28256 35 570
π ε π ε
, . , . , V 
 
 
 
 
 
 
 
 
3.5.5 
The problem is sketched below. Applying superposition and equation (3.37) yields 
V Q Q
o o
=
+
−





 +
+
−





 = − + =1 2 2
2
2 24
1
5 3
1
3 4
1
5 2
1
2
7 28256 28 287 42 21
π ε π ε
, . , . ,005V . 
 
 
(0, 5, 5)
–
+
y = 2
z = 3
V
Q2 = 5 �C
Q1 = 10 �C
z
y
(5)2 +
 (5 – 3
)2
29
=
(5 – 2)2 + (5)2 34=
(0, 0, 5)
–
+
y = 3y = –2
V
Q2 = 5 �CQ1 = 10 �C
z
y
52 + 22 52 + 32
3-12 
 
 
 
 
 
 
 
 
3.5.6 
The problem is sketched below. Applying superposition and equation (3.38) yields 
V
o o
=
+





 +
+





 = − + =
ρ
π ε
ρ
π ε
1
2 2
2
2 22
2
5 2 2
3
5 3
89145 119 30 477Vln ln , ,622 , . 
 
 
 
 
 
 
 
 
3.5.7 
The problem is sketched below. By the law of cosines, the distance from each charge to 
the center of the triangle is related to the side length as ( )l d d d o2 2 2 22 120= + − cos so 
that d l=
1732.
. Applying superposition and equation (3.42) yields 
V Q
l
Q
l
o
= 



= × =3
4
1732
4 677 10 31177V10
π ε
.
. , .
y
z
x
x = 3
Q1 = 5 �C
Q2 = –10 �C
(0, 5, 0)
z = 2
52 + 22
52 + 32
–
+
y = 2
z = 5
y = –3
V
�l = –10 �C/m �1 = 5 �C/m
z
y
52 + 32 52 + 22
3-13 
 
 
 
 
 
 
 
 
3.5.8 
The problem is sketched below. Using (3.44) gives 
V
ad
z a
a
z a
l
o
l
o
=
+
∫ =
+=
ρ φ
π ε
ρ
εφ
π
4 22 20
2
2 2
. The potential is only a function of z and hence 
the gradient is 
( ) ( )E a a a= − = − = − + = +
−
gradientV V
z
a
z
z a
a z
z a
z
l
o
z
l
o
z
∂
∂
ρ
ε
∂
∂
ρ
ε2 2
2 2
1
2
2 2
3
2
 which 
agrees with the results of Problem 3.2.4. 
 
 
 
 
 
 
 
 
 
 
120°
120°120°
l
ll
Q
Q Q
dd
d
R
a
y
z
x
�l ad�C
3-14 
3.5.9 
The problem is sketched below. Using (3.44) gives 
V rdrd
z r
z a z
r
a
s
o
s
o
= ∫
+
∫ = + −


= =0 2 20
2 2 2
4 2
ρ φ
π ε
ρ
εφ
π
. The potential is only a function of z 
and hence the gradient is 
E a a a= − = − = − + −

 = − +






gradientV V
z z
z a z z
z a
z
s
o
z
s
o
z
∂
∂
ρ
ε
∂
∂
ρ
ε2 2
12 2
2 2
 which 
agrees with the results of Problem 3.2.5. 
 
 
 
 
 
 
 
 
3.5.10 
(a) 
( ) ( ) ( )
E a a a
a a a
= − = − − −
=
+ +
+
+ +
+
+ +
gradientV V
x
V
y
V
z
x
x y z
y
x y z
z
x y z
x y z
x y z
∂
∂
∂
∂
∂
∂
2 2 2
3
2 2 2 2
3
2 2 2 2
3
2
 
(b) 
E a a a
a a a
= − = − − −
= − + +− − −
gradientV V
r r
V V
z
e e re
r z
z
r
z z
z
∂
∂
∂
∂ φ
∂
∂
φ φ φ
φ
φ
1
cos sin cos
 
R
�s rdrd�
�s
a
r y
z
x
3-15 
E a a a
a a a
= − = − − −
= − +
gradientV V
r r
V
r
V
r r r
r
r
∂
∂
∂
∂ θ θ
∂
∂ φ
θ φ θ φ φ
θ φ
θ φ
1 1
2 3 3 3
sin
sin cos cos cos sin
 
3 61. . 
E V
d
= = 10 kV
m
, D Eo r= =ε ε
µ0 478. C
m2
, P D Eo= − =ε
µ0 389. C
m2
, 
C A
d
pFr o= =ε ε 4775. . 
3 6 2. . 
( )C A
do o
= = =
−
ε ε
π 01
10
277 8
2
3
.
. pF . 
3 6 3. . 
There are two capacitors in series: C A
dr o1 1 1
= ε ε and C A
dr o2 2 2
= ε ε . Capacitors in 
series add like resistors in parallel so that C C C
C C
A d d
d d
o
r r
r r
=
+
=
+
1 2
1 2
1 2
1 2
1
1
2
2
ε
ε ε
ε ε
. The total free 
charge on the upper (and lower) plate is Q CVf = . Using Gauss’ law and surrounding 
the upper plate with a closed surface yields DA Q f= . Therefore D
Q
A
CV
A
f
= = . In the 
upper dielectric, E D CV
Ar o r o
1
1 1
= =
ε ε ε ε
 and in the lower dielectric, 
E D CV
Ar o r o
2
2 2
= =
ε ε ε ε
. Evaluating these gives C = 70 74. pF , E1 4000=
V
m
, 
E2 2000=
V
m
. 
3 6 4. . 
There are two capacitors in parallel: C A
dr o1 1
1
= ε ε and C A
dr o2 2
2
= ε ε . Capacitors in 
parallel add like resistors in series so that ( )C C C A A
do
r r
= + =
+
1 2
1 1 2 2ε
ε ε
. Using 
Gauss’ law and surrounding each portion of the upper plate with a closed surface yields 
D A Q f1 1= and D A Q f2 2= . Therefore E
D C V
Ar o r o
1
1
1
1
1 1
= =
ε ε ε ε
 and 
 
(c) 
3-16 
E D C V
Ar o r o
2
2
2
2
2 2
= =
ε ε ε ε
. Evaluating these gives C1 88 42= . pF , C2 707 4= . pF , 
C = 7958. pF , E1 5000=
V
m
, E2 5000=
V
m
. Observe that V E d E d= = =1 2 10V and 
Q Q Q CVf f f= + = =1 2 7 96. nC . 
3 6 5. . 
We observe that there are essentially two spherical capacitors in series. First we obtain 
the capacitance of a spherical capacitor with a homogeneous dielectric filling the interior. 
Using Gauss’ law and surroundingthe inner sphere with a sphere of radius r gives 
D
Q
r
f
=
4 2π
. Hence the electric field is radially directed and is E
Q
r
f
=
4 2π ε
. The voltage 
between the spheres is V
Q
r
dr
Q
a b
f
b
a f
= − ∫ = −

4 4
1 1
2π ε π ε
. Hence the capacitance is 
( )C
Q
V
ab
b a
f
= =
−
4π . This result was derived in Exercise Problem 3.8. Hence the 
capacitances are ( )C
ar
r a1
1
1
4
=
−
π and ( )C
br
b r2
1
1
4
=
−
π . Since these are in series and capacitors 
in series add like resistors in parallel we obtain 
C C C
C C
r b a r
o
r r
=
+
=
−



 + −




1 2
1 2
2 1 1 1
4 1
1 1 1 1 1 1
π ε
ε ε
. Substituting the values gives 
0.303pF. 
3 6 6. . 
The per-unit-length capacitance for a coaxial cable filled with a homogeneous dielectric 
was obtained in Example 3.15 as c
b
a
= 



2π ε
ln
F
m
. For this problem we observe that there 
ae two such capacitors in series: c
r
a
o r
1
1
1
2
= 



π ε ε
ln
 and c
b
r
o r
2
2
1
2
= 



π ε ε
ln
. Since capacitors in 
3-17 
series add like resistors in parallel, c c c
c c b
r
r
a
o r r
r r
=
+
= 


 +




1 2
1 2
1 2
1
1
2
1
2π ε ε ε
ε εln ln
. Evaluating this 
for the given dimensions yields 82 06. pF m . 
3.7.1 
Converting the radius from mils to meters gives 
16 254cm 1 4 064 10 4mils 1inch
1000mils 1inch
meter
100cm
m× × × = × −. . . Hence the resistance is 
( )R A= × × = × =−
1000
4 064 10
3323
4 2
m
= 5.8 107σ π .
. Ω . 
3.7.2 
The electric field is E V
d
= . Hence the current density is J E V
d
= =σ σ . The total 
current is I JA A V
d
= = σ . Hence the resistance is R V
I
d
A
= =
σ
Ω . Evaluating this 
gives 50mΩ. 
3.7.3 
The electric field intensity at a radius r was determined in Problem 3.6.5 as E
Q
r
f
=
4 2π ε
. 
The voltage between the spheres was also determined as 
V
Q
r
dr
Q
a b
f
b
a f
= − ∫ = −

4 4
1 1
2π ε π ε
. The current flowing between the two spheres is 
I JA EA
Q
r
r
Qf f
= = = =σ σ
π ε
π σ
ε4
42
2 . Hence the resistance is 
R V
I a b
= = −




1
4
1 1
π σ
. Evaluating this for the given dimensions yields 106.1Ω. 
3.7.4 
The voltage between the inner wire and the shield was determined in Example 3.15 in 
terms of the charge distribution on the inner wire of ρ l
C
m
 as V b
a
l
=




ρ
π ε2
ln . 
Similarly, the electric field in this region was determined in Example 3.7 to be 
3-18 
E
r
l
=
ρ
π ε2
. Hence the current flowing from one cylinder to the other (per-unit-of line 
length) is I JA
r
rl l= = =σ
ρ
π ε
π σ
ρ
ε2
2 . Thus the resistance per unit length is 
r V
I
b
a
= =




1
2π σ
ln /Ω m . Evaluating this for the given dimensions yields 01. /Ω m 
3.7.5 
From the results of Example 3.16, the magnetic flux density vector at a perpendicular 
distance from the midpoint of the current element is B a=
+
µ
π φ
o I
r
L
r L
2
2
4
2
2
. (a) At 
(0,3m,0) the field is 0.2858 Taφ µ . (b) Off the ends of the curent element d Rl a× is 
zero and hence the field is zero off the ends. 
3.7.6 
Utilizing the result obtained in Example 3.16 for an infinite current, the magnetic flux 
density at a distance r is B I
r
o
=
µ
π2
Wb
m2
. Hence the total magnetic flux penetrating the loop 
is ψ = •∫ B sd . By the right-hand rule the B field is directed into the page and hence the 
dot product can be removed. However, the B field depends on distance away from the 
current and cannot be removed from the integral. Hence 
ψ µ
π
µ
π
= ∫ ∫ = 


= =z
l
o
r r
r
oI
r
drdz Il r
r0
2
12 21
2
ln Wb . For the given parameters we obtain 
ψ µ= 0139. Wb . 
3.7.7 
The problem is sketched below. Using the results of Example 3.16 we may superimpose 
the contributions from the four identical sides. Observe that the contributions from 
opposite sides cause the net B from those two to be directed in the +z direction. similarly 
the other two sides cause a similar result. Hence B a=
+
4
2
2
4
2 2
µ
π
αo z
I
R
l
R l
cos where 
3-19 
R z l= +2
2
4 and cosα =
l
R
2 . Combining these gives 
B a=
+

 +
2 4
4 2
2
2 2 2 2
µ
π
o
z
I l
z l z l
. For l = 2m and I = 10A , at the center of the 
loop, z = 0 , we obtain B a= 566. µ T z . 
 
 
 
 
 
 
 
 
 
 
 
 
3.7.8 
The problem is sketched below. Using the results of Example 3.16 we may superimpose 
the contributions from the three identical sides. The radial distance from the center of 
each side to the center of the triangle is r l= 0 577
2
. . Using (3.60) of Example 3.16 we 
obtain B I
r
l
r l
I
l
o o
=
+
=3
2
2
4
9
22 2
µ
π
µ
π
 into the page. For l = 5cm and I = 4A we obtain 
B = 144µ T . 
 
R
�
��
�
R
I
I
I
I
l
2
( ,l
2
, 0)l
2
(– ,l
2
, 0)l
2
l
2
y
z
x
3-20 
 
 
 
 
 
 
 
3.7.9 
The problem is sketched below. According to the Biot Savart law, the magnetic field off 
the ends of a current is zero since d Rl a× is zero. Hence there are no contributions to the 
magnetic field at point P due to sides DA and BC. For side CD d rdRl a× = θ and is out 
of the page. Hence the Biot Savart law gives for this contribution 
( )
µ
π
θ µ
π
θ
θ
θ
o oI
r
r d I
r4 42
2 2
0 2=
∫ = . Similarly the contribution from the segment AB is µ
π
θo I
r4 1
 
which is into the page. Hence the total is B I
r r
o
= −




µ θ
π4
1 1
1 2
 which is directed into the 
page since r r2 1> . 
 
 
 
 
 
 
 
 
I
30°
l
l l
r
rr
I
I
�
I
AD
C
B
P
I
I
I
r1
r2
3-21 
3.9.1 
The problem is sketched below. Use the result for the magnetic field from an infinitely 
long current filament obtained in Example 3.19: H I
r
=
2π
. (a) At the center, the H fields 
add giving H I
d
I
d
= 



=2
2
2
2
π
π
. (b) At a distance D from the center and in a plane 
containing the currents we superimpose the fields to give 
( ) ( )H
I
D d
I
D d
I d
D d
=
−
−
+
=
−



2 2 2 2
2
4
2 2π π π
. 
 
 
 
 
 
 
 
 
3.9.2 
The problem is sketched below. Place the strip on the z axis centered about the origin. 
Treat the linear current density as filaments of current Kdz A . Superimpose the magnetic 
fields using equation (3.81) of Example 3.19. The magnetic field intensity vector is 
H a= ∫
=
2
20
2 Kdz
Rz
w
zπ
αcos where R d z= +2 2 and cosα = d
R
. Hence 
( )H
Kd
d z
dz K z
d
K w
dz
w
z
w
=
+
∫ = 

 =
=
−
=
−
π π π
1
22 20
2 1
0
2 1tan tan . For an infinite strip as 
w → ∞ , H K→
2
 which agrees with the result for an infinite strip in Example 3.21. 
 
d
2
d
2
II
D
3-22 
 
 
 
 
 
 
 
3.9.3 
For (b) a r b< < , H rd NIφ
φ
π
φ
=
∫ =
0
2
. Hence H NI
rφ π= 2 . For (a) r a< and (c) r b> the 
net current penetrating this loop of radius r is zero and hence H=0. 
3.9.4The problem is sketched below. Construct a rectangular contour as shown. By symmetry, 
the magnetic field is directed along the solenoid and is directed right to left. Applying 
Ampere’s law to this yields H l•∫ = = =d HL I InLenclosed . Hence H nI= and 
B H nIr o r o= =µ µ µ µ . 
 
 
 
 
 
3.11.1 
The magnetic flux density along the axis of the solenoid was obtained in Problem 3.9.4 as 
B nIr o= µ µ . Since this is uniform over the cross section of the core and is axially 
directed, the flux is ψ π= •∫ =B sd B a
core
cross section
2 . The flux linkages per unit length are 
�
��
z
d
R
R
Kdz
y
z =
w
2
z = _
w
2
L
C
H H
II
3-23 
Λ = nψ . Hence the per-unit-length self inductance is l
I
n ar o= =
Λ µ µ π2 2 H
m
. For 
the given dimensions, ( )l = × × × 

 =
−1000 4 10 2000 0 01 1587
2
2π π
turns
m
H
m
. . . 
3.11.2 
Treating this as a long parallel wire line of length l and separation w and neglecting the 
contribution from the end segments gives, using the result of Example 3.25 for an infinite 
line, L l w
a
o≅ 


µ
π
ln . 
3.11.3 
The problem is sketched below. Assuming the plate width is much greater than the plate 
separation we can use the result in Example 3.18 for the magnetic flux density from an 
infinite plate carrying a linear current density of K A
m
. The magnetic flux density is 
parallel to the plate and opposite to the direction of the current: B Ko= µ
2
. 
Superimposing the fields due to both plates gives B Ko= µ which is constant across the 
cross section between the plates. The flux penetrating the surface between the plates is 
ψ µ= •∫ = =B sd Bs l Ks lo∆ ∆ . The total current on each plate is I Kw= . Hence the 
flux is ψ µ= o I l
s
w
∆ . Thus the inductance per unit length is l I
l
s
wo
= =
ψ
µ
∆
. 
 
 
 
 
 
 
 
 
 
s
w
B
3-24 
3.11.4 
First we solve the basic problem shown below. The magnetic field intensity due to the 
infinitely long current is B I
r
o
=
µ
π2
. The flux through the loop is obtained by integration 
as ψ µ
π
µ
π
= ∫ = 


=
l I
r
dr Il b
a
o
r a
b
o
2 2
ln . Applying this to the original problem and 
superimposing the fluxes due to each current gives 
( )
( )
( )
( )ψ
µ
π
µ
π2 2
2 2
2 2
2
2 2
2 2
=
− +
− −





 −
+ +
+ −





o o
Il D s w
D s w
I D s w
D s w
ln ln . Hence the mutual inductance 
is 
( )( )
( )( )M I
l D s w D s w
D s w D s w
o
= =
− + + −
− − + +






ψ µ
π
2
2
2 2 2 2
2 2 2 2
ln . 
 
 
 
 
 
 
3.12.1 
The particle is traveling with constant velocity so that F ma= = 0 . In order for the 
vertical forces to balance so that the particle passes through the hole we must have the 
electric force, F qEe = , equal the magnetic force, F qvBm = . Solving gives the critical 
velocity of the particle as v E
B
= . Evaluating this for the given conditions yields 
v =
×
= ×
−
2000
1 10
2 103
6 m
s
. 
3.12.2 
The magnetic flux density vector is radially directed about each wire and is given by 
B I
r
o
=
µ
π2
. The magnetic flux density at wire 2 due to the current of wire 1 is therefore 
b
I
a l
3-25 
B I
s
o
21
1
2
=
µ
π
 and is perpendicular (into the page) to current I2 . The force exerted on a ∆l 
section of wire 2 is, according to the Lorentz force equation, F I lB I l I
s
o
21 2 21 2
1
2
= =∆ ∆ µ
π
. 
Hence the force per unit length exerted on wire 2 is f F
l
I I
s
o
21
21 1 2
2
= =
∆
µ
π
N
m
. 
3.12.3 
From the previous problem, the force exerted on the left side of the loop is 
F I I
a
wo1 1 22
=
µ
π
, and the force exerted on the right side of the loop is F I I
b
wo3 1 22
=
µ
π
. 
Along the upper and lower segments of the loop, the B field varies along the wire. Hence 
the force along the upper segment is F I I
r
dr I I b
a
Fo
r a
b
o
2
1 2 1 2
42 2
= ∫ = 

 =
=
µ
π
µ
π
ln . 
3.12.4 
The sliding bar cuts the magnetic field resulting in a voltage source , Bvw , inserted in it 
as shown below. Hence the current is I Bvw
R
= − . 
 
 
 
 
 
3.12.5 
The vertical side of the loop cuts the magnetic field so that a voltage is induced in it. The 
horizontal sides have no inserted source since v B× is perpendicular to the wire. Shown 
below is a view in the xy plane. The tangential velocity is v l= ω and 
( )v B× = l B tω ωsin assuming that the loop starts at the x axis at t=0. The voltage 
source is inserted as shown below so that the current through the resistor is 
( )I wl B t
R
= −
ω ωsin
. 
I
R
Bvw+
–
3-26 
 
 
 
 
 
 
 
 
 
 
B
B
l
v
�t
�
x
y
I
R
l� B sin �t+
–
4-1 
Chapter 4 
Problem Solutions 
411. . 
The flux in the left loop is ψ1 301 1 0 2 10= × × = × −B t. . Wb , and the flux in the right 
loop is ψ 2 301 0 5 01 10= × × = × −B t. . . Wb . The induced sources are shown below: 
V d
dt1
1 0 2= =ψ . mV and V d
dt2
2 01= =ψ . mV . Solving the resulting circuit gives 
V = − −
+
= −01 100
100 50
0 2 0 233. . .mV mV mV . 
 
 
 
 
 
 
412. . 
The flux in the inner loop is ψ1 301 0 5 01 10= × × = × −B t. . . Wb , and the flux in the 
outer loop is ψ 2 30 3 1 0 6 10= × × = × −B t. . Wb . The induced sources are shown below: 
V d
dt1
1 01= =ψ . mV and V d
dt2
2 0 6= =ψ . mV . Solving the resulting circuit gives 
V = −
+
=0 6 100
100 50
01 0533. . .mV mV mV . 
 
 
 
 
 
 
 
–
V
+
50 100
0.1 mV
+ –
0.2 mV
+ –
–
V
+
50 100
0.1 mV
0.6 mV
+ –
+ –
4-2 
413. . 
The flux in the left loop is ( )ψ π1 305 0 2 10 2 60= × × = −B t. . sin Wb , and the flux in the 
right loop is ( )ψ π2 303 0 2 0 6 10 2 60= × × = × −B t. . . sin Wb . The induced sources are 
shown below: ( )V d
dt
t1 1
377
2 60 2 60= = ×ψ π πcos1 2444 3444 mV and 
( )V d
dt
t2 2
266
2 60 0 6 2 60= = × ×ψ π π. cos1 24444 34444 mV . Solving the resulting circuit gives 
( ) ( ) ( )V t t t= × +
+
× = ×226 2 60 50
200 50
377 2 60 3016 2 60cos cos . cosπ π πmV mV mV . 
 
 
 
 
 
 
 
414. . 
(b) and (d) are not correct. In (b) a current must be established that will induce a 
secondary current I that will oppose the change in the original magnetic field. For this 
case the induced current must be counterclockwise to keep the field from decreasing. 
Similarly, in (d) a current must be established that will induce a secondary current I that 
will oppose the change in the original magnetic field. For this case the induced current 
must be clockwise to keep the field from increasing. 
415. . 
The flux in the loop is 
( ) ( )ψ π π= × = × × × ×B t t t tArea mWb = mWb2 2 60 05 10 10 2 60cos . cos . The induced 
–
V
+
200 50
V1 V2
+– +–
4-3 
voltage is ( ) ( )[ ]V ddt t t t= = × − ×
ψ
π π10 2 60 3770 2 60cos sin mV . Hence the current is 
( ) ( )[ ]I V t t t= = × − ×100 01 2 60 37 7 2 60. cos . sinπ π mA . 
 
 
 
 
 
 
416. . 
The position of the bar is ( )L d tt= ∫ =100 10 10 10
0
cos sin( )τ τ m . The flux in the loop is 
( ) ( )ψ = × = × ×B t tArea mWb = 0.05 Wb10 05 10 10 10. sin sin . Hence the induced voltage 
is ( )V d
dt
t= =ψ 05 10. cos V . Hence the currentis ( ) ( )I t t= =05 10
100
5 10
. cos
cos mA . 
 
 
 
 
 
 
 
417. . 
A view in the xy plane is shown below. The flux through the loop is 
( ) ( )ψ ω ω= •∫ = × × =B sd B t tArea mWbcos cos1 . Hence an induced voltage in the 
loop is ( ) ( )V d
dt
t t= = −ψ ω ω ωsin sinmV = -5 mV with polarity shown below. Hence 
the current is ( )I V t= = −
2
2 5. cos ω mA . 
I
V
100
50 cm
v
+ –
I
100
50 cm
v
+–
4-4 
 
 
 
 
 
 
 
 
418. . 
The magnetic flux density a distance r from an infinitely long current was obtained in the 
previous chapter as B t I t
r
o( ) ( )= µ
π2
 and is circumferentially directed about the wire. 
Hence the flux through the loop formed by the household power wiring is 
ψ µ
π
= •∫ = ∫ ∫ = × 

 = ×
=
− −B sd I t
r
drdz I t I t
z 0
3 7 8
2
6 10 109 517 10
m
o
r=1km
1.09km km
1km
( ) ln . ( ) . ( ) . 
The induced voltage is V d
dt
dI t
dt
= = × −
ψ 517 10 8. ( ) and is sketched below. Hence 
V = 2 585, V for 0 1< <t µ s and is V = −287V for 1 10µ µs s< <t . 
 
 
 
 
 
 
 
419. . 
The magnetic flux density threading the loop is B I
r
o
=
µ
π2
. Assuming that the loop starts 
at t=0 barely touching the wire, at some time t the flux through the loop is (downward) 
I
V
2
+ –
B
B
�t
x
y
2585 V
–287 V
V (t)
1 �s
10 �s
t
4-5 
ψ µ
π
µ
π
= •∫ = ∫ ∫ = +


= =
+
B sd I
r
drdz Il vt w
vtz
l
o
r vt
vt w
o
0 2 2
ln . The induced voltage in the loop 
(tending to push current counter clockwise) is 
( )V
d
dt
d
dt
Il vt w
vt
Il vt
vt w
w
vt
Ilw
vt w t
o o o
= =
+


 = +



 −



 = − +
ψ µ
π
µ
π
µ
π2 2 22
ln . Hence 
( )I
V
R
Ilw
R vt w t
o
= − =
+
µ
π2
. 
4110. . 
A sketch of the problem looking down on the plane of rotation is shown below. The 
further we go out along the bar length, the greater the velocity cutting the magnetic field 
lines. The linear velocity of a section of the bar at a radius r is v r= ω . The voltage 
induced in the bar is ( )v B l× •∫ = ∫ =
=
d B rdr B l
r
l
ω
ω
0
2
2
. By the right-hand rule, the 
rotating end is the positive end. 
 
 
 
 
 
 
 
 
4111. . 
Forming ∇ × = −

 + −



 + −



E a a a
∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
E
y
E
z
E
z
E
x
E
x
E
y
z y
x
x z
y
y x
z . From the 
form of E this reduces to 
( ) ( )∇ × = − + = − + −E a a a a∂∂
∂
∂ β α ω β α α ω β
E
z
E
x
E x t z E x t zy x
y
z m x m zsin cos cos sin
l
B
�
4-6 
. From Faraday’s law: ∇ × = −E Hµ ∂∂o t we obtain 
( ) ( )H a a= − − + −E x t z E x t zm
o
x
m
o
z
β
ω µ
α ω β α
ω µ
α ω βsin sin cos cos . 
4112. . 
Forming ∇ × = −

 + −



 + −



E a a a
∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
E
y
E
z
E
z
E
x
E
x
E
y
z y
x
x z
y
y x
z . From the 
form of E this reduces to ( )∇ × = =E a a∂∂ β β ω
E
z
E z tx y m ycos cos . From Faraday’s 
law: ∇ × = −E Hµ ∂∂o t we obtain ( )H a= −
E z tm
o
y
β
ω µ
β ωcos sin . 
4 21. . 
Above a frequency for which σ
ω ε εr o
= 1 the displacement current dominates the 
conduction current. This occurs for f > 45kHz . 
4 2 2. . 
Forming ∇ × = −

 + −



 + −



H a a a
∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
H
y
H
z
H
z
H
x
H
x
H
y
z y
x
x z
y
y x
z . From 
the form of H this reduces to 
( ) ( )∇ × = −

 = − + −H a a
∂
∂
∂
∂ β α α ω β
H
z
H
x
H H x t zx z y x z ysin cos . From Ampere’s 
law: ∇ × =H Eε ∂∂o t we obtain 
( ) ( )E a= − + −β α
ωε
α ω βH H x t zx z
o
ysin sin 
4 2 3. . 
Forming ∇ × = −

 + −



 + −



H a a a
∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
H
y
H
z
H
z
H
x
H
x
H
y
z y
x
x z
y
y x
z . From 
the form of H this reduces to ( )∇ × = − =H a a∂∂ β β ω
H
z
H z ty x m xsin sin . From 
Ampere’s law: ∇ × =H Eε ∂∂o t we obtain ( )E a= −
β
ωε
β ωH z tm
o
xsin cos 
4-7 
Gauss’ law becomes, in rectangular coordinates, ( )∇ • = = + +ε ∂∂
∂
∂
∂
∂o
x y zE
x
E
y
E
z
E 0 . 
But for the given electric field, ( )E a= −E x t zm ysin sinα ω β , it has only a y component 
which is independent of y. Hence, the divergence is zero. 
4 3 2. . 
Gauss’ law becomes, in rectangular coordinates, ( )∇ • = = + +ε ∂∂
∂
∂
∂
∂o
x y zE
x
E
y
E
z
E 0 . 
But for the given electric field, ( )E a= E z tm xsin cosβ ω , it has only an x component 
which is independent of x. Hence, the divergence is zero. 
4 3 3. . 
Gauss’ law becomes, in rectangular coordinates, ( )∇ • = = + +µ ∂∂
∂
∂
∂
∂o
x y zH
x
H
y
H
z
H 0 . 
For the given magnetic field, 
( ) ( )H a a= − + −H x t z H x t zx x z zsin sin cos cosα ω β α ω β , so that 
( ) ( )∂∂
∂
∂ α α ω β β α ω β
H
x
H
z
H x t z H x t zx z x x z z+ = − + −cos sin cos sina a . Hence the 
divergence is zero and Gauss’ law is satisfied only if α βH Hx y+ = 0 . 
4 3 4. . 
Gauss’ law becomes, in rectangular coordinates, ( )∇ • = = + +µ ∂∂
∂
∂
∂
∂o
x y zH
x
H
y
H
z
H 0 . 
But for the given magnetic field, ( )H a= H z tm ycos sinβ ω , the y component is 
independent of y. Hence, the divergence is zero. 
4 61. . 
The Poynting vector is S E H a= × = − −

 +








−715 1
2
2 8
4
1
2 4
8. cos cose t zz zω
π π W
m2
. 
The Poynting vector averaged over one cycle is S aaverage 2
W
m
=




−
715
2 4
8. cose z z
π . This 
is perpendicular only to the sides at z=0 and z=1m. Hence the average power exiting the 
cube is 
4 31. .
4-8 
( ) ( )
P S S
e e
z z
z z
average average average m
m
Area+ Area
= - = -25.27W
= − × ×



 +




= =
− = − =
0 1
8 0 8 1715
2 4
715
2 4
. cos . cosπ π
 
4 71. . 
The components of E that are tangential to the boundary must be continuous. Hence 
E a a2, tan = +β γy z . The components of D that are normal to the boundary must be 
continuous. Hence D E a D E1 1 1 1 2 2 2, , , , norm norm norm norm= = = =ε ε α εx so that 
E a2 1
2
, norm =
ε
ε
α x . Therefore, E E E a a a2 0 2 2
1
2
x x y z= = + = + +, , norm tan
ε
ε
α β γ . 
4 7 2. . 
The components of B that are normal to the boundary must be continuous. Hence 
B a2, norm = α x . The components of H that are tangential to the boundary must be 
continuous. Hence H B a a H B1
1
1
1 1
2
2
2
1 1
, , , , tan tan tan tan= = + = =µ
β
µ
γ
µ µy z
 so that 
B a a2 2
1
2
1
, tan = +
µ
µ
β µ
µ
γy z . Therefore, 
B B B a a a2 0 2 2
2
1
2
1
x x y z= = + = + +, , norm tan α
µ
µ
β µ
µ
γ . 
4 7 3. . 
The components of D that are normal to the boundary must be continuous. Hence 
D a2, norm = α x . The components of E that are tangential to the boundary must be 
continuous. Hence E D a a E D1
1
1
1 1
2
2
2
1 1 1 1
, , , , tan tan tan tan= = + = =ε ε
β
ε
γ
εy z
 so that 
D a a2 2
1
2
1
, tan = +
ε
ε
β ε
ε
γy z . Therefore, 
D D D a a a2 0 2 2
2
1
2
1
x x y z= = + = + +,tan ,norm α β
ε
ε
γ ε
ε
. 
4 7 4. . 
The componentsof H that are tangential to the boundary must be continuous. Hence 
H a a2, tan = +β γy z . The components of B that are normal to the boundary must be 
continuous. Hence B H a B H1 1 1 1 2 2 2, , , , norm norm norm norm= = = =µ µ α µx so that 
4-9 
B a a2 2
1
2
1
, tan = +
µ
µ
β µ
µ
γy z . Therefore, 
B B B a a a2 0 2 2
2
1
2
1
x x y z= = + = + +, , norm tan α
µ
µ
β µ
µ
γ . 
4 7 5. . 
The tangential components of E must be zero at the surface of a perfect conductor. Hence 
γ = 0 . Also, the x and y components of E must be equal in order that there be no 
tangential component from these two components and hence α β= . the normal 
components of B must be zero at the surface of a perfect conductor. Hence the x and y 
components must form a resultant that is tangent to the surface so that σ δ= − . 
4 81. . 
The components tangent to the plane are 2 3a ax y− . Reversing these gives − +2 3a ax y . 
Similarly reversing the z component gives −4a z . Hence the image current is 
I a a aimage A= − + −2 3 4x y z at (0,0,-2). 
4 91. . 
To convert a phasor quantity to the time domain we simply multiply by e ej t j tω π= ×2 10
6
 
and take the real part of the result. Hence, 
(a) 
( ) ( )E E a a
a a
= = − + −




= − × +



 + × +




Re $ Re Re
cos cos
e j e
j
e
t t
j t j t
x
j t
y
x y
ω ω ω
π
π
π
π
30 10
30 2 10
2
10 2 10
2
6 6
, 
(b) 
( )H H a a= =   = × + Re $ Re cose e e tj t j t j z zω ω π π π10 10 2 10 434 3 6 , 
(c) 
( ) ( ) ( )B B a a= = = × −− − −Re $ Re cose e e e e t zj t z j t j z x z xω ω π π π4 4 2 10 42 4 2 6 , 
(d) $E a= − −10 3 3e ex j z yπ , (e) ( )$ sinB a= −5 3 6z e j z x . 
4-10 
The phasor form of the field is $ sinE a= −E xem j z yα β . Faraday’s law in phasor form is 
∇ × = −$ $E Hj oω µ . Writing out the curl gives 
∇ × = −





 + −



 + −





$
$ $ $ $ $ $
E a a a∂∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
E
y
E
z
E
z
E
x
E
x
E
y
z y
x
x z
y
y x
z . From the form of 
the field this reduces to 
∇ × = − + = +− −$
$ $
sin cosE a a a a
∂
∂
∂
∂ β α α α
β βE
z
E
x
j E xe E xey x
y
z m
j z
x m
j z
z . Hence 
( )$ $ $ sin cos
sin cos
H E E a a
a a
= − ∇ × = ∇ × = +
= − +
− −
− −
1 1 1
j
j j j E xe E xe
E xe j E xe
o o o
m
j z
x m
j z
z
m
o
j z
x
m
o
j z
z
ω µ ω µ ω µ
β α α α
β
ω µ
α
α
ω µ
α
β β
β β
The time-domain field is 
( )
( )
( ) ( )
H H
a a
a a
=
= − − + − +


= − − − −
Re $
sin cos cos cos
sin cos cos sin
e
E x t z E x t z
E x t z E x t z
j t
m
o
x
m
o
z
m
o
x
m
o
z
ω
β
ω µ
α ω β α
ω µ
α ω β π
β
ω µ
α ω β α
ω µ
α ω β
2
 
4 9 3. . 
Consider two field vectors, ( )E Ee= Re $ j tω and ( )H He= Re $ j tω . The instantaneous 
power density or Poynting vector is S E H= × . Rewriting the field vectors as 
( ) ( )E Ee Ee E e= = + ∗ −Re $ $ $j t j t j tω ω ω12 and substituting into S E H= × yields 
( ) ( )S E H E H E H E H E H= × = × + × + × + ×∗ ∗ ∗ ∗ −14 14 2 2$ $ $ $ $ $ $ $e ej t j tω ω . But since 
( ) ( )$ $ $ $E H E H× = ×∗ ∗ ∗ and ( )$ $ Re $A A A+ =∗ 2 , this may be written as 
( ) ( )S E H E H= × + ×∗12 12 2Re $ $ Re $ $ e j tω . The time average of this is S Sav = ∫1 0T dt
T
 and T 
is the period of the sinusoid. The first term of our result is a constant and the second term 
is sinusoidal and hence averages to zero giving the desired result. 
4 9 2. . 
4-11 
The magnetic flux density in the core due to a current I is, from Chapter 3, B I
r
r oφ
µ µ
π
=
2
 
where the flux is circumferentially directed and r is approximately the mean radius of the 
core. The total flux through the windings is approximately ψ = BA where A is the area of 
the core cross section. According to Faraday’s law a voltage will be generated in the turns 
of wire of V N d
dt
NA
r
dI
dt
r o
= =
ψ µ µ
π2
. But dI
dt
j I⇔ ω $ . Hence $
$
$Z
V
I
NA
rT
r o
= =
ω µ µ
π2
. 
Substituting numerical values gives 
( )( ) ( )( )
( )$ . .Z
f N A
rT
r o
=
= = = = ×
=
=
−ω π µ µ
π
2 200 10 4 10
2 0 02
50 3
4
Ω Ω= 34dB 
 
 
4101. . 
 5-1 
Chapter 5 
Problem Solutions 
511. . 
The time derivative is related to the phasor form by j
t
ω
∂
∂⇔ . Hence 
− ⇔ −j H z
H z t
ty
y
ω µ µ
∂
∂
$ ( )
( , )
 and − ⇔ −j E z E z t
tx
xω ε ε
∂
∂
$ ( ) ( , ) . 
51 2. . 
The time-domain equations are ( ) ( )E z t E t z E t zx m m( , ) cos cos= − + ++ −ω β ω β and 
( ) ( )H z t E t z E t zy m m( , ) cos cos= − − +
+ −
η
ω β
η
ω β . Substituting into the first equation 
gives ( ) ( )∂ ∂ β ω β β ω β
E z t
z
E t z E t zx m m
( , ) sin sin= − − ++ − and 
( ) ( )− = − − ++ −µ ∂ ∂ µω η ω β µω η ω β
H z t
t
E t z E t zy m m
( , )
sin sin . Matching the 
corresponding coefficients of the sin terms gives the requirement that β µω
η
= . But 
substituting β ω µ ε= and η µ
ε
= shows this to be true. Similar results are shown for 
the second equation of Problem 5.1.1. 
51 3. . 
(a) pvc, ( )ε r = 35. β ω µ ε= =vo r r 0392.
rad
m
, η η µ
ε
= =o
r
r
202 Ω , 
v vo
r r
= = ×
µ ε
16 108. m
s
, λ = =v
f
16 m (b) Teflon, ( )ε r = 21. 
β ω µ ε= =
vo
r r 0304.
rad
m
, η η µ
ε
= =o
r
r
260 Ω , v vo
r r
= = ×
µ ε
2 07 108. m
s
, 
λ = =v
f
20 7. m . (c) Mylar, ( )ε r = 5 β ω µ ε= =vo r r 0 468.
rad
m
, η η µ
ε
= =o
r
r
169 Ω , 
v vo
r r
= = ×
µ ε
134 108. m
s
, λ = =v
f
134. m . (d) Polyurethane ( )ε r = 7 
5-2 
β ω µ ε= =
vo
r r 0554.
rad
m
, η η µ
ε
= =o
r
r
142 Ω , v vo
r r
= = ×
µ ε
113 108. m
s
, 
λ = =v
f
113. m . 
51 4. . 
A sketch is shown below. The phase constant is β ω= =
vo
0105. rad
m
 and the intrinsic 
impedance is η η= =o 377 Ω . The electric field intensity vector is given by 
$ .E a= 10 0105e j y z or ( )E a= × +10 10 10 01056cos .π t y z . In order for the power flow 
E H× to be in the -y direction, the magnetic field intensity vector must be in the -x 
direction so that $ . .H a= −0 0265 0105e j y x or ( )H a= − × +0 0265 10 10 01056. cos .π t y x . 
 
 
 
 
 
 
 
 
515. . 
A sketch is shown below. Since the wavelength is λ = v
f
, the frequency of the wave is 
800MHz. Also the velocity of propagation is v vo
r
=
ε
. Hence ε r = 2 25. . The phase 
constant is β ω πλ= = =v
2 251. rad
m
 . The electric field is 
( )E a= × −100 16 10 2518cos .π t x z . The intrinsic impedance is η η
ε
= =
o
r
251Ω . In 
order that E H× be in the +x direction, the magnetic field must be directed in the -y 
direction. Hence ( )H a= − × −0 398 16 10 2518. cos .π t x y 
Ez
Hx
y
z
x
5-3 
 
 
 
 
 
 
 
 
51 6. . 
The frequency of the wave is 40MHz. The phase constant is β ω ε= =
vo
r 29.
rad
m
. The 
intrinsic impedance is η η
ε
= =
o
r
109Ω . The problem is sketched below. The wave is 
traveling in the +y direction. From this E must be in the z direction in order that E H× be 
in the +y direction. The magnitude of the electric field is the product of the magnitude of 
the magnetic field and the intrinsic impedance. Hence, the electric field is 
( )E a= × −10 9 8 10 2 97. cos .π t y z . 
 
 
 
 
 
 
 
 
Ez
Hy
x
y
zEz
Hx
y
z
x
5-4 
51 7. . 
The problem is sketched below. The phase constant is β ω ε µ= =
vo
r r 251
rad
m
. The 
intrinsic impedance is η η µ
ε
= =o
r
r
565Ω . The wave is traveling in the -z direction and 
the magnetic field is in the y direction. Hence the magnetic field intensity vector is given 
by ( )H a= × +0 02 4 10 2519. cos π t z y From this E must be in the -x direction in order that 
E H× be in the -z direction. The magnitude of the electric field is the product of the 
magnitude of the magnetic field and the intrinsic impedance. Hence, the electric field is 
( )E a= − × +113 4 10 2519. cos π t z x . 
 
 
 
 
 
 
 
 
51 8. . 
Since λ
ε
=
v
f
o
r
 we have that ε λλr
o
= = 2 . Hence ε r = 4 . 
5 21. . 
The frequency is 500MHz. The propagation constant is 
( )$γ ω µ µ σ ω ε ε π π π
π
= + = × × × × + × × × ×




− −j j j jo r o r 10 10 4 10 4 1 10 10
1
36
10 368 7 8 9
This evaluates to $ . .γ = ∠ = +149 67 5 57 2 138o j . Hence we identify α = 57 2. and 
β = 138 rad
m
. The intrinsic impedance is 
Hy
Ex
z
x
y
5-5 
( )$η
ω µ µ
σ ω ε ε
π π
π
π
=
+
=
× × × ×
+ × × × ×




−
−
j
j
j
j
o r
o r
10 10 4 10 4
1 10 10 1
36
10 36
8 7
8 9
. This evaluates to 
$ .η = ∠106 22 5o . The problem is sketched below. The wave is traveling in the +x 
direction and the electric field is in the z direction. From this H must be in the -y direction 
in order that E H× be in the +x direction. Hence the magnetic field vector is 
( )H a= − × − −−0 946 10 10 138 22 557 2 8. cos ..e t xx o yπ . 
 
 
 
 
 
 
 
 
5.2.2 
(a) 60Hz. The propagation constant is 
( )$ .γ ω µ µ σ ω ε ε π π π
π
= + = × × + × × ×




− −j j j jo r o r 120 4 10 0 01 120
1
36
10 157 9 
This evaluates to $ . . .γ = × ∠ = × + ×− − −218 10 45 154 10 154 103 3 3o j . Hence we identify 
α = × −154 10 3. and β = × −154 10 3. rad
m
. The velocity of propagation is 
v = = ×ωβ 2 45 10
5. m
s
. The intrinsic impedance is 
( )$ .
η ω µ µ
σ ω ε ε
π π
π
π
=
+
=
× ×
+ × × ×




−
−
j
j
j
j
o r
o r
120 4 10
0 01 120 1
36
10 15
7
9
 which evaluates to 
$ .η = ∠0 22 45o . 
Ez
Hy
x
y
z
5-6 
(b) 1MHz. The propagation constant is 
( )$ .γ ω µ µ σ ω ε ε π π π
π
= + = × × × + × × × ×




− −j j j jo r o r 2 10 4 10 0 01 2 10
1
36
10 156 7 6 9
This evaluates to $ . . . .γ = ∠ = +0 281 47 4 019 0 21o j . Hence we identify α = 019. and 
β = 0 21. rad
m
. The velocity of propagation is v = = ×ωβ 303 10
7. m
s
. The intrinsic 
impedance is ( )$ .
η ω µ µ
σ ω ε ε
π π
π
π
=
+
=
× × ×
+ × × × ×




−
−
j
j
j
j
o r
o r
2 10 4 10
0 01 2 10 1
36
10 15
6 7
6 9
.which evaluates to 
$ . .η = ∠28 05 42 62o . 
(c) 100MHz. The propagation constant is 
( )$ .γ ω µ µ σ ω ε ε π π π
π
= + = × × × + × × × ×




− −j j j jo r o r 2 10 4 10 0 01 2 10
1
36
10 158 7 8 9
This evaluates to $ . . . .γ = ∠ = +814 86 58 0 49 813o j . Hence we identify α = 0 49. and 
β = 813. rad
m
. The velocity of propagation is v = = ×ωβ 7 73 10
7. m
s
. The intrinsic 
impedance is ( )$ .
η ω µ µ
σ ω ε ε
π π
π
π
=
+
=
× × ×
+ × × × ×




−
−
j
j
j
j
o r
o r
2 10 4 10
0 01 2 10 1
36
10 15
8 7
8 9
.which evaluates to 
$ . .η = ∠96 99 342o . 
(d) 10GHz. The propagation constant is 
( )$ .γ ω µ µ σ ω ε ε π π π
π
= + = × × × + × × × ×




− −j j j jo r o r 2 10 4 10 0 01 2 10
1
36
10 1510 7 10 9
This evaluates to $ . . . .γ = ∠ = +81116 89 97 0 49 81116o j . Hence we identify α = 0 49. and 
β = 8112. rad
m
. The velocity of propagation is v = = ×ωβ 7 75 10
7. m
s
. The intrinsic 
impedance is ( )$ .
η ω µ µ
σ ω ε ε
π π
π
π
=
+
=
× × ×
+ × × × ×




−
−
j
j
j
j
o r
o r
2 10 4 10
0 01 2 10 1
36
10 15
10 7
10 9
.which evaluates 
to $ . .η = ∠97 34 0 03o . 
5-7 
The propagation constant is 
( )$γ ω µ µ σ ω ε ε π π π
π
= + = × × × × + × × × ×




− −j j j jo r o r 2 10 4 10 16 2 2 10
1
36
10 99 7 9 9
This evaluates to $ . . . .γ = ∠ = +510 33 52 02 314 06 402 25o j . Hence we identify α = 314 06. 
and β = 402 25. rad
m
. The velocity of propagation is v = = ×ωβ 156 10
7. m
s
. The intrinsic 
impedance is ( )$η
ω µ µ
σ ω ε ε
π π
π
π
=
+
=
× × × ×
+ × × × ×




−
−
j
j
j
j
o r
o r
2 10 4 10 16
2 2 10 1
36
10 9
9 7
9 9
.which evaluates to 
$ . .η = ∠247 55 37 98o . 
5 2 4. . 
The frequency is 10GHz. Forming the propagation constant as 
$γ π π σ π
π
ε= + = × × × + × × × ×


− −200 300 2 10 4 10 2 10 1
36
1010 7 10 9j j j r . Squaring this 
gives $γ π ε π π σ2 4 4 2 4 10 75 10 12 10 16
36
10 2 10 4 10= − × + × = − × + × × × −j jr . Solving gives 
ε r = 114. and σ = 152.
S
m
. The intrinsic impedance can now be computed from 
( )$ . .
η ω µ µ
σ ω ε ε
π π
π
π
=
+
=
× × ×
+ × × × ×




−
−
j
j
j
j
o r
o r
2 10 4 10
152 2 10 1
36
10 114
10 7
10 9
.which evaluates to 
$ . .η = ∠218 98 3369o . The problem is sketched below. The wave is traveling in the +y 
direction and the magnetic field is in the x direction. From this E must be in the +z 
direction in order that E H× be in the +y direction. Hence the electric field vector is 
( )E a= × − +−219 2 10 300 3369200 10. cos .e t yy o zπ . 
 
 
 
 
 
j
5 2 3. . 
Ez
Hx
y
z
x
5-8 
5 31. . 
The surface is sketched below and has a surface area of ( ) ( )3 1 2 1 12− − × − − =( ) ( ) m2 . 
The intrinsic impedance of the medium is η η= =o
5
168 6. Ω . Hence the average power 
density is ( )1
2
10
168 6
0 3
2
.
.= W
m2
. The wave is propagating perpendicular to the surface and is 
uniform over it so that the total average power crossing the surface is 
P d
s
AV AV= •∫ =S s 356W. . 
 
 
 
 
 
 
 
 
 
 
5 3 2. . 
The surface is sketched below and has a surface area of 3 5 15m× = 2 . The intrinsic 
impedance of the medium is η η= =o
9
125 66. Ω . The magnitude of the electric field 
intensity is ( )E = × =0 2 125 66 2513. . . V
m
. Hence the average power density is 
( )1
2
2513
12566
2 51
2.
.
.= W
m2
. The wave is propagating perpendicular to the surface and is 
uniform over it so that the total average power crossing the surface is 
P d
s
AV AV W= •∫ =S s 37 7. . 
 
z
x
y
(3, –1, 2) m
(–1, –1, 2) m
(–1, 2, 2) m
(3, 2, 2) m
5-9 
 
 
 
 
 
 
 
 
 
 
5 3 3. . 
The frequency is 500MHz. The propagation constant is 
( )$γ ω µ µ σ ω ε ε π π π
π
= + = × × × × + × × × ×




− −j j j jo r o r 10 10 4 10 4 1 10 10
1
36
10 368 7 8 9
This evaluates to $ . .γ = ∠ = +149 67 5 57 2 138o j . Hence we identify α = 57 2. and 
β = 138 rad
m
. The intrinsic impedance is 
( )$η
ω µ µ
σ ω ε ε
π π
π
π
=
+
=
× × × ×
+ × × × ×




−
−
j
j
j
j
o r
o r
10 10 4 10 4
1 10 10 1
36
10 36
8 7
8 9
. This evaluates to 
$ .η = ∠106 22 5o . The problem is sketched below. The wave is propagating in the +x 
direction and is perpendicular to the side of area 2m3m = 6m2× . The average power 
density is ( ) ( )S a aAV 2Wm= =− × −12 100106 22 5 436
2
2 57 2 114 4e ex o y
x
y
. .cos . . . Hence the 
power dissipated is ( )436 1 6 235114 4 20. .− × =− ×e mm 2m W . 
 
 
 
 
z
x
y
(3, 0, 0) m
(3, 5, 0) m
(0, 5, 0) m
(0, 0, 0)
5-10 
 
 
 
 
 
 
 
 
 
5 3 4. . 
The propagation constant is 
( )$γ ω µ µ σ ω ε ε π π π
π
= + = × × × × + × × × ×




− −j j j jo r o r 2 10 4 10 16 2 2 10
1
36
10 99 7 9 9
This evaluates to $ . . . .γ = ∠ = +510 33 52 02 314 06 402 25o j . Hence we identify α = 314 06. 
and β = 402 25. rad
m
. The intrinsic impedance is 
( )$η
ω µ µ
σ ω ε ε
π π
π
π
=
+
=
× × × ×
+ × × × ×




−
−
j
j
j
j
o r
o r
2 10 4 10 16
2 2 10 1
36
10 9
9 7
9 9
.which evaluates to 
$ . .η = ∠247 55 37 98o . The wave is perpendicular to the surface of area 100cm 0 012 2m= . . 
Hence the average power density is 
( ) ( )S e ez o zAV 2Wm= = ×− × − −12 124755 3798 159 10
2
2 314 06 3 62812
.
cos . .. . . Hence the power 
dissipated is ( )159 10 1 0 01 15 23 62812 5. . ..× − × =− − ×e mm 2m Wµ . 
5 41. . 
We need to compute the attenuation constant from 
( )$γ α β ω µ µ σ ω ε ε ω µ µ ε ε σ
ω ε ε
= + = + = − −j j j jo r o r o r o r
o r
1 . The 
y
x
z
(20 mm, 2 m, 0)(0, 2, 0) m
(20 mm, 0, 0)
(20 mm, 0, 3 m)
(0, 2 m, 3 m)
20 mm
2 m
3 m
5-11 
attenuation of the amplitude varies as e d−α . In dB this is 
( )20 20 8 6910 10log log ( ) .e d e dd− = − = −α α α . For an attenuation of 80dB, d = 808 69. α . 
(a) 1kHz. σ
ω ε εr o f
=
×
= ×
0 889 10 8 89 10
9
5. . . Good conductor so 
α π µ µ σ≅ = × =−f fr o 397 10 01263. . . Hence d=73.3m. 
(b) 10kHz. σ
ω ε εr o f
=
×
= ×
0 889 10 8 89 10
9
4. . and is a good conductor. Thus α = 0 397. 
and d=23.2m. 
(c) 100kHz. σ
ω ε εr o f
=
×
= ×
0 889 10 8 89 10
9
3. . and is a good conductor. Thus α = 126. 
and d=7.33m. 
(d) 1MHz. σ
ω ε εr o f
=
×
= ×
0 889 10 8 89 10
9
2. . and is a good conductor. Thus α = 397. 
and d=2.32m. 
(e) 10MHz. σ
ω ε εr o f
=
×
=
0 889 10 88 9
9. . and is a good conductor. Thus α = 12 6. and 
d=0.733m. 
(f) 100MHz. σ
ω ε εr o f
=
×
=
0 889 10 8 89
9. . and is a good conductor. Thus α = 39 7. and 
d=23.2cm. 
5 4 2. . 
The attenuation of the amplitude varies as e d−α . In dB this is 
( )20 20 8 6910 10log log ( ) .e d e dd− = − = −α α α . For an attenuation of 20dB, d = 208 69. α . 
The attenuation constant was calculated for these cases in Problem 5.2.2. (a) 60Hz 
α = × −154 10 3. . Hence, d=1.49km. (b) 1MHz. α = 019. . Hence, d=12.1m. (c) 100MHz. 
α = 0 49. . Hence, d=4.7m. (d) 10GHz. α = 0 49. . Hence, d=4.7m. 
5-12 
At these frequencies, sea water may be considered a good conductor. The attenuation 
constant was calculated in Problem 5.4.1 as α = × −397 10 3. f . Hence the skin depth is 
δ
α
= =
1 252
f
. The intrinsic impedance is 
( )$ .η
ω µ µ
σ ω ε ε σ δ σ= + ≅ ∠ = ∠ = × ∠
−
j
j
f
fo r
o r
o o o2 45
2
252
45 14 10 453 . The average 
power dissipated is 
( ) [ ]P e Area f e fn
d
AV
2 W= ∠ −





 = − × =
×−
−
1
2
1
1 253 1
0 865
5m 109 10
2 2 2
3
η
θ δcos
.
.
124 34
. (a) 1kHz. 
The average power dissipated is 34.5W. (b) 10kHz. The average power dissipated is 
10.9W. (c) 100kHz. The average power dissipated is 3.45W. 
5 4 4. . 
The propagation constant is ( ) ( )$ tanγ α β ω µ σ ω ε ω µ ε φ= + = + = − −j j j j2 1 . 
Squaring both sides and equating real and imaginary parts yields α β ω µ ε2 2 2− = − and 
2 2α β ω µ ε φ= tan . Solving gives the desired result. 
5 51. . 
The intrinsic impedances of each region are η η µ
ε
η1 1
1
1
2
188= = =o r
r
o Ω and 
η η µ
ε
η2 2
2
2
3
251= = =o r
r
o Ω . The reflection and transmission coefficients are 
Γ =
−
+
=
−
+
=
η η
η η
η η
η η
2 1
2 1
2
3
1
2
2
3
1
2
1
7
o o
o o
 and T
o
o o
=
+
=
+
=
2
4
3
2
3
1
2
8
7
2
2 1
η
η η
η
η η
. Observe as a 
check that 1+ =Γ T . Since the phase constant in medium 1 is 
β ω µ µ ε ε ω µ ε π1 1 1 0 1 1 6= = =r o r r r
ov
 the frequency of the wave is 450MHz. Hence 
the phase constant in medium 2 is β ω µ µ ε ε ω µ ε π2 2 2 0 2 2 18= = =r o r r r
ov
. The 
electric fields can now be written as ( )E ai xt z= × −100 9 10 68cos π π , 
5 4 3. . 
5-13 
( )E ar xt z= × +1007 9 10 68cos π π , ( )E at xt z= × −8007 9 10 188cos π π . The magnetic 
fields can be found by dividing the electric fields by the intrinsic impedance of the 
appropriate medium and ensuring that the sign is such that E H× is in the correct 
direction for the particular wave. Hence, ( )H ai yt z= × −100188 9 10 68cos π π , 
( )H ar yt z= −
×
× +
100
7 188
9 10 68cos π π , ( )H at yt z=
×
× −
800
7 251
9 10 188cos π π . The 
average power transmitted through a 2m2 area of the surface is 
P
Et
AV,trans
2m= × =1
2
2 52W
2
2η
. 
5 5 2. . 
The intrinsic impedances of each region are η η µ
ε
η1 1
1
2 754= = =o r
r
o Ω and 
η η µ
ε
η2 2
2
1
3
126= = =o r
r
o Ω . The reflection and transmission coefficients are 
Γ =
−
+
=
−
+
= −
η η
η η
η η
η η
2 1
2 1
1
3
2
1
3
2
5
7
o o
o o
 and T
o
o o
=
+
=
+
=
2
2
3
2 1
3
2
7
2
2 1
η
η η
η
η η
. Observe as a 
check that 1+ =Γ T . Since the phase constant in medium 1 is 
β ω µ µ ε ε ω µ ε π1 1 1 0 1 1 83= = =r o r
r r
ov
 the frequency of the wave is 50MHz. Hence 
the phase constant in medium 2 is β ω µ µ ε ε ω µ ε π2 2 2 0 2 2= = =r o r r r
ov
. The electric 
fields can now be written as E ai xt z= × −



10 10 10
8
3
7cos π π , 
E ar xt z= − × +




50
7
10 10 8
3
7cos π π , ( )E at xt z= × −207 10 107cos π π . The magnetic 
fields can be found by dividing the electric fields by the intrinsic impedance of the 
appropriate medium and ensuring that the sign is such that E H× is in the correct 
direction for the particular wave. Hence, H ai yt z= × −




10
754
10 10 8
3
7cos π π , 
5-14 
H ar yt z= ×
× +




50
7 754
10 10 8
3
7cos π π , ( )H at yt z=
×
× −
20
7 126
10 107cos π π . The 
average power transmitted through a 5m2 area of the surface is 
P
Et
AV,trans
2 mW= × =1
2
5m 162
2
2η
. 
5 5 3. . 
The intrinsic impedances of each region are η η µ
ε
η1 1
1
1
3
126= = =o r
r
o Ω and 
η η µ
ε
η2 2
2
1
2
188= = =o r
r
o Ω . The reflection and transmission coefficients are 
Γ =
−
+
=
−
+
=
η η
η η
η η
η η
2 1
2 1
1
2
1
3
1
2
1
3
1
5
o o
o o
 and T o
o o
=
+
=
+
=
2
1
2
1
3
6
5
2
2 1
η
η η
η
η η
. Observe as a 
check that 1+ =Γ T . Since the phase constant in medium 1 is 
β ω µ µ ε ε ω µ ε π1 1 1 0 1 1 2= = =r o r r r
ov
 the frequency of the wave is 100MHz. Hence 
the phase constant in medium 2 is β ω µ µ ε ε ω µ ε π2 2 2 0 2 2 163= = =r o r
r r
ov
. The 
electric fields can now be written as ( )E ai yt z= × −5 2 10 28cos π π , 
( )E ar yt z= × +1 2 10 28cos π π , E at yt z= × − 6 2 10 1638cos π π . The magnetic fields 
can be found by dividing the electric fields by the intrinsic impedance of the appropriate 
medium and ensuringthat the sign is such that E H× is in the correct direction for the 
particular wave. Hence, ( )H ai xt z= − × −5126 2 10 28cos π π , 
( )H ar xt z= × +1126 2 10 28cos π π , H at xt z= − × − 6188 2 10 1638cos π π . The average 
power transmitted through a 5m2 area of the surface is 
P
Et
AV,trans
2m mW= × =1
2
4 383
2
2η
. 
5-15 
The intrinsic impedances of each region are η η µ
ε
η1 1
1
2
3
251= = =o r
r
o Ω and 
η η µ
ε
η2 2
2
4 1508= = =o r
r
o Ω . The reflection and transmission coefficients are 
Γ =
−
+
=
−
+
=
η η
η η
η η
η η
2 1
2 1
4 2
3
4 2
3
5
7
o o
o o
 and T o
o o
=
+
=
+
=
2 8
4 2
3
12
7
2
2 1
η
η η
η
η η
. Observe as a 
check that 1+ =Γ T . Since the phase constant in medium 1 is 
β ω µ µ ε ε ω µ ε π1 1 1 0 1 1 8= = =r o r r r
ov
 the frequency of the wave is 200MHz. Hence 
the phase constant in medium 2 is β ω µ µ ε ε ω µ ε π2 2 2 0 2 2 163= = =r o r
r r
ov
. The 
electric fields can now be written as ( )E ai yt z= − × −2513 4 10 88. cos π π , 
( )E ar yt z= − × +17 95 4 10 88. cos π π , E at yt z= − × − 4308 4 10 1638. cos π π . The 
magnetic fields can be found by dividing the electric fields by the intrinsic impedance of 
the appropriate medium and ensuring that the sign is such that E H× is in the correct 
direction for the particular wave. Hence, ( )H ai xt z= × −01 4 10 88. cos π π , 
( )H ar xt z= − × × +57 01 4 10 88. cos π π , H at xt z= × × − 27 01 4 10 1638. cos π π . The 
average power transmitted through a 3m2 area of the surface is 
P
Et
AV,trans
2m W= × =1
2
3 185
2
2η
. . 
5 5 5. . 
The intrinsic impedance and phase constant of the first medium are η η π1 120= =o and 
β β ω= = =o ov 0 063. . Medium 2 is a good conductor as evidenced by 
σ
ω ε ε
2
2
615 10
o r
= ×. . Hence the attenuation and phase constants can be computed as 
α β δ π µ σ2 2 2 2 2
1 108 83= = = =f . . The intrinsic impedance is 
5 5 4. . 
5-16 
$ .η
σ δ2 2 2
2 45 0154 45= ∠ = ∠o o . The reflection and transmission coefficients are 
$ $
$ . .Γ =
−
+
= ∠ ≅ −
η η
η η
2
2
09999 179 97 1o
o
o and $
$
$ .T o
o
=
+
= × ∠−
2
816 10 452
2
4η
η η
. The fields 
in medium 1 are ( )E ai xt z= × −10 6 10 0 0636cos .π , ( )E ar xt z= − × +10 6 10 0 0636cos .π , 
( )H ai yt z= × −10120 6 10 0 0636π πcos . , ( )H ar yt z= × +10120 6 10 0 0636π πcos . . The fields 
in medium 2 are ( )E at z o xe t z= × × − +− −816 10 6 10 108 83 453 108 83 6. cos .. π and 
( )H at z o o ye t z= × × − + −− −816 100154 6 10 108 83 45 45
3
108 83 6.
.
cos .. π . Hence the average 
power dissipated in the volume is 
( )P T E eiAV 2m W= −  × =− × −12 1 2 59 9
2 2
2
2 10
2
2
3
$
$
cos .
η
θ µη α . 
5 5 6. . 
The intrinsic impedance and phase constant of the first medium are η η π1 120= =o and 
β β ω= = =o ov 20 94. . Medium 2 (stainless steel) is not a good conductor as evidenced 
by σ
ω ε ε
2
2
036
o r
= . . Hence we must compute the attenuation and phase constants directly 
as 
( )$ .γ ω µ µ σ ω ε ε π π π
π2
9 7 9 92 10 4 10 500 0 02 2 10 1
36
10= + = × × × × + × × ×


− −j j j jo r o r
This evaluates to $ . . .γ 2 482 81 801 83 475 62= ∠ = +o j . Hence we identify α 2 83= and 
β 2 47562= . radm . The intrinsic impedance is 
( )$ .
η ω µ µ
σ ω ε ε
π π
π
π
2
9 7
9 9
2 10 4 10 500
0 02 2 10 1
36
10
=
+
=
× × × ×
+ × × ×




−
−
j
j
j
j
o r
o r
.which evaluates to 
$ . .η 2 8176 83 9 9= ∠ o . The reflection and transmission coefficients are 
$ $
$ . .Γ =
−
+
= ∠
η η
η η
2
2
0 91 091o
o
o and $
$
$ . .T o
o
=
+
= ∠
2
191 0 432
2
η
η η
. The fields in medium 1 are 
( )E ai xt z= × −100 2 10 20 949cos .π , ( )E ar o xt z= × + +91 2 10 20 94 0 919cos . .π , 
5-17 
( )H ai yt z= × −100120 2 10 20949π πcos . , ( )H ar o yt z= − × + +91120 2 10 20 94 0 919π πcos . . . 
The fields in medium 2 are ( )E at z o xe t z= × − +−191 2 10 475 62 0 4383 9cos . .π and 
( )H at z o o ye t z= × − + −−1918176 83 2 10 47562 0 43 9 983 9. cos . . .π . In the stainless steel a 
skin depth is δ
π µ µ σ
= =
1 5 03
f o r
. mm . Hence the average power dissipated in the 
volume is ( )P T E eiAV 2m W= −  × =− × × −12 1 2 2 49
2 2
2
2 5 03 10
2
2
3
$
$
cos ..
η
θη α . 
5 5 7. . 
The intrinsic impedance and phase constant of the first medium are η η µ
ε
π1 40= =o r
r
 
and β ω µ ε= =
vo
r r 314. . Medium 2 is a good conductor as evidenced by 
σ
ω ε ε
2
2
720
o r
= . Hence the attenuation and phase constants can be computed as 
α β δ π µ σ2 2 2 2 2
1 198 69= = = =f . . The intrinsic impedance is 
$ .η
σ δ2 2 2
2 45 14 05 45= ∠ = ∠o o . The reflection and transmission coefficients are 
$ $
$ . .Γ =
−
+
= ∠
η η
η η
2
2
0 854 1709o
o
o and $
$
$ . .T o
o
=
+
= ∠
2
0 207 40 812
2
η
η η
. The fields in medium 
1 are ( )E ai xt z= × −5 10 10 3148cos .π , ( )E ar o xt z= × + +4 27 10 10 314 170 98. cos . .π , 
( )H ai yt z= × −540 10 10 3148π πcos . , ( )H ar o yt z= − × + +4 2740 10 10 314 170 98. cos . .π π . 
The fields in medium 2 are ( )E at z o xe t z= × − +−104 10 10 198 69 40 81198 69 8. cos . .. π and 
( )H at z o o ye t z= × − + −−10414 05 10 10 198 69 4081 45198 69 8.. cos . .. π . Hence the average 
power dissipated in the volume is 
( )P T E eiAV 2m W= −  × =− × −−12 1 10 2 64
2 2
2
2 10 4
2
2
2
$
$
cos .
η
θ µη α . 
 
5-18 
5 5 8. . 
The ocean is not a good conductor as evidenced by σ
ω ε εo r
= 0127. . Hence we must 
calculate the intrinsic impedance directly as 
( )$ . .η
ω µ µ
σ ω ε ε
π π
π
π
2
9 7
9 9
2 7 10 4 10
4 2 7 10 1
36
10 81
4172 362=
+
=
× × × ×
+ × × × × ×




= ∠
−
−
j
j
j
j
o r
o r
o . Hence the 
reflection coefficient is $
$
$ . .Γ =
−
+
= ∠
η η
η η
2
2
0 801 17919o
o
o . The reflected power is 
proportional to the square of the magnitude of the reflection coefficient. Hence the 
portion of the incident power that is reflected is 64.2% and the incident power that is 
dissipated in the ocean is 35.8%. 
5 5 9. . 
The total electric field is approximately zero at a distance of one-half wavelength from the 
surface of a good conductor. Hence λo = 2m . Therefore the lowest possible frequency 
of the wave is f vo= =
2
150 MHz . 
5 61. . 
The problem solution is sketched below. From Snell’s law 
( )sin sin cosψ θ θ= − =1 90 1
n n
. Similarly, sin
sin
θ
ψ
t n
=
1
. Thus ( )
sin
sin
sin
sin
ψ
θ
θ
ψ90 1− =
t . 
Thus the direction of the beam as it exits the material is the same as the incident beam. 
The angle φ θ ψ= − −90 and r t=
cosψ . Hence ( )d r
t
= = +sin
cos
cosφ ψ θ ψ . We 
need to eliminate ψ from this expression by writing it in terms of θ . We have the 
identity ( )cos cos cos sin sinθ ψ θ ψ θ ψ+ = − . Also we have Snell’s law sin cosψ θ=
n
. 
Substituting gives d t t
n
= −
−
cos sin cos
cos
θ θ θ
θ2 2
. 
 
 
5-19 
 
 
 
 
 
 
 
 
 
5 6 2. . 
The critical angle is given by sin
.
θ c n
= =
1 1
15
 giving θ c o= 4181. . Hence, the ray strikes 
the back face with an angle of incidence of 45o which is greater than the critical angle 
and hence the ray is completely reflected. According to Snell’s law it will be reflectedalso with an angle of 45o and will strike the bottom face normal to it. The transmission 
coefficient at the front face is T n
no
1
2 2
1 1
0 8=
+
=
+
=
η
η η
. . At the bottom face, the 
transmission coefficient is T
n
o
o
1
2 2
1 1
12=
+
=
+
=
η
η η
. . Hence the net transmission 
coefficient is T T1 2 0 96= . . The incident power density is S
Ei i
o
=
2
2η
 and the transmitted 
power density is S T T Et
i
o
=
1
2
2
2 2
2η
. Hence the ratio of the transmitted to incident power 
densities is ( )T T1 2 2 0922= . . 
t
d
r
n
�t�
�
90 – �
�
�
5-20 
 
 
5 6 3. . 
The problem is sketched below. Snell’s law requires that sin sinθ θt in= . Hence 
( )tan
tan
90 6
20
1
− = =θ
θt t
 so that θ t o= 733. and θ i o= 39 7. . Thus 
D i= =10 83tan .θ ft . Thus the fish is at a distance of 28.3 ft from the boat. 
D
n = 1.5
6 ft.
10 ft.
20 ft.
�t
�i
 6-1 
Chapter 6 
Problem Solutions 
611. . 
The circuits are shown below. 
(a) Relating the voltages gives ( ) ( ) ( )V z z t V z t l z I z z t
t
+ − = −
+
∆ ∆
∆
, ,
,∂
∂ . Dividing both 
sides by ∆z and letting ∆z → 0 gives the first transmission line equation. Similarly, 
relating the currents gives ( ) ( ) ( )I z z t I z t c z V z t
t
+ − = −∆ ∆, ,
,∂
∂ . Dividing both sides by 
∆z and letting ∆z → 0 gives the second transmission line equation. 
(b) Relating the voltages gives ( ) ( ) ( ) ( )V z z t V z t l z
t
I z t c z
V z t
t
+ − = − −



∆ ∆ ∆, , ,
,∂
∂
∂
∂
1
3
. 
Dividing both sides by ∆z gives ( ) ( ) ( ) ( )V z z t V z t
z
l
t
I z t c z
V z t
t
+ −
= − −




∆
∆
∆
, ,
,
,∂
∂
∂
∂
1
3
. 
Letting ∆z → 0 gives the first transmission line equation. Similarly, relating the currents 
gives ( ) ( ) ( ) ( )I z z t I z t c z V z t
t
c z
V z z t
t
+ − = − −
+
∆ ∆ ∆
∆
, ,
, ,1
3
2
3
∂
∂
∂
∂ . Dividing both sides 
by ∆z gives ( ) ( ) ( ) ( )I z z t I z t
z
c
V z t
t
c
V z z t
t
+ −
= − −
+∆
∆
∆, , , ,1
3
2
3
∂
∂
∂
∂ . Letting ∆z → 0 
gives the second transmission line equation. 
(c) Relating the voltages gives 
( ) ( ) ( ) ( )V z z t V z t l z I z t
t
l z
I z z t
t
+ − = − −
+
∆ ∆ ∆
∆
, ,
, ,1
4
3
4
∂
∂
∂
∂ . Dividing both sides by ∆z 
gives ( ) ( ) ( ) ( )V z z t V z t
z
l
I z t
t
l
I z z t
t
+ −
= − −
+∆
∆
∆, , , ,1
4
3
4
∂
∂
∂
∂ . Letting ∆z → 0 gives the 
first transmission line equation. Similarly, relating the currents gives 
( ) ( ) ( )I z z t I z t c z
t
V z t l z I z t
t
+ − = − −



∆ ∆ ∆, , ,
( , )∂
∂
∂
∂
1
4
. Dividing both sides by ∆z gives 
( ) ( ) ( )I z z t I z t
z
c
t
V z t l z I z t
t
+ −
= − −




∆
∆
∆
, ,
, ( , )∂∂
∂
∂
1
4
. Letting ∆z → 0 gives the second 
transmission line equation.
6-2 
 
 
+
(a)
V(z, t)
I(z, t) I(z +∆z, t)
V(z +∆z, t)
l∆z
c∆z
c∆z
�V(z, t)
�t–
+
–
�V(z, t)
�t
+
(b)
V(z, t)
I(z, t) I(z +∆z, t)
V(z +∆z, t)
l∆z
c∆z
c∆z
–
+
–
c∆z
�V(z + ∆z, t)
�t
1
3
1
3
2
3
c∆z23
+
(c)
V(z, t)
I(z, t) I(z +∆z, t)
V(z +∆z, t)
l∆z
–
+
–
l∆z
c∆z
3
4
1
4
V(z, t) �I(z, t)
�t
�
�t
1
4c∆z l∆z–
6-3 
61 2. . 
Substitute into (6.3) and (6.4). Exact: 27.33pF/m, 0.4065µH/m, Approximate: 24.38pF/m, 
0.4558µH/m. The ratio of wire separation to wire radius is only 3.13. Evidently this is 
not sufficient for the approximate relations in (6.4) to be valid. The closeness of the wires 
means that proximity effect cannot be ignored and the charge is not uniformly distributed 
around the wire peripheries as is assumed by the approximate results in (6.4). 
61 3. . 
Substitute into (6.5) and (6.6). Exact: 42.18pF/m, 0.2634µH/m, Approximate: 40.07pF/m, 
0.2773µH/m. Observe that the ratio of wire height above ground to wire radius is only 2.0 
which is not sufficient for the approximate results to be valid although the error is only 
about 5%. 
61 4. . 
Using (6.7) we obtain a per-unit-length capacitance of 53.83pF/m and a per-unit-length 
inductance of 0.3µH/m. The velocity of propagation relative to free space is 
v
vo r
= =
1 0 83
ε
. . 
61 5. . 
From (6.10b) w
s
w
s
e
= =
1
2
. Substituting into (6.10a) and (6.10c) yields l=0.334µH/m and 
c=156.4pF/m. 
616. . 
The width to height ratio is w
h
= 0156. . Using (6.11a) gives the per-unit-length 
inductance of 0.7873µH/m. From (6.11b) the effective relative permittivity is ′ =ε r 3079. . 
Using (6.11c) gives the per-unit-length capacitance of 43.46pF/m. 
61 7. . 
The parameter k is k s
s w
=
+
=
2
1
3
 which is less than 1
2
. Hence we must compute 
′ = − =k k1 09432 . . Substituting into (6.12a) gives a per-unit-length inductance of 
6-4 
0.8038µH/m. From (6.12c) the effective relative permittivity is ′ =ε r 2825. . Substituting 
into (6.12d) gives a per-unit-length capacitance of 39.06pF/m. 
6 21. . 
The assumed solutions are given in (6.13). Define s t z
v
+
= − and s t z
v
−
= + . Hence 
{ {
∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
V
z
V
s
s
z
V
s
s
z
v v
= +
+
+
+
−
−
−
−
+
1 1
 and 
{ {
∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
I
t Z
V
s
s
t Z
V
s
s
tC C
= −
+
+
+ −
−
−1 1
1 1
. 
Substituting into the first transmission line equation given in (6.1a) yields 
∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
V
z v
V
s v
V
s
l I
t
l
Z
V
s
l
Z
V
sC C
= − + = = − = − +
+
+
−
−
+
+
−
−
1 1 ? . Comparing terms 
requires for an equality that 1
v
l
ZC
= . But equations (6.14) and (6.15) that define v and 
ZC confirm this equality. By a virtually identical method we can prove that (6.13) satisfy 
the second transmission line equation given in (6.1b). 
6 2 2. . - 6 2 7. . 
Z l
cC
= , v
lc
vo
r
= =
′
1
ε
. 
6 2 8. . 
Z l
cC
= so that Z l
cC
2
= . v
lc
=
1 so that v
lc
2 1
= . Solving gives Z vlC = and 
Z
vcC
=
1 . Hence l Z
v
C
= and c
vZC
=
1 . 
6 2 9. . 
First sketch the load voltage. The one-way time delay is T
v
= = 1ns . The source 
reflection coefficient is ΓS = −
1
4
, and the load reflection coefficient is ΓL =
1
2
. The 
initially sent out voltage is Vinit = .25V= ×
50
80
10V 6 . The individual incident and 
reflected voltages are sketched below as is the total. The steady-state load voltage is 
150
180
10 8333× = . V . 
6-5 
V(�, t)
V(�, t)
1 2 3 4 5 6 7 8 9 10 11
6.25
3.125
0.098
0.049
–0.006
–0.012–0.391
–0.781
1 2 3 4 5
9.375
8.203
8.350
8.331
8.334
6 7 8 9 10 11
t(ns)
t(ns)
0.002
0.001
6-6 
Next sketch the input current to the line. The source reflection coefficient is ΓS =
1
4
, and 
the load reflection coefficient is ΓL = −
1
2
. The current reflection coefficients are the 
negatives of the corresponding voltage reflection coefficients. The initially sent out 
current is Iinit A = .125A=
10
80
0 . The individual incident and reflected currents are 
sketched below as is the total. The steady-state input current is 10
180
0 056= . A . 
I (0, t)
I (0, t)
1 2 3 4 5 6 7 8 9 10 11
0.125
0.008
0.002
–2.44 × 10–4
3.052 × 10–5
–0.001–0.016
–0.063
1 2 3 4 5
0.125
0.047
0.057
0.055
0.056
6 7 8 9 10 11
t(ns)
t(ns)
1.221 × 10–4
6-7 
6 210. . 
First sketch the input voltage to the line. The one-way time delay is T
v
= = 2ns . The 
source reflection coefficient is ΓS = −
1
3
, and the load reflection coefficient is ΓL = +1. 
The initially sent out voltage is Vinit V = .33V= ×
100
150
5 3 . The individual incident and 
reflected voltages are sketched below as is the total. The steady-state load and input 
voltages are 5V. 
V(0, t)
V(0, t)
2 4 6 8 10 12 14 16 18 20 22
3.33
0.37
0.37
–0.123
–0.123
–1.111
–1.111
2 4 6 8 10
3.33
5.556
4.815
5.062
4.979
12 14 16 18 20 22
t(ns)
t(ns)
3.33
0.041
6-8 
Next sketch the load voltage. The individual incident and reflected voltages are sketched 
below as is the total. 
V(�, t)
V(�, t)
2 4 6 8 10 12 14 16 18 20 22
3.33
0.37
0.37
–0.123
–0.123
–1.111
–1.111
2 4 6 8 10
6.667
4.444
5.185
4.938
5.021
12 14 16 18 20 22
t(ns)
t(ns)
3.33
0.041
0.041
6-9 
6 211. . 
First sketch the input voltage to the line. The one-way time delay is T
v
= = 1µ s . The 
source reflection coefficient is ΓS = −
1
3
, and the load reflection coefficient is ΓL = −1. 
The initially sent out voltage is Vinit == ×
100
150
100V 66.67V . The individual incident 
and reflected voltages are sketched below as is the total. The steady-state input voltage is 
0V. 
V(0, t)
V(0, t)
1 2 3 4 5 6 7 8 9 10 11
22.22
7.407
2.469
–2.469
–7.407
–22.22
–66.67
1 2 3 4 5
66.67
22.22
7.407
2.469
0.823
6 7 8 9 10 11
t(�s)
t(�s)
66.67
0.823
6-10 
Next sketch the load current. For the currents, the reflection coefficients are the negative 
of the voltage reflection coefficients. The source reflection coefficient is ΓS =
1
3
, and the 
load reflection coefficient is ΓL = +1. The initially sent out current is 
Iinit = .667A=
100
150
0 . The individual incident and reflected currents are sketched below 
as is the total. 
I (�, t)
I (�, t)
1 2 3 4 5 6 7 8 9 10 11
1 2 3 4 5
1.333
1.778
1.926
1.975
1.992
6 7 8 9 10 11
t(�s)
t(�s)
0.222
0.222
0.667
0.667
0.074
0.074 0.025
0.025 0.008
0.008
6-11 
6 212. . 
First sketch the input voltage to the line. The one-way time delay is T
v
= = 4ns . The 
source reflection coefficient is ΓS =
1
3
, and the load reflection coefficient is ΓL = −
1
3
. 
The initially sent out voltage pulse is Vinit == ×
100
300
30V 10V . The individual incident 
and reflected voltages are sketched below as is the total. The steady-state load and input 
voltages are 0V since the input pulse only lasts for 12ns. 
V(0, t)
V(0, t)
4 8 12 16 20 24 28 32 36 40 44
4 8 12 16 20
10
5.556
–4.444
–3.951
0.494 0.439
–0.055
24 28 32 36 40 44
t(ns)
t(ns)
10/27
10/8
–10/9
–10/3
–0.041
–0.014
10
6-12 
Next sketch the load voltage. The individual incident and reflected currents are sketched 
below as is the total. 
V(�, t)
V(�, t)
4 8 12 16 20 24 28 32 36 40 44
4 8 12 16 20
6.667
5.926
–0.741
–0.658
0.082
0.073
24 28 32 36 40 44
t(ns)
t(ns)
10
0.37
0.123
0.005
–0.014
–0.041
–1.111
–3.333
6-13 
6 213. . 
The problem is sketched below. Sketch the individual incident and reflected waves as we 
would if we knew the numerical values as shown below. Comparing the total to the given 
result for Vin we obtain 
VS
2
100= giving VS = 200V and 100 20ΓL = giving ΓL =
1
5
. 
Hence ΓL L
L
R
R
= =
−
+
1
5
50
50
 or RL = 75Ω . Next we identify 10
2 16+ =
v
 giving 
v
= 3µ s . 
But v = × = ×3 10
21
2 07 10
8
8
.
. . Hence, = 621059. m . 
10
50
ZC = 50
RL = ?
v = 2.0702 × 108
VS
VS (t)
–
VS (t)
+
–
Vin
+
� = ?
t(�s)
2 � /v 10 + 2 � /v10
Vin
VS
2
t(�s)
VS
2
�L
(1 + �L)
VS
2
6-14 
6 214. . 
The problem is sketched below. Sketch the individual incident and reflected waves as we 
would if we knew the numerical values as shown below. Comparing the total to the given 
result for Iin we obtain 
12 150
ZC
= mA giving ZC = 80Ω . Also ( )1 2 12 10− = −ΓL
CZ
mA 
giving ΓL = 0 533. . Hence ΓL L
L
R
R
= =
−
+
0 533 80
80
. or RL = 2629. Ω . 
 
 
 
 
 
 
 
 
6 215. . 
The source reflection coefficient is ΓS = −1 and the load reflection coefficient is 
ΓL = +1 . The resulting sketches are shown below. 
 
 
 
t4 T
2 T
12/ZC
Iin
2 (12/ZC)�L
(12/ZC)–�L
�S = –1 �L = +1�r
5 V
VS (t)
VS (t)
–
VL
+
t
+
–
6-15 
(a) τ r T=
1
10
 
 
 
 
 
 
 
 
 
 
 
 
(b) τ r T= 2 
 
 
 
 
 
 
 
 
10
5
VL
3T 5T 7T 9T 11T tT
10
VL
3T 5T 7T 9T 11T tT
6-16 
(c) τ r T= 3 
 
 
 
 
 
 
 
 
 
 
 
(d) τ r T= 4 
3.335
6.65
VL
3T 5T 7T 9T 11T tT
5
VL
3T 5T 7T 9T 11T tT
6-17 
6 216. . 
The source reflection coefficient is ΓS = −
2
3
 and the load reflection coefficient is 
ΓL =
2
3
. The initially sent out voltage is V Z
Z Z
C
C C
init V = 4.167V=
+
×1
5
5 . The load 
voltage is sketched below. 
VL
VL
T 2T 3T 4T 5T 6T 7T 8T 9T 10T 11T
2.778
0.823
0.549
–0.244
–0.366
–1.235
–1.852
T 2T 3T 4T 5T
6.994
3.858
5.23
4.62
6T 7T 8T 9T 10T 11T
t
t
4.167
6-18 
6 217. . 
For currents, the source reflection coefficient is ΓS = −
1
3
 and the load reflection 
coefficient is ΓL =
3
7
. The initially sent out current is Iinit = 40mA= +
×
1
100 50
6V . 
The input current to the line is sketched below. 
Iin
Iin
1 2 3 4 5
1 mA9.8 mA
9.796 mA
17.14 mA
40 mA
40 mA
40 mA
11.4 mA
11.43 mA
1 2 3 4 5
51.43 mA
51.43 mA
–5.7 mA
–2.45 mA
–1.633 mA
–1.633 mA
t(�s)
t(�s)
6-19
6 218. . 
The SPICE circuit with nodes labeled is shown below. The PSPICE program is 
PROBLEM 6.2.18 
VS 1 0 PWL(0 0 0.01N 10) 
RS 1 2 30 
T 2 0 3 0 Z0=50 TD=1N 
RL 3 0 150 
.TRAN 0.01N 10N 0 0.01N 
.PROBE 
.END 
The PSPICE outputs are shown on the next pages. 
 
 
10 V
0.01 ns
VS
t
30
150
T = 1 ns
I(0, t)
VS
–
V(�, t)
+
+
–
321
0
ZC = 50 Ω
6-20
 
V(3) Time
0ns 2ns 4ns 6ns 8ns 10ns
0V
2V
4V
6V
8V
10V
Date/Time run: 07/01/02 10:18:48 Temperature: 27.0
PROBLEM 6.2.18
I(RS) Time
0ns 2ns 4ns 6ns 8ns 10ns
0V
20mA
40mA
60mA
80mA
100mA
120mA
140mA
Date/Time run: 07/01/02 10:18:48 Temperature: 27.0
PROBLEM 6.2.18
6-21
6 219. . 
The SPICE circuit with nodes labeled is shown below. The PSPICE program is 
PROBLEM 6.2.19 
VS 1 0 PWL(0 0 0.01N 5) 
RS 1 2 50 
T 2 0 3 0 Z0=100 TD=2N 
RL 3 0 1E8 
.TRAN 0.01N 20N 0 0.01N 
.PROBE 
.END 
The PSPICE output is shown on the next page. 
5 V
0.01 ns
VS
t
50
ZC = 100 Ω
108
T = 2 ns
I(0, t)
VS
–
V(�, t)
+
–
V(0, t)
+
+
–
321
0
6-22
 
V(2) Time
0ns 5ns 10ns 15ns 20ns
0V
1.0V
2.0V
3.0V
4.0V
5.0V
6.0V
Date/Time run: 07/01/02 10:23:55 Temperature: 27.0
PROBLEM 6.2.19
V(3) Time
0ns 5ns 10ns 15ns 20ns
0V
1.0V
2.0V
3.0V
4.0V
5.0V
7.0V
6.0V
Date/Timerun: 07/01/02 10:23:55 Temperature: 27.0
PROBLEM 6.2.19
6-23
6 2 20. . 
The SPICE circuit with nodes labeled is shown below. The PSPICE program is 
PROBLEM 6.2.20 
VS 1 0 PWL(0 0 0.01U 100) 
RS 1 2 50 
T 2 0 3 0 Z0=100 TD=1U 
RL 3 0 1E-8 
.TRAN 0.01U 10U 0 0.01U 
.PROBE 
.END 
The PSPICE outputs are shown on the next pages. 
100 V
0.01 �s
VS
t
50
ZC = 100 Ω
10–8
T = 1 �s
+
–
321
0
6-24 
 
V(2) Time
0us 2us 4us 6us 8us 10us
0V
10V
20V
30V
40V
50V
70V
60V
Date/Time run: 07/01/02 10:29:50 Temperature: 27.0
PROBLEM 6.2.20
I(RL) Time
0us 2us 4us 6us 8us 10us
0.0A
0.4A
0.8A
1.2A
1.6A
2.0A
Date/Time run: 07/01/02 10:29:50 Temperature: 27.0
PROBLEM 6.2.20
6-25
6 2 21. . 
The SPICE circuit with nodes labeled is shown below. The PSPICE program is 
PROBLEM 6.2.21 
VS 1 0 PWL(0 0 0.01N 30 12N 30 12.01N 0) 
RS 1 2 200 
T 2 0 3 0 Z0=100 TD=4N 
RL 3 0 50 
.TRAN 0.01N 32N 0 0.01N 
.PROBE 
.END 
The PSPICE outputs are shown on the next pages. 
30 V
0.01 ns 12 ns 12.01 ns
VS
VS
t
200
ZC = 100 Ω
50
T = 4 ns
+
–
321
0
6-26
 
V(2) Time
0ns 4ns 12ns 20ns 28ns 32ns
–5V
0V
5V
10V
Date/Time run: 07/01/02 10:35:59 Temperature: 27.0
PROBLEM 6.2.21
8ns 16ns 24ns
V(3) Time
0ns 4ns 12ns 20ns 28ns 32ns
–2.0V
0.0V
2.0V
4.0V
6.0V
8.0V
Date/Time run: 07/01/02 10:35:59 Temperature: 27.0
PROBLEM 6.2.21
8ns 16ns 24ns
6-27
6.2.22 
The SPICE circuit with nodes labeled is shown below. The PSPICE program is 
PROBLEM 6.2.22 
VS 1 0 PWL(0 0 0.01U 200 10U 200 10.01U 0) 
RS 1 2 50 
T 2 0 3 0 Z0=50 TD=3U 
RL 3 0 75 
.TRAN 0.01U 20U 0 0.01U 
.PROBE 
.END 
The PSPICE output is shown on the next page. 
200 V
0.01 �s 10 �s 10.01 �s
VS
VS
t
50
ZC = 50 Ω
75
T = 3 �s
+
–
321
0
6-28
 
V(2) Time
0us 20us
0V
20V
40V
60V
100V
80V
120V
Date/Time run: 07/01/02 10:46:35 Temperature: 27.0
PROBLEM 6.2.22
5us 10us 15us
6-29
6 2 23. . 
The SPICE circuit with nodes labeled is shown below. The PSPICE program is 
PROBLEM 6.2.23 
VS 1 0 PWL(0 0 0.01U 12) 
RS 1 2 1E-8 
T 2 0 3 0 Z0=80 TD=2.25U 
RL 3 0 262.9 
.TRAN 0.01U 6U 0 0.01U 
.PROBE 
.END 
The PSPICE output is shown on the next page. 
12 V
0.01 �s
VS
VS
t
262.9
ZC = 80 Ω
10–8
T = 2.25 �s
+
–
321
0
6-30
 
I(RS) Time
0.0us 1.0us 2.0us 3.0us 4.0us 5.0us 6.0us
–50mA
0mA
50mA
100mA
150mA
Date/Time run: 07/01/02 10:55:51 Temperature: 27.0
PROBLEM 6.2.23
6-31
6 2 24. . 
The SPICE circuit with nodes labeled is shown below. Arbitrarily choose ZC = 100Ω 
and the one-way time delay T = 10ns . For (a) choose a rise time of τ r T= =
1
10
1ns . For 
(b) choose a rise time of τ r T= =2 20ns . For (c) choose a rise time of τ r T= =3 30ns . 
For (d) choose a rise time of τ r T= =4 40ns . The PSPICE program for (a) is 
PROBLEM 6.2.24 
VS 1 0 PWL(0 0 1N 5) 
RS 1 2 1E-8 
T 2 0 3 0 Z0=100 TD=10N 
RL 3 0 1E8 
.TRAN 0.1N 100N 0 0.1N 
.PROBE 
.END 
The PSPICE outputs are shown on the next pages. 
 
5 V
1 ns
VS
VS
t
ZC = 100 Ω
10–8
108
T = 10 ns
+
–
321
0
6-32
 
V(3) Time
0ns 20ns 40ns 60ns 80ns 100ns
0V
2V
4V
6V
8V
10V
Date/Time run: 07/01/02 11:05:54 Temperature: 27.0
PROBLEM 6.2.24(a)
V(3) Time
0ns 20ns 40ns 60ns 80ns 100ns
0V
2V
4V
6V
8V
10V
Date/Time run: 07/01/02 11:09:56 Temperature: 27.0
PROBLEM 6.2.24(b)
6-33
 
V(3) Time
0ns 20ns 40ns 60ns 80ns 100ns
0.0V
1.0V
2.0V
3.0V
4.0V
5.0V
6.0V
7.0V
Date/Time run: 07/01/02 11:11:53 Temperature: 27.0
PROBLEM 6.2.24(c)
V(3) Time
0ns 20ns 40ns 60ns 80ns 100ns
0.0V
1.0V
2.0V
3.0V
4.0V
5.0V
6.0V
Date/Time run: 07/01/02 11:13:23 Temperature: 27.0
PROBLEM 6.2.24(d)
6-34 
6 2 25. . 
The SPICE circuit with nodes labeled is shown below. The PSPICE program is 
PROBLEM 6.2.25 
VS 1 0 PWL(0 0 0.001N 5) 
RS 1 2 10 
T 2 0 3 0 Z0=50 TD=1N 
RL 3 0 250 
.TRAN 0.001N 10N 0 0.001N 
.PROBE 
.END 
The PSPICE output is shown on the next page. 
5 V
0.001 ns
VS
VS
t
250
ZC = 50 Ω
10
T = 1 ns
+
–
321
0
6-35
 
V(3) Time
0ns 2ns 4ns 6ns 8ns 10ns
0.0V
1.0V
2.0V
3.0V
4.0V
5.0V
7.0V
Date/Time run: 07/01/02 11:18:44 Temperature: 27.0
PROBLEM 6.2.25
6.0V
6-36
6 2 26. . 
The SPICE circuit with nodes labeled is shown below. The PSPICE program is 
PROBLEM 6.2.26 
VS 1 0 PWL(0 0 0.001U 6 3U 6 3.001U 0) 
RS 1 2 100 
T 2 0 3 0 Z0=50 TD=1U 
RL 3 0 20 
.TRAN 0.001U 5U 0 0.001U 
.PROBE 
.END 
The PSPICE output is shown on the next page. 
 
VS 20
ZC = 50 Ω
100
T = 1 �s
+
–
321
0
6 V
0.001 �s 3 �s 3.001 �s
VS
t
6-37
 
I(RS) Time
0.0ms 1.0ms 2.0ms 3.0ms 4.0ms 5.0ms
60mA
50mA
40mA
30mA
20mA
10mA
0mA
Date/Time run: 07/01/02 11:25:40 Temperature: 27.0
PROBLEM 6.2.26
6-38
6 2 27. . 
The SPICE circuit with nodes labeled is shown below. The PSPICE program is 
PROBLEM 6.2.27 
VS 1 0 PWL(0 0 1N 5) 
RS 1 2 30 
T 2 0 3 0 Z0=100 TD=0.5N 
CL 3 0 10P 
.TRAN 0.01N 10N 0 0.01N 
.PROBE 
.END 
The PSPICE output is shown on the next page. 
 
5 V
1 ns
VS
VS
t
10 pF
ZC = 100 Ω
30
T = 0.5 ns
+
–
321
0
6-39
 
V(3) Time
0ns 2ns 4ns 6ns 8ns 10ns
0.0V
2.0V
4.0V
6.0V
8.0V
Date/Time run: 07/01/02 09:40:09 Temperature: 27.0
PROBLEM 6.2.27
6-40
6 31. . 
The phasor transmission line equations are ( ) ( )dV z
dz
j l I z
$ $
= − ω and ( ) ( )dI z
dz
j c V z
$ $
= − ω . 
The solutions are $( ) $ $V z V e V ej z j z= ++ − −β β and $( )
$ $
I z V
Z
e V
Z
e
C
j z
C
j z
= −
+
−
−β β . 
Differentiating the voltage expresion with respect to z gives 
dV z
dz
j V e j V e j l V
Z
e V
Z
ej z j z
C
j z
C
j z$( ) $ $ $ $
= − + = − −





+ − −
+
−
−
β β ωβ β β β . Matching 
corresponding exponentials and using the relations Z l
cC
= and β ω= lc shows an 
equality for the first transmission line equation. Equality in the second transmission line 
equation can be similarly shown. 
6 3 2. . 
The time-domain transmission line equations are ( ) ( )∂ ∂
∂
∂
V z t
z
l
I z t
t
, ,
= − and 
( ) ( )∂
∂
∂
∂
I z t
z
c
V z t
t
, ,
= − . The time-domain solutions are 
( ) ( ) ( )V z t V t z V t z, cos cos= − + + + ++ + − −ω β θ ω β θ and 
( ) ( ) ( )I z t VZ t z VZ t zC C, cos cos= − + − + +
+
+
−
−ω β θ ω β θ . Substitution into the first 
transmission line equation gives 
( ) ( ) ( )
( ) ( ) ( )
∂
∂ β ω β θ β ω β θ
∂
∂ ω ω β θ ω ω β θ
V z t
z
V t z V t z
l
I z t
t
l V
Z
t z l V
Z
t z
C C
,
sin sin
,
sin sin
= − + − + + =
− = − + − + +
+ + − −
+
+
−
−
 
Using the relations Z l
cC
= and β ω= lc shows an equality for the first transmission 
line equation. Equality in the second transmission line equation can be similarly shown. 
6-41
6 3 3. . 
The line length as a fraction of a wavelength is = kλ . Hence k f
v
= = 13. . The load 
reflection coefficient is $
$
$ . .ΓL
LC
L C
oZ Z
Z Z
=
−
+
= ∠0 9338 9866 . The exponential is 
e j o− = ∠ −2 1 936β . Hence the input reflection coefficient is 
( )$ $ . .Γ Γ0 0 9338 15392= = ∠−L j oe β . The input impedance is 
( )[ ]
( )[ ]$
$
$ . .Z Zin C
o
=
+
−
= ∠
1 0
1 0
1173 8116
Γ
Γ
Ω . The phasor input voltage to the line is 
( )$ $$ $ $ . .V
Z
Z Z
Vin
S in
S
o0 20 55 1213=
+
= ∠ . The undetermined constant in the solution is 
( )
( )[ ]$
$
$ . .V
V o+
=
+
= ∠
0
1 0
46 51 52 79
Γ
. Hence the phasor load voltage is 
( ) [ ]$ $ $ . .V V e j L o= + = ∠ −+ − β 1 89 6 50 45Γ . The average power delivered to the load is 
( )
P
V
ZL
ZLAV,load = =
1
2
2 77W
2$
cos .θ . This can be confirmed by determining the 
average power delivered to the input of the line: 
( )
P
V
Zin
ZinAV,load = =
1
2
0
2 77W
2$
cos .θ 
which should equal the power delivered to the load since the line is lossless. The VSWR 
is VSWR =
+
−
=
1
1
29 21
$
$ .
Γ
Γ
L
L
. 
6 3 4. . 
The line length as a fraction of a wavelength is = kλ . Hence k f
v
= = 14. . The load 
reflection coefficient is $
$
$ . .ΓL
L C
L C
oZ Z
Z Z
=
−
+
= ∠ −0 8521 126 5 . The exponential is 
e j o− = ∠ −2 1 1008β . Hence the input reflection coefficient is 
( )$ $ . .Γ Γ0 08521 5452= = ∠ −−L j oe β . The input impedance is 
( )[ ]
( )[ ]$
$
$ .Z Zin C
o
=
+
−
= ∠ −
1 0
1 0
192 7883
Γ
Γ
Ω . The phasor input voltage to the line is 
6-42 
( )$ $$ $ $ . .V
Z
Z Z
Vin
S in
S
o0 9 25 4633=
+
= ∠ . The undetermined constant in the solution is 
( )
( )[ ]$
$
$ . .V
V o+
=
+
= ∠
0
1 0
5 613 7123
Γ
. Hence the phasor load voltage is 
( ) [ ]$ $ $ .V V e j L o= + = ∠ −+ − β 1 4 738 127Γ . The average power delivered to the load is 
( )
P
V
ZL
ZLAV,load mW= =
1
2
43
2$
cosθ . This can be confirmed by determining the 
average power delivered to the input of the line: 
( )
P
V
Zin
ZinAV,load mW= =
1
2
0
43
2$
cosθ 
which should equal the power delivered to the load since the line is lossless. The VSWR 
is VSWR =
+
−
=
1
1
1252
$
$ .
Γ
Γ
L
L
. 
6 3 5. . 
The line length as a fraction of a wavelength is = kλ . Hence k f
v
= = 0 7. . The load 
reflection coefficient is $
$
$ .ΓL
L C
L C
oZ Z
Z Z
=
−
+
= ∠ −1 64 01 . The exponential is 
e j o− = ∠ −2 1 504β . Hence the input reflection coefficient is 
( )$ $Γ Γ0 1 1522= = ∠−L j oe β . The input impedance is 
( )[ ]
( )[ ]$
$
$ .Z Zin C
o
=
+
−
= ∠
1 0
1 0
2494 90
Γ
Γ
Ω . The phasor input voltage to the line is 
( )$ $$ $ $ . .V
Z
Z Z
Vin
S in
S
o0 3901 38 72=
+
= ∠ . The undetermined constant in the solution is 
( )
( )[ ]$
$
$ . .V
V o+
=
+
= ∠ −
0
1 0
8 059 37 27
Γ
. Hence the phasor load voltage is 
( ) [ ]$ $ $ . .V V e j L o= + = ∠+ − β 1 1367 38 72Γ . The average power delivered to the load is 
( )
P
V
ZL
ZLAV,load = =
1
2
0W
2$
cosθ . This can be confirmed by determining the average 
power delivered to the input of the line: 
( )
P
V
Zin
ZinAV,load = =
1
2
0
0W
2$
cosθ which should 
6-43
equal the power delivered to the load since the line is lossless. The VSWR is 
VSWR =
+
−
= ∞
1
1
$
$
Γ
Γ
L
L
. 
6 3 6. . 
The line length as a fraction of a wavelength is = kλ . Hence k f
v
= = 159. . The load 
reflection coefficient is $
$
$ . .ΓL
L C
L C
oZ Z
Z Z
=
−
+
= ∠ −08668 2549 . The exponential is 
e j o− = ∠ −2 1 1145β . Hence the input reflection coefficient is 
( )$ $ . .Γ Γ0 08668 90 292= = ∠ −−L j oe β . The input impedance is 
( )[ ]
( )[ ]$
$
$ . .Z Zin C
o
=
+
−
= ∠ −
1 0
1 0
74 62 8184
Γ
Γ
Ω . The phasor input voltage to the line is 
( )$ $$ $ $ . .V
Z
Z Z
Vin
S in
S
o0 17 71 1937=
+
= ∠ . The undetermined constant in the solution is 
( )
( )[ ]$
$
$ . .V
V o+
=
+
= ∠
0
1 0
1341 60 41
Γ
. Hence the phasor load voltage is 
( ) [ ]$ $ $ . .V V e j L o= + = ∠ −+ − β 1 24 43 1638Γ . The average power delivered to the load is 
( )
P
V
ZL
ZLAV,load = =
1
2
0 298W
2$
cos .θ . This can be confirmed by determining the 
average power delivered to the input of the line: 
( )
P
V
Zin
ZinAV,load = =
1
2
0
0 298W
2$
cos .θ 
which should equal the power delivered to the load since the line is lossless. The VSWR 
is VSWR =
+
−
=
1
1
14 02
$
$ .
Γ
Γ
L
L
. 
6 3 7. . 
The line length as a fraction of a wavelength is = kλ . Hence k f
v
= = 0 36. . The load 
reflection coefficient is $
$
$ . .ΓL
L C
L C
oZ Z
Z Z
=
−
+
= ∠ −05423 139 4 . The exponential is 
e j o− = ∠ −2 1 259 2β . . Hence the input reflection coefficient is 
6-44
( )$ $ . .Γ Γ0 05423 38 62= = ∠ −−L j oe β . The input impedance is 
( )[ ]
( )[ ]$
$
$ . .Z Zin C
o
=
+
−
= ∠ −
1 0
1 0
6571 4379
Γ
Γ
Ω . The phasor input voltage to the line is 
( )$ $$ $ $ . .V
Z
Z Z
Vin
S in
S
o0 99 2 6127=
+
= ∠ − . The undetermined constant in the solution is 
( )
( )[ ]$
$
$ . .V
V o+
=
+
= ∠
0
1 0
67 79 7 24
Γ
. Hence the phasor load voltage is 
( ) [ ]$ $ $ . .V V e j L o= + = ∠ −+ − β 1 465 1533Γ . The average power delivered to the load is 
( )
P
V
ZL
ZLAV,load = =
1
2
5 406W
2$
cos .θ . This can be confirmed by determining the 
average power delivered to the input of the line: 
( )
P
V
Zin
ZinAV,load W= =
1
2
0
5 405
2$
cos .θ 
which should equal the power delivered to the load since the line is lossless. The VSWR 
is VSWR =
+
−
=
1
1
337
$
$ .
Γ
Γ
L
L
. 
6 3 8. . 
The SPICE circuit with nodes labeled is shown below. The source impedance is 
represented by a 20Ω resistor in series with a 1.061nF capacitor, and the load impedance 
is represented by a 200Ω resistor in series with a 15.92µH inductor. The SPICE (PSPICE) 
program is 
PROBLEM 6.3.8 
VS 1 0 AC 50 0 
RS 1 2 20 
CS 2 3 1.061N 
T 3 0 4 0 Z0=50 TD=260N 
RL 4 5 200 
LL 5 0 15.92U 
.AC DEC 1 5E6 5E6 
6-45
.PRINT AC VM(3) VP(3) VM(4) VP(4) 
.END 
 
 
 
 
 
 
 
The SPICE result is ( )$ . .V o0 2056 1213= ∠ and ( )$ . .V o= ∠ −89 61 50 45 which compares 
well with the hand calculation. 
6 3 9. . 
The SPICE circuit with nodes labeled is shown below. The source impedance is 
represented by a 50Ω resistor, and the load impedance is represented by a 10Ω resistor in 
series with a 15.92pF capacitor. The SPICE (PSPICE) program is 
PROBLEM 6.3.9 
VS 1 0 AC 10 60 
RS 1 2 50 
T 2 0 3 0 Z0=100 TD=7N 
RL 3 4 10 
CL 4 0 15.92P 
.AC DEC 1 2E8 2E8 
.PRINT AC VM(2) VP(2) VM(3) VP(3) 
.END 
0°50
20 Ω 1.061 nF
200 Ω
2 3 4
5
1
0
+
–
ZC = 50 Ω
T = 260 ns
15.92 �H
6-46
 
 
 
 
 
 
The SPICE result is ( )$ . .V o0 9 25 4633= ∠ and ( )$ .V o= ∠ −4 737 127 which compares well 
with the hand calculation. 
6 310. . 
The SPICE circuit with nodes labeled is shown below. The source impedance is 
represented by a 20Ω resistor, and the load impedance is represented by a 0.9947pF 
capacitor. The SPICE (PSPICE) program is 
PROBLEM 6.3.10 
VS 1 0 AC 5 0 
RS 1 2 20 
T 2 0 3 0 Z0=100 TD=0.7N 
CL 3 0 0.9947P 
.AC DEC 1 1E9 1E9 
.PRINT AC VM(2) VP(2) VM(3) VP(3).END 
 
 
 
 
 
 
 
60°10
50 Ω
10 Ω
2 3
4
1
0
+
–
ZC = 100 Ω
T = 7 ns
15.92 pF
0°5
20 2 31
0
+
–
ZC = 100 Ω
T = 0.7 ns
0.9947 pF
6-47
The SPICE result is ( )$ . .V o0 3901 38 72= ∠ and ( )$ . .V o= ∠1367 38 72 which compares well 
with the hand calculation. 
6 311. . 
The SPICE circuit with nodes labeled is shown below. The source impedance is 
represented by a 30Ω resistor, and the load impedance is represented by a 100Ω resistor in 
series with a 0.8842pF capacitor. The SPICE (PSPICE) program is 
PROBLEM 6.3.11 
VS 1 0 AC 20 40 
RS 1 2 30 
T 2 0 3 0 Z0=75 TD=2.65N 
RL 3 4 100 
CL 4 0 0.8842P 
.AC DEC 1 6E8 6E8 
.PRINT AC VM(2) VP(2) VM(3) VP(3) 
.END 
 
 
 
 
 
 
 
The SPICE result is ( )$ . .V o0 17 71 19 37= ∠ and ( )$ . .V o= ∠ −24 43 1638 which compares 
well with the hand calculation. 
6 312. . 
The SPICE circuit with nodes labeled is shown below. The source impedance is 
represented by a 50Ω resistor in series with a 7.958µH inductor, and the load impedance is 
40°20
30
100 Ω
2 3
4
1
0
+
–
ZC = 75 Ω
T = 2.65 ns
0.8842 pF
6-48
represented by a 100Ω resistor in series with a 1.592nF capacitor. The SPICE (PSPICE) 
program is 
PROBLEM 6.3.12 
VS 1 0 AC 100 0 
RS 1 2 50 
LS 2 3 7.958U 
T 3 0 4 0 Z0=300 TD=0.36U 
RL 4 5 100 
CL 5 0 1.592N 
.AC DEC 1 1E6 1E6 
.PRINT AC VM(3) VP(3) VM(4) VP(4) 
.END 
 
 
 
 
 
 
 
The SPICE result is ( )$ . .V o0 99 2 6128= ∠ − and ( )$ . .V o= ∠ −465 1533 which compares 
well with the hand calculation. 
6 313. . 
The line length as a fraction of a wavelength is = kλ . Hence k f
v
= = 1385. . The load 
reflection coefficient is $
$
$ . .ΓL
L C
L C
oZ Z
Z Z
=
−
+
= ∠0 6152 162 9 . The exponential is 
e j o− = ∠ −2 1 996 9β . . Hence the input reflection coefficient is 
( )$ $ .Γ Γ0 0 6152 1142= = ∠ −−L j oe β . The input impedance is 
0°100
50 Ω 7.958 �H
100 Ω
2 4
5
1 3
0
+
–
ZC = 300 Ω
T = 0.36 �s
1.592 nF
6-49
( )[ ]
( )[ ]$
$
$ .Z Zin C
o
=
+
−
= ∠ −
1 0
1 0
205 6105
Γ
Γ
Ω . The phasor input voltage to the line is 
( )$ $$ $ $ . .V
Z
Z Z
Vin
S in
S
o0 8 785 1081=
+
= ∠ − . The undetermined constant in the solution is 
( )
( )[ ]$
$
$ . .V
V o+
=
+
= ∠
0
1 0
9 379 26 05
Γ
. Hence the phasor load voltage is 
( ) [ ]$ $ $ . .V V e j L o= + = ∠ −+ − β 1 4 221 88 71Γ . The average power delivered to the load is 
( )
P
V
ZL
ZLAV,load mW= =
1
2
9114
2$
cos .θ . This can be confirmed by determining the 
average power delivered to the input of the line: 
( )
P
V
Zin
ZinAV,load mW= =
1
2
0
9114
2$
cos .θ 
which should equal the power delivered to the load since the line is lossless. The VSWR 
is VSWR =
+
−
=
1
1
4 2
$
$ .
Γ
Γ
L
L
. With the line removed, the voltage at the input to the antenna 
is $
$
$ $ . .V
Z
Z Zin
L
S L
o o
=
+
∠ = ∠10 0 6 49 1115 . Hence the average power delivered to the 
antenna (and hence radiated) is P
V
Z
in
L
ZLAV,ant mW= =
1
2
21553
2$
cos .θ 
6 314. . 
Because the frequency of the source is 300MHz and the velocity of propagation on each 
line is v = ×3 108 m s , a wavelength is 1m. Hence each transmission line is a multiple of 
a half wavelength long. Therefore the input impedance replicates. Hence the source sees 
an impedance of ( ) ( )$ . . . .Z j j jin = + + = +73 42 5 73 425 36 5 2125 Ω . Thus the input voltage 
to the entire transmission setup is $
$
$ $ . .V
Z
Z Zin
in
S in
o o
=
+
∠ = ∠10 0 4 742 16 41 . The average 
power delivered by the source is P
V
Z
in
in
ZinAV,source mW= =
1
2
230
2$
cosθ . Because all 
lines are assumed to be lossless, all this power is delivered to the two antennas. Because 
6-50 
they have identical input impedances, the power is divided equally so that the average 
power to each antenna is 115W. 
6 315. . 
For an open-circuit load, the load reflection coeffiecient is $ΓL = +1. Hence the reflection 
coefficient at the input to the line is ( )$ $Γ Γ0 2 2= =− −L j je eβ β . Hence the input 
impedance is 
( )[ ]
( )[ ]
[ ]
[ ]
[ ]
[ ] ( )$
$
$ tanZ Z Z
e
e
Z
e e e
e e e
jZin C C
j
j C
j j j
j j j C
=
+
−
=
+
−
=
+
−
= −
−
−
− −
− −
1 0
1 0
1
1
1
2
2
Γ
Γ
β
β
β β β
β β β β and 
we have used the fact that [ ] ( )e ej jβ β β+ =− 2 cos and [ ] ( )e e jj jβ β β− =− 2 sin . 
Similarly, for a short-circuit load, the load reflection coefficient is $ΓL = −1. Hence the 
reflection coefficient at the input to the line is ( )$ $Γ Γ0 2 2= = −− −L j je eβ β . Hence the 
input impedance is 
( )[ ]
( )[ ]
[ ]
[ ]
[ ]
[ ] ( )$
$
$ tanZ Z Z
e
e
Z
e e e
e e e
jZin C C
j
j C
j j j
j j j C
=
+
−
=
−
+
=
−
+
=
−
−
− −
− −
1 0
1 0
1
1
2
2
Γ
Γ
β
β
β β β
β β β β . 
6 316. . 
For a quarter-wavelength line, − = − = − = −2 4 1
4
180β π π o . Hence the reflection 
coefficient at the input to the line is ( )$ $ $Γ Γ Γ0 2= = −−L j Le β . Hence the input impedance 
is 
( )[ ]
( )[ ]
[ ]
[ ]$
$
$
$
$Z Z Zin C C
L
L
=
+
−
=
−
+
1 0
1 0
1
1
Γ
Γ
Γ
Γ
. But the input impedance at the load is the load 
impedance: 
[ ]
[ ]$
$
$Z ZL C
L
L
=
+
−
1
1
Γ
Γ
. Hence 
[ ]
[ ]
1
1
−
+
=
$
$ $
Γ
Γ
L
L
C
L
Z
Z
. Substituting gives $ $Z
Z
Zin
C
L
=
2
 for a 
quarter-wavelength line. If the load is an open circuit, $ZL = ∞ then $Zin = 0 and the line 
looks like a short circuit at its input. If the load is short circuit, $ZL = 0 then $Zin = ∞ and 
the line looks like an open circuit at its input. 
6-51 
The normalized load impedance is $z jL = +4 10 which is located at 0.237λ on the TG 
scale or ∠7o on the angle scale. The line length as a fraction of a wavelength is = kλ . 
Hence k f
v
= = 13. . Rotating 1.3λ on the TG scale (clockwise) yields the normalized 
input impedance as $ .z jin = +0 04 12 located at 1.537λ=0.037λ on the TG scale. 
Multiplying this by the characteristic impedance gives the input impedance as 
$ $Z Z z jin C in= = +2 12 . Using the compass and transferring to one of the lower scales 
gives ( )$ .Γ 0 092 154= ∠ o , $ .ΓL o= ∠092 7 , and VSWR=30. 
6 4 2. . 
The normalized load impedance is $ . .z jL = −01 05 which is located at 1.4λ on the TG scale 
or ∠ −126o on the angle scale. The line length as a fraction of a wavelength is = kλ . 
Hence k f
v
= = 14. . Rotating 1.4λ on the TG scale (clockwise) yields the normalized 
input impedance as $ . .z jin = −038 188 located at 1.826λ=0.326λ on the TG scale. 
Multiplying this by the characteristic impedance gives the input impedance as 
$ $Z Z z jin C in= = −38 188 . Using the compass and transferring to one of the lower scales 
gives ( )$ .Γ 0 0842 55= ∠ − o , $ .ΓL o= ∠ −0 842 126 , and VSWR=12. 
6 4 3. . 
The normalized load impedance is $ .z jL = − 16 which is located at 0.339λ on the TG scale 
or ∠ − 64o on the angle scale. The line length as a fraction of a wavelength is = kλ . 
Hence k f
v
= = 0 7. . Rotating 0.7λ on the TG scale (clockwise) yields the normalized 
input impedance as $ .z jin = 0 25 located at 0.039λ on the TG scale. Multiplying this by 
the characteristic impedancegives the input impedance as $ $Z Z z jin C in= = 25 . Using the 
compass and transferring to one of the lower scales gives ( )$Γ 0 1 152= ∠ o , $ΓL o= ∠ −1 64 , 
and VSWR= ∞ . 
6 4 4. . 
The normalized load impedance is $ .z jL = −133 4 which is located at λ on the TG scale or 
∠ − 91o on the angle scale. The line length as a fraction of a wavelength is = kλ . 
6 4 1. . 
6-52 
Hence k f
v
= = 159. . Rotating 1.59λ on the TG scale (clockwise) yields the normalized 
input impedance as $ . .z jin = −013 097 located at 1.876λ=0.376λ on the TG scale. 
Multiplying this by the characteristic impedance gives the input impedance as 
$ $ . .Z Z z jin C in= = −9 75 72 75 . Using the compass and transferring to one of the lower 
scales gives ( )$ .Γ 0 087 91= ∠ − o , $ .ΓL o= ∠ −087 26 , and VSWR=15. 
6 4 5. . 
The normalized load impedance is $ . .z jL = −033 033 which is located at 0.444λ on the TG 
scale or ∠ −139o on the angle scale. The line length as a fraction of a wavelength is 
= kλ . Hence k f
v
= = 036. . Rotating 0.36λ on the TG scale (clockwise) yields the 
normalized input impedance as $ . .z jin = −158 15 located at 0.804λ=0.304λ on the TG 
scale. Multiplying this by the characteristic impedance gives the input impedance as 
$ $Z Z z jin C in= = −474 450. Using the compass and transferring to one of the lower scales 
gives ( )$ .Γ 0 054 39= ∠ − o , $ .ΓL o= ∠ −0 54 139 , and VSWR=3.4. 
6 4 6. . 
(a) The normalized input impedance is $ .z jin = −0 6 2 which is located at 0.18λ on the TL 
scale. The line length is 0.4λ. Rotating 0.4λ on the TL scale (counter-clockwise) yields 
the normalized load impedance as $ . .z jL = −014 054 located at 0.58λ=0.08λ on the TL 
scale. Multiplying this by the characteristic impedance gives the input impedance as 
$ $Z Z z jL C L= = −7 27 . Using the compass and transferring to one of the lower scales 
gives $ .ΓL o= ∠ −08 122 , and VSWR=9. 
(b) The normalized input impedance is $ .z jin = +0 667 0 which is located at 0λ on the TL 
scale or ∠180o on the angle scale. The line length is 1.3λ. Rotating 1.3λ on the TL scale 
(counter-clockwise) yields the normalized load impedance as $ . .z jL = +135 034 located at 
1.3λ=0.3λ on the TL scale. Multiplying this by the characteristic impedance gives the 
input impedance as $ $ . .Z Z z jL C L= = +10125 255. Using the compass and transferring to 
one of the lower scales gives $ .ΓL o= ∠0 21 36 , and VSWR=1.5. 
6-53 
(c) The normalized input impedance is $ . .z jin = +15 23 which is located at 0.2985λ on the 
TL scale. The line length is 0.6λ. Rotating 0.6λ on the TL scale (counter-clockwise) 
yields the normalized load impedance as $ . .z jL = +0 275 0 705 located at 0.899λ=0.399λ 
on the TL scale. Multiplying this by the characteristic impedance gives the input 
impedance as $ $ . .Z Z z jL C L= = +275 705 . Using the compass and transferring to one of 
the lower scales gives $ . .ΓL o= ∠0 7 106 5 , and VSWR=5.5. 
(d) The normalized input impedance is $ .z jin = 25 which is located at 0.3105λ on the TL 
scale. The line length is 0.8λ. Rotating 0.8λ on the TL scale (counter-clockwise) yields 
the normalized load impedance as $ .z jL = − 083 located at 1.11λ=0.11λ on the TL scale. 
Multiplying this by the characteristic impedance gives the input impedance as 
$ $Z Z z jL C L= = − 83 . Using the compass and transferring to one of the lower scales gives 
$ΓL o= ∠ −1 100 , and VSWR= ∞ . 
6 4 7. . 
(a) The normalized impedances are $ .z jin = − 0 2 located at 0.031λ on the TL scale, and 
$ .z jL = 0 5 located at 0.4265λ on the TL scale. The line length is obtained by rotating 
from $zin toward $zL (counterclockwise) on the TL scale giving a line length of 0.4265-
0.031=0.396λ. The VSWR is obtained by transferring the compass length to a lower scale 
to yield VSWR=∞. Similarly the load reflection coefficient is $ΓL o= ∠1 127 . 
(b) The normalized impedances are $ .z jin = −05 2 located at 0.179λ on the TL scale, and 
$ . .z jL = −012 0 5 located at 0.074λ on the TL scale. The line length is obtained by rotating 
from $zin toward $zL (counterclockwise) on the TL scale giving a line length of 0.5-
(0.179-0.074)=0.395λ. The VSWR is obtained by transferring the compass length to a 
lower scale to yield VSWR=10. Similarly the load reflection coefficient is 
$ .ΓL o= ∠ −0 83 126 . 
(c) The normalized impedances are $ . .z jin = +0 3 0 5 located at 0.422λ on the TL scale, and 
$z jL = +2 2 located at 0.2915λ on the TL scale. The line length is obtained by rotating 
from $zin toward $zL (counterclockwise) on the TL scale giving a line length of 0.5-
6-54 
(0.422-0.2915)= 0.37λ. The VSWR is obtained by transferring the compass length to a 
lower scale to yield VSWR=4.2. Similarly the load reflection coefficient is 
$ . .ΓL o= ∠0 62 29 5 . 
(d) The normalized impedances are $ .z jin = +18 0 located at 0.25λ on the TL scale, and 
$ . .z jL = −08 0 5 located at 0.384λ on the TL scale. The line length is obtained by rotating 
from $zin toward $zL (counterclockwise) on the TL scale giving a line length of 0.5-(0.25-
0.116)=0.366λ. The VSWR is obtained by transferring the compass length to a lower 
scale to yield VSWR=1.8. Similarly the load reflection coefficient is $ .ΓL o= ∠ −0 285 96 . 
6 4 9. . 
With an open circuit (the load removed), the input impedance, normalized is 
$ .$z jin ZL =∞ = − 0 8 which is located at 0.107λ on the TL scale. Rotating from this to the 
open-circuit load $zL = ∞ (counter clockwise) which is located at 0.25λ on the TL scale 
gives the line length as 0.25-0.107=0.143λ. Now we repeat this knowing the line length. 
Plotting the normalized input impedance with the load attached, $ . .z jin = +0 3 0 4 which is 
located at 0.435λ on the TL scale, and rotating 0.435+0.143=0.578λ=0.078λ on the TL 
scale and reading off the result gives the normalized unknown load impedance as 
$ . .z jL = −032 0 49 . Hence the unknown load impedance is ( )$Z jL = −32 49 Ω . 
6 410. . 
At 1GHz, the impedance of a 10pF capacitor is − j1592. Ω which is the desired input 
impedance of this short-circuited line. Normalizing this with the 50Ω characteristic 
impedance of the line gives a desired normalized input impedance of − j0318. . This is 
plotted on the Smith chart at 0.048λ on the TL scale. Rotating this counter clockwise to 
the load of $zL = 0 at 0.5λ on the TL scale gives a line length of 0.5-0.048=0.452λ. The 
wavelength of 1GHz in air is 30cm. Hence the physical length of the line is 13.6cm. 
6 51. . 
At 30MHz in air, a wavelength is 10m. Hence the line length of 1m is 110 λ and is thus 
approximately electrically short so that a lumped-Pi model should give sufficient accuracy. 
6-55
The exact input impedance is 
[ ]
[ ]$
$
$Z Z
e
e
in C
L
j
L
j
=
+
−
−
−
1
1
2
2
Γ
Γ
β
β where e e
j j o− −
= = .∠2
4
10 1 72β
π
Substituting Z C = 50Ω and $Z jL = −200 200 gives the exact value of 
 ( )$ . .Z jin = −12 89 5149 Ω . To prepare the SPICE lumped-Pi model we need the 
total line inductance and capacitance. The per-unit-length values are 
l Z
v
C
= = 01667. µ H
m
 and c
vZC
= =
1 66 67. pF
m
. Hence the totals are L H= 01667. µ and 
C pF= 66 67. . The SPICE circuit is shown below. The SPICE code is 
PROBLEM 6.5.1 
VS 1 0 AC 1 0 
RS 1 2 1 
C1 2 0 33.33P 
L 2 3 0.1667U 
C2 3 0 33.33P 
RL 3 4 200 
CL 4 0 26.53P 
.AC DEC 1 3E7 3E7 
.PRINT AC VM(2) VP(2) IM(RS) IP(RS) 
.END 
 
 
 
 
 
 
 
0°1
1 Ω 0.1667 �F
200 Ω
2 3
4
1
0
+
–
26.53 pF
33.33 pF 33.33 pF
6-56
The results are $( ). .V o2 0 9952 1021= ∠ − and $( ) . .I oRS = × ∠−1842 10 74 232 . The input 
impedance is the ratio $
$( )
$( )
. . . .Z V
I
jin
o
= = ∠ − = −2 54 028 75251 1376 52 25
RS
Ω which is 
close to the exact value. 
6 5 2. . 
At 4MHz with v = ×2 108 m s , a wavelength is 50m. Hence the line length of 5m is 
1
10 λ and is thus approximately electrically short so that a lumped-Pi model should give 
sufficient accuracy. The exact values for the line input and output voltages are computed 
from the results of Section 6.3.3 (or from a SPICE model) as $( ) . .V o0 7954 6578= ∠ − , and 
$( ) . .V o= ∠ −10 25 336 . To prepare the SPICE lumped-Pi model we need the total line 
inductance and capacitance. The per-unit-length values are l Z
v
C
= = 0 5. µ H
m
 and 
c
vZC
= =
1 50 pF
m
. Hence the totals are L H= 2 5. µ and C pF= 250 . The SPICE circuit is 
shown below. The SPICE code is 
PROBLEM 6.5.2 
VS 1 0 AC 10 0 
RS 1 2 25 
C1 2 0 125P 
L 2 3 2.5U 
C2 3 0 125P 
RL 3 4 150 
CL 4 0 795.8P 
.AC DEC 1 4E6 4E6 
.PRINT AC VM(2) VP(2) VM(3) VP(3) 
.END 
6-57
 
 
 
 
 
 
 
 
The results are $( ) . .V o0 7959 5906= ∠ − , and $( ) . .V o= ∠ −10 27 35 02 . which are close to 
the exact values. 
6 61. . 
In all cases, e− =2 0 327α . and e f
v
j o− ≅ ∠ − = ∠ − = ∠ −2 1 4 1 89 4 1 252β π π. . 
(a) For a short-circuit load, $ΓL = −1 so that ( )$ $ .Γ Γ0 0327 2522 2= = − ∠ −− −L j oe eα β . 
Hence the input impedance is {
( )[ ]
( )[ ]$ $
$
$ . .Z Zin C
o
=
+
−
= ∠ −
75
1 0
1 0
90 209 34 86
Γ
Γ
 
(b) For a open-circuit load, $ΓL = +1 so that ( )$ $ .Γ Γ0 0 327 2522 2= = ∠ −− −L j oe eα β . 
Hence the input impedance is {
( )[ ]
( )[ ]$ $
$
$ . .Z Zin C
o
=
+
−
= ∠
75
1 0
1 0
62 355 34 06
Γ
Γ
 
(c) For a 300Ω resistive load, $ .ΓL = 0 6 so that ( )$ $ .Γ Γ0 0196 2522 2= = ∠ −− −L j oe eα β . 
Hence the input impedance is {
( )[ ]
( )[ ]$ $
$
$ . .Z Zin C
o
=
+
−
= ∠
75
1 0
1 0
66 7 212
Γ
Γ
 
6 6 2. . 
If the cable is matched, the phasor voltage and current on the line are 
( )$ $V z V e ez j z= + − −α β and ( )$ $$I z
V
Z
e e
C
z j z
=
+
− −α β . The average power delivered to the 
cable at any z along it is ( ) ( ) ( )[ ]P z V z I z VZ eC z ZCAV = =∗
+
−
1
2 2
2
2Re $ $
$
cosα θ . The power 
0°10
25 2.5 �H
150 Ω
2 3
4
1
0
+
–
795.8 pF
125 pF 125 pF
6-58
loss is defined as the ratio ( )( )Power Loss
AV
AV
=
=
=
=
P z
P z
e
0 2α . In decibels, 
( ) ( ) ( )Power Loss Power LossdB = = =10 10 20
8 686
10 10
2
10log log log
.
e eα α1 24 34 . It is important 
to realize that this power loss as specified by cable manufacturers is valid only if the cable 
is matched. If the cable is mismatched, this loss specification has nothing whatever to do 
with the cable loss. 
 7-1 
Chapter 7 
Problem Solutions 
711. . 
Using the complete equations in (7.1): 
(a) 10cm. At 100MHz a wavelength is 3m. Hence 10cm is 1
30
λo so the fields are in the 
near field of the dipole. The magnetic field is 
( )$ . . . . .H j eo j ooφ = × ∠ + = ∠− −2 468 10 30 4 77 22 797 0575 2982 12 Am . The electric fields are 
$ . .Er o= ∠ −2069 67 6017
V
m
 and $ . .E oθ = ∠ −9914 59 64
V
m
. The ratios are 
$
$ .
E
Er
θ
= 0 479 
and 
$
$ .
E
H
θ
φ
= 1724 65 Ω . 
(b) 1m. At 100MHz a wavelength is 3m. Hence 1m is 1
3
λo so the fields are in the border 
between the near field and far field of the dipole. The magnetic field is 
$ . .H oφ = × ∠ −−1306 10 2552
A
m
. The electric fields are $ . .Er o= ∠ −4 701 1155
V
m
 and 
$ . .E oθ = ∠ −4 033 3174
V
m
. The ratios are 
$
$ .
E
Er
θ
= 0858 and 
$
$ .
E
H
θ
φ
= 3088 Ω . 
(c) 10m. At 100MHz a wavelength is 3m. Hence 10m is 3333. λo so the fields are in the 
far field of the dipole. The magnetic field is $ . .H oφ = × ∠ −−118 10 2 73
A
m
. The electric 
fields are $ . .Er o= × ∠ −−4 247 10 92 732
V
m
 and $ . .E oθ = ∠ −0 444 2 74
V
m
. The ratios are 
$
$ .
E
Er
θ
= 10 45 and 
$
$ .
E
H o
θ
φ
η= ≅376 08 Ω . Since 10m is in the far field of the dipole it is 
expected that the fields should approach the far-field approximations. 
 
 
 
7-2 
712. . 
R dlrad
o
=



 = ×
−80 8 773 102
2
3π λ . Ω , P I Rradiated rms radAV mW,
$ .= =2 438 64 where 
$Irms o= ∠
10
2
30 . 
713. . 
The field points are in the far field of the dipole and at electrical distances of 166 7. λo and 
1667λo . Hence the electric field and magnetic field are $ , $E H
e
r
j r
o
θ φ
π λ
∝
−



2
. Hence 
the electric field at 1000m is 1/10 of the field at 100m or 10 mV
m
. The magnetic field at 
100m is the ratio of the electric field and the intrinsic impedance of free space: 
$ $ .H E
o
= = × −
η
2 65 10 4 A
m
. The magnetic field at 1000m is $
$
.H E
o
= = × −
η
2 65 10 5 A
m
. The 
phase angle at a point is −2π λ
r
o
. Hence the difference in phase angles between the two 
points is ( )2
0 6
1000 100 1500 2 0π π
.
− = × = o . Hence the fields at 100m and 1000m are 
exactly in phase. The fields at 1000m lag those at 100m because of the minus sign in the 
exponential. The average power densities are those of plane waves: S
E
o
AV =
1
2
2$
η
. 
Hence the average power density at 100m is 1326. µ W
m2
 and at 1000m is 01326. µ W
m2
. 
Observe that the electric and magnetic fields decay inversely with distance, whereas the 
power densities decay inversely with the square of the distance. 
7 21. . 
( )F θ = 1, ( )E I
r
F I
ro
m m
θ η π
θ= = =
2
60 60 mV
m
, H E
o
φ θη
µ
= = 15915. A
m
. The power 
density in the wave is S
E
o
AV 2
W
m
= =
1
2
4 775
2$
.θ
η
µ , and the total average power radiated 
is P I Rrad m rms radAV, ,$= =
2
365mW . 
7-3 
7 2 2. . 
The input impedance to the monopole is ( )36 5 2125. .+ j Ω . Attaching the source, we 
determine the input current to the antenna as $
. .
. .I
jm
o
=
+ +
= ∠ −100
50 36 5 2125
1123 1388 A . 
Hence the total average power radiated is P Irad mAV W, $ .= =
1
2
365 23
2
. Broadside to he 
antenna E I
r
m
θ = =60 0 674.
V
m
 which is perpendicular to the ground plane (hence 
satisfying the boundary condition on the electric field at the surface of a perfect 
conductor). The average power density in the wave is S
E
o
AV 2
mW
m
= =
1
2
0 602
2$
.θ
η
. 
7 2 3. . 
The input impedance to the 1
5
λo monopole is ( )20 50− j Ω . Attaching the source, we 
determine the input current to the antenna as $ . .I
jm
o
=
+ −
= ∠100
50 20 50
1162 3554 A . 
Hence the total average power radiated is P Irad mAV W, $ .= =
1
2
20 1351
2
 and the radiation 
resistance for this antenna is the real part of its input impedance since it is assumed 
lossless. 
7 2 4. . 
The input impedance to the 1
10
λo monopole is ( )4 180− j Ω . Attaching the source, we 
determine the input current to the antenna as $ . .I
jm
o
=
+ −
= ∠100
50 4 180
0532 733 A . Hence 
the total average power radiated is P Irad mAV, $ .= =
1
2
4 0 566W
2
 and the radiation 
resistance for this antenna is the real part of its input impedance since it is assumed 
lossless. 
7 2 5. . 
The load on the transmission line is the input impedance to the antenna, 
$ .ZjL = +73 425 Ω . The input impedance to the transmission line can be found using the 
methods of Chapter 6. The load reflection coefficient is 
7-4 
$ .
.
. .ΓL
oj
j
=
+ −
+ +
= ∠73 425 50
73 42 5 50
0 3713 4252 . The reflection coefficient at the input to the line 
is $ ( ) $ . . . .Γ Γ0 0 3713 42 52 1 216 0 3713 17348
4
= = ∠ × ∠ − = ∠ −
−
L
j o o oe
π λ . The input 
impedance is 
( )[ ]
( )[ ]$
$
$ . . . .Z Z jin C
o
=
+
−
= ∠ − = −
1 0
1 0
2309 5585 2298 2 25
Γ
Γ
. Hence the input 
current to the line is $
. .
. .I
j
o
input to line = + −
= ∠100
50 22 98 2 25
137 176 . Hence the power 
delivered to the line input is P IAV to line input to line W, $ . .= =
2
2298 431 . But since the line 
is lossless this is also the power delivered to the antenna input. 
7 31. . 
Since E d
o
∝ +



cos cos
π
λ φ
α
2
, we can determine the location of maxima and minima by 
differentiating this expression with respect to φ and setting the result to zero: 
d
d
d d d
o o oφ
π
λ φ
α π
λ φ
α π
λ φcos cos sin cos sin+






= − +



 × −



 =2 2 0 . The sin φ = 0 
condition results in maxima or minima at φ = 0 and φ = 180o . This condition is a direct 
result of the pattern being symmetrical with respect to a line through the two antennas. 
The other condition yields πλ φ
αd
o
o ocos , , ,+ = ± ±
2
0 180 360 L . 
7 3 2. . 
E d
o
∝ +



cos cos
π
λ φ
α
2
 (a) d o o= =λ α2 90, , E ∝ +



cos cos
π φ π
2 4
. Nulls at 
π φ π π π π
2 4 2
3
2
5
2
cos , ,+ = ± ± ± or cosφ = 1
2
 or 60o . Maxima and minima at 
sin cosπ φ π
2 4
0+

 = or 
π φ π π
2 4
0cos ,+ = ± or cosφ = − 1
2
 giving ±120 180o o,0, . The 
pattern is sketched below. 
 
 
 
0.707
0.707
60° 120°
1
7-5 
 
 
 
(b) d o o= =5 8 45
λ α, , E ∝ +

cos cos
5
8 8
π φ π . Nulls at 
5
8 8 2
3
2
5
2
π φ π π π πcos , ,+ = ± ± ± or cos . ,φ = −0 6 1 or φ = ±5313 180. ,o o . Maxima and 
minima at sin cos5
8 8
0π φ π+

 = or 
5
8 8
0π φ π πcos ,+ = ± giving φ = ±0 180 10154, , .o o . 
The pattern is sketched below. 
 
 
 
 
 
 
 
 (c) d o
o
= =λ α, 180 , E ∝ +

cos cosπ φ
π
2
. Nulls at π φ π π π πcos , ,+ = ± ± ±
2 2
3
2
5
2
 or 
cos , ,φ = − +0 1 1 or φ = 90 180o o o,270 , ,0 . Maxima and minima at sin cosπ φ π+

 =2 0 or 
π φ π πcos ,+ = ±
2
0 giving φ = ± ±120 60o o, . The pattern is sketched below. 
 
 
 
 
 
 
0.707
53.13°
101.54°
1.00.92
60° 120°
1.01.0
7-6 
 
(d) d o o= =λ α4 180, , E ∝ +



cos cos
π φ π
4 2
. Nulls at π φ π π π π
4 2 2
3
2
5
2
cos , ,+ = ± ± ± 
or cos ,φ = −0 4 or φ = ±90o . Maxima and minima at sin cosπ φ π
4 2
0+

 = or 
π φ π π
4 2
0cos ,+ = ± giving φ = 0 180o o, . The pattern is sketched below. 
 
 
 
 
 
 
 
 
7 3 3. . 
d = =164ft 50m , f = ×1500 103 so that λo = 200m . Therefore d o=
λ
4
. 
E d
o
∝ +



 = +



cos cos cos cos
π
λ φ
α π φ π
2 4
3
8
. Nulls at π φ π π π π
4
3
8 2
3
2
5
2
cos , ,+ = ± ± ± or 
cos . , .φ = −0 5 35 or φ = ±60o . Maxima and minima at sin cosπ φ π
4
3
8
0+

 = or 
π φ π π
4
3
8
0cos ,+ = ± giving φ = 0 180o o, . The pattern is sketched below. 
 
 
 
 
 
 
 
0.7070.707
0.920.38
60°
7-7 
 
7 3 4. . 
E d
o
∝ +



 = +



cos cos cos cos
π
λ φ
α
π φ π
2 4
. Nulls at π φ π π π πcos , ,+ = ± ± ±
4 2
3
2
5
2
 or 
cos . , .φ = −0 25 0 75 or φ = ± ±7552 138 58. , .o o . Maxima and minima at 
sin cosπ φ π+

 =4 0 or π φ
π
πcos ,+ = ±
4
0 giving φ = ± ±0 180 104 48 4141o o o o, , . , . . The 
pattern is sketched below. 
 
 
 
 
 
 
 
 
7 41. . 
S
r
AV ∝
1
2 , 
( ) ( )S S r rAV 5000ft AV max max max ft = miles, , .× = × ⇒ = ×5000ft 3535 10 6702 2 6 . 
7 4 2. . 
P P G G
dR T R T
o
=




λ
π4
2
, GT = 12dB = 15.85 , 
P P dR T o= = = = ×
−10 01 3 3844 109 8, . , , .λ m . 
G P
P G
d
R
R
T T o
=



 = × =
4 1636 10 9214dB
2
9π
λ . . . 
0.707
0.707 1
0.707
41.41°
75.52° 104.48°
138.58°
7-8 
45dB 31,622.78⇒ , P dR o= = =−10 01 303, . ,λ m miles = 48,280m . 
P P
G G
d
T
R
R T o
=



 =
4 36 81
2
π
λ . W . 
7 4 4. . 
$E P G
d
T T
=
60
, P G dT T= × = =
−5 10 12dB 23, = 15.85, miles = 3218.7m . 
$ .E = 0 68 V
m
. 
7 4 5. . 
G = ⇒215. dB 1.64 . $
.
. .I
jant
o
=
+
= × ∠ −−10
123 42 5
7 68 10 1912 , 
P Irad ant= =
1
2
73 0 216W
2$ . , S G P
r
rad
rad
= = × −
4
282 102
10
π
. W
m2
, 
S
E
Erad
o
= ⇒ =1
2
0 461
2$
$ .
η
mV
m
. $ .E P G
d
T T
= =
60
0 461 mV
m
. 
By equation (7.14) ( ){$ .E
I
r
Fo m= =η
π
θ
2
4 608
1
mV
m
. Equation (7.16) gives the same result. 
7 4 3. . 
	Solucionario
	Solução cap. 01 - Electromagnetismo para Engenheiros. Clayton R. Paul
	Solução cap. 02 - Electromagnetismo para Engenheiros. Clayton R. Paul
	Solução cap. 03 - Electromagnetismo para Engenheiros. Clayton R. Paul
	Solução cap. 04 - Electromagnetismo para Engenheiros. Clayton R. Paul
	Solução cap. 05 - Electromagnetismo para Engenheiros. Clayton R. Paul
	Solução cap. 06 - Electromagnetismo para Engenheiros. Clayton R. Paul
	Solução cap. 07 - Electromagnetismo para Engenheiros. Clayton R. Paul