Sadiku & Kulkarni
Principles of Electromagnetics, 6e
INSTRUCTOR’S SOLUTIONS MANUAL FOR
PRINCIPLES
OF
ELEMENTS OF
ELECTROMAGNETICS
ASIAN EDITION
INTERNATIONAL
SIXTH EDITION
Matthew N. O. Sadiku
Prairie View A&M University
Sudarshan R. Nelatury
Pennsylvania State University
S.V. Kulkarni
IITYork
Bombay
New
Oxford
Oxford University Press
Copyright © 2015 by Oxford University Press
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Sadiku & Kulkarni
Principles of Electromagnetics, 6e
Copyright © 2015 by Oxford University Press
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Sadiku & Kulkarni
Principles of Electromagnetics, 6e
1
CHAPTER 1
P. E. 1.1
A + B = (1,0,3) + (5,2,−6 ) = (6,2,−3)
(a)
A + B = 36 + 4 + 9 = 7
(b)
5 A − B = (5,0,15) − (5,2,−6) = (0,−2,21)
(c)
The component of A along ay is
(d)
Ay = 0
3 A + B = (3,0,9 ) + (5,2,−6 ) = (8,2,3)
A unit vector parallel to this vector is
(8,2,3)
a11 =
64 + 4 + 9
= ±(0.9117a x + 0.2279a y + 0.3419a z )
P. E. 1.2 (a) rp = a x − 3a y + 5a z
rR = 3a y + 8a z
(b)
The distance vector is
rQR = rR − rQ = (0,3,8) − (2, 4, 6) = −2a x − a y + 2a z
(c)
The distance between Q and R is
| rQR |= 4 + 1 + 4 = 3
P. E. 1.3
Consider the figure shown on the next page:
40
uZ = uP + uW = −350a x +
( −a x + a y )
2
= −378.28a x + 28.28a y km/hr
or
uz = 379.3∠175.72 km/hr
Where up = velocity of the airplane in the absence of wind
uw = wind velocity
uz = observed velocity
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Sadiku & Kulkarni
Principles of Electromagnetics, 6e
22
N
y
up
x
uW
uz
W
E
S
P. E. 1.4
Using the dot product,
A B
−13
13
=
=−
AB
50
10 65

θ AB = 120.66
cos θ AB =
P. E. 1.5
(a) E F = ( E ⋅ a F )a F =
(E ⋅ F )F
F
2
=
− 10(4,−10,5)
141
= − 0.2837a x + 0.7092a y − 0.3546a z
ax
(b) E × F = 0
4
a y az
3
4 = (55,16,−12 )
− 10 5
a E ×F = ± (0.9398,0.2734,−0.205)
P. E. 1.6
a + b + c = 0 showing that a, b, and c form the sides of a triangle.
a ⋅ b = 0,
hence it is a right angle triangle.
1
1
1
a×b = b×c = c×a
2
2
2
1
1 4 0 −1 1
a×b =
= (3,−17,12)
2
21 3 4
2
Area =
Area =
1
9 + 289 + 144 = 10.51
2
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Principles of Electromagnetics, 6e
3
P. E. 1.7
( x2 − x1 ) + ( y2 − y1 ) + ( z2 − z1 )
2
(a) P1 P2 =
2
2
= 25 + 4 + 64 = 9.644
(
(b) rP = rP1 + λ rP2 − rP1
)
= (1,2,−3) + λ (− 5,−2,8)
= (1 − 5λ ,2 − 2λ ,−3 + 8λ ).
(c) The shortest distance is
d = P1 P3 sin θ = P1 P3 × a P1P2
1
=
−3 5
6
93 − 5 − 2 8
1
(− 14,−73,−27 ) = 8.2
=
93
P.E. 1.8
1.3 a) A – 3B
Prob.1.1
r = 4a= 4a
− 5xa– 2a
+ ay + 6az – 3(12ax + 18ay – 8az)
OP
x
y
z
56a
=
y + 30az
r –32ax –(4,
−5,1)
arOP = OP =
= 0.6172a x − 0.7715a y + 0.1543a z
b) | (2A
rOP | + 5B)/|B|
(16 + 25 + 1)
[2(4ax – 2ay + 6az) + 5(12ax + 18ay – 8az)]
=
2
Prob. 1.2
(12
+ 182 + 82)1/2
r = (−3, 2, 2) − (2, 4, 4) = (−5, −2, −2)
68ax + 86ay – 28az
r = (−5, −2, −2)
ar = =
23.06 = −0.8704a x − 0.3482a y − 0.3482a z
r
25 + 4 + 4
= 2.94ax + 3.72ay – 1.214az
c) 1.3
ax × A
Prob.
rMN = rN= −axrM× =(4a
(3,5,
2a1)y −+(1,
6a−z)4, −2) = 2a x + 9a y + a z
x–−
= 4(ax × ax) – 2(ax × ay) + 6(ax × az)
Prob. 1.4
= 0 – 2az – 6ay = –6ay – 2az
A
−
2 B = (4, −6,3) − 2(−1,8,5) = (4, −6,3) − ( −2,16,10)
(a) d) (B × ax) ⋅ ay
−22, −7)
((12a =+(6,
18a
– 8a ) × a ) ⋅ a
x
y
z
x
y
(12(ax × ax) + 18(ay × ax) – 8(az × ax)) ⋅ ay
(b) A B = (4, −6,3)(−1,8,5) = −4 − 48 + 15 = −37
(0 – 18az – 8ay) ⋅ ay = 0 – 8 = –8
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−2 4 4 3 −462−6 1
C
= =A ×
B=
= (−30, −18,12)
Let
Q
P
×
(c)
as expected. 2 −1 2 −20= ( −54,12, −10 )
Sadiku & Kulkarni
Principles of Electromagnetics, 6e
−30, −
PC⋅ R= =±( −(4,12,
−18,12)
= −4 a y + 0.3244a z )
10 ) ⋅ ( −1,1,
2 )(−=0.8111
4 + 12a−x 20
a⊥ =Q±×1.14
=±
− 0.4867
Prob.
2
2
2
4
|C |
+
+
30
18
12
(a) Using the fact that
−1 1 2
×B
× )P×⋅CR == (RA⋅⋅QC×)BP −=(B4⋅ C )3A, 2 = − ( −6 + 2 ) − ( −8 − 4 ) + 2 ( −4 − 6 ) = −4
or ( A Q
get
P.E.we
1.9
Prob.
1.12
2 −1 −2 7
(Q
)×=−R6,5)
)5A=−7−)(13
A×
×PQ
A) ×
B(2,
−)(=A(×(0,3,1)
× A==0)(−
A
⋅ A+)B120 + 70 = 206
(d) ( P
4,B−)12,10
4,⋅ A
⋅=( Q
⋅B
−+10,
= 16
(18
(b) A × ( A × ( A ×2B )−) 6= A45× −( 12
A ⋅ B10
) A- ( A+ ⋅6Aa ) B 
y a +z 12a + 8a
(e) ( P ×PQ×)Q
× (=Q0× R3) = 1 = −21a x - 2a
= 16
x
y
z
= ( 4A ⋅ B−)10
- ( A7× A) − ( A ⋅ A) - ( A × B )
P ⋅Q
13
2−
2A−×1 −B=4) )−0.51 −7⎯⎯
cos θ P=⋅ R = =− A( −
→ θ PQ = 120.66o
(
(f) cos θ PR =PQ
=
=
=
−0.9526
PQ
10 65
since AxA = 0P R
4 +1+ 4 1+1+ 4 3 6

Prob. 1.13 θ PR = 162.3
P.E.
Prob.1.10
1.15
If A and B are parallel, then B = kA and A x B = 0. It is evident that k = -2 and that
1
1 4 1 −5 1
Area = a x| D a×yE |=a z
= | (3 + 10)a x + (5 − 12)a y + (8 + 1)a z |
P
×
Q
16 +21443+ 100
−
1
2
2
2 = 260
3 = 0.998
(g)
sin=θ PQ
A× B
1 = −2 3 = = 0
P Q
31 16 + 9 + 4
3 29
1
4 −7,9)
− 6 |=
169 + 49 + 81 = 8.646
= −2| (13,
2 = 86.45
2
P. E. 1.7θ PQ
as expected.
2
2
2
(a) P1 P2 = ( x2 − x1 ) + ( y2 − y1 ) + ( z2 − z1 )
Prob.1.11
1.11
P.E.
Prob.
1.14
= the
25fact
+ 4 +that
64 = 9.644
(a) Using
A B = (4, −6,1)(2, 0,5) = 8 − 0 + 5 = 13
( A × B ) × C = ( A ⋅ C )B − (B ⋅ C )A,
rP1
(b)
(a) r|P B=|2r=P1 2+2 λ+ 5rP22 =−29
we get
(1,2,−3B))|+2==λ−13
(−( A+5,×2−2B×,829
AA=B
) ×) A= 71
× (+A2×| B
= (B ⋅ A)A − ( A ⋅ A)B
= (1 − 5λ ,2 − 2λ ,−3 + 8λ ).
(b) A × ( A × ( A × B ) ) = A × ( A ⋅ B ) A- ( A ⋅ A) B 
(b)
(c)
The shortest distance is
A× B
A ⋅PB )×-a( A × A) − ( A ⋅ A) - ( A × B )
a⊥ = ± d = P1 P3 sin θ ==( P
1 3
P1 P2
| A× B |
= − A2 ( A × B )
1 6 4 −−63 15
==0
since
Ax=A
A ×93
B =− 5 − 2 8 = (−30, −18,12)
Let C
2 0 5
1
Prob. 1.15
C=
30,,−−73
18,12)
((−14
,−27 ) = 8.2
a⊥ = ± 1 = 93
±
4
1 −5= ±(1−0.8111a x − 0.4867a y + 0.3244a z )
1
2
2
Area = | C || D × E30
|= + 18 + 122 = | (3 + 10)a x + (5 − 12)a y + (8 + 1)a z |
2
2 −1 2 3
2
(
)
1
1
169 + 49 + 81 = 8.646
= | (13, −7,9) |=
Prob. 1.12
2
Prob.1.1 2
P Q = (2, −6,5)(0,3,1) = 0 − 18 + 5 = −13
rOP = 4a x − 5a y + a z
arOP =
rOP
(4, −5,1)
=
= 0.6172a x − 0.7715a y + 0.1543a z
| rOP |
(16 + 25 + 1)
Prob. 1.2
r = (−3, 2, 2) − (2, 4, 4) = (−5, −2, −2)
r (−5, −2, −2)
ar = =
= −0.8704a x − 0.3482a y − 0.3482a z
r
25 + 4 + 4
Prob. 1.3
rMN = rN − rM = (3,5, −1) − (1, −4, −2) = 2a x + 9a y + a z
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Prob. 1.4
A − 2 B = (4, −6,3) − 2(−1,8,5) = (4, −6,3) − ( −2,16,10)
(a)
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ar =
Sadiku & Kulkarni
r (−5, −2, −2)
=
= −0.8704a x − 0.3482a y − 0.3482a z
r
25 + 4 + 4
Principles of Electromagnetics, 6e
Prob. 1.3
5
rMN = rN − rM = (3,5, −1) − (1, −4, −2) = 2a x + 9a y + a z
Prob. 1.3
1.4
A − 2 B = (4, −6,3) − 2(−1,8,5) = (4, −6,3) − ( −2,16,10)
(a)
= (6, −22, −7)
4
(b) A B = (4, −6,3)(−1,8,5) = −4 − 48 + 15 = −37
(c)
A× B =
4
−6 3
−1
8
5
= (−30 − 24)a x + (−3 − 20)a y + (32 − 6)a z
= −54a x −23a y + 26a z
1.5
Prob. 1.4
3 5 1
= (−35 − 1)a x + (0 + 21)a y + (3 − 0)a z
B×C =
0 1 −7
= −36a x + 21a y + 3a z
A( B × C ) = (4, 2,1)(−36, 21,3) = −144 + 42 + 3 = −99
Prob. 1.6
1.5
(a)
B×C =
1 1 1
0 1 2
= a x − 2a y + a z
A( B × C ) = (1, 0, −1)(1, −2,1) = 1 + 0 − 1 = 0
(b)
A× B =
1 0 −1
1 1
1
= a x − 2a y + a z
( A × B )C = (1, −2,1)(0,1, 2) = 0 − 2 + 2 = 0
(c) A × ( B × C ) =
1 0 −1
= −2a x − 2a y − 2a z
1 −2 1
(d) ( A × B ) × C =
1 −2 1
= −5a x − 2a y + a z
0 1 2
Prob.1.7
Prob. 1.6
(a) T = (3, -2, 1) and S = (4, 6, 2)
(b) rTS = rs – rt = (4, 6, 2) – (3, -2, 1) = ax + 8ay + az
(c) distance = |rTS| = 1 + 64 + 1 = 8.124 m
Prob. 1.8
(a) If A and B are parallel, A=kB, where k is a constant.
(α ,3, −2) = k (4, β ,8)
Equating coefficients gives
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Prob.1.7
(a) T = (3, -2, 1) and S = (4, 6, 2)
Sadiku & Kulkarni
Principles of Electromagnetics, 6e
(b) rTS = rs – rt = (4, 6, 2) – (3, -2, 1) = ax + 8ay + az
6
(c) distance = |rTS| = 1 + 64 + 1 = 8.124 m
Prob. 1.8
1.7
(a) If A and B are parallel, A=kB, where k is a constant.
(α ,3, −2) = k (4, β ,8)
Equating coefficients gives
−2 = 8k
⎯⎯
→
5
k =−
1
4
α = 4k = −1
3 = βk
⎯⎯
→ β = 3 / k = −12
This can also be solved using A X B = 0.
(b) If A and B are perpendicular to each other,
A• B = 0
⎯⎯
→ 4α + 3β − 16 = 0
Prob. 1.9
1.8
(a) A ⋅ B = AB cos θ AB
A × B = ABsin θ AB an
( A⋅ B)
2
+ A × B = ( AB ) ( cos 2 θ AB + sin 2 θ AB ) = ( AB )
2
2
2
(b) a x ⋅ (a y × a z ) = a x ⋅ a x = 1. Hence,
a y × az
a x ⋅ a y × az
=
ax
= ax
1
ay
az × a x
=
= ay
1
a x ⋅ a y × az
ax × ay
a x ⋅ a y × az
=
az
= az
1
1.9
Prob. 1.10
(a) P + Q = ( 6, 2, 0 ) , P + Q − R = ( 7,1, −2 )
P + Q − R = 49 + 1 + 4 = 54 = 7.3485
2 −1 −2
(b) P .Q × R = 4 3 2 = 2 ( 6 − 2 ) + ( 8 + 2 ) − 2 ( 4 + 3) = 8 + 10 − 14 = 4
−1 1 2
Q×R =
4 3 2
= ( 4, −10, 7 )
−1 1 2
P .Q × R = ( 2, −1, −2 ) ⋅ ( 4, −10, 7 ) = 8 + 10 − 14 = 4
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Sadiku & Kulkarni
Principles of Electromagnetics, 6e
Prob. 1.10
(a) P + Q = ( 6, 2, 0 ) , P + Q − R = ( 7,1, −2 )
7
P + Q − R = 49 + 1 + 4 = 54 = 7.3485
2 −1 −2
(b) P .Q × R = 4 3 2 = 2 ( 6 − 2 ) + ( 8 + 2 ) − 2 ( 4 + 3) = 8 + 10 − 14 = 4
−1 1 2
Q×R =
4 3 2
= ( 4, −10, 7 )
−1 1 2
P .Q × R = ( 2, −1, −2 ) ⋅ ( 4, −10, 7 ) = 8 + 10 −614 = 4
(c) Q × P =
4 3 2
= ( −4,12, −10 )
2 −1 −2
Q × P ⋅ R = ( −4,12, −10 ) ⋅ ( −1,1, 2 ) = 4 + 12 − 20 = −4
or
−1
Q × P ⋅ R = R ⋅Q × P = 4
2
1 2
3 2 = − ( −6 + 2 ) − ( − 8 − 4 ) + 2 ( − 4 − 6 ) = − 4
−1 −2
(d) ( P × Q ) ⋅ ( Q × R ) = ( 4, −12,10 ) ⋅ ( 4, −10, 7 ) = 16 + 120 + 70 = 206
(e) ( P × Q ) × ( Q × R ) =
4 −12 10
= 16ax + 12a y + 8az
4 −10 7
( −2 − 1 − 4 ) = −7 = −0.9526
P⋅R
=
P R
4 +1+ 4 1+1+ 4 3 6
7
θ PR = 162.3
(f) cos θ PR =
5 + 144 + 100
P ×2Q −6 16
a z = 0.998
= −21a x - 2a=y + 6260
(g) sin θPPQ× Q
= =
=
0
3
1
P Q
3 16 + 9 + 4
3 29
P ⋅ Q
−13
θcos
=PQ86.45
=
=
= −0.51 ⎯⎯
→ θ PQ = 120.66o
PQ θ
PQ
10 65
Prob. 1.11
1.13
Prob. 1.10
If A and B are parallel, then B = kA and A x B = 0. It is evident that k = -2 and that
A B = (4, −6,1)(2, 0,5) = 8 − 0 + 5 = 13
ax a y az
(a) | B |2 = 22 + 52 = 29
A × B = 1 −2 3 = 0
A B + 2 | B |2 = 13 + 2 × 29 = 71
−2 4 −6
as expected.
(b)
A× B
Prob.
a⊥ = ±1.14
| Athe
× Bfact
| that
(a) Using
( A × B ) × C = (4A ⋅−C6)B 1− =(B(−⋅ C30,)A−,18,12)
Let C = A × B =
2 0 5
we get
30,
( A−×18,12)
A ×C( A × B ) (=−−
B ) × A = (B ⋅ A)A − ( A ⋅ A)B
a⊥ = ±
=±
= ±(−0.8111a x − 0.4867a y + 0.3244a z )
2
2
2
|C |
+
+
30
18
12
(b) A × ( A × ( A × B ) ) = A × ( A ⋅ B ) A- ( A ⋅ A) B 
= ( A ⋅ B ) - ( A × A) − ( A ⋅ A ) - ( A × B )
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Prob. 1.12
= − A2 ( A × B )
Q==0(2, −6,5)(0,3,1) = 0 − 18 + 5 = −13
since AP
xA
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If A and B are parallel, then B = kA and A x B = 0. It is evident that k = -2 and that
ax a y az
Sadiku & Kulkarni
Principles of Electromagnetics, 6e
A × B = 1 −2 3 = 0
−2 4 −6
8
as expected.
1.11
Prob. 1.14
(a) Using the fact that
( A × B ) × C = ( A ⋅ C )B − (B ⋅ C )A,
we get
A × ( A × B ) = −( A × B ) × A = (B ⋅ A)A − ( A ⋅ A)B
(b) A × ( A × ( A × B ) ) = A × ( A ⋅ B ) A- ( A ⋅ A) B 
= ( A ⋅ B ) - ( A × A) − ( A ⋅ A ) - ( A × B )
since AxA = 0
= − A2 ( A × B )
8
Prob.
1.15
Prob. 1.12
1.16
P2
1
1 4 1 −5 1
Area = | D × E |=
= | (3 + 10)a x + (5 − 12)a y + (8 + 1)a z |
2
2 −1 2 a 3
2
=
1
1
| (13, −7,9) |=
169 + 49 + 81 = 8.646
2
2
b
P1
c
P3
a = rp 2 − rp1 = (1, −2, 4) − (5, −3,1) = ( −4,1,3)
(a) b = rp 3 − rp 2 = (3,3,5) − (1, −2, 4) = (2,5,1)
c = rp1 − rp 3 = (5, −3,1) − (3,3,5) = (2, −6, −4)
Note that a + b + c = 0
⎯⎯
→ perpendicular
a ⋅ b = −8 + 5 + 3 = 0
b ⋅ c = 4 − 30 − 4 ≠ 0
c ⋅ a = −8 − 6 − 12 ≠ 0
Hence P2 is a right angle.
1
1 −4 1 3 1
| a × b |=
= | (1 − 15)a x + (6 + 4)a y + (−20 − 2)a z |
2
2 2 5 1 2
Area =
(b)
=
1
1
| (−14,10, −22) |=
196 + 100 + 484 = 13.96
2
2
Prob. 1.17
Given rP = (−1, 4,8),
rQ = (2, −1,3),
rR = (−1, 2,3)
(a) | PQ |= 9 + 25 + 25 = 7.6811
(b) PR = −2a y − 5a z
 QP QR 
o
∠PQR = cos −1 
 = 42.57
 | QP || QR | 
(d) Area of triangle PQR = 11.023
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(e) Perimeter = 17.31
(c)
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Hence P2 is a right angle.
Sadiku & Kulkarni
Area =
(b)
Principles of Electromagnetics, 6e
1
1 −4 1 3 1
| a × b |=
= | (1 − 15)a x + (6 + 4)a y + (−20 − 2)a z |
2
2 2 5 1 2
9
1
1
196 + 100 + 484 = 13.96
= | (−14,10, −22) |=
2
2
Prob. 1.13
1.17
Given rP = (−1, 4,8),
rQ = (2, −1,3),
rR = (−1, 2,3)
(a) | PQ |= 9 + 25 + 25 = 7.6811
(b) PR = −2a y − 5a z
 QP QR 
o
∠PQR = cos −1 
 = 42.57
 | QP || QR | 
(d) Area of triangle PQR = 11.023
(e) Perimeter = 17.31
(c)
Prob.1.18
Prob. 1.14
Let R be the midpoint of PQ.
1
rR = {(2, 4, −1) + (12,16,9)} = (7,10,94)
2
OR = 49 + 100 + 16 = 165 = 12.845
OR 12.845
t=
=
= 42.82 ms
300
v
Prob. 1.15
1.19
Ax
A ⋅ ( A× B ) = Bx
Ay
By
Az
Ax
Bz , ( A× B ) ⋅ C = Bx
Ay
By
Az
Bz
Cx
Cy
Cz
Cy
Cz
Cx
Hence, A ⋅ ( A× B ) = ( A× B ) ⋅ C
Also, each equals the volume of the parallelopiped formed by the three vectors as sides.
Prob. 1.16
1.20
(a) Let P and Q be as shown below:
y
Q
θ2
P
θ1
x
P = cos 2 θ 1 + sin 2 θ 1 = 1, Q = cos 2 θ 2 + sin 2 θ 2 = 1,
Hence P and Q are unit vectors.
(b) P ⋅ Q = (1)(1)cos(θ 2 -θ1 )
But P ⋅ Q = cos θ1 cos θ 2 + sin θ1 sin θ 2 . Thus,
cos(θ 2 − θ1 ) = cos θ1 cos θ 2 + sin θ1 sin θ 2
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Let P1 = P = cos θCopyright
1 a x + sin θ 1 a y and
POESM_Ch01.indd 9
Q1 = cos θ 2 a x − sin θ 2 a y .
9/14/2015 3:29:31 PM
θ2
Sadiku & Kulkarni
P
θ1
Principles of Electromagnetics, 6e
x
P = cos 2 θ 1 + sin 2 θ 1 = 1, Q = cos 2 10
θ 2 + sin 2 θ 2 = 1,
Hence P and Q are unit vectors.
(b) P ⋅ Q = (1)(1)cos(θ 2 -θ1 )
But P ⋅ Q = cos θ1 cos θ 2 + sin θ1 sin θ 2 . Thus,
cos(θ 2 − θ1 ) = cos θ1 cos θ 2 + sin θ1 sin θ 2
Let P1 = P = cos θ 1a x + sin θ 1a y and
Q1 = cos θ 2 a x − sin θ 2 a y .
P1 and Q1 are unit vectors as shown below:
y
P1
θ1
θ1+θ2
θ2
10
x
Q1
P1 ⋅ Q1 = (1)(1) cos(θ 1 + θ 2 )
But P1 ⋅ Q1 = cosθ 1 cosθ 2 − sin θ 1 sin θ 2 ,
cos(θ 2 + θ 1 ) = cosθ 1 cosθ 2 − sin θ 1 sin θ 2
Alternatively, we can obtain this formula from the previous one by replacing
θ2 by -θ2 in Q.
(c)
1
1
| P − Q |= | (cos θ1 − cos θ 2 ) ax + (sin θ1 − sin θ 2 ) a y
2
2
=
1
cos 2 θ1 + sin 2 θ1 + cos 2 θ 2 + sin 2 θ 2 − 2 cos θ1 cos θ 2 − 2sin θ1 sin θ 2
2
1
1
2 − 2(cos θ1 cos θ 2 + sin θ1 sin θ 2 ) =
2 − 2 cos(θ 2 − θ1 )
2
2
Let θ 2 − θ1 = θ , the angle between P and Q.
=
1
1
| P − Q |=
2 − 2 cos θ
2
2
But cos 2A = 1 – 2 sin 2A.
1
1
| P − Q |=
2 − 2 + 4sin 2 θ / 2 = sin θ / 2
2
2
Thus,
θ −θ
1
| P − Q |=| sin 2 1 |
2
2
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POESM_Ch01.indd 10
Prob. 1.21
9/14/2015 3:29:32 PM
1
1
2 − 2(cos θ1 cos θ 2 + sin θ1 sin θ 2 ) =
2 − 2 cos(θ 2 − θ1 )
2
2
Sadiku & Kulkarni
Let θ 2 − θ1 = θ , the angle between P and Q.
=
Principles of Electromagnetics, 6e
11
1
1
| P − Q |=
2 − 2 cos θ
2
2
But cos 2A = 1 – 2 sin 2A.
1
1
| P − Q |=
2 − 2 + 4sin 2 θ / 2 = sin θ / 2
2
2
Thus,
θ −θ
1
| P − Q |=| sin 2 1 |
2
2
Prob. 1.17
1.21
w(1,−2,2)
= (1,−2,2), r = rp − ro = (1,3,4) − ( 2,−3,1) = ( −1,6,3)
3
1 −2 2
u = w×r =
= ( −18,−5,4)
−1 6 3
w=
u = −18a x − 5a y + 4az
1.22
Prob. 1.18
r1 = (1,1,1),
cos θ =
r2 = (1, 0,1) − (0,1, 0) = (1, −1,1)
r1 ⋅ r2 (1 − 1 + 1) 1
=
=
r1r2
3
3 3
⎯⎯
→ θ = 11
70.53o
Prob. 1.19
1.23
T ⋅ S ( 2, −6,3) ⋅ (1, 2,1) −7
=
=
= −2.8577
S
6
6
(S ⋅ T )T = − 7(2,−6,3)
(b) S T = ( S ⋅ a T )a T =
72
T2
= − 0.2857a x + 0.8571a y − 0.4286a z
(a) Ts = T ⋅ as =
(c) sin θ TS =
T×S
=
T S
2 − 6 3 (− 12,1,10)
245
=
=
= 0.9129
1 2 1
7 6
7 6
 θ TS = 65.91
1.24
Prob. 1.20
Let A = AB + AB ⊥
AB = ( A ⋅ a B )a B =
Hence,
A⋅ B
B
B⋅B
AB⊥ = A − AB = A −
A⋅ B
B
B⋅B
Prob.1.25
(a) H (1,3, −2) = 6a x + a y + 4a z
aH =
POESM_Ch01.indd 11
(6,1, 4)
= 0.8242a x + 0.1374a y + 0.5494a z
36 + 1 + 16Copyright © 2015 by Oxford University Press
(b) | H |= 10 = 4 x 2 y 2 + ( x + z ) 2 + z 4
9/14/2015 3:29:32 PM
Let
A
Sadiku & Kulkarni B
A = AB + AB ⊥
= ( A ⋅ a B )a B =
Hence,
A⋅ B
B
B⋅B
AB⊥ = A − AB = A −
Principles of Electromagnetics, 6e
12
A⋅ B
B
B⋅B
Prob.
1.21
Prob.1.25
(a) H (1,3, −2) = 6a x + a y + 4a z
(6,1, 4)
= 0.8242a x + 0.1374a y + 0.5494a z
36 + 1 + 16
aH =
(b) | H |= 10 = 4 x 2 y 2 + ( x + z ) 2 + z 4
or
100 = 4 x 2 y 2 + x 2 + 2 xz + z 2 + z 4
1.26
Prob. 1.22
C = 5a x + a z
(a) B × C =
1 1 0
= a x − a y − 5a z
5 0 1
A( B × C ) = (4, −1,1)(1, −1, −5) = 4 + 1 − 5 = 0
(b) AB = ( AaB )a B =
( A ⋅ B ) B (4 − 1)(1,1, 0)12
=
= 1.5a x + 1.5a y
| B |2
1+1
1.23
Prob. 1.27
(a) At (1, -2, 3), x = 1, y = -2, z = 3.
G = a x + 2a y + 6a z ,
H = −6a x + 3a y − 3a z
G = 1 + 4 + 36 = 6.403
H = 36 + 9 + 9 = 7.348
(b) G  H = −6 + 6 − 18 = −18
(c)
GH
−18
=
= −0.3826
6.403 × 7.348
GH
= 112.5o
cos θGH =
θGH
Prob. 1.28
1.24
rPQ = rQ − rP = (−2,1, 4) − (1, 0,3) = ( −3,1,1)
At P, H = 0a x − 1a z = −a z
The scalar component of H along rPQ is
D = H arPQ =
H • rPQ
| rPQ |
=
−1
= −0.3015
9 +1+1
Prob. 1.29
(a) At P, x = -1, y = 2, z = 4
D = 8a x − 4a y - 2a z ,
E = −10a x + 24a y + 128a z
Copyright
© 2015
C = D + E = −2a x + 20
a y + 126
a z by Oxford University Press
POESM_Ch01.indd 12
C a x
−2
9/14/2015 3:29:32 PM
rPQ = rQ − rP = (−2,1, 4) − (1, 0,3) = ( −3,1,1)
At P, H = 0a x − 1a z = −a z
Sadiku & Kulkarni
Principles of Electromagnetics, 6e
The scalar component of H along rPQ is
D = H arPQ =
H • rPQ
| rPQ |
−1
= −0.301513
9 +1+1
=
1.25
Prob. 1.29
(a) At P, x = -1, y = 2, z = 4
D = 8a x − 4a y - 2a z ,
E = −10a x + 24a y + 128a z
C = D + E = −2a x + 20a y + 126a z
(b)
C a x = C cos θ x
⎯⎯
→ cos θ x =
C a x
−2
=
= −0.01575
C
22 + 202 + 1262
θ x = 90.9o
13
Prob. 1.26
1.30
(a) At (1,2,3), E = (2,1,6)
E = 4 + 1 + 36 = 41 = 6.403
(b) At (1,2,3), F = (2,-4,6)
E F = ( E ⋅ aF ) aF =
( E ⋅ F )F
F
2
=
36
( 2,−4,6)
56
= 1.286a x − 2.571a y + 3.857az
(c) At (0,1,–3), E = (0,1,–3), F = (0,–1,0)
E×F =
a E ×F = ±
0 1 −3
= (−3,0,0)
0 −1 0
E×F
= ± ax
E×F
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Sadiku & Kulkarni
Principles of Electromagnetics, 6e
14
14
CHAPTER 2
P. E. 2.1
(a) At P(1,3,5),
ρ=
x = 1,
2
x + y
2
y = 3,
10 ,
=
z =5,
z = 5,
φ = tan −1 y / x = tan −1 3 = 71.6o
P( ρ , φ , z ) = P( 10, tan −1 3,5) = P (3.162, 71.6o ,5)
Spherical system:
r=
x2 + y2 + z2 =
35 = 5.916
θ = tan −1 x 2 + y 2 z = tan −1 10 5 = tan −1 0.6325 = 32.31°
P (r ,θ , ϕ ) = P (5.916,32.31°, 71.57°)
At T(0,-4,3),
x=0
y =-4,
z =3;
ρ = x + y = 4, z = 3, ϕ = tan y / x = tan − 4 / 0 = 270°
T ( ρ, ϕ , z) = T (4,270° ,3).
2
−1
2
−1
Spherical system:
r=
x 2 + y 2 + z 2 = 5,θ = tan −1 ρ / z = tan −1 4 / 3 = 5313
. °.
T (r ,θ , ϕ ) = T (5,5313
. ° ,270° ).
At S(-3-4-10),
x =-3, y =-4, z =-10;
 −4 
ρ = x 2 + y 2 = 5, φ = tan −1   = 233.1°
 −3 
S ( ρ , φ , z ) = S (5, 233.1,− 10).
Spherical system:
r = x 2 + y 2 + z 2 = 5 5 = 11.18.
5
= 153.43°;
−10
S (r , θ , φ ) = S (11.18,153.43°, 233.1°).
θ = tan −1 ρ z = tan −1
(b)
In Cylindrical system,
Qx =
ρ
ρ + z2
2
;
ρ = x2 + y 2 ;
yz = z ρ sin φ ,
z ρ sin φ
Qy = 0;
Qz = −
ρ 2 + z2
;
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Sadiku & Kulkarni
Principles of Electromagnetics, 6e
15
 Qρ   cos φ
  
 Qφ  =  − sin φ
 Qz   0
Qρ = Qx cos φ =
ρ cos φ
ρ +z
2
0  Qx 
 
0  0  ;
1  Qz 
sin φ
cos φ
0
2
,
Qφ = −Qx sin φ =
− ρ sin φ
ρ 2 + z2
Hence,
Q=
ρ
ρ + z2
2
(cos φ a ρ − sin φ a φ − z sin φ a z ).
In Spherical coordinates:
r sin θ
Qx =
= sin θ ;
r
1
Qz =− r sin φ sin θ r cosθ = − r sin θ cosθ sin φ .
r
Qr   sin θ cos φ sin θ sin φ cosθ  Qx 
  
 
Qθ  = cosθ cos φ cosθ sin φ − sin θ   0  ;
Qφ   − sin φ
cos φ
0  Qz 
 
Qr = Qx sin θ cos φ + Qz cosθ = sin 2 θ cosφ − r sin θ cos2 θ sin φ .
Qθ = Qx cosθ cos φ − Qz sin θ = sin θ cosθ cos φ + r sin 2 θ cosθ sin φ .
Qφ =− Qx sin φ = − sin θ sin φ.
∴ Q = sin θ ( sin θ cosφ − r cos2 θ sin φ ) a
At T :
r
+ sin θ cosθ (cosφ + r sinθ sin φ )aθ − sinθ sin φ aφ .
4
12
a x + a z = 0.8a x + 2.4a z ;
5
5
4
Q ( ρ , φ , z ) = (cos 270° a ρ − sin 270° aφ − 3sin 270°a z
5
= 0.8aφ + 2.4 a z ;
Q ( x, y , z ) =
4
45
4 3
20
4
Q (r ,θ , φ ) = (0 − (−1))ar + ( )(0 + (−1))aθ − (−1)aφ
5
25
5 5
5
5
36
48
4
= a r − aθ + aφ = 1.44ar − 1.92aθ + 0.8 aφ ;
25
25
5
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Sadiku & Kulkarni
Principles of Electromagnetics, 6e
16
16
Note, that the magnitude of vector Q = 2.53 in all 3 cases above.
P.E. 2.2 (a)
 Ax  cosφ −sinφ 0  ρz sinφ 
 A  = sinφ cosφ 0  3ρ cosφ 
 y 


 Az   0
0
1 ρ cosφ sinφ
A = (ρz cosφ sinφ −3ρ cosφ sinφ) ax + (ρz sin2 φ +3ρ cos2 φ) ay + ρ cosφ sinφ az.
y
x
y
But ρ = x2 + y2 , tanφ = , cosφ =
, sinφ =
;
2
2
2
2
x
x +y
x +y
Substituting all this yields:
1
A=
[(xyz −3xy)ax + (zy2 + 3x2 ) ay + xy az ].
2
2
x +y
 Bx  sin θ cos φ
 B  =  sin θ sin φ
 y 
 Bz   cos θ
cos θ cos φ
cos θ sin φ
− sin θ
Since r = x 2 + y 2 + z 2 , tan θ =
and sin θ =
and sin φ =
x2 + y 2
2
2
x +y +z
y
x2 + y2
,
2
− sin φ 
cos φ 
0 
 r2 


 0 
sin θ 


x2 + y2
y
, tan φ = ;
z
z
, cos θ =
cos φ =
z
2
x + y2 + z2
x
x2 + y2
;
;
y
1
= ( r 2 x − y ).
r
r
x
1
B y = r 2 sin θ sin φ + sin θ cos φ = ry + = (r 2 y + x ).
r
r
1
Bz = r 2 cos θ = r z = ( r 2 z ).
r
Bx = r 2 sin θ cos φ − sin θ sin φ = rx −
Hence,
B=
1
2
2
x +y +z
2
[{x ( x 2 + y 2 + z 2 ) − y} a x + { y ( x 2 + y 2 + z 2 ) + x} a y + z ( x 2 + y 2 + z 2 )a z ].
Copyright © 2015 by Oxford University Press
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Sadiku & Kulkarni
Principles of Electromagnetics, 6e
17
P.E.2.3 (a) At:
(1, π / 3, 0),
H = (0, 0.06767,1)
1
a x = cos φ a ρ − sin φ aφ = (a ρ − 3 aφ )
2
H • a x = −0.0586.
(b)
At:
(1, π / 3, 0),
aθ = cos θ a ρ − sin θ a z = − a z .
aρ
aφ
az
0.06767 1 = − 0.06767 a ρ .
0
1
H × az = 0
0
(c)
(d)
( H • a ρ ) a ρ = 0 aρ .
H × az =
aρ
aφ
az
0
0.06767
1
0
0
1
= 0.06767 a ρ .
H × a z = 0.06767
P.E. 2.4
(a)
A • B = (3, 2, − 6) • ( 4, 0,3) = − 6.
(b)
A× B =
3 2 −6
= 6 ar − 33aθ − 8aφ .
4 0 3
Thus the magnitude of A × B = 34.48.
(c)
At (1, π / 3, 5π / 4), θ = π / 3,
a z = cos θ a r − sin θ aθ =
1
3
ar −
aθ .
2
2
3 
3
1
( Aa z )a z =  − 3   ar −
aθ = −0.116ar + 0.201aθ
2
2
2 
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Sadiku & Kulkarni
Principles of Electromagnetics, 6e
18
18
P.E. 2.5
In spherical coordinates, the distance between two points is given by eq. 2.33:
By solving the above equation, we obtain P.E. 2.6
In Cartesian system the dot product of two vectors    and    is given by
(1)
  
Now using the eq. 2.21 to represent spherical coordinates in Cartesian system and the dot
product is determined by using Eq. (1)
 ­  Dot product of the given vectors is 3
Using the above equation
 € € € € €
Hence by using the above derived equation we can directly calculate dot product of
vectors in spherical system
33
33
P.E. 2.7
2.7
Prob.
2.24
P.E.
Prob.
2.24
At
P
(0,
2,−−5),
5),
90°°;;
At P(0, 2,
φφ ==90
cosφφ −−sin
sinφφ 00
BBxx cos



 BB  ==  sin
sinφφ cos
cosφφ 00
y
y


  

BBzz  00
00
11
00

== 11
00
−−11
00
00
00

00
11
BB== −−aaxx −−55aayy −−33aazz
BBρρ
BB 
 φφ 
BBzz 
−−55
 11 
 
−−33
(2,4,10)
4,10)++((−−1,1,−−5,
5,−−3)
3)
((aa)) AA++ BB == (2,
== aaxx−−aayy ++77aazz..
−52
A• B
cosθθAB
= A • B == −52
((bb)) cos
AB =
AB
4200
AB
4200
−52
= cos
cos−−11(( −52 )) == 143.36
143.36°°..
θθAB
AB =
4200 © 2015 by Oxford University Press
Copyright
4200
Chapter02.indd 18
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 Bz 
Sadiku & Kulkarni
 0
0
1   Bz 
0 −1 0   −5
= 1 0 0   1 
0 0 1   −3
Principles of Electromagnetics, 6e
19
B = −a x − 5a y − 3a z
(a) A + B = (2, 4,10) + ( −1, −5, −3)
= ax − a y + 7 az .
−52
A• B
=
AB
4200
−52
= cos −1 (
) = 143.36°.
4200
(b) cos θ AB =
θ AB
(c) AB = A • a B =
A• B
52
=−
= −8.789.
B
35
18
Prob. 2.24
Prob.At2.1(1, 60o , −1),
ρ = 1, φ = 60o , z = −1,
2
+ sin
y 2 60
= o )4a+ρ 36
(a) ρA== (−x2 −
+ (4=+6.324
2 cos 60o )aφ − 3(1)( −1)a z
1 y
−1 6
3a z o
ρ + 5aφ=+71.56
(a) φ = =
tan−−2.866
= atan
x
2
o
o
P is 60
(6.324,
71.56
B = 1cos
a ρ + sin
60o ,a−φ 4)
+ a z = 0.5aφ + 0.866aφ + a z
A B = −1.433 + 4.33 + 3 = 5.897
r = x 2 +2 y 2 + z 2 = 4 + 36 + 16 = 7.485
AB = 2.866 + 26 + 9 0.25 + 1 + 0.8662 = 9.1885
2
+ y2
−1B x 5.897
−1 6.324
o
−1 6.324
o
A
==tan
90→
+ tan
(b)
cos θθAB==tan =
0.6419 =⎯⎯
θ AB = 50.07=o 147.69
z
−4
4
AB 9.1885
o
o
P is (7.483,147.69 , 71.56 )
Let D = A × B. At (1,90o , 0),
ρ = 1, φ = 90o , z = 0
(b)
A = − sin 90o a ρ + 4aφ = −a ρ + 4aφ
Prob.2.2
(a)
Given P(1,-4,-3), convert to cylindrical and spherical values;
B = 1cos 90o2 a ρ +2 sin 90o a2 φ + a z =2 aφ + a z
ρ = x + y = 1 + (−4) = 17 = 4.123.
y
−4
= tan −1
= 284.04°.
x
1
∴ P ( ρ , φ , z ) = (4.123, 284.04°, − 3)
φ = tan −1
Spherical :
r = x 2 + y 2 + z 2 = 1 + 16 + 9 = 5.099.
ρ
4.123
= 126.04°.
−3
z
P (r , θ , φ ) = P (5.099, 126.04°, 284.04°).
θ = tan −1
(b)
= tan −1
y
0
= tan −1 = 0o
x
3
o
Q( ρ , φ , z ) = Q(3, 0 ,5)
ρ = 3,
φ = tan −1
r = 9 + 0 + 25 = 5.831,
θ = tan −1
Q(r , θ , φ ) = Q(5.831,30.96 , 0 )
o
(c)
POESM_Ch02.indd 19
o
ρ
z
= tan −1
3
= 30.96o
5
6 Universityo Press
Copyright © 2015 by−1Oxford
ρ = 4 + 36 = 6.325,
φ = tan
o
−2
= 108.4
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Spherical :
Sadiku & Kulkarni
Principles of Electromagnetics, 6e
r = x 2 + y 2 + z 2 = 1 + 16 + 9 = 5.099.
ρ
4.123
= 126.0420
°.
−3
z
P (r , θ , φ ) = P (5.099, 126.04°, 284.04°).
θ = tan −1
(b)
= tan −1
y
0
= tan −1 = 0o
3
x
o
Q( ρ , φ , z ) = Q(3, 0 ,5)
ρ = 3,
φ = tan −1
ρ
θ = tan −1
r = 9 + 0 + 25 = 5.831,
z
Q(r , θ , φ ) = Q(5.831,30.96 , 0 )
o
(c)
ρ = 4 + 36 = 6.325,
o
φ = tan −1
R ( ρ , φ , z ) = R(6.325,108.4o , 0)
r = ρ = 6.325,
θ = tan −1
ρ
= tan −1
3
= 30.96o
5
6
= 108.4o
−2
= tan −1
z
R (r , θ , φ ) = R(6.325,90 ,108.4o )
o
6.325
= 90o
0
19
Prob. 2.3
(a)
x = ρ cos φ = 2 cos 30° = 1.732;
y = ρ sin φ = 2sin 30° = 1;
z = 5;
P1 ( x, y, z ) = P1 (1.732,1, 5).
(b)
x = 1cos 90° = 0;
y = 1sin 90° =1;
z = − 3.
P2 ( x, y, z ) = P2 (0, 1, − 3).
(c)
(d)
x = r sin θ cos φ = 10sin(π / 4) cos(π / 3) = 3.535;
y = r sin θ sin φ = 10sin(π / 4) sin(π / 3) = 6.124;
z = r cos θ
= 10 cos(π / 4)
= 7.0711
P3 ( x, y, z ) = P3 (3.535, 6.124, 7.0711).
x = 4sin 30° cos 60° =1
y = 4sin 30° sin 60° = 1.7321
z = r cos θ = 4 cos 30° = 3.464
P4 ( x, y, z ) = P4 (1,1.7321,3.464).
Copyright © 2015 by Oxford University Press
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Principles of Electromagnetics, 6e
21
20
Prob. 2.4
x = ρ cos φ = 5cos120o = −2.5
y = ρ sin φ = 5sin120o = 4.33
(a)
z =1
Hence Q = (−2.5, 4.33,1)
r = x 2 + y 2 + z 2 = ρ 2 + z 2 = 25 + 1 = 5.099
(b)
θ = tan −1
x2 + y2
ρ
5
= tan −1 = tan −1 = 78.69o
z
z
1
φ = 120o
Hence Q = (5.099, 78.69o ,120o )
Prob. 2.5
T (r ,θ , φ )
⎯⎯
→
r = 10, θ = 60o , φ = 30o
x = r sin θ cos φ = 10sin 60o cos 30o = 7.5
y = r sin θ sin φ = 10sin 60o sin 30o = 4.33
z = r cos θ = 10 cos 60o = 5
T ( x, y, z ) = (7.5, 4.33,5)
ρ = r sin θ = 10sin 60o = 8.66
T ( ρ , φ , z ) = (8.66,30o ,5)
Prob. 2.6
(a)
x = ρ cos φ ,
y = ρ sin φ ,
V = ρ z cos φ − ρ sin φ cos φ + ρ z sin φ
2
(b)
U = x2 + y 2 + z 2 + y 2 + 2z 2
= r 2 + r 2 sin 2 θ sin 2 φ + 2r 2 cos 2 θ
= r 2 [1 + sin 2 θ sin 2 φ + 2 cos 2 θ ]
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Principles of Electromagnetics, 6e
22
21
Prob. 2.7
(a)
 Fρ   cos φ
 F  =  − sin φ
 φ 
 Fz   0
Fρ =
Fφ =
Fz =
_
F =
1
ρ +z
2
[ ρ cos 2 φ + ρ sin 2 φ ] =
2
1
ρ 2 + z2
4
ρ + z2
2
1
ρ + z2
2
sin φ 0 
cos φ 0 
0
1 


x
 2

 ρ + z2 


y


 ρ 2 + z2 


4


 2
2 
 ρ +z 
ρ
ρ + z2
2
;
[− ρ cos φ sin φ + ρ cos φ sin φ ] = 0;
;
( ρ aρ + 4 az )
In Spherical:
 Fr 
sinθ cosφ sinθ sinφ cosθ 
 


Fθ  = cosθ cosφ cosθ sin φ −sinθ 
 
cosφ
0 
 −sinφ
Fφ 
 x
r 
 
 y
r 
4
 
 r 
r
r
4
4
Fr = sin2 θ cos2 φ + sin2 θ sin2 φ + cosθ
= sin2 θ + cosθ ;
r
r
r
r
4
4
Fθ =sinθ cosθ cos2 φ + sinθ cosθ sin2 φ − sinθ = sinθ cosθ − sinθ ;
r
r
Fφ =− sinθ cosφ sinφ + sinθ sinφ cosφ = 0;
_
4
4
∴ F = (sin2 θ + cosθ ) ar + sinθ (cosθ − )aθ
r
r
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Principles of Electromagnetics, 6e
23
22
(b)
Gρ   cos φ
 G  =  − sin φ
 φ 
 Gz   0
ρ2
Gρ =
sin φ
0
cos φ 0 
0
1 
 xρ 2 
 2

 ρ + z2 


2
 yρ

 ρ 2 + z2 


 zρ 2 
 2

2
 ρ + z 
[ ρ cos φ + ρ sin φ ] =
2
ρ 2 + z2
ρ3
2
ρ 2 + z2
;
Gφ = 0;
zρ 2
Gz =
ρ 2 + z2
ρ2
G=
ρ 2 + z2
;
( ρ aρ + z az )
Spherical :
G=
Prob. 2.8
B = ρ ax +
ρ2
y
ρ
( xa x + ya y + za z ) =
r
r 2 sin 2 θ
rar = r 2 sin 2 θ ar
r
a y + za z
 Bρ   cos φ
 B  =  − sin φ
 φ 
 Bz   0
Bρ = ρ cos φ +
sin φ
cos φ
0
y
ρ
Bφ = − ρ sin φ +
0  ρ 
0   y / ρ 
1   z 
sin φ
y
ρ
cos φ
Bz = z
But y = ρ sin φ
Bρ = ρ cos φ + sin 2 φ , Bφ = − ρ sin φ + sin φ cos φ
Hence,
B = ( ρ cos φ + sin 2 φ )a ρ + sin φ (cos φ − ρ )aφ + za z
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Principles of Electromagnetics, 6e
24
23
Prob. 2.9
 Ax  cos φ
 A  =  sin φ
 y 
 Az   0
− sin φ
cos φ
0
At P, ρ = 2,
φ = π / 2,
0 2
0   3 
1   4 
z = −1
Ax = 2 cos φ − 3sin φ = 2 cos 90o − 3sin 90o = −3
Ay = 2sin φ + 3cos φ = 2sin 90o + 3cos 90o = 2
Az = 4
Hence, A = −3a x + 2a y + 4a z
Prob. 2.10
(a)
 Ax  cos φ
  
 Ay  =  sin φ
 
 Ay   0
− sin φ
cos φ
0
0   ρ sin φ 
0   ρ cos φ 
1   −2 z 
Ax = ρ sin φ cos φ − ρ cos φ sin φ = 0
Ay = ρ sin 2 φ + ρ cos 2 φ = ρ = x 2 + y 2
Hence,
Az = −2 z
A = x 2 + y 2 a y − 2 za z
(b)
 Bx 
sin θ cos φ cos θ cos φ − sin φ 
 B  =  sin θ sin φ cos θ sin φ cos φ 
 y


 Bz 
 cos θ
− sin θ
0 
Bx = 4r sin θ cos 2 φ + r cos θ cos φ
By = 4r sin θ sin φ cos φ + r cos θ sin φ
 4r cos φ 
 r 


 0 
Bz = 4r cos θ cos φ − r sin θ
But
2
2
sin θ =
2
r= x +y +z ,
sin φ =
y
2
x + y2
,
cos φ =
x2 + y2
,
r
x
cos θ =
z
r
x2 + y 2
Copyright © 2015 by Oxford University Press
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Principles of Electromagnetics, 6e
25
24
x2
zx
+
2
2
2
x +y
x + y2
xy
zy
+
By = 4 x 2 + y 2 2
2
2
x +y
x + y2
Bx = 4 x 2 + y 2
x
Bz = 4 z
B=
2
x +y
1
2
x +y
2
2
− x2 + y2
 x(4 x + z )a x + y (4 x + z )a y + (4 xz − x 2 − y 2 )a z 
Prob. 2.11
 Gx  cos φ
G  =  sin φ
 y 
 Gz   0
− sin φ 0   ρ sin φ 
cos φ 0   − ρ cos φ 
0
1   ρ 
Gx = ρ cos φ sin φ + ρ sin φ cos φ = 2 ρ sin φ cos φ
G y = ρ sin 2 φ − ρ cos 2 φ = ρ (1 − cos 2 φ ) − ρ cos 2 φ = ρ − 2 ρ cos 2 φ
Gz = ρ
ρ = x 2 + y 2 , cos φ =
Gx = 2 x 2 + y 2
But
x
ρ
=
xy
=
x + y2
2
Gy = x 2 + y 2 − 2 x 2 + y 2
x
2
x +y
2
,sin φ =
y
ρ
=
y
2
x + y2
2 xy
x2 + y2
x2
2 x2
2
2
=
+
−
x
y
x2 + y 2
x2 + y 2
Gz = x 2 + y 2
Thus,
G=

2x2
ax +  x2 + y 2 −

x2 + y 2
x2 + y2

2 xy

 a y + x2 + y 2 az


Copyright © 2015 by Oxford University Press
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Principles of Electromagnetics, 6e
26
25
Prob. 2.12
 H x  sin θ cos φ cos θ cos φ − sin φ  cos θ 
 H  =  sin θ sin φ cos θ sin φ cos φ   sin θ 
 y 


 H z   cos θ
− sin θ
0   0 
H x = sin θ cos θ cos φ + sin θ cos θ cos φ = 2sin θ cos θ cos φ
H y = cos θ sin θ sin φ + sin θ cos θ sin φ = 2sin θ cos θ sin φ
H z = cos 2 θ − sin 2 θ
But,
x2 + y 2
sin θ =
,
r
y
sin φ =
,
2
x + y2
Hx =
2z x2 + y2
x2 + y2 + z 2
Hy =
2z x2 + y2
x2 + y 2 + z 2
z
cos θ = ,
r
cos φ =
x
x2 + y2
y
x2 + y2
x
2
x + y2
=
2 xz
x + y2 + z2
=
2 yz
x + y2 + z2
2
2
z 2 − x2 − y2
x2 + y 2 + z 2
1
H= 2
2 xza x + 2 yza y + [ z 2 − x 2 − y 2 ]a z )
2
2 (
x +y +z
Hz =
Prob. 2.13
x = ρ cos φ
(a)
B = ρ cos φ a z
x = r sin θ cos φ
(b) B = r sin θ cos φ a z ,
Bx = 0 = By , Bz = r sin θ cos φ
 Br   sin θ cos φ sin θ sin φ cos θ  
0

  



0
 Bθ  = cos θ cos φ cos θ sin φ − sin θ  

 
cos φ
0   r sin θ cos φ 
 Bφ   − sin φ
Br = r sin θ cos θ cos φ = 0.5r sin(2θ ) cos φ
Bθ = −r sin 2 θ cos φ ,
Bφ = 0
B = 0.5r sin(2θ ) cos φ ar − r sin 2 θ cos φ aθ
Copyright © 2015 by Oxford University Press
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Principles of Electromagnetics, 6e
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26
Prob. 2.14
(a)
a x • a ρ = (cos φ a ρ − sin φ aφ ) • a ρ = cos φ
a x • aφ = (cos φ a ρ − sin φ aφ ) • aφ = − sin φ
a y • aρ = (sin φ a ρ + cos φ aφ ) • aρ = sin φ
_
_
a y • aφ = (sin φ a ρ + sin φ aφ ) • aφ = cos φ
(b) and (c)
In spherical system :
a x = sin θ cos φ ar + cos θ cos φ aθ − sin φ aφ .
a y = sin θ sin φ a r + cos θ sin φ aθ − cos φ aφ .
a z = cos θ a x − sin θ aθ .
Hence,
a x • a r = sin θ cos φ ;
a x • aθ = cos θ cos φ ;
a y • a r = sin θ sin φ ;
a y • aθ = cos θ sin φ ;
_
_
a z • a r = cos θ ;
_
_
a z • aθ = − sin θ ;
Prob. 2.15
(a)
r=
x2 + y2 + z2 =
ρ
θ = tan −1 ;
z
or
ρ=
ρ 2 + z2 .
φ = φ.
x 2 + y 2 = r 2 sin 2 θ cos2 φ + r 2 sin 2 θ sin 2 φ .
= r sin θ ;
z = r cosθ ;
φ = φ.
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Principles of Electromagnetics, 6e
28
27
(b) From the figures below,
cosθ aρ
z
z
aρ
aρ
az
ar
-az
θ
θ
sin θ aρ
cos θ a z
sin θ ( − az )
ρ
a r = sin θ a ρ + cos θ a z ;
ρ
aθ = cos θ a ρ − sin θ a z ;
aφ = aφ .
Hence,
 
 
 a r   sin θ 0 cos θ  a ρ 
  
 
 aθ  = cos θ 0 − sin θ   aφ 
   0
1
0   
 aφ  
az 
 
 
From the figures below,
a ρ = cos θ aθ + sin θ a r ; a z = cos θ ar − sin θ aθ ; aφ = aφ .
z
sin θ ar
z
az
cosθ aθ
sin θ ( − aθ )
−aθ
aρ
cosθ ar
ar
θ
θ
aθ
ρ
ar
ρ
Copyright © 2015 by Oxford University Press
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Principles of Electromagnetics, 6e
29
28
 
a ρ   sin θ
  
 aφ  =  0
   cos θ
 az  
 
cos θ
0
− sin θ
0
1 
0 
 
ar 
 
 aθ 
 
 az 
 
Prob. 2.16
If A and B are perpendicular to each other, AB = 0
A B = ρ 2 sin 2 φ + ρ 2 cos 2 φ -ρ 2
=ρ 2 (sin 2 φ + cos 2 φ )-ρ 2
=ρ 2 − ρ 2
=0
As expected.
Prob. 2.17
(a ) A + B = 8a ρ + 2aφ − 7a z
(b) A B = 15 + 0 - 8 = 7
(c ) A × B =
3 2
1
5 0 −8
=-16a ρ + (5 + 24)aφ − 10a z
=-16a ρ + 29aφ − 10a z
A⋅ B
7
7
=
=
AB
9 + 4 + 1 25 + 64
14 89
=0.19831
(d ) cosθ AB =
θ AB =78.56o
Prob. 2.18 (a)
 Ax 
cos φ
 A  =  sin φ
 y

 Az 
 0
− sin φ 0 
cos φ 0 
0
1 
 ρ cos φ 


0


 ρ z 2 sin φ 
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29
Ax = ρ cos 2 φ =
x2 + y 2
Ay = ρ sin φ cos φ =
Az = ρ z 2 sin φ = ρ z 2
1
A =
2
x +y
2
x2
=
x2 + y2
x2 + y 2
y
ρ
x2
x2 + y2
xy
=
x + y2
xy
2
x2 + y 2
= yz 2
[ x 2 a x + xya y + yz 2 x 2 + y 2 a z ]
At (3,-4,0)
x=3, y=-4, z=0;
1
A = [9a x − 12 a y ]
5
A=3
(b)
 Ar   sin θ cos φ
  
 Aθ  =  − cos θ cos φ
 
 Aφ   − sin φ
sin θ sin φ
cos θ sin φ
cos φ
cos θ 
− sin θ 
0 
x = r sin θ cos φ , y = r sin θ sin φ ,
Ar =
 x2 
 ρ 
 
 xy 
 
 ρ 
 yz 2 
 
 
z = r cos θ , ρ = r sin θ .
r sin θ cos φ
r sin θ cos φ sin φ
sin θ cos φ +
sin θ sin φ +
r sin θ
r sin θ
r 3 sin θ cos 2 θ sin φ cos θ
2
2
2
2
2
= r sin 2 θ cos3 φ + r sin 2 θ sin 2 φ cos φ + r 3 sin θ sin φ cos3 θ
Aθ = r sin θ cos 2 φ cos θ cos φ + r sin θ cos φ sin φ cos θ sin φ − r 2 cos 2 θ sin φ sin θ
= r sin θ cos θ cos φ − r 2 sin θ cos 2 sin φ
= r sin θ cos θ [cos φ − r cos θ sin φ ]
Aφ = − r sin θ cos 2 φ sin φ + r sin θ cos φ sin φ cos φ = 0.
∴
A=
r sin θ [cos φ sin θ + sin θ cos φ sin 2 φ +r 2 cos 3 θ sin φ ] ar
+ r sin θ cos θ [cos φ − r 2 cos θ sin θ sin φ ] aθ
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30
At (3 − 4, 0), r = 5, θ = π / 2, φ = 306.83
cos φ = 3 / 5,
sin φ = −4 / 5.
3
A = 5[12 * + 5(0)( −4 / 5)] ar + 5(1)(0)aθ
5
= 3ar
A = 3.
Prob. 2.19
 Ax 
cos φ − sin φ 0  Aρ 
 A  =  sin φ cos φ 0  A 
 y

  φ
 Az 
 0
0
1  Az 
x
y


0
−
 2
2
2
2
x +y
 x +y



y
x
0
=
2
2
x2 + y 2
 x +y



0
0
1



 Ax 
sin θ cos φ
 A  =  sin θ sin φ
 y

 Az 
 cos θ
cos θ cos φ
cos θ sin φ

x

 x2 + y 2 + z 2


y

2
2
2
=
 x +y +z

z

 x2 + y 2 + z 2

− sin θ
 Aρ 
A 
 φ
 Az 
− sin φ 
cos φ 

0 
 Ar 
 
 Aθ 
 Aφ 
 
xz
x2 + y 2 x2 + y 2 + z 2
yz
x2 + y 2 x2 + y 2 + z 2
−
x2 + y 2
x2 + y 2 + z 2


x2 + y 2 


x
2
2 
x +y 


0


−y
 Ar 
 
 Aφ 
 Aφ 
 
Copyright © 2015 by Oxford University Press
POESM_Ch02.indd 31
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Sadiku & Kulkarni
Principles of Electromagnetics, 6e
32
31
Prob. 2.20 (a) Using the results in Prob.2.14,
Aρ = ρz sin φ = r 2 sin θ cosθ sin φ
Aφ = 3ρ cos φ = 3r sin θ cos φ
Az = ρ cos φ sin φ = r sin θ cos φ sin φ
Hence,
 Ar   sin θ
  
 Aθ  = cos θ
 Aφ   0
 
0 cos θ   r 2 sin θ cos θ sin φ 


0 − sin θ   3r sin θ cos φ 

1
0   r sin θ cos φ sin φ 
A( r , θ , φ ) = r sin θ sin φ cos θ ( r sin θ + cos φ ) ar + sin φ ( r cos 2 θ − sin θ cos φ ) aθ + 3cos φ aφ 
At (10, π / 2,3π / 4),
r = 10, θ = π / 2, φ = 3π / 4
A = 10(0ar + 0.5aθ −
(b)
Br = r 2 = ( ρ 2 + z 2 ),
3
aφ ) = 5aθ − 21.21aφ
2
Bθ = 0,
 Bρ   sin θ
B  =  0
 φ 
 Bz   cos θ
Bφ = sin θ =
cos θ
0
− sin θ
ρ
ρ 2 + z2
0  Br 
 
1  Bθ 

0  Bφ 


ρ
+
B( ρ , φ , z ) = ρ 2 + z 2  ρ aρ + 2
a
z
a

φ
z
ρ + z2


At (2, π / 6,1),
ρ = 2, φ = π / 6, z = 1
B = 5(2aρ + 0.4aφ + a z ) = 4.472a ρ + 0.8944aφ + 2.236a z
Prob. 2.21
(a) d =
(b)
(6 − 2) 2 + ( − 1 − 1) 2 + (2 − 5) 2 =
29 = 5.385
d 2 = 32 + 52 − 2(3)(5) cos π + ( − 1 − 5) 2 = 100
d = 100 = 10
Copyright © 2015 by Oxford University Press
POESM_Ch02.indd 32
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Sadiku & Kulkarni
Principles of Electromagnetics, 6e
33
32
(c)
d 2 = 102 + 52 − 2(10)(5) cos
= 125 − 100(cos
π
4
d = 99.12 = 9.956.
π
cos
6
π
4
cos
− sin
π
6
π
− 2(10)(5) sin
π
π
4
sin
π
6
cos(7
π
4
−
3π
)
4
sin ) = 125 − 100 cos 75o = 99.12
4
6
Prob. 2.22
We can convert Q to cylindrical system and then use equation 2.32
At Q, r = 4 θ =
π
φ=
π
2
2
o
ρ = r sin θ = 4sin 90 = 4
φ=
π
2
z=r cos θ = 4 cos 90o = 0
Q is (4, π / 2, 0).
d 2 = ρ 22 + ρ12 − 2 ρ1 ρ 2 cos(φ2 − φ1 ) + ( z2 − z1 ) 2
= 102 + 42 − 2(10)(4) cos(π / 4 − π / 2) + 0 = 59.431
d = 7.709
Prob. 2.23
(a)
An infinite line parallel to the z-axis.
(b)
Point (2,-1,10).
(c)
A circle of radius r sin θ = 5 , i.e. the intersection of a cone and a sphere.
(d)
An infinite line
(e)
A semi-infinite line parallel to the x-y plane.
(f)
A semi-circle of radius 5 in the y-z plane.
parallel to the z-axis.
Copyright © 2015 by Oxford University Press
POESM_Ch02.indd 33
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AB
θ AB = cos −1 (
Sadiku & Kulkarni
4200
−52
) = 143.36°.
4200
(c) AB = A • a B =
Prob. 2.24
At (1, 60o , −1),
Principles of Electromagnetics, 6e
A• B
52 34
=−
= −8.789.
B
35
ρ = 1, φ = 60o , z = −1,
(a) A = (−2 − sin 60o )a ρ + (4 + 2 cos 60o )aφ − 3(1)( −1)a z
= −2.866 a ρ + 5aφ + 3a z
B = 1cos 60o a ρ + sin 60o aφ + a z = 0.5aφ + 0.866aφ + a z
A B = −1.433 + 4.33 + 3 = 5.897
AB = 2.8662 + 26 + 9 0.25 + 1 + 0.8662 = 9.1885
A B 5.897
cos θ AB =
=
= 0.6419
⎯⎯
→ θ AB = 50.07o
AB 9.1885
Let D = A × B. At (1,90o , 0),
ρ = 1, φ = 90o , z = 0
(b) A = − sin 90o a ρ + 4aφ = −a ρ + 4aφ
34
B = 1cos 90o a ρ + sin 90o aφ + a z = aφ + a z
aρ aφ
D = A × B = −1 4
0 1
aD =
az
0 = 4a ρ + aφ − a z
1
D
(4,1, −1)
=
= 0.9428a ρ + 0.2357aφ − 0.2357a z
D
16 + 1 + 1
Prob. 2.25
At T (2,3, −4)
x2 + y 2
13
= tan −1
= 137.97
z
−4
−4
13
cos θ =
= −0.7428,sin θ =
= 0.6695
29
29
y
3
φ = tan −1 = tan −1 = 56.31
x
2
2
3
cos φ =
sin φ =
13,
13
θ = tan −1
a z = cos θ ar − sin θ aθ = −0.7428ar − 0.6695aθ .
ar = sin θ cos φ a x + sin θ sin φ a y + cos θ a z .
= 0.3714a x + 0.5571a y − 0.7428a z .
Prob. 2.26
G a y = G y = Gr sin θ sin φ + Gθ cos θ sin φ + 0
= 6r 2 sin θ sin 2 φ + r 2 cos θ sin φ
At (2, −3,1), x = 2, y = −3, z = 1
r 2 = x 2 + y 2 + z 2 = 4 + 9 +Copyright
1 = 14 © 2015 by Oxford University Press
POESM_Ch02.indd 34
ρ
x2 + y2
13
9/28/2015 12:25:57 PM
13,
Sadiku & Kulkarnia z
13
= cos θ ar − sin θ aθ = −0.7428ar − 0.6695aθ .
ar = sin θ cos φ a x + sin θ sin φ a y + cos θ a z .
= 0.3714a x + 0.5571a y − 0.7428a z .
Principles of Electromagnetics, 6e
35
Prob. 2.26
G a y = G y = Gr sin θ sin φ + Gθ cos θ sin φ + 0
= 6r 2 sin θ sin 2 φ + r 2 cos θ sin φ
At (2, −3,1), x = 2, y = −3, z = 1
r 2 = x 2 + y 2 + z 2 = 4 + 9 + 1 = 14
ρ
sin θ =
r
x2 + y2
=
x2 + y2 + z 2
z
1
=
,
r
14
cos θ =
13
14
=
sin φ =
y
ρ
=
−3
13
13  9 
1  −3 
35− 3.1132 = 52.925
  + 14

 = 56.04
14  13 
14  13 
G y = 6(14)
Prob. 2.27
_
G = cos 2 φ a x +
_
2r cos θ sin φ _
a y + (1 − cos 2 φ ) a z
r sin θ
_
_
_
= cos 2 φ a x + 2 cot θ sin φ a y + sin 2 φ a z
 Gr 
 sin θ cos φ
 

 Gθ  = cos θ cos φ
 
 − sin φ
 Gφ 
sin θ sin φ
cos θ sin φ
cos φ
cos θ 
− sin θ 
0 
 cos 2 φ 


 2 cot θ sin φ 
2


 sin φ 
Gr = sin θ cos3 φ + 2 cos θ sin 2 φ + cos θ sin 2 φ
= sin θ cos3 φ + 3cos θ sin 2 φ
Gθ = cos θ cos3 φ + 2 cot θ cos θ sin 2 φ − sin θ sin 2 φ
Gφ = − sin φ cos 2 φ + 2 cot θ sin φ cos φ
_
G = [sin θ cos3 φ + 3cos θ sin 2 φ ] ar
+ [cos θ cos3 φ + 2 cot θ cos θ sin 2 φ − sin θ sin 2 φ ]aθ
+ sin φ cos φ (2 cot θ − cos φ ) aφ
Prob. 2.28
(a) J z = ( J • a z )a z .
At (2, π / 2, 3π / 2), a z = cos θ ar − sin θ aθ = − aθ .
J z = − cos 2θ sin φ aθ = − cos π sin(3π / 2) aθ = − aθ .
(b) Jφ = tan
θ
2
ln r aφ = tan
π
4
ln 2 aφ = ln 2aφ = 0.6931a φ .
(c) J t = J − J n = J − J r = − aθ + ln 2 aφ = − aθ + 0.6931aφ .
(d )
_
_
POESM_Ch02.indd 35
_
_
_
J P = (J • ax ) ax
_
_ by Oxford University
_
_
Copyright © 2015
Press
a x = sin θ cos φ a r + cos θ cos φ aθ − sin φ aφ = aφ .
At (2, π / 2, 3π / 2),
9/28/2015 12:25:57 PM
Gφ = − sin φ cos 2 φ + 2 cot θ sin φ cos φ
_
Sadiku & Kulkarni G
= [sin θ cos3 φ + 3cos θ sin 2 φ ] ar
Principles of Electromagnetics, 6e
+ [cos θ cos3 φ + 2 cot θ cos θ sin 2 φ − sin θ sin 2 φ ]aθ
36
+ sin φ cos φ (2 cot θ − cos φ ) aφ
Prob. 2.28
(a) J z = ( J • a z )a z .
At (2, π / 2, 3π / 2), a z = cos θ ar − sin θ aθ = − aθ .
J z = − cos 2θ sin φ aθ = − cos π sin(3π / 2) aθ = − aθ .
(b) Jφ = tan
θ
2
ln r aφ = tan
π
4
ln 2 aφ = ln 2aφ = 0.6931a φ .
(c) J t = J − J n = J − J r = − aθ + ln 2 aφ = − aθ + 0.6931aφ .
(d )
_
_
_
_
J P = (J • ax ) ax
_
_
_
_
_
a x = sin θ cos φ a r + cos θ cos φ aθ − sin φ aφ = aφ .
At (2, π / 2, 3π / 2),
_
_
J P = ln 2 aφ .
Prob. 2.29
H a x = H x
 H x   cos φ
 H  =  sin φ
 y 
 H z   0
− sin φ
cos φ
0
0   ρ 2 cos φ 


0   − ρ sin φ 
36
1   0 
H x = ρ 2 cos 2 φ + ρ sin 2 φ
At P, ρ = 2, φ = 60o , z = −1
H x = 4(1/ 4) + 2(3 / 4) = 1 + 1.5 = 2.5
Prob. 2.30
(a) 5 = r ⋅ a x + r ⋅ a y = x + y
(b)
10 = rxa z =
a plane
x y z
=| ya x − xa y |= x 2 + y 2 = ρ
0 0 1
a cylinder of infinite length
Copyright © 2015 by Oxford University Press
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Sadiku & Kulkarni
Principles of Electromagnetics, 6e
37
CHAPTER 3
P. E. 3.1
(a) DH =
φ = 60°
φ
π
r sin θ dφ
= 45°
θ = 90 °
(b) FG =
rdθ
r=5
θ = 60 °
(c)
π
π
= 3(1)[ − ] =
= 0.7854.
r = 3,90
3 4
4
o
5π
π π
= 5( − ) =
= 2.618.
2 3
6
θ = 90 ° φ = 60 °
AEHD =
r
2
sin θ dθ dφ
θ = 60 ° φ = 45 °
r=3
φ = 60 °
90 °
= 9 ( − cos θ )|θθ == 60
° φ |φ = 45 °
1 π
3π
= 9 ( )( ) =
= 1178
. .
2 12
8
(d)
ABCD =
r = 5 θ = 90
rdθ dr =
r = 3 θ = 60
r2 r = 5 π π
4π
( − )=
= 4.189.
2 r =3 2 3
3
(e)
Volume =
r =5
r =3
=
φ = 60°
φ
= 45°
θ =90
θ
r 2 sin θ dr dθ dφ =
= 60
r3
3
r =5
r =3
( − cos θ )
θ = 90°
θ = 60°
φ
φ = 60°
φ = 45°
1
1 π
(98)( )
3
2 12
=
49π
= 4.276 .
36
P.E. 3.2
y
3
2
60o
1
x
Copyright © 2015 by Oxford University Press
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Sadiku & Kulkarni
Principles of Electromagnetics, 6e
38
38
A• dl = ( + L
1
2
+ ) A• dl = C1 + C2 + C3
3
2
Along (1), C1 = A• dl = ρ cos φ d ρ |φ =0 =
0
ρ2
20
= 2.
Along (2), dl = ρ dφ aφ , A• dl = 0, C2 = 0
0
Along (3), C3 = ρ cos φ d ρφ =60° = −
ρ2
2
2
0
2
A• dl = C
1
1
( ) = −1
2
+ C2 + C3 = 2 + 0 − 1 = 1
l
P.E. 3.3
(a)
∇U =
∂U
∂U
∂U
ax +
ay+
az
∂x
∂ y
∂z
= y (2 x + z ) a x + x( x + z ) a y + xy a z
(b)
∇V =
∂V
∂V
1 ∂V
aρ +
aφ +
az
∂ρ
ρ∂φ
∂z
= ( z sin φ + 2 ρ ) a ρ + ( z cos φ −
z2
ρ
sin 2φ ) aφ + ( ρ s inφ + 2 z cos 2 φ ) a z
(c)
∇f =
∂f
1∂ f
1 ∂f
ar +
aθ +
aφ
∂r
r ∂θ
r sin θ ∂φ
cos θ sin φ
sin θ sin φ ln r
(cos θ cos φ ln r + r 2 )
+ 2rφ )ar −
aθ +
aφ
r
r
r sin θ
sin θ sin φ ln r
cos θ sin φ
cot θ cos φ ln r
=
+ 2rφ a r −
aθ +
+ r cos ecθ aφ
r
r
r
=(
P.E. 3.4
∇Φ = ( y + z ) a x + ( x + z ) a y + ( x + y ) a z
At (1, 2,3), ∇Φ = (5, 4,3)
(2, 2,1) 21
=
= 7,
3
3
where (2, 2,1) = (3, 4, 4) − (1, 2,3)
∇Φ • a1 = (5, 4,3) •
Copyright © 2015 by Oxford University Press
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Sadiku & Kulkarni
Principles of Electromagnetics, 6e
39
P.E. 3.5
Let f = x 2 y + z − 3,
g = x log z − y 2 + 4,
∇ f = 2 xy a x + x 2 a y + a z ,
∇ g = log z a x − 2 y a y +
x
az
z
At P ( − 1 , 2,1),
∇f
(−4 a x + a y + a z )
∇g
(−4 a y − a z )
=±
,
=±
ng = ±
| ∇f |
| ∇g |
18
17
( − 5)
cos θ = n f . n g = ±
18 × 17
T ake positive value to get acute angle.
5
= 73.39 °
θ = cos − 1
17.49 3
nf = ±
P.E. 3.6
(a) ∇ • A =
At
∂ Ax ∂ Ay ∂ Az
+
+
= 0 + 4 x + 0 = 4 x.
∂x
∂y
∂z
(1, −2,3), ∇ • A = 4.
(b)
∇•B =
=
1 ∂
ρ ∂ρ
1
ρ
( ρ Bρ ) +
2 ρ z sin φ −
1
ρ
1 ∂ Bφ
ρ ∂φ
+
∂ Bz
∂ρ
3ρ z 2 sin φ = 2 z sin φ − 3 z 2 sin φ
= (2 − 3 z ) z sin φ .
At (5,
(c)
π
2
,1) ,
∇ • B = (2 − 3)(1) = −1.
1 ∂ 2
1 ∂
1 ∂ Cφ
r
C
C
θ
(
)
(
sin
)
+
+
r
θ
r2 ∂ r
r sin θ ∂θ
r sin θ ∂φ
1
= 2 6r 2 cos θ cos φ
r
= 6 cos θ cos φ
∇•C =
At (1,
π π
, ),
6 3
∇ • C = 6 cos
π
6
cos
π
3
= 2.598.
Copyright © 2015 by Oxford University Press
POESM_Ch03.indd 39
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Sadiku & Kulkarni
Principles of Electromagnetics, 6e
40
40
P.E. 3.7 This is similar to Example 3.7.
Ψ = D • dS = Ψ t + Ψ b + Ψ c
S
Ψ t = 0 = Ψ b since D has no z-component
Ψ c = ρ cos φρ dφ dz = ρ
2
2
3
φ = 2π
cos φ dφ
2
φ =0
z =1
dz ρ = 4
z =0
= (4)3 π (1) = 64π
Ψ = 0 + 0 + 64π = 64π
By the divergence theorem,
D • dS = ∇ • Ddv
S
V
∇• D =
1 ∂
ρ ∂ρ
= 3ρ cos 2 φ +
z
ρ
Ψ = ∇ • Ddv =
V
(3ρ cos 2 φ +
V
2π
1
0
0
0
= 3(
ρ ∂φ
z sin φ +
∂ Az
dz
cos φ .
4
= 3 ρ 2 d ρ
1 ∂
( ρ 3 cos 2 φ ) +
2
cos φ dφ dz +
z
ρ
cos φ ) ρ dφ dzd ρ
4
2π
0
0
d ρ cos φ dφ
1
zdz
0
3
4
) π (1) = 64π .
3
P.E. 3.8
(a)
∇ × A = a x (1 − 0) + a y ( y − 0) + a z (4 y − z )
= a x + y a y + (4 y − z ) a z
_
At (1, −2,3) , ∇ × A = a x − 2 a y −11 a z
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Sadiku & Kulkarni
Principles of Electromagnetics, 6e
41
(b)
∇ × B = a ρ (0 − 6 ρ z cos φ ) + aφ ( ρ sin φ − 0) + a z
1
ρ
(6 ρ z 2 cos φ − ρ z cos φ )
= −6 ρ z cos φ a ρ + ρ sin φ aφ + (6 z − 1) z cos φ a z
At (5,
π
2
, − 1) , ∇ × B = 5 aφ
(c)
_
1
2r cos θ sin φ 3 1/ 2 a φ
aθ
(r −1/ 2 cos θ − 0) +
(−
− r ) + (0 + 2r sin θ cos φ )
r sin θ
r
sin θ
2
r
3
= r −1/ 2 cot θ a r − (2 cot θ sin φ + r −1/ 2 ) aθ + 2sin θ cos φ aφ
2
∇ × C = ar
π π
, ), ∇ × C = 1.732 a r − 4.5 aθ + 0.5 aφ
6 3
At (1,
P.E. 3.9
A • dl = (∇ × A) • dS
L
S
But (∇ × A) = sin φ a z +
z cos φ
ρ
a ρ and
d S = ρ dφ d ρ a z
(∇ × A) • dS = ρ sin φ dφ d ρ
S
=
ρ2
2
|
2
0
60°
(− cos φ ) |
0
1
= 2(− + 1) = 1.
2
Copyright © 2015 by Oxford University Press
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Sadiku & Kulkarni
Principles of Electromagnetics, 6e
42
42
P.E. 3.10
ax
∂
∇ × ∇V =
∂x
∂V
∂x
=(
P.E. 3.11
(a)
∇ 2U =
ay
∂
∂y
∂V
∂y
az
∂
=
∂z
∂V
∂z
∂ 2V
∂ 2V
∂ 2V
∂ 2V
∂ 2V
∂ 2V
) ax + (
)ay +(
)az =0
−
−
−
∂ y∂ z ∂ y∂ z
∂ x∂ z ∂ z∂ x
∂ x∂ y ∂ y∂ x
∂
∂ 2
∂
(2 xy + yz ) +
( x + xz ) +
( xy )
∂x
∂y
∂z
= 2 y.
(b)
∇ 2V =
=
1 ∂
ρ ∂ρ
1
ρ
ρ ( z sin φ + 2 ρ ) +
( z sin φ + 4 ρ ) −
= 4 + 2 cos 2 φ −
(c)
∇2 f =
2z2
ρ2
ρ2
ρ
2
(− ρ z sin φ − 2 z 2
∂
∂
sin φ cos φ ) +
( ρ sin φ + 2 z cos 2 φ )
∂ρ
∂z
( z ρ sin φ + 2 z 2 cos 2φ ) + 2 cos 2 φ .
cos 2φ .
1 ∂ 21
1
∂
[r cos θ sin φ + 2r 3φ ] + 2
[− sin 2 θ sin φ ln r ]
2
r ∂r
r
r sin θ ∂ θ
+
=
1
1
1
[− cos θ sin φ ln r ]
r sin 2 θ
2
1
cos θ sin φ (1 − 2 ln r − csc2 θ ln r ) + 6φ
2
r
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Sadiku & Kulkarni
Principles of Electromagnetics, 6e
43
P.E. 3.12
If B is conservative , ∇ × B = 0 must be satisfied.
∇× B =
ax
ay
az
∂
∂x
y + z cos xz
∂
∂y
x
∂
∂z
x cos xz
47
Prob. 3.11
= 0 a x + (cos xz − xz sin xz1− cos
xz + xz sin2 xz ) a y + (1 − 1) a z = 0
2
z 2
zdxdz
dx
zdz
(1)
ψ
= ABdSis=aconservative
=
=
=2
Hence
field.
0
2
S
0
0
P.E. 3.13
ψ = B dS, dS = dxdya z
Prob.S 3.1
(a) 2 1 2
x3 1 y 2 2
ψ = d S3x= ρydxdy
dφ dz= 3 3 0 2 0 = (1)(2) = 2
y =0 x =0
5
Prob. 3.12
S = d S = ρ dφ dz = 2 dz
0
(a)
= dxdydz
P.E.dv
3.14
π
2
π
π dφ
π
= 2(5)[ − ] =
2 3
10π
= 5.236
6
3
(b)
2 1 1
1
1
z
����
dSxydxdydz
d=φ� xdx
cylindrical,
= ρ d�ρ ��
Inxydv
Cartesian
coordinates
��� �
=
ydy
dz� � � � 3 � 1�.
v
z =0 y =0 z =0
3
2
2
π
0
0
0
3
π ρin2 cylindrical
For representing
this vector
coordinates, we use eq. (2.13):
2 = field
( ) = 3.142
S = d S x= 1 ρ yd ρ1 dφ
4
z �= (1/
0.5� 0 3� � � ��
1� =s��
4 2)(1/
2cos2)(2)
2 10 2 0 0 0 �
�� θs��
�
�
(c) In spherical,
d �S��=� r�2 sin
dφ�dθcos � 0� �
0
0
1
��
0
2π
2π
(b)
�
�
π
2
3
s�� �� � �cos ������
� � �cos � �3� � ��� � s�� � ������ � ��3� � �����
ρ dφ dz sin θ dθ dφ = 100 (2π )(− cos θ ) 3| = 200
π (0.5 + 0.7071) = 758.4
S = dvd=Sρ=d 100
=
π
π
π
π
Using
the following
relationships,
from
eq.zdz
(2.8),dφ�� � � cos ��,
ρ
zdv
=
ρ
z
ρ
d
ρ
d
φ
dz
=
ρ
d
ρ
ρ
obtain φ
2
(d)
3
4
v
= 0 z = 0 =1
0
3
2
1
2
4
0
0
� � s�� ��, we
d S = r3dr3 dθ 2 2
ρ
z
1
=
π (π ) = (9 − )(2π ) = 54.45
�
� �cos ��3��
��� � �s�� ���� cos ��� ��
4 2 ��
0 2 � ��� s��
3 1cos
r 2 4 3π π
8π
S = dS = rdr dθ = |0�( − ) =
= 4.189
�0 ���3��
cos2�� �
s�� ��� � �cos ���� cos ��� ��
2 ���
3 s�� �����
6
π
Prob. 3.13 � �
3
������� � �3� cos � � �� s�� � cos � � � s�� � cos ����
Let
I = Adv = r�sin φ�ar dv
�
�
v
� ��3�
cos � s�� � � ��s�� � � �cos ����
v
ar = sin θ cos φ a x + sin θ sin φ a y + cos θ a z
� in cylindrical coordinates is �5, 53.13°, 5�
Ar = r sin θ sin φ cos φ a x + r sin θ sin 2 φ a y + r cos θ sin φ a z
dv = r 2 sin θ1dθ∂dφ dr
��� �
POESM_Ch03.indd 43
1 ∂�� ∂��
�
� ∂�
� ∂�
∂z
1
������������ ���� cos� � � �� s�� � cos ��
�
1
� ���� cos �s��� � � �3�� cos� ��
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� �� s�� � cos � � �� s�� � cos ��
���� � �
9/27/2015 9:06:17 PM
Sadiku & Kulkarni
� �cos ��3�� cos ��� � ��� s�� ��� � �s�� ���� cos ��� ��
Principles of Electromagnetics, 6e
� ���3�� cos ��� � ��� s�� ����� s�� ��� � �cos ���� cos ��� ��
������� � �3�� cos� � � �� s�� � cos � � � s�� � cos ����
44
� ��3�� cos� � s�� � � ��s��� � � �cos� ����
� in cylindrical coordinates is �5, 53.13°, 5�
1 ∂�� ∂��
1 ∂
���� � �
�
� ∂�
� ∂�
∂z
1
������������ ���� cos� � � �� s�� � cos ��
�
1
� ���� cos �s��� � � �3�� cos� ��
�
� �� s�� � cos � � �� s�� � cos ��
� 1�.00 (after substituting the values of � and �)
��� �
For representing the given vector field in spherical coordinates we use eq. (2.27):
��
s�� � cos � s�� � s�� �
��� � � �cos � cos � cos � s�� �
��
� s�� �
cos �
cos � 3� � � ��
�
� s�� �� �
�
0
0
� � ��s�� � cos ���3� � � ��� � �s�� � s�� ������
� ��cos � cos ���3� � � ��� � �cos � s�� ������
� ���� s�� ���3� � � ���� � ��cos ����� ��
Using the following relationships, from eq. (2.22), � � � s�� � cos ��, � �
� s�� � s�� ��, we obtain
� � ��3� � sin� �cos� �� � �3�sin� � cos � sin �����
� ��3� � cos �sin� �cos� �� � ��� sin � cos � cos � sin ��
�
�
�
�
�3�sin
� � ��3� sin �
�cos
����cos
sin �����
�� sin
� cos��cos
sin �
����
�
�
�
� ��3� � cos
�cos
�� � ��� sin � cos � cos
� sin ��
� �sin
�
�
� ���3� sin �cos � sin �� � ���� sin �sin� ��
� �� sin � cos � cos � sin �����
� �� sin �� cos�� ������
� ���3� sin �cos � sin �� � ���� sin �sin� ��
� �� sin � cos� �����
� in spherical system is �7.071, 45° , 53.13° �
1 ���
1 ∂ system is �7.071,
�
1
� in spherical
45° , 53.13° �
�� sin � �
� � � � � �� � �� � �
� sin � ��
� ��
� sin � ��
�
�
�
�
�
�
�
�
�1�� 1sin∂ �cos � � �� 1sin � cos � sin ��+ 1 ��
��3�
sin� �cos� � �
�
�� sin � � � ��� �
�� �� � � � �� � �� � �
�sin��
��� � cos� � �� 3�sin
sin � ��
�
�
��
cos��� sin
� cos � sin � ��sin
��
�cos� � cos � sin �� �
�
�
�
�
�
�
�
��3� � sin� �cos� � �
��1�� sin
�cos � � �� sin � cos � sin ��+
��
� ���
��� � sin� �sin� � cos � � 3� � sin� �cos� � �
4��sin � cos � sin � �
� ��� �
�� � cos� � sin� � cos�� � � 3�sin� � cos � sin � � �� sin �cos� � cos � sin �� �
���sin � cos � sin � � 1� ������ s��s�i���in� ��� �����s o� �, � �n� ��
��� � sin� �sin� � cos � � 3� � sin� �cos� � � 4� sin � cos � sin � �
� ��� �
Note:
this
is clear
that the divergence
of ao�vector
field
��
sin From
� cos �
sinexample,
�� � 1�it������
s��s�i���in�
��� �����s
�, � �n�
��is the
same irrespective of the coordinate system used.
Note: From this example, it is clear that the divergence of a vector field is the
same irrespective of the coordinate system used.
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POESM_Ch03.indd 44
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1 ∂ �
1 ���
1
�
�� �� � �
�� sin � �
�
� ��
� sin � ��
� sin � ��
Principles of Electromagnetics, 6e
�
�
�
�
�
�
43��+ � ��3� � sin� �cos� � �
�1�� sin �cos � � �� sin � cos � sin
�
��� �
Sadiku & Kulkarni
�
� ��� �
�� � cos� � sin� � cos� � � 3�sin� � cos � sin � � �� sin �cos� � cos � sin �� �
�
45
P.E. 3.12
��� � sin� �sin� � cos � � 3� � sin� �cos� � � 4� sin � cos � sin � �
� ��� �
If B is conservative
� , ∇ × B = 0 must be satisfied.
�� sin � cos � sin � � 1� ������ s��s�i���in� ��� �����s o� �, � �n� ��
ax
ay
az
y + z cos xz
x
x cos xz
Note: From this example,
it∂is clear∂that the divergence of a vector field is the
∂
∇ ×irrespective
B=
same
of
the
coordinate
∂x
∂y
∂system
z
50used.
P.E. 3.15
∇ F = a x − 2a y + a z
= 0 a x + (cos xza −−xz2asin+xza − cos xz + xz sin xz ) a y + (1 − 1) a z = 0
∇F
x
y
z
0.4082a x − 0.8165a y + 0.4082a z
aHence
n =
B is=a conservative=field.
| ∇F |
1+ 4 +1
Prob. 3.18
Method 1:
Prob. ∂3.1
T
1 ∂T
1 ∂T
∇T =
ar +
aθ +
aφ
(a)
∂dr S = ρrd∂φθdz
r sin θ ∂φ
= sin θ cos φ ar + cos θ cos φ aθ − sinπφ aφ
5
2
o
π π
10π
At P, Sr == 2,dθ S= =60ρo , φ d=φ30
dz
=
dz
2
o o
0o π dφo = 2(5)[ o2 − 3 ] = 6 = 5.236
∇T = sin 60 cos 30 ar + cos 60 cos 330 aθ − sin 30 aφ
(b)
= 0.75ar - 0.433aθ − 0.5aφ
In cylindrical, dS = ρ d ρ dφ
2
| ∇T |= 0.752 + 0.433
+ 0.52 = 1
π
3
2
3
π itsρ direction
The magnitude of T4is 1 and
is along ∇T.
S = d S = ρ d ρ dφ = ( ) = 3.142
4 2 1
1
0
Method
2:
(c) In spherical,
d S = r 2 sin θ dφ dθ
T = r sin θ cos φ = x
∇T = a x
S = d S = 100
| ∇T |= 1
2π
3
2π
π
0
2π
3
sin θ dθ dφ = 100 (2π )(− cos θ ) π|
4
= 200π (0.5 + 0.7071) = 758.4
4
(d) 3.19
Prob.
∂fd S = ∂rfdr dθ ∂f
∇f = a x + a y + πa z = (2 xy − 2 y 2 )a x + ( x 2 − 4 xy )a y + 3 z 2 a z
∂x
∂y 4 ∂z 2
r2 4 π π
8π
= dS = x =rdr
= z =|0−(3 − ) =
= 4.189
At pointS(2,4,-3),
2,yd=θ 4,
2 2 3
6
π
0
∇f = (16 − 32)a x + (4 − 32)
3 a y + 27a z = −16a x − 28a y + 27 a z
a x + 2a y − a z
1
(1, 2, −1)
1+ 4 +1
6
The directional derivative is
1
99
∇f a = (−16, −28, 27)
(1, 2, −1) = −
= −40.42
6
6
a=
=
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Sadiku & Kulkarni
Principles of Electromagnetics, 6e
46
44
Prob. 3.2
(a)
dl = ρ dφ ;
ρ=3
π
2
3π
π π
L = dl = 3 dφ = 3( − ) =
= 2.356
2 4
4
π
4
(b)
dl = r sin θ dφ;
r = 1, θ = 30° ;
π
3
L = dl = r sin θ
(c)
π
dφ = (1) sin 30° [( ) − 0] =
3
0
0.5236.
dl = rdθ
π
L=
π π
4π
d l = r dθ = 4( − ) =
= 4.189
2 6
3
π
2
6
Prob. 3.3
2
S = dS =
π /2
φ π
L = dl =
L
π /6
ρ = 10
z =0 = / 4
2
π /2
0
π /4
= 10 dz Prob. 3.4
ρ dφ dz
ρ dφ
φ =0
dφ = 10(2)(π / 2 − π / 4) = 5π = 15.71
= 4(π / 6) = 2.094
ρ =4
Prob. 3.5
(a ) dV = dxdydz
1
2
3
0
1
−3
V = dxdydz = dx dy dz = (1) (2 − 1)(3 − − 3) = 6
(b) dV = ρ dφ d ρ dz
5
4
π
2
−1
π
V = ρ d ρ dz dφ =
ρ2
|
2
5
2
π
1
2π
(4 − −1)(π − ) = (25 − 4)(5)( ) = 35π = 110
3
2
3
3
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Sadiku & Kulkarni
Principles of Electromagnetics, 6e
47
45
(c) dV = r 2 sin θ drdθ dφ
π
3
π
2
π
2
π
6
2
3
π /3 π
r3 3
π
(
−
cos
θ
)
(
−
)
|
|
π /2 2
3 1
6
V = r 2 dr sin θ dφ =
1
=
1
1 π
26 π
(27 − 1)( )( ) =
= 4.538
3
2 3
18
Prob. 3.6
H •dl = ( x dx + y
2
2
dy )
L
But on L, y = x 2 dy = 2 xdx
1
2
4
H •dl = ( x + x .2 x)dx =
L
Prob. 3.7
(a)
0
1
F • dl = ( x
y =0
2
− z )dy|
x3
x6 1 1 1
+ 2 | = + = 0.6667
3
6 0 3 3
2
x = 0, z = 0
+
x=2
2 xydx |
x =0
y =1, z = 0
+
z =3
(−3xz
z =0
2
)dz|
x = 2, y =1
x2 2
z3 3
−
3(2)
|
|
2 0
3 0
= 0 + 4 − 54 = − 50
= 0 + 2(1)
(b)
Let x = 2t. y = t , z = 3t
dx = 2dt , dy = dt , dz = 3dt ;
1
F • dl = (8t
2
− 5t 2 − 162 t 3 ) dt
0
1
= (t 3 − 40.5t 4 ) = −39.5
0
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Sadiku & Kulkarni
Principles of Electromagnetics, 6e
48
46
Prob. 3.8
W = F • dl =
L
π /4
z ρ dφ
+
z = 0, ρ = 2
φ =0
3
ρ cos φ dz ρ
= 2,φ =π / 4
z =0
= 0 + 2 cos(π / 4)(3) = 6 cos 45o = 4.243 J
Prob. 3.9
H • dl =
0
( x − y )dx
x =1
y = 0, z = 0
+ ( x 2 + zy )dy + 5yzdz
0
2
1
0
= xdx + ( y −
+
1
5 yzdz x = 0, y = 0
z =0
x = 0, z = 1 − y / 2
0
y2
)dy + (10 z − 10 z 2 )dz
2
1
= −1.5
Prob. 3.10
Method 1:
B ⋅ dl = −
L
1
y =0
But z = y
B ⋅ dl
yzdy
1
xzdz
z =0
x =1
+ (− yzdy + xzdz )
x =1
⎯⎯
→ dz = dy on the last segment (or integral).
= 0+
L
=
z=0
+
0
z2 1
1
y3 y 2 0
+ (− y 2 + y )dy = + (− + )
2 0 y =1
2
3
2 1
1 1 1 1
+ − = = 0.333
2 3 2 3
Method 2:
B ⋅ dl = ∇ × B dS
L
S
∂
∇ × B = ∂x
xy
∇ × B dS =
S
∂
∂y
-yz
1
∂
∂z = ya x − za y − xa z ,
xz
y
y =0 z =0
1
ydzdy = y 2 dy =
0
dS = dydza x
y3 1 1
= = 0.333
3 0 3
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POESM_Ch03.indd 48
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Sadiku & KulkarniProb.
3.11
Principles of Electromagnetics, 6e
1
2
0
0
ψ = AdS = zdxdz = dx zdz = (1)
S
z 2
=2
2 0 47
49
2
P.E. 3.13
Prob. 3.11
ψ = B dS, dS = dxdya1z 2
z2 2
ψ = S AdS = zdxdz = dx zdz = (1)
=2
2 1
3
2
0
2
1
2
S
0
0
x
y
ψ = 3x 2 ydxdy = 3
= (1)(2) = 2
0
0
3
2
y =0 x =0
P.E. 3.13
ψ
= B
dS, dS = dxdya z
Prob.
3.12
S
(a) dv
= dxdydz
2 1
ψ=
xydv = y =0 x =0
v
x3 1 y 2 2
= (1)(2)
=2
z
3 0 12 0 1
xydxdydz = xdx ydy dz
2
=3
23x1 ydxdy
1
z =0 y =0 z =0
0
0
0
1
1
z
Prob. 3.122
1 y2 1 2
(a) dv == xdxdydz
z = (1/ 2)(1/ 2)(2) = 0.5
2 0 2 0 0
2
1
1
v xydv = z=0 y=0 z=0 xydxdydz = 0 xdx 0 ydy 0 dz
(b)
dv = ρ d ρ dφ dz
x 2 1π y22 13 2
3
2
π
=
z = (1/ 2)(1/ 2)(2) 2= 0.5
ρ
zdv
=
ρ
z
ρ
d
ρ
d
φ
dz
=
ρ
d
ρ
zdz
2 0 2 0 0
dφ
φ = 0 z = 0 ρ =1
v
(b)
1
0
0
ρ 3 3 z2 2
1
=
(π ) = (9 − )(2π ) = 54.45
dv = ρ d ρ d3φ dz
1 2 0
3
π
2
3
3
2
π
zdv = ρ z ρ d ρ dφdz = ρ d ρ zdz dφ
v ρ3.13
Prob.
φ = 0 z = 0 ρ =1
1
0
0
Let
I = A3dv = 2r sin φ ar dv
ρ 3 zv 2
1
=v
(π ) = (9 − )(2π ) = 54.45
2 θ0 sin φ a y + cos
3 θ az
ar = sin θ cos φ3a x1+ sin
2
Ar = r sin θ sin φ cos φ a x + r sin θ sin 2 φ a y + r cos θ sin φ a z
Prob. 3.13
dv = r 2 sin
dφ =
dr r sin φ a dv
θdv
Let
I =θdA
r
v
v
ar = sin θ cos φ a x + sin θ sin φ a y + cos θ a z
Ar = r sin θ sin φ cos φ a x + r sin θ sin 2 φ a y + r cos θ sin φ a z
dv = r 2 sin θ dθ dφ dr
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Sadiku & Kulkarni
Principles of Electromagnetics, 6e
50
48
I = Adv = r 3 sin 2 θ sin φ cos φ dθ dφ dra x
v
+ r 3 sin 2 θ sin 2 φ dθ dφ dra y
+ r 3 sin θ cos θ sin φ dθ dφ dra z
= ax
1
π
r dr sin
3
r =1
0
+a y
1
r =1
+a z
1
2
θ dθ sin φ cos φ dφ
2π
0
0
π
4
2π
r dr sin θ cos θ dθ sin φ dφ
3
0
= 0a x + 0a z + a y
ay
0
π
r 3 dr sin 2 θ dθ sin 2 φ dφ
r =1
=
2π
0
2π
1
r 1 1
(1
cos
2
)
(1 − cos 2φ )dφ
d
−
θ
θ
4 00 2
2
0
(π / 2)(π ) =
π
4
π2
8
a y = 1.234a y
Prob. 3.14
∂V1
∂V
∂V
a x + 1 a y + 1 az
∂x
∂y
∂z
= (6 y − 2 z )a x + 6 xa y + (1 − 2 x )a z
(a) ∇V1 =
∂V2
∂V
1 ∂V2
aρ +
aφ + 2 a z
∂ρ
∂z
ρ ∂φ
= (10 cos φ − z )a ρ − 10sin φ aφ − ρ a z
(b) ∇V2 =
∇V3 =
(c)
∂V3
1 ∂V3
1 ∂V3
ar +
aθ +
aφ
∂r
r ∂θ
r sin θ ∂φ
2
1 2
cos φ ar + 0 +
− sin φ aφ
2
r
r sin θ r
2
2sin φ
aφ
= − 2 cos φ ar − 2
r
r sin θ
=−
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Sadiku & Kulkarni
Principles of Electromagnetics, 6e
51
49
Prob. 3.15
(a)
∇U =
∂U
∂U
∂U
ax +
ay +
az
∂x
∂y
∂z
= e x + 2 y cosh za x + 2e x + 2 y cosh za y + e x + 2 y sinh za z
∂T
1 ∂T
∂T
aρ +
aφ +
az
ρ ∂φ
∂ρ
∂z
3z
3z
3
= − 2 cos φ a ρ − 2 sin φ aφ + cos φ a z
∇T =
(b)
ρ
∇W =
(c)
ρ
ρ
∂W
1 ∂W
1 ∂W
ar +
aθ +
aφ
∂r
r ∂θ
r sin θ ∂φ
5sin φ 1
5cos θ
= −
+ 4r sin φ ar + 2r 2 cos φ −
aφ
2
r
r r sin θ
Prob. 3.16
r = x2 + y 2 + z 2 ,
r n = ( x 2 + y 2 + z 2 ) n /2
Method 1:
∇r n =
∂r n
∂r n
∂r n
n
ax +
ay +
a z = ( x 2 + y 2 + z 2 ) n /2−1 (2 x)a x + ∂x
∂y
∂z
2
= n( x 2 + y 2 + z 2 )
n−2
2
( xa x + ya y + za z ) = nr n − 2 r
Method 2:
∂r n
r
n
∇r =
ar = nr n −1 = nr n − 2 r
∂r
r
Prob. 3.17
∇ T = 2x a x + 2 y a y − a z
At
(1,1, 2 ) , ∇ T = (2, 2, −1).
The mosquito should move in the direction of
2ax + 2a y − az
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Sadiku & KulkarniP.E.
3.15
∇ F = a x − 2a y + a z
an =
Principles of Electromagnetics, 6e
a − 2a y + a z
∇F
52 a + 0.4082a
= x
= 0.4082a x − 0.8165
y
z
| ∇F |
1+ 4 +1
Prob. 3.18
Method 1:
∂T
1 ∂T
1 ∂T
∇T =
ar +
aθ +
aφ
∂r
r ∂θ
r sin θ ∂φ
= sin θ cos φ ar + cos θ cos φ aθ − sin φ aφ
At P, r = 2,θ = 60o , φ = 30o
∇T = sin 60o cos 30o ar + cos 60o cos 30o aθ − sin 30o aφ
= 0.75ar - 0.433aθ − 0.5aφ
| ∇T |= 0.752 + 0.4332 + 0.52 = 1
The magnitude of T is 1 and its direction is along ∇T.
Method 2:
T = r sin θ cos φ = x
∇T = a x
| ∇T |= 1
Prob. 3.19
∂f
∂f
∂f
∇f = a x + a y + a z = (2 xy − 2 y 2 )a x + ( x 2 − 4 xy )a y + 3 z 2 a z
∂x
∂y
∂z
At point (2,4,-3), x = 2, y = 4, z = −3
∇f = (16 − 32)a x + (4 − 32)a y + 27a z = −16a x − 28a y + 27a z
a x + 2a y − a z
1
(1, 2, −1)
1+ 4 +1
6
The directional derivative is
1
99
(1, 2, −1) = −
= −40.42
∇f a = (−16, −28, 27)
6
6
a=
=
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Principles of Electromagnetics, 6e
53
51
Prob. 3.20
(a) Let f = ax + by + cz – d = 0
∇f = aa x + ba y + ca z
aa + ba y + ca z
∇f
= x
| ∇f |
a 2 + b2 + c2
Let g = α x + β y + γ z − δ
∇g α a x + β a y + γ a z
an 2 =
=
| ∇g |
α2 + β2 +γ 2
an1 =
cos θ = an1 an 2 =
θ = cos −1
aα + bβ + cγ
a 2 + b2 + c 2 α 2 + β 2 + γ 2
aα + bβ + cγ
(a + b 2 + c 2 )(α 2 + β 2 + γ 2 )
2
a = 1, b = 2, c = 3
(b) α = 1, β = 1, γ = 0
θ = cos −1
1+ 2 + 0
(12 + 22 + 32 )(12 + 12 + 02 )
= cos −1
3
= cos −1 0.5669 = 55.46o
28
Prob. 3.21
∂Ax ∂Ay ∂A z
+
+
= 3y − x
∂x
∂y
∂z
1
1
∇ ⋅ B = 2 ρ z 2 + ρ 2sin φ cos φ + 2 ρ sin 2 φ
(a) ∇ ⋅ A =
(b)
ρ
ρ
= 2 z + sin 2φ + 2 ρ sin 2 φ
2
(c) ∇ ⋅ C =
1 2
3r + 0 = 3
r2
Prob. 3.22
∂Ax ∂Ay ∂Az
+
+
= 2 xy + 0 + 2 y = 2 y (1 + x)
∂x
∂y
∂z
(a) At (−3, 4, 2), x = −3, y = 4
∇ A = 2(4)(1 − 3) = −16
∇ A =
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Principles of Electromagnetics, 6e
54
52
∇ B =
1 ∂
1 ∂Bφ ∂Bz 1 ∂
ρ Bρ +
3ρ 2 sin φ + 0 + 8 z cos 2 φ
+
=
ρ ∂ρ
ρ ∂φ
ρ ∂ρ
∂z
(
)
(
)
= 6sin φ + 8 z cos 2 φ
(b)
At
(5,30o ,1), z = 1, φ = 30o
∇ B = 6sin 30o + 8(1) cos 2 30o = 3 + 6 = 9
1 ∂ 2
1 ∂Cφ
1 ∂ 4
( r Cr ) + 0 +
(r cos θ ) + 0 = 4r cos θ
= 2
2
r ∂r
r sin θ ∂φ
r ∂r
(c) At (2, π / 3, π / 2), r = 2, θ = π / 3
∇C = 4(2)cos(π / 3) = 4
∇C =
Prob. 3.23
2
∇• H = k ∇• ∇ T = k ∇ T
2
∇ T=
∂ 2T ∂ 2T
πx
π y π2 π2
h
50sin
cos
(−
+
=
+ )= 0
2
2
4
4
∂ x2 ∂ y
2
Hence, ∇• H = 0
Prob. 3.24
We convert A to cylindrical coordinates; only the ρ-component is needed.
Aρ = Ax cos φ + Ay sin φ = 2 x cos φ − z 2 sin φ
But x = ρ cos φ ,
Aρ = 2 ρ cos 2 φ − z 2 sin φ
Ψ = A ⋅ dS = Aρ ρ dφ dz = 2 ρ 2 cos 2 φ − ρ z 2 sin φ dφ dz
S
= 2(2) 2
π /2
0
1
1
1
(1 + cos 2φ )dφ dz − 2 z 2 dz
2
0
0
π /2
sin φ dφ
0
π /2
π /2
1
z3 1
(− cos φ )
= 4(φ + sin 2φ )
−2
= 2π − 2 / 3 = 5.6165
0
0
2
3 0
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Principles of Electromagnetics, 6e
55
53
Prob. 3.25
z
Z=1
y
Z=-1
x
(a)
D • dS = [ z =−1
+ +
z =1
ρ
]D • dS
=5
= − ρ 2 cos 2 φ dφ d ρ + ρ 2 cos 2 φ dφ d ρ + 2 ρ 2 z 2 dφ dz|
= 2(5) 2
2π
1
0
−1
z
dφ z dz = + 50(2π ) ( 3
2
3
ρ =5
|1−1 )
200 π
= 209.44
3
1 ∂
(2 ρ 2 z 2 ) = 4 z 2
(b) ∇ • D =
=
ρ ∂ρ
1
5
2π
−1
0
0
2
2
∇ • Ddv = 4 z ρ d ρ dφ dz = 4 z dz ρ d ρ dφ
= 4x
3
z
3
1
−1
ρ
5
2
2
(2π ) =
0
200π
= 209.44
3
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Principles of Electromagnetics, 6e
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54
Prob. 3.26
H dS =
S
π / 2 2π
10 cosθ r
θ φ
2
sin θ dθ dφ
=0 =0
2π
π /2
0
0
= 10(1)2 dφ
sin θ
= 20π 2
2
r =1
π /2
sin θ cos θ dθ = 10(2π )
sin θ d (sin θ )
0
π /2
= 10π = 31.416
0
Prob. 3.27
H ⋅ d S = ∇ ⋅ Hdv
S
v
H ⋅ dS = − 2 xydydz x = 0 + 2 xydydz x = 1 − ( x
2
+ z 2 )dxdz
S
+ ( x 2 + z 2 )dxdz
y=2
− 2 yzdxdy
2
3
1
2
1
2
1
−1
0
1
0
1
z = −1
+ 2 yzdxdy
y =1
z =3
= 0 + 2 ydy dz + 2 dx ydy + 6 dx ydy
= 12 + 3 + 9 = 24
∇⋅H =
∂H x ∂H y ∂H z
+
+
= 2y + 0 + 2y = 4y
∂x
∂y
∂z
1
2
3
0
1
−1
∇ ⋅ Hdv = 4 ydxdydz = 4 dx ydy dz
V
= 4(1)
y 2
(3 + 1) = 24
2 1
2
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Principles of Electromagnetics, 6e
57
55
Prob. 3.28
ψ2
Side 1:
ψ3
ψ1
ψ = DdS =ψ 1 +ψ 2 +ψ 3
S
= 0+
2π
3
10 z × ρ dφ d ρ
φ =0 ρ =0
z=4
+
4
2π
5ρ × ρ dφ dz ρ = 3
z =0 φ =0
ρ2 3
= 10(4)(2π ) + 5(9)(2π )(4) = 360π + 360π = 2261.95
2 0
Side 2:
ψ = ∇ Bdv,
1 ∂
(5 ρ 2 ) + 0 + 10 = 10 + 10 = 20
ρ ∂ρ
∇ B =
v
2π
ψ = 20dv = 20 4
3
φ =0 z =0 ρ =0
ρ2 3
2 0
ρ dφ d ρ dz = 20(2π )(4) = 720π = 2261.95
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Principles of Electromagnetics, 6e
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56
Prob. 3.29
Let ψ = AdS = ∇ Adv
v
S
∂Ay
∂A
∂Ax
+
+ z = 2( x + y + z )
∂x
∂y
∂z
∇ A = 2( ρ cos φ + ρ sin φ + z ), dv = ρ d ρ dφ dz
∇ A =
ψ = 2( ρ cos φ + ρ sin φ + z ) ρ d ρ dφ dz
2π
2ρ 2 1 z 2 4
1
(2π ) = (16 − 4)(2π )
= 0 + 0 + 2 ρ d ρ zdz dφ =
2 0 2 2
2
0
2
0
1
4
= 12π = 37.7
Prob. 3.30
∇• A =
1 ∂
1 ∂
(r 4 ) +
(r sin 2 θ cos φ )
2
r ∂r
r sin θ ∂θ
= 4 r + 2 cos θ cos φ
∇ • Adv = 4r
3
sin θ dθ dφ dr + 2r 2 sin θ cos θ cos φ dθ dφ dr
π
π /2 π
r4 3
2r 3 3 cos 2 θ π / 2
(
cos
)
(
)
(
)
sin
θ
φ
−
+
−
|
|0 2 3 |0 2 |0
|02
4 0
1
π
= 81(1)( ) + 18(0 + )(1 − 0)
2
2
81π
=
+ 9 = 136.23
2
=4
z
y
x
A • dS = [φ +φ π + +θ π
=0
= /2
r =3
] A • dS
= /2
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Principles of Electromagnetics, 6e
59
57
Since A has no φ − component, the first two integrals on the right hand side vanish.
A • dS =
π /2
π /2
φ
θ
=0
=0
r 4 sin θ dθ dφ|
π
3
+
r =3
r =0
π /2
π /2
φ
=0
r 2 sin 2 θ cos φ drdφ|
θ =π / 2
π /2
= 81 ( ) (− cos θ )| + 9(1) sin φ|
0
0
2
81π
=
+ 9 = 136.23
2
Prob. 3.31
Let ψ =
F • dS = ψ
t
+ ψ b + ψ o +ψ i
where ψ t , ψ b , ψ o , ψ i are the fluxes through the top surface, bottom surface,
outer surface ( ρ = 3), and inner surface respectively.
For the top surface, dS = ρ dφ d ρ a z ,
z = 5;
F • dS = ρ 2 z dφ dz. Hence:
ψt =
3
ρ
=2
2π
ρ 2 z dφ dz|z =5 =
φ
=0
For the bottom surface,
190 π
= 198.97
3
z = 0, dS = ρ dφ d ρ (− a z )
F • dS = − ρ 2 z dφ d ρ = 0. Hence, ψ b = 0.
For the outer curved surface, ρ = 3, dS = ρ dφ dz a ρ
F • dS = ρ 2 sin φ dφ dz. Hence,
ψa =
5
2π
dz ρ φ sin φ dφ|ρ
3
z =0
=0
=3
=0
For the inner curved surface, ρ = 2, d S = ρ dφ dz ( −a ρ )
F • dS = − ρ 3 sin φ dφ dz. Hence,
ψa = −
5
z =0
ψ =
dz ρ 3
2π
sin φ dφ|ρ
φ =0
=2
=0
190 π
190 π
+ 0+0+ 0=
= 198.97
3
3
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Principles of Electromagnetics, 6e
60
58
ψ=
F • dS = ∇ • FdV
∇•F =
1 ∂
ρ∂ρ
= 3 ρ sin φ −
V
z
ρ
( ρ 3 sin φ ) +
1 ∂
ρ ∂φ
( z cos φ ) + ρ
sin φ + ρ
z
(3ρ sin φ − ρ sin φ + ρ ) ρ dφ d ρ dz
∇ • Fdv =
5
2π
3
0
0
2
= 0 + 0 + dz dφ ρ 2 d ρ
=
190 π
= 198.97
3
Prob. 3.32
ax
∂
(a) ∇xA =
∂x
xy
ay
∂
∂y
y2
az
∂
= za y − xa z
∂z
− xz
1
1
∇xB = 2 ρ z 2sin φ cos φ − 0 a ρ + (2 ρ z − 2 z sin 2 φ )aφ + ( 2 ρ sin 2 φ − 0 ) a
ρ
ρ
(b)
= 4 z sin φ cos φ a ρ + 2( ρ z − z sin 2 φ )aφ + 2sin 2 φ a z
= 2 z sin 2φ a ρ + 2 z ( ρ − sin 2 φ )aφ + 2sin 2 φ a z
∇xC =
(c)
1
r sin θ
1 ∂ 2
∂
2
2 ∂θ (r cos θ sin θ ar − r ∂r (r cos θ aθ
cos 2 θ
r
(2 cos θ )(− sin θ ) sin θ + cos θ (cos 2 θ ) ar −
(2r )aθ
r sin θ
r
(cos3 θ − 2sin 2 θ cos θ )
=
ar − 2 cos 2 θ aθ
sin θ
=
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Principles of Electromagnetics, 6e
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59
Prob. 3.33
(a)
ax
∂
∇× A =
∂x
x2 y
ay
az
∂
∂
=
∂y
∂z
y 2 z −2 xz
− y2 a x + 2z a y − x2 a z
∇ •∇× A = 0
(b)
∇× A = (
1 ∂ Az
ρ ∂φ
−
∂ Aφ
∂ A ∂ Az
1 ∂ ( ρ Aρ ) ∂ Aρ
−
) aρ + ( ρ −
) aφ + (
)az
∂z
∂z
∂ρ
ρ
∂ρ
∂φ
= (0 − 0) a ρ + ( ρ 2 − 3z 2 ) aφ +
1
ρ
(4 ρ 3 − 0) a z
= ( ρ 2 − 3z 2 ) aφ + 4 ρ 2 a z
∇•∇× A = 0
(c)
1 sin φ 1 −1 cos φ
1 ∂ cos φ ) − 0 aθ + (
− 0 aφ
0 − 2 ar + 2
r sin θ r r sin θ r
r ∂r
r
sin φ
cos φ
cos φ
ar + 3
aθ + 3 aφ
=− 3
r sin θ
r sin θ
r
∇•∇× A =
− sin φ
sin φ
+0+ 4
=0
4
r sin θ
r sin θ
∇× A =
∇ •∇× A= 0
Prob. 3.34
∇ × H = 0a ρ + 1aφ +
1
ρ
(2 ρ cos φ − ρ cos φ )a z = aφ + cos φ a z
1
1
1
1
∇ × ∇ × H = − sin φ − 0 a ρ + 0aφ + (1 − 0)a z = - sin φ a ρ + a z
ρ
ρ
ρ
ρ
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Principles of Electromagnetics, 6e
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60
Prob. 3.35
Method 1: We can express A in spherical coordinates.
a
r
A = 3 ar = 2r ,
r
r
−2
a 1
∇ × A = ∇ × 2r = ∇ 2 × ar = 3 ar × ar = 0
r
r r Method 2:
x
y
z
a + 3 a y + 3 az
3 x
r
r
r
∂
∂
∂
3
∂x ∂y ∂z 3
∇× A =
= − z ( x 2 + y 2 + z 2 ) −5/ 2 (2 y ) − − y ( x 2 + y 2 + z 2 ) −5 / 2 (2 z ) a x + ...
2
x
y
z 2
3
3
3
r
r
r
=0
A=
Prob. 3.36
y
1
1
2
3
0
1
2
x
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Principles of Electromagnetics, 6e
63
61
(a)
F • d l = (
L
+ + ) F • d l
1
2
3
_
_
_
For 1, y = x dy = dx, dl = dx a x + dy a y ,
1
F •d l = 1
x 3 dx − xdx = −
0
1
4
_
_
_
For 2, y = − x + 2, dy = − dx, dl = dx a x + dy a y ,
2
3
2
F • d l = (− x + 2 x − x + 2)dx =
2
1
17
12
For 3,
0
F •d l = x
3
2
2
1
F •d l = − 4
ydx|
+
L
y =0
=0
17
7
+ 0 =
12
6
(b)
∇ × F = − x2 a z ;
dS = dxdy (− a z )
1 x
2
(∇ × F ) • d S = − (− x )dxdy =
=
1
2
x y| dx +
x
0
0
2
x dydx +
0 0
2
2
x y|
1
− x+2
0
dx =
2
− x+2
1
y =0
x1
| +
4 0
x 2 dydx
2
x (− x + 2)dx =
2
1
7
6
(c) Yes
Prob. 3.37
A • dl =
=
1
ρ
=2
ρ sin φ d ρ
φ =0
+
π /2
φ
=0
ρ 2 ρ dφ
ρ =1
+
2
ρ
ρ sin φ d ρ
=1
φ = 90
+
o
0
φ π
= /2
ρ 3 dφ
ρ =2
π
1
π
+ (4 − 1) + 8(− ) = −9.4956
2 2
2
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Principles of Electromagnetics, 6e
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62
62
62
63
Prob.
Prob.
3.38
Prob.3.38
3.38
(c)
63
1 y∂yy
1
∂
cos θ cos φ
1
sin θ cos φ
∇ ⋅∇V = ∇ V = 2
(− sin θ cos φ ) + 2
(sin θ
) + 2 2 (−
)
r ∂r
r sin θ ∂θ
r
r sin θ
r
cos φ(c)
cos φ
= 0+ 3
(1 − 2sin 2 θ ) − 3
1 ∂
1
∂
cos θ cos φ
1
sin θ cos φ
r sin θ
∇ ⋅∇V = ∇ 2V = r2 sin (θ− sin θ cos φ ) + 2
(sin θ
) + 2 2 (−
)
o
o
o r ∂r
r sin θ ∂θ
r
r sin θ
r
2sin θ cos φ
45
45
45
=−
cos φ
cos φ
r3
= 0+ 3
(1 − 2sin 2 θ ) − 3
r sin θ 222
000
xxxr sin θ
2sin
θ
cos
φ
Prob. 3.41
=−
r3
r
2
2 22
0 00
r sin θ [(cos
φ − sin φ ) a xπ+π/4π/4/4(cos φ + sin φ ) a y ]
r sin θFF
+++ 333zzzsin
sin
+++ 222ρρρzd
zd
zdρρρ
sinφφφρρρdddφφφ
zdρρρ
F•••dldldl===0Prob.
0222ρρρzdzd
zzz===111 000
zzz===111
ρρρ===2,2,
3.41
2,zzz===111
2 22
0
= r (cos φ − sin φ ) a x + r (cos
+
sin
)
φ
φ
a
y
r
r sin θ [(cos φ − sin φ ) a x + (cos φ + sin φ ) a y ]
Q=
22 r sin θ
πππ/ //444 2 22000
2 22 2
Qr ==
sin
sin
θ===(4
ρρ θ +cos
φφφ)θ)) sin φ+++ρρρcos
cos
cos
1.757
++((−(φ
−−666cos
−−0)
=ρ
(4
+++6(6(
cos
(4−Q
0)
6(−−−cos
cosπππ/ //444+++1)1)
1)+++(0(0
(0−−−4)4)
4)===1.757
1.757
x0)
22 φ + sin φ ) a y
0 r (cos φ − sin0φ00) a x + r2(cos
00=
Qθ = 111cos θ cos φ cos θ sin φ − sin θ Qy ∇
F=== [3[3−
∇
+++......
sin
φ
0 Q ∇xxφF
xF
sinφφφ−−−000]]aaaz zcos
...
Q
ρρρ 3zzzsin
Qr z sin θ cos φ sin θ zsin
φ cos θ Qx 2 22 ππ/π4//44 //44y4
Qθ333zzz= cos θ cos φ cos θ sin φ − sin θ πππ/Q
x
x
d
d
d
d
d
d
φ
φ
ρ
ρ
φ
φ
ρ
ρ
φ
φ
(
(
)
)
sin
sin
3(2)(
3(2)(
cos
cos
)
)
∇
∇
F
F
⋅
⋅
S
S
=
=
=
=
−
−
θ
θ
sin
cos
Q
=
a
+
a
+
a
r
r
r
x
d
d
d
φ
ρ
φ
ρ
φ
(
)
sin
3(2)(
cos
)
∇
F
⋅
S
=
=
−
r
0=0φρρρθ −φsin φ zzz=== cos φ
0
0
0
0
Q
ρ ρ=ρ0==00 φφ=φQ
z
=0 Q=
(a)
1
,−−cos
30
1 θ a + ra
a6(
ρππ
=πQ
°θ=a2(+ )r=cos
dl = ρ d=φ==6(
r+1)
φ−
cos
++
1)
1.757
1.757
6(
cos
1)=r==sin
1.757
=sin
θ
φ
r 2
(a)30° = 3
z = r cos
Prob.
Prob.
3.39
Prob.3.39
3.39
1
2 −d−y−lyy=2 ρ dφ−a
ρ = r sin 30° = 2( ) = 1
− −y−=
yy
−y−yφy ,
=
+
ρ
Q
r
z
∇∇
xe
xe
16
xe
∇⋅ ⋅A
⋅ AA==
8xe
xe +++888xe
xe ===16
16xe
xe
φ=88
2
− −y−yy
− −y−yy
2
π
∇∇
((∇
16
16
xe
∇(∇
∇⋅ ⋅A
⋅A
A)))===16
16eee aa2ax xx−−−16
16xe
xe aaayy 30° = 3
2 z = r ycos
ρ
ρ
Q
l
•
=
+
d
z
dφ = 2(1)(2π ) = 4π
0
Qφ = r = ρ 2 + z 2
aaax xx
aaay yy
aaaz zz
(b)
2π
∂∂∂
∂l∂∂=
∂∂∂ 2 2 ρ d−φ−y−yy= 2(1)(2
− −y y
• dcos
∇∇
θ aφ ρ ===+((−(z−−16
∇
⋅A
((=∇
16
16
A))θ)===a r − 2aQθ +
∇x×x∇
∇(∇
∇⋅cot
⋅A
xQ
16eee +++16
16eee −)yπa))aa)z zz====400π0
∂∂∂xxx
∂∂∂yyy 0 ∂∂∂zzz
2 − −y−yy
(b)
r eeesin θ−d−−16
dxe
θ16
φ−a−y−yyr 000
For S1 , dS =16
16
xe
16
16xe
2∇
∇∇
×x∇
Q
cot
xsin
==
000θ. θ
Should
Should
be
expected
expected
since
..daθr d−φ|2 aθ + cos θ aφ
x∇
∇VθVV==cot
Shouldbe
be
expected
since
(∇
× Q ) • dSsince
= r∇
S1
r =2
S301 ,° dS = r sin θ dθ dφ a r
For
Prob.
Prob.
3.40
Prob.3.40
3.40
2π
sin
sin
cos
cos
cos
2φ
sinθθθcos
cosφφφ = 4 cos
cos
cos
sin
φsin θ cot θ dθ dφ
cos
=sin
dφθθθcos
) φ•φφdaθaSθaθdθ−=θ−−sin
r4φ2π
(a)
(a)
aaar rr+++(∇ ×2Q
aaaφ φφ
|r = 2
(a)∇∇
∇VVV===−−−
2 22
2
2
2
2
0
0
rrr
rrr
rrr
S1
(b(b
(b))) ∇∇
∇xx∇
x∇
∇VVV===000
2
2π
30°
0
0
= 4 dφ
cos θ dθ = 4π
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Principles of Electromagnetics, 6e
65
63
(c)
1 ∂
1
∂
cos θ cos φ
1
sin θ cos φ
(− sin θ cos φ ) + 2
(sin θ
) + 2 2 (−
)
2
r ∂r
r sin θ ∂θ
r
r sin θ
r
cos φ
cos φ
= 0+ 3
(1 − 2sin 2 θ ) − 3
r sin θ
r sin θ
2sin θ cos φ
=−
r3
∇ ⋅∇V = ∇ 2V =
Prob. 3.41
r
r sin θ [(cos φ − sin φ ) a x + (cos φ + sin φ ) a y ]
Q=
r sin θ
= r (cos φ − sin φ ) a x + r (cos φ + sin φ ) a y
Qr sin θ cos φ
Qθ = cos θ cos φ
Qφ − sin φ
sin θ sin φ
cos θ sin φ
cos φ
cos θ − sin θ 0 Qx Q y
Qz Q = r sin θ a r + r cos θ aθ + r a φ
(a)
dl = ρ dφ aφ ,
1
2
ρ = r sin 30° = 2( ) = 1
z = r cos 30° = 3
Qφ = r =
2π
Q • dl = ρ 2 + z2
ρ 2 + z 2 ρ dφ = 2(1)(2π ) = 4π
0
(b)
∇ × Q = cot θ a r − 2 aθ + cos θ aφ
For S1 , dS = r 2 sin θ dθ dφ a r
(∇ × Q ) • dS = r
2
S1
sin θ cot θ dθ dφ|
2π
30°
0
0
= 4 dφ
r =2
cos θ dθ = 4π
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Principles of Electromagnetics, 6e
66
64
(c)
dS = r sin θ dθ dr aθ
For S2 ,
(∇ × Q) • dS = − 2 r sin θ dφ dr|θ
S2
2
2π
0
0
= −2sin 30 rdr
= 30°
dφ
= − 4π
(d)
For S1 , dS = r 2 sin θ d φ dθ a r
S1
Q • dS = r 3 sin 2 θ dθ d φ|
2π
30
0
0
= 8 dφ
= 4π [
(e)
r=2
°
π
3
sin
−
2
θ dθ
3
] = 2.2767
2
_
For S 2 , dS = r sin θ dφ dr aθ
S2
Q • dS = r 2 sin θ cos θ dφ dr|
θ =30°
=
(f)
4π 3
= 7.2552
3
1 ∂ 3
r
∂
(r sin θ ) +
(sin θ cos θ ) + 0
2
r ∂r
r sin θ ∂ θ
= 2sin θ + cos θ cot θ
∇•Q =
∇ • Qdv = (2sin θ + cos θ cot θ )r
=
=
r3 2
(2π )
3 0
30
(1 + sin
2
2
sin θ dθ dφ dr
θ )dθ
0
4π
3
(π −
) = 9.532
3
2
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Sadiku & Kulkarni
Principles of Electromagnetics, 6e
67
65
Check :
∇ • Qdv = (
S1
+ Q • dS
S2
π
3
3
= 4π [ −
+
]
3 2
3
=
4π
3
[π −
]
3
2
(It checks!)
Prob. 3.42
Since u = ω × r ,
∇ × u = ∇ × ( ω × r ). From Appendix A.10,
∇ × ( A× B ) = A(∇ • B ) − B (∇ • A) + ( B • ∇) A− ( A• ∇) B
∇ × u = ∇ × (ω × r )
_
∇ × (ω × r ) = ω (∇ • r ) − r (∇ • ω ) + (r • ∇) ω − (ω • ∇) r
= ω (3) − ω = 2 ω
or ω =
1
∇ × u.
2
Alternatively, let
u=
x = r cos ω t ,
y = r sin ω t
∂x
∂ y
ax +
ay
∂t
∂t
= − ω r sin ω t a x + ω r cos ωt a y
= −ω y ax + ω x a y
∂
∂
∇×u = ∂ x ∂ y
−ω y ω x
i.e., ω =
∂
∂ z = 2ω a z = 2ω
0
1
∇×u
2
Note that we have used the fact that ∇ • ω = 0,
(r • ∇)ω = 0,
(ω • ∇)r = ω
Copyright © 2015 by Oxford University Press
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Principles of Electromagnetics, 6e
68
66
Prob. 3.43
∇ B =
1 ∂
1 ∂Bφ ∂Bz 1
( ρ Bρ ) +
+
= 2 ρ cos φ + ρ cos φ − 4
ρ ∂ρ
ρ ∂φ ∂z ρ
= (2 + ρ ) cos φ − 4
∂B 1 ∂Bz ∂Bφ ∂B ∂B 1 ∂
−
aρ + ρ − z aφ + ( ρ Bφ ) − ρ a z
(a) ∇ × B = ρ ∂ρ
∂z ∂ρ ∂φ ρ ∂φ
∂z
1
= 0a ρ + 0aφ + 3ρ 2 sin φ + ρ sin φ a z
ρ
= (3ρ + 1) sin φ a z
(b)
∇F =
1 ∂ 2
1 ∂Fφ 1 ∂ 4
1
(
)
0
( −2r sin φ )
r
F
+
+
= 2 (r sin θ ) +
r
2
r ∂r
r sin θ ∂φ r ∂r
r sin θ
2sin φ
= 4r sin θ −
sin θ
∂
∂Fθ
∂θ ( Fφ sin θ ) − ∂φ
∂F 1 ∂
+ (rFθ ) − r aφ
r ∂r
∂θ ∇×F =
1
r sin θ
1 1 ∂Fr ∂
ar + r sin θ ∂φ − ∂r (rFφ ) aθ
1 1 2
∂
∂θ (2r sin θ cos φ ) − 0 ar + r sin θ r cos φ − 4r cos φ aθ + 0aφ
r cos φ
= 2 cot θ cos φ ar + − 4 cos φ aθ
sin θ
=
1
r sin θ
Prob. 3.44
(a)
∂V
∂V
∂V ax + V
ay+ V
az ∇(V∇V)=∇ V
∂y
∂z ∂x
=
∂ ∂V
V
∂x ∂x
∂ ∂V
+ V
∂y ∂y
∂ ∂V + V
∂z ∂z 2
2
∂ 2V
∂ 2V
∂ 2V ∂V ∂V ∂V = V 2 +V 2 +V 2 + +
+
∂x
∂y
∂z
∂x ∂y ∂z = V ∇ 2V + | ∇V |2
2
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Principles of Electromagnetics, 6e
69
67
∂
∂
∂
∇ × VA = ∂x ∂y ∂z
VAx VAy VAz
(b)
∂
∂
∂
∂
∂
∂
= (VAz ) − (VAy ) a x + (VAx ) − (VAz ) a y + (VAy ) − (VAx ) a z
∂z
∂x
∂y
∂z
∂y
∂x
∂A ∂V
∂A
∂V
= Az
+ V z − Ay
− V y ax
∂y
∂z
∂z ∂y
∂A
∂V
∂A ∂V
+ Ax
+ V x − Az
−V z ay
∂z
∂x
∂x ∂z
∂A
∂V
∂A ∂V
+ Ay
+ V y − Ax
− V x az
∂x
∂y
∂y ∂x
∂A ∂A ∂A ∂A ∂A ∂A ∇ × VA = V z − y a x + x − z a y + y − x a z ∂z ∂x ∂y ∂z
∂x
∂y
∂V
∂V
∂V ∂V ∂V ∂V
+ Az
− Ay
− Az
− Ax
a x + Ax
az
a y + Ay
∂z ∂z
∂x ∂x
∂y ∂y
= V ∇ × A + ∇V × A
Prob. 3.45
(a)
∂B ∂B ∂B
∇ B = x + y + z = 2 xy + 1 + 1 = 2 + 2 xy
∂x
∂y
∂z
(b)
∂
∂
∂
∂y
∂z = (−1 + 0)a x + (0 − 0)a y + (4 x + x 2 )a z
∇ × B = ∂x
x 2 y (2 x 2 + y ) ( z − y )
= −a x + x(4 − x)a z
(c)
∇(∇B) = ∇(2 + 2 xy ) = 2 ya x + 2 xa y
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Principles of Electromagnetics, 6e
70
68
(d)
∂
∂
∇ × ∇ × B = ∂x ∂y
−1 0
∂
∂z
= 0a x - (4 - 2x)a y + 0a z
2
(4 x − x )
= 2( x − 2)a y
Prob. 3.46
(a)
V1 = x 3 + y 3 + z 3
∂ 2V1 ∂ 2V1 ∂ 2V1
∇ V1 =
+
+
∂x 2 ∂y 2 ∂z 2
∂
∂
∂
( 3x 2 ) +
3 y 2 ) + (3z 2 )
=
(
∂x
∂y
∂x
2
= 6 x + 6 y + 6z = 6( x + y + z)
(b)
V2 = ρ z 2 sin 2φ
∇ 2V2 =
=
1 ∂
ρ ∂ρ
z
2
ρ
=(
( ρ z 2 sin 2φ ) −
sin 2φ −
−3z 2
ρ
4z
2
ρ
4z2
ρ
sin 2φ +
∂
(2 ρ z sin 2φ )
∂z
sin 2φ + 2 ρ sin 2φ
+ 2 ρ ) sin 2φ
(c)
V3 = r 2 (1 + cos θ sin φ )
∇ 2V3 =
1 ∂
[2r 3 (1 + cos θ sin φ )]
2
r ∂r
1
∂
1
( − sin 2 θ sin φ )r 2 + 2 2 r 2 ( − cosθ sin φ )
r sinθ ∂θ
r sin θ
2 sinθ
cosθ sin φ
= 6(1 + cosθ sin φ ) −
cosθ sin φ −
sinθ
sin 2 θ
cosθ sin φ
= 6 + 4 cosθ sin φ −
sin 2 θ
+
2
Copyright © 2015 by Oxford University Press
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Principles of Electromagnetics, 6e
71
69
Prob. 3.47
(a)
U = x 3 y 2 e xz
∂
∂
∂ 4 2 xz
(3 x 2 y 2 e xz + x 3 y 2 ze xz ) +
(2 x 3 y e xz ) +
(x y e )
∂x
∂y
∂z
∇ 2U =
= 6 xy 2 e xz + 3 x 2 yze xz + 3 x 2 y 2 ze xz + x3 y 2 z 2 e xz + 2 x3e xz + x5 y 2 e xz
= e xz (6 xy 2 + 3 x 2 y 2 z + 3 x 2 y 2 z + x3 y 2 z 2 + 2 x3 + x5 y 2 )
At (1, −1,1),
∇ 2U = e1 (6 + 3 + 3 + 1 + 2 + 1) = 16e= 43.493
(b)
V = ρ 2 z (cos φ + sin φ )
1 ∂
∇ 2V =
[2 ρ 2 z (cos φ + sin φ )] − z (cos φ + sin φ ) + 0
ρ ∂ρ
= 4 z (cos φ + sin φ ) − z (cos φ + sin φ )
= 3 z (cos φ + sin φ )
π
At (5,
(c)
6
, − 2), ∇ 2V = −6(0.866 + 0.5)= −8.196
W = e − r sin θ cos φ
∇ 2W =
1 ∂
e− r
∂
2 −r
φ
(
−
r
e
sin
θ
cos
)
+
cos φ
(sin θ cos θ )
2
2
∂θ
r ∂r
r sin θ
−
e − r sin θ cos φ
r 2 sin 2 θ
1
(−2re − r sin θ cos φ ) + e − r sin θ cos φ
2
r
e − r cos φ
e − r cos φ
+ 2
(1 − 2sin 2 θ ) − 2
r sin θ
r sin θ
2 2
∇ 2W = e − r sin θ cos φ (1 − − 2 )
r r
=
At
(1, 60°,30°),
∇ 2W = e −1 sin 60 cos 30(1 − 2 − 2) = −2.25e −1 = − 0.8277
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Principles of Electromagnetics, 6e
72
70
Prob. 3.48
(a ) Let V = 1nr = 1n x 2 + y 2 + z 2
∂V 1 1
x
=
( 2 x ) ( x 2 + y 2 + z 2 )−1/2 = 2
∂x r 2
r
∇V =
∂V
ox
ax +
∂V
oy
ay +
∂V
oz
az =
x a x + ya y + z a z
r
2
=
r
r2
r 1
= ar in spherical coordinates.
r2 r
1 ∂ 2
1 ∂
∇ 2 (1nr ) = ∇ ∇ (1nr ) = ∇ A = 2
(r A r ) = 2
(r )
r
∂r
r ∂r
1
= 2
r
(b) Let ∇V = A =
Prob. 3.49
∂V
∂V
∂V
∇V =
ax +
ay +
az
∂x
∂y
∂z
= y 2 z 3a x + 2 xyz 3a y + 3xy 2 z 2 a z
At P(1,2,3,) x = 1, y = 2, z = 3
∇V = 4(27)a x + 2(2)(27)a y + 3(4)(9)a z
= 108( a x + a y + a z )
∂ 2V ∂ 2V ∂ 2V
∇V = 2 + 2 + 2
∂x
∂y
∂z
∂
∂
∂
= ( y 2 z 3 ) + ( 2 xyz 3 ) + ( 3xy 2 z 2 )
∂x
∂y
∂z
2
= 0 + 2 xz 3 + 6 xy 2 z
= 2 xz ( z 2 + 3 y 2 )
At P(1,2,3,) x = 1, y = 2, z = 3.
∇ 2V = 2(1)(3)(9 + 3 × 4) = 6(9 + 12)
= 126
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Principles of Electromagnetics, 6e
73
71
Prob. 3.50
∂V
1 ∂V
∂V
aρ +
aφ +
a z = 2 ρ z cos φ a ρ - ρ z sin φ aφ + ρ 2 cos φ a z
∇V =
ρ ∂φ
∂ρ
∂z
1 ∂ 2V ∂ 2V 1 ∂
1
+ 2 =
2 ρ 2 z cos φ ) − 2 ρ 2 z cos φ + 0
(
+ 2
2
∂z
ρ ∂ρ
ρ
ρ ∂φ
= ( 4 − 1) z cos φ = 3z cos φ
∇ 2V =
1 ∂ ∂V
ρ
ρ ∂ρ ∂ρ
Prob. 3.51
∂V
1 ∂V
1 ∂V
ar +
aθ +
aφ
∂r
r ∂θ
r sin θ ∂φ
10
5sin φ
= − 3 cos φ ar − 3
aφ
r
r sin θ
∇V =
(a)
1 ∂ 2 ∂V
1
∂
∂V
1
∂ 2V
r
(
)
+
(sin
)
+
θ
r 2 ∂r
r 2 sin θ ∂θ
r 2 sin 2 θ ∂φ 2
∂r
∂θ
1 ∂ 2 10 cos φ
1
5cos φ
r (−
= 2
) + 0 + 2 2 (−
)
3
r ∂r
r
r sin θ
r2
10 cos φ 5cos φ
∇∇V =
− 4 2
r4
r sin θ
∇∇V = ∇ 2V =
(b)
(c)
∇ × ∇V = 0, see Example 3.10.
Prob. 3.52
∂U
∂U
∂U
ax +
ay +
a z = 4 yz 2 a x + (4 xz 2 + 10 z )a y + (8 xyz + 10 y )a z
∇U =
∂x
∂y
∂z
∂
∂
∂
∇∇U = (∇U x ) + (∇U y ) + (∇U z ) = 0 + 0 + 8 xy = 8 xy
∂x
∂y
∂z
∇ 2U =
∂ 2U ∂ 2U ∂ 2U
+
+
= 0 + 0 + 8 xy = 8 xy
∂x 2 ∂y 2 ∂z 2
Hence, ∇ 2U = ∇∇U
Copyright © 2015 by Oxford University Press
POESM_Ch03.indd 73
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Principles of Electromagnetics, 6e
74
72
Prob. 3.53
Method 1
∇ 2G
ρ
= ∇ 2Gρ −
2 ∂Gφ Gρ
−
ρ 2 ∂φ ρ 2
1 ∂
2 ρ sin φ
8ρ sin φ 2 ρ sin φ
+0+
−
( 2 ρ sin φ ) −
2
ρ ∂ρ
ρ
ρ2
ρ2
2sin φ 2sin φ 8sin φ 2sin φ
6sin φ
=
−
+
−
=
=
ρ
∇ 2G
φ
ρ
= ∇ 2Gφ +
ρ
ρ
ρ
2 ∂Gρ Gφ
−
ρ 2 ∂φ ρ 2
1 ∂
1
4 ρ cos φ 4 ρ cos φ
−
(4 ρ cos φ ) − 4 ρ cos φ + 0 +
ρ
ρ2
ρ2
ρ ∂ρ
4 cos φ 4 cos φ 4 cos φ 4 cos φ
=
−
+
−
=0
=
ρ
∇ 2G
z
ρ
= ∇ 2 Gz =
=
1
ρ
ρ
ρ
1 ∂
∂
ρ ( z 2 + 1) + 0 + (2 z ρ )
ρ ∂ρ
∂z
( z 2 + 1) + 2 ρ
Adding the components together gives
∇ 2G =
6sin φ
ρ
1
a ρ + 2 ρ + ( z 2 + 1) a z
ρ
Method 2:
∇ 2G = ∇(∇G ) − ∇ × (∇ × G )
1 ∂
1
(2 ρ 2 sin φ ) + (−4 ρ sin φ ) + 2 z ρ = 2 z ρ
ρ ∂ρ
ρ
∇(∇G ) = ∇V = 2 za ρ + 2 ρ a z
Let V = ∇G =
Copyright © 2015 by Oxford University Press
POESM_Ch03.indd 74
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Principles of Electromagnetics, 6e
75
73
Let
1
1 ∂
A = ∇ × G = 0 − 0 a ρ + 0 − ( z 2 + 1) aφ + (4 ρ 2 cos φ ) − 2 ρ cos φ a z
ρ ∂ρ
ρ
2
= − ( z + 1)aφ + 6 cos φ a z
6
1 ∂
∇ × ∇ × G = ∇ × A = − sin φ + 2 z a ρ + (0 − 0)aφ + ( − ρ ( z 2 + 1)) − 0 a z
ρ ∂ρ
ρ
6
1
= 2 z − sin φ a ρ − ( z 2 + 1)a z
ρ
ρ
2
∇ G = ∇V − ∇ × A
6
1
= 2 zaρ + 2 ρa z − 2 z − sin φ a ρ + ( z 2 + 1)a z
ρ
ρ
6
1
= sin φ a ρ + 2 ρ + ( z 2 + 1) a z
ρ
ρ
Prob. 3.54
∂
∂
∂
∇ A = ( xz ) + ( z 2 ) + ( yz ) = z + y
∂x
∂y
∂z
∇(∇ A) = a y + a z
∇ 2 A = ∇ 2 Ax a x + ∇2 Ay a y + ∇2 Az a z = 0 + 2a y + 0 = 2a y
∇(∇ A) - ∇2 A = − a y + a z
∂
∇ × A = ∂x
xz
∂
∂y
z2
∂
∇ × ∇ × A = ∂x
−z
(1)
∂
∂z = − za x + xa y
yz
∂
∂y
x
∂
∂z = − a y + a z
0
(2)
From (1) and (2),
∇ × ∇ × A = ∇( ∇ A) - ∇ 2 A
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74
Prob. 3.55
∂A ∂A ∂A
∇ A = x + y + z = 1 + 1 + 1 = 3 ≠ 0
∂x
∂y
∂z
∂
∇ × A = ∂x
x
∇B =
∂
∂y
y
∂
∂z = 0
z
1 ∂
1 ∂Bφ ∂Bz
( ρ Bρ ) +
+
= 4 cos φ − 4 cos φ = 0
ρ ∂ρ
ρ ∂φ ∂z
∂B 1 ∂Bz ∂Bφ ∂B ∂B 1∂
∇× B = −
a ρ + ρ − z aφ + ( ρ Bρ ) − ρ a z
∂z ∂ρ ∂φ ρ ∂ρ
ρ ∂φ
∂z
1
= 0aρ + 0aφ + [ −8 ρ sin φ + 2 ρ sin φ ] a z = −6sin φ a z ≠ 0
ρ
1 ∂ 2
2sin θ
(r sin θ ) + 0 + 0 =
≠0
2
r ∂r
r
1 ∂
1
∂ 2
2
r
∇×C =
(
sin
θ
)
−
0
+
0
−
(r sin θ ) aθ
a
r
r sin θ ∂θ
r ∂r
1
+ [ 0 − cos θ ] aφ
r
cosθ
= 2 cos θ ar - 2sinθ aθ aφ ≠ 0
r
(a) B is solenoidal.
(b) A is irrotational.
∇C =
Prob. 3.56 (a)
ax
ay
∂
∂
∇×G =
∂x
∂y
16 xy − z 8 x 2
az
∂
∂z
−x
= 0 a x + (−1 + 1) a y + (16 x − 16 x) a z = 0
Thus, G is irrotational.
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Principles of Electromagnetics, 6e
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75
(b) Assume that ψ represents the net flux.
ψ = G • dS = ∇ • Gdv
∇ • G = 16 y + 0 + 0 = 16 y
1
1
1
0
0
0
ψ = 16 ydxdydz = 16 dx dz ydy = 16(1)(1)(
(c)
y2 1
)=8
2 0
y
1
0
G • dl
L
=
1
y =1
x =1
(16 xy − z )dx|
x =0
y =0
z =0
= 0 + 8(1) y| + 16(1)
1
0
= 8−8 = 0
+
8x dy|
y =0
2
x =1
z =0
+
x
y =0
x =0
(16 xy − z )dx|
x =1
y =1
z =0
+
8 x dy|
y =1
2
x =0
z =0
x2 0
| +0
2 1
This is expected since G is irrotational, i.e.
G • dl = (∇ × G ) • dS =
0
∇ • T = −6 + 0 = −6
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Principles of Electromagnetics, 6e
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76
Prob. 3.57
∇⋅F = 0
ax
∂
∇xF =
∂x
yz
ay
∂
∂y
xz
az
∂
= ( x − x )a x + ( y − y )a y + ( z − z )a z = 0
∂z
xy
Hence F is both solenoidal and conservative.
Prob. 3.58
∇× H = 0
H dl = (∇ × H)dS = 0
L
S
Prob. 3.59
From Appendix A.10,
∇( A × B ) = B (∇ × A) − A(∇ × B )
If A and B are irrotational,
∇× A = 0 = ∇× B
i.e. ∇( A × B ) = 0
which implies that A × B is solenoidal.
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77
CHAPTER 4
P. E. 4.1
5 ×10−9 [(1, −3, 7) − (2, 0, 4)]
[(1, −3, 7) − (2, 0, 4)]3
−9
1×10
−9
(a) F =
(
2
10
)[(1,
3,
7)
(
3,
0,5)]
−
×
−
−
−
10−9 +
3
4π [(1, −3, 7) − ( −3, 0,5)] 36π =[
45(−1, −3,3) 18(4, −3, 2)
] nN
−
193/ 2
293/ 2
= −1.004a x − 1.284a y + 1.4 a z nN
(b)
E=
F
= −1.004a x − 1.284a y + 1.4a z V/m
Q
P. E. 4.2
Let q be the charge on each sphere, i.e. q=Q/3. The free body diagram below helps us to
establish the relationship between various forces.
P
θ
T
A
F1
d/2
F2
mg
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78
At point A,
T sin θ cos 30° = F1 + F2 cos 60°
=
q2
4πε 0 d
2
+
q2
1
( )
4πε 0 d 2
2
3 q2
=
8πε 0 d 2
T cos θ = mg
Hence,
sin θ =
But
Thus,
or
tan θ cos 30° =
q2 =
h
d
tan θ =
=
l
3l
Q2 =
d
3
l2 −
d2
3
d
3
( )
3q 2
3 2 =
8πε 0 d 2 mg
d2
2
l −
3
4π ε0 d 3 m g
3 l2 −
but q =
3q 2
8πε 0 d 2 mg
Q
3
d2
3
⎯⎯
→
q2 =
Q
. Hence,
9
12 π ε 0 d 3 m g
d2
l −
3
2
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79
P.E. 4.3
_
d2 l
eE = m 2
dt
_
_
d 2 x _ d 2 y _ d 2z _
ax + 2 a y + 2 az )
dt 2
dt
dt
E0 = 200 kV / m
_
eE0 ( − 2a x + a y ) = m(
where
d 2z
=0
⎯⎯ →
z = ct + c2
dt 2
d2x
− 2eE0 t 2
m 2 = − 2eE0
⎯⎯ →
x=
+ c3 t + c4
2m
dt
d2y
eE0 t 2
m 2 = eE0
⎯⎯ →
y=
+ c5 t + c6
2m
dt
At t = 0, ( x, y, z ) = (0, 0, 0) c1 = 0 = c4 = c6
dx dy dz
, , ) = (0, 0, 0)
dt dt dt
⎯⎯
→
c1 = 0 = c3 = c5
At t = 0
Also, (
Hence,
( x, y ) =
i.e. 2 | y | = | x |
eE0t 2
(−2,1)
2m
Thus the largest value of is
80 cm = 0.8 m
P.E. 4.4
(a)
Consider an element of area dS of the disk.
The contribution due to dS = ρ dφ d ρ is
dE =
ρ s dS
ρ s dS
=
2
4πε 0 r
4πε 0 ( ρ 2 + h 2 )
The sum of the contribution along ρ gives zero.
ρs
Ez =
4π ε 0
a
ρ
=0
2π
hρ
h ρ d ρ dφ
φ =0 ( ρ 2 + h2 )3/ 2 = 2 ε 0s
a
ρ
−0
ρ dρ
( ρ + h 2 )3/ 2
2
a
a
h ρs
hρs
2
2 −3/ 2
2
2
2 −1/ 2
(
)
(
)
(
2(
)
|
=
+
=
−
+
ρ
h
d
ρ
ρ
h
0
4 ε 0 0
4ε 0
=
ρs
h
[1 − 2
]
2ε 0
(h + a 2 )1/ 2
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Principles of Electromagnetics, 6e
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80
(b)
As a
E=
⎯⎯
→
∞,
ρs
az
2ε 0
(c) Let us recall that if a/h <<1 then (1+a/h)n can be approximated by (1+na/h).
Thus the expression for Ez from (a) can be modified for a<<h as follows.
ρs 1
= ρs
Ez =
1−
2
2ε o a 2ε o
1+ 2 h 2
ρ πa Q
=
= s 2 2ε o 2πh 4πε o h 2
2
1 − 1 + a
h 2
1
− 2
ρs a2 ⎯a⎯
⎯ ⎯ ⎯⎯→
→0 , but ρ sπa 2 = Q
2ε o 2h 2 This is in keeping with original Coulomb’s law.
P. E. 4.5
QS =
ρ
2
S
dS =
2
12 | y| dx dy
−2 −2
2
= 12(4) 2 y dy = 192 mC
0
ρ dS
ρ dS | r − r' |
E = s 2 ar = s
4πε r
4πε o | r − r' |3
where r − r' = (0, 0,10) − ( x, y, z ) = (− x, − y,10).
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Principles of Electromagnetics, 6e
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81
E=
2
x =−2
2
12 | y |10−3 (− x, − y,10)
10−9 2 2
y =−2
4π (
)( x + y + 100)3/ 2
36π
2
− xdx dy a x
= 108(10 )[ | y | 2
+
( x + y 2 + 100)3/ 2 −2
−2
−2
6
2
2
+ 10 a z
2
2
(x
−2 −2
2
E = 108(107 ) a z
−2
= − 216(107 ) a z
2
2
− y | y | dy dx a y
2
+ y 2 + 100)3/ 2
(x
−2
| y | dx dy
]
+ y 2 + 100)3/ 2
1
d ( y2 )
2
[2 2
]dx
( x + y 2 + 100)3/ 2
0
2
2
[
−2
1
1
]dx
− 2
1/ 2
( x + 104)
( x + 100)1/ 2
= − 216 (107 ) a z ln |
2
x + x 2 + 104
x + x 2 + 100
|
2
−2
2 + 108
−2 + 108
) − ln(
))
2 + 104
−2 + 104
= − 216 (107 )a z (−7.6202 (10−3 ) )
= − 216 (107 )a z (ln(
E = 16.46 a z MV/m
P.E. 4.6
y = -3 plane
z
line
charge
O
y
P
x
x=2 plane
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Principles of Electromagnetics, 6e
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82
E1 and E2 remain the same as in Example 4.6.
E3 =
ρL
aρ
2πε 0 ρ
This expression, which represents the field due to a line charge,
is modified as follows. To get a ρ , consider the z = −1 plane.
ρ= 2
a ρ = a x cos 45° − a y sin 45°
1
(a x − a y )
2
10(10−9 ) 1
(a x − a y )
E3 =
10−9 2
2π (
)
36π
=
= 90 π (a x − a y ).
Hence,
E = E 1 + E 2 + E3
= − 180 π a x + 270 π a y + 90 π a x − 90 π a y
= −282.7 a x + 565.5a y V/m
P.E. 4.7
ρ
Q
a + s an
2 r
4π r
2
[(0, 4,3) − (0, 0, 0)] 10 × 10−9
+
ay
5
2
D = DQ + Dρ =
30 ×10−9
=
4π (5) 2
30
=
(0, 4,3) + 5 a y nC / m 2
500π
= 5.076a y + 0.0573a z nC / m 2
P.E. 4.8
(a) ρv = ∇ • D = 4 x
ρv (−1, 0,3) = −4 C/m3
1 1 1
(b) Ψ = Q = ρv dv = 4 xdxdydz
0 0 0
= 4(1)(1)(1/ 2) = 2 C
(c ) Q = Ψ = 2 C
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Principles of Electromagnetics, 6e
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83
P.E. 4.9
Q = ρ vdv = ψ =
For 0 ≤ r ≤ 10,
Dr ( 4π r 2 ) =
D • dS
2r
(r 2 ) sin θ dθ dr dφ
2r 4 r
) = 2π r 4
|
0
4
r2
E=
ar nV/m
2ε 0
Dr (4π r 2 ) = 4π (
Dr =
r2
2
E (r = 2) =
For r ≥ 10,
4(10−9 )
ar = 72π ar = 226 ar V/m
−9
2( 1036π )
Dr (4π r 2 ) = 2π r0 4 ,
Dr =
r0 4
2r 2
⎯⎯
→
r0 = 10m
E=
r0 4
ar nV/m
2ε 0 r 2
104 (10−9 )
E (r = 12) =
ar = 1250π ar
10−9
2(
)(144)
36π
= 3.927ar kV/m
P. E. 4.10
V (r ) = k =1
3
Qk
+C
4πε 0 | r − rk |
At V (∞) = 0,
C =0
| r − r1 | =| (−1,5, 2) − (2, −1,3) |= 46
| r − r2 |=| (−1,5, 2) − (0, 4, −2) |= 18
| r − r3 |=| (−1,5, 2) − (0, 0, 0) |= 30
10−6
−4
5
3
[
+
+
]
V (−1,5, 2) =
−9
10
46
18
30
4π (
)
36π
= 10.23 kV
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Principles of Electromagnetics, 6e
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84
P.E. 4.11
V=
Q
4πε 0 r
+C
If V (0, 6, −8) = V (r = 10) = 2;
5(10−9 )
+C
10−9
4π (
)(10)
36π
2=
(a)
5(10−9 )
VA =
10−9
4π (
)|( − 3,2,6) − (0,0,0)|
36π
= 3.929 V
(b)
45
VB =
(c)
7 2 + 12 + 52
C = −2.5
⎯⎯
→
− 2.5
− 2.5 = 2.696 V
V AB = VB − VA = 2.696 − 3.929 = − 1233
.
V
P.E. 4.12
(a)
−W
= E • dl = (3x 2 + y )dx + xdy
Q
=
2
2
(3x + y)dx | +
y =5
0
−1
x dy
5
|
x=2
= 18 − 12 = 6 kV
W = −6 Q = 12 mJ
(b)
dy = −3 dx
2
W
− = E • dl = (3x 2 + 5 − 3 x)dx + x(−3)dx
Q
0
2
= (3x 2 − 6 x + 5)dx = 8 − 12 + 10 = 6
0
W = 12 mJ
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Principles of Electromagnetics, 6e
87
85
P.E. 4.13
(a)
(0, 0,10)
V=
(r = 10, θ = 0, φ = 0)
⎯⎯
→
10−12
= 9 mV
10−9
4π (
)
36π
100 cos 0
(10−12 ) =
2
4πε 0 (10 )
100(10−12 )
[2 cos 0 ar + sin 0 aθ ]
10−9 3
4π (
)10
36π
=1.8 ar mV/m
E=
(b)
At (1,
π π
, ),
3 2
π
(10−12 )
3
V=
= 0.45 V
10−9
2
4π (
)(1)
36π
100 (10−12 )
π
π
E=
(2 cos ar + sin aθ )
−9
10
3
3
4π (
)(1) 2
36π
100 cos
= 0.9 ar + 0.7794aθ V/m
P.E. 4.14
After Q1 , W1 = 0
After Q2 , W2 = Q2V21 =
=
Q2 Q1
4 π ε 0 |(1,0 ,0 ) − (0 ,0 ,0 )|
1( − 2) (10 −18 )
= − 18 nJ
1
−9
4 π (10 )
36 π
After Q3 ,
W3 = Q3 (V31 + V32 ) + Q2V21
1
−2
= 3(9 )(10 −9 ) +
|(0 ,0 ,− 1) − (0 ,0 ,0 )| |(0 ,0 ,− 1) − (1,0 ,0 )|
2
= 27 (1 −
) − 18
2
= − 29.18 nJ
}
− 18 nJ
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Principles of Electromagnetics, 6e
88
86
After Q4 ,
W4 = Q4 (V41 + V42 + V43 ) + Q3 (V31 + V32 ) + Q2V21
1
3
−2
= − 4(9 )(10 −9 ) +
+
+ W3
|(0 ,0 ,1) − (0 ,0 ,0 )| |(0 ,0 ,1) − (1,0 ,0 ) |(0 ,0 ,1) − (0 ,0 ,− 1) 2 3
= − 36 (1 −
+ ) + W3
2 2
= − 39.09 − 29.18 nJ = − 68.27 nJ
P.E. 4.15
E = −∇V = −( y + 1)a x + (1 − x)a y − 2a z
At (1,2,3), E = −3a x − 2a z V/m
1 1 1
1
1
W = ε o E • Edv = ε o ( x 2 + y 2 − 2 x + 2 y + 6)dxdydz
2
2 −1 −1 −1
1
1
1
1
1 2
2
= ε o x dx dydz + y dy dxdz − 2 xdx dydz + 2 ydy dxdz +6(2)(2)(2) 2 −1
−1
−1
−1
80ε o
1 x3 1
= ε o 2
(2)(2) + 0 + 0 + 6(8) =
2 3 −1
3
= 0.2358 nJ
87
P.E. 4.16 (a)
E ( 5, 0, 6) =
Q1
[(5, 0, 6) − (4, 0, −3)]
Q2 [(5, 0, 6) − (2, 0,1)]
+
3
4 π ε 0 | (5, 0, 6) − (4, 0, −3) |
4 π ε 0 | (5, 0, 6) − (2, 0,1) |3
Q1 (1, 0,9)
Q2 (3, 0,5)
+
3
4 π ε 0 ( 82)
4 π ε 0 (34)3/ 2
Prob. 4.1 If Ez = 0, then
=
9 Q1
1
5 Q2
1
+
= 0
3/ 2
3/ 2
−12
4Qπ1Qε20(r(82)
(34)
)[(3,
2,1) − (−4, 0, 6)]
(7, 2, −5)
Q1 − r Q2 ) 4 π−ε20(10
0
FQ1 =
=
= − 180
× 10−3
3
−9
82
10
5
82
5
3
688.88
− r 3/ 2 = −4π 4 ( (3,
Q1 4=π ε− rQQ
)3/ 22,1)
nC− (−4, 0, 6)
1 2 ( Q2 )
9
34
9 36π34
= −8.3232 nC
= −1.8291a x − 0.5226 a y + 1.3065 a z mN
(b)
F (5, 0, 6) = qE (5, 0, 6)
If Fx = 0, then
qQ1
3qQ2
+
=0
3/ 2
4πε 0 (82)
4πε 0 (34)3/ 2
82 3/ 2
82
) = −12( )3/ 2 nC
34
34
Q1 = −44.945 nC
Q1 = −3Q2 (
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POESM_Ch04.indd 88
Prob. 4.2
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9 Q1
1
5 Q2
1
+
= 0
3/ 2
4 π ε 0 (82)
4 π ε 0 (34)3/ 2
5
82
5 82
Q1 = − Q2 ( )3/ 2 = − 4 ( )3/ 2 nC
9
34
9 34
= −8.3232 nC
89
Sadiku & Kulkarni
(b)
Principles of Electromagnetics, 6e
F (5, 0, 6) = qE (5, 0, 6)
If Fx = 0, then
qQ1
3qQ2
+
=0
3/ 2
4πε 0 (82)
4πε 0 (34)3/ 2
82 3/ 2
82
) = −12( )3/ 2 nC
34
34
95
Q1 = −44.945 nC
Q1 = −3Q2 (
P.E. 4.17
88
Prob.
4.2
22
'
e
Q [ (0, 0, 0) − (a, 0, 0)]
Q [ (0, 0, 0) − (−a, 0, 0)]
Fe = E = 2 arQ(r − rk ) =
−
'
3
3
4πε o | (0, 0, 0) − ( a, 0, 0) | 4πε o | (0, 0, 0) − ( −a, 0, 0) |3
(c) 4πε okr=1 4πε o | r − rk |
(a)
2 Q ( −Q
, 0, aQ) (−a(,a0,, 0,
, 0, a×)10
− −(19−a,20, 0) ]
a, [0,(a0)
0)10) ] Q Q [ (a1.6
Fe
eE
1
=
−
a
= =
= − −9 3 = 3
2 x a4)πε
− (oa, 0, 0)2| πε o4aπε
, 0,×a10
) −−31(−a, 0, 0) |3
44πε2 ooa|3(4aπ, 0,
o | ( a
Fg 4πε o Gm
9.1
×10
× 6.67 × 10−11
36π
Q(0, 0, a ) Q(2a, 0, a )
−Q
Q 1 a +
az
−
=
1−
= 4.17 ×=1042
3
2 3/ 2
2 2 x
πεao a, 0) − (4aπε
(5a] )
4πε a 5 5 Q [4(0,
, 0,o 0)
Q10
(0,5aπε
, 0)
[
o a− ( − a , 0, 0)o]
(b) E =
−
4πε o | (0, a, 0) − (a, 0, 0) |3 4πε o | (0, a, 0) − (− a, 0, 0) |3
Prob. 4.16
P.E. 4.18
QLet
(−aQ
, a1, be
0) located
Q (aat, athe
, 0) origin. −
Q the spherical surface of radius r,
At
(a) = 4πε (2a 2 )3/ 2 − 4πε (2a 2 )3/2 2 = 4 2πε a 2 a x
Q1o = DdS = ε Eo r (4π r )
o
QR1
QR2
=
−
E
Or
4πε o R13 4πε o R23
Q1
E point
=
by Gauss's
law
r x-axis
A
on athe
is (x,0,0).
4πε r 2
d)
( x, 0, 0)Q−2 is
(0,placed
0, d ) =on
( x,the
0, −spherical
R1 =charge
surface, Q2 experiences a force
If a second
3
2
2 3/ 2
R1 = ( x + d )
QQ
F
, 0,=0) −1(0,220,ar−d ) = ( x, 0, d )
R2 ==Q( x2 E
42πε3/ 2r
3
2
=
+
x
d
(
)
which isRCoulomb’s
law.
2
−2Qda z
Q
E=
( x, 0, −d ) − ( x, 0, d ) ] =
2
2 3/ 2 [
Prob. 4.17
4πε o ( x + d )
4πε o ( x 2 + d 2 )3/ 2
Q
For a point
charge,
D =z-axis 3isR(0, 0, z).
(b)
A point
along the
4π R
R1 = (0, 0, z ) − (0, 0, d ) = ( z − d )a z
For the given three point charges,
R 3 = ( z − d )3
1 1QR1 QR2 2QR3 + 3 −
D=
4π R13
R2
R33 R = (0, 0, z ) − (0, 0, −d ) = ( z + d )a z
R1 = (0, 0)2 − (−1, 0) = (1, 0), R1 = 1
3
= (1,
( z 0)
+ d=)3(−1, 0), R = 1
R = (0,R0)2 −
2
2
Q( z −= d(0,
)a−z 1), RQ=( z1+ d )a z
Qa z 1
1 0)=− (0,1)
R3 = (0,E
− 3
=
−
3
3
2
4πε o ( z − d ) 4πε o ( z +Qd )
4πε oQ ( z − d ) ( z + d ) 2 Q
(0, 2) =
ay
−1, 0) − 2(0, −1) ] =
D=
[(1, 0) + (Qdz
az
4π
4π
2π
=
πε o ( z 2 − d 2 )2
Prob. 4.3
F = qE = mg
POESM_Ch04.indd 89
Copyright © 2015mg
by Oxford
2 University
× 9.8 Press
⎯⎯
→ E=
q
=
4 × 10−3
= 4.9 kV/m
9/27/2015 9:07:31 PM
R1 = ( x, 0, 0) − (0, 0, d ) = ( x, 0, − d )
R13 = ( x 2 + d 2 )3/ 2
Sadiku & Kulkarni
93
Principles of Electromagnetics, 6e
R2 = ( x, 0, 0) − (0, 0, −d ) = ( x, 0, d )
1 R 3 = ( x 2 + d 2 )3/ 2
1
(−10 + 5 − 20)(a z )10 −9 = −18π (25)a z
[2-10(az ) + 5(az ) + 20(−az ] ×10−9 =
−9
10
2ε o
90
−2Qda z
Q
E=
−
−
x
d
x
(
,
0,
)
(
,20,×d36
) ]π=
[
2
2 3/ 2
4πε o ( x + d )
4πε o ( x 2 + d 2 )3/ 2
= −1.4137a z kV/m
(b)
A point along the z-axis is (0, 0, z).
For z > 2,
R1 = (0, 0, z ) − (0, 0, d ) = ( z − d )a z
1 3
3 a ) + 20a × 10 −9 = 18π ( −10 + 5 + 20)(a )
+)5(
E = R[1 -10(
= ( za−
z )d
z
z]
z
2ε o
E=
= 848.23a z V/m
R = (0, 0, z ) − (0, 0, −d ) = ( z + d )a z
Thus, 2
R 3 = ( z + d )3
−2848.23a z V/m, z <0
-1.979Qa( z kV/m,
− d )a z 0<z<1
Q ( z + d )a z
Qa z
E = E = 4πεz ( z − d )3 − 4πε ( z + d )3 = 4πε
o
o
o
−1.4137a z kV/m, 1 <z <2
Qdza z z >2
848.23
= a z V/m,
πε o ( z 2 − d 2 )2
Prob.
4.3
P.E. 4.19
F = qE = mg
⎯⎯
→ E=
y
Prob. 4.4
5
1
1 ( z − d )2 − ( z + d )2 mg 2 × 9.8
=
= 4.9 kV/m
q
4 × 10−3
5
Q = ρ L dl = 12 x 2 dx = 4 x3 | mC = 0.5 C
(a)
0
0
an
x
Let f ( x, y ) = x + 2 y − 5;
∇f = a x + 2 a y
∇f
(a x + 2 a y )
= ±
| ∇f |
5
Since point (−1, 0,1) is below the plane,
an = ±
_
an = −
(a x + 2 a y )
.
5
ρs
6(10−9 )
(a x + 2 a y )
E=
an =
(−
)
−9
2ε 0
2(10 / 36π )
5
= −151.7 a x − 303.5 a y V/m
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POESM_Ch04.indd 90
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4 π ε 0 (82)3/ 2
4 π ε 0 (34)3/ 2
5
82
5 82
Q1 = − Q2 ( )3/ 2 = − 4 ( )3/ 2 nC
9
34
9 34
88
= −8.3232 nC
91
Sadiku & Kulkarni
(c)
Principles of Electromagnetics, 6e
Q [ (a, 0, a ) − (a, 0, 0) ]
Q [ (a, 0, a ) − (−a, 0, 0) ]
(b)
E=
−
3
Prob. 4.1
πε6)o |=(aq, E
0,(5,
a ) 0,
− (6)
a, 0, 0) | 4πε o | (a, 0, a ) − (− a, 0, 0) |3
F (5,40,
If FxQ=(0,
0,0,then
a ) Q(2a, 0, a )
−Q
Q 1 = Q1Q2 (3rQ −− r Q2 )
= −12 )[(3, 2,1)2 −a(x −+4, 0, 6)]2 1 −
2
3/
2
a2,z −5)
−
20(10
(7,
3qQ
(5
4πε o a = − 180
πε
10
5
a
5
5
FQ1 =qQ41πε o a +1 4πε
=
× 10−3
2 a )
o
o
3
−9
0
=
688.88
4π 3/ε 2 rQ1 −4πε
rQ20 (34)3/ 24π 10 (3, 2,1) − (−4, 0, 6) 3
4πε 0 (82)
36π
82 3/ 2
82 3/ 2
P.E. 4.18
Q1 = −3Q2 ( ) = −12( ) nC
(a)
= −1.8291
34 a x − 0.522634a y + 1.3065 a z mN
Q
QR
R
Q1== −44.945
1
− nC 2 3
E
3
4πε o R1 4πε o R2
A point on the x-axis is (x,0,0).
Prob. 4.2
'
2
Q [ (0, 0, 0) − (−a, 0, 0)]
− d0)) − (a, 0, 0)]
( x,Q
0,(0)
0, d ) =Q( x[ (0,
, 0, 0,
R =
r −−r(0,
k)
E =13 2
=
−
' 3
4πε o | (0, 0, 0) − ( a, 0, 0) |3 4πε o | (0, 0, 0) − ( −a, 0, 0) |3
R1 k ==1 (4xπε+o |dr2 −)3/r2k |
(a)
, 0,0)0)− (0,Q0,
(a−, d0,)0)
= (−x,a0,
= (=x, 0,Q
d) a
R=2 Q
−
x
3
3
3
2
2
3/
2
4πε o a
2πε o a 2
o ad )
R2 = 4( xπε +
−2Qda z
Q
E=
( x, 0, −d ) − ( x, 0, d ) ] =
[
2
2
3/
2
2
Q [ (0, a, 0) − (a, 0, 0) ]
Q [ (0, a, 0) − 4(−
πεao, 0,
+] d 2 )3/ 2
( x0)
(b) E = 4πε o ( x + d )
−
3
3
4πε o along
| (0, a, the
0) −z-axis
(a, 0, 0)
4πε
(b)
A point
is |(0, 0,
z).o | (0, a, 0) − (− a, 0, 0) |
Q(−a, a, 0)
Q (a, a, 0)
−Q
= R1 = (0,2 0,3/ z2 )−− (0, 0, d ) 2= 3/( z2 −=d )a z
a
2 x
4πε
4 2πε88
oa
R13o =(2(az −) d )3 4πε o (2a )
(c)
R2 = (0, 0, z ) − (0, 0, −d ) = ( z + d )a z
Q [ (a, 0, a ) − (a, 0, 0) ]
Q [ (a, 0, a ) − (−a, 0, 0) ]
R23== ( z + d )3
E
−
3
4πε o | (a, 0, a ) − (a, 0, 0) | 4πε o | (a, 0, a ) − (− a, 0, 0) |3
Q ( z − d )a
Q ( z + d )a
Qa 1
1 −
E = Q(0, 0, a ) z 3 −Q(2a, 0, a ) z 3 = −Qz 1 2 Q
) 4πε o (2z +3/ 2d )= 4πε o (2z a−xd+) ( z +2 d1)−2 az
=4πε o ( z −3 d −
4πε o a
4πε o (5a )
4πε o a 5 5 10 5πε o a
Qdza z
=
πε o ( z 2 − d 2 )2
P.E. 4.18
(a)
Prob. 4.3
QR1
QR2
−
E=
mg 2 × 9.8
3
3
4πε=o Rmg
4πε⎯⎯
F = qE
E=
=
= 4.9 kV/m
1
2
o R→
q
4 × 10−3
A point on the x-axis is (x,0,0).
Prob. 4.4
R1 = ( x, 0, 0) −5 (0, 0, d ) = ( x, 0, −5 d )
(a)
Q3= ρ2L dl =2 3/122 x 2 dx = 4 x3 | mC = 0.5 C
R1 = ( x + d )
0
0
R2 = ( x, 0, 0) − (0, 0, −d ) = ( x, 0, d )
R23 = ( x 2 + d 2 )3/ 2
E=
(b)
−2Qda z
Q
( x, 0, −d ) − ( x, 0, d ) ] =
2 3/ 2 [
4πε o ( x + d )
4πε o ( x 2 + d 2 )3/ 2
2
A point along the z-axis is (0, 0, z).
R1 = (0, 0, z ) − (0, 0, d ) = ( z − d )a z
R13 = ( z − d )3
Copyright © 2015 by Oxford University Press
POESM_Ch04.indd 91
R2 = (0, 0, z ) − (0, 0, −d ) = ( z + d )a z
3
3
9/27/2015 9:07:33 PM
Sadiku & Kulkarni
Principles of Electromagnetics, 6e
92
89
(b)
Q = ρ S dS =
2π
4
z =0
z3 4
ρ z ρ dφ dz ρ |=3 = 9(2π ) 3 0| nC
φ =0
2
= 1.206 μ C
(c)
10
r 2 sin θ dθ dφ dr
r sin θ
2π
4
π
42
= 10 dφ dθ rdr =10(2π ) (π )
2
0
0
0
Q = ρV dV = = 1579.1 C
Prob. 4.5
x 2 a a3 ρo
dxdydz = ( a)( a) ρ o =
a
2
2a 0 a a a
ρo x
Q = ρ v dv = v
0 0 9
Prob. 4.6
Q = ρ v dv =
v
2
1
φ π
ρ
5ρ 2 z ρdφ d ρ dz mC
=0 z =0 = / 6
ρ 4 2 z2 1
=5
π /2
4 0 2 0
φ
π /2 5
10π
= (16)(1)(π / 2 − π / 6) =
π /6 8
3
Q = 10.472 mC
Prob. 4.7
Q = ρ s dS = 6 xydxdy
=
x
2
x =0
y =0
2
=6
6 xydxdy +
x =0
x
4
− x+4
x=2
y =2
6 xydxdy
y x
y2 −x + 4
dx + 6x
dx
0
2 0
2
x=2
4
2
2
4
x2
= 6 x( − 0)dx + 3 x (4 − x) 2 − 0 dx
2
x =0
x=2
2
= 6
0
4
x3
dx + 3 (16 x − 8 x 2 + x 3 )dx
2
2
Copyright © 2015 by Oxford University Press
POESM_Ch04.indd 92
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Sadiku & Kulkarni
Principles of Electromagnetics, 6e
93
90
=3
x4 2
8 x3 x 4 4
+ 6(8 x 2 −
+ )
4 0
3
4 2
= 12 + 3(128 − 32 −
= 12 + 3(96 −
Q = 32 C
512 64
+ + 64 − 4)
3
3
448
+ 60)
3
Prob. 4.8
Q = ρ v dv
v
=
1
1
1
6x
−1
2
y 2 dxdydz
−1 −1
1
1
1
−1
−1
−1
= 6 dz x 2 dx y 2 dy
= 6(2)
x 1 y3 1
3 −1 3 −1
3
12
48
(1 − −1)(1 − −1) =
nC
9
9
= 5.33 nC
=
Prob. 4.9
Q = ρ v dv = 4 ρ 2 z cos φρ d ρ dφ dz
nC
v
2
1
π /4
0
0
0
= 4 ρ 3 d ρ zdz
cos φdφ = ρ 4
2 z2 1
π /4
(sin φ )
0 2 0
0
= (16)(0.5)(sin π / 4) = 5.657 nC
Copyright © 2015 by Oxford University Press
POESM_Ch04.indd 93
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Sadiku & Kulkarni
Principles of Electromagnetics, 6e
94
91
Prob. 4.10
For 2 < r < 4cm
Q = ρ v dv = ρv
v
2π
π
4cm
φ θ
5r 2 sin θ drdθ dφ
= 0 = 0 r = 2cm
4cm
= 5(4π ) ρv
r 2 dr
2cm
= 20πρv
r3
3
4cm
2cm
1
= 20π × 5 × (43 − 23 ) ×10−6 mC
3
100π
(64 − 8)nC
3
= 5864.3 nC
Q = 5.864 μ C
=
Prob. 4.11
ρ dl
E = L 2 aR
4πε o R
L
R = − aa ρ + ha z , R =| R |= a 2 + h 2 , dl = adφ
E=
ρ L ( − aa ρ + ha z )
adφ
4πε o (a 2 + h 2 )3/ 2
Due to symmetry, the ρ-component cancels
ρ 2π haa dφ
ρ L haa z
(2π )
E = L 2 z 2 3/ 2 =
4πε o 0 (a + h )
4πε o (a 2 + h 2 )3/ 2
out.
4 ×10−3 × 12 ×10−6 × 4 × 3a z
(2π ) = 260.58a z N
F = QE =
10−9 3
4π ×
×5
36π
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POESM_Ch04.indd 94
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Sadiku & Kulkarni
Principles of Electromagnetics, 6e
95
92
Prob. 4.12
(a) At P(5,-1,4),
ρ sk
10 × 10−6
−20 ×10−6
30 × 10−6
(
)
+
(
)
+
( −a z )
ank =
a
a
x
y
10−9
10−9
10−9
k =1 2ε o
2×
2×
2×
36π
36π
36π
3
= 36π (5, −10, −15) ×10 = 565.5a x − 1131a y −1696.5a z kV/m
3
E =
(b) At R(0,-2,1)
E = 36π 5(−a x ) − 10(a y ) + 15(−a z ) × 103 = −565.5a x − 1131a y −1696.5a z kV/m
(c) At Q(3,-4,10),
E = 36π 5a x −10(−a y ) + 15a z ×103 = 565.5a x + 1131a y +1696.5a z kV/m
Prob. 4.13
We apply E =
ρs
an
2ε o
For z<0,
1
E=
[ -10(-az ) + 5(-az ) + 20(−a z ] ×10−9 =
2ε o
= 18π ×15(−a z ) = −848.23a z V/m
For 0<z<1,
1
E=
[ -10(az ) + 5(-a z ) + 20(−az )] ×10−9 =
2ε o
= −18π × 35(a z ) = −1.979a z kV/m
For 1 < z < 2,
1
(−10 + 5 + 20)(−a z )10−9
−9
10
2×
36π
1
( −10 − 5 − 20)(a z )10−9
10−9
2×
36π
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POESM_Ch04.indd 95
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Sadiku & Kulkarni
Principles of Electromagnetics, 6e
96
93
E=
1
[ -10(az ) + 5(az ) + 20(−az ] ×10−9 =
2ε o
= −1.4137a z kV/m
1
(−10 + 5 − 20)(a z )10 −9 = −18π (25)a z
−9
10
2×
36π
For z > 2,
1
E=
[ -10(az ) + 5(az ) + 20az ] ×10−9 = 18π (−10 + 5 + 20)(a z )
2ε o
= 848.23a z V/m
Thus,
−848.23a z V/m, z <0
-1.979a kV/m, 0<z<1
z
E=
−1.4137a z kV/m, 1 <z <2
848.23a z V/m, z >2
94
Prob. 4.14
P.E. 4.19
y
− ρs
ρs
an
x=0
x=a
x
x
Let f ( x, y ) = x + 2 y − 5; ∇f = a x + 2 a y
(a) For x < 0,
∇f
ρ (a x + 2 a y()− ρ s )
aEn == E
±1 + E2 == ±s (−a x ) +
( −a x ) = 0
| ∇f | 2ε o
5 2ε o
Since
−1,a,.0,1) is below the plane,
(b)
Forpoint
0 < x(<
ρ
(_− ρ s )
ρ
(−a x ) = s a x
E = s ax +
2ε (a x + 2 a2yε)o
εo
an = − o
.
5
(c) For x > a,
ρ
(− ρ s )
E = ρs a x +
(a x−9) )= 0 (a + 2 a )
6(10
x
y
E = 2εso a n = 2ε o −9
(−
)
2ε 0
2(10 / 36π )
5
Prob. 4.15
= −151.7 a x − 303.5 a y V/m
ρ s dS
R, R = ρ (-a ρ ) + ha z , dS = ρ dφ d ρ
4πε o R 3
S
ρ
ρ dφ d ρ
E= s 2
(- ρ a ρ + ha z )
4πε o S ( ρ + h 2 )3/ 2
Due to symmetry, the ρ -component vanishes.
E=
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POESM_Ch04.indd 96
2π
ρ ha b
E = s z ρ ( ρ 2 + h 2 ) −3/ 2 d ρ dφ
4πε
9/27/2015 9:07:35 PM
(b) For 0 < x < a,.
ρ
ρ
(− ρ s )
E = s ax +
(−a x ) = s a x
2ε o
2ε o
εo
Sadiku & Kulkarni
(c) For x > a,
ρ
(− ρ s )
E = s ax +
(a x ) = 0
2ε o
2ε o
Principles of Electromagnetics, 6e
97
Prob. 4.15
ρ s dS
R, R = ρ (-a ρ ) + ha z , dS = ρ dφ d ρ
4πε o R 3
S
ρ
ρ dφ d ρ
E= s 2
(- ρ a ρ + ha z )
4πε o S ( ρ + h 2 )3/ 2
Due to symmetry, the ρ -component vanishes.
E=
E=
2π
ρ s ha z b
2
2 −3/ 2
ρ
ρ
ρ
+
(
)
h
d
0 dφ
4πε o a
95
ρ 2 + h 2 , du = 2 ρ d ρ
Let u=
P.E.
4.17
ea2
−1/ 2
b
ρ
ρ s ha z h
1
F
a 1 u −3/ 2 du = ρ s ha z 1/ 2u
=
s
2 πr )
Ee = 4πε zr (2
=
−
2
4πε oo
2ε o −1/ 2
2ε o ρ2 2 + h 2 a 1.6 ×10−19 Fe
e2
1
1
= ρs h =
1
1
−11 9.1× 10 −31
E
Fg = 4πε o Gm22 42 π−×102−9 2 ×a6.67
z
×
10
4πε o a + h
b +36
hπ = 4.17 × 1042
Prob. 4.16
Let Q1 be located at the origin. At the spherical surface of radius r,
Q1 = DdS = ε Er (4π r 2 )
Or
Q1
ar by Gauss's law
4πε r 2
If a second charge Q2 is placed on the spherical surface, Q2 experiences a force
E=
Q1Q2
ar
4πε r 2
which is Coulomb’s law.
F = Q2 E =
Prob. 4.17
Q
R
4π R 3
For the given three point charges,
For a point charge,
D=
1 QR1 QR2 2QR3 + 3 −
R2
R33 4π R13
R1 = (0, 0) − (−1, 0) = (1, 0), R1 = 1
D=
R2 = (0, 0) − (1, 0) = (−1, 0), R2 = 1
R3 = (0, 0) − (0,1) = (0, −1), R3 = 1
D=
Q
Q
Q
[(1, 0) + (−1, 0) − 2(0, −1)] = (0, 2) = a y
4π
4π
2π
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POESM_Ch04.indd 97
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Prob. 4.17
Sadiku & Kulkarni
Q
R
4π R 3
For the given three point charges,
Principles of Electromagnetics, 6e
D=
For a point charge,
98
QR1 QR2 2QR3 3 + 3 −
R2
R33 R1
R1 = (0, 0) − (−1, 0) = (1, 0), R1 = 1
1
D=
4π
R2 = (0, 0) − (1, 0) = (−1, 0), R2 = 1
R3 = (0, 0) − (0,1) = (0, −1), R3 = 1
D=
Q
Q
Q
[(1, 0) + (−1, 0) − 2(0, −1)] = (0, 2)96= a y
4π
4π
2π
Prob. 4.18
(a) Assume for now that the ring is placed on the z=0 plane.
z
(0,0,h)
R
y
a
ρ L dl
x
D=
ρ L dl R
4π R
ρL
D=
4π
φ = 2π
φ
=0
3
, R = − a aρ + h az
adφ ( − a a ρ + h a z )
(a 2 + h 2 )3/ 2
Due to symmetry, the ρ component vanishes.
ρ L a (2 π h) a z
ρL a h a z
=
2
2 3/ 2
4 π (a + h )
2(a 2 + h 2 )3/ 2
a = 2, h = 3, ρ = 5 μ C/m
D=
L
_
Since the ring is actually placed in x = 0, a z becomes a x .
D=
(6)(5) a x
= 0.32 a x μ C/m 2
2(4 + 9)3/ 2
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Principles of Electromagnetics, 6e
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97
(b)
Q [(3, 0, 0) − (0, −3, 0)]
Q [(3, 0, 0) − (0,3, 0)]
+
3
4 π | (3, 0, 0) − (0, −3, 0) |
4 π | (3, 0, 0) − (0,3, 0) |3
Q(3,3, 0) Q(3, −3, 0)
6 Q (1, 0, 0)
=
+
=
3/ 2
3/ 2
4π (18)
4π (18)
4 π (18)3/ 2
D = DR + DQ = 0
DQ =
6Q
=0
4 π (18)3/ 2
0.32(10−6 ) +
∴ Q = − 0.32(4π )(183/ 2 )10−6
1
= −51.182μ C
6
Prob. 4.19
∂Dx ∂Dy ∂Dz
+
+
= 8 y C/m 2
∂x
∂y
∂z
1 ∂
1 ∂Dφ ∂Dz
+
(b) ρ v = ∇D =
( ρ Dρ ) +
ρ ∂ρ
ρ ∂φ
∂z
= 8sin φ − 2sin φ + 4 z
ρv = ∇D =
(a)
= 6sin φ + 4 z C/m3
1 ∂ 2
1
∂
(r Dr ) +
(Dθ sinθ ) + 0
2
r ∂r
r sin θ ∂θ
2
2cosθ
= − 4 cosθ +
r
r4
= 0
ρ v =∇ D =
(c)
Prob. 4.20
(a)
ρ v = ∇ D =
∂Dx ∂Dy ∂Dz
+
+
= 2(1 + z 2 ) + 0 + 2 x 2
∂x
∂y
∂z
= 2(1 + x 2 + z 2 ) nC/m3
(b)
ψ = DdS =
S
=2
3
2
2 x 2 zdxdy
y =0 x =0
z =1
2
3
0
0
= 2(1) x 2 dx dy
x 2
(3) = 16 nC
3 0
3
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Principles of Electromagnetics, 6e
100
98
Prob. 4.21
Gaussian surface
r
a
b
Apply Gauss’s law,
DdS = Q
enc
Dr 4π r 2 = ρ s1 4π a 2 + ρ s 2 4π b 2 = 8 × 10−9 × 4π (1) 2 + (−6 × 10−3 ) × 4π (2)2 = −0.3016
Dr =
−0.3016 = 0.3016
=
= −0.0027
4π r 2
4π (3) 2
D = -2.7ar mC/m 2
Prob. 4.22
For r <a.
DdS = Q
enc
S
= ρv dv
v
π
2π
r
0
0
0
Dr (4π r ) = 5r r sin θ dθ dφ dr = 5 sin θ dθ dφ r 5/ 2 dr
2
1/ 2 2
= 5(2)(2π )
r 40π r
r
=
7/2 0
7
7/ 2
7/2
40π r 7 / 2
10
7
Dr = ε o Er =
= r 3/ 2
2
4π r
7
10 3/ 2
Er =
r ,0 < r < a
7ε o
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Principles of Electromagnetics, 6e
101
99
For r > a,
40 7 / 2
πa
7
40 7 / 2
πa
7
Dr = ε o Er =
4π r 2
10a 7 / 2
,r > a
Er =
7ε o r 2
Thus,
10 3/ 2
r ar , 0 < r < a
7ε o
E=
7/2
10a a , r > a
7ε o r 2 r
Dr (4π r 2 ) =
Prob. 4.23
(a) ρ v = ∇ • D =
∂Dx ∂Dy ∂Dz
+
+
= 2 y C/m3
∂x
∂y
∂z
(b) Ψ = D ⋅ dS = x 2 dxdz
1
y =1
1
= x 2 dx dz =
0
0
1
1
1
0
0
0
1
C
3
(c) Q = ρv dv = 2 ydxdydz = 2 dx ydy dz = 1 C
v
Prob. 4.24
(a)
ρV = ∇ • D =
1 ∂
( ρ Dρ ) +
1 ∂ Dφ
ρ ∂ρ
ρ ∂φ
ρV = 4 ( z + 1) cos φ − ( z + 1) cos φ + 0
+
∂ Dz
∂z
ρV = 3( z + 1) cos φ μ C/m3
(b)
Qenc = ρV dv = 3( z + 1) cos φ ρ dφ d ρ dz
2
4
π /2
0
0
0
= 3 ρ d ρ ( z + 1) cos φ dφ = 3(2)(
π /2
4
z2
+ z )| (sin φ| )
0
0
2
= 6(8 + 4)(1 − 0) = 72μ C
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Principles of Electromagnetics, 6e
102
100
(c)
Let ψ = ψ 1 +ψ 2 +ψ 3 +ψ 4 +ψ 5 = D • dS
where ψ 1 , ψ 2 , ψ 3 , ψ 4 , ψ 5 respectively correspond witn surfaces S1 ,S2 ,S3 ,S4 ,S4
(in the figure below) respectively.
y
S3
S4
S5
S1
y
x
S2
For S1 ρ = 2, dS = ρ dφ dza ρ
4
π /2
0
0
ψ 1 = 2 ρ ( z + 1) cos φ dS|ρ = 2 = 2(2)2 ( z + 1)dz
cos φ dφ
= 8(12)(1) = 96
For S 2 , z = 0, dS = ρ dφ d ρ (− a z )
2
π /2
0
0
ψ 2 = − ρ 2 cos φ ρ dφ d ρ = − ρ 3d ρ
ρ
4
cos φ dφ
| (1) = −4
4 0
For S3, z = 4, dS = ρ dφ d ρ a z , ψ 3 = +4
=−
2
For S4 , φ = π / 2, dS = d ρ dzaφ
2
4
0
0
ψ 4 = − ρ ( z + 1) sin φ d ρ dz|d =π / 2 = (11 ρ d ρ ( z + 1)dz
=−
ρ2
2
|
2
0
(12) = −(2)(12) = −24
For S5 , φ = 0, dS = d ρ dz (−aφ ) ,ψ 5 = ρ ( z + 1) sin φ d ρ dz|
φ =0
=0
ψ = 96 − 4 + 4 − 24 + 0 = 72μ C
This is exactly the answer obtained in part (b).
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Principles of Electromagnetics, 6e
103
101
Prob. 4.25
e
F = eE
e
ρ0 =
4π
3e
4 π R3
=
3
R
3
ρ 0,
0<r<R
0,
elsewhere
ρV = D • dS = Qenc = ρV dV =
Er =
R
3 e 4π r 3
= Dr (4π r 2 )
3
4π R
3
3e r
12πε 0 R 3
F = eE =
Prob. 4.26
(a)
ψ = Qenc
e2 r
4π ε 0 R 3
at r = 2
10
Qenc = ρV dV =
= 10
2
r
2π
2
r 2 sin θ dθ dr dφ
π
sin θ dθ drdφ
r =1 φ = 0 θ = 0
= 10 (1) (2π ) (2) = (40 π ) mC
Thus, ψ = 125.7 mC
At r = 6;
Qenc. = 10
4
r =1
dr
2π
φ =0
dφ
π
sin θ dθ
θ =0
= 10 (3)(2π ) (2) = 120 π mC
ψ = 377 mC
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Principles of Electromagnetics, 6e
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102
(b)
ψ = Qenc
But ψ =
At
r = 1,
D • dS =
⎯⎯
→ D=0
Qenc = 0
At r = 5,
Dr =
Dr dS = Dr (4π r 2 )
Qenc = 120 π
Qenc
120 π
=
= 1.2
2
4π r
4 π (5) 2
D = 1.2 ar mC/m 2
Prob. 4.27
ρv =
Q
Q
3Q
=
=
3
volume 4π a / 3 4π a 3
For r < a,
D dS = Q
enc
= ρv dv
3Q 4π r 3 Qr 3
= 3
4π a 3 3
a
For r > a, DdS = Q
Dr 4π r 2 =
Dr 4π r 2 = Q
Hence,
⎯⎯
→
Qr
4π a 3 ar ,
D=
Q a,
4π r 2 r
Dr =
⎯⎯
→
Dr =
Qr
4π a 3
Q
4π r 2
r<a
r>a
Prob. 4.28
VP =
10−9 2
4
−
−9 10 | (1, −2,3) − (1, 0,3) | | (1, −2,3) − (−2,1,5) | 4πε o r1 4πε o r2
4π ×
36π
4
2
= 9 −
= 1.325 V
9+9+4
2
Q1
+
Q2
=
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Principles of Electromagnetics, 6e
105
103
Prob. 4.29
Q
,
V =4
4πε o r
V=
r = a 2 + a 2 + h 2 = 22 + 22 + 32 = 17 cm
4 × 8 × 10−9
= 6.985 kV
10−9
−2
× 17 × 10
4π ×
36π
Prob. 4.30 (a)
Q/2
2
V=
4π ε0 r
=
(b)
(c)
Q
2
=
Q/2
Q
4π ε0 r
60(10−6 )
= 135 kV
10−9
4π ×
x4
36π
Q
3( )
3 = 135 kV
V=
4π ε0 r
Q
2π (4)
ρ dl
Q
V= L
= 8π
= 135 kV
4π ε0 r
4π ε 0 r 4π ε0 r
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Principles of Electromagnetics, 6e
106
104
Prob. 4.31
(a)
VP =
Qk
4π | r p − r k |
10−3
3(10−3 )
−2 (10−3 )
4 π ε oV p =
+
+
| (−1,1, 2) − (0, 0, 4) |
| (−1,1, 2) − (−2,5,1) |
| (−1,1, 2) − (3, −4, 6) |
1
2
3
1
2
3
−
+
=
−
+
4 π ε 0 (103 ) V p =
| (−1,1, −2) | | (1, −4,1) | (−4,5, −4) |
6
18
57
10−9
(103 ) V p = 0.3542
36 π
4π
∴V p = 3.008 × 106 V
(b)
VQ =
Qk
4 πε o | r p − r k |
10−3
3(10−3 )
−2 (10−3 )
+
+
| (1, 2,3) − (0, 0, 4) |
| (1, 2,3) − (−2,5,1) |
| (1, 2,3) − (3, −4, 6) |
1
2
3
1
2
3
−
+
=
−
+
4 π ε 0 (103 ) V p =
| (1, 2, −1) | | (3, −3, 2) | (−2, 6, −3) |
6
22 7
4 π ε oVQ =
10−9
(103 ) V p = 0.410
36π
VQ = 3.694 (106 )V
4π
∴VPQ = VQ − VP = 0.686 (106 ) = 686 kV
Prob. 4.32
V=
ρ S dS
;
S 4 πε r
0
ρS =
1
1
ρ
; dS = ρ dφ d ρ ;
( ρ dφ d ρ )
1
ρ
=
V=
2
2 1/ 2
4 πε 0 ( ρ + h )
4 πε 0
1
2π
a
dρ
( ρ + h2 )
=0
dφ ρ
0
r = ρ 2 + h2
2
=
a
2π
1
ln( ρ + ρ 2 + h 2 )| =
[ln(a + a 2 + h 2 ) − ln h]
ρ =0
4 πε 0
2ε 0
=
1
a + a 2 + h2
ln
2ε 0
h
Copyright © 2015 by Oxford University Press
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Principles of Electromagnetics, 6e
107
105
Prob. 4.33
(a)
∂V
∂V
∂V
E = −(
ax +
ay +
az )
∂x
∂ y
∂z
= − 2 xy ( z + 3)a x − x 2 ( z + 3) a y − x 2 y a z
x = 3, y = 4, z = −6,
At (3, 4, −6),
E = − 2(3)(4)(−3) a x − 9 (−3)a y − 9(4) a z
= 72 a x + 27 a y − 36a z V/m
(b)
ρV = ∇ • D = ε 0∇ • E = − ε 0 (2 y ) ( z + 3)
ψ = Qenc =
ρ
V
1
= − 2ε 0 dx
0
dV = − 2ε 0 y ( z + 3)dx dy dz
1
1
0
0
y dy ( z + 3)dz = − 2ε 0 (1)(1/ 2)(
1
z2
+ 3z ) |
0
2
1
−7 10−9
= − ε 0 ( + 3) =
(
)
2
2 36 π
Qenc = − 30.95 pC
Prob. 4.34
(a)
r2 Q = ρ v dv = ρ o 1 − 2 r 2 sin θ dθ dφ dr
a v
2π
π
a
2 r4 a 3 a 3 8π 3
= ρo dφ sin θ dθ r − 2 dr = ρo (2π )(2) − =
a ρo
a 3 5 15
0
0
0
(b) Outside the nucleus, r >a,
Qenc
→
E=
ar
S DdS = Qenc ⎯⎯
4πε o r 2
8π a 3 ρ o
2a 3 ρ o
E = 15 2 ar =
ar
4πε o r
15ε o r 2
V = − E dl = − Er dr =
Since
V=
2a 3 ρ o
+ C1
15ε o r
V (∞) = 0, C1 = 0.
2a 3 ρ o
15ε o r
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Principles of Electromagnetics, 6e
108
106
(c) Inside the nucleus, r <a
r3 r5 Qenc = 4πρo − 2 3 5a ρo r r 3 Qenc
E=
a
=
−
a
r
4πε o r 2
ε o 3 5a 2 r
V = − Er dr = −
ρo r 2
r4 −
+C
ε o 6 20a 2 2
2a 2 ρ o ρ o a 2 a 2 = − + C2
V (r = a ) =
15ε o
ε o 20 6 2a 2 ρ o 7 a 2 ρ o a 2 ρ o
+
=
C2 =
15ε o
60ε o
4ε o
V=
ρo r 4
r 2 a 2 ρo
−
+
ε o 20a 2 6 4ε o
(d) E is maximum when
ρo 1 3r 2 dE
=0=
−
dr
ε o 3 5a 2 r=
⎯⎯
→ 9 r 2 = 5a
5
a = 0.7454a
3
We are able to say maximum because
2
d 2E
6r
= − 2 < 0.
2
dr
5a
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Principles of Electromagnetics, 6e
109
107
Prob. 4.35
z
h
R
b
ρ
y
a
x
D=
ρs =
D=
ρ s dSR
,
4π R 3
R = − ρ a ρ + ha z ,
R =| R | = ρ 2 + h 2 , dS = ρ dφ d ρ
Q
Q
=
2
S π (b − a 2 )
ρ s ρ d φ d ρ ( − ρ a ρ + ha z )
4π
( ρ 2 + h 2 )3/ 2
Due to symmetry, the component along aρ vanishes.
b
ρ s h b 2π ρ d φ d ρ
ρs h
=
(2π ) ( ρ 2 + h 2 ) −3/ 2 ρ d ρ
Dz =
2
2 3/ 2
4π ρ = a φ =0 ( ρ + h )
4π
a
−1 b ρ s h 1
1
=
−
2
2 ρ + h 2 a
2 a2 + h2
b2 + h2 1
1
Qh
−
D=
az
2π (b 2 − a 2 ) a 2 + h 2
b2 + h2 =
ρs h Copyright © 2015 by Oxford University Press
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Principles of Electromagnetics, 6e
110
108
Prob. 4.36
∂Dx ∂Dy ∂Dz
+
+
= 4 − 20 y + 2 z
∂x
∂y
∂z
At P(1,2,3), x=1, y= 2, z=3
ρ v = ∇ D =
ρv = 4 − 20(2) + 2(3) = −30 C/m3
Prob. 4.37
r=3cm
r=5cm
For r < 3cm, Qenc = 0
For 3 < r < 5cm,
DdS = Q
enc
⎯⎯
→ D=0
= 10 nC
Dr 4π r 2 = 10 nC
⎯⎯
→ Dr =
For r > 5 cm,
DdS = Q
enc
Thus,
= 10 -5 = 5nC
10
nC/m 2
2
4π r
⎯⎯
→ Dr =
5
nC/m 2
2
4π r
r < 3 cm
0,
10
D=
a nC/m 2 , 3 < r < 5 cm
2 r
4π r
5
2
r > 5 cm
4π r 2 ar nC/m ,
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Principles of Electromagnetics, 6e
111
109
Prob. 4.38
1 ∂ ε o Eo ρ 2 ρ ∂ρ a ρv = ∇ D = ∇ε o E =
=
2ε o Eo
,0 < ρ < a
a
Prob. 4.39
Let us choose the following path of two segments.
(2,1, −1) → (5,1, −1) → (5,1, 2)
W = − q E • dl
5
W
− = E • dl = 2 xyzdx
q
x=2
2
+
z = −1, y = 1
x 2 ydz
z =−1
x = 5, y = 1
2
x2 5
= 2(1)(−1)
+ (5)2 (1) z
= −21 + 75 = 54
−1
2 2
W = −54q = −108 μ J
Prob. 4.40
(a)
From A to B, dl = rdθ aθ ,
WAB = −Q
(b)
90°
|
10 r cos θ r dθ
θ =30°
= − 1250 nJ
r =5
From A to C , dl = dr a r ,
WAC = − Q
(c)
10
20 r sin θ dr θ |
= −3750 nJ
= 30°
r =5
From A to D, dl = r sin θ dφ aφ ,
WAD = −Q 0(r sin θ ) dφ = 0 J
(d)
WAE = WAD + WDF + WFE
where F is (10,30o , 60o ). Hence,
10
90
WAE = − Q 20 r sin θ dr | + 10 r cos θ r dθ | θ =30°
r =10 r =5
θ =30o
75 100
] nJ = − 8750 nJ
= − 100[ +
2
2
o
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Principles of Electromagnetics, 6e
112
110
Prob. 4.41
B
5
10
dr
r2
1
VAB = − E dl = − A
=
10
r
5
1
1
= 10( − 1) = −8V
5
Prob. 4.42
A = (2,3, −1)
↓ dl = dxa x
dl = dya
A '(8,3, −1)
B(8, 0, −1)
y
⎯⎯⎯→
W = − Q E dl
8
W
− = E dl = 2 xy 2 dx
Q
x =2
= 2(3) 2
x2
2
8
2
+ 2(82 + 1)
y =3
+
0
2y
2
( x 2 + 1) dy
x =8
y =3
y3
3
0
3
2
= 9(64 − 4) + 65(0 − 81)
3
= 540 − 3510 = −2970
W = 2970Q = 5940 nJ
= 5.94 μ J
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Principles of Electromagnetics, 6e
113
111
Prob. 4.43
Method 1:
W = −Q E dl ,
dl = ρ dφ aφ
L
Eρ cos φ sin φ 0 Ex E = − sin φ cos φ 0 E φ y
Ez 0
0
1 Ez Eφ = − Ex sin φ + E y cos φ = −20 x sin φ + 40 y cos φ
x = ρ cos φ , y = ρ sin φ
Eφ = −20 ρ cos φ sin φ + 40 ρ sin φ cos φ = 20 ρ cos φ sin φ
W = −Q E dl = -2 ×10-3 20 ρ cos φ sin φρ dφ
L
π
= −2(20)(2)2 sin φ d (sin φ ) mJ = 160
0
ρ =2
sin 2 φ π / 2
= −80 mJ
0
2
Method 2:
W
− = E dl = 20xdx + 40ydy
Q L
y = 2 − x, dy = −dx
−
0
W
= 20xdx + 40(2 - x)(-dx) = (60x - 80)dx
Q x=2
=
0
60 x 2
− 80 x = 40
2
2
W = −40Q = −80 mJ
Method 3:
∂
∂
∂
∂y
∂z = 0
∇ × E = ∂x
20 x 40 y −10 z
V = − E dl = -10x 2 − 20 y 2 + 5 z 2 + C
L
W = Q(V2 − V1 ) = Q( −20 × 4 + 10 × 4) = −40Q
W = −40Q = −80 mJ
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POESM_Ch04.indd 113
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Principles of Electromagnetics, 6e
114
112
Prob. 4.44
W = −Q E dl
L
ρ
ρ
E = s an = s a x ,
2ε o
2ε o
dl = dxa x
Qρs 1
Qρs
Qρs
W =−
dx = −
(−2) =
εo
2ε o 3
2ε o
=
10 ×10−6 × 40 × 10−9
= 400 × 36π ×10−6 = 45.24 mJ
10−9
36π
Prob. 4.45
(a) E = −∇V = −
∂V
∂V
∂V
ax −
ay −
a y = −4 xa x − 8 ya y
∂x
∂y
∂z
ρv = ∇ D = ε o∇ E = ε o (−4 − 8) = −12ε o = −106.25 pC/m3
∂V
1 ∂V
∂V
aρ −
aφ −
az
ρ ∂φ
∂ρ
∂z
= −(20 ρ sin φ + 6 z )a ρ − 10 ρ cos φ aφ − 6 ρ a z
(b) E = −∇V = −
1
ρ v = ∇ D = ε o ∇ E = ε o −
ρ
[ 40 ρ sin φ + 6 z ] + 10sin φ 6z = − 30sin φ + ε o C/m3
ρ ∂V
1 ∂V
1 ∂V
ar −
aθ −
aφ
∂r
r ∂θ
r sin θ ∂φ
= −10r cos θ sin φ ar + 5r sin θ sin φ aθ − 5r cot θ cos φ aφ
(c) E = −∇V = −
ρ v = ∇ D = ε o ∇ E
ρv 1
5r sin φ
5r cot θ sin φ
= 2 (−30r 2 cos θ sin φ ) +
2sin θ cos θ +
εo r
r sin θ
r sin θ
ρv = ε o (5sin φ csc 2 θ cos θ − 20 cos θ sin φ ) C/m3
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113
Prob. 4.46
V = −r −3 sin θ cos φ
∂V
1 ∂V
1 ∂V
ar +
aθ +
aφ
∂r
r ∂θ
r sin θ ∂φ
− E = ∇V =
r −4 sin θ
(− sin φ )aφ
sin θ
3
1
sin φ
E
= 4 sin θ cos φ ar − 4 cos θ cos φ aθ + 4 aφ
r
r
r
o
o
o
o
At (1,30 , 60 ), r = 1,θ = 30 , φ = 60
= −3r −4 sin θ cos φ ar + r −4 cos θ cos φ aθ +
= 3sin 30o cos 60o ar − cos 30o cos 60o aθ + sin 60o aφ
E
= 0.75ar − 0.433aθ + 0.866aφ
10−9
(0.75ar − 0.433aθ + 0.866aφ )
36π
= 6.635ar − 3.829aθ + 7.657aφ pC/m 2
D = εo E =
Prob. 4.47
For a < r < b, we apply Gauss’s law.
DdS = Q
enc
=Q
S
Dr (4π r 2 ) = Q
b
Vab = − E dl = −
a
⎯⎯
→ Er =
Dr
εo
=
Q
4πε o r 2
Q 1b
Q 1 1
1
=
=−
dr
− 2
4πε o a r
4πε o r a
4πε o a b Q
b
Prob. 4.48
π /2
ρ s dS
ρ 2π π / 2 r 2 sin θ dθ dφ
ρs
= s =
(2π ) sinθ dθ
4πε o r 4πε o φ =0 θ =0
4πε o r
r
S
0
V =
π / 2
ρs − cos θ
0 r=a
2ε o r ρ
= s
2ε o a
=
V
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Principles of Electromagnetics, 6e
116
114
Prob. 4.49
∇× E = 0
∇× D = 0
⎯⎯
→
∂D 1 ∂Dz ∂Dφ ∂D ∂D 1 ∂
∇× D = −
aρ + ρ − z aφ + ( ρ Dφ ) − ρ a z
∂z ∂φ ρ ρ ∂ρ
ρ ∂φ
∂z
1
= 0 a ρ − 0aφ − 2 ρ cos φ a z ≠ 0
ρ
Hence D is not a genuine EM field.
ψ = DdS =
S
= −2 cos φ
π /4 1
π /4
1
0
0
2 ρ sin φρ dφ dz = 2 sin φ dφ dz ρ
φ
=0 z =0
π /4
0
2
ρ =1
(1)(1) 2 = −2(cos π / 4 − 1) = 0.5858 C
Prob. 4.50
(a)
d2y
m 2 = eE ; divide by m , and integrate once, one obtains :
dt
dy
eEt
u=
=
+ c0
dt
m
e E t2
+ c0t + c1
2m
"From rest" implies c1 = 0 = c0
y =
(1)
V
or V = E d .
d
Substituting this in (1) yields :
2m d
t2 =
eE
Hence :
At t = t0 ,
y = d, E =
eE
m
u=
that is, u α
or
(b)
2md
eE
=
2 e Ed
2eV
=
m
m
V
u= k V
k =
2e
=
m
2 (1.603) 10−19
9.1066 (10−31 )
= 5.933 × 105
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Principles of Electromagnetics, 6e
117
115
(c)
1
9(1016 )
u2 m
100 = 2.557 k V
V=
=
2e
2 (1.76) (1011 )
Prob. 4.51
(a)
This is similar to Example 4.3.
eEt
, u x = u0
uy =
m
e E t2
, x = u0 t
y =
2m
t=
x 10 (10−2 )
=
= 10 ns
107
u0
Since x =10 cm when y =1cm,
2m y
2 (10−2 )
=
= 1.136 kV/m
E=
1.76 (1011 ) (10−16 )
et2
E = − 1.136 a y kV/m
(b)
u x = u0 = 107 ,
uy =
2000
(1.76)1011 (10 −8 ) = 2(106 )
1.76
u = (a x + 0.2a y ) (107 ) m/s
Prob. 4.52
1 ∂ 20 cos θ
∇ E = 2 r ∂r r
1
∂ 10sin 2 θ
+
3
r sin θ ∂θ r
20 cos θ 20 cos θ
=−
+
=0
r4
r4
1 ∂ 10sin θ 20sin θ aφ
∇× E = +
r ∂r r 2 r 3 +0
1 20sin θ 20sin θ aφ = 0
= −
+
r
r3
r 3 At r = 0, ∇E and ∇ × E are not defined. So they are zero every where except at the origin.
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118
116
Prob. 4.53
The dipole is oriented along y − axis.
p•r
V=
; p • r = Q d a y • a r = Qd sin θ sin φ
4πε 0 r 2
V=
Qd sin θ sin φ
4π ε0 r2
E = − ∇V = −
=
E=
Qd
4π ε0
∂V
1∂V
1 ∂V
ar −
aθ −
aφ
∂r
r ∂θ
r sin θ ∂ φ
2sin θ sin φ
cos θ sin φ
cos φ
ar −
aθ − 3 a φ 3
3
r
r
r
Qd
4π ε 0r 3
_
(2sin θ sin φ a r − cos θ sin φ aθ − cos φ aφ )
Prob. 4.54
V=
p cos θ
k cos θ
=
2
r2
4π ε0 r
At (0, l nm),
θ = 0,
r = 1 nm, V = 9;
k (1)
, ∴ k = 9(10−18 )
−18
1(10 )
cos θ
V = 9(10−18 ) 2
r
θ = 45°,
At (1,1) nm, r = 2 nm,
that is,
9=
9(10−18 ) cos 45°
9
=
= 3.182 V
V=
−18
2
10 ( 2)
2 2
Prob. 4.55
W = W1 + W2 = 0 + Q2V21 = Q2
Q1
4πε o | (2, 0, 0) − (0, 0,1) |
40 ×10−9 × ( −50) ×10 −9 40 × 9 × ( −50) ×10 −9
=
10−9
4 +1
4π ×
| (2, 0, −1) |
36π
= −8.05 μ J
=
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Principles of Electromagnetics, 6e
119
117
Prob. 4.56
E = −∇V = −
∂V
∂V
ax −
a y = −4 xa x − 12 ya y V/m
∂x
∂y
1
1
1
1
1
W = ε o | E |2 dv = ε o (16 x 2 + 144 y 2 )dxdydz
2 v
2 z =−1 y =−1 x =−1
x3 1
y 3 1 1 10−9
1 1
(160)(4) (1 + 1)
= ε o 16(4)
+ 144(4)
=
2 3 −1
3 −1 2 36π
3
= 1.886 nJ
Prob. 4.57
Given that E = 2r sin θ cos φ ar + r cos θ cos φ aθ − r sin φ aφ
E 2 = 4r 2 sin 2 θ cos 2 φ + r 2 cos 2 θ cos 2 φ + r 2 sin 2 φ
= r 2 cos 2 φ ( 4sin 2 θ + cos 2 θ ) + r 2 sin 2 φ
r 2 cos 2 φ + 3r 2 cos 2 φ sin 2 θ + r 2 sin 2 φ
=
= r 2 (1 + 3cos 2 φ sin 2 θ )
W=
ε
2 =
ε
2
E 2 r 2 sin θ drdθ dφ
π π
r dr (1 + 3cos
2
θ φ
4
2
φ sin 2 θ ) sin θ dθ dφ
=0
0
π
=
16ε
3π
(π sin θ +
sin 2θ )dθ
5 0
2
=
16 10−9
16
x
(4π ) =
nJ= 0.36 nJ
5 36π
45
Prob. 4.58
Method 1:
W=
But
W=
1
V
1
ρ sVdS = ρ s dS = QV
2S
2S
2
V=
Q
Q
4πε o a
2
8πε o a
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120
118
Method 2:
1
1
W = D Edv = ε o E 2 dv
2v
2 v
2
Q 2
1
r sin θ dθ drdφ
= ε o 2 2
4πε o r 2π
π
∞
εo
Q2
Q2
1
dr = (2π )(2)
= ε o dφ sin θ dθ 2 0
16π 2ε o2 r 2
2
16π 2ε o2 a
0
r =a
W=
Q2
8πε o a
Prob. 4.59
W=
=
1
ε o | E |2 dv
2 v
εo
2
3
2
1
(9 x
2
+ 25 z 2 )dxdydz
z =0 y =0 x =0
3
2
1
3
2
1
1 10−9 2
2
=
9 dz dy x dx + 25 z dz dy dx 2 36π 0 0 0
0
0
0
=
1 x3 1
z3
+ 25(1)(2)
9(3)(2)
72π 3 0
3
3
1
468
(18 + 50 × 9) =
nJ
nJ=
0 72π
72π
W=2.069 nJ
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Principles of Electromagnetics, 6e
121
CHAPTER 5
dS = ρ dφ dzaρ
P. E. 5.1
I =  J • dS =
2π
5
10 z sin


φ
2
φρ dzdφ |ρ = 2
= 0 z =1
z2
= 10(2)
2
5 2π
1
1
 2 (1 − cos 2φ )dφ = 240π
0
I = 754 A
P. E. 5.2
I = ρ s wu = 0.5 ×10−6 × 0.1×10 = 0.5μ A
V = IR = 1014 × 0.5 × 10−6 = 50 MV
P. E. 5.3 σ = 5.8 ×107 S/m
J
8 × 106
J =σE
⎯⎯
→
E= =
= 0.138 V/m
σ 5.8 ×107
8 × 106
= 4.42 ×10−4 m/s
10
ρv 1.81×10
P. E. 5.4 The composite bar can be modeled as a parallel combination of resistors as
shown below.
J = ρvu
⎯⎯
→
u=
J
=
RL
For the lead,
RL =
l
,
σ L SL
SL = d 2 − π r 2 = 9 −
Rc
π
4
cm2
RL = 0.974 m Ω
For copper,
Rc =
l
,
σ c Sc
Sc = π r 2 =
π
4
cm2
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Principles of Electromagnetics, 6e
12
Rc =
4
5.8 × 107 ×
π
×10−4
= 0.8781 m Ω
4
RL Rc
0.974 x0.8781
=
= 461.7 μΩ
R=
RL + Rc 0.974 + 0.8781
ρ Ps = P • a x = ax 2 + b
P. E. 5.5
ρ ps
ρ ps
x=0
x=L
= P • ( − a x ) x = 0 = −b
= P • ax
x=L
= aL2 + b
Qs =  ρ ps dS = −bA + (aL2 + b) A = AaL2
ρ pv = −∇ • P = −
ρ pv
x=0
= 0,
d
(ax 2 + b) = −2ax
dx
ρ pv
x=L
= −2aL
L
Qv =  ρ pv dv =  (−2ax) Adx = − AaL2
0
Hence,
QT = Qv + Qs = − AaL2 + AaL2 = 0
P. E. 5.6
V
103
ax =
a x = 500a x kV/m
d
2 x10−3
10−9
P = χ eε o E = (2.55 − 1) x
x0.5 x106 a x = 6.853a x μC / m 2
36π
ρ ps = P • a x = 6.853μ C/m 2
E=
P. E. 5.7 (a) Since P = ε o χ e E ,
χe =
Px = ε o χ e Ex
Px
3x10−9 1
=
x36π x109 = 2.16
ε o Ex
10π 5
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123
(b) E =
P
χ eε o
=
36π ×109 1
(3, −1, 4)10−9 = 5ax − 1.67 a y + 6.67az V/m
2.16 10π
(c)
D = ε oε r E =
ε r P 3.16  1
=
χ e 2.16  10π

2
 (3, −1, 4) nC/m = 139.7ax −46.6a y + 186.3az

P. E. 5.8 From Example 5.8,
ρs 2S
2ε F
⎯⎯
→
F=
ρs 2 = o
2ε o
S
V
But
ρ s = ε o E = ε o d . Hence
d
2ε o F ε o 2Vd 2
=
S
d2
ρs2 =
⎯⎯
→
Vd 2 =
2 Fd 2
ε oS
i.e.
Vd = V1 − V2 =
2 Fd 2
εoS
as required.
P. E. 5.9 (a) Since
D1n = 12ax ,
E2t = E1t
an = ax ,
D1t = −10ax + 4az ,
⎯⎯
→
D2t =
D2 n = D1n = 12ax
ε 2 D1t
1
=
(−10a y + 4az ) = −4a y + 1.6az
ε1
2.5
D2 = D2 n + D2t = 12ax − 4a y + 1.6az nC/m2.
tan θ 2 =
(−4) 2 + (1.6) 2
D2t
=
= 0.359
D2 n
12
θ 2 = 19.75o
⎯⎯
→
(b) E1t = E2t = E2 sin θ 2 = 12sin 60o = 10.392
E2
E 2t
θ2
E2n
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x
pC/m 2
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Principles of Electromagnetics, 6e
12
E1n =
εr2
1
12cos 60o = 2.4
E2 n =
ε r1
2.5
E1 =
tan θ1 =
E1t 2 + E1n 2 = 10.67
ε r1
2.5
tan θ 2 =
tan 60o = 4.33
εr2
1
⎯⎯
→
θ1 = 77o
Note that θ1 > θ 2 .
P. E. 5.10
D = εoE =
10−9
(60,20, −30) x10−3 = 0.531ax + 0.177 a y − 0.265az pC/m2
36π
ρ s = Dn =| D |=
10−9
(10) 36 + 4 + 9(10−3 ) = 0.619 pC/m2
36π
Prob. 5.1
I =  J • dS ,
dS = dydza x
I =  e cos(4 y )dydz
−x
x=2
=e
−2
π /3
4
0
0
 cos(4 y)dy  dz
 sin 4 y π / 3  −2 
4π

= 4e−2 
 = e  sin( ) − 0  = −0.1172 A
0 
3


 4
Prob. 5.2
Method 1:
I =  J • dS = 
3
10 −103 t 2
e r sin dθ dφ
t = 2ms, r = 4m
r
= 10(4)e−10 ×2×10
−3
π

θ
=0
sin θ dθ
2π
 dφ = 40e
φ
−2
(2)(2π ) = 160π e−2
=0
= 68.03 A
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125
Method 2:
3
10 −103 t
10
e dS = e −10 t (4π r 2 )
r
r
since r is constant on the surface.
I =  J • dS = 
I=40π re −2 = 160π e −2
= 68.03 A
Prob. 5.3
dQ
= 3x10−4 e−3t
dt
I =−
I(t=2.5) = 0.3 e-7.5 = 166 nA
I(t=0) = 0.3 mA,
Prob. 5.4
a
I =  J • dS
=5
2π
 
ρ φ
=0 =0
2π
e −10 ρ ρ dφ d ρ = 5  dφ
0
a

ρ
ρ e−10 ρ d ρ
=0
 e−10 ρ
 a 10π −10 a
 e (−10a − 1) − 1(−0 − 1) 
(−10 ρ − 1)  =
= 5(2π ) 
 100
 0 100
=
π
π
e−0.04 (−0.04 − 1) + 1 = (0.00078) = 244.7 μ A
10 
10
Prob. 5.5
I =  J dS = 
10
ρ
5
π
0
0
sin φρ dφ dz = 10  dz  sin φ dφ
= 10(5)(− cos φ )
π
0
= 100 A
Prob. 5.6
l
l
2 × 10−2
R=
⎯⎯
→ σ=
= 6
= 3.978 × 10−4 S/m
−3 2
RS 10 (π )(4 × 10 )
σS
l
8 ×10−2
8
=
=
= 33.95mΩ
Prob. 5.7 (a) R =
4
−6
75π
σ S 3 ×10 π (25)10
(b) I = V / R = 9 ×
75π
= 265.1 A
8
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126
(c) P = IV = 2.386 kW
Prob. 5.8
If R and S are the same,


σ
⎯⎯
→ 1 =  2 1
R1 = 1 = R2 = 2
σ 1S
σ 2S
σ2
If 1 corresponds to copper and 2 to silver,
σ 1 = 5.8 ×107 S/m, σ 2 = 6.1×107 S/m
5.8
= 0.951 2
6.1
That is, the copper wire is shorter than silver wire or the silver wire is longer.
1 =  2
Prob. 5.9 (a) Si = π ri 2 = π (1.5) 2 x10−4 = 7.068 x10−4
So = π (ro 2 − ri 2 ) = π (4 − 2.25) ×10−4 = 5.498 ×10−4
RI =
Ro =
ρI l
SI
ρol
So
=
11.8 × 10−8 × 10
= 16.69 × 10−4
−4
7.068 X 10
=
1.77 × 10−8 × 10
= 3.219 × 10−4
−4
5.498 × 10
R = Ri // Ro =
(b)
Ri Ro
16.69 × 3.219 × 10−4
=
= 0.27mΩ
16.69 + 3.219
Ri + Ro
V = I i Ri = I o Ro
⎯⎯⎯
→
I i + I o = 1.1929 I o = 60 A
I o = 50.3 A
(copper),
I i Ro 0.3219
=
=
= 0.1929
I o Ri
1.669
I i = 9.7 A
(steel)
Alternatively, using the principle of current division,
I o = 60
Ri
= 50.3 A
Ri + Ro
I i = 60
Ro
= 9.7 A
Ri + Ro
(c)
10 × 1.77 ×10−8
R=
= 0.322mΩ
1.75π ×10−4
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127
Prob. 5.10
From eq. (5.16),
ρ ρ
R1 = 1 = 1 ,
S1
ab
R2 =
ρ2
S2
=
ρ2
ac
ρ1 ρ 2
R1 R2
ρ1 ρ 2  2
= ab ac =
R = R1  R2 =
R1 + R2 ρ1 + ρ 2  ac ρ1 + abρ 2 
ab ac
ρ1 ρ 2 
R=
a (c ρ1 + bρ 2 )
Prob. 5.11
| P |= n | p |= nQd = 2ned = χ eε o E
χe =
(Q = 2e)
2ned 2 × 5 × 1025 × 1.602 × 10−19 × 10−18
=
= 0.000182
10−9
εoE
4
×10
36π
ε r = 1 + χ e = 1.000182
Prob. 5.12
N
P=
 qi di
i =1
v
| P |=
N
=
p
i
i =1
v
N
| p |= 2 ×1019 ×1.8 ×10−27 = 3.6 ×10−8
v
P =| P | a x = 3.6 × 10−8 a x C/m 2
But
P = χ eε o E
or
P
3.6 × 36π × 109 × 10−18
χe =
=
= 0.0407
εoE
105
ε r = 1 + χ e = 1.0407
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Principles of Electromagnetics, 6e
128
Prob. 5.13
E=
Q
4πε oε r r 2
ar
χ eQ
3(10)10−3
P = χ eε o E =
a =
ar = 596.8ar μ C/m 2
2 r
2
4πε r r
4π (4)1
Prob. 5.14
P = χ eε o E
100 × 10−9
E=
=
a ρ = 2.261a ρ kV/m
10−9
χ eε o
2.5
(2)
36π
P
⎯⎯
→
D = ε oε r E = 3.5 ×
10−9
2.261×103 a ρ = 70a ρ nC/m 2
36π
Prob. 5.15
(a)
Qs1 =  P dS , dS = r 2 sin dθ dφ (−ar )
S
= −  4r r 2 sin θ dθ dφ
r = 1.2cm
2π
= −4(1.2) (10−6 )  dφ  sin θ dθ (10−12 )
3
0
= −6.912(2π )(2) × 10−18
= −86.86 × 10−18 C
(b)
Qs 2 =  P dS , dS = r 2 sin dθ dφ (−ar )
S
= −  4r r 2 sin θ dθ dφ
r = 2.6cm
2π
= −4(2.6) (10−6 )  dφ  sin θ dθ (10−12 )
3
0
= −4(2.6) (2π )(2) ×10−18 = 883.5 ×10−18 C
3
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Principles of Electromagnetics, 6e
129
(c)
ρ pv = −∇P = −
1 ∂
(4r 3 ) pC/m3 = −12pC/m3
r 2 ∂r
π
2π
2.6
0
0
1.2
Qv =  ρ pv dv = −12 dv = −12 sin θ dθ  dφ  r 2 dr (10−18 )
v
= −12(2)(2π )
r 2.6 −18
(10 ) = −16π (2.63 − 1.23 )(10−18 )
3 1.2
3
= −796.61×10−18 C
Prob. 5.16
10−9
(6,12, −20) = 0.1114a x + 0.2228a y − 0.3714a z nC/m 2
36π
10−9
(6,12, −20) = 0.0584a x + 0.1167a y − 0.1945a z nC/m 2
P = χ eε o E = 1.1x
36π
D = ε oε r E = 2.1x
Prob. 5.17
 ∂V
1 ∂V
∂V 
E = −∇V = − 
aρ +
aφ +
az 
ρ ∂φ
∂z 
 ∂ρ
= (10zsinφ a ρ + 10 z cos φ aφ + 10 ρ sin φ a z )
D = ε E = 5ε o E
= −50ε 0 ( zsinφ a ρ + z cos φ aφ + ρ sin φ a z ) C/m 2
Prob. 5.18
 ∂V
∂V
∂V 
(a) E = −∇V = − 
ax +
ay +
a z  = −20 xyza x − 10 x 2 za y − 10( x 2 y − z )a z V/m
∂y
∂z 
 ∂x
(b) D = ε E = 5ε o E = −0.8842 xyza x − 0.4421x 2 za y − 0.4421( x 2 y − z )a z nC/m 2
(c) P = χ eε o E = 4ε o E = −0.7073xyza x − 0.3537 x 2 za y − 0.3537( x 2 y − z )a z nC/m 2
(d) ρ v = −ε∇ 2V
∂
∂
∂
(20 xyz ) + (10 x 2 z ) + (10 x 2 y − 10 z ) = 20 yz − 10
∂x
∂y
∂z
ρv = −5ε o10(2 yz − 1) = −0.8854 yz + 0.4427 nC/m3
∇ 2V =
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Principles of Electromagnetics, 6e
130
Prob. 5.19
P = χ e ε o E =χ e ε o
D
ε oε r
=
(ε r − 1) D
εr
=
1.4
× 450a x nC/m 2
2.4
= 262.5a x nC/m 2
P
Prob. 5.20 (a) Applying Coulomb’s law,
Q
 Dr
 ε = 4πε r 2 , r > b
o
Er =  o
 Dr = Q , a < r < b
 ε
4πε r 2
ε −1
P= r
D
(= D − ε O E )
εr
Hence
Pr =
εr −1 Q
.
,
ε r 4π r 2
a<r <b
ρ pv = −∇ • P = −
(b)
(c)
ρ ps = P • (−ar ) = −
ρ ps = P • (ar ) = −
1 d 2
(r Pr ) = 0
r 2 dr
Q εr −1
(
),
4π a 2 ε r
Q εr −1
(
),
4π b 2 ε r
r=a
r =b
Prob. 5.21
F1 =
Q1Q2
Q1Q2
= 2.6 nN,
= 1.5 nN
F2 =
2
4πε o d
4πε oε r d 2
F1 2.6
=
= ε r = 1.733
F2 1.5
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Principles of Electromagnetics, 6e
131
Prob. 5.22
(a) By Gauss’s law,
 DdS = Q
enc
⎯⎯
→ Dr =
S
Dr
Er =
ε
=
Q
4π r 2
Q
4πε r 2
1
W =  ε | E |2 dv,
2
v
dv = r 2 sin θ drdφ dθ
2π π ∞
1
Q2
Q2
2
sin
θ
φ
θ
=
W= ε   
r
drd
d
2 φ =0 θ r = a 16π 2ε 2 r 4
8πε a
(b) Dr remains the same but
D
Q
Q
=
Er = r =
2
2
ε
4πε o ( r + a )
 a
4π r 2ε o 1 + 
 r
2π
π
∞
1
1
Q 2 r 2 sin θ drdθ dφ  r + a 
εo 
W =  ε | E |2 dv =   

2
2 φ =0 θ =0 r = a 16π 2ε 2 ( r + a )4
 r 
v
=
W=
∞
1 ∞  Q2 1
Q2
dr
Q2 
=
−
(4
)
π
a ( r + a )2 8πε o  r + a a  = 8πε o 2a
32π 2ε o
Q2
16aπε o
Prob. 5.23
(a)
ρ ,
ρv =  o
 0,
0<r <a
r>a
4π r 3
3
⎯⎯
→
V = −  E • dl = −
ρo r 2
+ c1
6ε
For r < a, ε Er (4π r 2 ) = ρo
For r > a, ε o Er (4π r 2 ) = ρ o
4π a 3
3
V = −  E • dl =
As r ⎯⎯
→∞,
At r = a,
⎯⎯
→
Er =
ρo r
3ε
Er =
ρo a3
3ε o r 2
ρo a3
+ c2
3ε o r
V = 0 and c2 = 0
V(a+) = V(a-)
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2
Sadiku & Kulkarni
Principles of Electromagnetics, 6e
132
−
ρo a 2
ρ a2
+ c1 = o
6ε oε r
3ε o
V(r=0) = c1 =
⎯⎯⎯
→
c1 =
ρo a 2
(2ε r + 1)
6ε oε r
ρo a 2 (2ε r + 1)
6ε oε r
ρo a 2
(b) V (r = a ) =
3ε o
Prob. 5.24
 Dx 
4 1 1  1 
 D  = ε E 1 3 1   1 
o o 
 y
 
 Dz 
1 1 2   −1
Dx = ε o Eo (4 + 1 − 1) = 4ε o Eo
Dy = ε o Eo (1 + 3 − 1) = 3ε o Eo
Dz = ε o Eo (1 + 1 − 2) = 0
D = ε o Eo (4a x + 3a y ) C/m 2
Prob. 5.25
Since
∂ρv
= 0,
∂t
∇ • J = 0 must hold.
(a)
∇ • J = 6 x2 y + 0 − 6 x2 y = 0
(b)
∇ • J = y + ( z + 1) ≠ 0
(c)
∇•J =
(d)
∇•J =
1 ∂
ρ ∂ρ
⎯⎯
→
⎯⎯
→
( z 2 ) + cos φ ≠ 0
1 ∂
(sin θ ) = 0
r2 ∂ r
This is possible.
This is not possible.
⎯⎯
→
⎯⎯
→
This is not possible.
This is possible.
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Sadiku & Kulkarni
Principles of Electromagnetics, 6e
133
Prob. 5.26
∂J
∂J
∂ρ
∂J
∇ J = x + y + z = 2e −2 y cos 2 x − 2e −2 y cos 2 x + 1 = 1 = − v
∂x
∂y
∂z
∂t
∂ρ v
= −1 C/m3 s
Hence,
∂t
Prob. 5.27
(a) ∇ • J =
−
1 ∂ 100
100
(
)=− 3
ρ ∂ρ
ρ
ρ
∂ρv
100
= ∇•J = − 3
∂t
ρ
I =  J • dS = 
(b)
⎯⎯
→
100
ρ2
∂ρ v 100
= 3 C/m3 .s
∂t
ρ
ρ dφ dz ρ = 2
100
=
2
2π
1
0
0
 dφ  dz = 100π = 314.16 A
Prob. 5.28
ε
2.5 × 10−9
Tr = =
= 4.42 μ s
σ 5 ×10−6 × 36π
Q
1
=
= 29.84 kC / m3
π
4
V
×10−6 × 8
3
ρ v = ρ vo e − t / Tr = 29.84e −2 / 4.42 = 18.98 kC/m3
ρvo =
Prob. 5.29
−
∂ρv
∂J
= ∇ • J = x = 0.5π cos π x
∂t
∂x
At P(2,4,-3), x= 2
∂ρv
= −0.5π cos(2π ) = −0.5π = −1.571 C/m 2 s
∂t
Prob. 5.30
(a)
ε
=
σ
(b)
3.1×
10
10−9
36π = 2.741×104 s
−15
10−9
ε
36π = 5.305 × 104 s
=
10−15
σ
6×
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Principles of Electromagnetics, 6e
134
ε
=
σ
(c)
10−9
36π = 7.07 μ s
10−4
80 ×
Prob. 5.31
∂ρ
∇ J = − v
⎯⎯
→ ρv = −  ∇ Jdt
∂t
1 ∂ 2
1 ∂
0.5
(r J r ) = 2
(0.5r exp( −104 t )) = 2 exp(−104 t )
∇ J = 2
r ∂r
r ∂r
r
0.5
ρ v = −  ∇ Jdt = 2 4 exp(−104 t ) + C
r 10
If ρ v → 0 as t → ∞, C = 0.
ρv =
50
exp(−104 t ) μ C/m 3
2
r
Prob. 5.32
P1 = χ e1ε o E1 = χ e1ε o
D1
4 −1
3
ε oε r1 = 4 D1 = 4 D1
= 12a x + 22.5a y − 15az nC/m 2
D2 n = D1n = −20a z
E2t = E1t
D2t =
⎯⎯
→
D2t
ε2
=
D1t
ε1
6.5ε o
ε2
(16a x + 30a y ) = 26a x + 48.75a y
D1t =
ε1
4ε o
D2 = D2 n + D2t = 26a x + 48.75a y − 20a z nC/m 2
Prob. 5.33
Let x > 0 be region 1 and x < 0 be region 2.
D1n = 50a x ,
D1t = 80a y − 30a z
D2 n = D1n = 50a x
E2t = E1t
D2t =
⎯⎯
→
D2t
ε2
=
D1t
ε1
ε2
7.6
(80a y − 30a z ) = 289.5a y − 108.6a z
D1t =
2.1
ε1
D2 = D2t + D2 n = 50a x + 289.5a y − 108.6a z nC/m 2
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Principles of Electromagnetics, 6e
135
Prob. 5.34
f(x,y)= 4x +3y –10=0
−(4a x + 3a y )
∇f
=
= −0.8a x − 0.6a y
| ∇f |
5
The minus sign is chosen for an because it is directed toward the origin.
∇f = 4a x + 3a y
⎯⎯
→
an = −
D1n = ( D1 ⋅ an )an = (1.6 − 2.4)an = −0.64a x − 0.48a y
D1t = D1 − D1n = 2.64a x − 3.52a y + 6.5a z
D2 n = D1n = −0.64a x − 0.48a y
E2t = E1t
D2t =
⎯⎯
→
D2t
ε2
=
D1t
ε2
ε2
2.5
D1t =
(2.64, −3.52, 6.5) = (6.6, −8.8,16.25)
1
ε1
D2 = D2 n + D2t = 5.96a x − 9.28a y + 16.25a z nC/m 2
θ 2 = cos −1
D2 an
= 87.66o
| D2 |
Prob. 5.35
(a) P1 = ε o χ e1 E1 = 2 ×
(b)
E1n = −6a y ,
D2 n = D1n
or
E2 n =
10−9
(10, −6,12) = 0.1768a x − 0.1061a y + 0.2122a z nC/m 2
36π
E2t = E1t = 10a x + 12a z
⎯⎯
→
ε 2 E2 n = ε1 E1n
3ε o
ε1
E1n =
(−6a z ) = −4a y
ε2
4.5ε o
E2 = 10a x − 4a y + 12a z V/m
E2t
102 + 122
tan θ 2 =
=
= 3.905
E2 n
4
⎯⎯
→
θ 2 = 75.64o
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Principles of Electromagnetics, 6e
136
(c) wE =
1
1
D•E = ε | E | 2
2
2
1
1
10−9
wE1 = ε1 | E1 |2 = x3x
(102 + 62 + 122 ) = 3.7136 nJ/m3
2
2
36π
1
1
10−9
wE 2 = ε 2 | E2 |2 = x 4.5 x
(102 + 42 + 122 ) = 5.1725 nJ/m3
2
2
36π
Prob. 5.36 (a) D2 n = 12a ρ = D1n ,
E2t = E 2t
D1t =
⎯⎯
→
D1t
ε1
=
D2t = −6aφ + 9a z
D2t
ε2
3.5ε o
ε1
D2t =
(−6aφ + 9a z ) = −14aφ + 21a z
ε2
1.5ε o
D1 = 12a ρ − 14aφ + 21a z nC/m 2
(12, −14, 21) ×10−9
= 387.8a ρ − 452.4aφ + 678.6a z
10−9
3.5 ×
36π
0.5ε o
D
(b) P2 = ε o χ e 2 E2 = 0.5ε o 2 =
(12, −6,9) = 4aρ − 2aφ + 3a z nC/m 2
ε 2 1.5ε o
E1 = D1 / ε1 =
ρv 2 = ∇ • P2 = 0
(c)
wE1 =
wE 2 =
1
1 D1 • D1 1 (122 + 142 + 212 ) x10−18
D1 • E1 =
=
= 12.62 μ J/m 2
−9
10
2
2 ε oε r 1
2
3.5 x
36π
1 D2 • D2 1 (122 + 62 + 92 ) x10−18
=
= 9.839 μ J/m 2
−9
10
2 ε oε r 2
2
1.5 x
36π
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Principles of Electromagnetics, 6e
137
Prob. 5.37
Gaussian surface
Q =  D.dS = ε1 Er
4π r 2
4π r 2
+ ε 2 Er
= 2π r 2 (ε1 + ε 2 ) Er
2
2
Q

,
r>a

E r =  2π (ε1 + ε 2 )r 2

0,
r<a

Prob. 5.38
(a) The two interfaces are shown below
oil
glass
1
glass
2
2
oil-glass
E1n = 2000,
⎯⎯
→
3
glass-air
E1t = 0 = E2t = E3t
D1n = D2 n = D3n
air
ε1 E1n = ε 2 E2 n = ε 3 E3n
E2 n =
ε1
3.0
E1n =
(2000) = 705.9 V/m, θ 2 = 0o
8.5
ε2
E3n =
ε1
3.0
E1n =
(2000) = 6000 V/m, θ 3 = 0o
1.0
ε3
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Principles of Electromagnetics, 6e
138
(b)
glass
oil
glass
θ2
air
θ 1 = 75 o
1
2
2
3
E1n = 2000cos 75o = 517.63,
E1t = 2000sin 75o = E2t = E3t = 1931.85
ε
3
ε
3
E2 n = 1 E1n =
E3n = 1 E1n = (517.63) = 1552.89
(517.63) = 182.7,
ε2
ε3
8.5
1
E2 = E2 n 2 + E2t 2 = 1940.5,
θ 2 = tan −1
E2t
= 84.6o ,
E2 n
E3 = E3n 2 + E3t 2 = 2478.6,
θ3 = tan −1
E3t
= 51.2o
E3n
Prob. 5.39
10−9
2900
302 + 402 + 202 × 10−3 =
pC/m 2
36π
36π
= 0.476 pC/m 2
ρ s = Dn = ε o E =
10−9
Prob. 5.40 (a) ρ s = Dn = ε o En =
152 + 82 = 0.1503 nC/m 2
36π
(b)
Dn = ρ s = −20 nC
D = Dn an = ( − 20 nC)(-a y ) = 20 a y nC / m2
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Principles of Electromagnetics, 6e
139
Prob. 5.41
At the interface between ε o and 2ε o ,
E1n = Eo cos 30o ,
E1t = Eo sin 30o
E2t = E1t = 0.5Eo
D2 n = D1n
E2 n =
⎯⎯
→
ε1 E1n = ε 2 E2 n
ε
ε1
E1n = o (0.866 Eo ) = 0.433Eo
ε2
2ε o
The angle E makes with the z-axis is
E
0.5
θ1 = tan −1 2t = tan −1
= 49.11o
E2 n
0.433
At the interface between 2ε o and 3ε o ,
E3t = E2t = 0.5 Eo
D3n = D2 n
⎯⎯
→
E3n =
2ε
ε2
E2 n = o (0.433Eo ) = 0.2887 Eo
ε3
3ε o
The angle E makes with the z-axis is
E
0.5
= 60o
θ 2 = tan −1 3t = tan −1
E3n
0.2887
At the interface between 3ε o and ε o ,
E4t = E3t = 0.5 Eo
D4 n = D3n
⎯⎯
→
E4 n =
ε3
3ε
E3n = o (0.2887 Eo ) = 0.866 Eo
ε4
εo
The angle E makes with the z-axis is
E
0.5
= 30o
θ 3 = tan −1 4t = tan −1
E4 n
0.866
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Principles of Electromagnetics, 6e
140
138
CHAPTER 6
P. E. 6.1
∇ 2V = −
ρ
ε
ρ x
d 2V
=− o
2
dx
εa
⎯⎯
→
ρo x3
+ Ax + B
6ε a
 ρo x 2

dV
− A  ax
E=−
ax = 
dx
 2ε a

V =−
If E = 0 at x =0, then
0=0− A
⎯⎯
→
A=0
If V= 0 at x =a, then
ρo a3
0=−
+B
6ε a
Thus
ρo a 2
B=
6ε
⎯⎯
→
ρ
V = o (a 3 − x 3 ),
6ε a
P. E. 6.2
ρo x 2
E=
ax
2ε a
V1 = A1 x + B1 ,
V2 = A2 x + B2
V1 ( x = d ) = Vo = A1d + B1
⎯⎯
→
B1 = Vo − A1d
V1 ( x = 0) = 0 = 0 + B2
⎯⎯
→
B2 = 0
V1 ( x = a) = V2 ( x = a)
⎯⎯
→
aA1 + B1 = A2 a
D1n = D2 n
⎯⎯
→
A1a + Vo − A1d =
or
A1 =
Hence
ε1 A1 = ε 2 A2
ε1
aA1
ε2
E1 = − A1a x =
A2 =
−Vo ax
,
d − a + ε1a / ε 2
A2 =
ε1
A1
ε2

ε 
Vo = A1  − a + d + 1 a 
ε2 

⎯⎯
→
Vo
,
d − a + ε1a / ε 2
⎯⎯
→
ε1
ε1Vo
A1
ε 2 ε 2 d − ε 2 a + ε1a
E2 = − A2a x =
−Vo a x
a + ε 2 d / ε1 − ε 2 a / ε1
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10/13/2015 9:12:14 PM
Sadiku & Kulkarni
Principles of Electromagnetics, 6e
141
139
P. E. 6.3 From Example 6.3,
V
E = − o aφ ,
D = εoE
ρφo
ρ s = Dn (φ = 0) = −
Voε
ρφo
The charge on the plate φ = 0 is
Q =  ρ s dS = −
C=
Voε
φo
L
b
 ρ
z =0
1
=a ρ
dzd ρ = −
Voε
φo
L ln(b / a )
|Q | εL b
ln
=
Vo
φo a
4mm
a
45o
a
a sin
45o
=2
2
⎯⎯
→
a=
2
= 5.226 mm
sin 22.5o
10−9
1.5 ×
36π 5ln 1000 = 444 pF
C=
π
5.226
4
Q = CVo = 444 × 10−12 × 50 C = 22.2 nC
P. E. 6.4 From Example 6.4,
Vo = 50,
θ 2 = 45o ,
θ1 = 90o ,
r = 32 + 42 + 22 = 29 ,
5
⎯⎯
→ θ = 68.2o ;
tan 45o = 1
2
50ln(tan 34.1o )
V=
= 22.125 V,
ln(tan 22.5o )
50aθ
= 11.35aθ V/m
E=
29 sin 68.2o ln(tan 22.5o )
θ = tan −1
ρ
z
=
tan −1
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140
P. E. 6.5
E = −∇V = −
=−
4Vo
b
∞
∂V
∂V
ax −
ay
∂x
∂y
1
 sinh nπ a/b cos(nπ x/b)sinh (nπ y/b) a
n = odd
x
+ sin(nπ x/b)cosh(nπ y/b)a y 
(a) At (x,y) = (a, a/2),
V=
400
π
(0.3775 − 0.0313 + 0.00394 − 0.000585 + ...) = 44.51 V
E = 0a x + (−115.12 + 19.127 − 3.9411 + 0.8192 − 0.1703 + 0.035 − 0.0074 + ...)a y
= −99.25a y V/m
(b) At (x,y) = (3a/2, a/4),
V=
400
π
(0.1238 + 0.006226 − 0.00383 + 0.0000264 + ...) = 16.50 V
E = (24.757 − 3.7358 − 0.3834 + 0.0369 + 0.00351 − 0.00033 + ...)a x
+ (−66.25 − 4.518 + 0.3988 + 0.03722 − 0.00352 − 0.000333 + ...)a y
= 20.68a x − 70.34a y V/m
P. E. 6.6
∞
V ( y = a ) = Vo sin(7π x / b) =  cn sin( nπ x / b)sinh(nπ a / b)
n =1
By equating coefficients, we notice that cn = 0 for n ≠ 7 . For n=7,
Vo sin(7π x / b) = c7 sin(7π x / b)sinh(7π a / b)
Hence
V ( x, y) =
⎯⎯
→
c7 =
Vo
sinh(7π a / b)
Vo
sin(7 π x / b) sinh(7 π y / b)
sinh(7 π a / b)
P. E. 6.7 Let V (r ,θ , φ ) = R(r )F (θ )Φ (φ ).
Substituting this in Laplace’s equation gives
RΦ d 
dF 
RF d 2Φ
ΦF d  2 dR 
r
+
+
=0
sin
θ




r 2 dr  dr  r 2 sin θ dθ 
dθ  r 2 sin 2 θ dφ 2
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141
Dividing by RF Φ / r 2 sin 2 θ gives
sin 2 θ d 2
sin θ d
1 d 2Φ
r R ') +
= λ2
( sin θ F ') = −
(
R dr
F dθ
Φ dφ 2
Φ ''+ λ 2Φ = 0
1 d 2
1
d
r R ') +
( sin θ F ') = λ 2 / sin 2 θ
(
R dr
F sin θ dθ
1 d 2
λ2
1
d
r
R
'
=
−
( sin θ F ') = μ 2
(
)
2
sin θ F sin θ dθ
R dr
2rR '+ r 2 R '' = μ 2 R
or
R ''+
or
2
μ2
R '− 2 R = 0
r
r
sin θ d
( sin θ F ') − λ 2 + μ 2 sin 2 θ = 0
F dθ
F ''+ cos θ F '+ ( μ 2 sin θ − λ 2 cs cθ ) F = 0
P. E. 6.8 (a) This is similar to Example 6.8(a) except that here 0 < φ < 2π instead of
0 < φ < π / 2 . Hence
b
ln
2π tVoσ
Vo
a
I=
and
=
R=
ln(b / a )
I 2π tσ
(b) This to similar to Example 6.8(b) except that here 0 < φ < 2π . Hence
I=
Voσ
t
and R =
b 2π
  ρ d ρ dφ =
a 0
Voσπ (b 2 − a 2 )
t
Vo
t
=
2
I σπ (b − a 2 )
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142
P. E. 6.9 From Example 6.9,
J1 =
σ 1Vo
ρ ln
b
a
J2 =
,
σ 2Vo
ρ ln
b
a
2π
π

Vl
ρ
φ
+
J
d
J 2 ρ dφ dz = o [πσ 1 + πσ 2 ]

z = 0 φ= 0 1

b

φ =π

ln
a
b
ln
Vo
a
R=
=
I π l [σ 1 + σ 2 ]
I =  J • dS =
P. E. 6.10 (a)
L
C=
4πε
,
1 1
−
a b

10  2.5
C1 = 4π x

36π  103 103
−

3
 2
−9
C=
C1 and C2 are in series.


 = 5 / 3 pF,




10  3.5
C2 = 4π x

36π  103 103
−

 1
2
−9


 = 7 / 9 pF



C1C2
(5 / 3)(7 / 9 )
=
= 0.53 pF
C1 + C2 (5 / 3) + (7 / 9 )
2πε
, C1 and C2 are in parallel.
1 1
−
a b



−9 
−9 

10
2.5
10
3.5
= 5 / 24 pF, C2 = 2π ×
C1 = 2π ×
 3
 3
3 
36π  10 10 
36π  10 103
−
−



3 
3
 1
 1
(b)
C=


 = 7 / 24 pF



C = C1+ C2 = 0.5 pF
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P. E. 6.11 As in Example 6.8, assuming V ( ρ = a) = 0,
V = Vo
ln ρ / a
,
ln b / a
Q =  ε E • dS =
C=
E = −∇V = −
L
V ( ρ = b) = Vo ,
Vo
aρ
ρ ln b / a
2π
Voε
1
V 2πε L
dz ρ dφ = o


ln b / a z = 0 φ = 0 ρ
ln b / a
Q
2πε L
=
Vo ln b / a
P. E. 6.12
(a) Let C1 and C2 be capacitances per unit length of each section and CT be the total
capacitance of 10m length. C1 and C2 are in series.
2πε r1ε o 2π x 2.5 10−9
=
= 342.54 pF/m,
ln b / c
ln 3 / 2 36π
2πε r 2ε o 2π x3.5 10−9
=
= 280.52 pF/m
C2 =
ln c / a
ln 2 36π
C1 =
C1C2
342.54 x 280.52
=
= 154.22 pF
C1 + C2 342.54 + 280.52
CT = Cl = 1.54 nF
C=
(b) C1 and C2 are in parallel.
C = C1 + C2 =
πε r1ε o
ln b / a
CT = Cl = 1.52 nF
+
πε r 2ε o
ln b / a
=
π (ε r1 + ε r 2 )ε o
ln b / a
=
6π 10−9
= 151.7 pF/m
ln 3 36π
P. E. 6.13 Instead of Eq. (6.31), we now have
a
a
Qdr
Qdr
Q
= −
=−
ln b / a
2
10ε o 2
πε
4πε r
40
o
b
b 4π
r
r
40 π 10 −9
Q
C=
=
= 113
. nF
|V | ln 4 / 1.5 36 π
V = −
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144
P. E. 6.14 Let
F = F1 + F2 + F3 + F4 + F5
where Fi ,
F =−
−
=
Q2
4πε o r
i = 1,2,...,5 are shown on in the figure below.
a +
2 y
Q 2 (a x sin 30o + a y cos30o )
4πε o (2r cos30o ) 2
−
Q 2 (a x cos30o + a y sin 30o )
4πε o (2r ) 2
+
Q 2a x
4πε o (2r cos30o ) 2
Q 2 (a x cos30o − a y sin 30o )
4πε o r 2

3a y
1 a
 −a y +  x +
2
4πε o r 
3  2
2

Q2
 1  3a
ay  1
3a x a y 
x
+  + ax −
+ 
 − 
 4
2
2
3
2
2




 1 5 3
 −5
3 
= 9 x10−5 a x  −
+
 + a y 
  = −52.4279a x − 30.27a y μ N
2
8
8
6


 
 
| F |= 60.54 μ N
Note that the force tends to pull Q toward the origin.
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P.E. 6.15
(a) and (b) and E
Dielectric�1 Dielectric�2
ε1>ε2
ρ
E
Dielectric�1 Dielectric�2
ε1<ε2
ρ
Figure. Electric field intensity distribution in a multi-dielectric coaxial capacitor
It can be noted from this practice exercise that, in a non-uniform field, the dielectric with
higher permittivity should be placed in the high field region and the one with lower
permittivity in the low field region (this is case-a). The difference in the average stress
levels of the two dielectrics reduces. If the placement is done the other way round, the
difference increases further (case-b). The electric field intensities in the two cases are
plotted as a function of radius in the figure above.
P.E. 6.16
(a)
(b)
, breakdown voltage , breakdown occurs
, breakdown voltage , breakdown does not occur
Note: One has to properly size the thickness of dielectric layers with lower permittivity,
as in case-b in this practice exercise.
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164
P.E. 6.17
(a) For the parallel-plate capacitor,
V
E = − o ax
d
From Example 6.11,
1
1
V2
ε
εS
C = 2  ε | E |2 dv = 2  ε o2 dv = 2 Sd =
Vo
Vo
d
d
d
(b) For the cylindrical capacitor,
E=−
Vo
aρ
ρ ln b / a
From Example 6.8,
b
1
εVo2
2πε L
dρ
2πε L
C = 2 
ρ d ρ dφ dz =
=
2
2 
Vo
( ρ ln b / a )
( ln b / a ) a ρ ln b / a
(c) For the spherical capacitor,
E=
Vo
ar
r (1/ a − 1/ b)
2
From Example 6.10,
C=
b
1
εVo2
ε
dr
4πε
r 2 sin θ dθ drdφ =
4π  2 =
2
2
2  4
Vo
r (1 / a − 1 / b )
(1 / a − 1 / b ) a r 1 − 1
a b
Prob. 6.31
C=
εS
d
⎯⎯
→
2 ×10−9 × 10−6 2
S=
=
m = 0.5655 cm 2
−9
ε oε r 4 ×10 / 36π
Cd
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145
Prob. 6.1
(a)
20
10
1 10
cos θ sin φ ar − 3 sin θ sin φ aθ +
cos θ cos φ aφ
3
r
r
rsinθ r 2
At P(1,60o ,30o ), r = 1, θ = 60o , φ = 30o
− E = ∇V = −
20
10
1 10 cos 60o cos 30o
o
o
o
o
+
−
a
a
aφ
cos
60
sin
30
sin
60
sin
30
θ
r
13
13
sin60o
13
= 5ar + 4.33aθ − 5aφ V/m
E=
(b)
1 ∂  −20 cos θ sin φ 
1
∂  −10sin 2 θ sin φ 
+



 2
r 2 ∂r 
r
r2
 r sin θ ∂θ 

1
10 cos θ sin φ
− 2 2
r sin θ
r2
20 cos θ sin φ 20sin θ cos θ sin φ 10 cos θ sin φ
=
−
−
r4
r 4 sin θ
r 4 sin 2 θ
10 cos θ sin φ
=−
r 4 sin 2 θ
ρ
10ε o cos θ sin φ
∇ 2V = − v
⎯⎯
→ ρ v = −ε ∇ 2V =
ε
r 4 sin 2 θ
At P, r =1, θ =60o , φ =30o
∇ 2V =
ρ v =10 ×
10-9 cos60o sin 30o
=29.47 pC/m3
36π 14 sin 2 60o
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Prob. 6.2
(a)
∂V
∂V 
 ∂V
E = −∇V = − 
ax +
ay +
az 
∂z
∂z 
 ∂x
= −(15 x 2 y 2 za x + 10 x 3 yza y + 5 x 3 y 2 a z )
At P, x=-3, y=1, z=2,
E = −15(9)(1)(2)a x + 10(−27)(1)(2)a y − 5(−27)(1)a z = −270a x + 540a y + 135a z V/m
ρv = ∇ • D
(b)
or
ρv = −ε∇ 2V
∂ 2V ∂ 2V ∂ 2V
∂
∂
∂
∇ V = 2 + 2 + 2 = (15 x 2 y 2 z ) + (10 x 3 yz ) + (5 x 3 y 2 )
∂x
∂y
∂y
∂x
∂y
∂z
2
= 30 xy 2 z + 10 x 3 z
At P,
ρv = −ε∇ 2V = −2.25 ×
Prob. 6.3
(a) E = −∇V = −
10−9
[30(−3)(1)(2) + 10(−27)(2)] = 14.324 nC/m3
36π
∂V
1 ∂V
sin 3φ
3cos 3φ
aρ −
aφ =
aρ −
aφ
2
ρ ∂φ
ρ
ρ2
∂ρ
At A, ρ =1, φ =20o , z=4,
E=
sin 60o
3cos 60o
a
aφ = 0.866a ρ − 1.5aφ V/m
−
ρ
12
12
10−9
(0.866a ρ − 1.5aφ )
36π
= 13.783a ρ − 23.87aφ pC/m 2
P = χ eε o E = 1.8 ×
 1 ∂  sin 3φ  9sin 3φ 
(b) ρ v = ∇ D = ε∇ E = ε 


+
ρ3 
 ρ ∂ρ  ρ 
ρv − sin 3φ 9sin 3φ
=
+
3
3
ε
ρv =
ρ
ρ
8ε sin 3φ
ρ3
10−9 sin 60o
= 171.52 pC/m 3
At A, ρv = 8 × 2.8 ×
3
36π 1
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Prob. 6.4
∇ 2V = −
ρv
ε
d 2V
y 10−9
y 10−9
=−
⋅
=−
⋅
= −2.25 y
dy 2
4π 4ε o
4π 10−9
4
36π
2
dV
y
= −2.25 + B
dy
2
3
V = −0.375 y + By + C
V (1) = 0 = −0.375 + B + C
V (3) = 50 = −10.125 + 3B + C
From (1) and (2), B=29.875 and C=-29.5
V = −0.375 y 3 + 29.875 y − 29.5
V (2) = 27.25 V
(1)
(2)
Prob. 6.5
ρv
d 2V
ρz
∇V =−
⎯⎯
→
=− o
2
ε
εd
dz
2
ρz
dV
=− o +A
2ε d
dz
ρ z3
V = − o + Az + B
6ε d
2
z = 0,V = 0
z = d ,V = Vo
⎯⎯
→
⎯⎯
→
0 = 0 + B,
Vo = −
Vo ρ o d
+
6ε
d
Hence,
i.e. B = 0
ρo d 2
+ Ad
6ε
A=
V =−
ρo z 3  Vo ρo d 
+ +
z
6ε d  d
6ε 
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148
Prob. 6.6
d 2V
ρv
50(1 − y 2 ) x10−6
=
−
=
−
= − k (1 − y 2 )
2
ε
ε
dx
−6
50 ×10
= 600π ×103
k=
−9
10
3×
36π
∇ 2V =
where
dV
= − k ( y − y 3 / 3) + A
dy
 y2 y4 
V = −k  −  + Ay + B = 50π .103 y 4 − 300π .103 y 2 + Ay + B
 2 12 
When y=2cm, V=30X103,
30 ×103 = 50π ×103 ×16 × 10−6 − 300π × 103 × 4 ×10−4 + Ay + B
or
30,374.5 = 0.02A + B
When y=-2cm,
(1)
V=30x103,
30,374.5 = −0.02A + B
From (1) and (2), A=0, B=30,374.5.
(2)
Thus,
V = 157.08 y 4 − 942.5 y 2 + 30.374 kV
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149
Prob. 6.7
ρ
∇ 2V = − v =
ε
−
10
ρ
3.6 ×
Let α = 0.1π .
∇ 2V = −
−α =
× 10−12
−9
10
36π
=−
0.1π
ρ
α 1 d  dV 
=
ρ
ρ ρ d ρ  d ρ 
d  dV 
ρ

dρ  dρ 
dV
= −αρ + A
dρ
dV
A
= −α +
ρ
dρ
V = −αρ + A ln ρ + B
ρ
At ρ =2, V=0
⎯⎯
→ 0 = −2α + A ln 2 + B
⎯⎯
→ 60 = −5α + A ln 5 + B
At ρ =5, V=60
Subtracting (1) from (2),
60 = −3α + A ln 5 / 2
⎯⎯
→
A=
(1)
(2)
60 + 3α
= 66.51
ln 2.5
From (1),
B = 2α − A ln 2 = −45.473
dV
A
66.51
)a ρ
E=−
a ρ = (α − )a ρ = (0.3142 −
dρ
ρ
ρ
Prob. 6.8
∂ 2U ∂ 2U ∂ 2U
∇ U = 2 + 2 + 2 = 0 + 0 − 2 = −2 ≠ 0
∂x
∂y
∂z
Does not satisfy Laplace's equation.
2
Prob. 6.9
∂ 2U ∂ 2U ∂ 2U
+
+
= 6 xy + 0 + 2c = 0
∂x 2 ∂y 2 ∂z 2
c = −3xy
∇ 2U =
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150
Prob. 6.10
V = V ( ρ ),
∇ 2V =
1 d  dV 
ρ
=0
ρ d ρ  d ρ 
dV
=A
⎯⎯
→
dρ
V = A ln ρ + B
dV A
=
dρ ρ
ρ
When ρ =4mm=a, V=0
⎯⎯
→ 0=Alna+B or B = -Alna
When ρ =12mm=b, V=Vo
E = −∇V = −
At ρ =8mm,
dV
A
V
aρ = − aρ = − o aρ
b
ρ
dρ
ρ ln
a
−6aρ kV/m = −
Prob. 6.11
∇ 2V =
⎯⎯
→ Vo =Alnb-Alna or A =
Vo
12
8 × 10 ln
4
d 2V
=0
dz 2
−3
⎯⎯
→
aρ
⎯⎯
→
Vo = 48ln 3 = 53.73
V = Az + B
When z=0, V = 0
B=0
When z=d, V = Vo
Vo=Ad or A = Vo/d
Hence,
V=
Vo
b
ln
a
Vo z
d
dV
V
az = − o az
dz
d
V
D = ε E = −ε oε r o a z
d
Since Vo = 50 V and d = 2mm,
E = −∇V = −
V = 25z kV, E = - 25az kV/m
10−9
D=−
(1.5)25 ×103 a z = −332a z nC/m 2
36π
ρ s = Dn = ± 332 nC / m2
The surface charge density is positive on the plate at z=d and negative on the plate at z=0.
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Prob. 6.12 From Example 6.8, solving
∇ 2V = 0 when V = V ( ρ ) leads to
Vo ln ρ / a
ln(a / ρ )
= Vo
ln b / a
ln(a / b)
Vo
Vo
E = −∇V = −
aρ =
aρ ,
ρ ln b / a
ρ ln a / b
V=
ρs = Dn = ±
D =εE =−
ε oε rVo
aρ
ρ ln b / a
ε oε rVo
ρ ln b / a ρ = a ,b
In this case, Vo=100 V, b=5mm, a=15mm, ε r = 2. Hence at ρ =10mm,
V=
100ln(10 / 15)
= 36.91 V
ln(5 / 15)
100
a ρ = 9.102a ρ kV/m
10 x10−3 ln 3
10−9
D = 9.102 x103 x
2aρ = 161a ρ nC/m 2
36π
E=
10−9
105
ρ s ( ρ = 5mm) =
= 322 nC/m 2
(2)
36π
5ln 3
ρ s ( ρ = 15mm) = −
10−9
105
= −107.3
(2)
nC/m 2
152
36π
15ln 3
Prob.
Prob.6.14
6.13
(a) 2
1 d V 1 ∂  ∂V  d 21V ∂ 2V
dV
=0
⎯⎯
→ + 2 = 0 + 0⎯⎯
→
=A
2
2=
ρ
∇
V
dφ
ρ dφ ρ ∂ρ  ∂ρ  dρ
φ 2 ∂φ 2


V = Aφ + B
1 ∂
4
2
−2
−2
c1ρ 2 −
( 2→
0 = 0 +=Bρ ∂ρ⎯⎯
B 2=c02 ρ ) sin 2φ − ρ 2 (c1ρ + c2 ρ )sin 2φ
100
−4
c2 ρ→
4c2 ρ −4 )sin 2φ = 0
− A4c=1 −
50 = A=π (4
/ 2c1 + 4⎯⎯
π
(b)
Ao
100
o 1 dV
P(2,45
EAt
a
= −∇
V = −,1), ρ = a1,φφ==−45 aφ = −
ρ dφ o
ρ
πρ φ
50 = (c1 + c2 ) sin 90 = c1 + c2
(1)
− E = ∇V =
∂V
1 ∂V
aρ +
a +0
∂ρ
ρ ∂φ φ
= (2c1 ρ − 2c2 ρ −3 ) sin 2φ a ρ + (c1 ρ + c2 ρ −3 )(2) cos 2φ aφ
At P,
− E = (2c1 − 2c2 )(1)aρ + 0
| E |= 100 = 2c1 − 2c2
From (1) and (2),
POESM_Ch06.indd 155
Prob. 6.15
(a)
⎯⎯
→
50 = c1 − c2
(2)
c1 = 50, c2 = 0
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10/13/2015 9:12:20 PM
(b)
100ln(10 / 15)
= o ,1), ρ = 1, φ = 36.91
At V
P(2,45
45o V
ln(5
/
15)
Sadiku & Kulkarni
50 = (c1 + c2 ) sin 90o = c1 + c2
(1)
Principles of Electromagnetics, 6e
∂V
1 ∂V
100
a φ +a0ρ kV/m 156
− EE= =∇V = −3 a ρ +a ρ = 9.102
152
∂ρ ln 3 ρ ∂φ
10 x10
−3−9
= 9.102
(2c1 ρ x−10
2c3 2xρ10
) sin
(c1 ρnC/m
+ c2 ρ2−3 )(2) cos 2φ aφ
ρ +a
D=
2a2ρ φ=a161
ρ
Prob. 6.13
36π
At P,
(a)
− E = (2
1 c1∂− 2c2∂)(1)
10−9∂ 2V 105
V aρ +10
ρ
∇ 2V = ρ s ( ρ =
+
+ 0 = 322 nC/m 2
(2)
c5mm)
 = 2⎯⎯
2
=
2
2
(2)
| E |= 100
−
→
c
ρ ∂ρ  1 ∂ρ 2 36
ρ π∂φ 5ln50
3 = c1 − c2
From (1)
c =0 4
1 and
∂ (2), 2 c1 = −50,
ρ −29) sin2 2φ10−5 2 (c1ρ 2 + c2 ρ −2 )sin
2c1ρ − 2c210
2φ
=
(
ρ = −107.3 nC/m 2
ρρs ( ∂ρρ= 15mm) = −
(2)
36π
15ln 3
(4c1 + 4c2 ρ −4 − 4c1 − 4c2 ρ −4 )sin 2φ = 0
Prob.=6.15
(a)
(b)
Prob. 6.14 2
∂At
V P(2,45o ,1),
a
= 1,
φ = 45o
φ
= Vo (1 + 2 )ρsin
ρd=2V(c + c ρ) sin 90o = dc 2V+ c
∂150
dV
1= 0 2
⎯⎯
→ 1 2 =2 0
⎯⎯
→(1)
=A
2
2
d∂φV
dφ
ρ ∂dVφ
a
∂
1
V
ρ ==∇VVo (=ρ + a) sin
φ
∂ρAφ + B ∂ρρ ρ + ρ ∂φ a φ +0
=
V− E
2
−3
V c1 ρ⎯⎯
0∂= 0 +=∂B(2
0
− 2→
c2−ρ a−B3 )=)sin
=
ρ
(1
sin2φφ a ρ + (c1 ρ + c2 ρ )(2) cos 2φ aφ
V

 o
2
∂ρ  ∂ρ 
ρ
100
At=P,Aπ / 2
⎯⎯
→ A=
50
2 π
1− E∂ =(2c∂1 V− 2c )(1)a1 + a0
ρ−
) sin φ 100
o(
ρ
 1=2 VdV
3 A
ρ
∂
ρ
∂
ρ
ρ
ρ


E
a = − = caφ− c
− →
V=
| E=|=−∇
100
=−
2cρ1 −d2φc2aφ = ⎯⎯
(2)
1
2
ρ φ 50πρ
2
2
∂V
a
−Voand
( ρ −(2), ) sin φc1 = 50, c2 = 0
From= (1)
ρ
∂φ 2
1 ∂ 2V
1 a2
= −Vo ( − 3 ) sin φ
Prob.
6.15
2
2
ρ
φ
ρ ρ
∂
(a)
2
∂V2V = 1 ∂ a 2ρ ∂V  + ∂ V = 0
∇
= Vo (1 +  2 ) sin φ
ρ ∂ρ ρ ∂ρ  ∂φ 2
∂ρ
ρ
∂V
a2
= Vo ( ρ + ) sin φ
ρ
∂ρ
∂  ∂V 
a2
ρ
 = Vo (1 − 2 ) sin φ
∂ρ  ∂ρ 
ρ
1 ∂  ∂V 
1 a2
ρ
=
− ) sin φ
(
V
ρ ∂ρ  ∂ρ  o ρ ρ 3
∂ 2V
a2
ρ
=
−
−
(
) sin φ
V
o
ρ
∂φ 2
1 ∂ 2V
1 a2
=
−
− ) sin φ
(
V
o
ρ 2 ∂φ 2
ρ ρ3
∇ 2V =
1 ∂  ∂V
ρ
ρ ∂ρ  ∂ρ
 ∂ 2V
+ 2 = 0
 ∂φ
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Principles of Electromagnetics, 6e
157
153
(b)
2
If ρ 2  a 2 , then a
E = −∇V = −
ρ2
 1 and V  Vo ρ sin φ
∂V
1 ∂V
aρ −
a = −Vo sin φ aρ − Vo cos φ aφ
ρ ∂φ φ
∂ρ
Prob. 6.16
d 2V
∇ 2V = 2 = 0
⎯⎯
→
V = Ax + B
dx
At x = 20 mm = 0.02 m, V = 0
0 = 0.02 A + B
⎯⎯
→ (1)
E=−
From (1)
Then
At x = 0
dV
ax
dx
⎯⎯
→ A = 110 ⎯⎯
→ (2)
B = −0.02 A = −2.2
V = 110 x − 2.2
V = −2.2V
At x = 50 mm = 0.05 m,
V = 110 × 0.05 − 2.2 = 3.3V
Prob. 6.17
∇ 2V = 0
⎯⎯
→
V = −A / r + B
At r=0.5, V=-50
Or
-50 = -A/0.5 + B
-50 = -2A + B
At r = 1, V =50
(1)
50 = -A + B
(2)
From (1) and (2), A = 100, B = 150, and
V =−
100
+ 150
r
E = −∇V = −
A
100
a = − 2 ar V/m
2 r
r
r
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Principles of Electromagnetics, 6e
158
154
Prob. 6.18 From Example 6.4,
 tan θ / 2 
Vo ln 

tan θ1 / 2 

V=
 tan θ 2 / 2 
ln 

 tan θ1 / 2 
Vo = 100, θ1 = 30o , θ 2 = 120o , r = 32 + 02 + 42 = 5,
θ = tan −1 ρ / z = tan −1 3 / 4 = 36.87 o
 tan 18.435 o 
ln

 tan 15 o 
. V
V = 100
= 117
 tan 60 o 
ln

 tan 15 o 
E=
−Vo aθ
−100aθ
=
= −17.86aθ V/m
 tan θ 2 / 2  5sin 36.87o ln 6.464
r sin θ ln 

 tan θ1 / 2 
Prob. 6.19
(a)
1 ∂  ∂V
ρ
∇ 2V =
ρ ∂ρ  ∂ρ
V ( ρ = b) = 0

=0

⎯⎯
→
V ( ρ = a) = Vo
⎯⎯
→
V =−
⎯⎯
→
V = A ln ρ + B
0 = A ln b + B
⎯⎯
→
Vo = A ln a / b
⎯⎯
→
Vo
V ln b / ρ
ln ρ / b = o
ln b / a
ln b / a
V ( ρ = 15mm)=70
B = − A ln b
Vo
A=−
ln b / a
ln2
= 12.4 V
ln50
(b) As the electron decelerates, potential energy gained = K.E. loss
e[70 − 12.4] =
1
m[(107 ) 2 − u 2 ]
2
u 2 = 1014 −
⎯⎯
→
1014 − u 2 =
2e
× 57.6
m
2 × 1.6 × 10−19
× 57.6 = 1012 (100 − 20.25)
9.1× 10−31
u = 8.93 ×106 m/s
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Principles of Electromagnetics, 6e
159
155
Prob. 6.20 This is similar to case 1 of Example 6.5.
X = c1 x + c2 ,
X (0) = 0
But
Y = c3 y + c4
⎯⎯
→
0 = c2 ,
Y (0) = 0
0 = c4
⎯⎯
→
Hence,
V ( x, y ) = XY = ao xy,
V ( xy = 4) = 20
Also,
Thus,
ao = c1c3
⎯⎯
→
20 = 4ao
⎯⎯
→
ao = 5
V ( x, y ) = 5 xy and E = −∇V = −5 ya x − 5 xa y
At (x,y) = (1,2),
V = 10 V, E = −10a x − 5a y V/m
Prob. 6.21
(a) As in Example 6.5, X ( x) = A sin(nπ x / b)
For Y,
Y ( y ) = c1 cosh(nπ y / b) + c2 sinh(nπ y / b)
Y (a) = 0
⎯⎯
→
0 = c1 cosh(nπ a / b) + c2 sinh(nπ a / b) ⎯⎯
→
c1 = −c2 tanh(nπ a / b)
∞
V =  an sin(nπ x / b) [sinh(nπ y / b) − tanh(nπ a / b)cosh(nπ y / b)]
n =1
∞
V ( x, y = 0) = Vo = − an tanh(nπ a / b)sin(nπ x / b)
n =1
 4Vo
b
, n = odd
2

− an tanh(nπ a / b) =  Vo sin(nπ x / b)dx =  nπ
b0
 0, n = even
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Principles of Electromagnetics, 6e
160
156
Hence,
V =−
=−
4Vo
 sinh(nπ y / b)
∞
 sin(nπ x / b)  n tanh(nπ a / b) −
π

n = odd
cosh( nπ y / b) 

n

sin(nπ x / b)
[sinh(nπ y / b)cosh(nπ a / b) − cosh(nπ y / b)sinh(nπ a / b)]
π n = odd n sinh(nπ a / b)
4Vo
∞

=
sin(nπ x / b)sinh[nπ (a − y ) / b]
n sinh(nπ a / b)
π n = odd
4Vo
∞

Alternatively, for Y
Y ( y ) = c1 sinh nπ ( y − c2 ) / b
Y (a) = 0
0 = c1 sinh[nπ (a − c2 ) / b]
⎯⎯
→
⎯⎯
→
c2 = a
∞
V =  bn sin(nπ x / b)sinh[ nπ ( y − a) / b]
where
n =1
4Vo

,
−
bn =  nπ sinh(nπ a / b)

0,

n = odd
n = even
(b) This is the same as Example 6.5 except that we exchange y and x. Hence
V ( x, y ) =
sin(nπ y / a )sinh(nπ x / a )]
n sinh(nπ b / a)
π n = odd
4Vo
∞

(c) This is the same as part (a) except that we must exchange x and y. Hence
V ( x, y ) =
sin(nπ y / a )sinh[nπ (b − x ) / a ]
π n = odd
n sinh(nπ b / a)
4Vo
∞

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Principles of Electromagnetics, 6e
161
157
Prob. 6.22 (a) X(x) is the same as in Example 6.5. Hence
∞
V ( x, y ) =  sin( nπ x / b) [ an sinh( nπ y / b) + bn cosh( nπ y / b) ]
n =1
At y=0, V = V1
∞
V1 =  bn sin(nπ x / b)
⎯⎯
→
n =1
At y=a, V = V2
 4V1
 nπ ,

bn = 
 0,


n = odd
n = even
∞
V2 =  sin(nπ x / b) [ an sinh(nπ a / b) + bn cosh(nπ a / b) ]
n =1
 4V2
 nπ , n = odd

an sinh( nπ a / b) + bn cosh(nπ a / b) = 
 0, n=even


or
4V2

 nπ sinh(nπ a / b) (V2 − V1 cosh(nπ a / b) ) , n = odd

an = 

0, n=even


Alternatively, we may apply superposition principle.
V2
y
0
0 0
V
V1
0
V2
x
0 0
VA
0
0
VB
V1
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Principles of Electromagnetics, 6e
162
158
i.e. V = V A + VB
VA is exactly the same as Example 6.5 with Vo = V2 , while VB is exactly the same
as Prob. 6.19(a). Hence
4 ∞ sin(nπ x / b)
V=
V sinh[nπ (a − y ) / b] + V2 sinh(nπ y / b)

π n = odd n sinh(nπ a / b) 1
[
(b)
]
V ( x, y ) = (a1e −α x + a2e+α x )(a3 sin α y + a4 cos α y )
lim V ( x, y ) = 0
⎯⎯
→
x ⎯⎯
→∞
a2 = 0
V ( x, y = 0) = 0
⎯⎯
→
a4 = 0
V ( x, y = a ) = 0
Hence,
⎯⎯
→
α = nπ / a,
n = 1, 2,3,...
∞
V ( x, y ) =  an e− nπ x / a sin(nπ y / a)
n =1
∞
V ( x = 0, y ) = Vo =  an sin( nπ y / a)
⎯⎯
→
n =1
V ( x, y ) =
 4Vo
, n = odd

an =  nπ
0,
n = even
sin(nπ y / a )
exp(− nπ x / a)
π n = odd
n
4Vo
∞

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Principles of Electromagnetics, 6e
163
159
(c) The problem is easily solved using superposition theorem, as illustrated below.
y
V3
a
V2
V
V4
x
0
0
b
V1
V2
0
0
VI
VII
0
0
V1
0
0
V3
VIV
0
VIII
0
V4
0
0
Therefore,
V = VI + VII + VIII + VIV
 sin(nπ x / b)
[V1 sinh(nπ (a − y) / b) + V3 sinh(nπ y / b)] 

4
1  sinh(nπ a / b)

=  

π n =odd n  sin(nπ x / a)
+
[V sinh(nπ y / a) + V4 sinh(nπ (b − x) / a)]
 sinh(nπ b / a) 2

∞
where
VI =
VII =
sin( nπ x / b)sinh[nπ ( a − y ) / b]
n sinh(nπ a / b)
π n = odd
4V1
∞

sin( nπ x / a)sinh( nπ y / a)
n sinh(nπ b / a )
π n = odd
4V2
∞

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Principles of Electromagnetics, 6e
164
160
VIII =
VIV =
sin( nπ x / b)sinh( nπ y / b)
n sinh(nπ a / b)
π n = odd
4V3
∞

sin(nπ y / a )sinh[nπ (b − x ) / a ]
n sinh(nπ b / a)
π n = odd
4V4
∞

Prob. 6.23
E = −∇V = −
∂V
∂V
ax −
ay
∂x
∂y
nπ sin(nπ y / a )
exp(−nπ x / a )
n
π n =odd a
4V ∞ nπ cos(nπ y / a )
Ey = − o 
exp(−nπ x / a )
n
π n =odd a
4V ∞
E = o  exp(− nπ x / a ) sin(nπ y / a )a x − cos(nπ y / a )a y 
a n =odd
Ex =
4Vo
∞

Prob. 6.24
This is similar to Example 6.5 except that we must exchange x and y. Going through the
same arguments, we have
 nπ x   nπ y 
V ( x, y ) =  cn sinh 
 sin 

 b   b 
Applying the condition at x=a, we get
π y 
 nπ a   nπ y 
Vo sin 
 =  cn sinh 
 sin 

 b 
 b   b 
This yields
 nπ a  Vo ,
cn sinh 
=
 b   0,
Hence,
n =1
n ≠1
πx  π y 
sinh 
 sin 

b   b 

V ( x, y ) = Vo
πa 
sinh 

 b 
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Principles of Electromagnetics, 6e
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161
Prob. 6.25
∇ 2V =
1 ∂  ∂V
ρ
ρ ∂ρ  ∂ρ
 1 ∂ 2V
=0
+ 2
2
 ρ ∂φ
If we let V ( ρ ,φ ) = R ( ρ )Φ (φ ),
Φ ∂
ρ ∂ρ
or
( ρ R ') +
1
ρ2
RΦ '' = 0
ρ ∂
Φ ''
( ρ R ') = −
=λ
R ∂ρ
Φ
Hence
Φ ''+ λΦ = 0
163
and
Prob. 6.29
∂
λR
( ρ R ') −
=0
∂ρ
ρ
This is the same as Problem 6.30 except that α = π . Hence,
or
1
1 1 1
1 1
R=
 − =
 − 
2πσ (1 −Rcos
' πλ)R a b  4πσ  a b 
R ''+ − 2 = 0
ρ
ρ
Prob. 6.26
6.30
For a spherical capacitor, from Eq. (6.38),
1 ∂  2 ∂V 
1
∂
∂V
(sinθ
)=0
1 r 1  + 2
2
r ∂ r  − ∂ r  r sinθ ∂θ
∂θ
a b
If V (r ,Rθ )== R(r ) F (θ ), r ≠ 0,
4πσ
∇2V =
d hemisphere,
R Rd ' = 2 R since the sphere consists of two hemispheres in parallel.
ForF the
(r 2 R ') +
(sin θ F ') = 0
sin θ dθ
As dr
b
⎯⎯
→ ∞,
Dividing through by RF gives
1 1
2 − 
1 Rd' = 2lim  a 1b  =d 1
(r R ') = −
(sin θ F ') = λ
sin θ dθ2π aσ
R dr b ⎯⎯→ ∞ 4Fπσ
G = 1 / R ' = 2π aσ
Hence,
sin θ F ''+ cos θ F '+ λ F sin θ = 0
or
F ''+ cot θ F '+ λ F = 0
Also,
POESM_Ch06.indd 165
d 2
(r R ') − λ R = 0
dr
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d
1 d 2
1
(r R ') = −
(sin θ F ') = λ
R dr
F sin θ dθ
Sadiku & Kulkarni
Hence,
Principles of Electromagnetics, 6e
166
sin θ F ''+ cos θ F '+ λ F sin θ = 0
or
F ''+ cot θ F '+ λ F = 0
Also,
162
Alternatively,
for an isolated sphere, C = 4πε a. But
d 2
(r R ') − λ R = 0
dr
1
ε
or
⎯⎯
→
RC =
R=
4π aσ
σ
2R ' λ
R ''+
− 21R = 0
r
or G = 2π aσ
R ' = 2R = r
2π aσ
Prob. 6.27 If the centers at φ = 0 and φ = π / 2 are maintained at a potential difference
of Vo, from Example 6.3,
Eφ =
Hence,
2Vo
πρ
,
I =  J • dS =
and
R=
J =σE
2Voσ
π
b
t
 
ρ
=a z =0 ρ
d ρ dz =
2Voσ t
π
ln(b / a )
Vo
π
=
I 2σ t ln(b / a )
Prob. 6.28 If V (r = a ) = 0,
E=
1
Vo
,
r (1 / a − 1 / b)
V (r = b) = Vo , from Example 6.9,
J =σE
2
Hence,
α
2π
Voσ
1 2
2π Voσ
I =  J • dS =
r sin θ dθ dφ =
( − cosθ ) |0α
2


1 / a − 1 / b θ =0 φ =0 r
1 / a −1 / b
1 1
−
Vo
a
b
=
R=
I 2πσ (1 − cos α )
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POESM_Ch06.indd 166
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R ∂ρ
Hence
Φ
Sadiku & Kulkarni
Principles of Electromagnetics, 6e
Φ ''+ λΦ = 0
167
163
164
and
∂
λR
=0
Prob. 6.29( ρ R ') −
∂ρ
ρ
P.E. 6.17
or
(a)
thesame
parallel-plate
capacitor,
ThisFor
is the
as Problem
6.30 except that α = π . Hence,
Vo
E = −1 R 'a x λ R 1 1 
1 1 1
d −  = −0  =
R=
 − 
R
''
+
2πσ (1 − ρ
cos πρ) 2 a b  4πσ  a b 
From Example 6.11,
1
1
V2
ε
εS
Prob.C6.30
= 2  ε | E |2 dv = 2  ε o2 dv = 2 Sd =
Prob.
6.26
For
a spherical
Vo capacitor, from
Vo Eq.
d (6.38),
d
d
(b) For the cylindrical capacitor,
1 ∂ 1 1∂ V 
1
∂
∂V
(sinθ
)=0
∇2V = 2  −r 2
+ 2

Vrao b∂ r  r sinθ ∂θ
r
∂
∂θ
=
R
E=−
aρ
/(ar ) F (θ ), r ≠ 0,
If V (r ,θρ)ln= b4Rπσ
For
theExample
hemisphere,
From
6.8, R ' = 2 R since the sphere consists of two hemispheres in parallel.
AsF d (r 2 R ') + R d (sin θ F ') = 0
b dr⎯⎯
→ sin
,θ dθ
b
1
ε∞
Vo2
2πε L
dρ
2πε L
C = 2 
ρ
d
ρ
d
φ
dz
=
=
1
1
2 
2 

Vo through
Dividing
/ a−)RF gives
( ρ ln2 b by
( ln b / a ) a ρ ln b / a
1
a b 

R ' = lim
=
b ⎯⎯
→∞
4πσ1
d
d 2π aσ
1
2
(c) For (the
r Rspherical
') = − capacitor,
(sin θ F ') = λ 162
R dr
F sin θ dθ
G = 1 / R ' = 2π aσ
Vo
E
ar
=
Hence,
2
Alternatively,
r (1/ afor
− 1/anb)isolated sphere, C = 4πε a. But
sin θ F ''+ cos
ε θ F '+ λ F sin θ = 0 1
From Example
→
RC = 6.10,⎯⎯
R=
σ
4π aσ
or
b
2
1
εVo
ε
dr
4πε
2
C = 2  4
r
sin
θ
d
θ
drd
φ
4
π
=
=
1
2
2
2

1 1
r
F V''+o cot
θrF2(R'1+/=λaF− =1 /0b ) or G = 2π aσ (1 / a − 1 / b )
R' =
a
−
2π aσ
a b
Also,
Prob.
6.27 If the centers at φ = 0 and φ = π / 2 are maintained at a potential difference
Prob.
d 6.31
r 2 R ') Example
− λ R = 0 6.3,
of Vo,(from
dr
Cd 2 ×10−9 × 10−6 2
orC = ε S
⎯⎯
S=
=
m = 0.5655 cm 2
2Vo→
−9
,
J = σεEoε r 4 × 10 / 36π
dEφ =
2πρ
R' λ
− 2 R=0
Hence, R ''+
r
r
I =  J • dS =
and
R=
2Voσ
π
b
 
ρ
POESM_Ch06.indd 167
Hence,
1
=a z =0 ρ
d ρ dz =
2Voσ t
π
ln(b / a )
Vo
π
=
I 2σ t ln(b / a )
Prob. 6.28 If V (r = a ) = 0,
E=
t
V (r = b) = Vo , from Example 6.9,
Vo
J =σE
,
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r (1 / a − 1 / bCopyright
)
2
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Sadiku & Kulkarni
Principles of Electromagnetics, 6e
168
165
Prob. 6.32
This can be regarded as three capacitors in parallel.
C1
C = C1 + C2 + C3 = 
εo
C2
C3
ε oε rk S k
dk
3 × 15 × 10−2 × 20 × 10−2 + 5 × 15 × 10−2 × 20 × 10−2 + 8 × 15 × 10−2 × 20 × 10−2 
2 ×10
10−9 15 × 10−2 × 20 × 10−2
[3 + 5 + 8] = 2.122 nF
=
×
36π
2 × 10−3
=
−3
Prob. 6.33
This may be regarded as three capacitors in series.
C1
C2
C3
3
1
1
1
1
d
1×10−3
= +
+
=  k = −9
Ceq C1 C2 C3 k =1 ε k Sk 10
× 80 × 10−4
36π
Ceq =
 1 1 1  36π
9
 3 + 5 + 8  = 8 × 10 × 0.6583
8
nF=0.1076 nF
36π x0.6583
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Sadiku & Kulkarni
Principles of Electromagnetics, 6e
169
166
Prob. 6.34
A
C1
C1
d
C3
C3
C2
C2
From the figure above,
C=
here
C1 =
C=
C1C2
+ C3
C1 + C2
εo A / 2
d/2
=
εo A
d
,
C2 =
ε oε r A
d
,
C3 =
εo A
2d
ε o 2ε r A2 / d 2
ε A ε A 1
ε  10−9 10 × 10−4  1 6 
+ o = o  + r =
 +  ≅ 6 pF
ε o (ε r + 1) A / d 2d
d  2 ε r + 1  36π 2 × 10−3  2 7 
Prob. 6.35
C=
εoS
d
⎯⎯
→S =
Cd
εo
−3
1 × 1 × 10
= 36π × 106
−9
10 / 36π
S = 1.131 × 108 m 2
S=
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Sadiku & Kulkarni
Principles of Electromagnetics, 6e
170
167
Prob. 6.36
Fdx = dWE
⎯⎯
→
F=
dWE
dx
1
1
1
WE =  ε | E |2 dv = ε oε r E 2 xad + ε o E 2 da(1 − x)
2
2
2
where E = Vo / d .
dWE 1 Vo 2
= ε o 2 (ε r − 1)da
dx
2 d
⎯⎯
→
F=
ε o (ε r − 1)Vo 2 a
2d
1
WE = CVo 2 , where
2
ε ε ax ε ε (1 − x)
C = C1 + C2 = o r + o r
d
d
2
dWE 1 Vo a
(ε r − 1)
= εo
2
dx
d
Alternatively,
F=
ε o (ε r − 1)Vo 2 a
2d
Prob.6.37
(a)
10−9
4πε
36π = 25 pF
=
C=
1 1
1
1
−
−
−2
a b 5 x10
10 x10−2
4π × 2.25 ×
(b)
Q = C Vo= 25x80 pC
ρs =
Q
25 × 80
=
pC/m 2 = 63.66 nC/m 2
2
4π r
4π × 25 × 10−4
Prob. 6.38
εεS
ε S
C1 = o r , C2 = o
d
d
C1
56μ F
= εr
⎯⎯
→ εr =
= 1.75
32μ F
C2
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POESM_Ch06.indd 170
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Sadiku & Kulkarni
Principles of Electromagnetics, 6e
171
168
Prob. 6.39
d 2V
=0
⎯⎯
→ V = Az + B
dz 2
At z=0, V=40
⎯⎯
→ 40=B
At z=2mm=d, V=0
V =−
⎯⎯
→ 0 = Ad + 40
⎯⎯
→
A = −40 / d
40
40 z
+ 40 = −20 × 103 z + 40 V
z + 40 = −
−3
2 × 10
d
Prob. 6.40
10−9
ε S 6.8 × 36π × 0.5
(a) C =
=
= 7.515 nF
d
4 × 10−3
Q
Q
, C=
⎯⎯
→ Q = CV
S
V
(b)
7.515 ×10−9 × 9
CV
ρs = ±
=±
= ±135.27 nC/m 2
0.5
S
ρs =
Prob. 6.41
Q2
WE =
2C
ε oε r S
C=
d
Q2d
WE =
2ε 0ε r S
When the plate spacing is doubled
WE =
Q 2 (2d ) Q 2 d
=
2ε 0ε r S ε 0ε r S
When the plate spacing is halved
2
Q 2 (d )
2 = Qd
WE =
2ε 0ε r S 4ε 0ε r S
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Sadiku & Kulkarni
Principles of Electromagnetics, 6e
172
169
Prob.6.42
This can be treated as three capacitors in series
C1 =
C2 =
ε1S
=
ε 2S
=
d1
d2
ε 3S
ε o (2.5)(0.4)
1.5 ×10−3
ε o (5.6)(0.4)
1.5 × 10−3
= 666.67ε o
= 1493.33ε o
ε o (8.1)(0.4)
= 1620ε o
d3
2 ×10−3
1 1
1
1 0.002787
= +
+
=
C C1 C2 C3
ε0
C3 =
=
10−9
C = 358.81ε o = 358.81×
36π
C = 3.1726 nF
Prob. 6.43
(a)
C=
εoS
d
=
10−9 200 × 10−4
= 59 pF
36π 3 × 10−3
(b) ρ s = Dn = 10−6 nC/m 2 . But
Dn = ε En =
or
Vo =
(c)
ε oVo
d
= ρs
ρs d
= 10−6 × 3 × 10−3 × 36π × 109 = 339.3 V
εo
2
Q2
ρ s S 10−12 × 200 ×10−4 × 36π ×109
F=
=
=
= 1.131 mN
2Sε o
2ε o
2
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POESM_Ch06.indd 172
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Sadiku & Kulkarni
Principles of Electromagnetics, 6e
173
170
Prob. 6.44
C1
b
c
C2
d
C3
a
1
1
1
1
=
+
+
C C1 C2 C3
4πε 3
where C1 =
,
1 1
−
b a
C2 =
4πε 2
,
1 1
−
c b
C3 =
4πε1
,
1 1
−
d c
4π 1/ b − 1/ a 1/ c − 1/ b 1/ d − 1/ c
=
+
+
C
ε3
ε2
ε1
4π
C=
ε1
ε
ε
+ 2 + 3
1 1 1 1 1 1
−
−
−
d c c b b a
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POESM_Ch06.indd 173
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Sadiku & Kulkarni
Principles of Electromagnetics, 6e
174
171
Prob. 6.45
We may place a charge Q on the inner conductor. The negative charge –Q is on the outer
surface of the shell. Within the shell, E = 0, i.e. between r=c and r=b. Otherwise,
Q
E=
ar
4πε o r 2
The potential at r=a is
a
Va = −  E dl = −
−∞
=−
C=
Q
4πε o
Q
=
Va
c
dr
r
−∞
2
c

−∞
b
a
Er dr −  Er dr −  Er dr
c
−0−
Q
4πε o
b
a
dr
r
b
2
=
Q
+
4πε o c
Q 1 1
 − 
4πε o  a b 
1
1
1 1 1
+
 − 
4πε o c 4πε o  a b 
Prob. 6.46
10−9
× 3 × 103
2π × 2.5 ×
2πε L
36π
=
= 0.8665 μ F
C=
ln(b / a)
ln(8 / 5)
Prob. 6.47
Let the plate at φ=0 be 0, i.e. V(0)=0 and let the plate at φ=π/4 be Vo , i.e. V(π/4)=Vo.
1 d 2V
dV
∇ 2V = 2
=0
⎯⎯
→
=A
⎯⎯
→ V = Aφ + B
2
ρ dφ
dφ
V (0) = 0
⎯⎯
→ 0 = 0+ B
V (π / 4) = Vo
⎯⎯
→ B=0
⎯⎯
→ Vo = Aπ / 4
⎯⎯
→
A=
4V
1 dV
A
aφ = − aφ = − o aφ
ρ dφ
ρ
πρ
4ε Vo
D =εE = −
aφ
4Vo
π
E = −∇V = −
πρ
4ε Vo
ρ s = Dn = −
πρ
Q =  ρ s dS = −
C=
b
L
 
ρ
= a z =0
4ε Vo
πρ
d ρ dz = −
4ε Vo
π
L ln(b / a)
| Q | 4ε L
=
ln(b / a)
Vo
π
Copyright © 2015 by Oxford University Press
POESM_Ch06.indd 174
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Principles of Electromagnetics, 6e
175
172
Prob. 6.48
C=
2πε L
=
ln(b / a )
2π ×
10−9
× 4.2 × 400 × 10−3
36π
= 74.5 pF
ln(3.5 /1)
Prob. 6.49
10−9
× 100 × 10−6
2π ×
2πε o L
36π
C=
=
= 1.633 × 10−15 F
ln(b / a )
ln(600 / 20)
V = Q/C =
50 ×10−15
= 30.62 V
1.633 ×10−15
Prob. 6.50
2πε1
2πε 2
, C2 =
C1 =
ln(b / a)
ln(c / b)
Since the capacitance are in series, the total capacitance per unit length is
2πε1ε 2
CC
C= 1 2 =
C1 + C2 ε 2 ln(b / a ) + ε1 ln(c / b)
Prob. 6.51
E=
Q
4πε r 2
ar
Q
Vo
0
W=
1
Q2
2
ε
E
=
ε r 2 sin θ dθ dφ dr
|
|
dv
2 2 4


2
32π ε r
b
Q2
dr Q 2  1 1 
=
π
(2
)(2)
c r 2 = 8πε  c − b 
32π 2ε
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POESM_Ch06.indd 175
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Sadiku & Kulkarni
Principles of Electromagnetics, 6e
176
173
Q 2 (b − c)
W=
8πε bc
Prob. 6.52
ρs
(−a x ) , where ρ s is to be determined.
ε
d
d
−ρ
1 d
ρ
Vo = −  E • dl = −  s dx = ρ s 
dx = s d ln( x + d )
0
ε
ε d+x
ε
0 o
E=
(a) Method 1:
Vo = ρ s d ln
E=−
2d
d
⎯⎯
→
ρs =
Voε o
d ln 2
ρs
Vo
ax = −
ax
( x + d )ln 2
ε
Method 2: We solve Laplace’s equation
∇ • (ε∇V ) =
d
dV
(ε
)=0
dx dx
⎯⎯
→
ε
dV
=A
dx
dV A
Ad
c
= =
= 1
dx ε ε o ( x + d ) x + d
V = c1 ln( x + d ) + c2
V ( x = 0) = 0
⎯⎯
→
V ( x = d ) = Vo
c1 =
⎯⎯
→
0 = c1 ln d + c2
⎯⎯
→
c2 = −c1 ln d
Vo = c1 ln 2d − c1 ln d = c1 ln 2
Vo
ln 2
V = c1 ln
E=−
x + d Vo
x+ d
=
ln
d
d
ln 2
dV
Vo
ax = −
ax
( x + d )ln 2
dx
Copyright © 2015 by Oxford University Press
POESM_Ch06.indd 176
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Principles of Electromagnetics, 6e
177
174
(b)
ε oVo
ε o xVo
 x+d 
P = (ε r − 1)ε o E = − 
− 1
ax = −
ax
 d
 ( x + d ) ln 2
d ( x + d ) ln 2
(c)
x=d
x=0
ρ ps |x = 0 = P • ( − a x ) |x = 0 = 0
ρ ps |x = d = P • a x |x = d = −
(d)
E=
ε oVo
2d ln 2
ρs
Q
Q
ax =
ax =
ax
x
ε
εS
ε o (1 + ) S
d
Q
dx
Q
V = −  E dl = −
d ln 2
=

ε o S a (1 + x ) ε o S
d
ε S
Q
C= = o
V d ln 2
d
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POESM_Ch06.indd 177
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Principles of Electromagnetics, 6e
178
175
Prob. 6.53
We solve Laplace’s equation for an inhomogeneous medium.
d  dV 
dV
∇(ε∇V ) =  ε
⎯⎯
→ ε
=A
=0
dx  dx 
dx
2
dV A
A  x 
1
= =
+
   
dx ε 2ε o   d  
A
x3
(x + 2 ) + B
V=
2ε o
3d
When x=d, V=Vo ,
Vo =
A
d
(d + ) + B
2ε o
3
Vo =
⎯⎯
→
0=−
When x = -d, V=0,
A
d
0=
(−d − ) + B
2ε o
3
Vo = 2 B
Adding (1) and (2),
2 Ad
+B
3ε o
⎯⎯
→
2 Ad
+B
3ε o
(1)
(2)
⎯⎯
→ B = Vo / 2
From (2),
B=
2 Ad Vo
=
3ε o
2
A=
⎯⎯
→
3ε oVo
4d
  x 2 
1 +   
2
−3Vo   x  
3ε oVo   d  
dV
A
E = −∇V = −
ax = − ax = −
ax = −
1 +    a x
dx
4d
2ε o
8d   d  
ε
ρ s = Dan = ε E ax
x=d
Q =  ρ s dS = ρ s S = −
S
= −A = −
3ε oVo
4d
3Sε oVo
4d
| Q | 3ε o S
=
C=
Vo
4d
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POESM_Ch06.indd 178
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Principles of Electromagnetics, 6e
179
176
Prob. 6.54
Method 1: Using Gauss’s law,
Q =  D • dS = 4π r 2 Dr
Q
E = D/ε =
4πε o k
V = −  E • dl = −
C=
D=
⎯⎯
→
Q
ar ,
4π r 2
ε=
ε ok
r2
ar
Q
a
Q
(b − a )
dr = −
4πε k 
4πε k
o
o
b
Q 4π ε o k
=
|V | b − a
Method 2: Using the inhomogeneous Laplace’s equation,
∇ • (ε∇V ) = 0
dV
= A'
dr
V (r = a ) = 0
ε ok
⎯⎯
→
⎯⎯
→
⎯⎯
→
V (r = b) = Vo
E=−
⎯⎯
→
Vo = Ab + B = A(b − a)
dV
V
ar = − Aar = − o ar
dr
b−a
ρ s = Dn = −
⎯⎯
→
A=
Vo
b−a
Vo ε o k
|r = a , b
b − a r2
Q =  ρ s dS = −
C=
1 d  ε o k 2 dV 
r

=0
r 2 dr  r 2
dr 
dV
= A or V = Ar + B
dr
0 = Aa + B
⎯⎯
→
B = − Aa
Voε o k 1 2
Vε k
r sin θ dθ dφ = − o o 4π
2

b−a r
b−a
| Q| 4π ε o k
=
Vo
b− a
Prob. 6.55
10−9
C = 4πε o a = 4π ×
× 6.37 × 106 = 0.708 mF
36π
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177
Prob.6.56
Q
C=
V
Q
D=
aρ
2πρ L
D
Q
E= =
ε 2πρ Lε o (3)(1 + ρ )
Q
V = −  E dl =
6π Lε o
b
dρ
 ρ (1 + ρ )
a
1
A
B
= +
ρ (1 + ρ ) ρ 1 + ρ
Using partial fractions
Let
A=1, B= -1
b dρ b dρ 
−


6πε 0 L  a ρ a 1 + ρ 
Q
=
[ ln ρ − ln(1 + ρ )]
6πε 0 L
Q
V=
b
a
Q
b
a 

− ln
ln

6πε 0 L  1 + b
1 + a 
If a=1 mm, and b=5 mm
=
C=
Q
|V |
=
6πε o
b
a
− ln
1+ b
1+ a
−9
10
6π ×
36π
=
5
1
ln − ln
6
2
1
× 10−9
1
6
=
= × 1.9591 nF
ln 0.8333 − ln 0.5 6
C = 0.326 nF
ln
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178
Prob. 6.57
o
y=2
y=4
y=8
y=-4
y=0
y=2
y=4
y=8
At P(0,0,0), E=0 since E does not exist for y<2.
At Q(-4,6,2), y=6 and
ρs
10−9
E =
an =
(−30a y + 20a y − 20a y − 30a y ) = 18π (−60)a y
2ε o
2 x10−9 / 36π
= −3.4a y kV/m
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179
Prob. 6.58
4nC
-2
4
3nC
-3nC
-1
0
3
1
2
-4nC
2
1
(a) Qi = -(3nC – 4nC) = 1nC
(b) The force of attraction between the charges and the plates is
F = F13 + F14 + F23 + F24
| F |=
10−18
4π × 10−9 / 36π
 9 2(12) 16 
 22 − 32 + 42  = 5.25 nN
Prob. 6.59
We have 7 images as follows: -Q at (-1,1,1), -Q at (1,-1,1), -Q at (1,1,-1),
-Q at (-1,-1,-1), Q at (1,-1,-1), Q at (-1,-1,1), and Q at (-1,1,-1). Hence,
(2a x + 2a y + 2a z ) (2a y + 2a z ) 
 2
2
2
− 3 ax − 3 a y − 3 az −
+


3/ 2
Q
2
2
2
12
83/ 2


F=
4πε o  (2a x + 2a y ) (2a x + 2a z )

+
 +

3/ 2
3/ 2
8
8
2
1
1 
 1
= 0.9(a x + a y + a z )  − −
+
 = −0.1092(a x + a y + a z ) N
 4 12 3 4 2 
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180
Prob. 6.60
 360 o 
N=
− 1 = 7

 45 o
Prob. 6.61
(a)
ρL  aρ1 aρ 2  16 × 10−9  (2, −2,3) − (3, −2,4)
(2, −2,3) − (3, −2, −4) 
−
−
E = E+ + E − =

=
−9 
2
10  | (2, −2,3) − (3, −2,4) | | (2, −2,3) − (3, −2, −4) |2 
2πε o  ρ1 ρ2 
2π ×
36π
 (−1,0, −1) (−1,0,7) 
= 18 x16 
−
= −138.2a x − 184.3a y V/m
2
50 

(b) ρ s = Dn
D = D+ + D− =
=
ρL  aρ1 aρ 2  16 x10−9  (5, −6,0) − (3, −6,4)
(5, −6,0) − (3, −6, −4) 
−
−

=


2
2π  ρ1 ρ2 
2π  | (5, −6,0) − (3, −6,4) | | (5, −6,0) − (3, −6, −4) |2 
8  (2, 0, −4) (2, 0, 4) 
−
nC/m 2 = −1.018a z nC/m 2
π  20
20 
ρ s = −1.018 nC/m 2
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181
Prob. 6.62
The images are shown with proper sign at proper locations. Figure does not show the
actual direction of forces but they are expressed a follows:
z
+Q (0, 0, a)
F2
F1
F3
-Q(a, 0, a)
y
O
- Q (0, 0, -a)
x
F1 =
Q2
4πε o
PEC
+Q (a, 0, -a)
 −a x 
 a 2 


Q 2  aa x + 2 aa z 
F2 =
4πε o  a 2 + 4a 2 3 




2
Q  −a z 
F3 =
4πε o  4a 2 
(
Ftotal
=
Q2
=
4πε o a 2
)
 1
1 

 2
− 1 a x + 
−  az 

 5 5 4 
 5 5 
Q2
 −0.91a x − 0.071a y  N
4πε o a 2 
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182
CHAPTER 7
P.E. 7.1
z
α1
α2
2
27
−a x + a y
 −a − a y 
aφ = al × a ρ =  x
 × az =
2 
2

ρ = 5, cos α1 = 0, cos α 2 =
H3 =
  −a + a y 
10  2
− 0   x

 = −30.63a x + 30.63a y mA/m
4π (5)  27
2 

P.E. 7.2
2 
3 
1+

 az = 0.1458az A/m
4π (2) 
13 
12
ρ = 32 + 42 = 5, α 2 = 0, cos α1 = − ,
(b)
13
(a) H =
 3a − 4a z  4a x + 3a z
aφ = −a y ×  x
=
5
5


2  12   4a x + 3az 
1
H=
( 4ax + 3az )
1 + 
=
4π (5)  13 
5
 26π
= 48.97a x + 36.73a z mA/m
P.E. 7.3
(a) From Example 7.3,
Ia 2
H=
az
2(a 2 + z 2 )3/ 2
At (0,0,-1cm), z = 2cm,
50 × 10−3 × 25 ×10−4
H=
a z = 400.2a z mA/m
2(52 + 22 )3/ 2 × 10−6
(b) At (0,0,10cm), z = 9cm,
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183
H=
50 × 10−3 × 25 ×10−4
a z = 57.3a z mA/m
2(52 + 92 )3/ 2 × 10−6
P.E. 7.4
NI
2 × 103 × 50 × 10−3 (cosθ 2 − cosθ1 )a z
( cosθ 2 − cosθ1 ) a z =
2L
2 × 0.75
100
=
( cosθ 2 − cosθ1 ) az
1.5
0.75
(a) At (0,0,0), θ = 90o , cosθ 2 =
0.752 + 0.052
θ1
= 0.9978
100
H=
( 0.9978 − 0 ) az
1.5
H=
= 66.52 az A/m
(b) At (0,0,0.75), θ 2 = 90o ,cosθ1 = −0.9978
100
( 0 + 0.9978 ) a z
H=
1.5
= 66.52az A/m
−0.5
(c) At (0,0,0.5), cosθ1 =
= −0.995
0.52 + 0.052
0.25
cosθ1 =
= 0.9806
0.252 + 0.052
100
( 0.9806 + 0.995) a z
H=
1.5
= 131.7az A/m
P.E. 7.5
H=
(a)
(b)
θ2
θ1
θ2
θ1
θ2
1
K × an
2
1
H (0, 0, 0) = 50a z × (−a y ) = 25a x mA/m
2
1
H (1,5, −3) = 50a z × a y = −25a x mA/m
2
P.E. 7.6
 NI
, ρ − a < ρ < ρ + a, 9<ρ < 11

H =  2πρ

0,
otherwise

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184
(a) At (3,-4,0), ρ = 32 + 42 =5cm ‹ 9cm
H =0
(b) At (6,9,0), ρ = 62 + 92 = 117 ‹ 11
H =
103 × 100 ×10−3
= 147.1 A/m
2π 117 × 102
P.E. 7.7
(a) B = ∇ × A = (−4 xz − 0)ax + (0 + 4 yz )a y + ( y 2 − x 2 )az
B (−1, 2,5) = 20a x + 40a y + 3a z Wb/m2
(b) ψ =  B.dS =
4
1
 
( y 2 − x 2 )dxdy =
y =−1 x = 0
4

−1
1
y 2 dy − 5 x 2 dx
0
1
5
= (64 + 1) − = 20 Wb
3
3
Alternatively,
1
4
0
0
−1
1
ψ =  A.dl =  x 2 (−1)dx +  y 2 (1)dy +  x 2 (4)dx + 0
5 65
=− +
= 20 Wb
3 3
P.E. 7.8
z
h
R
y
k
dS
x
H =
kdS × R
,
4π R 3
dS = dxdy, k = k y a y ,
R = (− x, − y, h),
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188
185215
2
2π
−40
(b) k × IR==(Jhadx S+ =xa z )k y ,  ρ dφ d ρ , dS = ρ dφ d ρ a z
ρ =0 φ =0
k y (ha x + xaμzo)dxdy
H = 2
2π
3 2
2 2ρ
−40 4π ( x 2 + y 2 + −h40
2
)
d
d
=
=
ρ
ρ
φ
0 (2π )


μko ha0 ∞ ∞ 0
μo 2
k y az ∞ ∞
dxdy
xdxdy
= y x   −6
+
3
3


−80 × 2 × 10 2 + 2 + 2 2 4π
2
2
2 2
(
)
x
y
h
−∞ −∞ ( x + y + h )
= 4π −∞ −∞
=
−
400
A
4π × 10−7
The integrand in the last term is zero because it is an odd function of x.
Prob. 7.51
H = − ∇V
→ V = − H ⋅ dl = −mmf
m
m

k y h2π a x ∞ 2 2 2 − 3 2 d ( ρ 2 )
=
H=
( ρIa+ h )
3
  Example
H 4=
2
2 7.3,
2
π 0 2 2 3 2 a z 2
4π From
φ =0 ρ =0 ( ρ + h )
2 (z + a )


k h
k
−1
3
 ∞0 = yIaa2x
− Iz
= y ax 
2
2 − 2
1
2 m 2 =
dz =
−
z
+
a
+ c
 ( ρ 2 + hV
(
)
2
2
1

)


2
2 ( z2 + a 2 ) 2
1
Similarly, for
point
(0,0,-h),
H
=
−
k y ax
As z → ∞, Vm = 0 , 2i.e.
I
I
Hence,
0 = − + c → c =
1
2
k a ,
z >2 0

y
x
2
H =Hence,
1
z I< 0
 k a ,

z
 2 y Vxm =
1 −

2 
z2 + a 2 
P.E. 7.9
I
H =
aφ
2πρ
Prob. 7.1
But
H = − ∇Vm ( J = 0 )
(a) See text
I H = Hy + Hz 1 ∂Vm
I
(b) Let
aφ = −
aϕ → Vm = −
φ + C
ρ ∂φ
2πρ
2π
I
=
a
ρ = (π−3) 2 + 42 = 5
For H
I π
o z
φ =φ 60o =
⋅
+ C
At (10, 60 , 7 ) , 2πρ
, Vm = 0 → 0 = −
3
2π 3
(−3a +I 4a y ) (3a y − 4ax )
aφ =or−az ×C = x
=
56
5
I
I
20Vm = −
+
Hz =
(4ax + 32aπy )φ= 0.5093
6 ax + 0.382a y
2π (25)
π
At ( 4, 30o , − 2 ) , φ = 30o =
,
I
2
6
For H y =
aφ , ρ = (−3) + 52 = 34
2πρ
I π
I
12
I
Vm = −
⋅
+
=
=
12
12
(−3a + 5a )2π 36a + 5a x6
aφ = a y × Vm x = 1 zA = z
34
34
k y ha x
2π
∞
ρ dφ d ρ
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185
P.E. 7.10
∂
∂
k × R = (ha x + xa z )k y , ∂
∂y
∂z
(a ) B = ∇k y×(hAa x =
+ xa z )dxdy∂x
H =
2 3
xy 2 − xz3 − 6 xy + 2z 2 y 2
2 yz
4π ( x 2 + y 2 +2x
h 2 )y +
2
∞ + 4 x 2 y + 3 xz 2 )a + y + 6yz-4xy
∞ ∞
B k=y ha( x− ∞6xz
a y + y 2 − z 3 − 2 x 2 − z a z Wb/m 2
x k y az
dxdy
xdxdy
=
 −∞ ( x 2 + y 2 + h2 ) 3 2 + 4π −∞ −∞ ( x2 + y 2 + h2 ) 3 2
4
π
−∞
(b)
(
ψ =
)
(
)
x y + 3 xz ) dy dz
( −6 xz +in4the
 integrand
The
last term is zero because it is an odd function of x.
2
2
2
2
z=0 y=0
x =1
2
2
= k ha ( −
2π6 xz
∞ ) dy dz + 4  x y dy
∞ dz + 3  xz2 dy dz
k
h
2
π
a
3
0
0
0
ρ
φ
ρ
ρ
(
)
d
d
d
−
y
x
x
= y
H=
(2ρ 2 +2 h 2 ) 2
3



2 4π
4π 2 =0 ρ =20 ( ρ 2 + h 2 ) 22
2
= − 6 φdz
dy + 4 dz  y dy + 03 dy  z 2 dz

0
0
0
0
0
0
 ∞ ky
kyh 
−1

2
2
ax 
a
=
=
2
3
0
x
2y 
z 
 ( ρ 2 + h2 ) 12 
2
 ( 2) + 4(2)

= − 6(2)

 +3(2) 
 = -24+16+16
 2 0 1
 3 0
Similarly, for point (0,0,-h), H = − k y a x
2
ψ = 8 Wb
Hence,
∂A y
∂Ax 1
∂A
(c ) ∇ ⋅ A =
+
+z > 0 z = 4xy + 2xy − 6 xy = 0
,
k
a
∂x y x ∂y
∂z
H =2
∇ ⋅ B = − 6 z + 8 xy +1 3kz 3a+ ,6 z − 8zxy<+01 − 3z 3 − 1 = 0
2 y x
As a matter of mathematical necessity,
∇ • B = ∇ • (∇ × A) = 0
Prob. 7.1
7.43
∂
∂
(a) See∂ text
∂y
∂z = (cos x + sin y )a z
B1 = ∇ × A1 = ∂x
(b) Let H = Hy + Hz
0 (sin x + x sin y ) 0
I
For ∂H z = ∂ aφ∂ ρ = (−3) 2 + 42 = 5
2πρ
∂y ∂z = (cos x + sin y )a z
B2 = ∇ × A2 = ∂x
sin(−x3ax 0+ 4a y ) = (3a y − 4ax )
aφcos
= −yaz ×
5
5
B =B =B
1
2
20 the same B.
Hence, A 2 and A 2 give
Hz =
(4ax + 3a y ) = 0.5093ax + 0.382a y
2π (25)
∇ B = 0
showing that B is solenoidal.
I
For H y =
aφ , ρ = (−3)2 + 52 = 34
2πρ
aφ = a y ×
(−3a x + 5a z ) 3a z + 5a x
=
34
34
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186
Hy =
10
(5a x + 3a z ) = 0.234a x + 0.1404a z
2π (34)
H = Hy + Hz
= 0.7433ax + 0.382ay + 0.1404az A/m
Prob. 7.2
Idl × R
4π R 3
(a) At (1,0,0), R=(1,0,0) - (0,0,0) = (1,0,0)
4a × a
dH = x 3x = 0
4π (1)
(b) At (0, 1,0), R = ay
dH =
4a x × a y
dH =
4π (1)3
(c) At (0,0,1), R =az
dH =
= 0.3183a z A/m
4a x × a z
= −0.3183a y A/m
4π (1)3
(d) At (1,1,1), R=(1,1,1)
dH =
4a x × ( a x + a y + a z )
4π (3)3/ 2
= 61.26(−a y + a z ) mA/m
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187
Prob. 7.3
Let H = H1 + H 2
where H1 and H 2 are respectively due to the lines located at (0,0) and (0,5).
H1 =
I
2πρ
aφ ,
ρ = 5,
aφ = a × a ρ = a z × a x = a y
ay
10
ay =
2π (5)
π
I
ρ = 5 2, aφ = a × a ρ , a = −a z
H2 =
aφ ,
2πρ
5a x − 5a y a x − a y
=
aρ =
5 2
2
 a x − a y  -a x − a y
aφ = −a z × 
=
2 
2

10  -a x − a y  1
H2 =
( -a x − a y )

=
2π 5 2 
2  2π
H1 =
H = H1 + H 2 =
Prob. 7.4
I
H=
aφ ,
2πρ
ay
π
+
1
(-a x − a y ) = −0.1592a x + 0.1592a y
2π
ρ = 5, I = 12
 3a + 4a y  4
3
aφ = a × a ρ = −a z ×  x
 = ax − a y
5
5

 5
12  4
3 
H=
 a x − a y  = 0.3056a x − 0.2292a y
2π (5)  5
5 
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188
Prob. 7.5
b
I
α2
a
α1
y
x
y
x
H=
I
4πρ
(cos α 2 − cos α1 )aφ
ρ = x 2 + y 2 , cos α1 =
a
a2 + ρ 2
, cos α 2 =
b
b2 + ρ 2
aφ = al × aρ = a z × a ρ = aφ . Hence,
H=

b
a
−

x2 + y 2 + a2
4π x 2 + y 2  x 2 + y 2 + b 2
I

aφ

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Prob. 7.6
y
1 A
α2
6A
P
ρ
B
O
H =
I
4πρ
α1
x
1
( cos α 2 − cos α1 ) aφ
1
2
2 =
2
2
 −a x + a y 
 −a x − a y 
1 -1 1 0
= 
×
=
= az



2 -1 -1 0

2
2
α 1 = 135o , α 2 = 45o , ρ =
aφ = al × a ρ
6
H =
4π
( cos 45
2
o
)
− cos135o a z =
3
π
az
2
H ( 0, 0, 0) = 0.954a z A/m
Prob. 7.7
(a) At (5,0,0), ρ = 5,
H=
aφ = a y ,
cos α 1 = 0,
cos α 2 =
2
10
(
)a y = 28.471a y mA/m
4π (5) 125
(b) At (5,5,0), ρ = 5 2,
aφ =
H=
cos α1 = 0,
cos α 2 =
−a x + a y
10
125
10
150
2
2
10  −a x + a y
(
)
4π (5 2) 150 
2

 = 13(−a x + a y ) mA/m

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(c) At (5,15,0), ρ = 250 = 5 10,
aφ =
cos α1 = 0,
cos α 2 =
5a y - 15a x
10
350
5 10
2
10  −15a x + 5a y 
(
)
 = −5.1a x + 1.7a y mA/m
4π (5 10) 350 
5 10

d) At (5,-15,0), by symmetry,
H=
H = 5.1a x + 1.7a y mA/m
Prob. 7.8
z
α1
x
A (2, 0, 0)
C (0, 0, 5)
y
α2
B (1, 1, 0)
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191
(a)
Consider the figure above.
AB
=
AC
=
(1, 1, 0)
( 0, 0, 5)
AB ⋅ AC
=
(
=
BA =
i.e.
)
al × a ρ
H2 =

10
0+

4π 27 
2
2
29 
2
=
( −1,
=
ρ =
×
( −1,
2 27
2
29
− 1, 5)
=
0
27
− 1, 5)
5
=
( 5, 5, 2)
=
27
( 5, 5, 2)
cos α 1 = −
→
2 29
−1 + 1
BC BA
=
( −1, − 1, 5) ,
( −1, 1, 0)
=
aφ =
2π 29
54
⋅
( 5, 5, 2)
27
A/m
27.37 a x + 27.37a y + 10.95 a z mA/m
H = H1 + H 2 + H 3 =
+
=
=
AB ⋅ AC
AB AC
=
BC ⋅ BA
BC BA
=
BC = ρ =
=
(b)
− ( 2, 0, 0 )
( 0, 0, 5) − (1, 1, 0)
(1, − 1, 0)
cos α 2
( −1, 1, 0)
( −2, 0, 5)
=
2, i.e AB and AC are not perpendicular.
cos 180o − α 1
BC
− ( 2, 0, 0)
( 0,
− 59.1, 0) +
( −30.63, 30.63, 0)
( 27.37,
27.37, 10.95)
− 3.26 a x − 1.1 a y + 10.95a z mA/m
Prob. 7.9
y
(a)
Let H = H x + H y = 2H x
Hx =
I
4πρ
( cos α 2 − cos α1 ) aφ
O
α2
5A
α1
x
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where aφ =
5
( cos 45
4π ( 2 )
Hx =
− a z , α1 = 180o , α 2 = 45o
− ax × a y =
o
− cos 180o
) ( −a )
z
= −0.6792 a z A/m
(b)
H = Hx + H y
5
where H x =
4π ( 2 )
(1 − 0)
aφ , aφ = − a x × −a y = a z
= 198.9a z mA/m
H y = 0 since α 1 = α 2 = 0
H = 0.1989 a z A/m
(c )
H = Hx + H y
where H x =
Hy =
5
(1 − 0) ( −ax × az )
5
(1 − 0)
4π ( 2 )
4π ( 2 )
(a
y
× az
)
= 198.9 a y mA/m
= 198.9 a x mA/m
H = 0.1989 a x + 0.1989 a y A/m.
Prob. 7.10
3
Let H = H1 + H 2 + H 3 + H 4
4
where H n is the contribution by side n.
(a)
H = 2H1 + H 2 + H 4 since H1 = H 3
I
1
10  6
1 
+

 az
4π ( 2)  40
2
H1 =
( cos α 2 − cos α1 ) aφ
4πρ
H2 =
10 
2 
2×

 az , H 4 =
4π ( 6 ) 
40 
=
2
10 
1 
2⋅

 az
4π ( 2 ) 
2
 5  3
1 
5
5 
= 
+
+
+
 a z = 1.964a z A/m


2
6π 10
2π 2 
 2π  10
At ( 4, 2, 0) , H = 2 ( H1 + H 4 )
H
(b)
H1 =
10
4π ( 2)
H
2 5  1
1 +  a =
π  4 z
=
8
az , H 4 =
20
10
4π ( 4 )
4
az
20
1.78a z A/m
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(c )
(d )
At
( 4, 8, 0) ,
H = H1 + 2H 2 + H 3
H1 =
10 
4 
2⋅
az , H 2 =

4π ( 8)  4 5 
H3 =
10  2 
4π ( 4)  2 
H
5
8π
=
( az )
10  8
1 
−

 az
4π ( 4 )  4 5
2
( −a z )
4
4 
 1
+
−


5
5
2
=
−0.1178a z A/m
At ( 0, 0, 2) ,
H1 =
10  8

− 0


4π ( 2)  68
H2 =
10
 4

 2a − 8a x 
− 0 a y ×  z




4π 68
84
68 
( ax × az )
=
−
10
ay
π 68
5 ( a x + 4a z )
=
17π 84
 2a x − 4a y 
10
 8

− 0  ax × 
 =
−
4π 20 
84
20 


−5a x
10 
4 
=
0 +
 ( −a y × a z ) =
π 20
4π 2 
20 
H3 =
H4
H
a y + 2a z
π 21
5
5 
10 
2 

 1
 20
= 
−
−
+
 az
 ax + 
 ay + 
 34π 21 π 21 
 34π 21 π 20 
 π 21 π 68 
= −0.3457 ax − 0.3165 a y + 0.1798 az A/m
Prob. 7.11
For the side of the loop along y-axis,
I
H1 =
( cos α 2 − cos α1 ) aφ
4πρ
where aφ =
− a x , ρ = 2 tan 30o =
5
3
cos 30o − cos 150o
4π 2
H = 3H1 = − 1.79a x A/m
H1 =
(
2
, α 2 = 30o , α 1 = 150o
3
) ( −a )
x
=
−
15
ax
8π
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Prob. 7.12
H = 4 H1 = 4
I
4πρ
(cos α 2 − cos α1 )aφ
ρ = a = 2cm, I = mA, α 2 = 45o , α1 = 90o + 45o = 135o
aφ = a × a ρ = a y × (−a x ) = a z
H=
1
2I
2 × 5 ×10−3
I 1
(
−−
)a z =
az =
a = 0.1125a z
πa 2
πa
π × 2 ×10−2 z
2
Prob. 7.13
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195
(a) Consider one side of the polygon as shown. The angle subtended by the Side At the
center of the circle
360° 2π
=
n
n
The filed due to this side is
I
H1 =
(cos α 2 − cos α1 )
4π ρ
where ρ = r ,
π
π
cos α 2 = cos(90 − ) = sin
n
n
π
cos α1 = − sin
H1 =
n
I
2 sin
4π r
π
π
nI
sin
2π r
n
π
3I
For n = 3, H =
sin
2π r
3
2
r cot 30o = 2 → r =
3
n
H = nH 1 =
(b)
H =
3× 5
2π 2
For n = 4, H =
⋅
3
3
2
45
8π
=
4I
π
sin
2π r
4
=
= 1.79 A/m.
4×5
⋅
2π ( 2 )
1
2
= 1.128 A/m.
(c)
As
n → ∞,
nI
π
sin
=
n →∞ 2π r
n
From Example 7.3, when h = 0,
H = lim
nI
π
⋅
2π r
n
=
I
2r
I
2r
which agrees.
H =
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Prob. 7.14
4
β
1
Let H =
3
α2
2
H1 + H 2 + H 3 + H 4
I
10
az =
a z = 62.5 a z
4a
4 × 4 × 10−2
I
4
cos α 2 − cos 90o ) a z , α 2 = tan −1
= H4 =
−2 (
4π × 4 × 10
100
= 19.88 a z
H1 =
H2
H3 =
I
4π (1)
2 cos β a z , β = tan −1
100
4
= 2.29o
= 87.7o
10
2 cos 87.7 oa z = 0.06361 a z
4π
= ( 62.5 + 2 × 19.88 + 0.06361) a z
=
H
= 102.32 a z A/m.
Prob. 7.15
From Example 7.3, H due to circular loop is
H1 =
(a)
Iρ 2
az
2 ρ2 + z2
(
H ( 0, 0, 0) =
)
(
5 × 22
2 22 + 02
)
3
az +
2
(
5 × 22
2 22 + 42
)
3
2
az
= 1.36 a z A/m
(b)
H ( 0, 0, 2) = 2
(
5 × 22
2 22 + 22
)
3
2
az
= 0.884 a z A/m
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Prob. 7.16
α2
H
=
nI
( cos θ 2 − cos θ1 )
2

cos θ 2 = -cos θ1 =
H
=
θ2
(
nI 
2
2 a + 
2
4
)
(
1
2
2
a + 
2
2
=
4)
1
2
0.5 × 150 × 2 × 10−2
2 × 10−3 × 42 + 102
= 69.63 A/m
(b)
α1
α 1 = 90o , tan θ 2 =
H
=
nI
cos θ 2 =
2
α2
a
4
=
= 0.2 → θ 2 = 11.31o
b
20
150 × 0.5
cos 11.31o = 36.77 A/m
2
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Prob. 7.17
e
y
• P (4, 3, 2)
x
H = Hl + H p
Let
Hl =
ρ =
aρ =
1
2πρ
aφ
( 4, 3, 2)
− (1, -2, 2) = (3, 5, 0),
3a x + 5a y
34
=
34
=
3a y − 5a x
34
20π  −5a x + 3a y 
x10-3 = ( − 1.47a y + 0.88a y ) mA/m


2π 
34

1
1
K × an =
100 × 10 −3 a z × ( -a x ) = − 0.05a y A/m
2
2
= H l + H p = −1.47a x − 49.12 a y mA/m
Hp =
H
ρ
, al = az
 3a x + 5a y 
aφ = a l × a ρ = a z × 

34 
Hl =
ρ =
(
)
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Prob. 7.18
(a) See text
(b)
I
a
b
For ρ < a,
 H ⋅ dl
For a < ρ < b,
= Ienc = 0 → H = 0
Hφ ⋅ 2πρ =
Hφ =
For ρ > b,
Iπ ( ρ 2 − a2 )
π ( b2 − a 2 )
 ρ 2 − a2 


2πρ  b2 − a2 
I
Hφ ⋅ 2πρ = I →
Hφ =
I
2πρ
Thus,
Hφ

 0,

 I
= 
 2πρ
 I

 2πρ
ρ <a
 ρ 2 − a2 
 2 2 ,
 b −a 
,
a < ρ <b
ρ >b
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Prob. 7.19
x
y
-1
1
1
H =  K × an
2
1
1
= (20a x ) × (−a y ) + (−20a x ) × a y
2
2
= 10(−a z ) − 10(a z )
= −20a z A/m
Prob. 7.20
1
1
k × an = 10a x × a z = −5a y
2
2
HP =
HL =
I
2πρ
aφ =
I
I
(a x × −a z ) =
ay
2π (3)
6π
H P + H L = −5a y +
I
ay = 0
6π
⎯⎯
→ I = 30π = 94.25 A
Prob. 7.21
(a)
Applying Ampere's law,
πρ 2
→
π a2
Iρ
aφ
2π a 2
Hφ ⋅ 2πρ = I ⋅
i.e
H
=
Hφ = I ⋅
ρ2
2π a 2
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201
(b)
From Eq. (7.29),
 Iρ
 2π a 2 , ρ < a
Hφ = 
 I , ρ >a
 2πρ
At ( 0, 1 cm, 0 ) ,
3 × 1× 10−2
=
Hφ =
2π × 4 × 10−4
H = 11.94 aφ A/m
At
300
8π
( 0, 4 cm, 0 ) ,
3
2π × 4 × 10−2
11.94 aφ A/m
Hφ =
H =
=
300
8π
Prob. 7.22
For 0 < ρ < a

L
H dl = I enc =  J dS
H φ 2πρ = 
2π
φ =0
ρ

ρ
=0
Jo
ρ
ρ dφ d ρ
= J o 2πρ
Hφ = J o
For ρ > a
2π
a
 Hdl =  J dS = φ ρ
=0
Jo
=0
ρ
ρ dφ d ρ
H ρ 2πρ = J o 2π a
Hφ =
Joa
ρ
 J o , 0<ρ <a

Hence H φ =  J o a
 ρ , ρ >a

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Prob. 7.23
2k
1 d
1 d
ρ2
( ρ H φ )a z =
( ko
)a z = o a z
a
a
ρ dρ
ρ dρ
(a) J = ∇ × H =
(b) For ρ>a,
 H ⋅ dl = I
2π
2 ko
2ko
ρ2 a
(2π )
ρ d ρ dφ =
=  J ⋅ dS =  
a
2 0
ρ =0 φ =0 a
a
enc
H φ 2πρ = 2π ko a
⎯⎯
→ Hφ =
a
H = ko   aφ ,
ρ
ρ >a
ko a
ρ
Prob. 7.24
∂
J = ∇ × H = ∂x
y2
∂
∂y
x2
∂
∂z = (2 x − 2 y )a z
0
At (1,-4,7), x =1, y = -4, z=7,
J = [ 2(1) − 2(−4) ] a z = 10a z A/m 2
Prob. 7.25
(a)
J = ∇× H =
1 ∂
1 ∂
( ρ Hφ )a z =
(103 ρ 3 )a z
ρ ∂ρ
ρ ∂ρ
= 3ρ ×103 a z A/m
2
(b)
Method 1:
2
2π
0
0
I =  J dS =  3ρ ρ dφ d ρ103 = 3 × 103  ρ 2 d ρ  dφ
S
= 3 × 103 (2π )
Method 2:
ρ 2
3
3 2
= 16π × 103 A = 50.265 kA
2π
I =  H dl =103  ρ 2 ρ dφ = 103 (8)(2π ) = 50.265 kA
L
0
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Prob. 7.26
Let H = H1 + H 2
where H1 and H 2 are due to the wires centered at x = 0 and x = 10cm respectively.
For H1 , ρ = 50 cm, aφ = al × a ρ = a z × a x = a y
(a)
H1 =
5
50
ay =
a
−2
π y
2π ( 5 × 10 )
For H 2 , ρ = 5 cm, aφ = − a z × −a x = a y , H 2 = H1
H = 2H1 =
100
π
ay
= 31.83 a y A/m
2a y − a x
 2a + a y 
For H1 , aφ = a z ×  x
 =
5 
5

 −a x + 2a y 
5
H1 =
 = − 3.183a x + 6.366a y
−2 
2π 5 5 ×10 
5

For H 2 , a ρ = − a z × a y = a x
(b)
H2 =
5
2π ( 5 )
a x = 15.915a x
H = H1 + H 2
= 12.3 a x + 6.366a y A/m
Prob. 7.27
(a) B =
μo I
aφ
2πρ
At (-3,4,5), ρ=5.
B=
4π × 10−7 × 2
aφ = 80aφ nW/m 2
2π (5)
μI
Ψ =  B • dS = o
2π
(b)

d ρ dz
ρ
6 4
4π ×10−7 × 2
=
ln ρ z
2 0
2π
= 16 ×10−7 ln 3 = 1.756 μ Wb
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Prob. 7.28
(a) I =  J dS
=
2π
a
 
ρ2
J o (1 −
a2
φ =0 ρ =0
) ρ d ρ dφ
 ρ2 ρ4 
= 2π J o 
− 2
 2 4a 
1
= π a2 Jo
2
(b)
 H dl = I
enc
a
0
2π
a
0
0
= J o  dφ  ( ρ −
ρ3
a2
)d ρ
2π  2 a 2 
=
Jo  a − 
2
2 

=  J dS
For ρ < a,
H φ 2πρ =  J dS
 ρ2 ρ4 
= 2π J o 
− 2
 2 4a 
H ρ 2πρ = 2π J o
Hρ =
ρ2 
ρ2 
2
−


4 
a2 
Jo ρ 
ρ2 
2
−


4 
a2 
For ρ > a,
 Hdl = 
J o dS = I
1
H φ 2πρ = π a 2 J o
2
2
a Jo
Hφ =
4ρ
 Jo ρ 
ρ2 
2
−


, ρ < a
a2 
 4 
Hence Hφ = 
aJ o

, ρ >a

4ρ
Copyright © 2015 by Oxford University Press
POESM_Ch07.indd 208
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Sadiku & Kulkarni
Principles of Electromagnetics, 6e
209
205
Prob. 7.29
B =
μ0 I
aφ
2πρ
ψ =

=
Prob. 7.30
B ⋅ dS =
d+a
μ0 I
dρ dz
z = 0 2πρ
ρ 
=d
b
μ0 Ib d + a
In
2π
d
For a whole circular loop of radius a, Example 7.3 gives
H=
Ia 2 a z
2  a 2 + h 2 
3/2
→0
Let h ⎯⎯
I
az
2a
For a semicircular loop, H is halfed
I
H=
az
4a
μI
B = μo H = o a z
4a
H=
Prob. 7.31
∂Bx ∂By ∂Bz
+
+
=0
∂x
∂y
∂z
showing that B satisfies Maxwell’s equation.
(a) ∇ • B =
(b)
dS = dydza x
4
Ψ =  B • dS = 
1

z =1 y = 0
y 2 dydz =
4
y3 1
( z ) = 1 Wb
1
3 0
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Sadiku & Kulkarni
Principles of Electromagnetics, 6e
210
206
(c) ∇ × H = J
J = ∇×
⎯⎯
→
B
μo
∂
∂
∂
∇ × B = ∂x ∂y ∂z = −2 za x − 2 xa y − 2 ya z
y 2 z 2 x2
2
J = − ( za x + xa y + ya z ) A/m 2
μo
Prob. 7.32
h
( ρ − a)
6
where H1 and H 2 are due to the wires centered at x = 0 and x = 10cm respectively.
On the slant side of the ring, z =
ψ =
 B.dS
μo I
 2πρ
=
h
( ρ −a)
b
dρ dz
=
μo I
2π
=
μo Ih 
a + b
 b − a ln
 as required.
a 
2π b
a+b
ρ 
dz dρ
ρ
z=0
=a
=
μo Ih
2π b
 a
1−
dρ
=a 
 ρ 
a+b
ρ
If a = 30 cm, b = 10 cm, h = 5 cm, I = 10 A,
ψ =
4π × 10 −7 × 10 × 0.05 
4
 0.1 − 0.3 ln 
−2
3
2π 10 × 10
(
)
= 1.37 × 10 −8 Wb
Prob. 7.33
ψ =
 BdS
= μo 
0.2

50o
z=0 φ =0
106
ρ
sin 2φ ρ dφ dz
 cos 2φ 
ψ = 4π × 10 × 10 ( 0.2)  −


2 
−7
(
50o
6
=
0.04π 1 − cos 100
=
0.1475 Wb
o
)
0
Copyright © 2015 by Oxford University Press
POESM_Ch07.indd 210
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Sadiku & Kulkarni
Principles of Electromagnetics, 6e
211
207
Prob. 7.34
π /4 2
ψ =  B  dS = 

ρ
20
φ = 0 =1
S
= 20(1)
π /4

0
ρ
2
π /4
1
0
sin φ ρ d ρ dφ = 20 d ρ
2
 sin
2
φ dφ
π /4
1
1
(1 − cos 2φ )dφ = 10(φ − sin 2φ )
0
2
2
π
1
= 10( − ) = 2.854 Wb
4 2
Prob. 7.35
ψ =  B dS , dS = r 2sinθ dθ dφ ar
S
ψ = 
2π
2
cos θ r 2sinθ dθ dφ
= 2  dφ
r =1
r3
0
π /3
= 2(2π )  sin θ d (sin θ ) = 4π
0
π /3
 cos θ sin θ dθ
0
sin 2 θ π / 3
= 2π sin 2 (π / 3)
0
2
= 4.7123 Wb
Prob. 7.36
B = μo H =
μo J × R
dv
4π v R 3
Since current is the flow of charge, we can express this in terms of a charge moving with
velocity u. Jdv = dqu.
μo
4π
 qu × R 
 R 3 
In our case, u and R are perpendicular. Hence,
B=
μo qu 4π ×10−7 1.6 ×10−19 × 2.2 ×106
1.6 ×10−20
=
×
=
B=
4π R 2
4π
(5.3 ×10−11 ) 2
(5.3) 2 ×10 −22
= 12.53 Wb/m 2
Copyright © 2015 by Oxford University Press
POESM_Ch07.indd 211
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Sadiku & Kulkarni
Principles of Electromagnetics, 6e
212
208
Prob. 7.37
(a ) ∇A = − ya sin ax ≠ 0
∂
∂x
∇×A =
∂
∂y
y cos ax
∂
∂z
y + e-x
0
= a x + e − x a y − cos axa z ≠ 0
A is neither electrostatic nor magnetostatic field
1 ∂
1 ∂
ρ Bρ =
( 20) = 0
ρ ∂ρ
ρ ∂ρ
∇× B = 0
B can be E-field in a charge-free region.
(b)
∇⋅ B =
(c )
∇⋅ C =
(
)
1
∂ 2
(r sinθ ) = 0
r sin θ ∂φ
1
∂
1∂ 3
∇×C =
r 2 sin 2 θ ar (r sinθ )aθ ≠ 0
r sin θ ∂θ
r ∂r
C is possibly H field.
(
)
Prob. 7.38
(a) ∇⋅ D = 0
∂
∂x
∇× D =
y2 z
∂
∂y
∂
∂z
2(x + 1)yz -(x + 1)z 2
= 2(x + 1)ya x + . . . ≠ 0
D is possibly a magnetostatic field.
(b)
∇⋅ E =
∇× E =
∂  sin φ 
1 ∂
=0
( ( z + 1) cos φ ) + 
∂z  ρ 
ρ ∂ρ
1
ρ2
cos θ a ρ + . . . ≠ 0
E could be a magnetostatic field.
(c )
∇⋅ F =
1 ∂
1 ∂  sinθ 
( 2cosθ ) +

 ≠ 0
2
r ∂r
rsinθ ∂θ  r 2 
1  ∂
2 sin θ 
−1
aθ ≠ 0
r
sin
θ
+
r  ∂r
r 2 
F can be neither electrostatic nor magnetostatic field.
∇×F =
(
)
Copyright © 2015 by Oxford University Press
POESM_Ch07.indd 212
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Sadiku & Kulkarni
Principles of Electromagnetics, 6e
213
209
Prob. 7.39
A=
μo Idl μo ILaz
=
4π r
4π r
This requires no integration since L << r.
∂A
1 ∂Az
B = ∇× A =
a ρ − z aφ
∂ρ
ρ ∂φ
But r = ρ 2 + z 2
μo ILa z
4π ( ρ 2 + z 2 )1/ 2
μ IL 1
∂Az μo IL ∂
( ρ 2 + z 2 )1/ 2 = o (− )( ρ 2 + z 2 ) −3/ 2 (2 ρ )
=
4π ∂ρ
4π
2
∂ρ
μo IL ρ aφ
μo IL ρ aφ
B=
=
4π ( ρ 2 + z 2 )3/ 2
4π r 3
A=
Prob. 7.40
y
2
a
1
I
3
0
a
dl
R
P
2a
x
4
Divide the loop into four segments as shown above. Due to segment 1,
Copyright © 2015 by Oxford University Press
POESM_Ch07.indd 213
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Sadiku & Kulkarni
Principles of Electromagnetics, 6e
214
210
A1 = 
μo Idl
,
4π R
dl = dya y , R =
y2 + a2
)
(
a
a
μo I
μI
dy
= o a y ln( y + y 2 + a 2
A1 =
ay 
−a
4π
y 2 + a 2 4π
y =− a
=
 2 + 1  μo I
μo I
a y ln 
ln( 2 + 1) a y
 =
4π
 2 − 1  2π
By symmetry, the contributions due to sides 2 and 4 cancel. For side 3,
μ Idl
A3 =  o ,
dl = dy (−a y ), R = y 2 + (−3a) 2
4π R
A3 =
=
)
(
 10 + 1 
a μo I
μo I
(-a y ) ln( y + y 2 + 9a 2
(-a y ) ln 
=

−a 4π
4π
 10 − 1 
μo I  10 + 1 
ln 
 (-a y )
2π  3 
A = A1 + A2 + A3 + A4
=
=
μo I
ln( 2 + 1) a y −
2π
μo I  10 + 1 
ln 
 ay
2π  3 
μo I  3( 2 + 1) 
ln 
 ay
2π  10 + 1 
Prob. 7.41
∂
B = ∇ × A = ∂x
0
=−
π
2
sin
∂
∂y
0
πx
2
sin
∂
∂A
∂A
∂z
= z ax − z a y
∂y
∂x
Az ( x, y )
πy
2
ax −
π
2
cos
πx
2
cos
πy
2
ay
Prob. 7.42
∂
B = μ o H = ∇ × A = ∂x
0
∂
∂y
0
∂
∂A
∂A
∂z
= z a x − z a z = −2 μo kya x + 2 μo kxa y
∂y
∂x
Az ( x, y )
H = −2kya x + 2kxa y
Copyright © 2015 by Oxford University Press
POESM_Ch07.indd 214
10/14/2015 11:46:10 AM
(c )
Sadiku & Kulkarni ∇ ⋅ B
∇⋅ A =
x
∂x
+
y
∂y
+
=
z
∂z
4xy + 2xy − 6 xy = 0
= − 6 z + 8 xy + 3 z 3 + 6 z − 8 xy + 1 − 3z 3 − 1 =
As a matter of mathematical necessity,
∇ • B = ∇ • (∇ × A) = 0
Prob. 7.43
∂
B1 = ∇ × A1 = ∂x
0
∂
∂y
(sin x + x sin y )
∂
∂
∂y
B2 = ∇ × A2 = ∂x
cos y sin x
0
Principles of Electromagnetics, 6e
215
∂
∂z = (cos x + sin y )a z
0
∂
∂z = (cos x + sin y )a z
0
B1 = B2 = B
Hence, A 2 and A 2 give the same B.
∇ B = 0
showing that B is solenoidal.
212
Prob. 7.44
1 ∂Az
∂A
a ρ − z aφ
ρ ∂φ
∂ρ
B = ∇× A =
=
15
ρ
π
e − ρ cos φ a ρ + 15 e − ρ sin φ aφ
1
1


B  3, , -10  = 5 e −3
a ρ + 15 e −3
aφ
2
2
 4

B
107 15 −3  1

H =
e  a ρ + aφ 
=
μo
4π
2
3

H =
(14 a
ψ =
 B ⋅ dS
= 15 z
ρ
10
0
+ 42 aφ ) ⋅10 4 A/m
=
15
 ρ
π
( sin φ ) 0 2
e − ρ cos φ ρ dφ dz
e −5 = 150 e −5

ψ = 1.011 Wb
Prob. 7.45
 ∂
∂Aθ 

1  1 ∂Ar ∂
 ∂θ ( Aφ sin θ ) − ∂φ  ar + r  sin θ ∂φ − ∂r (rAφ )  aθ




1 ∂
∂A 
+  (rAθ ) − r  aφ
r  ∂r
∂θ 
1 10
1 ∂
2sin θ cos θ ar −
(10) sin θ aθ + 0aφ
=
r sin θ r
r ∂r
20
B = 2 cos θ ar
r
At (4, 60o , 30o ), r = 4, θ =60o
B = ∇× A =
H=
B
μo
=
1
r sin θ
1
 20

cos 60o ar  = 4.974 × 105 ar A/m
−7  2
4π × 10  4

Copyright © 2015 by Oxford University Press
POESM_Ch07.indd 215
10/14/2015 11:46:11 AM
o
(14 a
H =
Sadiku & Kulkarni
ψ =
ρ
+ 42 aφ ) ⋅10 4 A/m
15
 B ⋅ dS =
= 15 z
10
0
 ρ
π
( sin φ ) 0 2
Principles of Electromagnetics, 6e
e − ρ cos φ ρ dφ dz
216
e −5 = 150 e −5

ψ = 1.011 Wb
Prob. 7.45
 ∂
∂Aθ 

1  1 ∂Ar ∂
 ∂θ ( Aφ sin θ ) − ∂φ  ar + r  sin θ ∂φ − ∂r (rAφ )  aθ




1 ∂
∂A 
+  (rAθ ) − r  aφ
r  ∂r
∂θ 
1 10
1 ∂
2sin θ cos θ ar −
(10) sin θ aθ + 0aφ
=
r sin θ r
r ∂r
20
B = 2 cos θ ar
r
At (4, 60o , 30o ), r = 4, θ =60o
B = ∇× A =
H=
B
μo
=
1
r sin θ
1
4π × 10−7
 20

o
5
 42 cos 60 ar  = 4.974 × 10 ar A/m
213
Prob. 7.46
Applying Ampere's law gives
H φ ⋅ 2πρ = J o ⋅ πρ 2
Jo
ρ
2
Hφ =
ρ
Bφ = μo H φ = μo
But B
=
−
∂AZ
∂ρ
∇× A
=
=
−
1
μ Jo ρ
2
or
a
Jo ρ
2
∂AZ
aϕ + . . .
∂ρ
⎯⎯
→
AZ =
− μo
Jo ρ 2
4
1
A = - μo J o ρ 2 a z
4
Prob. 7.47
∂
∂x
B = μo H = ∇ × A =
10sin π y
H=
∂
0
POESM_Ch07.indd 216
∂
∂z
= π sin π xa y − 10π cos π ya z
4 + cos π x
π 

sin π xa y − 10 cos π ya z 

μo 

π
∂x
J = ∇× H =
μo
J=
∂
∂y
0
∂
∂
π 

∂y
∂z
10π sin π ya x + π cos π xa z 
=

μo 

sin π x −10 cos π y
π2
(10sin π ya x + cos π xa z )
μo
Copyright © 2015 by Oxford University Press
10/14/2015 11:46:11 AM
But B
=
∂A
− Z
∂ρ
Sadiku & Kulkarni
∇× A
=
=
−
1
μ Jo ρ
2
or
∂ρ
aϕ + . . .
⎯⎯
→
AZ
=
− μo
Jo ρ 2
4
Principles of Electromagnetics, 6e
1
A = - μo J o217
ρ 2 az
4
Prob. 7.47
∂
∂x
B = μo H = ∇ × A =
10sin π y
H=
∂
∂z
= π sin π xa y − 10π cos π ya z
4 + cos π x
π 

sin π xa y − 10 cos π ya z 

μo 

∂
π
∂x
J = ∇× H =
μo
0
J=
∂
∂y
0
∂
∂
π 

∂y
∂z
10π sin π ya x + π cos π xa z 
=

μo 

sin π x −10 cos π y
π2
(10sin π ya x + cos π xa z )
μo
214
Prob. 7.48
1
∂
1 ∂
( Aφ sin θ )ar −
(rAφ )aθ
r sin θ ∂θ
r ∂r
A
1 Ao
=
(2sin θ cos θ )ar − o sin θ ( −r −2 )aθ
2
r sin θ r
r
A
= 3o (2 cos θ ar + sin θ aθ )
r
B = ∇× A =
Prob. 7.49
∂
∂
∂
∂z = (−2 yz − x 2 )a x + (2 xz − 2 xy )a z
(a) J = ∇ × H = ∂x ∂y
xy 2 x 2 z − y 2 z
At (2,-1,3), x=2, y=-1, z=3.
J = 2a x + 16a z A/m 2
(b) −
∂ρ v
= ∇ • J = 0 − 2x + 2x = 0
∂t
At (2,-1,3),
∂ρv
= 0 C/m3s
∂t
Prob. 7.50
(a) B = ∇ × A
POESM_Ch07.indd 217
∂Aρ 
 1 ∂Az ∂Aφ 
 ∂Aρ ∂Az 
1 ∂
a
a
az
=
−
+
−
+
(
A
)
−
ρ
ρ
φ
φ
 ∂z
∂z 
∂ρ 
∂φ 
ρ  ∂ρ
 ρ ∂φ

∂A
2
= − z aφ = 20 ρ aCopyright
© 2015 by Oxford University Press
φ μ Wb/m
∂ρ
B −20 ρ
10/14/2015 11:46:12 AM
(b) −
Sadiku & Kulkarni
∂ρ v
= ∇ • J = 0 − 2x + 2x = 0
∂t
At (2,-1,3),
Principles of Electromagnetics, 6e
∂ρv
= 0 C/m3s
∂t
218
Prob. 7.50
(a) B = ∇ × A
∂A 
 1 ∂Az ∂Aφ 
 ∂A ∂A 
1 ∂
a ρ +  ρ − z  aφ +  ( ρ Aφ ) − ρ  a z
=
−

∂z 
∂ρ 
∂φ 
ρ  ∂ρ
 ρ ∂φ
 ∂z
∂A
= − z aφ = 20 ρ aφ μ Wb/m 2
∂ρ
B −20 ρ
=
H=
aφ μ A/m
μo
μo
1 ∂
( ρ Aφ )a z
ρ ∂ρ
−40
1
=
(−40 ρ )a z =
a z μ A/m 2
μo ρ
μo
215
J = ∇× H =
I =  J dS =
(b)
2
−40
μo
2π
 
ρ dφ d ρ , dS = ρ dφ d ρ a z
ρ =0 φ =0
2π
2
−40 ρ 2
=
ρ d ρ  dφ =
μo 0
μo 2
0
−40
=
2
0
(2π )
−80π × 2 × 10−6
= −400 A
4π × 10−7
Prob. 7.51
H =
− ∇Vm
→
Ia 2
From Example 7.3, H =
Vm =
−
Ia 2
2
 (z
−  H ⋅ dl = −mmf
Vm =
2 (z + a
2
+ a2 )
2
−3
2
2
)
3
2
az
dz =
− Iz
2 (z + a
2
2
)
1
2
+ c
As z → ∞, Vm = 0 , i.e.
0 = −
Hence,
Vm =
P.E. 7.9
H =
2πρ
I
2πρ
aφ =
−
I 
1 −
2 
→
c =
I
2


z +a 
z
2
2
aφ
H = − ∇Vm
But
POESM_Ch07.indd 218
I
I
+ c
2
( J = 0)
1 ∂Vm
I
aϕ → Vm = −
φ + C
Copyright
ρ ∂φ© 2015 by Oxford University Press
2π
At (10, 60o , 7 ) , φ = 60o =
π
, V
= 0→
0 = −
I
⋅
π
+ C
10/14/2015 11:46:12 AM
Sadiku & Kulkarni
Principles of Electromagnetics, 6e
219
216
Prob. 7.52
∇'
1
R
=
=
=
1
R
=
=
=
(
)
1
2
2
2
2
= ( x − x') + ( y − y ') + z − z ' 


1
 ∂
∂
∂  
2
2
2 − 2

x
x'
y
y
'
z
z'
a
+
a
+
a
−
+
−
+
−
(
)
(
)
(
)
y
z
 ∂x x

∂y
∂z  
r − r'
R =
∇
1
 ∂
 
∂
∂
2
2
2 − 2

+
+
−
+
−
y
+
−
x
x'
y
'
z
z'
a
a
a
(
) (
) (
)

x
y
z 
∂y '
∂z '  
 ∂x '
3
2
2
2 − 2
 1
 −  ( −2 ) ( x − x' ) a x ( x − x' ) + ( y − y' ) + ( z − z')  + a y and a z terms
 2
R
R3
3
1
2
2
2 − 2
2 ( x − x') a x ( x − x') + ( y − y') + ( z − z')  + a y and az terms


2
− ( x − x') a z + ( y − y ') a y + ( z − z') a z 
R
= − 3
3
R
R
−
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Principles of Electromagnetics, 6e
217220
CHAPTER 8
P.E. 8.1
∂u
= QE = 6a z N
∂t
∂u
∂
(b)
= 6a z = (u x , u y , u z ) 
∂t
∂t
∂u x
= 0 → ux = A
∂t
∂u y
= 0 → uy = B
∂t
∂u z
= 6 → u z = 6t + C
∂t
A=B=C=0
Since u ( t = 0 ) = 0 ,
(a) F = m
ux = 0 = uy, uz = 6t
∂x
ux =
=0→ x= A
∂t
∂y
uy =
=0→ y= B
∂t
∂z
uz =
= 6t → z = 3t 2 + C1
∂t
At t = 0, (x,y,z) = (0,0,0) → A1 = 0 = B1 = C1
Hence , (x,y,z) = (0,0,3t2),
u = 6ta z at any time. At P(0,0,12), z = 12 =3t2 → t =2s
t =2s
(c) u = 6ta z = 12a z m/s .
∂u
a=
= 6a z m 2
s
∂t
(d) K .E =
1
1
2
m u = (1)(144 ) = 72 J
2
2
P.E. 8.2
(a)
ma = eu × B = (eBouy, -eBoux, 0)
d 2 x eBo dy
dy
=
=ω
2
dt
m dt
dt
(1)
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Principles of Electromagnetics, 6e
221
218
d2y
eBo dx
dx
=−
= −ω
2
dt
m dt
dt
(2)
d 2z
dz
= 0; 
= C1
2
dt
dt
(3)
From (1) and (2),
d 3x
d2y
dx
= ω 2 = −ω 2
3
dt
dt
dt
(D2 + w2 D)x = 0 → Dx = (0, ±jω)x
x = c2 + c3cosωt +c4sinωt
dy 1 d 2 x
=
= −c3ω cos ωt − c4ω sin ωt
dt ω dt 2
At t = 0, u = (α , 0, β ) . Hence,
c1 = β , c3 = 0, c4 =
α
ω
dx
dy
dz
= α cos ωt , = −α sin ωt , = β
dt
dt
dt
(b)
Solving these yields
a
α
x = sin ωt , y = cos ωt , z = β t
ω
ω
The starting point of the particle is (0,
(c)
x2 + y2 =
α
,0)
ω
α2
, z=βt
ω2
showing that the particles move along a helix of radius α
ω placed along the z-axis.
P.E. 8.3
(a)
From Example 8.3, QuB = QE regardless of the sign of the charge.
E = uB = 8 x 106 x 0.5 x 10-3 = 4 kV/m
(b)
Yes, since QuB = QE holds for any Q and m.
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Principles of Electromagnetics, 6e
222
219
P.E. 8.4
By Newton’s 3rd law, F12 = F21 , the force on the infinitely long wire is:
μIIb 1
1
Fl = − F = o 1 2 ( −
)a
2π
ρo ρo + a ρ
=
4π × 10−7 × 50 × 3  1 1 
 −  a ρ = 5a ρ μ N
2π
 2 3
P.E. 8.5
m = ISan = 10 × 10−4 × 50
(2, 6, −3)
7
= 7.143 x 10-3 (2, 6, -3)
= (1.429a x + 4.286a y − 2.143a z ) × 10−2 A-m 2
P.E. 8.6
(a)
(b)
T = m×B =
= 0.03a x − 0.02a y − 0.02a z N-m
T = ISB sin θ →
| T |max =
P.E. 8.7
(a)
μr =
(b)
H=
(c)
10 × 10−4 × 50 2 6 −3
6 4 5
7 × 10
T
max
= ISB
50 × 10 -3
| 6a x + 4a y + 5a z |= 0.04387 Nm
10
μ
= 4.6, χ m = μ r − 1 = 3.6
μo
10 × 10−3 e − y
a z A / m = 1730e − y a z A/m
−7
μ 4π ×10 × 4.6
M = χ m H = 6228e − y a z A/m
B
=
P.E. 8.8
3a x + 4a y 6a x + 8a y
an =
=
5
10
(6 + 32)(6a x + 8a y )
B1n = ( B1 • an )an =
1000
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Principles of Electromagnetics, 6e
223
220
= 0.228a x + 0.304a y = B 2 n
B1t = B1 − B1n = −0.128a x + 0.096a y + 0.2a z
B2t =
μ2
B = 10 B1t = −1.28a x + 0.96a y + 2a z
μ1 1t
B2 = B2 n + B2t = −1.052a x + 1.264a y + 2a z Wb/m2
P.E. 8.9
(a)
B1n = B2 n → μ1 H1n = z μ2 H 2 n
or μ1 H1 • an 21 = μ2 H 2 • an 21
(6 H 2 x − 10 − 12)
(60 + 2 − 36)
= 2μo
μo
7
7
35 = 6 H 2 x
H 2 x = 5.833 A/m
(b)
K = ( H1 − H 2 ) × an12 = an 21 × ( H1 − H 2 )
= an 21 ×  (10,1,12) − (35 , −5, 4) 
6


=
2 −3
1 6
25
7 6 6 8
K = 4.86a x − 8.64a y + 3.95a z A/m
(c)
Since B = μ H , B1 and H1 are parallel, i.e. they make the same angle with the
normal to the interface.
H •a
26
cos θ1 = 1 n 21 =
= 0.2373
H1
7 100 + 1 + 144
θ1 = 76.27 o
cos θ 2 =
H 2 • an 21
13
=
= 0.2144
H2
7 (5.833) 2 + 25 + 16
θ 2 = 77.62o
P.E. 8.10
(a)
L ' = μo μr n 2 S = 4π × 10−7 × 1000 × 16 × 106 × 4 × 10−4
= 8.042 H/m
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Principles of Electromagnetics, 6e
224
221
(b)
Wm ' = 1 L' I 2 = 1 (8.042)(0.5 2 ) = 1.005 J/m
2
2
P.E. 8.11 From Example 8.11,
Lin =
μo l
8π
Lext =
2 wm 1
= 2
I2
I
=
=
1
4π
2
l
2π
0
0
μI 2
 4π 2 ρ 2 ρ d ρ dφ dz
b
2 μo
 dz  dφ  (1 + ρ ) ρ dρ
a
b
1
2μo
1 
dρ
• 2π l   −
2
ρ (1 + ρ ) 
4π
a 
1+ b 
μ ol  b
ln − ln

π  a
1 + a 
μ l μ l b
1+ b
L = Lin + Lext = o + o ln − ln
8π
π  a
1 + a 
=
P.E. 8.12
(a)
L’in =
μo
4π × 10−7
=
= 0.05 μH/m
8π
8π
L’ext = L’ – L’in = 1.2 – 0.05 = 1.15 μH/m
(b)
L’ =
ln
μo
2π
d − a
1
 4 + ln a 


d − a 2π L '
2π ×1.2 ×10−6
=
− 0.25 =
− 0.25
a
μo
4π x10−7
= 6 − 0.25 = 5.75
d −a
= e 5.75 = 314.19
a
d − a = 314.19a = 314.19 ×
d = 407.9mm = 40.79cm
2.588 × 10 −3
= 406.6mm
2
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Principles of Electromagnetics, 6e
225
222
P.E. 8.13
This is similar to Example 8.13. In this case, however, h=0 so that
μ I a 2b
A1 = o 1 3 aφ
4b
μ I a2
μ πI a 2
φ12 = o 12 • 2πb = o 1
4b
φ12
m12 =
=
μoπ a
2b
I1
= 2.632 μH
P.E. 8.14
Lin =
2b
2
=
4π ×10−7 × π × 4
2×3
μo
μ 2πρo
4π × 10−7 × 10 × 10−2
l= o
=
8π
8π
4
= 31.42 nH
P.E. 8.15
(a) From Example 7.6,
Bave =
μo NI
l
φ = Bave • S =
or I =
=
μo NI
2πρ o
μo NI
• πa 2
2πρ o
2 ρ oφ 2 × 10 × 10−2 × 0.5 × 10−3
=
μa 2 N
4π × 10− 7 × 10− 4 × 103
= 795.77A
Alternatively, using circuit approach
l
2πρo
2πρo
R=
=
=
μ S μo S μoπ a 2
φℜ 2 ρ oφ
ℑ = NI =
=
, as obtained before.
N
μa 2 N
ℜ=
2 ρo
2 × 10 × 10−2
=
= 1.591× 109
μa 2 4π × 10− 7 × 10− 4
ℑ = ϕℜ = 0.5x10-3x1.591x109=7.9577x105
ℑ
I = = 795.77 A as obtained before.
N
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c
ψ
Sadiku & Kulkarni
F
(b)
226
223
Principles of Electromagnetics, 6e
Ra
If μ=500μo,
795.77
I=
= 1.592 A
F = NI500
= 500 x 0.2 = 100 A.t
P.E. 8.16
l
42 × 10−2
42 × 106
Rc = 2c =
=
3 −4
−4
B S (1.5) 2 ×−710 × 10
22500
16
ℑ = μ Sa =4π × 10 × 10 −7× 4 ×=10
= π895.25N
2 μo
2 × 4π × 10
8π241
la
10−3
108
=
=
Ra =
−7
−4
16π
P.E. 8.17 μo S 4π ×10 × 4 × 10
μ NI
We may approximate the longer
solenoid as infinite so that B1 = o 1 1 . The flux linking
8
l1
1.42 × 10
Ra + Rc =
the
second
is:π
Prob.
8.1 solenoid 16
μ NI
ψ 2 =2 N 2 B1S1 = o−31 1 1 • π r1216 N22
16πl ×100 16)π (0.4 ×10 −10 ) = 14.576 nN
F =ψ
mω= r =F9.11×
μ Wb
=10 1 × (2 8×10
=
R
1.42
ψa2 + Rμc o N11.42
N 2 × 102
M =
=
• πr1
l1
Prob. 8.2 I1
−6
(a) we assume
ψ air-core
16π × 10
Here
solenoids.
Ba = =
= 88.5 mWb/m 2
−4 6
10
−
2
× 4 ×10
−3
F = Q(u × B )S= 101.42
= 10−3 (−50a x − 250a y )
0 0 25
Prob.
8.45
P.E. 8.18
= −0.05a x −I 0.25a y N
H = R =a ρ 
2πρ μ S
2
(b) Constant 1velocity2implies
that
a = 0.−2
1
1I acceleration
−2
w
|
|
=
μ
H
=
μ
πρ
π
π
2
2
(5
6)10
11

=
=
×
+
=
×10
m
2
2
F = ma = 0 =2Q( E + uo× B
2 ) 4π2 ρ
2
E = -u × B = S =
501.5
a x ×+10
250
a y −IV/m
1−2 (6
5)10−2 = 1.5 × 10−4 1
μ I 2 L ln(b / a)
W =  wm dv =  μ 2 2 ρ dφ d ρ dz =
2 4π ρ
4π ψ 
→ μ=
Prob. 8.3 1 F = NI = ψ−7R = ψ μ S −6 ⎯⎯
NIS
At P, x==4π
2, ×y4=×5,4πz ×=10
-3−3 (625 × 10 −2 )3ln(18 /12) = 304.1 pJ
12 × 10 (11π × 10 2 )
μa=x + (2) 2 (−3)a y + (2)
= 27.65 × 10−3 H/m
E = 2(2)(5)(−3)
−4 (5)a z = −60a x − 12a y + 20a z
500(2)1.5
10
×
Alternatively,
B =ψ
−3)−32 a y + 22 a z = 25a x + 9a y + 4a z
(5) 2 a12
x +×(10
B= =
= 80 Wb
F =Q
+ u××10B−)4 1 2 1 μ L b 2 μ I 2 L b
S ( E1.5
W = LI =
ln × I =
ln
1.4 3.2 −21
2 2π a
4π
a
u× B =
= 21.8a x − 30.6a y − 67.4a z
25 9
4
E + u × B = (−60, −12, 20) + (21.8, −30.6, −67.4) = ( −38.2, −42.6, −47.4)
F = Q( E + u × B ) = 4( E + u × B ) mN
= −152.8a x − 170.4a y − 189.6a z mN
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ℑ=
Sadiku & Kulkarni
B 2 a S (1.5) 2 × 10 × 10−4 22500
=
=
= 895.25N
2 μo
2 × 4π × 10−7
8π
Principles of Electromagnetics, 6e
227
Prob. 8.1
F = mω 2 r = 9.11×10−31 × (2 ×1016 ) 2 (0.4 ×10 −10 ) = 14.576 nN
Prob. 8.2
(a)
F = Q(u × B ) = 10−3
10 −2
6
0
25
0
= 10−3 (−50a x − 250a y )
= −0.05a x − 0.25a y N
(b) Constant velocity implies that acceleration a = 0.
F = ma = 0 = Q( E + u × B )
E = -u × B =
50a x + 250 a y V/m
Prob. 8.3
At P, x = 2, y = 5, z = -3
E = 2(2)(5)(−3)a x + (2) 2 (−3)a y + (2) 2 (5)a z = −60a x − 12a y + 20a z
B = (5) 2 a x + (−3) 2 a y + 22 a z = 25a x + 9a y + 4a z
F = Q( E + u × B)
u× B =
1.4 3.2 −1
= 21.8a x − 30.6a y − 67.4a z
25 9
4
E + u × B = (−60, −12, 20) + (21.8, −30.6, −67.4) = ( −38.2, −42.6, −47.4)
F = Q( E + u × B ) = 4( E + u × B ) mN
= −152.8a x − 170.4a y − 189.6a z mN
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Principles of Electromagnetics, 6e
228
224
Prob. 8.4
F = qE = ma = m
du
dt
du qE 10 ×10−3
=
=
(30, 0, 0) ×103
dt
m
2
d
(u x , u y , u z ) = (150, 0, 0)
dt
Equating components gives
du x
= 150
⎯⎯
→ u x = 150t + c1
dt
du y
=0
⎯⎯
→ u y = c2
dt
du z
=0
⎯⎯
→ u z = c3
dt
At t = 0, u =(2,5,0) × 103 .
2000 = 0 + c1
⎯⎯
→ c1 = 2000
5000 = c2
0 = c3
Hence, u= (150t+2000,5000,0)
At t = 4s,
u = (2600,5000, 0) m/s
dx
= 150t + 2000
⎯⎯
→ x = 75t 2 + 2000t + c4
dt
dy
uy =
= 5000
⎯⎯
→ y = 5000t + c5
dt
dz
uz =
=0
⎯⎯
→ z = +c6
dt
At t=0, (x,y,z)=(0,0,0)
⎯⎯
→ c4 = 0 = c5 = c6
ux =
Hence,
( x, y, z ) = (75t 2 + 2000t ,5000t , 0)
At t = 4s, x=9,200, y=20,000, z=0.
i.e. ( x, y, z ) = (9200, 20000, 0)
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Principles of Electromagnetics, 6e
229
225
Prob. 8.5
ma = Qu × B
10−3 a = −2 × 10−3
ux
0
uy
6
uz
0
d
(u x , u y , u z ) = (12u z ,0,−12u x )
dt
i.e.
du y
du x
= 12u z
dt
(1)
(2)
= 0 → u y = A1
dt
du z
= −12u x
dt
(3)
From (1) and (3),
ux = 12u z = −144u x
or
ux + 144u x = 0 → u x = c1 cos12t + c 2 sin 12t
From (1), uz= - c1sin12t + c2cos12t
At t=0,
Hence,
ux=5, uy=0, uz=0 → A1=0=c2, c1=5
u = (5cos12t , 0, −5sin12t )
u(t = 10s ) = (5cos120, 0, −5sin120) = 4.071a x − 2.903a z m/s
dx
= 5 cos12t → x = 5 sin 12t + B1
12
dt
dy
uy =
= 0 → y = B2
dt
dz
uz =
= −5 sin 12t → z = 5 cos12t + B3
12
dt
ux =
At t=0, (x, y, z) = (0, 1, 2) → B1=0, B2=1, B3=
19
12
19 
5
 5
( x , y , z) =  sin 12t ,1, cos 12t + 
 12
12 
12
(4)
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Principles of Electromagnetics, 6e
230
226
At t=10s,
19 
5
 5
( x , y , z) =  sin 120 ,1, cos 120 +  = (0.2419, 1, 1.923)
 12
12 
12
By eliminating t from (4),
x 2 + ( z − 19 ) 2 = ( 5 ) 2 , y = 1 which is a circle in the y=1 plane with center at
12
12
(0,1,19/12). The particle gyrates.
Prob. 8.6
(a)
ma = −e(u × B )
−
u
m d
(u x , u y , u z ) = x
0
e dt
uy
0
uz


= u y Bo ax − Bo u x a y
Bo
du z
= 0 → uz = c = 0
dt
du x
Be
Be
= −u y o = −u y w , where w = o
m
dt
m
du y
= ux w
dt
Hence,
ux = − wu y = − w2u x
or ux + w2u x = 0 → u x = A cos wt + B sin wt
uy = −
u x
= A sin wt − B cos wt
w
At t=0, ux = uo, uy = 0 → A = uo, B=0
Hence,
u
dx
→ x = o sin wt + c1
dt
w
dy
u
u y = uo sin wt =
→ y = − o cos wt + c2
dt
w
u
At t=0, x = 0 = y → c1=0, c2= o . Hence,
w
uo
uo
x = sin wt , y = (1 − cos wt )
w
w
u x = uo cos wt =
2
u 2o
u
u 
(cos 2 wt + sin 2 wt ) =  o  = x 2 + ( y − o ) 2
2
w
w
w
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Principles of Electromagnetics, 6e
231
227
showing that the electron would move in a circle centered at (0,
uo
). But since the field
w
does not exist throughout the circular region, the electron passes through a semi-circle
and leaves the field horizontally.
(b)
d = twice the radius of the semi-circle
=
2u o 2u o m
=
w
Bo e
Prob. 8.7
F =  Idl × B =
0.2
 2dy(−a
y
) × (4a x − 8a z )
0
ax
ay
az
(−a y ) × (4a x − 8a z ) = 0
−1
0 = 8a x + 4a z
0
−8
4
F = 2(8a x + 4a z )(0.2) = 3.2a x + 1.6a z N
Prob. 8.8
qE
o
mg = qE
mg
mg 0.4 × 10−3 × 9.81
⎯⎯
→ q=
=
= 26.67 nC
1.5 ×105
E
Prob. 8.9
μ I I a ×a
F
= I1al × B2 = o 1 2 l φ
L
2πρ
−7
a × (−a y )4π ×10 (−100)(200)
F21 = z
= 4a x mN/m (repulsive)
2π
ℑ = IL × B → ℑ =
(a)
(b)
F12 = − F21 = −4a x mN/m (repulsive)
(c)
4
3
3
4
al × aφ = a z × (− a x + a y ) = − a x − a y , ρ = 5
5
5
5
5
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Principles of Electromagnetics, 6e
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228
4π ×10−7 (−3 ×104 )  3
4 
F31 =
 − ax − a y 
2π (5)
5 
 5
= 0.72a x + 0.96a y mN/m (attractive)
(d)
F3 = F31 + F32
4π ×10−7 × 6 × 104 )
( az × a y ) = −4ax mN/m(attractive)
2π (3)
F3 = −3.28a x + 0.96a y mN/m
F32 =
(attractive due to L2 and repulsive due to L1)
Prob. 8.10
F=
μo I1 I 2 4π ×10−7 (10)10
=
= 100 μ N
2πρ
2π (20 ×10−2 )
Prob. 8.11
F =  Ldl × B = 3(2a z ) × cos φ
W = −  F • dl ,
F = 6 cos φ
2π
3

aφ N
W = −  6 cos φ
0
= -1.8sin
3
ρo dφ = −6 ρo × 3sin φ 3
2π
0
3
aφ
J
2π
= -1.559 J
3
Prob. 8.12
6
(a)
μo I1 I 2
4π × 10−7
d ρ a ρ × aφ =
F1 = 
(2)(5) ln 6 a z
2
πρ
π
2
2
ρ =2
= 2 ln 3a z μ N = 2.197a z μ N
(b)
F2 =  I 2 dl2 × B1
μo I1 I 2
2π
μII
= o 1 2
2π
=
1
 ρ  d ρ aρ + dza  × aφ
1
z
 ρ d ρ a
z
− dzaρ 
But ρ = z+2, dz=dρ
2
4π ×10−7
1
(5)(2)   d ρ a z − dza ρ 
F2 =
2π
ρ =4 ρ
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Principles of Electromagnetics, 6e
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229
2 ln 2 (a z − a ρ ) μ N = 1.386aρ − 1.386a z a z μ N
4
μo I1 I 2 1
 d ρ a z − dza ρ 
F3 =
2π  ρ 
But z = -ρ + 6, dz = -dρ
4
4π × 10−7
1
(5)(2)   d ρ a z − dza ρ 
F3 =
2π
ρ =6 ρ
2 ln 4 (a z + a ρ ) μ N = −0.8109a ρ − 0.8109a z μ N
6
F = F1 + F2 + F3
= aρ (ln 4 + ln 4 − ln 9) + a z (ln 9 − ln 4 + ln 4 − ln 9)
= 0.575a ρ μ N
Prob. 8.13
A
From Prob. 8.7,
f =
C
μo I1 I 2
aρ
2πρ
60o
30
o
fBC
fAC
B
f = f AC + f BC
4π ×10−7 × 75 ×150
| f AC |=| f BC |=
= 1.125 × 10−3
2π × 2
o
f = 2 × 1.125cos 30 a x mN/m
= 1.949a x mN/m
Prob. 8.14
The field due to the current sheet is
B=
μ
2
K × an =
μo
2
10a x × (−a z ) = 5μ o a y
L
F = I 2  dl2 × B = 2.5 dxa x × (5μo a y ) = 2.5L × 5μo (a z )
0
F
= 12.5 × 4π × 10−7 (a z ) = 15.71a z μ N/m
L
Prob. 8.15
F =  Idl × B = IL × B = 5(2a z ) × 40a x 10−3 = 0.4a y N
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Principles of Electromagnetics, 6e
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230
Prob. 8.16
T = m × B = [ 0.4(0.6)(3)a x ] × (0.5a x + 0.8a y ) = 0.72(0.8)a z
= 0.576a z Nm
Prob. 8.17
F =  Idl × B
⎯⎯
→ F = IB = 520 × 0.4 × 10 −3 × 30 × 10 −3
F = 6.24 mN
Prob. 8.18
m = IS
I=
⎯⎯
→ I=
m
m
= 2
S πr
8 × 1022
= 6.275 × 108 = 627.5 MA
3 2
π (6370 ×10 )
Prob. 8.19
Let F = F1 + F2 + F3
0
F1 =  Idl × B =  2dxa x × 30a z mN
5
=-60a y x
0
= 300a y mN
5
5
F2 =  2dya y × 30a z mN
0
=60a x y
5
0
= 300a x mN
5
F3 =  2(dxa x + dza z ) × 30a z mN
0
=60(-a y ) x
5
= −300a y mN
0
F = F1 + F2 + F3 = 300a y +300a x -300a y mN=300a x mN
1
T = m × B = ISan × B = 2( )(5)(5)a y × 30a z 10−3 = 0.75a x N.m
2
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Principles of Electromagnetics, 6e
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231
Prob. 8.20
For each turn, T = m × B, m = ISan
For N turns,
T = NISB = 50 × 4 × 12 × 10 −4 × 100 × 10 −3 = 24 mNm
Prob. 8.21
f ( x, y, z ) = x + 2 y − 5 z − 12 = 0
an =
∇f = a x + 2a y − 5a z
a x + 2a y − 5a z
∇f
=
| ∇f |
30
m = NISan = 2 × 60 × 8 ×10−4
Prob. 8.22
M = χmH = χm
B
μo μr
=
(a x + 2a y − 5a z )
30
M = χm H = χm
B
μo μ
4999
1.5
×
5000
4π × 10 −7
M =
= 17.53a x + 35.05a y − 87.64a z mAm
χm B
μo (1 + χ m )
Prob. 8.23
(a)
⎯⎯
→
=
1.193 × 106 A/m
N
(b)
M =
m
k =1
k
Δv
If we assume that all mk align with the applied B field,
M =
Nmk
Δv
→
mk =
Nmk
N
Δv
=
1.193 × 106
8.5 × 1028
m k = 1.404 × 10 −23 A ⋅ m 2
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Principles of Electromagnetics, 6e
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232
Prob. 8.24
μr = χ m + 1 = 6.5 + 1 = 7.5
M = χm H
⎯⎯
→ H=
M
χm
=
24 y 2
az
6.5
At y = 2cm,
24 × 4 × 10−4
a z = 1.477a z mA/m
6.5
∂
∂
∂
∂x ∂y
∂z
48 y
=
ax
J = ∇× H =
2
6.5
24 y
0
0
6.5
At y=2cm,
H=
J=
48 × 2 ×10−2
a x = 0.1477a x A/m 2
6.5
Prob. 8.25
(a)
(b)
(c)
(d)
χ m = μr − 1 = 3.5
4y a z × 10 −3
= 707.3y a z A/m
μ
4π × 10 −7 × 4.5
M = χ m H = 2.476y a z kA/m
H =
B
=
Jb = ∇ × M =
∂
∂x
0
∂
∂
∂y
∂z
M z (y)
0
=
dM z
ax
dy
= 2.476ax kA/m 2
Prob. 8.26
When H = 250,
2H
2(250)
B=
=
= 1.4286 mWb/m 2
100 + H 100 + 250
But B=μo μr H
μr =
B
1.4286 × 10−3
=
= 4.54
μo H 4π ×10−7 × 250
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Principles of Electromagnetics, 6e
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233
Prob. 8.27
 H ⋅ dl
=
Ienc
πρ 2
Hφ ⋅ 2πρ =
⋅I
π a2
M = χm H =
Jb = ∇ × M =
→
Iρ
aφ
2π a 2
( μr − 1)
1 ∂
ρ ∂ρ
Iρ
2π a 2
Hφ =
( ρM ) a
φ
z
=
( μr − 1)
I
a
π a2 z
Prob. 8.28
(a)
From H1t – H2t = K and M = χmH, we obtain:
M 1t
χ m1
−
M 2t
χm2
=K
Also from B1n – B2n = 0 and B = μH = (μ/χm)M, we get:
μ1 M 1n μ 2 M 2 n
=
χ m1
χ m2
(b)
From B1cosθ1 = B1n = B2n = B2cosθ2
B sin θ1
B sin θ 2
= H1t = K + H 2t = K + 2
and 1
μ1
μ2
(1)
(2)
Dividing (2) by (1) gives
tan θ1
k
tan θ 2 tan θ 2 
kμ 2 

1 +
=
+
=
μ1
B2 cosθ 2
μ2
μ 2  B2 sin θ 2 
i.e.
tan θ 1 μ1 
kμ 2
1 +
=
tan θ 2 μ 2  B2 sin θ 2



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Principles of Electromagnetics, 6e
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234
Prob. 8.29
B2 n = B1n = 1.8a z
H 2t = H1t
B2t =
⎯⎯
→
B2t
=
μ2
B1t
μ1
4 μo
μ2
B1t =
(6a x − 4.2a y ) = 9.6a x − 6.72a y
μ1
2.5μo
B2 = B2 n + B2t = 9.6a x − 6.72a y + 1.8a z mWb/m 2
H2 =
B2
μ2
=
10−3 (9.6, −6.72,1.8)
4 × 4π ×10−7
= 1,909.86 a x − 1,336.9a y + 358.1a z A/m
z
B2n
θ2
B2t
B2 n
1.8
=
= 0.1536
B2t
9.62 + 6.722
θ 2 = 8.73o
tan θ 2 =
Prob. 8.30
(a)
B1n = B 2n = 15aφ
B1t
H1t = H 2t →
B1t =
=
μ1
B2t
μ2
μ1
2
B2t =
10a ρ − 20a z
μ2
5
(
)
= 4a ρ − 8 a z
Hence,
B1 = 4aρ + 15aφ − 8a z mWb/m 2
(b)
w m1 =
1
B1 ⋅ H1 =
2
B12
2 μ1
w m1 = 60.68 J / m3
w m2 =
B22
2 μ2
=
(10
2
=
(4
2
)
+ 152 + 82 × 10 −6
2 × 2 × 4π × 10 −7
)
+ 152 + 202 × 10 −6
2 × 5 × 4π × 10
−7
= 57.7 J / m3
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Principles of Electromagnetics, 6e
239
235
Prob. 8.31
B2 n = B1n = 40a x
B2 n = μ2 H 2 n
H 2t = H1t
B2t =
⎯⎯
→ H 2n =
⎯⎯
→
B2t
μ2
=
40a x
μ2
=
40a x
50 μo
B1t
μ1
μ2
B
μ1 1t
B2t
H 2t =
μ2
=
B1t
μ1
=
H 2 = H 2 n + H 2t =
(−30a x + 10a y )
μo
1 40
10−3
( , −30,10) ⋅10−3 =
(0.8, −30,10)
μo 50
4π ×10−7
H 2 = 0.6366a x − 23.87a y + 7.957a z kA/m
Prob. 8.32
H 2t = H1t = α a x + δ a z
B2 n = B1n
⎯⎯
→ μ2 H 2 n = μ1 H1n
μ1
μ
H1n = r1 β a y
μ2
μr 2
μ
H = α a x + r1 β a y + δ a z
μr 2
H 2n =
Prob. 8.33
B2 n = B1n = 0.6a y
H 2t = H1t
B1t =
⎯⎯
→
B2t
μ2
=
B1t
μ1
μ
μ1
B2t = o (1.4a x − 2a z ) = 0.1167a x − 0.1667a z
12μo
μ2
B1 = B1n + B1t = 0.1167a x + 0.6a y − 0.1667a z Wb/m 2
H1 =
B1
μ1
=
10−3 (0.1167, 0.6, −0.1667)
4π ×10−7
= (0.0929 a x + 0.4775a y − 0.1327a z ) ⋅106 A/m
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Principles of Electromagnetics, 6e
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236
Prob. 8.34
f ( x, y , z ) = x − y + 2 z
∇f = a x − a y + 2a z
an =
(a)
∇f
1
=
(a x − a y + 2a z )
| ∇f |
6
H1n = ( H1 an )an = (40 − 20 − 60)
(a x − a y + 2a z )
6
= −6.667a x + 6.667a y − 13.333a z A/m
(b)
H 2 = H 2 n + H 2t
⎯⎯
→
B 2 n = B1n
μ2 H 2 n = μ1 H1n
B2 = μ2 H 2 = μ2 H 2 n + μ 2 H 2t = μ1 H1n + μ 2 H 2t = μo (2 H1n + 5 H 2t )
But
= 4π × 10−7 [ (−13.333,13.333, −26.667) + (233.333, 66.666, −83.333]
= 4π × 10−7 (220,80, −110)
= 276.5a x + 100.5a y − 138.2a z μ Wb/m 2
Prob. 8.35
an = a ρ
B2 n = B1n = 22 μo a ρ
H 2t = H1t
B2t =
⎯⎯
→
B2t
μ2
=
B1t
μ1
μ2
μo
B1t =
(45μo aφ ) = 0.05625μo aφ
μ1
800 μo
B2 = μo (22a ρ + 0.05625aφ ) Wb/m 2
Prob. 8.36
r = a is the interface between the two media.
B2 n = B1n
⎯⎯
→
Bo1 (1 + 1.6) cos θ ar = Bo 2 cos θ ar
2.6 Bo1 = Bo 2
(1)
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Principles of Electromagnetics, 6e
241
237
H 2t = H1t
⎯⎯
→
B2t
μ2
=
B1t
μ1
μ2 B1t = μ1 B2t
μ2 Bo1 (−0.2) sin θ aθ = μo Bo 2 (− sin θ )aθ
μB
(2)
μ2 = o o 2
0.2 Bo1
Substituting (1) into (2) gives
μ2 =
μo
0.2
(2.6) = 13μo
Prob. 8.37
(a)
H = 1 K × an = 1 (30 − 40)a x × (−a z ) = −5a y A/m
2
2
B = μo H = 4π × 10−7 (−5a y ) = −6.28a y μ Wb/m2
(b)
H = 1 (−30 − 40)a y = −35a y A/m
2
B = μo μr H = 4π ×10−7 (2.5)( −35a y ) = −110a y μ Wb/m 2
(c)
H = 1 (−30 + 40)a y = 5a y
2
B = μo H = 6.283a y μ Wb/m2
Prob. 8.38
H1n = −3a z ,
H1t = 10a x + 15a y
H 2t = H1t = 10a x + 15a y
H 2n =
μ1
1
H1n =
(−3a z ) = −0.015a z
μ2
200
H 2 = 10a x + 15a y − 0.015a z
B2 = μ2 H 2 = 200 × 4π × 10−7 (10,15, −0.015)
B2 = 2.51a x + 3.77a y − 0.0037a z mWb/m2
tan α =
B2 n
B2t
or α = tan −1
0.0037
2
2.51 + 3.77
2
= 0.047o
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Principles of Electromagnetics, 6e
242
238
Prob. 8.39
(a)
The square cross-section of the toroid is shown below. Let (u,v) be the local
coordinates and ρ o =mean radius. Using Ampere’s law around a circle passing
through P, we get
v
u
(0, ρ o )
H (2π )( ρ o + v) = NI
⎯⎯
→
H=
The flux per turn is
Ψ=
a/2

a/2

u =− a / 2 v =− a / 2
L=
(b)
Bdudv =
NI
2π ( ρo + v)
μo NIa  ρo + a / 2 
ln 

2π
 ρo − a / 2 
N Ψ μo N 2 a  2 ρo + a 
=
ln 

2π
I
 2 ρo − a 
The circular cross-section of the toroid is shown below. Let (r,θ) be the local
coordinates. Consider a point P( r cos θ , ρ o + r sin θ ) and apply Ampere’s law
around a circle that passes through P.
H (2π )( ρ o + r sin θ ) = NI
H=
⎯⎯
→
r
NI
NI  r sin θ 
≈
1 −

2π ( ρ o + r sin θ ) 2πρo 
ρo 
θ
(0, ρ o )
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Principles of Electromagnetics, 6e
243
239
a 2π

Flux per turn Ψ =
r =0 θ
L=
μ NI  r sin θ 
μ NI a 2
(2π )
1 −
rdrdθ =
2πρ o 
ρo 
2πρ o 2
N Ψ μ N 2a2
=
2 ρo
I
Or from Example 8.10,
L = L' l =
μo N 2lS
l2
=
μo N 2πa 2 μo N 2a 2
=
2πρ o
2 ρo
Prob. 8.40
1
2
a = 2 cm
ρo = (3 + 5) = 4cm
L=
μo N 2 a  2 ρ o + a 
ln 

2π
 2 ρo − a 
N2 =
2π L
2π (45 × 10−6 )
=
= 22, 023.17
 2 ρo + a 
8+ 2
−7
−2
μo a ln 
 4π ×10 (2 ×10 ) ln  8 − 2 
a
ρ
−
2
 o

N = 148.4 or 148
Prob. 8.41
μ
L= o
8π
⎯⎯
→
L 4π × 10−7
=
= 50 nH/m

8π
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Principles of Electromagnetics, 6e
244
240
Prob. 8.42
Lin =
μo 
,
8π
μo 
ln(b / a )
2π
μo  μo 
⎯⎯
→
=
ln(b / a)
8π
π
Lext =
If Lin = 2L ext
1
b
= e1/ 8 = 1.1331
8
a
b = 1.1331a = 7.365 mm
ln(b / a ) =
Prob. 8.43
L' =
μ
2π
b  4π ×10−7
1
ln
[0.25 + ln(6 / 2.5)] = 225 nH
+
=
 4
a 
2π
Prob. 8.44
For N = 1,
M 12 =
=
Nψ 12
I
ψ 12
I1
=
b
μo I
μ Ib a + ρo
dzd ρ = o ln
ρo
2πρ
2π
=
z =0
N μo b a + ρ o
ln
=
ρo
2π
ψ 12 =  B1 • dS =
M 12 =
ρo + a
 
ρ ρ
o
μo b a + ρ o
ln
ρo
2π
4π × 10−7
(1) ln 2 = 0.1386 μ H
2π
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ψ 2 = N 2 B1S1 =
Sadiku & Kulkarni
M =
ψ2
=
I1
μo N1 I1
μo N1N 2
l1
l1
2
• π r1  N 2
Principles of Electromagnetics, 6e
2
r1
•π
245
Here we assume air-core solenoids.
Prob. 8.45
H=
I
2πρ
wm =
aρ
1
1
I2
μ | H |2 = μ 2 2
2
2 4π ρ
1
I2
1
μ I 2 L ln(b / a)
W =  wm dv =  μ 2 2 ρ dφ d ρ dz =
2 4π ρ
4π
1
=
× 4 × 4π × 10−7 (625 × 10−6 )3ln(18 /12) = 304.1 pJ
4π
Alternatively,
1 2 1 μ L b 2 242
μI 2L b
W = LI =
ln × I =
ln
2
2 2π a
4π
a
Prob. 8.46
μr = χ m + 1 = 20
1
1
μH ⋅ H
B1 ⋅ H1 =
2
2
1
μ 25x 4 y 2 z 2 + 100x 2 y 4 z 2 + 225x 2 y 2z 4
=
2
wm =
(
Wm =
)
 w dv
m
1
2
2
1
2
2
1 
μ  25 x 4 dx  y 2 dy  z 2 dz + 100 x 2 dx y 4 dy  z 2 dz
0
0
0
−1
−1
2  0
1
2
2
+ 225 x 2 dx  y 2 dy  zdz 

0
0
−1
1
2
2
1
2
2
25μ  x 5
y3
z3
x3
y5 z 3
=
+ 4

2  5 0 3 0 3 −1
3 0 5 0 3 −1

1
2
2
x3
y3 z 5 
+ 9

3 0 3 0 5 −1 

=
=
25μ  1 8 9
 ⋅ ⋅
2 5 3 3
=
25
3600
× 4π × 10 −7 × 20 ×
2
45
+
4 32 9
⋅
⋅
3 3 3
+
9 8 33 
⋅ ⋅

3 3 5
Wm = 25.13 mJ
Prob. 8.47
W=
POESM_Ch08.indd 245
1
1
2 University
−6
Copyright © 2015
Oxford
μ H 2 dv =  4.5
× 4π × 10−7by
+ 5002 ]10Press
dxdydz
[200

2v
2
10/14/2015 12:24:30 PM
25
3600
× 4π × 10 −7 × 20 ×
2
45
=
Sadiku & Kulkarni
Wm = 25.13 mJ
Principles of Electromagnetics, 6e
246
Prob. 8.47
1
1
μ H 2 dv =  4.5 × 4π × 10−7 [2002 + 5002 ]10−6 dxdydz

2v
2
6
= 2π (4.5)10 −7 (29 × 10 4 )10 −6 (2)(2)(2)10 −243
= 6.56 pJ
W=
Prob. 8.48
NI = Hl =
Bl
l
μ
ρo
1.5 × 0.6π
N=
=
μo μr I 4π ×10−7 × 600 ×12
= 313 turns
Bl
N
Prob. 8.49
F = NI = 400 x 0.5 = 200 A.t
Ra =
100
MAt/Wb,
4π
Fa =
R 1 = R2 =
6
MAt/Wb,
4π
R3 =
1.8
MAt/Wb
4π
Ra F
= 190.8 A.t
Ra + R3 + R1 // R2
Ha =
Fa
190.8
=
= 19080 A/m
l a 1 × 10 − 2
Prob. 8.50
Total F = NI = 2000 x 10 = 20,000 A.t
lc
(24 + 20 − 0.6) × 10 −2
=
Rc =
= 0.115 x 107 A.t/m
μ o μ r S 4π × 10 −7 × 1500 × 2 × 10 − 4
la
Ra =
=
0.6 × 10−2
= 2.387 x 107 A.t/m
−7
−4
4π × 10 (1) × 2 × 10
μo μ r S
R = Ra + Rc = 2.502 x 107 A.t/m
ψ=
ℑ
20,000
=ψa =ψc =
= 8 x 10-4 Wb/m2
R
2.502 × 107
Ra
2.387 × 20,000
ℑ=
= 19,081 A.t
2.502
R a + Rc
Rc
0.115 × 20,000
ℑc =
ℑ=
= 919 A.t
2.502
R a + Rc
ℑa =
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Sadiku & Kulkarni
Principles of Electromagnetics, 6e
247
244
Prob. 8.51
Rc
ψ
Ra
F
F = NI = 500 x 0.2 = 100 A.t
Rc =
lc
42 × 10−2
42 × 106
=
=
μ S 4π × 10−7 × 103 × 4 ×10−4
16π
Ra =
la
10−3
108
=
=
μo S 4π ×10−7 × 4 ×10−4 16π
Ra + Rc =
ψ=
Ba =
1.42 × 108
16π
F
16π ×100 16π
μ Wb
=
=
Ra + Rc 1.42 × 108 1.42
ψ
S
=
16π × 10−6
= 88.5 mWb/m 2
1.42 × 4 ×10−4
P.E. 8.18
R=

μS
1
 = 2πρo = 2π × (5 + 6)10−2 = 11π ×10−2
2
−2
S = 1.5 ×10 (6 − 5)10−2 = 1.5 × 10−4

ψ
F = NI = ψ R = ψ
⎯⎯
→ μ=
μS
NIS
μ=
B=
ψ
S
=
12 × 10−3 (11π × 10−2 )
= 27.65 × 10−3 H/m
−4
500(2)1.5 ×10
12 ×10−3
= 80 Wb
1.5 × 10−4
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Principles of Electromagnetics, 6e
248
245
Prob. 8.52
F=
B2S
ψ2
4 × 10−6
=
=
= 53.05 kN
2 μo 2μo S 2 × 4π × 10−7 × 0.3 ×10−4
Prob. 8.53
(a)
F = NI = 200 x 10-3 x 750 = 150 A.t.
l
10−3
Ra = a =
= 3.183 × 107
μo S 25 × 10− 6 μo
2π × 0.1
lt
=
Rt =
= 6.7 x 107
μo μ r S μo × 300 × 25 × 10− 6
ψ=
ℑ
150
= 7
= 15.23 × 10 −7
Ra + Rt 10 (3.183 + 20 / 3)
F=
B2S
ψ2
2.32 × 10−12
=
=
2 μo 2μo S 2 × 4π ×10−7 × 25 × 10−6
= 37 mN
(b)
Rt
ℑ
Ra
ℑ
150
=
Ra 3.183 × 107
If μt → ∞, Rt = 0, ψ =
F2 = I 2 dl2 • B1 = I 2 dl2
ψ
ψ1
S
=
2 × 10−3 × 5 × 10−3 × 150
3.183 × 107 × 25 × 10− 6
F2 = 1.885 μN
Prob. 8.54
ψ2
ψ2
ψ1
ℑ
Ra
ψ 1 = 2ψ 2 ,ψ 1 =
Ra
ℑ
3 R
2 a
ψ1
=
ℑ
Ra

Ra/2
Ra
ℑ
2ℑ
→ψ2 =
3Ra
3Ra
2
 ψ 2  ψ
3ψ 1
ℑ2
ℑ = 2 2  + 1 =
=
2
 2μo S  2 μo S 4 μo S 3Ra μo S
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POESM_Ch08.indd 248
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Sadiku & Kulkarni
Principles of Electromagnetics, 6e
249
246
=
μo S ℑ2
3la
2
4π × 10−7 × 200 ×10−4 × 9 × 106
=
3 ×10−6
= 24π × 10 3 = mg → m =
24π × 10 3
= 7694 kg
9.8
Prob. 8.55
ℑ
ℑ = NI
Rs
Ra
Rs

Rs/2
Ra
Since μ → ∞ for the core (see Figure) , Rc = 0.
a
Rs  ψ ( 2 + x)

ℑ = NI = ψ  Ra +  =
2 
μo S

ψ (2 x + a)
=
2μo S
ℑ=
=
N 2 I 2 4μo2 S 2
B2S
1
1
=ψ 2
=
•
2 μo
2 μo S 2 μo S
(a + 2 x) 2
2 N 2 I 2 μo S
(a + 2 x)2
F = − Fa x since the force is attractive, i.e.
F=
−2 N 2 I 2 μo Sa x
(a + 2 x) 2
Copyright © 2015 by Oxford University Press
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Principles of Electromagnetics, 6e
250
247
CHAPTER 9
P.E. 9.1
(a) Vemf =  ( u × B ) ⋅ dl = uBl = 8 ( 0.5 )( 0.1) = 0.4 V
(b) I =
Vemf
R
=
0.4
= 20 mA
20
(c) Fm = Il × B = 0.02 ( −0.1a y × 0.5a z ) = −a x mN
(d) P = FU = I 2 R = 8 mW
P=
or
Vemf
R
=
(0.4)2
20
= 8 mW
P.E. 9.2
(a) Vemf =  ( u × B ) ⋅ dl
where B = Bo a y = Bo ( sin φ a ρ + cos φ aφ ) , Bo = 0.05 Wb/m2
( u × B ) ⋅ dl = − ρω Bo sin φ dz = −0.2π sin (ωt + π 2 ) dz
Vemf =
0.03
 ( u × B ) ⋅ dl = −6π cos (100π t ) mV
0
At t = 1ms,
Vemf = −6π cos 0.1π = − 17.93 mV
Vemf
= −60π cos(100π .t ) mA
R
At t = 3ms, i = −60π cos 0.3π = −110.8 mA
i=
(b) Method 1:
ρ o zo
Ψ =  B ⋅ dS =  Bot ( cos φ a ρ − sin φ aφ ) ⋅ d ρ dzaφ = −   Bot sin φ d ρ dz = − Bo ρo zot sin φ
0 0
where Bo = 0.02 , ρ o = 0.04 , zo = 0.03
φ = ωt + π 2
Ψ = − Bo ρo zot cos ωt
Vemf = −
∂Ψ
= Bo ρ o zo cos ωt − Bo ρo zotω sin ωt
∂t
Copyright © 2015 by Oxford University Press
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Sadiku & Kulkarni
Principles of Electromagnetics, 6e
251
248
= ( 0.02 )( 0.04 )( 0.03) [ cos ωt − ωt sin ωt ]
= 24 [ cos ωt − ωt sin ωt ] μV
Method 2:
∂B
• dS +  (u × B ).dl
∂t
B = Bo ta x = Bot (cos φ a ρ − sin φ aφ ), φ = ωt + π
Vemf = − 
2
∂B
= Bo (cos φ aρ − sin φ aφ )
∂t
Note that only explicit dependence of B on time is accounted for, i.e. we make φ
= constant because it is transformer (stationary) emf. Thus,
ρo zo
0
0 0
zo
Vemf = − Bo   (cos φ a ρ − sin φ aφ )d ρ dzaφ +  − ρoω Bot cos φ dz
= Bo ρo zo (sin φ − ωt cos φ ), φ = ωt + π
At t = 1ms,
2
= Bo ρo zo (cos ωt − ωt sin ωt ) as obtained earlier.
Vemf = 24[cos18o − 100π × 10−3 sin 18o ]μV
= 20.5μV
At t = 3ms,
i = 240[cos54o − .03π sin 54o ]mA
= -41.93 mA
P.E. 9.3
dψ
dψ
, V2 = − N 2
dt
dt
V2 N 2
N
300 × 120
=
→ V2 = 2 V1 =
= 72V
V1 N 1
N1
500
V1 = − N 1
P.E. 9.4
(a)
Jd =
∂D
= −20ωε o sin(ωt − 50 x)a y A / m 2
∂t
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Principles of Electromagnetics, 6e
252
249
(b)
∂H z
a y = −20ωε o sin(ωt − 50 x)a y
∂x
20ωε o
or H =
cos(ωt − 50 x)a z
50
∇ × H = Jd → −
= 0.4ωε o cos(ωt − 50 x)a z A/m
(c)
∇ × E = − μo
∂E y
∂H
→
a z = 0.4 μoω 2ε o sin(ωt − 50 x)a z
∂t
∂x
1000 = 0.4μoε oω = 0.4
2
ω2
10
or ω = 1.5 x 10 rad/s
c2
P.E. 9.5
2
2

 1+ j 
2∠45o 
j 
=
−
j
= − j 2 ∠143.13o


o
5
2− j
 5∠ − 26.56 
= 0.24 + j0.32
3
(a)
(b)
(
)
o
6∠30o + j 5 − 3 + e j 45 = 5.196 + j 3 + j 5 − 3 + 0.7071(1 + j )
= 2.903 + j8.707
P.E. 9.6
(
)
P = 2sin(10t + x − π )a y = 2 cos 10t + x − π − π a y , w = 10
4
4
2
(
= Re 2e
i.e. Ps = 2e
j ( x − 3π )
4
j ( x −3π )
4
)
a y e jwt = Re ( Ps e jwt )
ay
Q = Re ( Qs e jwt ) = Re ( e j ( x + wt ) (a x − a z ) ) sin π y
= sin π y cos( wt + x)(a x − a z )
P.E. 9.7
−μ
∂H
1
∂
1 ∂
= ∇× E =
( Eφ sin θ )ar −
(rEφ )aθ
∂t
r sin θ ∂θ
r ∂r
Copyright © 2015 by Oxford University Press
POESM_Ch09.indd 252
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ψ = B  S = (0.2)2 π 40 × 10−3 sin104 t
∂ψ
Sadiku & Kulkarni
= −16π cos104 t
V =−
Principles of Electromagnetics, 6e
∂t
V 16π
cos104 t
i= =
253
250
4
R
= −12.57 cos104 t A
2 cos θ
β
cos(ωt − β r )ar − sin θ sin(ωt − β r )aθ
=
2
r
r
2 cos θ
β
Prob. 9.3
sin(ωt − β r )ar −
sin θ cos(ωt − β r )aθ
H =−
2
∂ψ μω
∂r
∂B μω r
Vemf = −
= −  B • dS = −  dxdya z
∂t
∂t ω 6 ×107
∂t
β= =
=
0.2
rad/m
0.1 0.8
8
c 3 ×π10
t − 3 y )dxdy mV
=   30π × 40sin(30
1
1
y = 0H
x = 0= −
cos θ sin(6 × 107 − 0.2r )ar −
sin θ cos(6 ×107 − 0.2r )aθ
2
0.8 12π
0.1r
120π r
= 1200π  dx  sin(30π t − 3 y )dy
P.E. 9.8
0
0
c
3  − 1 3cos(30
9 t×−1038y ) 0.1
π
π
= 1200
(0.8)

 × 10 8 rad/s
= −3
=
= 02.846
ω=

με  μ r ε r
10
π t − 0.3) − 6cos(30π t ) ] mV
= 320π [ cos(30
1
cos(ωt − 3 y )a x
E =  ∇ × Hdt = −
Vemf
Vεemf
320π ωε
=
=
I=
[ −2sin(30π t − 0.15) sin(−0.15)]
R
10 +=4
14 −6
cos(ωt − 3 y )a x
8
−9
9
×
10
10
= 143.62sin(30π t − 0.15)
• sin(0.15)
(5)
π
10
mA
I = 21.46sin(30π t − 0.15)36
E = −476.86 cos(2.846 × 108 t − 3 y )a x V/m
P.E. 9.9
V =−
∂ψ
∂
∂B
= −  B • dS = −
•S
∂t
∂t
∂t
Prob. 9.1
= 3770 sin377t x π(0.2)2 x 10-3
Measuring
the induced
direction,
= 0.4738
sin377temf
V in the clockwise253
Vemf =  (u × B )dl
0
1.2
P.E. 9.10
= (5a × 0.2a )dya y +  (15a x × 0.2a z )dya x
V =  (u × Bx)dl z
0
1.2
1.2
0
u = =ρω
, dyB− = B
o ady
z
- aφ(1)
(3)


 ω Bo  2
1
2
ρω+B1.2
ρ
ω
ρ
V == −1.2
d
B
=
=
×
3
=
−
1.2
+
3.6
o
o
0
2
2
ρ =0
= 2.4 V
30
= × 60 × 10−3 (8 × 10−2 ) 2 = 5.76 mV
2
 0
1.2
Prob. 9.8
Method 1:
We assume that the sliding rode is on − < z < 
 = x / 3 = 5t / 3

Vemf =  (u × B )dl =  5a x × 0.6a z • dya y = −3x  dy = −3 x × 25t 23 = −86.6025t t
−
POESM_Ch09.indd 253
Copyright © 2015 by Oxford University Press
Method 2:
The flux linkage is given by
10/14/2015 12:47:11 PM
•
(5)
10 36π
E = −476.86 cos(2.846 × 108 t − 3 y )a x V/m
Sadiku & Kulkarni
Principles of Electromagnetics, 6e
254
Prob. 9.1
Measuring the induced emf in the clockwise direction,
Vemf =  (u × B )dl
=
1.2
0
 (5ax × 0.2az )dya y +  (15ax × 0.2az )dyax
0
1.2
1.2
0
0
1.2
= -  (1) dy −  (3)dy
= −1.2 + 1.2 × 3 = −1.2 + 3.6
= 2.4 V
251
Prob. 9.2
ψ = B  S = (0.2)2 π 40 × 10−3 sin104 t
∂ψ
= −16π cos104 t
V =−
∂t
V 16π
cos104 t
i= =
4
R
= −12.57 cos104 t A
Prob. 9.3
∂ψ
∂
∂B
Vemf = −
= −  B • dS = −  dxdya z
∂t
∂t
∂t
=
0.1 0.8
  30π × 40sin(30π t − 3 y)dxdy
mV
y =0 x =0
= 1200π
0.8
0.1
0
0
 dx  sin(30π t − 3 y)dy
 1
0.1
= 1200π (0.8)  − cos(30π t − 3 y )

0 
 −3
= 320π [ cos(30π t − 0.3) − cos(30π t ) ] mV
Vemf
Vemf
320π
[ −2sin(30π t − 0.15) sin(−0.15)]
R
10 + 4
14
= 143.62sin(30π t − 0.15) sin(0.15)
I = 21.46sin(30π t − 0.15) mA
I=
=
P.E. 9.9
V =−
=
∂ψ
∂
∂B
= −  B • dS = −
•S
∂t
∂t
∂t
Copyright © 2015 by Oxford University Press
= 3770 sin377t x π(0.2)2 x 10-3
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Principles of Electromagnetics, 6e
255
252
Prob. 9.4
∂
∂B
• dS
Vemf = −  B • dS = − 
∂t
∂t
= -  (-4ω )sin ωt x 2 + y 2 dxdy = 4ω sin ωt  x 2 + y 2 dxdy
We change variables from Cartesian to cylindrical coordinates.
Vemf = 4ω sin ωt
2π
3
 
ρ ⋅ ρ d ρ dφ = 4ω sin ωt (2π )
φ =0 ρ =0
ρ3 3
3 0
= 72πω sin ωt = 226.2ω sin ωt V
Prob. 9.5
μI
B = o ( −a x )
2π y
μo I
2π
ψ =  B • dS =
Vemf = −
a ρ +a
 ρ
z =0 y =
dzdy μo Ia ρ + a
ln
=
2π
y
ρ
∂ψ
∂ψ ∂ρ
μ Ia d
=−
•
= − o uo
[ln( ρ + a ) − ln ρ ]
∂t
∂ρ ∂t
2π
dρ
μ o Ia  1
μ o a 2 Iu o
1
=−
uo 
− =
2π
 ρ + a ρ  2πρ ( ρ + a)
where ρ = ρo + u o t
Prob. 9.6
Vemf =
ρ +a
ρ 3a
z
×
μo I
3μ I ρ + a
aφ • d ρ a ρ = − o ln
2πρ
2π
ρ
4π × 10 −7
60
× 15 × 3 ln
= −9.888μV
2π
20
Thus the induced emf = 9.888μV, point A at higher potential.
=−
Prob. 9.7
Vemf = − N
∂ψ
∂
dS
= − N  B dS = − NB
∂t
∂t
dt
d
dφ
( ρφ ) = − NBρ
= − NBρω
dt
dt
= −50(0.2)(30 × 10−4 )(60) = −1.8V
= − NB
Copyright © 2015 by Oxford University Press
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u = ρω aφ ,
Sadiku & Kulkarni
B = Bo a z
 ω Bo  2
1
2
V =  ρω Bo d ρ = ω Bo ρ
=
0
2
2
ρ =0

=
Principles of Electromagnetics, 6e
256
30
× 60 × 10−3 (8 × 10−2 ) 2 = 5.76 mV
2
Prob. 9.8
Method 1:
We assume that the sliding rode is on − < z < 
 = x / 3 = 5t / 3

Vemf =  (u × B )dl =  5a x × 0.6a z • dya y = −3x  dy = −3 x × 25t 23 = −86.6025t t
−
Method 2:
The flux linkage is given by
ψ=
5t
x. 3
 
0.6 xdxdy = 0.6 ×
x = o y =− x / 3
Vemf = −
2
×125t 3 / 3 = 28,8675t 3
3
dψ
= −86.602t 2
dt
Prob. 9.9
Vemf = uB = 410 × 0.4 ×10−6 × 36 = 5.904 mV
Prob. 9.10
u
u
B
B
θ
Vemf =  (u × B ) ⋅ dl = uBl cos θ
 120 ×103

=
m / s  ( 4.3 × 10−5 ) (1.6 ) cos 65o
 3600

o
= 2.293cos 65 = 0.97 mV
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Principles of Electromagnetics, 6e
257
254
Prob. 9.11
dψ = 0.64 – 0.45 = 0.19, dt = 0.02
dψ
 0.19 
= 10 
 = 95V
dt
 0.02 
Vemf = N
I=
Vemf
 95 
=   = 6.33 A
R
 15 
Using Lenz’s law, the direction of the induced current is counterclockwise.
Prob. 9.12
V =  (u × B ) • dl , where u = ρω aφ , B = Bo a z
V=
ρ2
 ρω B d ρ =
ρ
o
ω Bo
2
1
V=
( ρ 2 2 − ρ 21 )
60 ×15
• 10−3 (100 − 4) • 10−4 = 4.32 mV
2
Prob. 9.13
J ds = jωDs → J ds
=
max
= ωεE s = ωε
10−9 2π × 20 × 106 × 50
×
36π
0.2 × 10− 3
Vs
d
= 277.8 A/m2
I ds = J ds • S =
1000
× 2.8 × 10 − 4 = 77.78 mA
3.6
Prob. 9.14
Jc = σ E,
∂D
∂E
=ε
∂t
∂t
| J d |= εω | E |
Jd =
| J c |= σ | E |,
If I c = I d , then | J c |=| J d |
ω = 2π f =
f =
σ
=
2πε
σ
ε
4
10−9
2π × 9 ×
36π
⎯⎯
→ σ = εω
= 8 GHz
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Jc
σE
σ
=
=
J d ωεE ωε
Prob. 9.15
(a)
σ
=
ωε
(b)
σ
=
ωε
σ
=
ωε
(c)
2 × 10−3
10−9
2π ×109 × 81×
36π
25
10−9
2π ×10 × 81×
36π
= 0.444 × 10−3
= 5.555
9
2 × 10−4
= 7.2 × 10−4
10−9
2π ×109 × 5 ×
36π
Prob. 9.16
J d ωε E ωε
=
=
=1
J
σE
σ
2π f = 12π × 105
σ
10−4
⎯⎯
→ ω= =
= 12π × 105
−9
10
ε
3×
36π
⎯⎯
→
f = 600 kHz
Prob. 9.17
J c = σ E = 0.4 cos(2π ×103 t )
E=
0.4
σ
cos(2π × 103 t )
∂E
0.4ε
=−
(2π ×103 ) sin(2π ×103 t )
∂t
σ
10−9
0.4 × 4.5 ×
36π (2π ×103 ) sin(2π ×103 t )
=−
−4
10
= −100sin(2π ×103 t ) A/m 2
Jd = ε
Prob. 9.18
(a)
∇ • Es =
ρs
ε ,∇ • Hs = 0
∇ × E s = jωμ H s ,
∇ × H s = (σ − jωε ) E s
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(b)
∂Dx ∂Dy ∂Dz
+
+
= ρv
∂x
∂y
∂z
∂B ∂By ∂Bz
∇•B = 0 → x +
+
=0
∂x
∂y
∂z
∂B
∂E ∂E
∂B
∇× E = −
→ z − y =− x
∂t
∂y
∂z
∂t
∇ • D = ρv →
(1)
(2)
(3)
∂By
∂Ex ∂Ez
−
=−
∂z
∂x
∂t
∂E y ∂E x
∂B
−
=− z
∂x
∂y
∂t
∇× H = J +
∂H x
∂z
∂H y
∂x
(4)
(5)
∂D
∂H z ∂H y
∂D
→
−
= Jx + x
∂t
∂y
∂z
∂t
∂D y
∂H z
= Jy +
−
∂t
∂x
∂H x
∂D z
= Jz +
−
∂t
∂y
(6)
(7)
(8)
Prob. 9.19
If J = 0 = ρv , then
∇•B = 0
∇ • D = ρv
∂B
∇× E = −
∂t
∂D
∇× H = J +
∂t
Since ∇ • ∇ × A = 0 for any vector field
(1)
(2)
(3)
(4)
A,
∂
∇•B = 0
∂t
∂
∇•∇× H = − ∇• D = 0
∂t
showing that (1) and (2) are incorporated in (3) and (4). Thus Maxwell’s equations can be
reduced to (3) and (4), i.e.
∇•∇× E = −
∇× E = −
∂B
∂D
, ∇× H =
∂t
∂t
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Prob. 9.20
∇ E = 0
⎯⎯
→
(1)
∇ H = 0
⎯⎯
→
(2)
∇ × E = −μ
∂H
∂t
∂
 ∂x

∇× E = 
0


=
∂E y
∂x
H =−
⎯⎯
→
∂
∂y
(3)
∂
∂z 


0 

E y ( x, t )
a z = − Eo sin x cos ta z
Eo
1
∇ × Edt =
μ
μ
∇× H = ε
∂E
∂t
∂
 ∂x

∇× H = 
0


sin× sin ta z
o
⎯⎯
→
∂
∂y
(4)
∂
∂z




0 H z ( x, t ) 

E
∂H z
a y = − o cos x sin ta y
=−
∂x
μo
E
=
Eo
1
∇ × Hdt =
cos x cos ta
ε
με
y
o
which is off the given E by a factor. Thus, Maxwell’s equations (1) to (3) are satisfied,
but (4) is not. The only way (4) is satisfied is for μoε = 1 which is not true.
Prob. 9.21
∇× E = −
∇×∇× E = −
∂B
∂t
∂
∂
∂J
∂2 E
∇ × B = −μ ∇ × H = −μ
− με 2
∂t
∂t
∂t
∂t
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But
∇ × ∇ × E = ∇(∇ • E ) − ∇ 2 E
∇(∇ • E ) − ∇ 2 E = − μ
∂J
∂2 E
− με 2 ,
∂t
∂t
J =σ E
In a source-free region, ∇ • E = ρ v / ε = 0 . Thus,
∂E
∂2 E
∇ E = μσ
+ με 2
∂t
∂t
2
Prob. 9.22
∇ • J = (0 + 0 + 3 z 2 )sin104 t = −
∂ρv
∂t
3z 2
ρv = −  ∇ • Jdt = −  3z sin10 tdt = 4 cos104 t + Co
10
2
4
If ρ v |z =0 = 0, then Co = 0 and
ρv = 0.3z 2 cos104 t mC/m3
Prob. 9.23
∂D
∂E 50ε o
4.421× 10−2
8
8
Jd =
= εo
=
(−10 ) sin(10 t − kz )a ρ = −
sin(108 t − kz )a ρ A/m
∂t
∂t
ρ
ρ
∇ × E = − μo
∇× E =
H =−
H=
∂Eρ
1
μo
∂z
2
∂H
∂t
aφ =
50k
ρ
sin(108 t − kz )aφ
1
 ∇ × Edt = 4π ×10
−7
50k
cos(108 t − kz )aφ
8
10 ρ
2.5k
cos(108 t − kz )aφ A/m
2πρ
∇× H = −
∂H φ
∂z
aρ = −
2.5k 2
sin(108 t − kz )a ρ
2πρ
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∇ × H = Jd
k2 =
⎯⎯
→
−
2π
× 4.421× 10−2
2.5
4.421x10−2
ρ
sin(108 t − kz )a ρ =
−2.5k 2
sin(108 t − kz )a ρ
2πρ
k = 0.333
⎯⎯
→
Prob. 9.24
∂D
∂E
1
= 0+ε
⎯⎯
→ E =  ∇ × Hdt
ε
∂t
∂t
∂
∂
∂
∂z
∇ × H = ∂x ∂y
= 10β sin(ωt + β x)a y
0
0 10 cos(ωt + β x)
∇× H = J +
E=
1
10 β sin(ωt + β x) dta
ε
But ∇ × E = − μ
∇× E =
∂
∂x
0
H =−
∂H
∂t
y
=
−10 β
ωε
⎯⎯
→ H =−
∂
∂y
−10 β
ωε
1 10β 2
μ  ωε
cos(ωt + β x)a y
cos(ωt + β x)
sin(ωt + β x)dta z =
∂
∂z
1
μ
=
0
10β 2
ω 2 με
∇ × Edt
10 β 2
ωε
sin(ωt + β x)a z
cos(ωt + β x)a z
Comparing this with the given H,
10 =
10 β 2
⎯⎯
→ β = ω με = 2π × 109 4π ×10−7 ×
ω 2 με
β = 60π = 188.5 rad/m
E=
−10β
ωε
10−9
× 81
36π
cos(ωt + β x )a y = −148cos(ωt + β x )a y V / m
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Prob. 9.25
D = ε o E = ε o Eo cos(ωt − β z )a x
∇× E = −
∂B
∂t
⎯⎯
→ B = −  ∇ × Edt
∂
∂x
∇× E =
Eo cos(ωt − β z )
B=
H=
∂
∂y
0
∂
∂z = − β Eo sin(ωt − β z )a y
0
β Eo
cos(ωt − β z )a y
ω
B
μo
β Eo
cos(ωt − β z )a y
μoω
=
Prob. 9.26
∂D
⎯⎯
→
D =  J d dt
dt
−60 ×10−3
cos(109 t − β z )a x = −60 × 10−12 cos(109 t − β z )a x C/m 2
D=
109
∂H
D
∂H
∇× E = μ
⎯⎯
→ ∇ × = −μ
ε
∂t
∂t
∂
∂
∂
∂x ∂y ∂z
1
D 1
∇× =
= (−60)(−1) ×10−12 sin(109 t − β z )a x
ε ε
ε
0
Dx 0
(a) J d =
=
H =−
1
60β
ε
× 10−12 sin(109 t − β z )a y
∇×
D
dt = −
1
(−1)
60 β 10−12
× 9 cos(109 t − β z )a y
ε
10
μ
ε
μ
60β
=
× 10−21 cos(109 t − β z )a y A/m
με
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(b)
∇ × H = J + Jφ = 0 + J d
∂
∂x
∂
∂y
∂
∂z
Jd = ∇ × H =
=
Hy
0
(− β )(−1)60β
με
0
× (10−21 ) sin(109 t − β z )a x
Equating this with the given J d
60 ×10−3 =
60 β 2
με
× 10−21
β 2 = με 1018 = 2 × 4π ×10−7 ×10 ×
β = 14.907 rad/m
Prob. 9.27
∇ × E = − μo
∂H
∂t
⎯⎯
→
10−9 2000
=
36π
9
H =−
1
μo
 ∇ × Edt
1 ∂
1 ∂
(rEθ )aφ =
[10sin θ cos(ωt − β r )] aφ
r ∂r
r ∂r
10β
sin θ cos(ωt − β r )aφ
=
r
10β
H =−
sin θ  sin(ωt − β r )dtaφ
μr
10β
sin θ cos(ωt − β r )aφ
=
ωμo r
∇× E =
Prob. 9.28
(a) ∇ • A = 0
∂
∂x
∂
∂y
∂
∂z
∇× A =
=−
0
0
E z ( x, t )
∂ E z ( x, t )
ay ≠ 0
∂x
Yes, A is a possible EM field.
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(b)
∇•B =0
∇× B =
1 ∂
ρ ∂ρ
[10 cos(ωt − 2ρ )] az ≠ 0
Yes, B is a possible EM field.
(c)
∇•C =
∇×C =
1 ∂
ρ ∂ρ
1 ∂
( 3ρ
ρ ∂ρ
3
)
cot φ sin ωt −
( cos φ sin ωt ) a z − 3ρ 2
sin φ sin ωt
ρ2
≠0
∂
(cot φ sin ωt )a z ≠ 0
∂φ
No, C cannot be an EM field.
1
∂
(d) ∇ • D = 2
sin(ωt − 5r ) (sin 2 θ ) ≠ 0
r sin θ
∂θ
∇× D = −
∂ Dθ
1 ∂
1
ar +
(rDθ )aφ = sin θ (−5) sin(ωt − 5r )aφ ≠ 0
∂φ
r ∂r
r
No, D cannot be an EM field.
Prob. 9.29
From Maxwell’s equations,
∂B
∇× E = −
(1)
∂t
∂D
∇× H = J +
(2)
∂t

Dotting both sides of (2) with E gives:
∂D
(3)
E • (∇ × H ) = E • J + E •
∂
t


But for any arbitrary vectors A and B ,
∇ • ( A × B ) = B • (∇ × A) − A • (∇ × B )
Applying this on the left-hand side of (3) by letting A ≡ H and B ≡ E , we get
∂
H • (∇ × E ) + ∇ • ( H × E ) = E • J + 1
( D • E ) (4)
2 ∂t
From (1),
 ∂B  1 ∂
H • (∇ × E ) = H •  −
 = 2 (B • H )
∂t
 ∂t 
Substituting this in (4) gives:
∂
∂
−1
(B • H ) − ∇ • (E × H ) = J • E + 1
(D • E)
2 ∂t
2 ∂t
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Rearranging terms and then taking the volume integral of both sides:
∂
 ∇ • ( E × H )dv = − ∂t 1 2  ( E • D + H • B)dv −  J • Edv
v
Using the divergence theorem, we get
 ( E × H ) • dS = −
s
v
v
∂W
− J • Edv
∂t v
∂W
or
= −  ( E × H ) • dS −  E • Jdv as required.
∂t
s
v
Prob. 9.30
∂B
= ∇ × E = β Eo sin(ωt + β y − β z )(a y + a z )
∂t
∂H
1
−μ
= ∇× E
⎯⎯
→ H = -  ∇ × Edt
∂t
μ
β Eo
H=
cos(ωt + β y − β z )(a y + a z ) A/m
−
μω
Prob. 9.31 Using Maxwell’s equations,
∇× H = σ E +ε
But
∂E
∂t
∇× H = −
E=
12sin θ
=−
εo
(σ = 0)
⎯⎯
→
E=
1
ε
 ∇ × Hdt
1 ∂ Hθ
1 ∂
12sin θ
(rHθ )aφ =
ar +
β sin(2π ×108 t − β r )aφ
r sin θ ∂φ
r ∂r
r
β  sin(2π ×108 t − β r )dtaφ
12sin θ
β cos(ωt − β r )aφ ,
ωε o r
ω = 2π ×108
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Prob. 9.32
With the given A, we need to prove that
∂2 A
∇ 2 A = με 2
∂t
2
∇ A = με ( jω )( jω ) A = −ω 2 με A
Let β 2 = ω 2 με , then ∇ 2 A = − β 2 A is to be proved. We recognize that
A=
μo jωt − jβ r
e e az
4π r
μ
e− jβ r
,
A = o e jωt ϕ a z
r
4π
1 ∂ 2
∂ϕ  1  ∂ 2  − j β 1  − jβ r 
∇ 2ϕ = 2
(r sin θ
) =
(r ) 
− 2 e


r sin θ  ∂r
dr  r 2  ∂r
r 
 r

− jβ r
1
e
= 2 ( − β 2 r + j β − j β ) e− jβ r = − β 2
= − β 2ϕ
r
r
∇2 A = −β 2 A
Therefore,
We can find V using Lorentz gauge.
−1
−1
∇ • Adt =
∇• A
V=

jωμoε o
μ oε o
ϕ=
Assume
=
∂  μo − j β r jωt 
−1
 − j β 1  − jβ r jωt
e e =
− 2  e e cos θ


jωμoε o ∂r  4π r
r 
 jωε o (4π )  r
−1
V=
cos θ 
1  j (ωt − β r )
 jβ +  e
j 4πωε o r 
r
Prob. 9.33
Take the curl of both sides of the equation.
∂
∇× A
∂t
But ∇ × ∇V = 0 and B =∇ × A. Hence,
∂B
∇× E = −
∂t
which is Faraday's law.
∇ × E = −∇ × ∇V −
Prob. 9.34
(a)
∇⋅ A =
Hence,
∂Az x
= ,
∂z c
∂V
= − xc,
∂t
∂V
∇ ⋅ A = − μoε o
∂t
− μ oε o
∂V
x
x
= 2c=
∂t c
c
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(b)
E = −∇V −
E
∂A
∂V 
 ∂V
= −
ax +
a z  + xa z = −( za x + xa z ) + xa z
∂t
∂z 
 ∂x
= − za x
Prob. 9.35
∇ A = 0 = − με
∂V
∂t
⎯⎯
→ V = constant
∂A
= 0 − Aoω cos(ωt − β z )a x
∂t
= − Aoω cos(ωt − β z )a x
(a) E = −∇V −
(b) Using Maxwell’s equations, we can show that
β = ω μ oε o
Prob. 9.36
(a)
z = 4∠30o − 10∠50o = 3.464 + 2 j − 6.427 − j 7.66 = −2.963 − j5.66
= 6.389∠ − 117.64o
z1/ 2 = 2.5277∠ − 58.82o
(b)
1 + j2
2.236 ∠ 63.43 o
2.236 ∠ 63.43 o
=
=
6 − j 8 − 7 ∠ 15 o 6 − j 8 − 7 .761 − j1.812 9.841∠ 265.57 o
= 0.2272∠ − 202.1o
(c)
(5∠ 53.13 o ) 2
25∠ 106 .26 o
z=
=
12 − j7 − 6 − j10 18.028 ∠ − 70.56 o
= 1.387 ∠ 176 .8 o
(d)
1.897 ∠ − 100 o
= 0.0349 ∠ − 68 o
(576
. ∠ 90 o )(9.434∠ − 122 o )
Prob. 9.37
(a) A = 5cos(2t + π / 3 − π / 2)a x + 3cos(2t + 30o )a y = Re( As e jωt ), ω = 2
o
o
As = 5e − j 30 a x + 3e j 30 a y
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(b) B =
100
ρ
Bs =
cos(ωt − 2π z − 90o )a ρ
100
ρ
o
e − j (2π z +90 ) a ρ
cos θ
cos(ωt − 3r − 90o )aθ
r
cos θ − j (3r +90o )
Cs =
e
az
r
(c) C =
(d) Ds = 10 cos(k1 x)e − jk2 z a y
267
Prob. 9.39
We can use Maxwell’s equations or borrow ideas from chapter 10.
μ
1 120π
η=
= ηo
=
ε
εr
9
Ho =
Eo
η
=
10 × 9
= 0.2387
120π
β = ω με =
ω
c
εr =
2π ×109
81 = 60π = 188.5 rad/m
3 × 108
Prob. 9.40
(a)
9
H = Re  40e j (10 t − β z ) a x  , ω = 109


= Re  40e − jβ z a x e jωt  = Re  H s e jωt 
H s = 40e− jβ z a x
(b)
∂
∂x
Jd = ∇ × H =
40 cos(109 t − β z )
= 40β sin(109 t − β z )a y A/m
Prob. 9.41
( jω ) 2 Y + 4 jωY + Y = 2∠0o ,
∂
∂y
0
∂
∂z
0
2
ω =3
Y (−ω 2 + 4 jω + 1) = 2
2
2
2
Y=
=
=
= −0.0769 − j 0.1154
2
−ω + 4 jω + 1 −9 + j12 + 1 −8 + j12
= 0.1387∠ − 123.7 o
y (t ) = Re(Ye jωt ) = 0.1387 cos(3t − 123.7 o )
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(b)
Sadiku & Kulkarni
∂
∂x
Jd = ∇ × H =
40 cos(109 t − β z )
= 40β sin(109 t − β z )a y A/m
Prob. 9.41
( jω ) 2 Y + 4 jωY + Y = 2∠0o ,
∂
∂y
0
∂
∂z
0
2
Principles of Electromagnetics, 6e
270
ω =3
Y (−ω 2 + 4 jω + 1) = 2
2
2
2
=
=
= −0.0769 − j 0.1154
Y=
2
−ω + 4 jω + 1 −9 + j12 + 1 −8 + j12
= 0.1387∠ − 123.7 o
y (t ) = Re(Ye jωt ) = 0.1387 cos(3t − 123.7 o )
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Principles of Electromagnetics, 6e
271
CHAPTER 10
P. E. 10.1 (a)
T=
2π
ω
=
2π
= 31.42 ns,
2 × 108
λ = uT = 3 × 108 × 31.42 × 10−9 = 9.425 m
k = β = 2π / λ = 0.6667 rad/m
(b) t1 = T/8 = 3.927 ns
(c)
H (t = t1 ) = 0.1cos(2 × 108
as sketched below.
π
8 × 108
− 2 x / 3)a y = 0.1cos(2 x / 3 − π / 4)a y
P. E. 10.2 Let xo = 1 + (σ / ωε ) 2 , then
α =ω
or
xo 2 =
μ oε o
2
μrε r ( xo − 1) =
xo − 1 =
ω 16
c
2
xo − 1
α c 1 / 3 × 3 × 108
1
=
=
108 8
8
ω 8
81
= 1 + (σ / ωε ) 2
64
xo = 9 / 8
σ
= 0.5154
ωε
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tan 2θη = 0.5154
θη = 13.63o
β
x +1
= o
= 17
α
xo − 1
β = α 17 =
(a)
17
= 1.374 rad/m
3
σ
= 0.5154
ωε
μ / ε 120π 2 / 8
(c) | η |=
=
= 177.72
(b)
xo
9/8
η = 177.72∠13.63o Ω
ω 108
u= =
= 7.278 × 107 m/s
β 1.374
(d)
(e) a H = ak × a E
H=
⎯⎯
→
a z × a x = aH
⎯⎯
→
aH = a y
0.5 − z / 3
e sin(108 t − β z − 13.63o )a y = 2.817e − z /3 sin(108 t − β z − 13.63o )a y mA/m
177.5
P. E. 10.3 (a) Along -z direction
(b) λ =
f =
2π
β
= 2π / 2 = 3.142 m
ω 108
=
= 15.92 MHz
2π 2π
β = ω με = ω μoε o μrε r =
or ε r = β c / ω =
ω
c
(1)ε r
3 × 108 × 2
=6
108
(c) θη = 0,| η |= μ / ε = μo / ε o 1 / ε r =
ak = a E × a H
−a z = a y × a H
ε r = 36
120π
= 20π
6
aH = ax
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273
50
sin(ωt + β z )a x = 795.8sin(108 t + 2 z )a x mA/m
20π
H=
P. E. 10.4 (a)
σ
=
ωε
10−2
10−9
10 π × 4 ×
36π
α ≅ω
β ≅ω
= 0.09
9
με 
2
 ω
1 σ 
σ
109 π
1
1
(2)(0.09) = 0.9425 Np/m
μ
ε
+
−
=
=


r r


2  2  ωε 
ωε 2 × 3 × 108
 2c
με 
2
 109 π
1 σ 
1
2[2 + 0.5(0.09) 2 ] = 20.965 rad/m
+
1 + 
=

8
2  2  ωε 
3
10
×

E = 30e −0.9425 y cos(109 π t − 20.96 y + π / 4)a z
At t = 2ns, y = 1m,
E = 30e −0.9425 cos(2π − 20.96 + π / 4)a z = 2.844a z V/m
(b) β y = 10o =
10π
rad
180
or
y=
π 1
π
=
= 8.325 mm
18 β 18 × 20.965
(c) 30(0.6) = 30 e−α y
y=
1
α
ln(1 / 0.6) =
1
1
ln
= 542 mm
0.9425 0.6
(d)
| η |≅
μ /ε
1
[1 + (0.09) 2 ]
4
=
60π
= 188.11 Ω
1.002
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2θη = tan −1 0.09
θη = 2.571o
a H = ak × a E = a y × a z = a x
H=
30 −0.9425 y
e
cos(109 π t − 20.96 y + π / 4 − 2.571o )a x
188.11
At y = 2m, t = 2ns,
H = (0.1595)(0.1518) cos(−34.8963rad )a x = −22.83a x mA/m
P. E. 10.5
w∞
w
∞
0 0
0
0
I s =   J xs dydz = J xs (0)  dy  e− z (1+ j ) / δ dz =
| I s |=
J xs (0) wδ
1+ j
J xs (0) wδ
2
P. E. 10.6 (a)
Rac
a a
1.3 × 10−3
π f μσ =
π × 107 × 4π × 10−7 × 3.5 × 107 = 24.16
=
=
Rdc 2δ 2
2
(b)
Rac 1.3 × 10−3
π × 2 × 109 × 4π × 10−7 × 3.5 × 107 = 341.7
=
Rdc
2
P. E. 10.7
E = Re[ E s e jωt ] = Re  Eo e jωt e − j β z a x + Eo e − jπ / 2e jωt e − j β z a y 
= Eo cos(ωt − β z )a x + Eo cos(ωt − β z − π / 2)a y
= Eo cos(ωt − β z )a x + Eo sin(ωt − β z )a y
At z = 0, Ex = Eo cos ωt , E y = Eo sin ωt
2
2
E  E 
→  x  + y  =1
cos ωt + sin ωt = 1 ⎯⎯
 Eo   Eo 
which describes a circle. Hence the polarization is circular.
2
2
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P. E. 10.8
1
Pave = η H o 2a x
2
(a) Let f(x,z) = x + y –1 = 0
an =
a + ay
∇f
,
= x
| ∇f |
2
Pt =  P.dS =P.San =
=
1
2 2
dS = dSan
a + ay
1
η H o 2a x . x
2
2
(120π )(0.2) 2 (0.1) 2 = 53.31 mW
(b) dS = dydzax , Pt =  P.dS =
1
η H o2S
2
1
Pt = (120π )(0.2) 2 π (0.05) 2 = 59.22 mW
2
P. E. 10.9
τ=
η1 = ηo = 120π ,η 2 =
μ ηo
=
ε
2
2η2
η −η
= 2 / 3, Γ = 2 1 = −1 / 3
η2 + η1
η2 + η1
Ero = ΓEio = −
Ers = −
10
3
10 j β1 z
e a x V/m
3
where β1 = ω / c = 100π / 3 .
20
Eto = τ Eio =
3
Ets =
20 − j β2 z
e
a x V/m
3
where β 2 = ω ε r / c = 2β1 = 200π / 3 .
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P. E. 10.10
α1 = 0, β1 =
σ2
=
ωε 2
ω
c
μrε r =
2ω
= 5 ⎯⎯
→ ω = 5c / 2 = 7.5 × 108
c
0.1
= 1.2π
10−9
7.5 × 10 × 4 ×
36π
8
α2 =
ω 4
1 + 1.44π 2 − 1 = 6.021
β2 =
ω 4
1 + 1.44π 2 + 1 = 7.826
c
| η 2 |=
c
2
2
60π
4

1 + 1.44π 2

= 95.445,η1 = 120π ε r1 = 754
tan 2θη2 = 1.2π ⎯⎯
→θη2 = 37.57o
η2 = 95.445∠37.57o
(a)
η2 − η1 95.445∠37.57 o − 754
Γ=
=
= 0.8186∠171.08o
o
η2 + η1 95.445∠37.57 + 754
τ = 1 + Γ = 0.2295∠33.56o
s=
(b)
1+ | Γ | 1 + 0.8186
=
= 10.025
1− | Γ | 1 − 0.8186
Ei = 50sin(ωt − 5 x)a y = Im( Eis e jωt ) , where Eis = 50e − j 5 x a y .
o
o
Ero = ΓEio = 0.8186e j171.08 (50) = 40.93e j171.08
o
Ers = 40.93e j 5 x + j171.08 a y
Er = Im( Ers e jωt ) = 40.93sin(ωt + 5 x + 171.1o )a y V/m
a H = ak × a E = −a x × a y = −a z
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Hr = -
40.93
sin(ωt + 5 x + 171.1o )a z = −0.0543sin(ωt + 5 x + 171.1o )a z A/m
754
(c)
o
o
Eto = τ Eio = 0.229e j 33.56 (50) = 11.475e j 33.56
o
Ets = 11.475e− j β2 x + j 33.56 e−α 2 x a y
Et = Im( Ets e jωt ) = 11.475e −6.021x sin(ωt − 7.826 x + 33.56o )a y V/m
a H = ak × a E = a x × a y = a z
Ht =
11.495 −6.021x
e
sin(ωt − 7.826 x + 33.56o − 37.57 o )a z
95.445
= 0.1202e−6.021x sin(ωt − 7.826 x − 4.01o )a z A/m
(d)
P1ave
Eio 2
Ero 2
1
=
ax +
( −a x ) =
[502 a x − 40.932 a x ] = 0.5469a x W/m2
2η1
2η1
2(240π )
P2ave =
Eto 2 −2α 2 x
(11.475) 2
e
cosθη2 a x =
cos37.57o e−2(6.021) x a x = 0.5469e−12.04 x a x W/m2
2 | η2 |
2(95.445)
P. E. 10.11 (a)
k = −2a y + 4a z ⎯⎯
→ k = 22 + 42 = 20
ω = kc = 3 × 108 20 = 1.342 × 109 rad/s ,
λ = 2π / k = 1.405m
(b) H =
ak × E
ηo
=
(−2a y + 4a z )
20(120π )
× (10a y + 5a z ) cos(ωt − k.r )
= −29.66cos(1.342 x109 t + 2 y − 4 z )a x mA/m
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(c) Pave =
| Eo |2
125 (−2a y + 4a z )
ak =
= −74.15a y + 148.9a z
2ηo
2(120π )
20
mW/m2
P. E. 10.12 (a)
y
ki
θi
kt
θr
z
kr
tan θi =
sin θt =
kiy
kiz
=
2
⎯⎯
→θi = 26.56 = θ r
4
μ1 ε1
1
→θt = 12.92o
sin θi = sin 26.56o ⎯⎯
μ2 ε 2
2
(b) η1 = ηo ,η 2 = ηo / 2
E is parallel to the plane of incidence. Since μ1 = μ2 = μo ,
we may use the result of Prob. 10.42, i.e.
Γ\ \ =
tan(θt − θi ) tan(−13.64o )
=
= −0.2946
tan(θt + θi ) tan(39.48o )
τ \\ =
2cos 26.56o sin12.92o
= 0.6474
sin 39.48o cos(−13.64o )
(c) kr = − β1 sin θ r a y − β1 cosθ r a z .
Once kr is known, Er is chosen such that
kr .Er = 0 or ∇.Er = 0. Let
Er = ± Eor (− cosθ r a y + sin θ r a z ) cos(ωt + β1 sin θ r y + β1 cosθ r z )
Only the positive sign will satisfy the boundary conditions. It is evident that
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Ei = Eoi (cos θi a y + sin θi a z ) cos(ωt + 2 y − 4 z )
Since θ r = θi ,
Eor cosθ r = Γ / / Eoi cosθi = 10Γ / / = −2.946
Eor sin θ r = Γ / / Eoi sin θi = 5Γ / / = −1.473
β1 sin θ r = 2, β1 cosθ r = 4
i.e.
Er = −(2.946a y − 1.473az ) cos(ωt + 2 y + 4 z )
E1 = Ei + Er = (10a y + 5a z )cos(ωt + 2 y − 4 z ) + (−2.946a y + 1.473a z ) cos(ωt + 2 y + 4 z )
V/m
(d) kt = − β 2 sin θt a y + β 2 cosθt a z .
Since kr • Er = 0 , let
Et = Eot (cosθt a y + sin θt a z ) cos(ωt + β 2 y sin θt − β 2 z cos θt )
β 2 = ω μ2ε 2 = β1 ε r 2 = 2 20
1
1
9
sin θt = sin θi =
,
cosθt =
2
2 5
20
19
β 2 cosθt = 2 20
= 8.718
20
19
Eot cosθt = τ / / Eoi cosθt = 0.6474 125
= 7.055
20
Eot sin θt = τ / / Eoi sin θt = 0.6474 125
1
= 1.6185
20
Hence
E2 = Et = (7.055a y + 1.6185a z ) cos(ωt + 2 y − 8.718 z ) V/m
(d) tan θ B / / =
ε2
= 2 ⎯⎯
→θ B / / = 63.43o
ε1
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P.E. 10.13
Si =
1 + 0.4 1.4
=
= 2.333
1 − 0.4 0.6
So =
1 + 0.2 1.2
=
= 1.5
1 − 0.2 0.8
Prob. 10.1 (a) Wave propagates along +ax.
(b)
2π
= 1μ s
ω 2π × 106
2π 2π
=
= 1.047m
λ=
β
6
T=
2π
=
ω 2π × 106
= 1.047 × 106 m/s
u= =
β
6
(c) At t=0, Ez = 25sin(−6 x) = −25sin 6 x
At t=T/8, Ez = 25sin(
2π T
π
− 6 x) = 25sin( − 6 x)
T 8
4
At t=T/4, Ez = 25sin(
At t=T/2, Ez = 25sin(
2π T
− 6 x) = 25sin(−6 x + 90o ) = 25cos 6 x
T 4
2π T
− 6 x) = 25sin(−6 x + π ) = 25sin 6 x
T 2
These are sketched below.
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Prob. 10.2
(a)
∂E
= − sin( x + ωt ) − sin( x − ωt )
∂x
∂2E
= − cos( x + ωt ) − cos( x − ωt ) = − E
∂x 2
∂E
= −ω sin( x + ωt ) − ω sin( x − ωt )
∂t
∂2E
= −ω 2 cos( x + ωt ) − ω 2 cos( x − ωt ) = −ω 2 E
2
∂t
2
∂2E
2 ∂ E
u
−
= −ω 2 E + u 2 E = 0
2
2
∂t
∂x
2
if ω = u 2 and hence, eq. (10.1) is satisfied.
(b) u = ω
Prob. 10.3
(a) ω = 108 rad/s
ω
108
= 0.333 rad/m
(b) β = =
c 3 × 108
2π
= 6π = 18.85 m
(c) λ =
β
(d) Along -ay
At y=1, t=10ms,
1
(e) H = 0.5cos(108 t × 10 × 10−9 + × 3) = 0.5cos(1 + 1)
3
= −0.1665 A/m
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Prob. 10.4
c 3 × 108
(a) λ = =
= 5 × 106 m
f
60
3 × 108
= 150 m
2 × 106
3 × 108
= 2.5 m
(c) λ =
120 × 106
3 × 108
= 0.125 m
(d) λ =
2.4 × 109
(b) λ =
Prob. 10.5 If
γ 2 = jωμ (σ + jωε ) = −ω 2 με + jωμσ
and γ = α + j β , then
| γ 2 |= (α 2 − β 2 ) + 4α 2 β 2 = (α 2 + β 2 ) 2 = α 2 + β 2
i.e.
α 2 + β 2 = ωμ (σ 2 + ω 2ε 2 )
Re(γ 2 ) = α 2 − β 2 = −ω 2 με
β 2 − α 2 = ω 2 με
(1)
(2)
Subtracting and adding (1) and (2) lead respectively to
α =ω
β =ω
με 
2

σ 

 1+ 

1
−
ωε 
2 




με 
2

σ 
 1 + 
+ 1
ωε 
2 




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(b) From eq. (10.25), E s ( z ) = Eo e −γ z a x .
∇ × E = − jωμ H s
Hs =
But H s ( z ) = H o e −γ z a y , hence H o =
η=
Eo
η
=−
jγ
ωμ
j
ωμ
∇ × Es =
j
ωμ
(−γ Eo e −γ z a y )
Eo
jωμ
γ
(c) From (b),
jωμ
η=
jωμ (σ + jωε )
4
jωμ
=
σ + jωε
μ /ε
σ
1− j
ωε
−1
μ /ε
| η |=
=
σ 
1+ 

 ωε 
2
σ
 ωε 
, tan 2θη = 
 =
ωε
σ 
Prob. 10.6 (a)
σ
=
ωε
8 × 10−2
10−9
2π × 50 × 10 × 3.6 ×
36π
=8
6
α =ω
β =ω
με 
2

2π × 50 × 106
σ 
 1 + 

=
−
1
3 × 108
ωε 
2 




2.1 × 3.6
[ 65 − 1] = 5.41
2
με 
2

σ 
 1 + 
 = 6.129
+
1
ωε 
2 




γ = α + j β = 5.41 + j 6.129 /m
(b)
λ=
2π
β
=
2π
= 1.025 m
6.129
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(c)
ω 2π × 50 × 106
=
= 5.125 × 107 m/s
β
6.129
u=
μ
ε
(d) | η |=
4
tan 2θη =
σ 
1+ 

 ωε 
2
=
2.1
3.6 = 101.4
4
65
120π
σ
= 8 ⎯⎯
→θη = 41.44o
ωε
η = 101.41∠41.44o Ω
(e)
H s = ak ×
Es
η
o
6
6
= a x × e −γ z a z = − e −γ z a y = −59.16e − j 41.44 e −γ z a y mA/m
η
η
Prob. 10.7
(a) tan θ =
σ
=
ωε
10−2
= 1.5
10−9
2π × 12 × 10 × 10 ×
36π
−4
10
(b) tan θ =
= 3.75 × 10−2
−9
10
2π × 12 × 106 × 4 ×
36π
(c) tan θ =
6
4
10−9
2π × 12 × 10 × 81 ×
36π
= 74.07
6
Prob. 10.8
(a)
α =ω
με 
2
 2π × 15 × 109 1 × 9.6
σ 

 1 + 9 × 10−8 − 1
 1+ 
=
1
−

8

2 
3 × 10
2 

 ωε 


1

= 100π 4.8  × 9 × 10−8  = 0.146
2

1
δ = = 6.85 m
α
(b) A = α  = 0.146 × 5 × 10−3 = 0.73 × 10−3 Np
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Prob. 10.9
The phase difference is θη
tan 2θη =
σ
=
ωε
2θη = 60.9o
8 × 10−3
10−9
2π × 20 × 10 × 4 ×
36π
⎯⎯
→ θη = 30.47o
= 1.8
6
Prob. 10.10
For silver, the loss tangent is
σ
6.1 × 107
=
= 6.1 × 18 × 108  1
−9
10
ωε
2π × 108 ×
36π
Hence, silver is a good conductor
For rubber,
σ
=
ωε
10−15
−9
=
18
× 10−14  1
3.1
10
36π
Hence, rubber is a poor conductor or a good insulator.
2π × 108 × 3.1 ×
Prob. 10.11
σ
4
=
= 9,000 >> 1
5
ωε 2π × 10 × 80 × 10−9 / 36π
α =β =
ωμσ
2
=
2π × 105
× 4π × 10−7 × 4 = 0.4π
2
(a)
2π × 105
= 5 × 105 m/s
u =ω / β =
0.4π
(b)
λ = 2π / β =
(c) δ = 1 / α =
2π
=5 m
0.4π
1
= 0.796 m
0.4π
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(d) η =| η | ∠θη ,θη = 45o
μ
ε
| η |=
4
η = 0.4443∠45o
σ 
1+ 

 ωε 
2
≅
4π × 10−7 × 2π × 105
μ ωε
=
= 0.4443
ε σ
4
Ω
Prob. 10.12
σ
=
ωε
1
10−9
2π × 10 × 4 ×
36π
=4.5
9
α =ω
με 
2

σ 
 1 + 
− 1
ωε 
2 




= 2π × 109
4π
10−9 
× 10−7 × 4 × 9 ×
1 + 4.52 − 1

2
36π 
= 20π 2[ 21.25 − 1] = 168.8 Np/m
β =ω
με 
2

σ 

 1+ 
 = 20π 2[ 21.25 + 1] = 210.5 rad/m
+
1
ωε 
2 




tan 2θη =
| η |=
σ
= 4.5
⎯⎯
→ θη = 38.73o
ωε
μ /ε
120π 9 / 4
σ 
4 1+


 ωε 
2
=
4
1 + 4.52
= 263.38
η = 263.38∠38.73o Ω
u=
ω 2π × 109
=
= 2.985 × 107 m/s
β
210.5
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287
Prob. 10.13
This is a lossy medium in which μ=μo.
2
σ 
Let x = 

 ωε 
o
jωμ j 2π × 109 × 4π
η=
=
= 35.31∠26.57
γ
100 + j 200
Eo = 0.05 × 35.31 = 1.765
a E = a H × a k = −a z
Thus, we obtain
E = -1.765cos (2π × 109 t − 200 x + 26.57 o )a z V/m
ε r (1 / 3) =
εr
α c 100 × 3 × 108 15
=
=
ω
2π × 109
π
1
= 4.776
3
tan 2θη =
⎯⎯
→
σ
4
=
ωε 3
ε r = 14.32
⎯⎯
→
θη = 26.57o
377
μ /ε
| η |= 4
= 14.32 = 77.175
1+ x
5/3
Eo =| η | H o = 77.175 × 50 × 10−3 = 3.858
a E = − (a k × a H ) = − ( a x × a y ) = − a z
E = −3.858e −100 x cos(2π × 109 t − 200 x + 26.57o )a z V/m
Prob. 10.14 (a)
T = 1 / f = 2π / ω =
(b) Let
σ 
x = 1+ 

 ωε 
2π
= 20 ns
π x108
2
1/ 2
α  x −1
=
β  x + 1 
But
α=
ω μrε r
c
2
x −1
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αc
0.1 × 3 × 108
=
= 0.06752 ⎯⎯
→ x = 1.0046
μrε r
π × 108 2
x −1 =
ω
2
1/ 2
1/ 2
 x +1
 2.0046 
β =
 α =
 0.1 = 2.088
 x −1
 0.0046 
λ = 2π / β =
(c)
| η |=
2π
= 3m
2.088
μ /ε
x
=
377
= 188.1
2 1.0046
2
σ 
x = 1+ 
 = 1.0046
 ωε 
σ
= 0.096 = tan 2θη ⎯⎯
→θη = 2.74o
ωε
η = 188.1∠2.74o
Ω
Eo = η H o = 12 × 188.1 = 2257.2
a E × a H = ak ⎯⎯
→ a E × a x = a y ⎯⎯
→ aE = az
E = 2.257e −0.1 y sin(π × 108 t − 2.088 y + 2.74o )a z kV/m
(d) The phase difference is 2.74o.
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Principles of Electromagnetics, 6e
289
Prob. 10.15
β = 6.5 = ω μoε o =
(a)
ω
c
ω = β c = 6.5 × 3 × 10 = 1.95 × 109 rad/s
8
2π
= 0.9666 m
β 6.5
(b) For z=0, Ez = 0.2cos ωt
λ=
2π
=
2π λ
) = −0.2cos ωt
λ 2
The two waves are sketched below.
For z=λ /2, Ez = 0.2cos(ωt −
300
z = 0
z = λ/2
Amplitude (mV / m)
200
100
0
-100
-200
-300
-3
-2
-1
0
1
Time t (ns)
2
(c) H = H o cos(ωt − 6.5 z )aH
Ho =
Eo
ηo
=
0.2
= 5.305 × 10−4
377
a E × a H = ak
⎯⎯
→ ax × aH = az
⎯⎯
→ aH = a y
H = 0.5305cos(ωt − 6.5 z )a y mA/m
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Prob. 10.16
c
u=
λ=
ε r μr
=
3 × 108
= 8.66 × 107 m/s
3× 4
u 8.66 × 107
=
= 1.443 m
f
60 × 106
μr
4
= 377
= 435.32 Ω
3
εr
η = ηo
Prob. 10.17 (a) Along -x direction.
(b) β = 6,
β = ω με =
ω = 2 × 108 ,
ω
μrε r
c
εr = βc / ω =
6 × 3 × 108
=9
2 × 108
ε r = 81
⎯⎯
→
10−9
ε = ε oε r =
× 81 = 7.162 × 10−10 F/m
36π
(c) η = μ / ε = μo / ε o μr / ε r =
120π
= 41.89 Ω
9
Eo = H oη = 25 × 10−3 × 41.88 = 1.047
a E × a H = ak ⎯⎯
→ a E × a y = −a x ⎯⎯
→ aE = az
E = 1.047sin(2 × 108 t + 6 x)a z V/m
σ
=
Prob. 10.18 (a)
ωε
10−6
2π × 107 × 5 ×
−9
10
36π
= 3.6 × 10−4 << 1
Thus, the material is lossless at this frequency.
(b) β = ω με =
2π × 107
5 × 750 = 12.83 rad/m
3 × 108
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2π
λ=
=
β
2π
= 0.49 m
12.83
(c) Phase difference = β l = 25.66 rad
(d) η = μ / ε = 120π
μr
750
= 120π
= 4.62 kΩ
5
εr
Prob. 10.19
(a) E = Re[ E s e jωt ] = (5a x + 12a y )e −0.2 z cos(ωt − 3.4 z )
At z = 4m, t = T/8, ωt =
2π T π
=
T 8 4
E = (5a x + 12a y )e −0.8 cos(π / 4 − 13.6)
| E |= 13e −0.8 | cos(π / 4 − 13.6) |= 5.662 V/m
(b) loss = αΔz = 0.2(3) = 0.6 Np. Since 1 Np = 8.686 dB,
loss = 0.6 x 8.686 = 5.212 dB
(c) Let
σ 
x = 1+ 

 ωε 
2
1/ 2
α  x −1
=
β  x + 1 
= 0.2 / 3.4 =
x −1
= 1 / 289
x +1
⎯⎯⎯
→
α = ω με / 2 x − 1 =
εr
αc
ω
c
1
17
x = 1.00694
εr / 2 x −1
0.2 × 3 × 108
=
=
= 7.2
2 ω x − 1 108 0.00694
| η |=
μo 1
.
εo εr
x
=
⎯⎯
→
ε r = 103.68
120π
= 36.896
103.68 × 1.00694
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Principles of Electromagnetics, 6e
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tan 2θη =
σ
= x 2 − 1 = 0.118
ωε
θη = 3.365o
⎯⎯
→
η = 36.896∠3.365o Ω
Prob. 10.20
This is a lossless material.
μ
μ
= 377 r = 105
ε
εr
η=
u=
(1)
ω
ω
c
=
=
= 7.6 × 107
β ω με
μrε r
(2)
From (1),
μr 105
=
= 0.2785
ε r 377
(1)a
From (2),
1
μrε r
=
7.6 × 107
= 0.2533
3 × 108
(2)a
Multiplying (1)a by (2)a,
1
= 0.2785 × 0.2533 = 0.07054
⎯⎯
→ ε r = 14.175
εr
Dividing (1)a by (2)a,
μr =
0.2785
= 1.0995
0.2533
Prob. 10.21
ax
ay
∂
∂
∇× E =
∂x
∂y
(a)
Ex ( z , t ) E y ( z , t )
az
∂E
∂
∂E
= − y ax + x a y
∂z
∂z
∂z
0
= −6 β cos(ωt − β z )a x + 8β sin(ωt − β z )a y
But ∇ × E = − μ
H=
6β
μω
∂H
∂t
⎯⎯
→ H =−
sin(ωt − β z )a x +
8β
μω
1
μ
∇ × Edt
cos(ωt − β z )a y
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2π f
2π × 40 × 106
4.5 =
4.5 = 1.777 rad/m
c
3 × 108
2π
2π
=
= 3.536 m
λ=
β 1.777
(b) β = ω με =
η=
u=
μ 120π
=
= 177.72 Ω
ε
4.5
c
3 × 108
=
=
= 1.4142 × 108 m/s
4.5
4.5
με
1
Prob. 10.22
0.4 Eo = Eo e −α z
Or
1
2
α = ln
1
= e2α
0.4
⎯⎯
→
1
= 0.4581
0.4
⎯⎯
→
δ = 1 / α = 2.183 m
λ = 2π / β = 2π / 1.6
u = f λ = 107 ×
2π
= 3.927 × 107 m/s
1.6
Prob. 10.23
(a)
ω
108 π
π
= = 1.0472 rad/m
β = ω με = =
8
c 3 × 10
3
(b)
E =0
⎯⎯
→ sin(108 π to − β xo ) = 0 = sin(nπ ), n = 1, 2,3,...
108 π to − β xo = π
108 π × 5 × 10−3 −
π
3
xo = π
⎯⎯
→
xo  5 × 105 m
(c)
H = H o sin(108 π t − β x)aH
50 × 10−3
Ho =
=
= 132.63 μ A/m
η
120π
a H = ak × a E = a x × a z = −a y
Eo
H = −132.63sin(108 π t − 1.0472 x)a y μ A/m
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Principles of Electromagnetics, 6e
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Prob. 10.24
α =β =
ωμσ
2
η =| η | ∠θη =
| η |=
2α 2
2 × 122
⎯⎯
→ σ=
=
= 36.48
ωμ 2π × 106 × 4π × 10−7
ωμ
∠45o
σ
2π × 106 × 4π × 10−7
ωμ
=
= 0.4652
36.48
σ
Eo =| η | H o = 0.4652 × 20 × 10−3 = 9.305 × 10−3
a E = a H × a k = a y × ( −a z ) = −a x
E = Eo e−α z sin(ωt + β z )aE
= −9.305e −12 z sin(2π × 106 t + 12 z + 45o )a x mV/m
Prob. 10.25 For a good conductor,
(a)
σ
=
ωε
10−2
10−9
2π × 8 × 10 × 15 ×
36π
σ
>> 1,
ωε
= 1.5
say
σ
> 100
ωε
⎯⎯⎯
→
lossy
6
No, not conducting.
(b)
σ
=
ωε
0.025
10−9
2π × 8 × 10 × 16 ×
36π
= 3.515
⎯⎯⎯
→
lossy
= 694.4
⎯⎯⎯
→
conducting
6
No, not conducting.
(c)
σ
=
ωε
25
10−9
2π × 8 × 10 × 81 ×
36π
Yes, conducting.
6
Prob. 10.26
α =ω
με 
2
 2π f
σ 

 1+ 
=
1
−
2 
ωε 
c




μrε r 
2π × 6 × 106 4
=
1.0049
1
−
× 2.447 × 10−3
8


2
α = 8.791 × 10−3
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3 × 10
2
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295
δ = 1 / α = 113.75 m
με 
2

σ 
4π
 1 + 
=
1
+

2 
 100
 ωε 


β =ω
u =ω / β =
4
1.0049 + 1 = 0.2515
2
2π × 6 × 106
= 1.5 × 108 m/s
0.2515
Prob. 10.27 (a)
Rdc =
l
l
600
=
=
= 2.287 Ω
2
7
5.8 × 10 × π × (1.2) 2 × 10−6
σ S σπ a
l
(b) Rac =
σ 2π aδ
(see Table 10.2).
Rac =
(c)
At 100 MHz, δ = 6.6 × 10−3 mm =6.6 × 10-6 m mm for copper
600
= 207.61 Ω
5.8 × 10 × 2π × (1.2 × 10−3 ) × 6.6 × 10−6
7
Rac
a
=
=1
Rdc 2δ
f =
.
⎯⎯
→
66.1 × 10−3
δ =a/2=
f
66.1 × 2 × 10−3 66.1 × 2
=
a
1.2
⎯⎯
→
f = 12.137 kHz
Prob. 10.28
(a) tan θ =
α =β =
σ
=
ωε
ωμσ
3.5 × 107
10−9
2π × 150 × 10 ×
36π
=
6
3.5 × 18 × 109
>> 1
15
= π f μσ = 150π × 106 × 4π × 10−7 × 3.5 × 107 = 143,965.86
2
γ = α + j β = 1.44(1 + j ) × 105 /m
(b)
δ = 1 / α = 6.946 × 10−6 m
(c) u =
ω 150 × 2π × 106
=
= 6547 m/s
β
1.44 × 105
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Prob. 10.29
σ
=
ωε
4
= 1.5
10−9
2π × 2 × 10 × 24 ×
36π
9
10−9
2
4π × 10−7 × 24 ×


με
σ 
9
36π  1 + (1.5 )2 − 1
 1 + 
α =ω
 − 1 = 2π × 2 × 10


2 
2

 ωε 


= 130.01 Np/m
10−5 Eo = Eo e −α d
Taking the log of both sides gives
5ln10 5ln10
=
= 0.0886 m
⎯⎯
→ d=
-5ln10 = −α d
α
130.01
Prob. 10.30
α = β =1/ δ
λ = 2π / β = 2πδ = 6.283δ
⎯⎯
→
δ = 0.1591λ
showing that δ is shorter than λ .
Prob. 10.31
t = 5δ =
5
5
=
= 2.94 × 10−6 m
9
−7
7
π f μσ
π × 12 × 10 × 4π × 10 × 6.1 × 10
Prob. 10.32
δ=
f =
1
π f μσ
⎯⎯
→
f =
1
δ πμσ
2
1
= 1.038 kHz
4 × 10 × π × 4π × 10−7 × 6.1 × 107
−6
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Principles of Electromagnetics, 6e
297
Prob. 10.33
(a) Linearly polarized along az
(b) ω = 2π f = 2π × 107
⎯⎯
→
β = ω με = ω μoε o ε r =
(c)
ω
c
f = 107 = 10 MHz
εr
β c 3 × 3 × 108
=
= 14.32
εr =
2π × 107
ω
Let H = H o sin(ωt − 3 y )a H
Ho =
Eo
η
,
η=
⎯⎯
→ ε r = 205.18
μ ηo 120π
=
=
= 26.33
ε
ε r 14.32
12
= 0.456
26.33
a H = ak × a E = a y × a z = a x
(d) H o =
H = 0.456sin(2π × 107 t − 3 y )a x A/m
Prob. 10.34
E = (2a y − 5a z )sin(ωt − β x)
The ratio E y / Ez remains the same as t changes. Hence the wave is linearly polarized
Prob. 10.35
(a)
Ex = Eo cos(ωt + β y ),
E y = Eo sin(ωt + β y )
Ex (0, t ) = Eo cos ωt
⎯⎯
→ cos ωt =
E y (0, t ) = Eo sin ωt
⎯⎯
→ sin ωt =
2
Ex (0, t )
Eo
E y (0, t )
Eo
2
 E   Ey 
cos ωt + sin ωt = 1 ⎯⎯
→  x  +   =1
 Eo   Eo 
Hence, we have circular polarization.
2
2
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Principles of Electromagnetics, 6e
298
(b)
Ex = Eo cos(ωt − β y ),
E y = −3Eo sin(ωt − β y )
In the y=0 plane,
Ex (0, t ) = Eo cos ωt
⎯⎯
→ cos ωt =
E y (0, t ) = Eo sin ωt
⎯⎯
→ sin ωt =
2
Ex (0, t )
Eo
− E y (0, t )
3Eo
2
 E  1  Ey 
cos ωt + sin ωt = 1 ⎯⎯
→  x  +   =1
 Eo  9  Eo 
Hence, we have elliptical polarization.
2
2
Prob. 10.36
(a) We can write
E = Re( Es e jωt ) = (40a x + 60a y ) cos(ωt − 10 z )
Since E x / E y does not change with time, the wave is linearly polarized.
(b) This is elliptically polarized.
Prob. 10.37
(a)
When φ = 0,
E ( y, t ) = ( Eo1a x + Eo 2a z )cos(ωt − β y )
The two components are in phase and the wave is linearly polarized.
(b)
When φ = π / 2,
Ez = Eo 2 cos(ωt − β y + π / 2) = − Eo 2 sin(ωt − β y )
We can combine Ex and Ez to show that the wave is elliptically polarized.
(c)
When φ = π ,
E ( y, t ) = Eo1 cos(ωt − β y )a x + Eo 2 cos(ωt − β y + π )a z
= ( Eo1 a x − Eo 2a y )cos(ωt − β y )
Thus, the wave is linearly polarized.
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Principles of Electromagnetics, 6e
299
Prob. 10.38
We can write Es as
E s = E1 ( z ) + E2 ( z )
where
1
Eo (a x − ja y )e − j β z
2
1
E2 ( z ) = Eo (a x + ja y )e− j β z
2
We recognize that E1 and E2 are circularly polarized waves. The problem is therefore proved.
E1 ( z ) =
Prob. 10.39
(a) The wave is elliptically polarized.
(b)
Let E = E1 + E2 ,
where E1 = 40cos(ωt − β z )a x ,
E2 = 60sin(ωt − β z )a y
H1 = H o1 cos(ωt − β z )aH 1
H o1 =
40
ηo
=
40
= 0.106
120π
a H 1 = ak × a E = a z × a x = a y
H1 = 0.106cos(ωt − β z )a y
H 2 = H o 2 sin(ωt − β z )a H 2
H o1 =
60
ηo
=
60
= 0.1592
120π
a H 2 = a k × a E = a z × a y = −a x
H 2 = −0.1592sin(ωt − β z )a x
H = H1 + H 2 = −159.2sin(ωt − β z )a x + 106cos(ωt − β z )a y mA/m
Prob. 10.40
Let E s = Er + jEi
and
H s = H r + jH i
E = Re(E s e jωt ) = Er cos ωt − Ei sin ωt
Similarly,
H = H r cos ωt − H i sin ωt
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Principles of Electromagnetics, 6e
300
1
P = E × H = Er × H r cos 2 ωt + Ei × H i sin 2 ωt − ( Er × H i + Ei × H r )sin 2ωt
2
T
T
T
T
1
1
1
1
2
2
Pave =  Pdt =  cos ωdt ( Er × H r ) +  sin ωdt ( Ei × Hi ) −
sin 2ωdt ( Ei × Hi + Ei × H r )
T0
T0
T0
2T 0
1
1
= ( Er × H r + Ei × H i ) = Re[( Er + jEi ) × ( H r − jH i )]
2
2
Pave =
1
Re(E s × H s* )
2
as required.
Prob. 10.41
(a)
β = ω με = ω μoε o ε r =
ω
c
εr
β c 8 × 3 × 108
=
= 2.4
ω
109
εr =
ε r = 5.76
μ
μo 1
377
=
=
= 157.1 Ω
ε
ε o ε r 2.4
(b) η =
(c) u =
ω 109
=
= 1.25 × 108 m/s
8
β
(d)
Let
H = H o cos(109 t + 8 x)a H
150
= 0.955
η 157.1
a H = ak × a E = −a x × a z = a y
Ho =
Eo
=
H = 0.955cos(109 t + 8 x)a y A/m
(e)
P = E × H = -150(0.955)cos 2 (109 t + 8 x)a x
= -143.25cos 2 (109 t + 8 x)a x W/m 2
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Principles of Electromagnetics, 6e
301
Prob. 10.42
P = E×H =
Ex
0
0
Hy
Ez
= − Ez H y a x + Ex H y a z
0
= − H o Eo 2 sin α x cos α x sin(ωt − β x) cos(ωt − β z )a x
+ H o Eo1 cos 2 α x cos 2 (ωt − β z )a z
1
− H o Eo 2 sin 2α x sin 2(ωt − β x)a x
4
+ H o Eo1 cos 2 α x cos 2 (ωt − β z )a z
=
Pave =
1T
1
Pdt = 0 + H o Eo1 cos 2 α xa z

2
T0
1
= Eo1H o cos 2 α xa z
2
Prob. 10.43
(a)
H
Let H s = o sin θ e − j 3r a H
r
10
1
E
=
Ho = o =
ηo 120π 12π
a H = ak × a E = ar × aθ = aφ
Hs =
1
sin θ e − j 3r aφ A/m
12π r
(b)
1
2
Pave = Re( E s × H s ) =
Pave =  Pave dS ,
10
sin 2 θ ar
2
2 × 12π r
dS = r 2 sin θ dθ dφar
S
10
Pave =
24π
π π /6
 r
φ θ
=0 =0
2
sin 3 θ dθ
r=2
=
5 5 3
−
= 0.007145
8 32
= 7.145 mW
Prob. 10.44
(a) Pave =
1
1
|E |
82 −0.2 z
Re( Es H s* ) = Re( s ) =
e
2
2
|η |
2 |η |
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Principles of Electromagnetics, 6e
302
α =ω
β =ω
Let
με 
2

σ 

 1+ 

−
1
ωε 
2 




με 
2

σ 
 1 + 

+
1
ωε 
2 




σ 
x = 1+ 

 ωε 
2
α
x −1
=
= 0.1 / 0.3 = 1 / 3
β
x +1
x −1 1
=
x +1 9
⎯⎯
→ x =5/4
5
σ 
= 1+ 

4
 ωε 
2
⎯⎯
→
μ
ε
| η |=
2
=
σ
=3/ 4
ωε
120π / 81
= 37.4657
5
4
σ 
1+ 

 ωε 
64
Pave =
e−0.2 z = 0.8541e −0.2 z W/m 2
2(37.4657)
4
(b) 20dB = 10log
P1
P2
⎯⎯
→
P1
= 100
P2
P2
1
= e −0.2 z =
⎯⎯
→ e0.2 z = 100
P1
100
z = 5log100 = 23 m
Prob. 10.45
(a) u = ω / β
η=
⎯⎯
→
ω = uβ =
βc
4.5
=
2 × 3 × 108
4.5
= 2.828 × 108 rad/s
120π
= 177.7Ω
4.5
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Principles of Electromagnetics, 6e
303
H = ak ×
E
η
=
az
η
×
(b) P = E × H =
Pave =
(c)
4.5
ρ2
40
ρ
9
ρ
2
sin(ωt − 2 z )a ρ =
0.225
ρ
sin(ωt − 2 z )aφ A/m
sin 2 (ωt − 2 z )a z W/m 2
a z , dS = ρ dφ dρ az
Pave =  Pave • dS = 4.5
3mm

2mm
dρ
ρ
2π
 dφ = 4.5ln(3/2)(2π ) = 11.46 W
0
Prob. 10.46
P=E×H =
Eo2
sin 2 θ sin 2 ω (t − r / c)ar
2
ηr
T
Pave =
Eo2
1
dt
=
sin 2 θ ar
P
2

T0
2η r
Prob. 10.47
β=
E=
ω
c
1
ε
⎯⎯
→
ω = β c = 40(3 × 108 ) = 12 × 109 rad/s
 ∇ × Hdt
∂
∂
∂
∂y
∂z
∇ × H = ∂x
0 10sin(ωt − 40 x) −20sin(ωt − 40 x)
= −800cos(ωt − 40 x)a y − 400cos(ωt − 40 x)a z
E=
1
800
sin(ωt − 40 x)a
∇ × Hdt = −
ε
ωε
800
y
400
ωε
sin(ωt − 40 x)a z
400
sin(ωt − 40 x)a z
10
10−9
9
12 × 10 ×
12 × 10 ×
36π
36π
= −7.539sin(ωt − 40 x)a y − 3.77sin(ωt − 40 x)a z kV/m
=−
9
−9
sin(ωt − 40 x)a y −
−
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Principles of Electromagnetics, 6e
304
P = E×H =
0 Ey
0 Hy
Ez
= ( E y H z − E z H y )a x
Hz
=  20(7.537)sin 2 (ωt − 40 x) + 37.7sin 2 (ωt − 40 x)  a x 103
1
[ 20(7.537) + 37.7] a x 103 = 94. 23 a x kW/m 2
2
Pave =
Prob. 10.48
P=
Eo2
2ηo
⎯⎯
→ Eo2 = 2ηo P = 2(120π )10 × 10−3 = 7.539
Eo = 2.746 V/m
Prob. 10.49
Let T = ωt − β z.
∂
∂B
−
= ∇ × E = ∂x
∂t
cos T
−μ
∂
∂y
sin T
∂
∂z
0
∂H
= β cos Ta x + β sin Ta y
∂t
H =−
β
β
β
 cos Ta x + sin Ta y dt = −
sin Ta x +
cos Ta y

μ
μω
μω
cosT
P = E×H =
=
sinT
β
sin T
−
μω
0
β
(cos 2 T + sin 2 T )a z
=
β
cos T 0 μω
μω
β
ε
az =
a
μω
μ z
which is constant everywhere.
Prob. 10.50
E2
P = o
2ηo
P= PS =
Eo2 S (2.4 × 103 ) 2 × 450 × 10−4
=
= 343.8 W
2ηo
2 × 377
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Sadiku & Kulkarni
Principles of Electromagnetics, 6e
305
Prob. 10.51
Vo I o
sin 2 (ωt − β z )a z
2πρ ln(b / a)
P = E×H =
2
T
Pave =
(a)
T
1
1
1
Vo I o
Vo I o
P dt =
sin 2 (ωt − β z )dta z =
az
2
2


2πρ ln(b / a ) T 0
2πρ ln(b / a) 2
T0
=
Vo I o
az
4πρ ln(b / a)
2
(b)
Pave =  Pave dS ,
dS = ρ d ρ dφ a z
S
2π
b
Vo I o
1
Vo I o
=
ρ d ρ dφ =
(2π ) ln(b / a )
2


4π ln(b / a) φ = 0 ρ = a ρ
4π ln(b / a)
1
= Vo I o
2
Prob. 10.52
E 2
(a) Pi , ave = io ,
2η1
R=
Pr , ave
Pi , ave
Pr , ave =
Ero 2
,
2η1
Pt , ave =
η −η 
E 2
= ro2 = Γ 2 =  2 1 
Eio
 η2 + η1 
Eto 2
2η2
2
2
 μo
μo 
−


ε2
ε 1   μ oε 1 − μ oε 2

=
R=
 μ
μo   μoε1 + μoε 2
o
+


ε1 
 ε2
Since n1 = c μ1ε1 = c μoε1 ,
n −n 
R= 1 2 
 n1 + n2 
T=
Pt , ave
Pi , ave
=




2
n2 = c μoε 2 ,
2
η1 Eto 2 η1 2 η1
4n1n2
= τ = (1 + Γ) 2 =
2
(n1 + n2 ) 2
η2 Eio η2
η2
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Principles of Electromagnetics, 6e
306
(b) If Pr , ave = Pt , ave ⎯⎯
→ RPi , ave = TPi.ave ⎯⎯
→R = T
i.e. (n1 − n2 ) 2 = 4n1n2
⎯⎯
→
n12 − 6n1n2 + n2 2 = 0
2
n 
n 
or  1  − 6  1  + 1 = 0, so
 n2 
 n2 
n1
0.1716
= 3 ± 8 = 5.828 or
n2
(Note that these values are mutual reciprocals, reflecting the inherent symmetry of the
problem.)
Prob. 10.53 (a) η1 = ηo ,
Γ=
μ
= ηo / 2
ε
η2 − η1 ηo / 2 − ηo
=
= −1 / 3,
3ηo / 2
η2 + η1
s=
(b)
η2 =
τ=
2η 2
ηo
=
=2/3
η2 + η1 3ηo / 2
1+ | Γ | 1 + 1 / 3
=
=2
1− | Γ | 1 − 1 / 3
1
Eor = ΓEoi = − × (30) = −10
3
Er = −10cos(ωt + z )a x V/m
Let H r = H or cos(ωt + z )a H
a E × a H = ak ⎯⎯
→ −a x × a H = −a z ⎯⎯
→ aH = a y
Hr =
10
cos(ωt + z )a y = 26.53cos(ωt + z )a y mA/m
120π
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Principles of Electromagnetics, 6e
307
Prob. 10.54
ω 2π × 108 2π
β1 = =
=
c
3 × 108
3
η1 = ηo , η 2 = 0
Γ=
η2 − η1 0 − ηo
=
= −1,
η 2 + η1 0 + ηo
τ =1+ Γ = 0
Et = 0
E r = −50sin(2π × 108 t + β1 x)a z V/m
Prob. 10.55 (a)
β =1=
(b) η1 = ηo ,
ω
c
Medium 1 is free space. Given that β = 1 ,
⎯⎯
→ ω = c = 3 × 108 rad/s
η 2 = ηo
μr
3
= ηo
= ηo / 2
εr
12
Γ=
η2 − η1
= −1 / 3,
η2 + η1
s=
1+ | Γ | 1 + 1 / 3
=
=2
1− | Γ | 1 − 1 / 3
τ =1+ Γ = 2 / 3
(c) Let H r = H or cos(ωt + z )a H , where
1
Er = − (30) cos(ωt + z )a y = −10cos(ωt + z )a y ,
3
H or =
10
ηo
=
10
120π
a E × a H = ak ⎯⎯
→ −a y × a H = −a z ⎯⎯
→ a H = −a x
Hr = −
(d)
10
cos(3 × 108 t + z )a x A/m= −26.53cos(3 × 108 t + z )a x mA/m
120π
Eot2
2
Pt =
az ,
Eot = τ Eoi = (30) = 20, η 2 = 60π
2η 2
3
202
Pt =
(a z ) = 1.061a z W/m 2
120π
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Principles of Electromagnetics, 6e
308
Prob. 10.56 η1 =
Γ=
η2 − η1
= 1 / 3,
η2 + η1
μ1
= ηo / 2,
ε1
τ =1+ Γ = 4 / 3
Eor = ΓEio = (1 / 3)(5) = 5 / 3,
β=
ω
c
μrε r =
η 2 = ηo
Eot = τ Eio = 20 / 3
108
4 =2/3
3 × 108
5
Er = cos(108 t − 2 y / 3)a z
3
(a)
E1 = Ei + Er = 5cos(108 t +
(b)
Pave1 =
(c) Pave2 =
2
5
2
y )a z + cos(108 t − y )a z V/m
3
3
3
Eio 2
E 2
25
1
(−a y ) + ro (+ a y ) =
(1 − )(−a y ) = −0.0589a y W/m 2
2η1
2η1
2(60π )
9
Eto 2
400
( −a y ) =
(−a y ) = −0.0589a y W/m 2
2η2
9(2)(120π )
Prob. 10.57
η1 = ηo
(a)
Ei = Eio sin(ωt − 5 x)a E
Eio = H ioηo = 120π × 4 = 480π
a E × a H = ak ⎯⎯
→ a E × a y = a x ⎯⎯
→ a E = −a z
Ei = −480π sin(ωt − 5 x)a z
η2 =
Γ=
μo 120π
=
= 60π
4ε o
4
η2 − η1 60π − 120π
=
= −1 / 3,
η2 + η1 60π + 120π
τ =1+ Γ = 2 / 3
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Principles of Electromagnetics, 6e
309
Ero = ΓEio = (−1 / 3)(−480π ) = 160π
Er = 160π sin(ωt + 5 x)a z
E1 = Ei + Er = −1.508sin(ωt − 5 x)a z + 0.503sin(ωt + 5 x)a z kV/m
(b) Eto = τ Eio = (2 / 3)(480π ) = 320π
Eto 2
(320π ) 2
ax =
a x = 2.68a x kW/m 2
2η 2
2(60π )
1+ | Γ | 1 + 1 / 3
(c) s =
=
=2
1− | Γ | 1 − 1 / 3
P=
Prob. 10.58
(a) In air, β1 = 1, λ1 = 2π / β1 = 2π = 6.283 m
ω = β1c = 3 × 108 rad/s
In the dielectric medium, ω is the same .
ω = 3 × 108 rad/s
β2 =
ω
c
2π
= 3.6276 m
β2
3
E
10
= 0.0265
(b) H o = o =
ηo 120π
a H = ak × a E = a z × a y = −a x
λ2 =
2π
ε r 2 = β1 ε r 2 = 3
=
H i = −26.5cos(ωt − z )a x mA/m
(c) η1 = ηo ,
Γ=
η 2 = ηo / 3
η2 − η1 (1 / 3) − 1
=
= −0.268,
η2 + η1 (1 / 3) + 1
τ = 1 + Γ = 0.732
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Principles of Electromagnetics, 6e
310
(d) Eto = τ Eio = 7.32,
Ero = ΓEio = −2.68
E1 = Ei + Er = 10cos(ωt − z )a y − 2.68cos(ωt + z )a y V/m
E2 = Et = 7.32cos(ωt − z )a y V/m
Pave1 =
1
1
(a z )[ Eio 2 − Ero 2 ] =
(a z )(102 − 2.682 ) = 0.1231a z W/m 2
2η1
2(120π )
Eto 2
3
=
(a z ) =
(7.32) 2 (a z ) = 0.1231a z W/m 2
2η2
2 × 120π
Pave2
Prob. 10.59
η1 = ηo = 120π
For seawater (lossy medium),
η2 =
Γ=
jωμo
=
σ + jωε
j 2π × 108 × 4
10−9
4 + j 2π × 10 × 81 ×
36π
= 10.44 + j 9.333
8
η2 − η1
= 0.9461∠177.16
η2 + η1
| Γ |2 = 0.8952, 1− | Γ |= 0.1084
Pr
Pt
= 89.51%,
= 10.84%,
Pi
Pi
η − η 7.924∠43.975 − 377
Γ= 2 1 =
= 0.9702∠178.2o
η 2 + η1 7.924∠43.975 + 377
The fraction of the incident power reflected is
Pr
=| Γ |2 = 0.97022 = 0.9413
Pi
The transmitted fraction is
Pt
= 1− | Γ |2 = 1 − 0.97022 = 0.0587
Pi
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Sadiku & Kulkarni
Principles of Electromagnetics, 6e
311
Prob. 10.60
(a)
μ 120π
η1 = 1 =
= 188.5,
ε1
4
μ2 120π
=
= 210.75
ε2
3.2
η − η 210.75 − 188.5
2η 2
2 × 210.75
Γ= 2 1 =
= 0.0557, τ =
=
= 1.0557
η2 + η1 210.75 + 188.5
η 2 + η1 210.75 + 188.5
Ero = ΓEio = (0.0557)(12) = 0.6684
Eto = τ Eio = 1.0557(12) = 12.668
β1 = ω μ1ε1 =
ω
c
η2 =
⎯⎯
→ ω=
4
β1 c
2
=
40π (3 × 108 )
= 6π × 109 rad/s
2
(b)
6π × 109 3.2
= 112.4
3 × 108
c
Er = Ero cos(ωt + 40π x)a z = 0.6684cos(6π × 109 t + 40π x)a z V/m
β 2 = ω μ2ε 2 =
ω
3.2 =
Et = Eto cos(ωt − β 2 x)a z = 12.668cos(6π × 109 t − 112.4 x)a z V/m
Prob. 10.61 (a) ω = β c = 3 × 3 × 108 = 9 × 108 rad/s
(b) λ = 2π / β = 2π / 3 = 2.094 m
(c)
σ
4
=
= 2π = 6.288
8
ωε 9 × 10 × 80 × 10−9 / 36π
tan 2θη =
σ
= 6.288
ωε
μ2 / ε 2
| η2 |=
4
σ 
1+  2 
 ωε 2 
2
=
⎯⎯
→
377 / 80
4
1 + 4π 2
θη = 40.47 o
= 16.71
η 2 = 16.71∠40.47o Ω
(d)
Γ=
η 2 − η1 16.71∠40.47o − 377
=
= 0.935∠179.7 o
η2 + η1 16.71∠40.47o + 377
Eor = ΓEoi = 9.35∠179.7 o
Er = 9.35sin(ωt − 3 z + 179.7)a x V/m
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Principles of Electromagnetics, 6e
312
α2 =

c
2
β2 =
τ=
2

 σ2 
9 × 108 80 

1+ 
1 + 4π 2 − 1 = 43.94 Np/m
 −1 =
8



 3 × 10
2
 ωε 2 


ω μr 2ε r 2 
9 × 108 80 
1 + 4π 2 + 1 = 51.48 rad/m
8


3 × 10
2
2η 2
2 × 16.71∠40.47o
=
= 0.0857∠38.89o
o
η 2 + η1 16.71∠40.47 + 377
Eot = τ Eo = 0.857∠38.89o
Et = 0.857e 43.94 z sin(9 × 108 t + 51.48 z + 38.89o )a x V/m
Prob. 10.62
Induced Currents on the surface
σ = 0 Standing waves of H
σ=∞
Zero fields
z
Curve 0 is at t = 0; curve 1 is at t = T/8; curve 2 is at t = T/4; curve 3 is at t = 3T/8, etc.
Prob. 10.63 Since μo = μ1 = μ2 ,
sin θt1 = sin θi
sin θt 2 = sin θt1
ε o sin 45o
=
= 0.3333
ε1
4.5
ε1 1 4.5
=
= 0.4714
ε 2 3 2.25
⎯⎯
→
⎯⎯
→
θt1 = 19.47 o
θt 2 = 28.13o
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Principles of Electromagnetics, 6e
313
Prob. 10.64
jk y
− jk y
20(e jkx x − e − jk x x ) (e y + e y )
Es =
az
j2
2
j (k x+ k y)
j (k x−k y)
− j (k x−k y)
− j (k x+ k y)
= − j 5 e x y + e x y − e x y − e x y  a z
which consists of four plane waves.
∇ × E s = − jωμo H s
Hs = −
j 20
⎯⎯
→
Hs =
j
ωμo
∇ × Es =
j  ∂ Ez
∂ Ez 
ax −
ay 

ωμo  ∂ y
∂x 
 k y sin(k x x)sin(k y y )a x + k x cos( k x x) cos( k y y )a y 
ωμo 
Prob. 10.65
η1 = ηo = 377Ω
For η 2 ,
σ2
=
ωε 2
4
10−9
2π × 1.2 × 10 × 50 ×
36π
= 1.2
9
tan 2θη2 =
σ2
= 1.2
ωε 2
μ /ε
| η 2 |=
4
σ 
1+  2 
 ωε 2 
2
=
⎯⎯
→ θη2 = 25.1o
120π 1 / 50
4
1 + 1.22
= 42.658
η 2 = 42.658∠25.1o
η − η 42.658∠25.1o − 377
Γ= 2 1 =
= 0.8146∠174.4o
η 2 + η1 42.658∠25.1o + 377
Prob. 10.66
(a)
Pt = (1− | Γ |2 ) Pi
s=
1+ | Γ |
1− | Γ |
⎯⎯
→ | Γ |=
s −1
s +1
2
Pt
4s
 s −1
=1− 
 =
( s + 1) 2
Pi
 s +1
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(b) Pi = Pr + Pt
⎯⎯
→
Pr
P  s −1
=1− t = 

Pi
Pi  s + 1 
2
Prob. 10.67
If A is a uniform vector and Φ( r ) is a scalar,
∇ × (ΦA) = ∇Φ × A + Φ (∇ × A) = ∇Φ × A
since ∇ × A = 0.
∇× E = (
∂
∂
∂
j (k x+ k
ax +
ay +
a z ) × Eo e
∂x
∂y
∂z
x
y y + k z z −ω t )
= j ( k x a x + k y a y + k z a z )e j ( k • r − ω t ) × E o
= jk × E o e j ( k • r − ω t ) = j k × E
Also,
−
∂B
= jωμ H .
∂t
Hence ∇ × E = −
∂B
becomes k × E = ωμ H
∂t
ak × a E = a H
From this,
Prob. 10.68
k =| k |= 1242 + 1242 + 2632 = 316.1
2π
λ=
= 19.88 mm
k
2π f
kc 316.1 × 3 × 108
k = ω με =
⎯⎯
→ f =
=
= 15.093 GHz
c
2π
2π
124
k • a x = k cos θ x
⎯⎯
→ cos θ x =
⎯⎯
→ θ x = 66.9o = θ y
316.1
θ z = cos −1
263
= 33.69o
361.1
Thus,
θ x = θ y =66.9o , θ z = 33.69o
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Prob. 10.69
k = −3.4a x + 4.2a y
kE = 0
⎯⎯
→ 0 = −3.4 Eo + 4.2
4.2
= 1.235
3.4
Eo =
k =| k |= β = ( −3.4) 2 + (4.2) 2 = 5.403
λ=
2π
β
=
2π
= 1.162
5.403
3 × 108
f = =
= 258 MHz
λ 1.162
c
Hs =
=
1
k × Es =
μω
1
k × Es
μ kc
0
−3.4 4.2
1
A
8
1 3 + j4 o
4π × 10 × 5.403 × 3 × 10 Eo
−7
Ao = e − j 3.4 x + 4.2 y
where
H s = 4.91Ao × 10 −4  4.2(3 + j 4)a x + 3.4(3 + j 4)a y + ( −3.4 − 4.2 Eo )a z 
= 0.491 (12.6 + j16.8)a x + (10.2 + j13.6)a y − 8.59a z  e − j 3.4 x + 4.2 y mA/m
Prob.10.70
∇•E = (
∂
∂
∂
j (k x + k
ax +
ay +
a z ) • Eo e
∂x
∂y
∂z
x
= jk • E o e j ( k • r − ω t ) = j k • E = 0
y y + k z z −ωt )
⎯⎯
→
= j ( k x a x + k y a y + k z a z )e j ( k • r −ω t ) • E o
k•E =0
Similarly,
∇ • H = jk • H = 0
⎯⎯
→
k•H =0
It has been shown in the previous problem that
∇× E = −
∂B
∂t
k × E = ωμ H
Similarly,
∇× H =
∂D
∂t
kxH = −εω E
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From k × E = ωμ H ,
ak × a E = a H
From k × H = −εω E ,
a k × a H = −a E
Prob. 10.71
ηo
η
,η 2 = o
ε r1
εr2
μo = μ1 = μ2 , η1 =
If
1
Γ\ \ =
εr2
1
εr2
cosθt −
cosθt +
and
1
ε r1
1
ε r1
ε r1 sin θi = ε r 2 sin θt
cos θi
cosθi
⎯⎯
→
ε r 2 sin θi
=
ε r1 sin θt
sin θi
cosθi
sin θt
sin θt cosθt − sin θi cosθi
Γ\ \ =
=
sin θi
cosθt +
cos θi sin θt cosθt + sin θi cos θi
sin θt
sin 2θt − sin 2θi sin(θt − θi )cos(θt + θi ) tan(θt − θi )
=
=
=
sin 2θt + sin 2θi cos(θt − θi )sin(θt + θi ) tan(θt + θi )
Similarly,
cos θt −
2
τ \\ =
εr2
1
εr2
cosθi
1
cosθt +
ε r1
cosθi
=
2cosθi
sin θi
cosθt +
cosθi
sin θt
2cosθi sin θt
sin θt cos θt (sin θi + cos 2 θi ) + sin θi cos θi (sin 2 θt + cos 2 θt )
2cosθi sin θt
=
(sin θi cos θt + sin θt cos θi )(cosθi cosθt + sin θi sin θt )
=
=
2
2cosθi sin θt
sin(θi + θt ) cos(θi − θt )
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1
Γ⊥ =
2
τ⊥ =
1
sin θi
cosθt
εr2
ε r1
sin θt
sin(θt − θi )
=
=
1
1
sin θi
cosθi +
cosθt cosθi +
cosθt sin(θt + θi )
θ
sin
εr2
ε r1
t
cosθi −
εr2
1
εr2
cosθt
cosθi −
cos θi
cos θi +
1
ε r1
cosθt
=
2cosθi
2cos θi sin θi
=
sin θi
sin(θt + θi )
cosθi +
cosθt
sin θt
Prob. 10.72
(a) n1 = 1, n2 = c μ2ε 2 = c 6.4ε o × μo = 6.4 = 2.5298
sin θt =
n1
1
sin θi =
sin12o = 0.082185
2.5298
n2
⎯⎯
→ θt = 4.714o
1
= 47.43π
6.4
Ero
η cosθt − η1 cosθi 47.43π cos 4.714o − 120π cos12o
=Γ= 2
=
Eio
η2 cosθt + η1 cos θi 47.43π cos 4.714o + 120π cos12o
η1 = 120π ,
η 2 = 120π
47.27 − 117.38
= −0.4258
47.27 + 117.38
Eto
2η2 cos θi
2 x 47.43cos12o 92.787
=τ =
=
=
= 0.5635
η2 cosθt + η1 cosθi 47.27 + 117.33 164.65
Eio
=
Prob. 10.73
(a) ki = 4a y + 3az
ki • an = ki cosθi
⎯⎯
→
cosθi = 4 / 5
⎯⎯
→
θi = 36.87o
(b)
Pave =
E2
1
( 82 + 62 ) 2 (4a y + 3a z )
Re( E s × H s* ) = o ak =
= 106.1a y + 79.58a z mW/m 2
2
2η
2 × 120π
5
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(c) θ r = θi = 36.87 o . Let
Er = ( Ery a x + Erz a z )sin(ωt − kr • r )
z
kt
Er
kr
Et
θr
ki
θt
θi
Ei
From the figure, kr = krz a z − kry a y . But
krz = kr sin θ r = 5(3 / 5) = 3,
Hence,
k r = ki = 5
kry = kr cos θ r = 5(4 / 5) = 4,
kr = −4a y + 3a z
sin θt =
c μ1ε1
3/5
n1
sin θi =
sin θi =
= 0.3
n2
4
c μ2ε 2
θt = 17.46, cosθt = 0.9539,
η1 =ηo = 120π ,η2 = ηo / 2 = 60π
ηo
(0.9539) − ηo (0.8)
Ero η 2 cos θt − η1 cosθi
2
Γ/ / =
=
=
= −0.253
Eio η2 cosθt + η1 cosθi ηo (0.9539) + η (0.8)
o
2
Ero = Γ / / Eio = −0.253(10) = −2.53
But
3
4
( Ery a y + Erz a z ) = Ero (sin θ r a y + cosθ r a z ) = −2.53( a y + a z )
5
5
Er = −(1.518a y + 2.024a z )sin(ωt + 4 y − 3z ) V/m
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Similarly, let
Et = ( Ety a y + Etz a z )sin(ωt − kt • r )
kt = β 2 = ω μ2ε 2 = ω 4μoε o
But
ki = β1 = ω μoε o
kt
=2
ki
⎯⎯
→
kt = 2ki = 10
kty = kt cos θt = 9.539,
ktz = kt sin θt = 3,
kt = 9.539a y + 3az
Note that kiz = krz = ktz = 3
τ \\ =
Eto
2η2 cosθi
ηo (0.8)
=
=
= 0.6265
Eio η2 cos θt + η1 cos θi ηo (0.9539) + η (0.8)
o
2
Eto = τ \ \ Eio = 6.265
But
( Ety a y + Etz a z ) = Eto (sin θt a y − cos θt a z ) = 6.256(0.3a y − 0.9539a z )
Hence,
Et = (1.879a y − 5.968a z )sin(ωt − 9.539 y − 3 z ) V/m
Prob. 10.74
c
(a) n = = μrε r = 2.1 × 1 = 1.45
u
(b) n = μrε r = 1 × 81 = 9
(c) n = ε r = 2.7 = 1.643
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Prob. 10.75
(a) From air to seawater,
ε r1 = 1,
ε r 2 = 81
tan θ B =
ε2
81
=
=9
ε1
1
⎯⎯
→ θ B = 83.66o
(b) From seawater to air,
ε r1 = 81,
εr2 = 1
ε2
1 1
=
=
ε1
81 9
tan θ B =
Prob. 10.76
(a)
k
1
tan θi = ix =
kiz
8
sin θt = sin θi
(b) β1 =
(c)
ω
c
⎯⎯
→
ε r1 1
= (3) = 1
εr2 3
ε r1 =
λ = 2π / β ,
⎯⎯
→
θt = 90o
⎯⎯
→
k = 3.333
λ1 = 2π / β1 = 2π / 10 = 0.6283 m
λ2 = 2π / β 2 = 2π × 3 / 10 = 1.885 m
(a x + 8a z )
3
9
= (23.6954a x − 8.3776a z ) cos(10 t − kx − k 8 z ) V/m
Ei = η1 H x × ak = 40π (0.2)cos(ωt − k • r )a y ×
(e) τ / / =
2cosθi sin θt
2cos19.47o sin 90o
=
=6
sin(θi + θt ) cos(θt − θi ) sin19.47 o cos19.47 o
Γ/ / = −
Let
θi = θ r = 19.47o
109
× 3 = 10 = k 1 + 8 = 3k
3 × 108
β 2 = ω / c = 10 / 3,
(d)
⎯⎯
→ θ B = 6.34o
cot19.47o
= −1
cot19.47o
Et = − Eio (cosθt a x − sin θt a z ) cos(109 t − β 2 x sin θt − β 2 z cosθt )
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where
Et = − Eio (cosθi a x − sin θi a z ) cos(109 t − β1 x sin θi − β1 z cosθi )
sin θ t = 1,
cos θ t = 0 ,
β 2 sin θ t = 10 / 3
Eto sin θt = τ \ \ Eio = 6(24π )(3)(1) = 1357.2
Hence,
Et = 1357 cos(109 t − 3.333x)a z V/m
Since Γ = −1,
θ r =θi
Er = (213.3a x + 75.4a z )cos(109 t − kx + k 8 z ) V/m
(f) tan θ B / / =
ε2
εo
=
=1/ 3
ε1
9ε o
⎯⎯
→
θ B / / = 18.43o
Prob.10.77
Microwave is used:
(1) For surveying land with a piece of equipment called the tellurometer. This radar
system can precisely measure the distance between two points.
(2) For guidance. The guidance of missiles, the launching and homing guidance of
space vehicles, and the control of ships are performed with the aid of microwaves.
(3) In semiconductor devices. A large number of new microwave semiconductor
devices have been developed for the purpose of microwave oscillator,
amplification, mixing/detection, frequency multiplication, and switching. Without
such achievement, the majority of today’s microwave systems could not exist.
Prob.10.78
(a) In terms of the S-parameters, the T-parameters are given by
T11 = 1/S21, T12 = -S22/S21, T21 = S11/S21, T22 = S12 - S11 S22/S21
(b) T11 = 1/0.4 = 2.5, T12 = -0.2/0.4,
T21 = 0.2/0.4, T22 = 0.4 - 0.2 x 0.2/0.4 = 0.3
Hence,
 2.5 −0.5
T= 

 0.5 0.3 
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Prob. 10.79
Since ZL = Zo , Γ L = 0.
Γ i = S11 = 0.33 – j0.15
Γ g = (Zg - Zo)/ (Zg + Zo) = (2 –1)/(2 + 1) = 1/3
Γ o = S22 + S12S21 Γ g /(1 - S11 Γ g )
= 0.44 – j0.62 + 0.56x0.56 x(1/3)/[1 – (0.11 – j0.05)]
= 0.5571 - j0.6266
Prob. 10.80 The microwave wavelengths are of the same magnitude as the circuit
components. The wavelength in air at a microwave frequency of 300 GHz, for example,
is 1 mm. The physical dimension of the lumped element must be in this range to avoid
interference. Also, the leads connecting the lumped element probably have much more
inductance and capacitance than is needed.
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CHAPTER 11
P.E. 11.1
Since Zo is real and α ≠ 0, this is a distortionless line.
Zo =
or
R
G
(1)
L C
=
R G
(2)
α = RG
(3)
β = ωL
G ωL
=
R Zo
(4)
(1) × (3) → R = α Z o = 0.04 × 80 = 3.2Ω / m ,
(3) ÷ (1) → G =
α
=
0.04
= 5 × 10−4 S / m
80
Zo
β Zo
1.5 × 80
L=
=
= 38.2 nH/m
2π × 5 × 108
ω
LG 12
0.04
1
C=
= × 10−8 ×
×
= 5.97 pF/m
80 0.04 × 80
R
π
P.E. 11.2
(a) Z o =
R + jω L
0.03 + j 2π × 0.1 × 10−3
=
G + jωC
0 + j 2π × 0.02 × 10−6
= 70.73 − j1.688 = 70.75∠ − 1.367o Ω
(b) γ =
( R + jω L )( G + jωC ) = ( 0.03 +
j 0.2π ) ( j 0.4 × 10−4 π )
= 2.121 × 10−4 + j8.888 × 10−3 / m
ω
2π × 103
= 7.069 × 105 m/s
(c) u = =
−3
β 8.888 × 10
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P.E. 11.3
(a) Z o = Z l → Z in = Z o = 30 + j 60Ω
(b) Vin = Vo =
V
Z in
Vg = g = 7.5∠0o Vrms
2
Z in + Z o
Vg
Vg
15∠0o
=
=
I in = I o =
Z g + Z in 2 Z o 2 ( 30 + j 60 )
= 0.2236∠ − 63.43o A
assuming that Zg = 0.
(c) Since Zo = Zr, Γ = 0 → Vo − = 0,Vo + = Vo
The load voltage is VL = Vs ( z = l ) = Vo + e −γ l
e
Vo +
7.5∠0o
=
=
1.5∠48o
o
VL 5∠ − 48
−γ l
eα l e j β l = 1.5∠48o
1
1
eα l = 1.5 → α = ln (1.5 ) = ln (1.5 ) = 0.0101
40
l
o
o
1 48
e j β l = e j 48 → β =
π rad = 0.02094
l 180o
γ = 0.0101 + j 0.02094 /m
assuming that Zg =0.
P.E. 11.4
(a) Using the Smith chart, locate S at s = 1.6. Draw a circle of radius OS. Locate P
where θ Γ = 300o . At P,
Γ =
OP 2.1cm
=
= 0.228
OQ 9.2cm
O
Γ = 0.228∠300o
Also at P, zL =1.15-j0.48,
S
R 
Q
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ZL =Zo zL =70(1.15-j0.48) = 80.5-j33.6Ω
 =0.6λ → 0.6 × 720o =432o =360o +73o
From P, move 432o to R. At R, zin = 0.68 − j 025
Z in = Z o Z in = 70(0.68 − j 0.25) = 47.6 − j17.5Ω
(b) The minimum voltage (the only one) occurs at θ Γ = 180o ; its distance from the load
180 − 60
λ
λ = = 0.1667λ
is
720
6
Values obtained using formulas are as follows:
s −1
Γ=
∠300o = 0.2308∠300o
s +1
Z L = 80.5755 − j 34.018 Ω
Z in = 48.655 − j17.63 Ω
These are pretty close.
P.E. 11.5
Z − Z o 60 + j 60 − 60
j
=
=
= 0.4472∠63.43o
(a) Γ = L
Z L + Z o 60 + j 60 + 60 2 + j
s=
1+ Γ
1− Γ
=
1 + 0.4472
= 2.618
1 − 0.4472
Let x = tan ( β l ) = tan
2π l
λ
 Z + jZ o tan ( β l ) 
Z in = Z o  L

 Z o + jZ L tan ( β l ) 
 60 + j 60 + j 60 x 
120 − j 60 = 60 

 60 + j ( 60 + j 60 ) x 
Or 2 − j =
1 + j (1 + x )
1 − x + jx
→ 1 − x + j ( 2x − 2) = 0
Or x = 1 = tan ( β l )
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π
4
i.e
+ nπ =
l=
λ
8
2π l
λ
(1 + 4n ) , n = 0,1, 2,3...
Z L 60 + j 60
=1+ j
Zo
60
Locate the load point P on the Smith chart.
(b) z L =
Γ =
OP 4.1cm
=
= 0.4457,θ Γ = 62o
OQ 9.2cm
Γ = 0.4457∠62o
Locate the point S on the Smith chart. At S, r = s = 2.6
Z
120 + j 60
Z in = in =
= 2 − j , which is located at R on the chart. The angle between
Zo
60
90λ λ
Q
= .
OP and OR is 64o-(-25o) = 90o which is equivalent to
64º
720 8
P
Hence l =
( Zin )max
λ
λ
λ
(1 + 4n ) , n = 0,1, 2........
8
2 8
= sZ o = 2.618 ( 60 ) = 157.08Ω
+n
=
S
O
( Z in ) min = Z o / s = 60 / 2.618 = 22.92 Ω
R
-26º
62o
l=
λ = 0.0851λ
720o
P.E. 11.6
ZL 57.6 º
Vmax
S = 1.8
λ
2
= 37.5 − 25 = 12.5cm or λ = 25cm
l = 37.5 − 35.5 = 2cm =
l = 0.08λ → 57.6o
2λ
25
z L = 1.184 + j 0.622
Z L = Z o z L = 50 (1.184 − j 0.622 )
= 59.22 + j31.11Ω
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P.E. 11.7
See the Smith chart
100 − j80
= 1.33 − j1.067
75
132o − 65
lA =
λ = 0.093λ
72
132o + 64o
lB =
= 0.272λ
720o
91
dA =
λ = 0.126λ
720
91º
zL =
d B = 0.5λ − d A
Ys = ±
lA
B´
1 + j0.97
Y = 1 + jb
A
yL
lB
zL
dB
A´ B
dA
-91º
= 0.374λ
1 - j0.97
j 0.95
= ± j12.67 mS
75
P.E. 11.8
1
Z − Zo
lim
(a) ΓG = , Γ L = Z L ⎯⎯
→0 L
= −1
3
Z L + Zo
lim
V∞ = zL ⎯⎯
→0
ZL
Vg = 0,
ZL + Zg
lim
I ∞ = zL ⎯⎯
→0
Vg
Zg + ZL
=
Vg
=
Zg
12
= 120mA
100
Thus the bounce diagrams for current and waves are as shown below.
ΓG =
ΓL = –1
1
3
4V
V=4
–4V
2t1
V = –1.33
– 43
V = 0.444
6t1
t1
V=0
3t1
4
3
4t1
V=0
4
9
V=0
5t1
− 94
V=0
− 274
ΓL = 1
ΓG= − 13
80mA
I = 80
2t1
I = 133.3
– 803
I = 115.6
6t1
(Voltage)
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I = 160
3t1
− 803
4t1
I=0
t1
80
80
9
80
9
− 80
27
(Current)
I = 106.27
5t1
I = 124.4
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V (l,t)
0V
t (μs)
V(0,t)
4V
4V
4
3
0
2
4
– 43
4
9
t (μs)
6
–4
I(l,t)
160mA
124.45
106.67
80
80
9
0
2
6
4
t (μs)
− 803
I(0,t)
133.33
115.5
80
80mA
80
9
0
2
4
6
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1
Z − Zo
lim
(b) ΓG = , Γ L = Z L ⎯⎯
→∞ L
=1
3
Z L + Zo
ZL
Vg = Vg = 12V ,
ZL + Zg
lim
V∞ = zL ⎯⎯
→∞
lim
I ∞ = zL ⎯⎯
→∞
Vg
ZL + Zg
=0
The bounce diagrams for current and voltage waves are as shown below.
ΓG =
ΓL = 1
1
3
4
V=4
t1
4
3
4t1
3t1
6t1
80
3
I = 8.89
6t1
(Voltage)
I=0
t1
I=0
3t1
− 803
4t1
V = 11.55
4
27
V = 11.7
I = 26.67
5t1
4
9
–80
2t1
V = 10.67
4
9
V = 11.11
I = 80
V=8
4
3
V = 9.33
80mA
V=0
4
2t1
ΓL =- 1
ΓG= − 13
80
9
I=0
5t1
− 809
(Current)
V(l,t)
10.67
11.55
12V
8V
4V
4
3
0
2
4
4
9
t (μs)
6
I (l,t)
0A
2
4
6
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9.333
V(0,t)
4V
4V
4
3
0
12V
11.11
2
4
9
8
6
4
t (μs)
I(0,t)
80
80mA
80
80
0
80
3
2
9
3
80
9
t (μs)
6
4
− 80 3
–80
P.E. 11.9
1
1
ΓG = − , Γ L = , t1 = 2μ s
2
7
(Vg )max = 10 = 100mA
( I o )max =
Z g + Z o 100
The bounce diagrams for maximum current are as shown below.
Γ=
Γ= −
1
2
100mA
– 100
7
2t1
t1
− 50 7
50
3t1
49
4t1
25
49
− 25 343
6t1
1
7
5t1
− 12.5 343
7t1
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I (0,t)mA
100
1.521
0
2
6
4
8
10
8
10
t (μs)
–21.43
I (l,t)mA
85.71
0
2
4
6
–6.122
P.E. 11.10
(a) For w / h = 0.8,
(b) Z o =
(c) λ =
ε eff
4.8 2.8  12 
1+
=
+
2
2  0.8 
−1
2
= 2.75
60
 8 0.8 
+
ln 
 = 36.18ln10.2 = 84.03Ω
2.75  0.8 4 
3 × 108
= 18.09 mm
1010 2.75
P.E. 11.11
π f μo
π × 20 × 109 × 4π × 10−7
Rs =
=
σc
5.8 × 107
= 3.69 × 10−2
R
8.686 × 3.69 × 10−2
α c = 8.685 s =
wZ o
2.5 × 10−3 × 50
= 2.564 dB/m
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P.E. 11.12
ZL
= 1 + j2
Zo
We locate z L on the Smith chart.
Z L = (1 + j 2) Z o
λ
⎯⎯
→ zL =
720o
= 180o
4
4
o
Sadiku & Kulkarni We move 180 toward the generator and locate point Q at which
z = 0.2 − j 0.4
⎯⎯
→
Z = zZ o = (0.2 − j 0.4) Z o
Prob.
11.33
P.E. 11.13
(a) Method 1:
366
At Y,
o
720

Z +
jZ o tan
λ
o β
Z in =→Z o  L = 180

4
 Z4o + jZ L tan β  
2π λ
= π,
tan π = 0
λ 2
Z
Z in = Z o L = Z L = 150 Ω
Zo
At X,
Z L ' = 150Ω
β =
65.274o
2π λ
= π / 2, tan π / 2 = ∞
λ 4
ZL ' 

YL
 jZ o + tan ZβL  Z 2 O(75) 2
o
Z in = lim Z o 
=
= 37.5Ω
=
tan β  →∞
 jZ '+ Z o  Z L ' 150
 L tan β  
A
β =
0.4093 λ
At A,
0.07225 λ
-52.02o
yin = 1 − j1.561
ystub = j1.5614
Position of the stub = 0.0723λ
Length of the stub = 0.4093λ
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Prob. 11.1
C=
πε l
−1
cosh (d / 2a )
≅
πε l
ln( d / a)
since (d/2a)2 = 11.11 >> 1.
C=
π×
10−9
× 16 × 10−3
36π
= 0.2342 pF
ln(2 / 0.3)
1
1
=
= 2.09 × 10−5 m << a
7
−7
7
π f μσ c
π × 10 × 4π × 10 × 5.8 × 10
δ=
16 × 10−3
Rac =
=
= 1.4 × 10−2 Ω
−3
−5
7
π aδσ c π × 0.3 × 10 × 2.09 × 10 × 5.8 × 10
l
Prob. 11.2
L=
μ
ln(b / a ),
2π
C=
2πε
ln(b / a )
2
 b
ln
L μ  a 
=
C 2π 2πε
L ln(b / a ) μ
Zo =
=
= 50
ε
C
2π
50 =
ln(b / a ) μo
ε oε r
2π
2
2
 b
 20 
 ln  η 2  ln  377 2
a
6 
ε r =  2  o2 = 
= 2.0874
4π 50
4π 2
502
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Prob. 11.3
Method 1:
Assume a charge per unit length Q on the surface of the inner conductor and –Q on the
surface of the outer conductor. Using Gauss’s law,
Q
, a<ρ <b
Eρ =
2περ
b
Q
ln(b / a )
2πε
V = −  E dl =
a
J =σE =
σQ
2περ
I =  J dS =
S
2π
σQ
σQ
(1) ρ dφ =
2περ
ε
=0

φ
σQ
ε
2πσ
I
=
=
Q
V
ln(b / a ) ln(b / a )
2πε
Method 2:
G=
Consider a section of unit length, Assume that a total current of I flows from inner
conductor to outer conductor. At any radius ρ between a and b,
I
J=
aρ , a < ρ < b
2πρ
J
I
E= =
aρ
σ 2πσρ
b
V = -  E dl =
a
G=
I
2πσ
ln(b / a )
2πσ
I
=
V ln(b / a)
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Prob. 11.4
1
1
=
= 7.744 × 10−6
6
−7
7
π f μcσ c
π × 80 × 10 × 4π × 10 × 5.28 × 10
δ=
1
1


+
−3
−3 

1 1 1
103 (1.25 + 0.3836)
0.8 × 10
2.6 × 10 

R=
+
=
=
= 0.6359 Ω /m
2πδσ c  a b  2π × 7.744 × 10−6 × 5.28 × 107
2569.09
μ b 4π × 10−7 2.6
ln =
ln
= 2.357 × 10−7 H/m
2π a
2π
0.8
L=
2πσ 2π × 10−5
G=
=
= 5.33 × 10−5 S/m
b
2.6
ln
ln
a
0.8
C=
2πε
=
b
ln
a
2π × 3.5 ∗
ln
2.6
0.8
10−9
36π = 1.65 × 10−10 F/m
Prob. 11.5
δ=
1
1
=
6
π f μσ c
π × 500 × 10 × 4π × 10−7 × 7 × 107
δ = 2.6902 × 10−6
R=
L=
2
=
wδσ c
μo d
=
w
2
= 0.0354Ω / m
0.3 × 2.6902 × 10−6 × 7 × 107
4π × 10−7 × 1.2 × 10−2
= 50.26 nH/m
0.3
εow
10−9
0.3
=
×
= 221 pF/m
C=
d
36π 1.2 × 10−2
Since σ = 0 for air,
G=
σw
d
=0
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Prob. 11.6
(a) Applying Kirchhoff’s voltage law to the loop yields
V ( z + Δz , t ) = V ( z , t ) − RΔzI1 − LΔz
But I1 = I ( z , t ) −
∂ I1
∂t
C ∂ V ( z, t ) G
Δz
− ΔzV ( z , t )
2
∂t
2
I ( z, t )
I1
Hence,
 ∂ I C ∂ 2V G ∂ V 
C ∂V G


− ΔzV  − LΔz  − Δz 2 − Δz
V ( z + Δz, t ) = V ( z, t ) − RΔz  I ( z, t ) − Δz
∂t 2
∂t
∂ t 
2
2


∂t 2
Dividing by Δz and taking limits as Δt → 0 give
Δz
lim
⎯⎯
→0
or −
V ( z + Δz, t ) − V ( z, t )
Δz

∂ I RC ∂ V RG
LC ∂ 2V LG ∂ V 
lim
=Δz ⎯⎯
→0 −RI − L +
Δz
+
ΔzV +
Δz 2 +
Δz 
∂t 2
∂t
∂t
∂t 
2
2
2

∂V
∂I
= RI + L
∂z
∂t
Similarly, applying Kirchhoff’s law to the node leads to
∂  V ( z, t ) + V ( z + Δz, t ) 
 V ( z , t ) + V ( z + Δz ) 
I ( z + Δz , t ) − I ( z , t ) = −GΔz 
 − C Δz 

2
∂t 
2



→ 0 , we get
Let Δz ⎯⎯
∂I
∂V
−
= GV + C
∂z
∂t
(b) Applying Kirchhoff’s voltage law,
V ( z, t ) = R
Δl
Δl ∂ I
I ( z, t ) + L
( z , t ) + V ( z + Δl / 2, t )
2
2 ∂t
or
V ( z + Δl / 2, t ) − V ( z, t )
∂I
−
= RI + L
∂t
Δl / 2
As Δl → 0,
−
∂V
∂I
= RI + L
∂z
∂t
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Here, we take Δl = Δz .
Applying Kirchhoff’s current law,
I ( z , t ) = I ( z + Δl , t ) + GΔlV ( z + Δl / 2, t ) + C Δl
or
−
I ( z + Δl , t ) − I ( z , t )
Δl
As Δl → 0, −
= GV ( z + Δl / 2, t ) + C
(
∂ V z + Δl 2 , t
∂t
)
∂ V ( z + Δl / 2, t )
∂t
∂ I ( z, t )
∂ V ( z, t )
= GV ( z , t ) + C
∂z
∂t
Prob. 11.7
(a)
γ = ( R + jω L)(G + jωC ) = jω LC (1 +
R
G
)(1 +
)
jω L
jωC
RG
R
G
+
+
2
ω LC jω L jωC
As R<< ωL and G<<ωC, dropping the ω2 term gives

R
G
R
G 
γ ≅ jω LC 1 +
+
≅ jω LC 1 +
+

jω L jωC
 2 jω L 2 jωC 
= jω LC 1 −
R C G L 
= 
+
 + jω LC
2 L 2 C
(b)
1+
R
1/ 2
R + jω L
L
L
R  
G 
jω L
Zo =
=
=
1+
1+




G + jωC
C 1+ G
C
jω L  
jωC 
jωC
−1/ 2
≅


L
R
G
L
R
G

+ ... 1 −
+ ...  =
+ j
+ ... 
1 +
1 − j
C  2 jω L
j 2ωC
C
2ω L
2ωC



≅
L
C

1 +

R 
 G
j
−

 2ωC 2ω L  
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Prob. 11.8
For a lossless line,
1
L
u=
, Zo =
C
LC
1
uZ o =
C
1
u=
Z oC
Prob. 11.9
R + jω L
G + jωC
Zo =
(1)
γ = α + j β = ( R + jω L)(G + jωC )
α = 0.04 dB/m =
(2)
0.04
Np/m = 0.00461 Np/m
8.686
Multiplying (1) and (2),
Z o (α + j β ) = R + jω L
⎯⎯
→ 50(0.00461 + j 2.5) = R + jω L
R = 50 × 0.00461 = 0.2305 Ω /m
L=
50 × 2.5
= 0.3316 μ H/m
2π × 60 × 106
Dividing (2) by (1),
α + jβ
= G + jωC
Zo
G=
C=
α
Zo
=
0.00461
= 92.2 μS/m
50
β
2.5
=
= 0.1326 nF/m
ω Z o 2π × 60 × 106 × 50
Prob. 11.10
L
μd d
d μ
=
.
=
c
w εw w ε
d
Z o = ηo = 78
w
Zo =
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Z o ' = ηo
d
= 75
w'
78 w'
= → w' = 1.04w
75 w
i.e. the width must be increased by 4%.
Prob. 11.11
(a) Z o =
R + jω L
6.8 + j 2π × 103 × 3.4 × 10−3
=
G + jωC
0.42 × 10−6 + j 2π × 103 × 8.4 × 10−9
= 103
6.8 + j 21.36
= 644.3 − j 97 Ω
0.42 + j 52.78
γ = ( R + jω L)(G + jωC ) = 10−3 (6.8 + j 21.36)(0.42 − j52.78)
= (5.415 + j 33.96) × 10−3 /mi
ω
2π × 103
=
= 1.85 × 105 mi/s
−3
β 33.96 × 10
2π
2π
(c) λ =
=
= 185.02 mi
β 33.96 × 10−3
(b)
u=
Prob. 11.12
Using eq. (11.42a),
Z in = − jZ o cot β 
c
3 × 108
=
= 0.75 m
f 400 × 106
2π
2π × 0.1
=
= 48o
β =
0.75
λ
Z in = − j (250)cot 48o = − j 225.1 Ω
Z o = 250,
λ=
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Prob. 11.13
Assume that the line is lossless.
L
Zo =
C
From Table 11.1,
L μ 1 b
ln 
= 
C ε  2π a 
μ b
ln ,
2π a
C=
2πε
b
ln
a
2
1 b
μ
ηo
L
b
ln ×
ln
=
=
ε 2π ε r a
C 2π a
Zo =
ln
L=
b
Z
75
= 2π ε r o = 2π 2.25
= 1.875
a
120π
ηo
b
= e1.875
a
⎯⎯
→ a = be −1.875 = 3e −1.875 mm = 0.46 mm
Prob. 11.14
(a) For a lossless line, R = 0 = G.
γ = jω LC
β = ω LC = ω μo co =
⎯⎯
→
ω
ω
1
u= =c=
β
LC
(b) For lossless line, R = 0 =G
L=
μ
πε
d
cosh −1 , C =
d
π
2a
cosh −1
2a
Zo =
=
120
εr
L
μ 1
d 120π
d
=
=
. cosh −1
cosh −1
π πε
C
2a π ε r
2a
cosh −1
d
2a
Yes, true for other lossless lines.
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Prob. 11.15
L=
μ
0.32
d
= 4 × 10−7 cosh −1
cosh −1
π
2a
0.12
L = 0.655 μ H/m
10−9
π×
× 3.5
πε
36π
=
C=
d
Cosh −1 2.667
Cosh −1
2a
C = 59.4 pF/m
Zo =
or
Zo =
L
0.655 × 10−6
=
= 105Ω
C
59.4 × 10−12
120
cosh −1 2.667 = 105Ω
3.5
Prob. 11.16
For a distortionless cable,
R G
=
⎯⎯
→
RC = LG
L C
(1)
L
= 60
(2)
C
1
4
ω

= =
(3)
u= =
−6
β
LC to 80 × 10
0.24
α  = 0.24dB =
Np = 0.0276
8.686
(4)
α = RG = 0.00069
From (2) and (3),
Zo =
1
60 × 4
=
C 80 × 10−6
From (2),
8 × 10−5
⎯⎯
→ C=
= 333.3 nF/m
240
L = (60) 2 C = 3600 × 333.3 × 10−9 = 1.20 mH/m
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From (1) and (4),
C
LG
0.000692
2
=
⎯⎯
→
=
G
G 0.000692
602
0.00069
= 11.51 μS/m
G=
60
From (4),
R=
0.000692
= 0.00069 × 60 = 0.0414 Ω /m
G
Prob. 11.17
R G
R
20 × 63 × 10−12
(a) = → G = C =
L C
L
0.3 × 10−6
G = 4.2 × 10−3 S/m
α = RG = 20 × 4.2 × 10−3 = 0.2898
β = ω LC = 2π × 120 × 106 0.3 × 10−6 × 63 × 10−12
= 3.278
γ = 0.2898 + j 3.278 /m
u=
ω 2π × 120 × 106
=
= 2.3 × 108 m/s
3.278
β
Zo =
L
0.3 × 10−6
=
= 69 Ω
C
63 × 10−12
(b) Let Vo be its original magnitude
Voe−α z = 0.2Vo → eα z = 5
z=
1
α
ln 5 = 5.554 m
(c) β l = 45o = π
4
→l =
π
4
=
= 0.3051 m
4 β 4 × 3.278
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Prob. 11.18
(a) α = 0.0025 Np/m, β = 2 rad/m,
ω 108
u= =
= 5 × 107 m/s
2
β
V
60 1
(b) Γ = o+ =
=
Vo
120 2
Z L − Zo
1 300 − Z o
→ =
→ Z o = 100Ω
Z L + Zo
2 300 + Z o
'
120 0.0025l '
60
I (l ' ) =
e
cos (108 + 2l ' ) − e−0.0025l cos (108 t − 2l ' )
Zo
Zo
But Γ =
= 1.2e0.0025l ' cos (108 + 2l ' ) − 0.6e −0.0025l cos (108 t − 2l ' ) A
'
Prob. 11.19
α = 10−3 , β = 0.01
γ = α + j β = 0.001 + j 0.01 = (1 + j10) × 10−3 /m
u=
ω 2π × 104
=
= 6.283 × 106 m/s
β
0.01
Prob. 11.20
L
0.6 × 10−6
=
= 85.54 Ω
C
82 × 10−12
RC
RC = LG
⎯⎯
→ G=
L
RC
R 10 × 10−3
=
=
= 1.169 × 10−4 Np/m
α = RG = R
L
Zo
85.54
Zo =
β = ω LC = 2π × 80 × 106 0.6 × 10−6 × 82 × 10−12
= 3.5258 rad/m
γ = 1.169 × 10−4 + j 3.5258 /m
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Prob. 11.21
R + jω L = 6.5 + j 2π × 2 × 106 × 3.4 × 10−6 = 6.5 + j 42.73
G + jωC = 8.4 × 10−3 + j 2π × 2 × 106 × 21.5 × 10−12 = ( 8.4 + j 0.27 ) × 10−3
Zo =
R + jω L
=
G + jωC
6.5 + j 42.73
(8.4 + j 0.27 ) × 10−3
Z o = 71.71∠39.75o = 55.12 + j 45.85Ω
γ=
( R + jω L )( G + jωC )
=
( 43.19∠81.34 )(8.4 × 10
o
= 0.45 + j0.4 /m
β
l
ω
t = , but u = ,
u
β
α
t=
βl
0.39 × 5.6
=
= 0.1783μ s
ω 2π × 2 × 106
Prob. 11.22
 Z + Z o tanh γ  
Z in = Z o  L

 Z 0 + Z L tanh γ  
γ  = α  + j β  = 1.4 × 0.5 + j 2.6 × 0.5 = 0.7 + j1.3
tanh γ  = 1.4716 + j 0.3984
 200 + (75 + j 60)(1.4716 + j 0.3984) 
Z in = (75 + j 60) 

 (75 + j 60) + 200(1.4716 + j 0.3984) 
= 57.44 + j 48.82 Ω
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−3
∠1.84o )
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Prob. 11.23
(a) For a lossy line,
 Z + Z o tanh γ  
Z in = Z o  L

 Z 0 + Z L tanh γ  
For a short-circuit, Z L = 0.
Z sc = Z in
tanh γ  =
ZL = 0
= Z o tanh γ 
Z sc 30 − j12
=
= 0.168 − j 0.276
Z o 80 + j 60
γ  = α  + j β  = tanh −1 (0.168 − j 0.276) = 0.1571 − j 0.2762
0.1571
= 0.0748 Np/m
2.1
0.2762
= 0.1316 rad/m
β=
2.1
α=
(b)
 (40 + j 30) + (80 + j 60)(0.168 − j 0.276) 
Z in = (80 + j 60) 

 (80 + j 60) + (40 + j30)(0.168 − j 0.276) 
= 61.46 + j 24.43 Ω
Prob. 11.24
V
ZLIL
2Z L I L
(a) TL = L+ =
=
1 (V + Z I ) Z L I L + Z o I L
Vo
o L
2 L
2Z L
=
Z L + Zo
1 + ΓL = 1 +
(b) (i) τ L =
Z L − Zo
2Z L
=
Z L + Zo Z L + Zo
2nZ o
2n
=
nZ o + Z o n + 1
lim
(ii) τ L =YL ⎯⎯
→0 =
lim
(iii) τ L = Z L ⎯⎯
→0 =
(iv) τ L =
2Z o
=1
2Z o
1+
2
Zo
=2
ZL
2Z L
=0
Z L + Zo
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Prob. 11.25
γ = ( R + jω L)(G + jωC ) = (3.5 + j 2π × 400 × 106 × 2 × 10−6 )(0 + j 2π × 400 × 106 × 120 × 10−12 )
= (3.5 + j 5026.55)( j 0.3016) = 0.0136 + j38.94
α = 0.0136 Np/m,
ω 2π × 400 × 106
=
= 6.452 × 107 m/s
β
38.94
u=
Zo =
β = 38.94 rad/m
R + jω L
3.5 + j 5026.55
=
= 129.1 − j 0.045 Ω
G + jωC
j 0.3016
Prob. 11.26
From eq. (11.33)
Z sc = Z in Z L = 0 = Z o tanh γ l
Z oc = Z in
Z L =∞
=
Zo
= Z o coth ( γ l )
tanh γ l
For lossless line, γ = j β , tan ( γ l ) = tanh ( j β l ) = j tan ( β l )
Z sc = jZ o tan ( β l ) , Z oc = − jZ o cot ( β l )
Prob. 11.27
1
( a ) β l = × 100 = 25 rad = 1432.4o = 352.4o

Z in = 60 

4
j 40 + j 60 tan 352.4o 
 = j 29.375Ω
60 − 40 tan 352.4o 
V ( z = 0) = Vo =
Z in
j 29.375(10∠0o
Vg =
Z in + Z g
j 29.375 + 50 − j 40
293.75∠90o
=
= 5.75∠102o
o
51.116∠ − 12
(b) Z in = Z L = j 40Ω.
Vo+ =
VL =
Vg
(e
jβ l
+ Γe − j β l )
Vg (1 + Γ)
(e
jβ l
+ Γe
− jβ l
)
(l is from the load)
= 12.62∠0o V
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1
( c ) β l ' = × 4 = 1rad = 57.3o
4
j 40 + j 60 tan 57.3o 
 = − j 3471.88Ω.
60 − 40 tan 57.3o 

Z in = 60 

Vg (e j + Γe− j )
= 22.74∠0o V
V = j 25
− j 25
( e + Γe )
(d) 3m from the source is the same as 97m from the load., i.e.
1
4
 j 40 + j 60 tan 309.42o 
Z in = 60 
= − j18.2Ω
o 
 60 − 40 tan 309.42 
V (e j 97 / 4 + Γe− j 97 / 4 )
= 6.607∠180o V
V = g j 25
− j 25
( e + Γe )
l ' = 100 − 3 = 97 m,
β l ' = × 97 = 24.25rad = 309.42o
Prob. 11.28
Z − Z o 120 − 50
(a) Γ = L
=
= 0.4118
Z L + Zo
170
For resistive load, s =
(b) Z in = Z o
βl =
ZL
= 2.4
Zo
Z L + jZ o tan ( β l )
Z o + jZ L tan ( β l )
2π λ
. = 60o
λ 6
120 + j 50 tan ( 60o ) 
 = 34.63∠ − 40.65o Ω
Z in = 50 
o
 50 + j120 tan ( 60 ) 
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= −0.4831 + j 0.5362
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Z in =
( 65 +
j 38 )( −0.2454 + j1.0548 )
−0.4831 + j 0.5362
Principles of Electromagnetics, 6e
348
= 113 + j 2.726Ω
Prob. 11.30
V1 = Vs ( z = 0) = Vo+ + Vo−
+V e
+
o
−
o
V2 = Vs ( z = l ) = V e
I1 = I s ( z = 0) =
(1)
+ −γ l
o
− γl
o
(2)
V
V
−
Zo Zo
I2 = −Is (z = l) = −
(3)
Vo+ −γ l Vo− γ l
e
e +
Zo
Zo
(4)
1
(1) + (3) → Vo+ = (V1 + Z o I1 )
2
1
(1) - (3) → Vo− = (V1 − Z o I1 )
2
+
Substituting Vo and Vo− in (2) gives
Sadiku & Kulkarni
1
1
V2 = (V1 + Z o I1 )e−γ l + (V1 − Z o I1 )eγ l
2
2
1
1
= (eγ l + e−γ l )V1 + Z o (e−γ l − eγ l ) I1
345
2
2
V2 = cosh γ lV1 − Z o sinh γ lI1
Principles of Electromagnetics, 6e
(5)
Substituting Vo+ and Vo− in (4),
I2 = −
=
I2 =
1
1
(V1 + Z o I1 )e −γ l +
(V1 − Z o I1 )eγ l
2Z o
2Z o
1 γl
1
(e − e −γ l )V1 + (eγ l + e−γ l ) I1
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2Z o
2
1
sinh γ lV1 − cosh γ lI1
Zo
(6)
From (5) and (6)
 cosh γ l
 V2  
 − I  =  − 1 sinh γ l
 2  Z
 o
But
 cosh γ l
 1
 − sinh γ l
 Z o
Thus
−1
− Z o sinh γ l 
 cosh γ l
 = 1
 sinh γ l
cosh γ l 

 Z o
 cosh γ l
V1  
 I  =  1 sinh γ l
 1 Z
 o
Prob. 11.31
− Z o sinh γ l 
 V1 
cosh γ l   I1 

Z o sinh γ l 

cosh γ l 

Z o sinh γ l 
  V2 
cosh γ l   − I 2 

(a)
za =
Z a 80
=
= 1.6
Z o 50
(b)
zb =
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Z b 60 + jCopyright
40
=
= 1.2 + j 0.8
Zo
50
 − sinh γ l
 Z o
cosh γ l 

 sinh γ l
 Z o
cosh γ l 

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Principles of Electromagnetics, 6e
 cosh γ l
V1  
 I  =  1 sinh γ l
 1 Z
 o
Prob. 11.31
Sadiku & Kulkarni
Z o sinh γ l 
  V2 
cosh γ l   − I 2 

349
(a)
za =
Z a 80
=
= 1.6
Z o 50
(b)
zb =
Z b 60 + j 40
=
= 1.2 + j 0.8
Zo
50
(c)
zc =
Z c 30 − j120
=
= 0.6 − j 2.4 346
Zo
50
Principles of Electromagnetics, 6e
The three loads are located on the Smith chart, as A, B, and C as shown next.
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Prob. 11.32
zL =
Z L 210
=
= 2.1 = s
Z O 100
Z L − Z O 110
=
,
Z L + Z O 310
1+ Γ
s=
= 2.1
1− Γ
V
But s = max → Vmax = sVmin
Vmin
Or Γ =
λ
λ
720o
= 120o
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4
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Hence the sending end will be Vmin , while the receiving end at Vmax
Since the line is
long,
→
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Principles of Electromagnetics, 6e
350
Prob. 11.32
Sadiku & Kulkarni
zL =
Z L 210
=
= 2.1 = s
Z O 100
Z L − Z O 110
=
,
Z L + Z O 310
1+ Γ
s
=
= 2.1
P.E.111.12
−Γ
Or Γ =
Principles of Electromagnetics, 6e
347
Z
Z L = (1 +Vmax
j 2) Z o
⎯⎯
→ zL = L = 1 + j 2
But s =
→ Vmax = sVmin
Zo
Vmin
We locate z L on the Smith chart. o
λ
λ
720
Since
the line 720
is o long, o →
= 120o
λ
⎯⎯
→
4 = 180 4
4
4
4 end will be V , while the receiving end at V
Hence
the sending
min
max
We move 180o toward the generator and locate point Q at which
Vmin = Vmax / s = 80 / 2.1 = 38.09
z = 0.2 − j 0.4
Vsending = 38.09∠90o
Z = zZ o = (0.2 − j 0.4) Z o
Prob. 11.33
(a) Method 1:
At Y,
 Z + jZ o tan β Copyright

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Z in = Z o  L

Z
jZ
β
+
tan

L
 o

2π λ
β =
= π,
tan π = 0
λ 2
Z
Z in = Z o L = Z L = 150 Ω
Zo
At X,
Z L ' = 150Ω
2π λ
= π / 2, tan π / 2 = ∞
λ 4
ZL ' 

 jZ o + tan β   Z 2 (75) 2
Z in = lim Z o 
= 37.5Ω
= o =
tan β  →∞
 jZ '+ Z o  Z L ' 150
 L tan β  
β =
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Method 2: Using the Smith chart,
Z ' 150
zL = L =
=3
50
Zo
Since  =λ /2, we must move 360o toward the generator. We arrive
at the same point. Hence,
Zin = Z L = 150
zL ' =
=
λ
150
=2
75
→ 180o
4
We move 180o toward the generator.
zin = 0.5
Z in = 75(0.6) = 37.5Ω
(b) From the Smith chart,
s = 3 for section XY
s = 2 for section YZ
Z − Z o 150 − 50
(c) Γ = L
=
= 0.5
ZL − Z o 150 + 50
Prob. 11.34
 Z + jZ o tan β  
Z in = Z o  L

 Z o + jZ L tan β  
c 3 × 108
= 0.75
λ= =
f 4 × 108
2π
2π × 3.2
=
= 26.81
β =
λ
0.75
 (30 − j 50) + j50 tan(26.81) 
Z in = 50 
 = 26.13 + j 44.23 Ω
 50 + j (30 − j 50) tan(26.81) 
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Prob. 11.35
Z
40 − j 25
zL = L =
= 0.8 − j 0.5
Zo
50
We locate this at point P on the Smith chart shown below
|Γ L |=
OP 2.4 cm
=
= 0.3,
OQ
8 cm
θ Γ = −96o
Γ L = 0.3∠ − 96o
At S,
s = r = 1.81
 = 0.27λ
⎯⎯
→
0.27 × 720o = 194.4o
From P, we move 194.4o toward the generator to G. At G,
z in = 1.0425 + j 0.6133
Z in = Z o z in = 50(1.0425 + j 0.6133) = 52.13 − j 30.66 Ω
69.323o
G
O
S=1.81
P
-96.277o
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Prob. 11.36
u=
λ=
c
εr
=
3 × 108
= 1.5767 × 108
3.62
u 1.5767 × 108
=
= 39.42 cm
400 × 106
f
 = 132 cm
⎯⎯
→
3.3485λ
⎯⎯
→ θ = 720o × 0.3485 = 250.92o
88.825o
ZL =196.03 -j109.82 Ω
G
O
P
zin =
Z in 40 + j 65
=
= 0.5333 + j 0.8666
Zo
75
We move toward the load from G (corresponding to Zin ) to
P (corresponding to the load). At P,
z L = 2.6137 − j1.4643
Z L = Z o z L = 75(2.6137 − j1.4643) = 196.03 − j109.82 Ω
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Prob. 11.37
(a)
0.12λ
⎯⎯
→ 0.12 × 720o = 86.4o
We draw the s=4 circle and locate Vmin . We move from that
location 86.4o toward the load.
0.37λ
Min
O
Max
s=4
P
-93.6o ZL =22.294 -j41.719 Ω
At P, z L = 0.45 − j 0.83
Z = 50(0.45 − j 0.83) = 22.3 − j 41.72 Ω
(b) the load is capacitive.
(c) Vmin and Vmax are λ /4 apart. Hence the first maximum
occurs at
0.12λ +0.25λ = 0.37λ
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Prob. 11.38
Z
75 + j 60
(a) z L = L =
= 1.5 + j1.2
Zo
50
| Γ |=
OP 3.8cm
=
= 0.475,
OQ
8cm
θ Γ = 42o
Γ = 0.475∠420
(Exact value = 0.4688 ∠ 41.76o)
(b) s=2.8
(Exact value = 2.765)
(c) 0.2λ → 0.2 x720o = 144o
zin = 0.55 − j 0.65
Z in = Z o zin = 50(0.55 + j 0.65) = 27.5 + j 32.5 Ω
(d) Since θ Γ = 42o , Vmin occurs at
42
λ = 0.05833λ
720
(e) same as in (d), i.e.. 0.05833λ
Prob. 11.39
If λ → 720o , then
zin = 0.35 + j 0.24
λ
6
→ 120o
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Prob. 11.40
Z 100 + j 60
=
= 2 + j1.2
z=
Zo
50
We locate z on the Smith chart. We move 180 o toward the generator
to reach point Q. At Q, y = 0.36-j0.22
Y = yYo =
1
(0.36 − j 0.22) = 7.4 − j 4.4 mS
50
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354
Prob. 11.41
u 0.5 × 3 × 108
=
= 0.9375 m
160 × 106
f
50 + j 30
Z
= 1 + j 0.6
zL = L =
50
Zo
λ=
We locate z at P on the Smith chart. We draw a circle that passes through P.
We locate point Q as the point where the circle crosses the Γ r − axis. At Q,
zin = 1.8
Z in = Z o zin = 50(1.8) = 90 Ω
The angular distance between P and Q is θ Γ = 73.3o.
If
=
λ → 720o ,
73.3o →
λ
720
o
73.3o
o
73.3
× 0.9375 = 0.0954 m
720o
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355
Prob. 11.42
Z
40 − j 30
zL = L =
= 0.8 − j 0.6
Zo
50
Locate this load at point P on the Smith chart.
λ
720o
= 180o
4
4
Draw a circle that passes through P and move 180o toward the generator.
At point Q,
zin = 0.8 + j 0.6
→
Z in = Z o zin = 50(0.8 + j 0.6) = 40 + j 30
Yin =
1
= 0.016 − j 0.012 = 16 − j12 mS
Z in
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356
Prob. 11.43
Z − Z o 0.5 − j − 1
(a) Γ = L
=
= 0.0769 − j 0.6154 = 0.6202∠ − 82.87 o
Z L + Z o 0.5 − j + 1
(b)
z −1
1 + Γ 1 + 0.4∠25o
Γ= L
⎯⎯
→ zL =
=
zL + 1
1 − Γ 1 − 0.4∠25o
z L = 1.931 + j 0.7771
Z L = (1.931 + j 0.7771) Z o
Prob. 11.44
Z
40 + j 25
zL = L =
= 0.8 + j 0.5
Zo
50
Locate this load at point P on the Smith chart. Draw a circle that passes through
P. Locate point Q where the negative Γ r − axis crosses the circle.
θ P = 97 o. The angular distance between P and Q is
θ =180 + θ P = 277o.
720o → λ
277 o →  =
λ
720
o
277 o = 0.3847λ
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Prob. 11.45
(a)
λ
2
= 120cm → λ = 2.4m
3 × 108
= 125MHz
2.4
λ
40λ λ
720°
(b) 40cm =
= →
= 120°
240 6
6
Z L = Z o z L = 150(0.48 + j 0.48
u = fλ → f =
u
=
= 72 + j 72 Ω
(Exact value = 73.308+j70.324 Ω)
(c)
s − 1 1.6
=
= 0.444,
s + 1 3.9
Γ = 0.444∠120°
Γ =
Prob. 11.46
(a)
Z
j 60
Z
j 40
zL = L =
= j 0.75,
zin = in =
= j 0.5
Zo
80
Zo
80
The two loads fall on the r=0 circle, the outermost resistance circle. The shortest
distance between them is
106.26o
126.87o
360o − (126.87 o − 106.26o )λ
= 0.4714λ
720o
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(b) s = ∞,
Γ L = 1∠106.26o
Prob. 11.47
(a) Z in =
Z in 100 − j120
=
= 1.25 − j1.5
Zo
80
u 0.8 × 3 × 108
=
= 20m
f
12 × 106
22λ
l1 = 22m =
= 1.1λ → 720° + 72°
20
28λ
l2 = 28m =
= 1.4λ → 720° + 72° + 216°
20
To locate P(the load), we move 2 revolutions
λ=
plus 72° toward the load. At P,
OP 5.1cm
=
= 0.5543
ΓL =
OQ 9.2cm
θ Γ = 72° − 47° = 25°
Γ L = 0.5543∠25°
(Exact value = 0.5624∠25.15o )
Z in , max = sZ o = 3.7(80) = 296Ω
(Exact value = 285.59 Ω)
Z
80
Z in , min = o =
= 21.622Ω
s 3.7
(Exact value =22.41 Ω)
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Z L = 2.3 + j1.55
(b) Also, at P,
Z L = 80(2.3 + j1.55) = 184 + j124Ω
(Exact value = 183.45+j128.25 Ω)
s = 3.7
At S,
To Locate Zin' , we move 216° from Zin toward the geneator.
At Zin' ,
zin' = 0.48 + j 0.76
Z in' = 80(0.48 + j 0.76) = 38.4 + j 60.8Ω
(Exact=37.56+j61.304 Ω)
(c) Between ZL and Zin , we move 2 revolutions and 72°. During
the movement, we pass through Zin, max 3 times and Zin,min twice.
Thus there are:
3 Z in ,max and 2 Z in ,min
Prob. 11.48
(a) From Eq. (11.43),
Z in1 =
Z in 2 =
Z o22
ZL
Z o21
Z2 Z2
= Z o , i.e. Z in 2 = o1 = o 2
Z in 2
Zo
ZL
Z o1 = Z o 2
(b) Also,
50
Zo
= 30
= 24.5Ω.
75
ZL
Zo  Zo2 
Zo Z L
=
 → Zo2 =
Z o1  Z L 
Z o1
(1)
2
Z o1  Z o 2 
3
2
=
Also,
 → ( Z o 2 ) = Z o1Z L
Zo2  Z L 
(2)
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From (1) and (2), ( Z o 2 )3 = Z o1Z L2 =
Z o3 Z L3
Z o31
or Z o1 = 4 Z o3 Z L = 4 (50)3 (75) = 55.33Ω
From (3), Z o 2 = 3 Z o1Z L2 = 3 (55.33)(75)2 = 67.74Ω.
Prob. 11.49
l1 =
λ
→ Z in1 =
4
200 + j150
= 20 + j15 mS
(100) 2
yin1 =
l2 =
λ
8
yin 2 =
→ Z in 2 = Z L


 = jZ o


7π

 Zi + jZo tan
4
= Zo 
π
7

 Zo + jZi tan
4



 =



Z o ( Zi − j Z o )
( Z o − jZ i )
yi = yin1 + yin2 = 20 + j5 mS
zi =
yin 3 =
π

 Z L + jZ o tan 4
lim
 0 Z o 
π
 Z o + jZ L tan

4
1
1
=
= − j10 mS
jZ o j100
7λ
→ Z in 3
l3 =
8
But
Zo2
Z
or yin1 = L2
ZL
Zo
1
yi
=
1000
= 47.06 − j11.76
20 + j5
Z o − jZ in
100-j47.06-11.76
=
Z o ( Z in − jZ o ) 100 ( 47.06-j11.76-j100 )
= 6.408 + j 5.189 mS
If the shorted section were open,
yin1 = 20 + j15 mS
yin 2 =
j tan π
1
4 = j = j10 mS
=
Z in 2
Zo
100
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7π 

Zi + jZo tan


Zo ( Zi − jZo )
7λ
4 
→ Z in 3 = Zo 
=
l3 =
7π 
8

( Z o − jZ i )
 Zo + jZi tan

4 

yi = yin1 + yin 2 = 20 + j15 + j10 = 20 + j 25 mS
Zi =
1
1000
=
= 19.51 − j 24.39Ω
yi 20 + j 25
yin 3 =
75.61-j19.51
Z o − jZ i
=
Z o ( Z i − jZ o ) 100 (19.51 − j124.39 )
= 2.461 + j 5.691 mS
Prob. 11.50
From the previous problem, Z in = 148Ω
I in =
Vg
Z g + Z in
=
120
= 0.5263 A
80 + 148
1
1
2
I in Rin = (0.5263) 2 (148) = 20.5W
2
2
Since the lines are lossless, the average power delivered to either antenna is 10.25W
Pave =
Prob. 11.51
2π λ π
(a) β l =
. = ,
4 4 2
tan β l = ∞
 ZL

+ jZ o 

 Z + jZ o tan β l 
 tan β l

Z in = Z o  L
 = Zo
 Zo

 Z o + jZ L tan β l 
 tan β l + jZ L 


As tan β l → ∞,
Z in =
Z o2 (50) 2
=
= 12.5Ω
200
ZL
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(b) If Z L = 0,
Z o2
=∞
(open)
0
25 × ∞
25
(c) Z L = 25 / / ∞ =
=
= 25Ω
25 + ∞ 1 + 25
∞
2
(50)
Z in =
= 200Ω
12.5
Z in =
Prob. 11.52
λ
74
1
= 1.48,
= 0.6756
zL
4
50
This acts as the load to the left line. But there are two such loads in parallel due to
→ 180°, z L =
the two lines on the right. Thus
 1 
 
Z
'
Z L = 50  L  = 25(0.6756) = 16.892
2
16.892
1
= 0.3378, z in = ' = 2.96
z L' =
zL
50
Z in = 50(2.96) = 148Ω.
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Prob. 11.53
zL =
Z L 60 − j 50
=
= 1.2 − j1
Zo
50
yL =
1
zL
0.08433 λ
65.039o
125.75o
A
YL
O
ZL
0.1307 λ
-94.096o
At A, y = 1 + j0.92, ys = -j0.92
− j 0.92
Ys = Yo ys =
= − j18.4 mS
50
Stub length = 0.1307λ
Stub position = 0.0843λ
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Prob. 11.54
d A = 0.12λ → 0.12 × 720o = 86.4o
l A = 0.3λ
(a)
→ 0.3 × 720o = 216o
From the Smith Chart below,
z L = 0.57 + j 0.69
Z L = 60 ( 0.57 + j 0.69 )
= 34.2 + j 41.4Ω
(b)
(c)
360o − 86.4o
λ = 0.38λ
dB =
720o
o
o
λ ( −62.4 − −82 )
lB = −
λ = 0.473λ
2
720o
s = 2.65
(Exact value = 2.7734)
0.4721 λ
ZL
O
YL
s = 2.773
A
0.38 λ
-421.97o
-82.033o
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Prob. 11.55
Z L 120 + j 220
=
= 2.4 + j 4.4
Zo
50
We follow Example 11.7. At A, ys=-j3 and at B, ys=+j3. The required stub
admittance is
± j3
Ys = Yo ys =
= ± j 0.06 S
50
The distance between the load and the stub is determined as follows. For A,value =
0.2308λ)
zL =
For B,
180 + 10 + 17
λ = 0.2875λ
720
The length of the stub line is determined as follows.
19
dA =
λ = 0.0264λ
720
(Exact value = 0.0515 λ)
360 − 19
dB =
λ = 0.4736λ
720
(Exact value = 0.4485λ)
lB =
0.2308 λ
33.863o
A
ZL
O
-159.96o
YL
0.05152 λ
-37.095o
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Prob. 11.56
90o
4V
V
=4
s = max =
Vmin 1V
Γ =
λ
2
→
s −1 3
= = 0.6
s +1 5
±180o
= 25 cm − 5 cm = 20 cm
Vmin
0o
s=4
λ = 40 cm
P
The load is l=5cm from Vmin, i.e.
l=
5λ λ
=
40 8
→
–90o
90o
On the s = 4 circle, move 90o from Vmin towards the load and obtain ZL = 0.46 – j0.88 at
P.
ZL = Zo zL = 60(0.46 – j0.88) = 27.6 – j52.8 Ω
(Exact value = 28.2353 –j52.9412 Ω)
θ Γ = 270o or -90o
Γ = 0.6∠-90o
Prob. 11.57
s=
λ
2
Vmax 0.95
=
= 2.11
Vmin 0.45
= 22.5 − 14 = 8.5
→
λ = 17 cm
3 × 108
f = =
= 1.764 GHz
0.17
λ
3.2
l = 3.2 cm =
λ → 135.5o
17
l
c
Vmin
•
O
•
S
P • ZL
Q
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At P,
z L = 1.4 − j 0.8
ZL = 50 (1.4 − j 0.8 ) = 70 − j 40Ω
(Exact value = 70.606-j40.496 Ω)
s − 1 1.11
Γ =
=
= 0.357,
s + 1 3.11
Γ = 0.357∠ − 44.5o
θ Γ = −44.5o
(Exact value = 0.3571∠-44.471o )
Prob. 11.58
Γs =
Rg − Ro
Rg + Ro
ΓL =
Prob. 11.59
ΓL =
Γg =
t1 =
=
0 − 50
= −1
0 + 50
RL − Ro 80 − 50
=
= 0.231
RL + Ro 80 + 50
Z L − Z o 0.5Z o − Z o
1
=
=−
1.5Z o
3
Z L + Zo
Z g − Zo
Z g + Zo
=
Zo 1
=
3Z o 3
l
Z
= 2 μ s, Vo = o ( 27 ) = 9 V,
3Z o
u
Io =
Vo
= 180 mA
Zo
0.5
ZL
Vg =
( 27 ) = 5.4 V,
2.5
Zg + ZL
I∞ =
V∞
= 216 mA
ZL
V∞ =
The voltage and current bounce diagrams are shown below
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ry = 1
3
z=0
ri = − 1 ry = − 1 3
3
z=0
z =1
9V
t1
–3V
2t1
60mA
2t1
–1V
–6.667mA
4t1
1 V
9
−1
27
V
3t1
2.222mA
5t1
0.741mA
6t1
t1
–20mA
3t1
1 V
3
4t1
180mA
ri = 1
3
z =1
5t1
6t1
–0.25mA
(Current bounce diagram)
(Voltage bounce diagram)
From the bounce diagrams, we obtain V(0,t) and I(0,t) as shown below:
V(0,t)
9V
5V
5.444V
5.395V
1
1
4
8
-1V
9
3
12
-3V
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I(0,t)
220mA
215.5mA
180mA
216.05mA
60mA
4
12
8
-20mA
Prob. 11.60
Using Thevenin equivalent at z = 0 gives
R g = Rs = 4 Z o = 200Ω
Vg = I s Rs = 10 × 200 × 10−3 = 2V
Γg =
ΓL =
Z g − Zo
Z g + Zo
=
4Z o − Z o 3
=
4Z o + Z o 5
Z L − Z o 2Z o − Z o 1
=
=
Z L + Z o 2Zo + Z o 3
10

=
= 50 ns
u 2 × 108
The bounce diagram is shown below.
t1 =
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Γg=0.6
ΓL=1/3
2V
t1
2/3
0.4
3t1
0.133
0.08
5t1
The load voltage is sketched below.
V(l,t)
3.4136
3.2
2..667
2
2/3
0
t1
2 t1
3 t1
4 t1
5 t1
V ( , t ) V (  , t )
=
ZL
100
To get I(I,t), we just scale down V(l,t) by 100.
I ( , t ) =
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Prob. 11.61
Z − Z o 32 − 75
Γg = g
=
= 0.4019
Z g + Z o 32 + 75
Z L − Z o 2 × 106 − 75
1
ΓL =
=
Z L + Z o 2 × 106 + 75
 50 × 10−2
=
= 2.5ns
2 × 108
u
The bounce diagram is shown below.
t1 =
At t =20 ns = 4t1 ,
V = 8 + 8 − 3.2152 − 3.2152 = 9.57 V
Γg =-0.4019
ΓL = 1
8V
t1
8V
2t1
-3.2152
3t1
-3.2152
4t1
1,293
5 t1
Prob. 11.62
(a)
l
150
=
= 0.5μ s,
u 3 × 108
Z − Z o 150 − 50 1
ΓL = L
=
= ,
Z L + Z o 150 + 150 2
t1 =
Vo =
Z oVg
Zo + Z g
=
50 (12 )
75
= 8V ,
Γg =
Io =
Z g − Zo
Z g + Zo
Vg
Z g + Zo
=
25 − 50
1
=− ,
75
3
=
12
= 160 mA
75
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Γ= − 13
0
8V
− 23
2μs
1.5μs
−
(Voltage)
40
9
− 209
3μs
1
27
1.5μs
40
3
2.5μs
3.5μs
0.5μs
– 803
2μs
2
9
1
9
3μs
–80
1μs
– 43
1
2
160
0.5μs
4
1μs
Γ= −
Γ=1/3
0
Γ= 12
2.5μs
20
− 27
(Current)
3.5μs
The bounce diagrams are for the leading pulse. The bounce diagrams for the second
pulse is delayed by 1µs and negated because of -12V.
(b) For each time interval, we add the contributions of the two pulses together.
For 0 < t < 1µs, V(0,t) = 8V
For 1 < t < 2µs, V(0,t) = -8 + 4 – 4/3 = -5.331V
For 2 < t < 3µs, V(0,t) = -(4-4/3)-2/3 + 2/9 = -2.667 -0.444 = -3.11V
For 3 < t < 4µs, V(0,t) = 0.444 + 1/9 -1/27 = 0.444+0.0741 = 0.518V
For 4 < t < 5µs, V(0,t)= -0.0741 – 0.0124 = -0.0864V
We do the same thing at the load end.
For 0 < t < 0.5 μ s, V (, t ) = 0
For 0.5 < t < 1.5μs, V(,t)=8+4 = 12
For 1.5 < t < 2.5μs, V(,t)=-12 + (-4/3 -2/3)=-12-2=-14
For 2.5 < t < 3.5μ s, V(,t)=2+2/9 +1/9=2.333
For 3.5 < t < 4.5μ s, V(,t)=-0.333-1/27-1/54=-0.3886V
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The results are shown below.
V(0,t)
8V
0.518V
0
1
2
3
-5.333V
V (, t )
-3.11V
4
5
-0.0864V
t (µs)
6
12V
2.333V
0
0.5
1.5
2.5
3.5
4.5
5.5
t (µs)
-0.3886V
-14V
Since I (, t ) =
V ( , t ) V (  , t )
=
, we scale V (, t ) by a factor of 1/150 as shown below.
ZL
150
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I (, t ) (mA)
80
15.55
0
0.5
1.5
2.5
3.5 -2.59
4.5
-93,3
Prob. 11.63
Vo = 8V =
Zo
50
Vg =
Vg
Zo + Z g
50 + 60
⎯⎯
→ Vg =
8 × 110
= 17.6 V
50

u
8
−6
 = ut1 = 3 × 10 × 2 × 10 = 600 m
2t1 = 4 μ s
⎯⎯
→ t1 = 2 μ s =
Prob. 11.64
t1 =
20
= 10−7 = 0.1μ s,
8
2 x10
Γg =
Z g − Zo
Z g + Zo
=
Vo =
Zo
50
Vg = (12) = 10V
Zo + Z g
60
10 − 50
= −2 / 3
10 + 50
Z L − Z o 0 − 50
=
= −1
Z L + Z o 0 + 50
The voltage bounce diagram is shown below
ΓL =
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Γ g = −2 / 3
Γ L = −1
10V
t1
-10V
2t1
6.667V
-6.667 V
3t1
From the bounce diagram, we obtain V(0,t) as shown. V(l,t) =0 due to the short circuit.
V(0,t)
10
6.667
4.44
2.963
0
0.1
0.2
0.3
0.4
0.5
-6.667
-10
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0.8
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Prob. 11.65
The initial pulse on the line is
Vo =
Zo
Vg
Zo + Z g
(1)
The reflection coefficient is
Z − Zo
Γ= g
(2)
Z g + Zo
The first reflected wave has amplitude ΓVo , while the second reflected wave
has amplitude Γ 2Vo , etc. For a very long time, the total voltage is
VT = Vo + ΓVo + Γ 2Vo + ...
 1 
= Vo (1 + Γ + Γ 2 + Γ3 + ...) = Vo 

1− Γ 
Substituting (1) and (2) into (3) gives


1
Zo
VT =
Vg 
Zg − Z o
Zo + Z g 
1−
Z g + Zo

Prob. 11.66
For w = 0.4 mm,
(3)


V
1
=ZV
= g
o g

2
Z g + Zo − Z g + Zo


w 0.4 mm
=
= 0.2 → narrow strip
h
2m
w
= 0.2, ε eff = 5.851, Z o = 91.53Ω
h
w
For
= 0.4, ε eff = 6.072, Z o = 73.24Ω
h
For
Hence,
73.24Ω < Z o < 91.53Ω
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Prob. 11.67
(a ) Let x = w h. If x < l ,
50 =
60
8
ln 
4.6  x

+ x

8

5 4.6 - 6ln 
+ x = 0
x

we solve for x (e.g using Maple) and get x = 2.027 or 3.945
which contradicts our assumptiom that x < 1. If x > 1,
120π
4.6 [ x + 1.393 + 0.667 ln( x + 1.444) ]
50 =
We solve this iteratively and obtain:
x = 1.8628, w = xh = 14.9024 mm
For this w and h,
ε eff = 3.4598
(b)
β=
ω ε eff
c
β  = 450 =
 =
π
4
=
ω ε eff
πc
4 ε eff 2π f
c
=
3 × 108
8 ×
3.4598 × 8 × 109
 = 0.00252 m
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Prob. 11.68
w = 1.5cm, h = 1cm,
(a)
εr - 1
0.6
 6 +1
= 1.6 +
 +
2 1 + 12h/w
1 + 12/1.5
 2 
ε eff = 
377
1.8 (1.5 + 1.393 + 0.667 In (2.944))
Z0 =
(b)
w
= 1.5
h
α c = 8.686
Rs =
1
281
3.613
= 77.77Ω
Rs
wZo
=
σc σ
=
= 1.8
μπ f
σc
=
19 × 2.5 × 109 × 4π ×10-3
1.1 ×107
= 2.995 × 10-2
αc =
u =
8.686 × 2.995 × 10−2
1.5 × 10−2 × 77.77
c
ε eff
u
c
=
f
f ε eff
=
3 × 108
2.5 × 109 1.8
0.8 ( 2.2 )
2 ×10−2
1.2 1.8 8.944 × 10−2
=
96.096
14.3996
→ λ =
α d = 27.3 ×
= 0.223dB/m
α d = 6.6735 dB/m
(c)
α = α c + α d = 6.8965 dB/m
α  = 20dB →  =
20
α
=
20
= 2.9 m
6.8965
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Prob. 11.69
0.1  5 × 1.2 
ln 
w ' = 0.5 +
 = 0.5 + 0.1279 = 0.6279
3.2  0.1 
Zo =

377
4 × 1.2  8 × 1.2

ln 1 +
+ 3.354  

2π 2  π × 0.6177  4 × 0.6279

= 42.43ln {1 + 2.49(3.822 + 3.354} = 42.43ln(20.255)
= 127.64 Ω
Prob. 11.70
Suppose we guess that w/h < 2
A =
w
h
75 3.3
60 2
=
+
1.3 
0.11 
 0.23 +
 = 1.117
3.3 
2.3 
8e A
24.44
=
= 3.331 → w = 3.331h = 4mm
2A
7.337
e - 2
If we guess that w/h > 2,
B=
60π 2
Zo ε r
=
60π 2
= 5.206
75 2.3
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w
2
1.3 
0.61 
=
4.266 − ln 9.412 +
 ln 4.206 + 0.39 −


h
4.6
2.3  
π
= 1.665 <2
w
Thus
= 3.331 > 2
h
ε eff =
3.3
+
2
3 × 108
u =
1.953
1.3
12
2 1 +
3.331
= 1.953
= 2.1467 × 108 m/s
Prob. 11.71
Z L − Z o 100 − 150
=
= −0.2
250
Z L + Zo
RL = −20log | Γ |= 13.98 dB
Γ=
Prob. 11.72
Γ=
Z L − Z o 120 − 50
=
= 0.4118
Z L + Zo
170
RL = −20log10 | Γ |= −20log10 (0.4118) = 7.706 dB
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CHAPTER 12
P. E. 12.1 (a) For TE10, fc = 3 GHz,
1 − ( f c / f ) 2 = 1 − (3 / 15) 2 = 0.96 , β o = ω / uo = 4π f / c
β=
4π f
c
0.96 =
4π × 15 × 109
0.96 = 615.6 rad/m
3 × 108
ω 2π × 15 × 109
=
= 1.531 × 108 m/s
β
615.6
60π
μ
= 192.4Ω
η'=
= 60π , ηTE =
ε
0.96
u=
(b) For TM11, fc = 3 7.25 GHz, 1 − ( f c / f ) 2 = 0.8426
4π f
4π × 15 × 109 (0.8426)
β=
(0.8426) =
= 529.4 rad/m
c
3 × 108
u=
ω 2π × 15 × 109
=
= 1.78 × 108 m/s
529.4
β
ηTM = 60π (0.8426) = 158.8Ω
P. E. 12.2 (a) Since Ez ≠ 0 , this is a TM mode
Ezs = Eo sin(mπ x / a)sin(nπ y / b)e − j β z
mπ
= 40π
a
i.e. TM21 mode.
Eo = 20,
m=2,
nπ
= 50π
b
n=1
u'
3 × 108
2
2
(b) f c =
( m / a ) + ( n / b) =
402 + 502 = 1.5 41 GHz
2
2
2π f
2π × 109
β = ω με 1 − ( f c / f ) 2 =
f 2 − fc 2 =
225 − 92.25 = 241.3 rad/m.
c
3 × 108
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(c)
− jβ
(40π )20cos 40π x sin 50π ye − j β z
h2
− jβ
E ys = 2 (50π )20sin 40π x cos50π ye− j β z
h
Ey
= 1.25 tan 40π x cot 50π y
Ex
Exs =
P. E. 12.3 If TE13 mode is assumed, fc and β remain the same.
fc = 28.57 GHz, β = 1718.81 rad/m, γ = j β
377 / 2
ηTE13 =
= 229.69 Ω
1 − (28.57 / 50) 2
For m=1, n=3, the field components are:
Ez= 0
H z = H o cos(π x / a )cos(3π y / b)cos(ωt − β z )
ωμ  3π 
  H o cos(π x / a )sin(3π y / b)sin(ωt − β z )
h2  b 
ωμ  π 
E y = 2   H o sin(π x / a)cos(3π y / b)sin(ωt − β z )
h a
β π 
H x = − 2   H o sin(π x / a)cos(3π y / b)sin(ωt − β z )
h a
β  3π 
H y = − 2   H o cos(π x / a )sin(3π y / b)sin(ωt − β z )
h  a 
Ex = −
Given that H ox = 2 = −
H oy = −
β
h2
β
h2
(π / a) H o ,
(3π / b) H o = 6a / b = 6(1.5) / 8 = 11.25
−2 × 14.51π 2 × 104 × 1.5 × 10−2
H oz = H o = −
=
= −7.96
βπ
1718.81π
ωμ  π 
2ωμ
Eoy = 2   H o = −
= 2ηTE = −459.4
β
h a
3a
Eox = − Eoy
= 459.4(4.5 / 0.8) = 2584.1
b
2h 2 a
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Ex = 2584.1cos(π x / a)sin(3π y / b)sin(ωt − β z ) V/m,
E y = −459.4sin(π x / a)cos(3π y / b)sin(ωt − β z ) V/m,
Ez= 0,
H y = 11.25cos(π x / a)sin(3π y / b)sin(ωt − β z ) A/m,
H z = −7.96cos(π x / a) cos(3π y / b)cos(ωt − β z ) A/m
P. E. 12.4
f c11 =
3 × 108
up =
ug =
u' 1
1 3 × 108 × 102
+
=
1 / 8.6362 + 1 / 4.3182 = 3.883 GHz
2
2
b
2 a
2
1 − (3.883 / 4)
2
= 12.5 × 108 m/s,
9 × 1016
= 7.2 × 107 m/s
8
12.5 × 10
P. E. 12.5 The dominant mode becomes TE01 mode
f c 01 =
c
= 3.75 GHz, ηTE = 406.7Ω
2b
From Example 12.2,
Ex = − Eo sin(3π y / b)sin(ωt − β z ) ,
Pave =

a
x =0
where Eo =
| Exs |2
Eo 2 ab
dxdy
=
y = 0 2η
4η
ωμb
Ho .
π
b
Hence Eo = 63.77 V/m as in Example 12.5.
Ho =
π Eo
π × 63.77
=
= 63.34 mA/m
10
ωμb 2π × 10 × 4π × 10−7 × 4 × 10−2
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P. E. 12.6 (a) For m=1, n=0, fc = u’/(2a)
σ
10−15
10−15
=
=
<< 1
ωε 2π × 9 × 109 × 2.6 × 10−9 / (36π ) 1.3
Hence,
u' ≅
αd =
1
με
fc =
= c / 2.6,
ση '
2 1 − ( fc / f )
2
=
3 × 108
2 × 2.4 × 10−2 2.6
10−15 × 377 / 2.6
2 1 − (2.2149 / 9)
2
= 2.2149 GHz
= 1.205 × 10−13 Np/m
For n = 0, m=1,
αc =
2 Rs
1 b
[ + ( f c / f )2 ]
bη ' 1 − ( f c / f ) 2 a
2
=
2 2.6 π × 9 × 109 × 1.1 × 107 × 4π × 10−7
−2
377 × 1.5 × 10 × 1.1 × 10
7
1 − (2.2149 / 9)
2
[0.5 + (2.4 / 1.5)(2.2148 / 9) 2 ] = 2 × 10−2 Np/m
(b) Since α c >> α d ,α = α c + α d ≅ α c = 2 x10−2
loss = α l = 2 × 10−2 × 0.4 = 0.8 × 10−2 Np = 0.06945 dB
P. E. 12.7 For TE11 , m = 1 = n,
H zs = H o cos(π x / a) cos(π y / b)e −γ z
jω
(π / b) H o cos(π x / a )sin(π y / b)e −γ z
2
h
jωμ
E ys = − 2 (π / a) H o sin(π x / a) cos(π y / b)e −γ z
h
jβ
H xs = 2 (π / a ) H o sin(π x / a) cos(π y / b)e −γ z
h
jβ
H ys = 2 (π / b) H o cos(π x / a)sin(π y / b)e −γ z
h
Ezs = 0
Exs =
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For the electric field lines,
dy E y
=
= (a / b) tan(π x / a ) cot(π y / b)
dx Ex
For the magnetic field lines
dy H y
=
= −(a / b)cot(π x / a) tan(π y / b)
dx H x
E H
Notice that ( y )( y ) = −1
Ex H x
showing that the electric and magnetic field lines are mutually orthogonal. The field
lines are as shown in Fig. 12.14.
P. E. 12.8
1
c
=
u' =
με
εr
1.5 × 1010
1 / 25 + 0 + 1 / 100 = 1.936 GHz
3
1
QTE101 =
, where
61δ
1
1
δ=
=
= 1.5 × 10−6
9
−7
7
π f101μσ c
π × 1.936 × 10 × 4π × 10 × 5.8 × 10
fTE101 =
QTE101 =
106
= 10,929
61 × 1.5
P. E. 12.9
(a) By Snell’s law, n1 sin θ 1 = n2 sin θ 2 . Thus
θ 2 = 90o
sin θ 2 = 1
sin θ 1 = n2/n1,
(b) NA =
θ 1 = sin –1 n2/n1 = sin –1 1.465/1.48 = 81.83o
n12 − n2 2 = 1.482 − 1.4652 = 0.21
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P. E. 12.10
α l = 10 log P(0)/P(l) = 0.2 X 10 = 2
P(0)/P(l) = 100.2, i.e. P(l) = P(0) 10-0.2 = 0.631 P(0)
i.e. 63.1 %
Prob. 12.1
2
2
u'  m   n 
3 × 108
fc =
+
=
   
2  a  b
2 2.25 × 10−2
1/ 2
 m  2  n  2 

 +
 
 2.28   1.01  
1/ 2
2
2
15  m   n  
=

 +
 
2.25  2.28   1.01  
GHz
Using this formula, we obtain the cutoff frequencies for the given modes as shown below.
Mode
f c (GHz)
TE01
TE10
TE11
TE02
TE22
TM11
TM12
TM21
9.901
4.386
10.829
19.802
21.658
10.829
20.282
13.228
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Prob. 12.2
2
2
2
2
u ' m 2 n 2 3 × 108
m n
m n
+ 2 =
× 102   +   = 15   +   GHz
2
2 a
b
2
 6  4
 6  4
15
f10 = = 2.5 GHz
6
15
f c 01 = = 3.75 GHz
4
2
f c 20 = 15 × = 5 GHz
6
1
1
f c11 = 15 2 + 2 = 15 × 0.3005 = 4.51 GHz
6
4
Possible modes are TE10 ,TE 01 ,TE 20 ,TE11 and TM11.
f cmn =
Prob. 12.3
(a) For TE10 mode, f c =
(b) f = 3 f c = 7.5 GHz
f cmn
u1 m 2 n 2
=
+
2 a 2 b2
2
u'
3 × 103
=
= 2.5 GHz
2a 2 × 6 × 10−2
2
3 × 102 × 102  m   n 
=

 + 
2
 6   4
2
2
m  n 
= 15 
 +   GHz
 6   4
2
f c 20 = 15 × = 5 GHz
6
f c 01 = 3.75 GHz,
f c 02 = 7.5 GHz
f c10 = 2.5 GHz,
f c 20 = 5.0 GHz
f c 21 = 6.25 GHz,
f c 30 = 7.5 GHz
2
2
2
2
1 2
f12 = 15   +   = 7.91 GHz
6 4
1 1
f11 = 15   +   = 4.507 GHz
6 4
The following modes are transmitted
TE01 , TE02 , TE10 , TE11 , TE20 , TE21 , TE30
TM 11 , TM 21
i.e. 7 TE modes and 2 TM modes
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Prob. 12.4
(a)
For TE10 mode,
2
u'  1 
3 × 108
=
= 6.25 GHz
 
2 a
2 × 2.4 × 10−2
For TE 01 mode,
fc =
2
u'  1 
3 × 108
=
= 12.5 GHz
 
2 b
2 × 1.2 × 10−2
For TE 20 mo de,
fc =
2
u'  2 
  = 2 × 6.25 = 12.5 GHz
2 a
For TE 02 mode,
fc =
2
fc =
u'  2
  = 2 × 12.5 = 25 GHz
2 b
(b) Since f = 12 GHz, only TE10 mode will propagate.
Prob. 12.5 (a) For TE10 mode, f c =
Or a =
u'
3 × 108
=
= 3 cm
2 f c 2 × 5 × 109
For TE01 mode, f c =
Or
b=
u'
2a
u'
2b
u'
3 × 108
=
= 1.25 cm
2 f c 2 × 12 × 109
(b) Since a > b, 1/a < 1/b, the next higher modes are calculated as shown below.
Mode
TE10
*TE20
TE30
TE40
*TE01
TE02
*TE11
TE21
fc (GHz)
5
10
15
20
12
24
13
15.62
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The next three higher modes are starred ones, i.e. TE20, TE01, TE11
1
(c) u ' =
με
=
c
2.25
= 2 × 108 m/s
For TE11 modes,
fc =
3 × 108
1
1
+
= 8.67 GHz
2
2
−2
2 × 10 2.25 3 1.25
For the dominant mode,
Prob. 12.6
fc =
c 3 × 108
=
= 18.75 MHz
2a
2×8
(a) It will not pass the AM signal, (b) it will pass the FM signal.
Prob. 12.7 a/b = 3
f c10 =
a = 3b
u'
2a
a=
u'
3 × 108
=
m = 0.833cm
2 f c10 2 × 18 × 109
A design could be a = 9mm, b = 3mm.
Prob. 12.8
2
2
 f 
 25 
Let F12 = 1 −  c12  = 1 −   = 0.7806
 40 
 f 
λ' =
λ12 =
c
3 × 108
=
= 0.0075 m = 7.5 × 10−3 m
9
f 40 × 10
λ'
F12
=
7.5 × 10−3 m
= 9.608 × 10−3 m
0.7806
u ' 3 × 108
=
= 3.843 × 108 m/s
u12 =
F12 0.7806
β12 =
2π
=
λ12
η'
ηTE12 =
F12
2π
= 653.95 rad/m
9.608 × 10−3
=
120π
= 482.95Ω
0.7806
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Prob. 12.9
u=
ω
u'
3 × 108
=
=
= 6.975 × 108 m/s
2
2
β
1 − ( fc / f )
1 − (6.5 / 7.2)
9 × 1016
ug =
= 1.2903 × 108 m/s
u
t=
2l
300
=
= 2.325 μs
u g 1.2903 × 108
Prob. 12.10
c
c 1.12 × 3 × 108
f = 1.12 f c10 = 1.12
⎯⎯
→ a = 1.12
=
= 4.2 cm
2a
2f
2 × 4 × 109
The next higher-order mode is fc01 =c/2b.
c
f = 0.85 f c 01 = 0.85
2b
0.85c 0.85 × 3 × 108
⎯⎯
→ b=
=
= 3.187 cm
2f
2 × 4 × 109
Prob. 12.11
u'
c
(a) f c10 =
=
2a 2a ε r
=
3 × 108
2 × 1.067 × 10−2 6.8
30
GHz
=
2 × 1.067 6.8
= 5.391 GHz
(b)
 f 
F = 1−  c 
 f 
2
 5.391 
= 1− 

 6 
= 0.439
2
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u'
c
3 × 108
=
=
= 2.62 × 108 m/s
F F ε r 0.439 × 6.8
uρ =
(c) λ =
λ'
u'
3 ×108
=
=
=
F
F
fF ε r 0.439 × 6 ×109 × 6.8
=
c
f
10−1
= 0.04368 m = 4.368 cm
2 × 0.439 6.8
Prob. 12.12
In evanescent mode,
2
 mπ   nπ 
k = ω με < 
 +

 a   b 
2
2
2
2
2
 mπ   nπ 
2
2
2
2
γ =α = 
 +
 − k = 4π με f c − ω με
a
b

 

β = 0,
 f 
α = με 4π f − 4π f = 2π με f c 1 −  
 fc 
2
2
c
2
2
2
Prob. 12.13
Ez ≠ 0 . This must be TM23 mode (m=2, n=3). Since a= 2b,
c
3 × 108
ω 1012
2
2
fc =
m + 4n =
4 + 36 = 15.81 GHz, f =
=
= 159.2 GHz
4b
4 × 3 × 10−2
2π
2π
ηTM = 377 1 − (15.81 / 159.2)2 = 375.1 Ω
P
ave
=
| Exs |2 + | E ys |2
2ηTM
az
β 2 Eo 2
(2π / a ) 2 cos 2 (2π x / a )sin 2 (3π y / b) + (3π / b) 2 sin 2 (2π x / a )cos 2 (3π y / b)  a z
= 4
2h ηTM
Pave =  Pave .dS =
=
a
b
 P
ave
dxdya z
x =0 y =0
β 2 Eo 2 ab  4π 2 9π 2  β 2 Eo 2 ab
+ 2 =

2h 4ηTM 4  a 2
b  8h 2ηTM
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Sadiku & Kulkarni
Principles of Electromagnetics, 6e
395
But
β=
ω
1 − ( f c / f )2 =
1012
1 − (15.81 / 159.2) 2 = 3.317 × 103
8
3 × 10
c
4π 2 9π 2 10π 2
2
h = 2 + 2 = 2 = 1.097 × 105
a
b
b
Pave =
(3.317) 2 × 106 × 52 × 18 × 10−4
= 1.5 mW
8 × (1.098 × 105 ) × 375.1
Prob. 12.14 (a) Since m=2 and n=1, we have TE21 mode
(b) β = β ' 1 − ( f c / f ) 2 = ω μoε o 1 − (ωc / ω ) 2
ωc 2 = ω 2 − β 2c 2
β c = ω 2 − ω 2c
ω
fc = c =
2π
(c) ηTE =
β 2c 2
144 × 9 × 1016
18
f −
= 36 × 10 −
= 5.973 GHz
4π 2
4π 2
2
η
1 − ( f c / f )2
=
377
1 − (5.973 / 6) 2
= 3978Ω
(d) For TE mode,
Ey =
ωμ
Hx =
h2
(mπ / a ) H o sin(mπ x / a) cos(nπ y / b)sin(ωt − β z )
−β
(mπ / a) H o sin(mπ x / a)cos(nπ y / b)sin(ωt − β z )
h2
β = 12, m = 2, n =1
ωμ
β
Eoy = 2 (mπ / a ) H o , H ox = 2 (mπ / a) H o
h
Eoy
h
ωμ 2π × 6 × 109 × 4π × 10−7
ηTE =
=
=
= 4π 2 × 100
β
12
H ox
H ox =
Eoy
ηTE
=
5
= 1.267 mA/m
4π × 100
2
H x = −1.267sin(mπ x / a ) cos(nπ y / b)sin(ωt − β z ) mA/m
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Sadiku & Kulkarni
Principles of Electromagnetics, 6e
396
Prob. 12.15 (a) Since m=2, n=3, the mode is TE23.
β = β ' 1 − ( fc / f )2 =
(b)
2π f
1 − ( fc / f )2
c
But
u'
3 × 108
2
2
( m / a ) + ( n / b) =
(2 / 2.86) 2 + (3 / 1.016) 2 = 46.19 GHz, f = 50 GHz
fc =
−2
2
2 × 10
β=
2π × 50 × 109
1 − (46.19 / 50) 2 = 400.68 rad/m
8
3 × 10
γ = j β = j 400.7 /m
(c) η =
η'
1 − ( fc / f )
2
=
377
1 − (46.19 / 50) 2
= 985.3Ω
Prob. 12.16 In free space,
η1 =
η1 =
η2 =
ηo
1 − ( fc / f )2
377
1 − (3 / 8) 2
,
c
3 × 108
=
= 3 GHz
fc =
2a 2 × 5 × 10−2
= 406.7Ω
η '1
1 − ( f c / f )2
,η ' =
120π
u'
c
,u ' =
= 80π , f c =
2a
2.25
εr
3 × 108
80π
=
2
GHz,
=
= 259.57Ω
η
2
2 × 5 × 10−2 2.25
1 − (2 / 8) 2
η −η
Γ = 2 1 = −0.2208
η2 + η1
fc =
s=
1+ | Γ |
= 1.5667
1− | Γ |
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Sadiku & Kulkarni
Principles of Electromagnetics, 6e
397
Prob. 12.17 Substituting Ez = RΦZ into the wave equation,
ΦZ d
RZ
( ρ R ') + 2 Φ ''+ RΦZ ''+ k 2 RΦZ = 0
ρ dρ
ρ
Dividing by RΦZ ,
1 d
Φ ''
Z ''
( ρ R ') +
+ k2 = −
= −k z 2
2
Rρ d ρ
Z
Φρ
i.e.
Z ''− k z 2 Z = 0
1 d
Φ ''
( ρ R ') +
+ (k 2 + k z 2 ) = 0
2
Rρ d ρ
Φρ
ρ d
Φ ''
( ρ R ') + (k 2 + k z 2 ) ρ 2 = −
= kφ 2
R dρ
Φ
or
Φ ''+ kφ 2Φ = 0
ρ
d
( ρ R ') + (k ρ 2 ρ 2 − kφ 2 ) R = 0 , where k ρ 2 = k 2 + k z 2 . Hence
dρ
ρ 2 R ''+ ρ R '+ (k ρ 2 ρ 2 − kφ 2 ) R = 0
Copyright © 2015 by Oxford University Press
Sadiku & Kulkarni
Principles of Electromagnetics, 6e
398
Prob. 12.18
(a)
For TE10 mode,
fc =
3 × 108
u'
=
= 2.083 GHz
2a 2 × 7.2 × 10−2
2
2
 f 
 2.083 
Let F = 1 −  c  = 1 − 
 = 0.942
 6.2 
 f 
β = ω με F =
ωF
=
2π × 6.2 × 109 × 0.942
= 122.32 rad/m
3 × 108
c
ω c 3 × 108
up = = =
= 3.185 × 108 m/s
β F 0.942
u g = u ' F = 3 × 108 (0.942) = 2.826 × 108 m/s
ηTE =
η'
F
=
377
= 400.21 Ω
0.942
(b)
1
u' =
fc =
με
=
c
εr
=
3 × 108
= 2 × 108
2.25
2 × 108
u'
=
= 1.389 GHz
2a 2 × 7.2 × 10−2
2
2
 f 
 1.389 
Let F = 1 −  c  = 1 − 
 = 0.9746
 6.2 
 f 
β = ω με F =
ωF εr
=
2π × 6.2 × 109 × 0.9746 × 1.5
= 189.83 rad/m
3 × 108
c
ω 2π × 6.2 × 109
= 2.052 × 108 m/s
up = =
β
189.83
u g = u ' F = 2 × 108 (0.9746) = 1.949 × 108 m/s
ηTE =
η'
F
=
377
= 257.88 Ω
1.5 × 0.9746
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Sadiku & Kulkarni
Principles of Electromagnetics, 6e
399
Prob. 12.19
1
f c10 =
με
u'
c
3 × 108
=
=
=
= 1.315 GHz
2a
2a
2a ε r 2 × 7.214 × 10−2 2.5
2
2
 f 
 1.315 
Let F = 1 −  c  = 1 − 
 = 0.9444
 4 
 f 
β = ω με F =
up =
ω εr F
c
c
3 × 108
ω
=
=
= 2.009 × 108 m/s
β F ε r 0.9444 × 2.5
cF
ug = u ' F =
εr
=
3 × 108 × 0.9444
2.5
= 1.792 × 108 m/s
Prob. 12.20
fc =
u'
2a
2
 f 
ug = u ' 1 −  c 
 f 
2
⎯⎯
→


2
2
8 

 ug 
 fc 
1.8 × 10
 = 0.208
  =1−   =1− 
8
 3 × 10 
 f 
 u' 


2.2 

f c = 0.208 f = 2.0523 GHz
a=
u'
3 × 108
=
= 4.927 cm
2 f c 2 2.2 × 2.053 × 109
Prob. 12.21
Let F = 1 − ( f c / f ) 2 = 1 − (16 / 24) 2 = 0.7453
u' =
1
με
=
3 × 108
= 2 × 108 ,
2.25
up =
u'
,
F
u g = u ' F = 2 × 108 × 0.7453 = 1.491 × 108
m/s
ηTE = η '/ F =
377
= 337.2Ω
1.5 x0.7453
Copyright © 2015 by Oxford University Press
Sadiku & Kulkarni
Principles of Electromagnetics, 6e
400
Prob. 12.22
3 × 108
(m / 0.025) 2 + (n / 0.01) 2 = 15 n 2 + (m / 2.5) 2 GHz
fc =
2
fc10 = 6 GHz, fc20 = 12 GHz, fc01 = 15 GHz.
Since fc20 , fc10 > 11 GHz, only the dominant TE10 mode is propagated.
u
1
1
=
= 1.193
(a) p =
2
u
1 − ( fc / f )
1 − (6 / 11) 2
(b)
ug
u
= 1 − (6 / 11) 2 = 0.8381
Prob. 12.23
For the TE10 mode,
 π x  − jβ z
H zs = H o cos 
e
 a 
jβ a
 π x  − jβ z
H xs =
H o sin 
e
π
 a 
 π x  − jβ z
H o sin 
e
π
 a 
Exs = 0 = Ezs = H ys
E ys = −
jωμ a
E s × H s* =
0
H
*
xs
0
E ys
0
jωμ a
H
*
zs
= E ys H zs* a x − E ys H xs* a z
ωμβ a 2 2  π x 
πx  πx 
H cos 
H o sin 
=−
 sin 
 ax +
 az
π
π2
 a   a 
 a 
ωμβ a 2 2 2  π x 
1
*
H o sin 
Pave = Re  E s × H s  =
 az
2
2π 2
 a 
2
2
o
Prob. 12.24
Pave =
| Exs |2 + | E ys |2
2η
ω 2 μ 2π 2 2 2
H o sin π y / ba z
az =
2η b 2 h 4
where η = ηTE10 .
Pave =  Pave .dS =
Pave =
ω 2 μ 2π 2 2 a b
H o   sin 2 π y / bdxdy
2 4
2η b h
x =0 y =0
ω 2 μ 2π 2 2
H o ab / 2
2η b 2 h 4
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Sadiku & Kulkarni
Principles of Electromagnetics, 6e
401
h 2 = (mπ / a ) 2 + (nπ / b) 2 =
But
Pave =
π2
b2
,
ω 2 μ 2 ab3 H o 2
4π 2η
Prob. 12.25
πμ f
π × 12 × 109 × 4π × 10−7
=
= 2.858 × 10−2
Rs =
7
σc
5.8 × 10
f c10 =
u'
3 × 108
=
= 4.651 GHz
2a 2 2.6 × 2 × 10−2
1/ 2
u' 1
1
f c11 =  2 + 2  = 10.4 GHz
2 a
b 
μ 377
=
= 233.81Ω
η'=
ε
2.6
(a) For TE10 mode, eq.(12.57) gives
α d + j β d = −ω 2 με + k x 2 + k y 2 + jωμσ d
= −ω 2 / u 2 +
π2
a2
+ jωμσ d
2
 2π × 12 × 109 
π2
(2.6)
= −
+
+ j 2π × 12 × 109 × 4π × 10−7 × 10−4

−2 2
8
(2 × 10 )
 3 × 10

= 0.012682 + j373.57
α d = 0.012682 Np/m
αc =
=
 1 b fc 2 
 + ( ) 
bη ' 1 − ( f c / f )  2 a f 
2 Rs
2
2 x 2.858 x10−2
 1 1 4.651 2 
)  = 0.0153 Np/m
 + (

10−2 (233.81) 1 − (4.651 / 12) 2  2 2 12
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Sadiku & Kulkarni
Principles of Electromagnetics, 6e
402
(b) For TE11 mode,
α d + j β d = −ω 2 / u 2 + 1 / a 2 + 1 / b 2 + jωμσ d
= −139556.21 +
π2
(10−2 ) 2
+ j 9.4748 = 0.02344 + j 202.14
α d = 0.02344 Np/m
αc =
 (b / a )3 + 1 
 (1 / 8) + 1 
2 × 2.858 × 10−2
=




2
bη ' 1 − ( f c / f ) 2  (b / a) + 1  10−2 (233.81) 1 − (10.4 / 12) 2  (1 / 4) + 1 
2 Rs
α c = 0.0441 Np/m
Prob. 12.26
ε c = ε '− jε '' = ε − j
Comparing this with
σ
ω
ε c = 16ε o (1 − j10−4 ) = 16ε o − j16ε o × 10−4
σ
= 16ε o x10−4
ε = 16ε o ,
ω
For TM21 mode,
1/ 2
u '  m2 n2 
fc =  2 + 2 
2 a
b 
= 2.0963 GHz,
f = 1.1 f c = 2.3059 GHz
σ = 16ε oω × 10−4 = 16 × 2π × 2.3059 × 109 ×
η'=
αd =
10−9
× 10−4 = 2.0525 × 10−4
36π
μ
= 30π
ε
ση '
2 1 − ( fc / f )2
Eo e −α d z = 0.8 Eo
=
4.1 × 10−4 × 30π
= 0.0231 Np/m
2 1 − 1 / 1.12
z=
1
αd
ln(1 / 0.8) = 9.66 m
Copyright © 2015 by Oxford University Press
Sadiku & Kulkarni
Principles of Electromagnetics, 6e
403
Prob. 12.27
For TM21 mode,
αc =
2 Rs
bη ' 1 − ( f c / f ) 2
π fμ
π × 2.3059 × 109 × 4π × 10−7
Rs =
=
=
= 0.0246
σ cδ
σc
1.5 × 107
1
αc =
2 × 0.0246
= 0.0314 Np/m
4π × 10−2 × 30π × 0.4166
Eo e − (αc +α d ) z = 0.7 Eo
z=
1
ln(1 / 0.7) = 6.5445 m
αc + αd
Prob. 12.28
u'
3 × 108
=
= 2.5 GHz
2a 2 × 6 × 10−2
377
η'
=
=
= 483
2
2
 f 
 2.5 
1− 
1−  c 

 4 
 f 
f c10 =
ηTE
From Example 12.5,
Pave =
Eo2 ab (2.2) 2 × 106 × 6 × 3 × 10−4
=
= 9.0196 mW
2η
2 × 483
Prob. 12.29 For TE10 mode,
u'
3 × 108
fc =
=
= 2.151 GHz
2a 2 2.11 × 4.8 × 10−2
(a) loss tangent =
σ
=d
ωε
σ = dωε = 3 × 10−4 × 2π × 4 × 109 × 2.11 ×
η'=
120π
= 259.53
2.11
10−9
= 1.4086 × 10−4
36π
Copyright © 2015 by Oxford University Press
Sadiku & Kulkarni
Principles of Electromagnetics, 6e
404
αd =
ση '
2 1 − ( fc / f )
2
=
1.4067 × 10−4 × 259.53
2 1 − (2.151 / 4)
2
= 2.165 × 10−2 Np/m
μ fπ
π × 4 × 109 × 4π × 10−7
(b) Rs =
=
= 1.9625 × 10−2
7
σc
4.1 × 10
αc =
3.925 × 10−2 (0.5 + 0.5 × 0.2892)
1 b
2
+
=
(
f
/
f
)
c


2.4 × 10−2 × 259.53 × 0.8431

bη ' 1 − ( f c / f ) 2  2 a
2 Rs
= 4.818 × 10−3 Np/m
Prob. 12.30
αc =
=
2 Rs

bη ' 1 − 

π fμ
2
 1 b  f 2 
σc
 +  c =
2
2
 fc 
f c   2 a  f  
bη ' 1 −  
f 
 f 
1
2 4π × 10−7 × π f ×
2

0.5 × 10−2 × (120π / 2.25) 5.8 × 107 1 − 

=
10−5 f

30 (5.8 / 2.25) 1 − 

 1 1  f 2 
 +  c 
 2 2  f  
  f 2 
1 +  c  
2
f c    f  
f 
  f 2 
1 +  c  
2
f c    f  
f 
The MATLAB code is shown below
k=10^(-5)/(30*sqrt(5.8/2.25));
fc=10^10;
for n=1:1000
f(n)=fc*(n/100+1);
fn=f(n);
num=sqrt(fn)*(1 +(fc/fn)^2);
den=sqrt(1- (fc/fn)^2);
alpha(n) =k*num/den;
end
plot(f/10^9,alpha)
xlabel('frequency (GHz)')
ylabel('attenuation')
grid
Copyright © 2015 by Oxford University Press
Sadiku & Kulkarni
Principles of Electromagnetics, 6e
405
The plot of attenuation versus frequency is shown below.
attenuation (Np/m)
0.3
0.25
0.2
0.15
0.1
0.05
0
0
10
20
30
40
50
60
frequency (GHz)
Prob. 12.31 (a) For TE10 mode,
fc =
u'
,
2a
c
2.11
u' =
3 × 108
fc =
= 4.589 GHz
2.11(2 × 2.25 × 10−2 )
(b)
α cTE10 =
1 b

+ ( f c / f )2 


bη ' 1 − ( f c / f ) 2  2 a
2 Rs
π fμ
π × 5 × 109 × 4π × 10−7
Rs =
=
= 3.796 × 10−2
7
σc
1.37 × 10
η' =
377
= 259.54
2.11
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70
80
90
100
Sadiku & Kulkarni
Principles of Electromagnetics, 6e
406
1.5
(4.589 / 5) 2 ]
2.25
= 0.05217 Np/m
−4
1.5 × 10 (259.54) 1 − (4.589 / 5) 2
2 × 3.796 × 10−2 [0.5 +
αc =
Prob. 12.32
(a)
f c10
u'
3 × 108
=
=
= 3.947 GHz
2a 2 × 3.8 × 10−2
2
 f 
u g = u ' 1 −  c  = 3 × 108 1 − (0.3947) 2 = 2.756 × 108 m/s
 f 
(b) α = α d + α c
α d = 0 since the guide is air-filled.
π fμ
π × 1010 × 4π × 10−7
=
= 2.609 × 10−2 Ω
σc
5.8 × 107
Rs =
2

b  fc  
αc =
 0.5 +   
2
a  f  
 f c  
bη ' 1 −  
 f 
2 Rs
=
2 × 2.609 × 10−2
1.6 × 10−2 (377) 1 − ( 0.3947 )
2
1.6
5.218 × 0.5656
2

0.5 + 3.8 ( 0.3947 )  =
554.23


= 5.325 × 10−3 Np/m
α c (dB) = 8.686 × 5.325 × 10−3 = 0.04626 dB/m
Prob. 12.33
f c10 =
β' =
ω
u'
=
u'
c
3 × 108
=
=
= 3.991 GHz
2a 2a ε r μr 2 × 2.5 × 10−2 2.26
2π f ε r
c
2
2
 f 
 3.991 
F = 1−  c  = 1− 
 = 0.8467
 7.5 
 f 
β = β 'F =
2π × 7.5 × 109 2.26
0.8467 = 199.94 rad/m
3 × 108
Copyright © 2015 by Oxford University Press
Sadiku & Kulkarni
Principles of Electromagnetics, 6e
407
αd =
αc =
Rs =
ση '
2F
=
σηo
10−4 (377)
=
= 1.481 × 10−2 Np/m
2 F ε r 2 × 0.8467 2.26
2 Rs

bη ' 1 − 

2

b  fc  
 0.5 +   
2
a  f  
f c  
f 
π fμ
π × 7.5 × 109 × 4π × 10−7
=
= 0.0519
σc
1.1 × 107
2

1.5  3.991  
2 × 0.0519 0.5 +

 
2.5  7.5   0.1038 × 0.6698

αc =
=
377
3.1848
−2
1.5 × 10 ×
× 0.8467
2.66
= 0.02183 Np/m
up =
u'
c
3 × 108
=
=
= 2.357 × 108 m/s
F F ε r 0.8467 2.26
3 × 108 × 0.8467
= 1.689 × 108 m/s
2.26
3 × 108
u'
c
λc = =
=
= 0.05 m = 5 cm(= 2a, as expected)
f c f c ε r 3.991 × 109 2.26
ug = u ' F =
Prob. 12.34
The cutoff frequency of the dominant mode is
u
3 × 108
f c10 =
=
= 6.56 GHz
2a 4.576 × 10−2
The surface resistance is
Rs =
π fμ
π × 8.4 × 109 × 4π × 10−7
=
= 23.91 × 10−3
5.8 × 107
σc
For TE10 mode,
αc =
2 Rs
2

b  fc  
 0.5 +   
2
a  f  
f c  

f 

bη ' 1 − 

f c 6.56
=
= 0.781,
f
8.4
η ' = ηo = 377
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2 × 23.91 × 10−3
1.016
2

0.5 +
0.781) 
(

2.286

1.016 × 10−2 × 377 1 − 0.7812 
−3
47.82 × 10 (0.5 + 0.2711)
=
= 15.42 × 10−3 Np/m
3.83 × 0.6245
=15.42 × 10−3 × 8.686 dB/m = 0.1339 dB/m
αc =
Prob. 12.35 For TE10 mode,
αc =
 1 b fc 2 
 + ( ) 
bη ' 1 − ( f c / f )  2 a f 
2 Rs
2
But a = b, Rs =
1
σ cδ
=
π fμ
σc
π fμ
σc
1
f 
k f  + ( c )2 
f 
1
f 
2
αc =
+ ( c )2  =
2 2
f 
aη ' 1 − ( f c / f ) 
1 − ( fc / f )2
2
where k is a constant.
f c 2 1/ 2 1 −1/ 2 3 2 −5/ 2 k 1 1/ 2
f
) ] [ f
] − [ f + f c 2 f −3/ 2 ](2 f c 2 f −3 )[1 − ( c ) 2 ]−1/ 2
− fc f
dα c
f
4
2
2 2
f
=
2
df
1 − ( fc / f )
dα c
For minimum value,
= 0 . This leads to f = 2.962 fc.
df
k[1 − (
Prob. 12.36
For the TE mode to z,
Ezs = 0, H zs = H o cos(mπ x / a ) cos(nπ y / b)sin( pπ z / c)
E ys = −
γ ∂ Ezs jωμ ∂ H zs
jωμ
+ 2
= − 2 (mπ / a ) H o sin(mπ x / a )cos(nπ y / b)sin( pπ z / c)
2
∂x
h ∂y
h
h
as required.
Exs = −
γ ∂ Ezs jωμ ∂ H zs
jωμ
− 2
= 2 (nπ / b) H o cos(mπ x / a )sin(nπ y / b)sin( pπ z / c)
2
h ∂x
h ∂y
h
From Maxwell’s equation,
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∂
− jωμ H s = ∇ × E s = ∂ x
∂
∂y
∂
∂z
Exs
E ys
0
H xs =
1 ∂ E ys
1
= − 2 (mπ / a )( pπ / c ) H o sin(mπ x / a ) cos(nπ y / b) cos( pπ z / c)
jωμ ∂ z
h
Prob. 12.37 Maxwell’s equation can be written as
jωε ∂ Ezs γ ∂ H zs
−
h2 ∂ y h2 ∂ x
For a rectangular cavity,
H xs =
h 2 = k x 2 + k y 2 = (mπ / a ) 2 + (nπ / b) 2
For TM mode, Hzs = 0 and
Ezs = Eo sin(mπ x / a )sin(nπ y / b) cos( pπ z / c)
Thus
jωε ∂ Ezs
jωε
= 2 (nπ / b) Eo sin(mπ x / a)cos(nπ y / b) cos( pπ z / c)
2
h ∂y
h
as required.
H xs =
H xs = −
jωε ∂ Ezs γ ∂ H zs
−
h2 ∂ x h2 ∂ y
jωε
(mπ / a ) Eo cos(mπ x / a )sin(nπ y / b) cos( pπ z / c)
h2
From Maxwell’s equation,
=−
∂
jωε E s = ∇ × H s = ∂ x
∂
∂y
∂
∂z
H xs
H ys
0
E ys =
1 ∂ H xs
1
= 2 (nπ / b)( pπ / c) Eo sin(mπ x / a )cos(nπ y / b)cos( pπ z / c)
jωε ∂ z
h
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Prob. 12.38
fr =
u'
( m / a ) 2 + ( n / b) 2 + ( p / c ) 2
2
where for TM mode to z, m = 1, 2, 3,…, n=1, 2, 3, …., p = 0, 1, 2, ….
and for TE mode to z, m = 0,1, 2, 3,…, n=0,1, 2, 3, …., p = 1, 2, 3, … , (m + n) ≠ 0 .
(a) If a < b < c, 1/a > 1/b > 1/c,
u' 1
1
+ 2
2
2 a
b
u' 1 1 u' 1
1
The lowest TE mode is TE011 with f r =
+ 2 <
+ 2
2
2
2 b
c
2 a
b
The lowest TM mode is TM110 with f r =
Hence the dominant mode is TE011.
(b) If a > b > c, 1/a < 1/b < 1/c,
u' 1
1
+ 2
2
2 a
b
u' 1
1 u' 1
1
The lowest TE mode is TE101 with f r =
+ 2 >
+ 2
2
2
2 a
c
2 a
b
The lowest TM mode is TM110 with f r =
Hence the dominant mode is TM110.
(c) If a = c > b, 1/a = 1/c < 1/b,
u' 1
1
+ 2
2
2 a
b
u' 1
1 u' 1
1
The lowest TE mode is TE101 with f r =
+ 2 <
+ 2
2
2
2 a
c
2 a
b
The lowest TM mode is TM110 with f r =
Hence the dominant mode is TE101.
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Prob. 12.39
(a) Since a > b < c, the dominant mode is TE101.
c 1
1 3 × 108
fr =
+
=
2 a2 c2
2
δ=
(b)
1
π f r101μoσ
=
1
1
+
= 9.014 GHz
−2 2
(2 × 10 )
(3 × 10−2 ) 2
1
π × 9.014 × 10 × 4π × 10−7 × 5.8 × 107
9
1
=
4π 2 × 5.8 × 9.014 × 109
(4 + 9)6 × 10−2
0.78
(a 2 + c 2 )abc
Q=
=
=
δ  2b(a 3 + c 3 ) + ac(a 2 + c 2 )  δ [ 2(8 + 27) + 6(4 + 9)] 148δ
=
0.78
4π 2 × 5.8 × 9.014 × 109 = 7571.5
148
Prob. 12.40
u' =
1
με
=
c
3 × 108
=
= 1.897 × 108
2.5
2.5
2
2
2
2
2
u '  m   n   p  1.897 × 108 × 102  m   n   p 
fc =
  +  +  =
  +  + 
2  a  b  c 
2
 1  2  3 
= 9.485 m 2 + 0.25n 2 + 0.111 p 2 GHz
f r101
= 9.485 1 + 0 + 0.111 = 10 GHz
f r 011
= 9.485 0 + 0.25 + 0.111 = 5.701 GHz
f r 012
= 9.485 0 + 0.25 + 0.444 = 7.906 GHz
f r 013
= 9.485 0 + 0.25 + 0.999 = 10.61 GHz
= 9.485 0 + 1 + 0.111 = 10 GHz
f r 021
Thus, the first five resonant frequencies are:
5.701 GHz(TE 011 )
7.906 GHz (TE 012 )
10 GHz (TE101 and TE 021 )
10.61 GHz (TE 013 or TM110 )
11.07 GHz (TE111 or TM111 )
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Prob. 12.41
Q=
(a 2 + c 2 )abc
δ  2b(a 3 + c3 ) + ac(a 2 + c 2 ) 
When
Q=
a = b = c,
a
2a 2 a 3
2a 5
=
=
4
3
2
2
3δ
δ  2a × 2a + a × 2a  6δ a
Prob. 12.42
(a) Since a > b < c, the dominant mode is TE101
f r101 =
u' 1
1 3 × 108 × 102
+
+
=
0
c2
2 a2
2
(b) QTE101 =
=
But
1 1
+ = 16.77 GHz
22 12
(a 2 + c 2 )abc
δ  2b(a 3 + c3 ) + ac(a 2 + c 2 ) 
(400 + 100)20 × 8 × 10 × 10−3
3.279 × 10−3
=
δ [16(8000 + 1000) + 200(400 + 100)]
δ
δ=
1
π f r101μoσ
QTE101 = 3.279 × 10−3
=
1
π 16.77 × 109 × 4π × 10−7 × 6.1 × 107
=
10−4
200.961
200.961
= 6589.51
10−4
Prob. 12.43
c
m2 + n2 + p 2
2a
The lowest possible modes are TE101, TE011, and TM110. Hence
fr =
fr =
c
2
2a
a=
c
fr 2
=
3 × 108
= 7.071 cm
2 × 3 × 109
a = b = c = 7.071 cm
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Prob. 12.44
(a) a = b = c
u'
fr =
m2 + n2 + p 2
2a
For the dominant mode TE101 ,
u'
c
1+1 =
2
2a
2a
c 2
3 × 108 2
a=
=
= 0.03788 m
2 f r 2 × 5.6 × 109
fr =
a = b = c = 3.788 cm
(b)
For ε r = 2.05,
a=
c
u' =
εr
c 2
0.03788
=
= 0.02646
2 fr ε r
2.05
a = b = c = 2.646 cm
Prob. 12.45
(a)
This is a TM mode to z. From Maxwell’s equations,
∇ × E s = − jωμ H s
Hs = −
1
jωμ
∇ × Es =
j
ωμ
∂
∂x
∂
∂y
∂
∂z
0
0
Ezs ( x, y )
=
j  ∂ Ezs
∂ Ezs 
ax −
ay 

ωμ  ∂ y
∂x

But
Ezs = 200sin 30π x sin 30π y,
1
ωμ
=
1
10−2
=
6 × 109 × 4π × 10−7 24π
j10−2
Hs =
× 200 × 30π {sin 30π x cos30π ya x − cos30π x sin 30π ya y }
24π
H = Re (Hs e jωt )
H = 2.5{− sin 30π x cos30π ya x + cos30π x sin 30π ya y } sin 6 × 109 π t A/m
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(b)
E = Ez a z ,
H = H xa x + H y a y
EH = 0
Prob. 12.46
(a) a = b = c
a=
⎯⎯
→
f r101 =
3 × 108
= 12 × 109
a 2
3 × 108
= 1.77 cm
2 × 12 × 109
(b) QTE101 =
=
a a π f r101μσ
=
3δ
3
1.77 × 10−2 π × 12 × 109 × 4π × 10−7 × 5.8 × 107
= 9767.61
3
Prob. 12.47
2
2
u'  m   n   p 
fr =
  +  + 
2  a  b  c 
f r101 =
3 × 108
2
2
1
1
+
= 44.186 MHz
2
(10.2)
(3.6) 2
f r 011 = 150
1
1
MHz = 45.093 MHz
+
2
(8.7)
(3.6) 2
f r111 = 150
1
1
1
MHz = 47.43 MHz
+
+
2
2
(10.2)
(8.7)
(3.6) 2
f r110 = 150
1
1
MHz = 22.66 MHz
+
2
(10.2)
(8.7) 2
f r102 = 150
1
4
MHz = 84.62 MHz
+
2
(10.2)
(3.6) 2
f r 201 = 150
4
1
MHz = 51 MHz
+
2
(10.2)
(3.6) 2
Thus, the resonant frequences below 50 MHz are
f r110 , f r101, f r 011 , and f r111
Prob. 12.48
n = c/um =
3 × 108
= 1.4286
2.1 × 108
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Prob. 12.49 When an optical fiber is used as the transmission medium, cable radiation is
eliminated. Thus, optical fibers offer total EMI isolation because they neither emit nor
pick up EM waves.
Prob. 12.50
(a) NA =
n12 − n2 2 =
1.622 − 1.6042 = 0.2271
(b) NA = sin θ a = 0.2271 or θ a = sin –1 0.2271 = 13.13o
(c) V =
πd
π × 50 × 10−6 × 0.2271
= 27.441
NA =
1300 × 10−9
λ
N = V2/2 6 modes
Prob. 12.51
π d 2 3 π × 2 × 5 × 10−6
1.482 − 1.462 = 5.86
V=
n1 − n2 =
−9
1300 × 10
λ
2
V
= 17.17 or 17 modes
N=
2
Prob. 12.52
(a) NA = sin θ a =
n12 − n2 2 = 1.532 − 1.452 = 0.4883
θ a = sin –1 0.4883 = 29.23o
(b) P(l)/P(0) = 10- α l / 10 = 10-0.4X5/10 = 0.631
i.e. 63.1 %
Prob. 12.53
P () = P(0)10−α  /10 = 10 × 10−0.5×0.85/10 = 9.0678 mW
Prob. 12.54 As shown in Eq. (10.35), log10 P1/P2 = 0.434 ln P1/P2 ,
1 Np = 20 log10 e = 8.686 dB or 1 Np/km = 8.686 dB/km,
or 1Np/m = 8686 dB/km. Thus,
α12 = 8686α10
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Prob. 12.55
α  = 10log10
Pin
1.2 × 10−3
= 10log10
= 30.792
1 × 10−6
Pout
0.4
Np/km
8.686
30.792 30.392 dB
=
= 76.98 km
=
0.4 dB/km
α
α = 0.4 dB/km =
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CHAPTER 13
P.E. 13.1
(a) For this case, r is at near field.
I dl sin θ  j β 1  − j β r
2π λ
. = 72o
+ 2 e
Hφ s = o
, βr =

λ 5
4π
r 
 r
2π c
2π × 3 × 108
λ=
=
= 6π ,
108
ω
Hφ s =
(0.25)
β=
2π
λ
=
1
3
6π
sin 30o 
 − j 72o
1
j1 / 3
100
e
+
= 0.2119∠ − 20.511o mA/m

2 
4π
 6π / 5 (6π / 5) 
H = Im ( Hφ s e jωt aφ )
Im is used since I = Io sin ωt
= 0.2119sin(108 − 20.5o )aφ mA/m
(b) For this case, r is at far field. β =
Hφ s =
j (0.25)(
2π
)
λ
λ
Sin60o e − j 0
λ 100
4π (6π × 200)
H = Im ( H φ s aφ e jωt )
2π
× 200λ = 0o
o
= 0.2871e j 90 μ Am
o
= 0.2871sin(108 + 90o )aφ
P. E. 13.2
(a) l =
λ
4
= 1.5m ,
(b) Io = 83.3mA
1
(c) Rrad = 36.56 Ω , Prad = (0.0833)2 36.56
2
= 126.8 mW.
(d) ZL = 36.5 + j21.25,
Γ=
36.5 + j 21.25 − 75
= 0.3874∠140.3o
36.5 + j 21.25 + 75
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s=
1 + 0.3874
= 2.265
1 − 0.3874
P.E. 13.3
D=
4π U max
Prad
(a) For the Hertzian monopole
U (θ ,φ ) = sin 2 θ , 0 < θ < π / 2,
π
Prad =
2 2π
  sin
θ φ
2
sin θ dθ dφ =
=0 =0
D=
(b) For the
λ
4
0 < φ < 2π , Umax = 1
4π
3
4π (1)
=3
4π
3
monopole,
π
cos 2 ( cosθ )
2
U (θ ,φ ) =
, Umax = 1
sin 2 θ
π
Prad
cos 2 ( cosθ )
2
=  
sin θ dθ dφ = 2π (0.609)
sin 2 θ
θ =0 φ =0
D=
4π (1)
= 3.28
2π (0.609)
π
2 2π
P. E. 13.4
(a) Prad = η r Pin = 0.95(0.4)
4π U max
4π (0.5)
=
= 16.53
D=
Prad
0.4 × 0.95
4π (0.5)
(b) D =
= 20.94
0.3
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P. E. 13.5
π
Prad =
2 2π
  sin θ sin θ dθ dφ =
θ =0 φ =0
D=
4π (1)
π2
π2
2
, Umax = 1
= 2.546
2
P. E. 13.6
1

(a) f (θ ) = cosθ cos  ( β d cosθ + α ) 
2

2π λ
. =π
where α = π , β d =
λ 2
1

f (θ ) = cosθ cos  (π cosθ + π ) 
2

unit pattern
group pattern
For the group pattern, we have nulls at
π (cosθ + 1) = ± π
2
2
θ = ±π 2
and maxima at
π (cosθ + 1) = 0, π
2
cosθ = −1,1
⎯⎯
→ θ = 0, π
Thus the group pattern and the resultant patterns are as shown in Fig.13.15(a)
1

(b) f (θ ) = cosθ cos  ( β d cosθ + α ) 
2

−
π
, βd = π / 2
where α =
2
1  π

f (θ ) = cosθ cos   cosθ − π  
2

2 2
unit pattern
group pattern
For the group pattern, the nulls are at
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π (cosθ − 1) = − π
4
2
θ = 180o
cos θ − 1 = 0
θ =0
and maxima at
Thus the group pattern and the resultant patterns are as shown in Fig.13.15(b)
P. E. 13.7
(a)
●
●
●
λ
●
●
●
●
●
●
λ
2
● 2 ●
λ ●●
2
1: 2 :1
1: 2 :1
x
x
λ
2
Thus, we take a pair at a time and multiply the patterns as shown below.
●
×
●
x
≡
x
(b) The group pattern is the normalized array factor, i.e.
1
N ( N − 1) i 2ψ N ( N − 1)( N − 2) i 3ψ
( AF ) n =
1 + Neiψ +
e +
e + ............ + ei ( N −1)ψ
2!
3!

where
N −1
N
i −1

=  i  = 1+ N +

N − 1 N ( N − 1)( N − 2)
+
+ ...........
2!
3!
= (1 + 1) N −1 = 2 N −1
( AF ) n =
1
2 N −1
1 + e jψ
N −1
=
1
2 N −1
e
jψ
2
e
− jψ
2
+e
jψ
N −1
2
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=
1
2 N −1
2cos
ψ
N −1
= cos
2
N −1
ψ
2
P. E. 13.8
λ2
c 3 × 108
Ae =
Gd , λ = =
=3m
f
108
4π
For the Hertzian dipole,
Gd = 1.5sin 2 θ
λ2
(1.5sin 2 θ )
4π
1.5λ 2 1.5 × 9
=
= 1.074 m 2
Ae,max =
4π
4π
Ae =
By definition,
Pr 3 × 10−6
Pave =
=
Ae
1.074
Pr = Ae Pave
= 2.793 μ W / m 2
P. E. 13.9
(a) Gd =
4π r 2 Pave
=
Prad
4π r 2
1 E2
2 η
Prad
=
2π r 2 E 2
η Prad
2π × 400 × 106 × 144 × 10−6
=
= 0.0096
120π × 100 × 103
G = 10log10 Gd = -20.18 dB
(b) G = η r Gd = 0.98 × 0.0096 = 9.408 × 10−3
P. E. 13.10
1
 λ 2Gd 2σ Prad  4
r=

3
 (4π ) Pr 
c 3 × 108
where λ = =
= 0.05m
f 6 × 109
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Ae = 0.7π a 2 = 0.7π (1.8) 2 = 7.125m 2
Gd =
4π Ae
λ
4π (7.125)
= 3.581 × 104
−4
25 × 10
=
2
 25 × 10−4 × (3.581) 2 × 108 × 5 × 60 × 103 
r=

(4π )3 × 0.26 × 10−3


1
4
= 1168.4m = 0.631 nm
At r =
P=
rmax
2
= 584.2m,
Gd Prad 3.581 × 104 × 60 × 103
=
= 501 W/m 2
4π r 2
4π (584.2) 2
Prob. 13.1
Using vector transformation,
Ars = Axs sin θ cos φ , Aθ s = Axs cosθ cos φ , Aφ s = − Axs sin φ
As =
50e− j β r
(sin θ cos φ ar + cosθ cos φ aθ − sin φ aφ )
r
∇ × As
μ
−
= Hs =
100cosθ sin φ − j β r
50
e ar − 2 (1 − j β r )sin φ e − j β r aθ
2
μ r sin θ
μr
50
cosθ cos φ (1 + j β r )e − j β r aφ
2
μr
At far field, only
1
term remains. Hence
r
j 50 − j β r
β e (sin φ aθ − cosθ cos φ aφ )
μr
− j 50 βη e − j β r
(sin φ aφ + cosθ cos φ aθ )
E s = −η ar × H s =
μr
Hs =
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H
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& Kulkarni
Kulkarni
= Re  H s e jωt  =
Prob. 13.2
(a)
λ=
−50
β sin(ωt − β r )(sin φ aθ − cosθ cos φ aφ ) A/m
μr
3 × 108
c
=
= 0.75 m
f 400 × 106
423
421
2
λ50)ηβ
2d 2 2(0.02
jω t
rEmin==Re  E=s e  =
sin(
r ωt − β r )(sin φ aφ + cosθ cos φ aθ ) V/m
μr
λ
λ
λ
λ
i.e. r is in the far field.
−50
jω t
 = − j β r β sin(ωt − β r )(sin φ aθ − cosθ cos φ aφ ) A/m
H = RejIoH
β dl
se
Hφ s =
sin θ e μ r
4π r
Prob. 13.2
2π
3 ×8
× 0.02λ × sin 90o
3
×
10
c
β
I
dl
λ= 0.75 m
= θ=
|(a)
H φ s |= λ o= sin
= 5 × 10−4 = 0.5 mA/m
6
4π rf 400 × 10
4π (60)
2
2
λ ) V/m
|Eθ s |= 2ηdo | H φ2(0.02
|= 0.1885
rmin =
= s
r
i.e.
inφ sthe
far field.
(b) r is| H
|= 0.5
mA/m|
jI o β dl
2
H =
e − j β r
sinθdl
2
(c)φ s R rad4π= r80π 2   = 80π 2 ( 0.02 ) = 0.3158 Ω
 λ  2π
3 × 1 × 0.02λ × sin 90o
1
2
β
I
dl
| I oθ| =R rad =λ (9)(0.3158) = 1.421
|(d)
H φ s |=Prado = sin
= 5W
× 10−4 = 0.5 mA/m
2 4π (60)
4π r2
|Eθ s |= ηo | H φ s |= 0.1885 V/m
Prob.
13.3
2
2
2  dl 
2  0.024λ 
(a)
(b) R|rad
H φ=s 80
|= π
0.5mA/m|
 = 80π 
 = 0.4548 Ω
 λ 2
 λ 
dl  − 502
2
2  0.4548
L − Zo
(c) Γ =
RZ
) = 0.3158 Ω
−0.982
(b)
rad = 80π =
  = 80π (=0.02
λ  + 50
ZL + Z o  0.4548
1
2
1+ | Γ | 1 1.982
(d)
= (9)(0.3158) = 1.421 W
s = Prad == 2 | I o | R rad= 110.11
2
1− | Γ | 1 − 0.982
Prob. 13.3
Prob. 13.4
2
2
8 
 dl
210
2  0.024λ 
c
3
×
(a) λR=rad ==80π  = 6= m
80π 
 = 0.4548 Ω
 λ 
f 50 × 10λ6 
−
Z 2− Z o
2(5 ×0.4548
10−3 ) 2 50 = −0.982
(b)Γ = 2dL
−6
rmin = Z +=Z = 0.4548 =+ 8.333
50 × 10 m  r = 15 cm
o
6
λL
1+ | Γ | far
1.982
i.e.
field. = 110.11
=
s = r is in the
1− | Γ | 1 − 0.982
Prob. 13.4
c
3 × 108
=6m
λ= =
f 50 × 106
2(5 × 10−3 ) 2 Copyright © 2015−6by Oxford University Press
= 8.333 × 10 m  r = 15 cm
6
λ
i.e. r is in the far field.
rmin =
2d 2
=
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ofElectromagnetics,
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6e
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ofElectromagnetics,
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6e
Principles
424
422
Hφ s =
jI o β dl
sin θ e − j β r
4π r
2π
× 5 × 10−3 × sin 30o
5
I o β dl
6
| H φ s |=
sin θ =
= × 10−2 = 2.778 mA/m
−2
4π r
4π × 15 × 10
18
−3
|Eθ s |= ηo | H φ s |= 377 × 2.778 × 10 = 1.047 V/m
2×
Prob. 13.5
e− jβ r
(a) Azs =
4π r
l
2
I
−l
o
(1 −
2z
l
2
)e j β z cos θ ∂ z
l

2
2z
2z
e − j β r  2
=
Io
(1 −
) cos( β z cosθ )dz + j  (1 −
)sin( β z cos θ ) dz 

l
l
4π r  −l
−l
 2

2
l
l
2
e− j β r
2z
=
2 I o  (1 − ) cos( β z cosθ )dz
l
4π r
0
I oe− jβ r
2
. 1 − cos( β l cosθ ) 
=
2
2
2

2π r β cos θ l 
E s = − jωμ As
Eθ s = jωμ sin θ Azs = j βη sin θ Azs
sin θ 1 − cos( β l cos θ ) 
2


2
β cos θ
βl
( cosθ ) 2
If β l  l , cos( β l cosθ ) = 1 − 2
.
2
2
2!
Hence
jη I o e − j β r
Eθ s =
π rl
jη I o
β le− j β r sin θ , H φ s = Eθ s / η
8π r
2
Eθ s
Pave =
, Prad =  Pave dS
2η
Eθ s =
Prad
2
2π π
n  I βl  1
=    o  2 sin 2 θ r 2 sin θ dθ dφ
0 0 2  8π  r
2
l
= 10π I o   = 1 I o 2 Rrad
2
λ
2
2
l
or Rrad = 20π 2  
λ
2
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423
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Kulkarni
l
(b) 0.5 = 20π 2  
λ
Principlesof
ofElectromagnetics,
Electromagnetics,6e
6e
Principles
2
l = 0.05λ
425
423
Prob. 13.6
c 3 × 108 l 2
 m
λ = 0.5== 20π 26= 100
l = 0.05λ
(b)
f 3 × 10  λ 
 
 = 4 m  λ . Hence the antenna is electrically short.
2
2
Prob. 13.6
1 2
 π I o dl 
 π × 3× 4 
8
Prad =c I 3o R×rad
10 = 40  λ  = 40  100  = 5.685 W
 m 


= 100
λ = 2=
f 3 × 106
=4m
Prob.
13.7 λ . Hence the antenna is electrically short.
Let us model this as a short Hertzian
dipole. 2
2
1 2
 π I o dl 
 π2 × 3 × 4 
Prad = I o Rrad = 40 
dl
 =40
 = 5.685 W
2
Rrad λ= 80π 2   =100
80π 2 (1 / 8) 2 = 12.34 Ω
λ
Prob. 13.7
Prob.
Let us 13.8
model this as a short Hertzian dipole.
2
40Ω
 dl 
Rrad = 80π 2   = 80π 2 (1 / 8)I2 = 12.34 Ω
λ
Prob. 13.8
24 V
+
-
40Ω
I
Zin = 73+j42
+
-
24 V
Zin = 73+j42
V
24
I=
=
= 0.1866 − j 0.0694
Rs + Z in 40 + 73 + j 42
1
| I |2 Rrad ,
Rrad = 73
2
1V
24
I rad
= = (0.1991)
= 2 × 73 = 1.447= W
0.1866 − j 0.0694
P
Rs 2+ Z in 40 + 73 + j 42
Prad =
1
| I |2 Rrad ,
Rrad = 73
2
1
= (0.1991)2 × 73 = 1.447 W
2
Prad =
Prad
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Principlesof
ofElectromagnetics,
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6e
Principles
426
424
Prob. 13.9
Change the limits in Eq. (13.16) to ± l
As =
2
i.e.
μ I oe j β zCosθ ( j β cosθ cos β t + β sin β t )
− β 2 cos 2 θ + β 2
4π r
μ I oe jβ r
1
=
2π r β sin 2 θ
l
2
−l
2
 βl
βl  βl
 βl


sin 2 cos  2 cosθ  − cosθ cos 2 sin  2 cosθ  





But B = μ H = ∇ × A
Hφ s =
1 ∂
∂A
( rAθ ) − r  ,

μr ∂ r
∂θ 
where Ao = − Az sin θ , Ar = Az cosθ
Hφ s =
I o e− j β r  j β   β l
βl  βl
Io − jβ r
 βl


e (......)

 sin cos  cosθ  − cosθ cos sin  cosθ  +
2
2π r β  sinθ  
2
2
 2

 2
 2π r
For far field, only the
Hφ s =
jI o − j β r
e
2π r
1
-term remains. Hence
r
 βl
βl  βl
 βl


sin 2 cos  2 cos θ  − cos θ cos 2 sin  2 cos θ  





sin θ
βl
 βl

cos  cosθ  − cos
2
 2

(b) f (θ ) =
sin θ
For l = λ , f (θ ) =
cos (π cosθ ) + 1
sin θ
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Principlesof
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Principles
427
425
 3π

cos  cosθ 
3λ
 2

For l =
, f (θ ) =
2
sin θ
For l = 2λ , f (θ ) =
cosθ sin ( 2π cosθ )
sin θ
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Principlesof
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Principles
428
426
Prob. 13.10
(a)
c
3 × 108
= 0.6667 m
λ= =
f 450 × 106
=
λ
2
= 0.333 m
(b)
σ
=
ωε
4
10−9
2π × 450 × 10 × 81 ×
36π
= 1.975
6
με 
2
 2π × 460 × 106 81
σ 
 1 + (1.975 )2 + 1
 1 + 
 + 1 =


c
2 
2

 ωε 


β =ω
2π × 460 × 106
× 11.4086 = 109.91
3 × 108
2π
λ=
= 0.0572
=
=
β
λ
2
= 28.58 mm
Prob. 13.11
(a)
c
3 × 108
= 260.8 m
λ= =
f 1.150 × 106
=
λ
4
= 65.22 m
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λ=
=
c
3 × 108
=
= 0.5 m
f 600 × 106
λ
4
Principlesof
ofElectromagnetics,
Electromagnetics,6e
6e
Principles
= 0.125 m
429
427
Prob. 13.12
(b)
This cis a monopole
3 × 108 antenna
λ = c = 3 × 1086 = 3.333 m
λ = f = 90 × 10 6 = 200
λf 1.5 × 10
 = = 0.8333 m
4
l λ ,hence it is a Hertzian monopole.
(c)
2
2
c
31× 1082  dl 
2 1 
λ =Rrad == 80π 6 = 3.75
 =m40π 
 = 9.87 mΩ
2 × 10  λ 
f 80
 200 
1 2
λ
Pt = m
I o Rrad
 = Prad==0.9375
2
4
(d)
8
c2 23P×t 108
= 810.54
−3
=
0.5
m
λ = I o == R = 9.87
6
10
×
rad
f 600 × 10
λ
I o== 28.47
= 0.125
A m
4
Sadiku & Kulkarni
Principles of Electromagnetics, 6e
Prob.
Prob. 13.13
13.12
8
c
3 × 10antenna
This is a monopole
428
= 30 m
(a) λ = = 8
c f3 × 10
10 × 106
= 200
λ= =
f 1.5 × 106
ηπ I o S
Emax rλ 2
Emax =
I
=
o
2
ηπ S
l λ ,hencerλit is a Hertzian monopole.
50 × 10−3 × 32× 302
2
2
=2 9.071
Io =1
 dl  2
 1 mA (S=Nπ r )
Rrad = 120
80π 22π (0.2)
40
9.87
m
Ω
 =100 π 
 =
2
 λ  Copyright©200
2015by Oxford University Press
1
Prad = Pt 320
= πI o42SR2rad 320π 4π 2 (0.2)4 × 104
(b) Rrad = 2 4
=
= 6.077Ω
304
λ
2P
8
I o 2 = t1 = 2
1 3 = 810.54
2
PradR=rad I o 9.87
Rrad ×=10−(9.071)
× 10−6 × 6.077
2
2
I o = 28.47 A
= 0.25 mW
Prob. 13.13
Prob. 13.14
c
3 × 108
= 30 m
(a) λ = =
6
f c 10 ×310
× 108
= 3.75 m
λ= =
f 80 × 106
S = Nπρ o2
Rrad =
N2 =
320π 4 S 2
λ4
=
320π 4 N 2π 2 ρo4
λ4
⎯⎯
→ N2 =
λ 4 Rrad
320π 6 (1.2 × 10−2 )4
(3.75) 4 × 8
=2015
248006
⎯⎯
→ Press
N  498
Copyright ©
by Oxford University
320π 6 (1.2 × 10−2 )4
1 2
1
I o Rrad = (9.071) 2 × 10−6 × 6.077
2
2
Prad =
Sadiku & Kulkarni
= 0.25 mW
Principles of Electromagnetics, 6e
430
Prob. 13.14
c
3 × 108
= 3.75 m
λ= =
f 80 × 106
S = Nπρ o2
Rrad =
N2 =
320π 4 S 2
320π 4 N 2π 2 ρo4
λ4
(3.75) 4 × 8
= 248006
320π 6 (1.2 × 10−2 )4
Prob. 13.15
Rrad =
(a)
λ4
=
⎯⎯
→ N2 =
λ 4 Rrad
320π 6 (1.2 × 10−2 )4
⎯⎯
→ N  498
320π 4 S 2
λ4
S = πρ o2 = π (0.4)2 = 0.5027 m 2
λ=
Sadiku & Kulkarni
c 3 × 108
=
= 50 m
f 6 × 106
Rrad =
(b)
(c)
320π 4 (0.5027)2
= 1.26 mΩ
(50) 4
1
1
Prad = I o 2 Rrad = (50)2 × 1.26 × 10−3429
= 1.575 W
2
2
R =
2π R μ f π
a
a 
a
R dc =
=
π fμσ =
2
2δ
2δ σ S 2σπ a
2π a
σ
0.4
4π × 10−7 × 6 × 106 × π
R μ fπ
=
= 63.91 mΩ
R =
σ
4 × 10−3
5.8 × 107
a
R rad
1.26
=
× 100% = 1.933%
η=
R rad + R 1.26 + 63.91
Prob. 13.16
Copyright © 2015 by Oxford University Press
π

cos  cosθ 
2

(a) f (θ ) =
sin θ
Copyright © 2015 by Oxford University Press
Principles of Electromagnetics, 6e
2δ
2δ σ S
2σπ a
2π a
σ
0.4
4π × 10 × 6 × 10 × π
R μ fπ
=
= 63.91 mΩ
−3
σ
4 × 10
5.8 × 107
a
R rad
1.26
=
× 100% = 1.933%
η=
431
R rad + R 1.26 + 63.91
Sadiku & Kulkarni R 
−7
=
6
Principles of Electromagnetics, 6e
Prob. 13.16
π

cos  cosθ 
2

(a) f (θ ) =
sin θ
(b) The same as for
λ
2
dipole except that the fields are zero for θ 
Sadiku & Kulkarni
π
2
as shown.
Principles of Electromagnetics, 6e
430
Prob. 13.17
Let Prad1 and Prad2 be the old and new radiated powers respectively.
Let Pohm1 and Pohm2 be the old and new ohmic powers respectively.
η r1 = 20% =
Prad 1
1
=
Prad 1 + Pohm1 5
But
⎯⎯
→
4 Prad 1 = Pohm1
(1)
1 2
I Rs Δz
2
1
= I 2 Rs 2Δz = 2Pohm1
2
Pohm1 =
Pohm2
(2)
2
1 2
1
2  Δz 
Prad 1 =
I o Rrad
= byI o2Oxford
× 80πUniversity
Copyright
© 2015
 Press
2
2
 λ 
2
1
1
 2Δz 
Prad 2 = I o2 Rrad = I o2 × 80π 2 
 = 4 Prad 1
2
2
 λ 
From (1) to (3),
η r2 =
4 Prad 1
Prad 2
P
=
= ohm1 = 33.3%
Prad 2 + Pohm 2 4 Prad 1 + 2 Pohm1 3Pohm1
Copyright © 2015 by Oxford University Press
Prob. 13.18
(3)
Pohm2 =
Sadiku & Kulkarni
1 2
I Rs 2Δz = 2Pohm1
2
(2)
2
1
1
 Δz 
Prad 1 = I o2 Rrad = I o2 × 80π 2  
2
2
 λ 
2
432
1 2
1 2
2  2 Δz 
Prad 2 = I o Rrad = I o × 80π 
 = 4 Prad 1
2
2
 λ 
From (1) to (3),
η r2 =
(3)
4 Prad 1
Prad 2
P
=
= ohm1 = 33.3%
Prad 2 + Pohm 2 4 Prad 1 + 2 Pohm1 3Pohm1
Prob. 13.18
(a) Let H s =
cos 2θ − j β r
e aH
ηo r
a E × a H = ak
Hs =
Principles of Electromagnetics, 6e
⎯⎯
→
aθ × a H = ar
⎯⎯
→ a H = aφ
cos 2θ − j β r
e aφ
120π r
| Es |2
cos 2 (2θ )
ar =
ar
(b) Pave =
2η
2η r 2
Sadiku & Kulkarni
Principles of Electromagnetics, 6e
π
1 cos 2 2θ 2
1
Prad =
r sin θ dθ dφ =
(2π )  cos 2 2θ sin θ dθ
2

r
2η
240π
0
431
2
2
2
But cos 2θ = cos θ − sin θ = 2cos θ − 1
Prad
π
1
(2cos 2 θ − 1) 2 d (cos θ )
=−

120 0
=−
π
1
(4cos 4 θ − 4cos 2 θ + 1) d (cos θ )
120 0
π
1  4cos5 θ 4cos3 θ
=−
−
+ cos θ 

120  5
3
0
1
4 4
4 4
1 14
[− + − 1 − + − 1] =
( )
120 5 3
5 3
120 15
= 7.778 mW
=−
(c)
Prad
1
=−
120
=−
120o

Copyright © 2015 by Oxford University Press
(2cos 2 θ − 1) 2 d (cos θ )
60o
 120o
1  4cos5 θ 4cos3 θ
−
+
θ
cos

 o
120  5
3
 60
1 4 1
4 1 1 4 1
4 1 1
1 1 1 1
[ (− ) − (− ) − − ( ) + ( ) − ] = [ + − ]
120 5 32 3 8 2 5 32 3 8 2 60 40 2 6
= 5.972 mW
5.972
= 0.7678 or 76.78%
7.778
=−
which is
Prob. 13.19
1
2
1
2
Pave = Re( E s × H s* ) = η | Hφ s |2 ar
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2 2
β Copyright
Io
1
2
Prad =  Pave dS = η 
θ cos 2 φ r 2 sin θ dθ dφ
sin
2
16π 2 r 2
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377 × 2502 × 1012
=
= 87.27Ω
3 × 9 × 1016
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433
Prob. 13.20
(a) Prad =  Prad ⋅ dS = Pave .2π r 2 (hemisphere)
Prad
200 × 103
=
= 12.73μW / m 2
2
6
2π r
2π (2500 × 10 )
Pave =
Pave = 12.73ar μ W/m 2 .
(b)
(E )
Pave = max
2η
2
Emax = 2η Pave = 240π × 12.73 × 10−6
= 0.098 V/m
Prob. 13.21
U
4π r 2 Pave 8π sin θ cos φ
=
=
Gd =
U ave  Pave .∂ s
 Pave .dS
But  Pave .dS =
π
π
2
  2sin θ cosφ sin θ dθ dφ
θ =0 φ =0
π
2
π
π
= 2  cos φ dφ  sin θ dθ = 2sin φ
0
0
2
2
0
π 
  =π
2
Gd = 8sin θ cos φ
D = Gd ,max = 8
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Prob. 13.22
3λ
2π 3λ
β =
= 3π
,
λ 2
2
  2π 1 3λ

 2π 1 3λ  
cos 
cos θ  − cos 

−β r 
jI e   λ 2 2

 λ 2 2 
Hφ s = o
2π r
sin θ
  3π

 3π  
cos  cosθ  − cos   
−β r 
−β r
jI e   2

 2   jI o e cos (1.5π cos θ )
= o
=
2π r
sin θ
2π r
sin θ
Hence, the normalized radiated field pattern is
From Prob. 13.11, set  =
f (θ ) =
cos (1.5π cosθ )
sin θ
which is plotted below.
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Prob. 13.23
The MATLAB code is shown below
N=20;
del= 2*pi/N;
sum=0;
for k=1:N
theta = del*k;
term = (1 – cos(theta))/theta;
sum = sum + term;
end
int = del*sum
When the program is run, it gives the value of 2.4335. The accuracy may be increased by
increasing N.
Prob. 13.24
jη I o β dl
sin θ e − j β r
4π r
2
 dl 
Rrad = 80π 2  
λ
1
4π r 2 .
Eθ s
2
4π r Pave
2η
Gd =
=
1 2
Prad
I o Rrad
2
(a) Eθ s =
2
2 2 2
2
4π r 2 1  λ  1 η I o β ( dl ) sin θ
=
.
  .
I o 2 80π 2  dl  η
16π 2 r 2
2
2
Gd = 1.5sin 2 θ
(b) D = Gd ,max = 1.5
(c) Ae =
λ2
1.5λ 2 sin 2 θ
Gd =
4π
4π
(d) Rrad
1
= 80π   = 3.084 Ω
 16 
2
2
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Prob. 13.25
Eφ s =
(a)
120π 2 I o S
sin θ e − j β r
2
λ
r
320π 4 S 2
Rrad =
4
λ
2
4π U (θ , φ ) 4π r 2 Pave 8π r 2 1 Eφ s
Gd =
=
= 2 .
1 2
Prad
I o 2η Rrad
I o Rrad
2
8π r 2 1
I 2 S2
λ2
= 2 . .14400π 4 o2 4 sin 2 θ
I o 2η
r λ
320π 4 S 2
Gd = 1.5sin 2 θ
(b)
D = 1.5
λ 2Gd λ 2
=
1.5sin 2 θ
4π
4π
(c)
Ae =
(d)
S = π a2 =
πd2
4
=
320π 6
(576) 2
Rrad = 0.927Ω
Prob. 13.26
(a)
c
3 × 108
= 250
λ= =
f 1.2 × 106
=
λ
= 62.5 m
4
(b) From eq. (13.30), Rrad = 36.5 Ω
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(c)
For λ /4-monopole,
π
cos( cosθ )
2
f (θ ) =
,
sin θ
0 <θ <π / 2
π
4π f 2 (θ )
Gd (θ ,φ ) =

f 2 (θ )d Ω
4π cos 2 ( cosθ )
2
sin 2 θ
=
2π π / 2

0
0
π
cos 2 ( cosθ )
2
dθ dφ
sin θ
π
π
4π cos 2 ( cosθ )
3.282cos 2 ( cosθ )
1
2
2
=
=
sin 2 θ
2π (0.6094)
sin 2 θ
D = Gd ,max = 3.282
Prob. 13.27
(a) Umax = 1
U ave =
Prad  Ud Ω
=
4π
4π
=
1
4π
  sin
2
2θ sin θ dθ dφ
π
1
2
=
(2π )  ( 2sin θ cos θ ) d ( − cosθ )
4π
0
π
= 2 ( cos 4 θ − cos 2 θ ) d ( cos θ )
0
 cos5 θ cos3 θ 
= 2
−

3 
 5
π
0
 2 2 8
= 2 − +  =
 5 3  15
U ave = 0.5333
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D=
(b)
U max
= 1.875
U ave
Umax = 4
1
4
U ave =
Ud Ω =

4π
4π
=
1π
π
π
sin θ
  sin
2
 dφ  cos ecθ dθ =
0
π
3
2
θ
dθ dφ
π
ln 3
π
Uave = 0.5493
D=
(c)
U max
16
=
= 9.7092
U ave 3ln 3
Umax = 2
1
1
Ud Ω =
2sin 2 θ sin 2 φ sin θ dθ dφ



4π
4π
π
π
1
2
=
sin φ dφ  (1 − cos 2 θ ) d ( − cosθ )

2π 0
0
U ave =
=

1 π  cos3 θ
. 
− cosθ 
2π 2  3

π
0
=
1 2
 1
− + 2 =

4 3
 3
Uave = 0.333
D=
U max
=6
U ave
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Prob. 13.28
Gd (θ ,φ ) =
(a)
U ave =
2π
U (θ , φ )
U ave
π
1
10sin θ sin 2 φ × sin θ dθ dφ
4π φ = 0 θ = 0
−
10
=
4π
2π
 sin
0
2
π
φ dφ  sin 2θ dθ
0
10 1 
sin 2θ  2π 1 
sin 2φ  π
θ −
 0 φ −

4π 2 
2 
2
2 0
10
5π
(2π − 0)(π − 0) =
=
16π
4
2
40sin θ sin φ
= 2.546sin θ sin 2 φ
Gd (θ , φ ) =
5π
D = Gd . max = 2.546
=
U ave
(b)
1
=
4π
π
π
  2sin
φ θ
2
θ sin 3 φ × sin θ dθ dφ
=0 =0
π
π
π

2
2 
3
3
2
=
=
φ
d
φ
θ
d
θ
sin
sin
  (1 − cos φ )d (− cos φ ) 


4π 0
4π  0
0

2
 cos3 φ
 π 1 4
16
−
=
=
φ
cos

 


 0  2π  3  18π
 3
18π
Gd (θ ,φ ) =
2sin 2 θ sin 3 φ = 2.25π sin 2 θ sin 3 φ
16
D = Gd . max = 7.069
=
1
2π
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1
=
4π
U ave
(c)
2π
π
  5(1 + sin
φ θ
2
θ sin 2 φ ) × sin θ dθ dφ
=0 =0
π
π
2π
2π

5 
3
2
=
  sin θ dθ  dφ +  sin θ dθ  sin dφ 
4π  0
0
0
0

π 4 φ sin 2φ 2π 
5 
) 
 2π (− cos θ ) + ( −
0 3 2
0
4π 
4
5 
4  20
=
4π + π  =

4π 
3  3
3
Gd (θ ,φ ) = 5(1 + sin 2 θ sin 2 φ ) = 0.75(1 + sin 2 θ sin 2 φ )
20
D = Gd . max = 1.5
=
Prob. 13.29
U max = 4
φ
1
1
Ud Ω =
4sin 2 θ sin sin θ dθ dφ


4π
4π
2
π
π
π
φ
φ π
1
1
=  sin 3 θ dθ  sin dφ =  (1 − cos 2 θ )d (− cosθ )(−2cos )
π0
π0
2
2 0
0
U ave =
1 4
8
( )(2) =
π 3
3π
U max
3π
= 4×
= 4.712
D=
U ave
8
=
Prob. 13.30
P
| Er |2
I 2 sin 2 θ
ar = o
ar
2η
2η r 2
ave =
Prad =
2π
I o2 sin 2 θ 2
I o2
φ
=
r
sin
θ
d
θ
d
(2π )  (1 − cos 2 θ )d (− cosθ )
2

2η
r
240π
0
=
π I2
I o2 cos3 θ
I2
− cos θ ) = o (−1 / 3 + 1 − 1 / 3 + 1) = o
(
0 120
120
3
90
I o2 = 90 Pave = 90 × 50 × 10−3
⎯⎯
→ I o = 2.121 A
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Prob. 13.31
Prad =
2π
π
π
| E |2 2
100 r 2
θ
θ
φ
=
r
sin
d
d
(2π )  sin 5 θ dθ
2


r 2ηo
φ = 0 θ = 0 2η o
0
=
2π
5
1
 5
π
100  − cosθ + cos3θ − cos5θ 
2 × 377
48
80
 8
0
=
100π  5 5 5
5
1
1 
−
+
+  = 0.889 W
 + −
377  8 8 48 48 80 80 
Prob. 13.32
This is similar to Fig. 13.10 except that the elements are z-directed.
E s = E s1 + E s 2 =
where r1 ≅ r −

jηβ I o dl 
e − j β r1
e − j β r2
+
θ
θ
sin
a
sin
aθ 2 

θ1
1
2
r1
r2
4π 

d
cosθ ,
2
r2 ≅ r +
d
cosθ ,
2
Es =
jηβ I o dl
sin θ aθ e j β d cos θ / 2 + e − j β d cos θ / 2 
4π
Es =
jηβ I o dl
1
sin θ cos( β d cosθ )aθ
2π
2
θ1 ≅ θ 2 ≅ θ ,
aθ 1 ≅ aθ 2 = aθ
Prob. 13.33
Equation (13.33) applies except that cosθ must be replaced by sinθ. Hence, the radiation
pattern is
1

f (θ ) = sin θ 2cos  ( β d cosθ + α ) 
2

2π λ
α = 0, d = 2(λ / 4) = λ / 2, β d=
=π
λ 2
1
f (θ ) = 2sin θ cos( π cosθ )
2
Prob. 13.34
1

(a) AF = 2cos  ( β d cos θ + α )  ,
2

α = 0,
βd =
2π
λ
AF = 2cos(π cos θ )
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(b) Nulls occur when
cos(π cosθ ) = 0
or
⎯⎯
→
π cosθ = ±π / 2, ±3π / 2,...
θ = 60o ,120o
(c) Maxima and minima occur when
df
=0
dθ
⎯⎯
→
i.e. sin θ = 0
cosθ = 0
or
sin(π cosθ )π sin θ = 0
⎯⎯
→
⎯⎯
→
θ = 0o ,180o
θ = 90o
θ = 0o ,90o ,180o
(d) The group pattern is sketched below.
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Prob. 13.35
1

f (θ ) = cos  ( β d cosθ + α ) 
2

2π
(a) α = π , β d =
.λ = 2π
2
λ
(
f (θ ) = cos π cosθ + π
4
Nulls occur at π cosθ + π
Maxima occur at
)
4
∂f
=0
∂θ
π
=±
π
2
,±
3π
,... or θ = 75.5o ,138.6o
2
sin θ = 0
θ = 0o ,180o

Or sin  π cos θ +  = 0
θ = 41.4o ,104.5o
4

With f max = 0.71,1 .
Hence the group pattern is sketched below.
θ = 41.0138o, abs(f) = 1,
θ = 104.6083o, abs(f) = 1,
abs(f) = 1/√2=0.707
θ = 180o,
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(b)
α=
3π
2π λ π
,βd =
. =
4
λ 4 2
3π 
π
f (θ ) = cos  cosθ +

8 
4
Nulls occur at
π
4
cosθ +
3π
π 3π
= ± , ± ,...
8
2
2
⎯⎯
→
θ = 60o
3π 
π
Minima and maxima occur at sin θ cos  cosθ +
=0
8 
4
i.e. θ = 0o ,180o → f (θ ) = 0.383,0.924
θ = 0, abs(f) = 0.3827
θ = 180, abs(f) = 0.9239
2π 3λ 3π
=
.
λ 4
2
 3π

f (θ ) = cos  cosθ 
 4

3π
π 3π
It has nulls at
cosθ = ± , ± ,... → θ = 48.2o ,131.8o
4
2
2
df
 3π

It has maxima and minima at
= 0 → sin θ sin  cos θ  = 0
dθ
 4

o
o
o
i.e. θ = 0 ,180 → f (θ ) = 0.71,1 ,
θ = ±90 , → f (θ ) = 1
(c) α = 0, β d =
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θ = 0 o,
θ = 90o,
θ = 180o,
Prob. 13.36
1

(a) For N = 2, f (θ ) = cos  ( β d cosθ + α ) 
2

α = 0, d =
λ
4
 1  2π λ

π

f (θ ) = cos   . cosθ + 0   = cos  cosθ 

4

2 λ 4
Maxima and minima occur at
d  π

cos  cosθ   = 0

dθ   4

π

sin θ sin  cosθ  = 0
4

sin θ = 0 → θ = π ,0 and f (θ ) = 0.707
π

sin  cosθ  → cos = 0 → θ = 90o , f (θ ) = 1
4

π
π 3π
Nulls occur as
cosθ = ± , ± ,... (No Solution)
4
2
2
The group pattern is sketched below.
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abs(f) = 1/√2=0.707,
abs(f) = 1,
abs(f) = 1/√2=0.707
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(b) For N = 4,
AF =
Now,
sin 2 ( β d cos θ + 0 )
1
sin ( β d cosθ + 0 )
2
sin 4θ 2sin 2θ cos 2θ
=
= 4cos 2θ cosθ
sin θ
sin θ
1

AF = 4cos ( β d cosθ ) cos  β d cos θ 
2

 2π λ

 1 2π λ 
f (θ ) = cos  . cosθ  cos 
 cosθ
 λ 4

2 λ 4
π

π

= cos  cosθ  cos  cosθ 
2

4

The plot is shown below.
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Prob. 13.37
The MATLAB code is shown below.
for n=1:180
phi=n*pi/180;
p(n)=n;
sn=sin(2*pi*cos(phi));
cn=cos(0.5*pi*cos(phi));
sd=sin(0.5*pi*cos(phi));
fun=sn*cn*cn/sd;
f(n)= abs(fun);
end
polar(p,f)
The polar plot and the xy plot are shown below.
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Prob. 13.38
(a) The resultant pattern is obtained as follows.
I ∠ 0o
I ∠ 0o
I ∠ 0o
I ∠ 0o
λ/2
λ/2
λ/2
λ
x
=
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(b) The array is replaced by
+
λ
∠0 o
4
+
∠π /2
where + stands for
∠0o
∠π
Thus the resultant pattern is obtained as shown.
I ∠ 0o
I ∠ 90o
I ∠ 270o
I ∠180o
x
λ/4
=
λ/4
λ/4
λ/2, π
x
=
Prob. 13.39
Gd (dB ) = 20dB = 10log10 Gd
λ=
⎯⎯
→
Gd = 102 = 100
c
3 × 108
=
= 3 × 10−2
9
f 10 × 10
λ2
9 × 10−4
Ae =
Gd =
100 = 7.162 × 10−2 m 2
4π
4π
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Prob. 13.40
Ae =
=
Pr
Pr
=
Pave | Er |2
=
2η
2η Pr
| Er |2
2 × 120π × 2 × 10−6 48π
=
= 0.6031
25 × 102 × 10−6
250
Prob. 13.41
Friis equation states that
Pr
 λ 
= Gr Gt 

Pt
 4π r 
2
3 × 108
c
= 1.5 m,
λ= =
f 200 × 106
r = 238,857 × 1.609 × 103 = 3.843 × 108
Gt (dB ) = 15dB = 10log10 Gt
⎯⎯
→ Gt = 1015/10 = 31.623
2
2
−9
8
 4π r  Pr  4π × 3.843 × 10  4 × 10
=
= 34.55 × 1010
Gr = 


−3
1.5
 λ  Pt 
 120 × 10
Gr (dB ) = 10log10 Gr = 10log10 34.55 × 1010 = 115.384 dB
Prob. 13.42
Using Frii’s equation,
2
 λ 
Pr = Gr Gt 
 Pt
 4π r 
2
 4π r  Pr
Pt = 

 λ  Gr Gt
c 3 × 108
λ= =
= 0.1,
f 3 × 109
r = 42 km
Gt (dB ) = 10log10 Gt = 25
⎯⎯
→ Gt = 102.5 = 316.23
Gr (dB ) = 10log10 Gr = 20
⎯⎯
→ Gt = 102 = 100
2
 4π × 42 × 103  3 × 10−6
Pt = 
= 2.642 kW

0.1

 31623
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450
Prob. 13.43
Gdt = 104 , Gdr = 103.2 = 1585
c
3 × 108
1
λ= =
= 0.02m =
9
f 15 × 10
50
2
2
0.02
 λ 


4
Pr = Gdr Gdt 
320
 Pt = 10 (1585 ) 
7 
 4π r 
 4π × 2.456741 × 10 
= 2.129 × 10−11 W = 21.29 pW
Prob. 13.44
Using Frii’s equation,
2
 λ 
Pr = G 2 
 Pt
 4π r 
2
 4π r  Pr
Pt = 
 2
 λ  G
⎯⎯
→ G = 102.5 = 316.23
G (dB) = 25
λ=
c
3 × 108
=
= 0.5
f 600 × 106
2
−3
 4π × 450  4 × 10
= 5.1163 W
Pt = 
2

 0.5  (316.23)
Prob. 13.45
30dB = log
Pt
P
→ t = 103 = 1000
Pr
Pr
But Pr = ( Gd )
2
2
3


 Gd 

 Pt = Pt 

 50 × 4π × 12 
 800π 
2
Pr
1
 1 
 Gd 
=

 = =
Pt 1000  10 10 
 800π 
or Gd =
2
2
800π
= 79.476
10 10
Gd = 10log 79.476 = 19 dB
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Prob. 13.46
(a) Pi =
E
2
2ηo
Ei =
=
Prad Gd
→ Ei =
4π r 2
240π Prad Gd
4π r 2
1
1
60 Prad Gd =
60 × 200 × 103 × 3500
r
120 × 103
= 1.708 V/m
2
Ei σ
(b)
Es =
(c)
Pc = Piσ =
(d)
4π r 2
=
1.7082 × 8
= 11.36 μ V/m
4π × 14400 × 106
1.7082
( 8) = 30.95 mW
240π
2
2
E
(11.36 ) × 10−12 = 1.712 × 10−13 W/m 2
=
Pi =
2ηo
240π
λ=
3 × 108
λ 2G 0.04 × 3500
=
0.2
m
,
A
=
=
2r
15 × 108
4π
4π
Pr = Pa Aer = 1.712 × 10−13 × 11.14 = 1.907 × 10−12
( λGd ) σ Prad
Pr =
3
( 4π ) r 4
2
or
( 0.2 × 3500 ) × 8 × 2 × 105
=
3
( 4π ) × 124 × 1016
2
= 1.91 × 10−12 W
Prob. 13.47
Pr =
(λGd ) 2 σ Prad
(4π )3 r 4
Gd (dB) = 30dB = 10log10 Gd
⎯⎯
→ Gd = 103
λ=
c 3 × 108
=
= 0.075 m
f 4 × 109
Pr =
(0.075 × 103 ) 2 × 12 × 80 × 103
= 272.1 pW
(4π )3 (10 × 103 ) 4
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Prob. 13.48
2
4π  4π r1r2  Pr
Prad =


Gdt Gdr  λ  σ
But Gdt = 36dB = 103.6 = 3981.1
Gdr = 20dB = 102 = 100
λ=
c 3 × 108
=
= 0.06
f 5 × 109
r1 = 3km , r2 = 5km
2
Prad
 4π × 15 × 106  8 × 10−12
4π
=


3981.1 × 100  6 × 10−2 
2.4
= 1.038 kW
Prob. 13.49
( λGd ) σ Prad
Pr =
3
( 4π ) r 4
2
λ=
4π ) r 4 Pr
(
→ Prad =
2
( λGd ) σ
3
c 3 × 108
1
=
=
<< r = 250m
9
f 6 × 10
20
40 = log10 Gd → Gd = 104
Prad =
( 4π )
3
( 0.25 × 10 )
3 4
2
× 2 × 10 −6
 1
4
 × 10  × 0.8
 20

= 77.52 W
Prob. 13.50
(a)
π fL π × 300 × 106 × 50 × 10−9
F=
=
= 2.356
R
20
IL = 10log10 (1 + F 2 ) = 10log10 (1 + 2.3562 ) = 8.164 dB
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(b)
F = π fRC = π × 300 × 106 × 10 × 103 × 60 × 10−12 = 180π = 565.5
IL = 10log10 (1 + F 2 ) = 10log10 (1 + 565.52 ) = 55.05 dB
Prob. 13.51
Zg
I1
I2
+
V1
Vg
-
By definition,
+
A B
C

  A DB 
 C D


V1 = AV2 – BI2
I1 = CV2 – DI2
V2
ZL
-
(1)
(2)
Let V2 and V2 be respectively the load voltages when the filter circuit is
present and when it is absent.
V2 = − I 2 Z L =
=
=
=
Vg Z L

V1 
 Z g +  ( CZ L + D )
I1 

=
Vg Z L

AV2 − BI 2 
 Zg +
 ( CZ L + D )
CV2 − DI 2 

Vg Z L

AZ L + B 
 Zg +
 ( CZ L + D )
CZ
D
+
L


( Z ( CZ
g
V2 =
I1Z L
CZ L + D
Vg Z L
L
+ D ) + AZ L + B )
Vg Z L
(Z
g
+ ZL )
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Ratio and modulus give
( Z g ( CZ L + D ) + AZ L + B )
V2
=
V2
Zg + ZL
Insertion loss =
IL =
20log10
V2
= 20log10
V2
( Z ( CZ
g
L
+ D ) + AZ L + B )
Zg + ZL
which is the required result
Prob. 13.52
SE = 20log10
Ei
6
= 20log10
= 20log10 (3 × 105 )
Eo
20 × 10−6
= 109.54 dB
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CHAPTER 14
P. E. 14.1
The program in Fig. 14.3 was used to obtain the plot in Fig. 14.5.
P. E. 14.2
For the exact solution,
(D2 + 1) y = 0
y (0) = 0
y(1) = 1
Thus,
y = A cos x + B sin x
A =0
1 = B sin 1 or B = 1/sin 1
y = sin x/sin 1
For the finite difference solution,
y ( x + Δ) − 2 y ( x) + y ( x − Δ)
+ y=0
Δ2
y’’ + y = 0
or
y ( x + Δ) + y ( x − Δ)
, y (0) = 0, y (1) = 1, Δ = 1 / 4
2 − Δ2
With the MATLAB program shown below, we obtain the exact result ye and FD
result y.
y ( x) =
y(1)=0.0;
y(5)=1.0;
del=0.25;
for n=1:20
for k=2:4
y(k)=(y(k+1) +y(k-1))/(2-del*del)
x=(k-1)*del;
ye=sin(x)/sin(1.0)
end
end
The results are listed below.
y(x)
N=5
y(0.25) 0.2498
y(0.5) 0.5242
y(0.75) 0.7867
N=10
N=15
N=20
Exact
ye(x)
0.2924
0.5682
0.8094
0.2942
0.5701
0.8104
0.2943
0.5702
0.8104
0.2940
0.5697
0.8101
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P. E. 14.3 By applying eq. (14.16) to each node as shown below, we obtain the
following results after 5 iterations.
0
0
25
0
0
0
0
10.01
9.82
9.35
8.19
5.56
4.69
0
28.3
28.17
27.06
25
19.92
18.95
0
12.05
11.87
11.44
10.30
7.76
2.34
0
28.3
28.17
27.85
27.06
25.06
19.92
0
44.57
44.46
44.26
43.76
42.48
37.5
0
50
50
10.01
9.82
9.35
8.19
5.56
4.69
0
28.3
28.17
27.85
27.06
25
19.92
0
50
0
25
P. E. 14.4 (a) Using the program in Fig. 14.16 with nx = 4+1=5 and ny = 8+1=9, we
obtain the potential at center as
V(3,5) = 23.796 V
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Sadiku & Kulkarni
(b) Using the same program with nx = 12+1=13 and ny = 24+1=25, the potential at the
Principles of Electromagnetics, 6e
center is
V(7,13) = 23.883 V
472
P. E.
E. 14.7
14.5 By
ܸ ൌcombining
ͲǤ͹ͻ (the solution
further
and
gets closer
to the expected
P.
the ideasimproves
in Figs. 14.23
and
14.27,
and dividing
each wirevalue
into
due to a finer grid).
C (in pF)
N segments, the results listedθin Table 14.2 is obtained.
Code: It is the same as in Example
following lines:
10 14.5 except the
8.5483
20
9.0677
P.
E. 15.2
grid_size=0.05;
3
8.893
(a)
nt=90; %% Top plate at 90th 30
grid row from the upper
boundary
8.606
nB=110; %% Bottom plate at40
110th grid row from
the upper boundary
50
13.004
Sadiku & Kulkarni ns1=90; %% Both plates start from 90th grid column
Principles of Electromagnetics, 6e
60
8.5505
ns2=110; %% Both plates ends at 110th grid column
2
70
9.3711 2
4 457 8.7762
80
v = V((nt×lx)+100)
1
90
8.665
100
8.665
(b)
Using
nx = 12+1=13
nyExample
= 24+1=25,
potential
at the
P. E.
14.6the
Thesame
codeprogram
remains with
essentially
the sameand
as in
14.6the
except
for boundary
110
10.179
center iswhich need to be120
conditions
explicitly defined 8.544
as explained at the end of Section 14.4
130
9.892
1
above Practice Exercise 14.6.
V(7,13) = 23.883
140V
8.7449
150
9.5106
P.
14.7 By
ideas
in
Figs.
14.23
and
14.27,
and A
dividing
each wire into
ForE.element
1, combining
local 1-2-3 the
corresponds
with
global
1-3-4
so that
1 = 0.35,
160
8.5488
P. E. 14.9 Consider Poisson’s170
equation as the partial
11.32differential equation and Dirac delta
N
results
listed
in
14.2
P
= 0.8, as
P2the
=the
0.6,
P3 =function.
-1.4,180
Q1Table
= -0.5,
Q2is=obtained.
0.5,
Q3 = 0
1 segments,
function
forcing
8.6278
P. E. 15.2
‫׏‬ଶ ‫ ܩ‬ൌwe
െ݃
3 the program in Fig. 14.23.
P.E. 14.8 To determine V and E at (-1,4,5),
use
(a)
 0.6357 0.1643 −0.8 
Green’s
function is the impulse
response of the given differential equation. So forcing
L
C (1) = ρ0.1643
0.4357 −0.6 
L dl
=  ݂ in, the
where
= 26 + (4is−Dirac-delta
y ') 2
Vfunction
aboveRexpression
function.
R
4
πε

−
0.8
−
0.6
1.4
o
0 


2
સଶ ‫ ܩ‬ൌ െ݃2
Δ N 2, localρ1-2-3
k
For
corresponds 4with global 1-2-3
so that A2 = 0.7,
1
V = element

By considering
axial
symmetry
the variation in the direction of ‫ ׎‬and ܼ is zero.
4πε k =1 26 + ( y − yk ) 2
P1 = 0.1, P2 = 1.4, P3 = -1.5, Q1 = -1, Q2 = 0, Q3 = 1
ͳ ߲
߲‫ܩ‬
L
સଶ ‫ ܩ‬ൌ
൬ߩ ൰
ρ L dlR
ߩ ߲ߩ ߲ߩ
0 4πε o R−30.4107 
 0.3607E = 0.05

1
(2)
=  0.05
−over
C
0.7
0.75the
Substituting
and integrating
 surface of ߩ and ‫׎‬
where R
r – r’ =−(-1,
5), R = |R|
 −=0.4107
0.754-y’,1.1607
For element 1, local 1-2-3 corresponds
global
1-3-4
so that A1 = 0.35,
ඵ સଶ ‫ ܩ‬with
ߩ݀ߩ݀‫׎‬
ൌඵ
െߜ ߩ݀ߩ݀‫׎‬
N
Δ
(−1) ρ k
The
coefficient matrix
is given by
E ≅global
P1x = 0.8,
0.6,
Q = 0.5, Q3 = 0
4πε Pk =21 =[26
) 2 ]3/ͳ2Q1߲= -0.5,
+ (4P3−=yk-1.4,
߲‫ ܩ‬2
߲‫ܩ‬
N
ඵ
ߩ
ߩ݀ߩ݀‫׎‬
ൌ ඵ ߲ ൬ߩ ൰ ݀‫ ׎‬ൌ െͳ
Δ
(4 − yk ) ρ k ߩ ߲ߩ ߲ߩ
߲ߩ
Ey ≅

2 3/ 2
4πε k =1 [26 + (4 − yk ) ]
 0.6357 0.1643 −0.8  ଶగ
߲‫ܩ‬
߲‫ܩ‬


(1)
න
߲
න
݀‫ ׎‬ൌ ߩ
ʹߨ ൌ െͳ
൬ߩ
൰
C
=
0.1643
0.4357
−
0.6
Ez = −5Ex
߲ߩ ଴
߲ߩ
  −0.8
−0.6
1.4 
For N = 20, Vo = 1V, L = 1m, a = 1mm, the program in Fig. 14.23 is modified. The
result
is:
For element
2, local 1-2-3
corresponds
with
global
1-2-3
so that A2 = 0.7,
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Copyright ©
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2015 by
by Oxford
Oxford University
University Press
Press
V = 12.47 mV, E = -0.3266 a + 1.1353a + 1.6331a mV/m
P1 = 0.1, P2 = 1.4, P3 = -1.5, Qx1 = -1, Q2 =y0, Q3 = 1 z
Code: It is the same as in Example 14.5 except the following lines:
where R = r – r’ = (-1, 4-y’, 5), R = |R|
grid_size=0.05;
Sadiku & Kulkarni
Principles of Electromagnetics, 6e
(−1)atρ90th
nt=90; Δ
%%NTop plate
grid row from the upper boundary
k
Ex ≅

2
nB=110;
Bottom plate2 at3/110th
grid row from the upper boundary
4πε%%
k =1 [26 + (4 − yk ) ]
ns1=90; %%
Both
plates
start
from
90th
grid column
459
Δ N
(4 − yk ) ρ k
ns2=110;
%%
Both plates ends
at 110th grid column
Ey ≅

2 3/ 2
4πε k =1 [26 + (4 − yk ) ]
v = V((nt×lx)+100)
E z = −5 E x
P.
remains
essentially
as in in
Example
14.6isexcept
for boundary
ForE.N14.6
= 20,The
Vocode
= 1V,
L = 1m,
a = 1mm,the
thesame
program
Fig. 14.23
modified.
The
result is: which need to be explicitly defined as explained at the end of Section 14.4
conditions
above Practice Exercise 14.6.
V = 12.47 mV, E = -0.3266 ax + 1.1353ay + 1.6331az mV/m
P. E. 14.9 Consider Poisson’s equation as the partial differential equation and Dirac delta
function as the forcing function.
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‫׏‬ଶ ‫ ܩ‬ൌ െ݃
Green’s function is the impulse response of the given differential equation. So forcing
function ݂ in the above expression is Dirac-delta function.
સଶ ‫ ܩ‬ൌ െ݃
By considering axial symmetry the variation in the direction of ‫ ׎‬and ܼ is zero.
સଶ ‫ ܩ‬ൌ
ͳ ߲
߲‫ܩ‬
൬ߩ ൰
ߩ ߲ߩ ߲ߩ
Substituting and integrating over the surface of ߩ and ‫׎‬
ඵ
ඵ સଶ ‫ ׎݀ߩ݀ߩ ܩ‬ൌ ඵ െߜ ߩ݀ߩ݀‫׎‬
ͳ ߲ ߲‫ܩ‬
߲‫ܩ‬
ߩ
ߩ݀ߩ݀‫ ׎‬ൌ ඵ ߲ ൬ߩ ൰ ݀‫ ׎‬ൌ െͳ
ߩ ߲ߩ ߲ߩ
߲ߩ
ଶగ
߲‫ܩ‬
߲‫ܩ‬
න ߲ ൬ߩ ൰ න ݀‫ ׎‬ൌ ߩ
ʹߨ ൌ െͳ
߲ߩ ଴
߲ߩ
is distance between the source and observation points.
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Prob. 14.1 (a) Using the Matlab code in Fig. 14.3, we input the data as:
>> plotit([-1 2 1], [-1 0; 0 2; 1 0], 1, 1, 0.01, 0.01, 8, 2, 5)
and the plot is shown below.
(b) Using the MATLAB code in Fig. 14.3, we input the required data as:
>> plotit([1 1 1 1 1], [-1 -1; -1 1; 1 –1; 1 1; 0 0], 1, 1, 0.02, 0.01, 6, 2, 5)
and obtain the plot shown below.
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Prob. 14.2
∇ 2V =
∂ 2V 1 ∂ V ∂ 2V
+
+
=0
∂ρ 2 ρ ∂ρ ∂ z 2
The equivalent finite difference expression is
V ( ρ o + Δρ , zo ) − 2V ( ρo , zo ) + V ( ρo − Δρ , zo ) 1 V ( ρ o + Δρ , zo ) − V ( ρ o − Δρ , zo )
+
(Δρ ) 2
2Δρ
ρo
+
V ( ρo , zo + Δz ) − 2V ( ρo , zo ) + V ( ρ o , zo − Δz )
=0
(Δz ) 2
If Δz = Δρ = h, rearranging terms gives
1
1
h
V ( ρo , zo ) = V ( ρ o , zo + h) + V ( ρo , zo − h) + (1 +
)V ( ρ + h, zo )
4
4
2 ρo
+(1 −
h
2 ρo
)V ( ρ − h, zo )
as expected.
Prob. 14.3 (a)
dV V ( x + Δx) − V ( x − Δx)
=
dx
2Δx
For Δx = 0.05 and at x = 0.15,
dV 2.0134 − 1.00
=
= 10.117
dx
0.05 X 2
d 2V V ( x + Δx) − 2V ( x) + V ( x − Δx) 2.0134 + 1.0017 − 2 x1.5056
=
=
= 1.56
dx 2
(Δx) 2
(0.05) 2
(b) V = 10 sinh x, dV/dx = 10 cosh x. At x = 0.15, dV/dx = 10.113
which is close to the numerical estimate.
d2V/dx2 = 10 sinh x. At x = 0.15, d2V/dx2 = 1.5056
which is slightly lower than the numerical value.
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Prob.14.4
Exact solution:
( D 2 + 4) y = 0
⎯⎯
→ y ( x) = A cos 2 x + B sin 2 x
y (0) = 0
⎯⎯
→ 0= A
y (1) = 10
⎯⎯
→ 10 = B sin 2
⎯⎯
→ B=
10
sin 2
sin 2 x
sin 2
sin 0.5
y (0.25) = 10
= 5.272
sin 2
Finite difference solution:
y ( x + Δ) − 2 y ( x) + y ( x − Δ)
+ 4 y ( x) = 0
Δ2
y ( x + Δ ) + y ( x − Δ) = 2 y ( x) − 4Δ 2 y ( x) = (2 − 4Δ 2 ) y ( x)
or
y ( x + Δ) + y ( x − Δ)
y ( x) =
, Δ = 0.25
(2 − 4Δ 2 )
y ( x) = 10
Using this scheme, we obtain the result shown below. The number of iterations is not
enough to get accurate result. The numerical results are compared with the exact
solution as shown in the figure below.
Iteration
0
1
2
3
4
5
0
0
0
0
0
0
0
0.25 0.5
0.75
0
0
0
0
0
5.7143
0
3.2653 7.5802
1.8659 5.398 8.7987
3.0844 6.7904 9.5945
3.8802 7.7904 10.1142
1.0
10
10
10
10
10
10
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Prob. 14.5
∂ 2V 1 ∂ V 1 ∂ 2V
∇V =
+
+
=0,
∂ρ 2 ρ ∂ρ ρ 2 ∂φ 2
(1)
∂ 2V
Vm +1n − 2Vm n + Vm +1n
=
,
∂ρ 2
(Δρ ) 2
(2)
∂ 2V
Vm n +1 − 2Vm n + Vm n −1
=
,
∂φ 2
(Δφ ) 2
(3)
∂V
∂ρ
(4)
2
m, n
V n m +1 − V n m −1
=
.
2Δρ
Substituting (2) to (4) into (1) gives
∇ 2V =
V n m +1 − V n m −1
V n − 2Vm n + Vm +1n
Vm n +1 − 2Vm n + Vm n −1
+ m +1
+
mΔρ (2Δρ )
(Δρ ) 2
(mΔρΔφ ) 2
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=
1
(Δρ ) 2

1
1
1
)Vm −1n − 2Vm n + (1 +
)Vm −1n +
(Vm n +1 − 2Vm n + Vm n −1
(1 −
2
2m
2m
(mΔφ )


)

as required.
Prob. 14.6
Iteration

V1
V2
V3
V4
0
1
2
3
4
5
0.0000
25.0000
35.6250
38.9063
39.7266
39.9316
0.0000
26.2500
32.8125
34.4531
34.8633
34.9658
0.0000
16.2500
22.8125
24.4531
24.8633
24.9658
0.0000
15.6250
18.9063
19.7266
19.9316
19.9829
Prob. 14.7
Vo =
V1 + V2 + V3 + V4 10 − 40 + 50 + 80
=
= 25V
4
4
Prob. 14.8
V
1
V1 = [ 20 + 20 + 40 + V4 ] = 20 + 4
4
4
V
1
V2 = [ −10 + 20 + 0 + V3 ] = 2.5 + 3
4
4
1
1
V3 = [ 0 + 20 + V2 + V4 ] = 5 + (V2 + V4 )
4
4
1
1
V4 = [ 0 + 40 + V1 + V3 ] = 10 + (V1 + V3 )
4
4
Using these relationships, we obtain the data in the table below.
Iteration
V1
V2
V3
V4
1st
20
2.5
5.625
16.406
2nd
24.1
3.91
10.08
18.54
3rd
24.64
5.02
10.89
18.88
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4th
24.72
5.22
11.03
18.94
5th
24.73
5.26
11.05
18.95
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Prob. 14.9
(a) We follow Example 6.5 with a=b.
 nπ x 
 nπ y 
sin 
sinh 


∞
4V
 a 
 a  + 4Vo
V = V1 + V2 = o 
n sinh(nπ )
π n = odd
π
 nπ y 
 nπ x 
sin 
 sinh 

 a 
 a 

n sinh(nπ )
n = odd
∞
(b) At the center of the region, finite difference gives
1
V
V (a / 2, a / 2) = (0 + 0 + Vo + Vo ) = o = 25 V
4
2
Prob. 14.10
h2 ρs
50 × 10−9
k=
= 10−4 ×
= 0.18π = 0.5655
10−9
ε
36π
At node 1,
1
V1 = [0 + V2 + V3 + k ]
⎯⎯
→ 4V1 − V2 − V3 = k
4
At node 2,
1
V2 = [0 + V1 + V4 + k ]
⎯⎯
→ 4V2 − V1 − V4 = k
4
At node 3,
1
V3 = [0 + 2V1 + V4 + k ]
⎯⎯
→ 4V3 − 2V1 − V4 = k
4
At nde 4,
1
V4 = [0 + 2V2 + V3 + k ]
⎯⎯
→ 4V4 − 2V2 − V3 = k
4
Putting (1) to (4) in matrix form,
 4 −1 −1 0  V1  0.5655
 −1 4 0 −1 V  0.5655

 2 = 

 −2 0 4 −1 V3  0.5655

  

 0 −2 −1 4  V4  0.5655
Using a calculator or MATLAB, we obtain
V1 = V2 = 0.3231 V, V3 = V4 = 0.4039 V
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Prob. 14.11
(a)
 −4 1 0 1 0 0  Va   −200 
 1 −4 1 0 1 0  V   −100 

  b 

 0 1 −4 0 0 1  Vc   −100 

  =

 1 0 0 −4 1 0  Vd   −100 
 0 1 0 1 −4 1  Ve   0 

  

 0 0 1 0 1 −4  V f   0 
[A]
[B]
(b)
 −4 1 0 1 0 0 0 0  V1   −30 

 V  

 1 −4 1 0 1 0 0 0   2   −15 
 0 1 −4 0 0 1 0 0  V3   −30 

  

 1 0 0 −4 1 0 1 0  V4  =  −7.5
 0 1 0 1 −4 1 0 1  V5   0 

  

 0 0 1 0 1 −4 0 0  V6   −7.5
 0 0 0 1 0 0 −4 1  V   0 

 7 

−
0
0
0
0
1
0
1
4

 V8   0 
[A]
[B]
Prob. 14.12 (a) Matrix [A] remains the same. To each term of matrix [B], we add
−h 2 ρv / ε .
(b) Let Δx = Δy = h = 0.25 so that nx = 5= ny.
ρv x( y − 1)10−9
=
= 36π x( y − 1)
ε
10−9 / 36π
Modify the program in Fig. 14.16 as follows.
H=0.25;
for I=1:nx –1
for J=1: ny-1
X = H*I;
Y=H*J;
RO = 36.0*pi*X*(Y-1);
V(I,J) = 0.25*(V(I+1,J) + V(I-1,J) + V(I,J+1) + V(I,J-1) + H*H*RO);
end
end
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This is the major change. However, in addition to this, we must set
v1 = 0.0;
v2 = 10.0;
v3 = 20.0;
v4 = -10.0;
nx = 5;
ny = 5;
The results are:
Va = 4.6095 Vb= 9.9440 Vc= 11.6577
Vd = -1.5061 Ve =3.5090 Vf= 6.6867
Vg= -3.2592 Vh = 0.2366 Vi = 3.3472
Prob. 14.13
1
1
V1 = (0 + 0 + V2 + V4 ) = (V2 + V4 )
4
4
1
1
V2 = (0 + 50 + V1 + V3 ) = (50 + V1 + V3 )
4
4
1
1
V3 = (0 + 100 + 50 + V2 ) = (150 + V2 )
4
4
1
1
V4 = (0 + 50 + V1 + V5 ) = (50 + V1 + V5 )
4
4
1
1
V5 = (0 + 0 + V4 + V6 ) = (V4 + V6 )
4
4
1
1
V6 = (0 + 50 + V5 + V7 ) = (50 + V5 + V7 )
4
4
1
1
V7 = (0 + 100 + V6 + 50) = (150 + V6 )
4
4
Initially set all free potentials equal to zero. Apply the seven formulas above iteratively
and obtain the results shown below.
n
V1
V2
V3
V4
V5
V6
V7
1
0
12.5
40.625
12.5
3.12
13.281
40.82
2
6.25
24.22
43.55
14.84
7.03
24.46
43.62
9.77
25.83
43.96
16.70
10.29
25.98
43.99
3
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10.63
26.15
44.04
17.73
10.93
26.23
44.06
4
10.97
26.25
44.06
17.97
11.05
26.28
44.07
5
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Prob. 14.14
j +1
j −1
j
Φ j m +1, n + Φ j m −1, n − 2Φ j m , n
1 Φ m , n + Φ m , n − 2Φ m , n
=
(Δt ) 2
(Δx) 2
c2
+
Φ j m , n +1 + Φ j m, n −1 − 2Φ j m , n
(Δz ) 2
If h = Δx = Δz , then after rearranging we obtain
Φ j +1m, n = 2Φ j m , n − Φ j −1m , n + α (Φ j m +1, n + Φ j m −1, n − 2Φ j m , n )
+α (Φ j m , n +1 + Φ j m , n −1 − 2Φ j m , n )
where α = (cΔt / h) 2 .
Prob. 14.15
∂ 2V ∂ 2V
V ( x + Δx, t ) − 2V ( x, t ) + V ( x − Δx, t )
= 2
⎯⎯
→
=
2
∂x
∂t
(Δx) 2
V ( x, t + Δt ) − 2V ( x, t ) + V ( x, t − Δt )
(Δt ) 2
2
 Δt 
V ( x, t + Δt ) =   [V ( x + Δx, t ) − 2V ( x, t ) + V ( x − Δx, t ] + 2V ( x, t ) − V ( x, t − Δt )
 Δx 
or
V (i, j + 1) = α [V (i + 1, j ) + v(i − 1, j ) ] + 2(1 − α )V (i, j ) − V (i, j − 1)
2
 Δt 
where α =   .
 Δx 
Applying the finite difference formula derived above, the following
programs was developed.
xd=0:.1:1;td=0:.1:4;
[t,x]=meshgrid(td,xd);
Va=sin(pi*x).*cos(pi*t);%Analytical result
subplot(211) ;mesh(td,xd,Va);colormap([0 0 0])
% Numerical result
N=length(xd);M=length(td);
v(:,1)=sin(pi*xd');
v(2:N-1,2)=(v(1:N-2,1)+v(3:N,1))/2;
for k=2:M-1
v(2:N-1,k+1)=-v(2:N-1,k-1)+v(1:N-2,k)+v(3:N,k);
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end
subplot(212);mesh(td,xd,v);colormap([0 0 0])
The results of the finite difference algorithm agree perfectly with the exact solution as
shown below.
Prob. 14.16
y
θ
x
h
To find C, take the following steps:
(1) Divide each line into N equal segments. Number the segments in the lower conductor
as 1, 2, …, N and segments in the upper conductor as N+1, N+2, …, 2N,
(2) Determine the coordinate (xk, yk) for the center of each segment.
For the lower conductor, yk = 0, k=1, …, N, xk = h + Δ (k-1/2), k = 1,2,… N
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For the upper conductor, yk = [h + Δ (k-1/2)] sin θ , k=N+1, N+2, …,2 N,
xk = [h + Δ (k-1/2)] cos θ ,
k = N+1,N+2,… 2N
where h is determined from the gap g as
g
h=
2sin θ / 2
(3) Calculate the matrices [V] and [A] with the following elements
 Vo , k = 1,..., N
Vk = 
−Vo , k = N + 1,...2 N
 Δ
,i ≠ j

Aij =  4πε Rij
2ln Δ / a, i = j

where Rij = ( xi − x j ) 2 + ( yi − y j ) 2
(4) Invert matrix [A] and find [ ρ ] = [A]-1 [V].
(5) Find the charge Q on one conductor
N
Q =  ρk Δ = Δ ρk
k =1
(6) Find C = |Q|/2Vo
Taking N= 10, Vo = 1.0, a program was developed to obtain the following result.
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θ
10
20
30
40
50
60
70
80
90
100
110
120
130
140
150
160
170
180
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C (in pF)
8.5483
9.0677
8.893
8.606
13.004
8.5505
9.3711
8.7762
8.665
8.665
10.179
8.544
9.892
8.7449
9.5106
8.5488
11.32
473 8.6278
Principles of Electromagnetics, 6e
P.E.
14.8
To determine
andideas
E at in
(-1,4,5),
we use in
theFigs.
program
Fig.14.27,
14.23.we develop a
Prob.
14.17
CombiningVthe
the programs
14.23inand
MATLAB
code which gives
L
ρ L dl
, where R = 26 + (4 − y ') 2
V =
= 20
C = 19.4 pF/m
oR
0 4πεN
V=
Δ N
ρk

N
=
40
4πε k =1 26 + ( y − yk ) 2
N = 100
E=
L
ρ L dlR
 4πε R
0
o
3
C = 13.55 pF/m
C = 12.77 pF/m
For the exact value, d/2a = 50/10 = 5
where R = r – r’ = (-1, 4-y’, 5), R = |R|
πε
π × 10−9 / 36π
C= Δ N
= (−1) ρ −1
= 12.12 pF/m
k
cosh
5
−1 d
Ex ≅ cosh
2 3/ 2
4πε k =1 [26
2a + (4 − yk ) ]
Δ N
(4 − yk ) ρ k
EProb.

y ≅
2 3/ 2 the program in Fig. 14.27 and obtain the result in the table
14.18
We
may
4πε k =1 [26 + (4
− ymodify
k) ]
below. Z o ≅ 100 Ω .
E z = −5 E x
N
Zo, in Ω
For
N
=
20,
Vo = 1V, L = 1m, a = 1mm, the program in Fig. 14.23 is modified. The
10
97.2351
result
20 is: 97.8277
30
98.0515
V
=
12.47
mV, E = -0.3266 ax + 1.1353ay + 1.6331az mV/m
40
98.1739
50
98.2524
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Prob. 14.19
We make use of the formulas in Problem 14.18.
2N
Vi =  Aij ρi
j =1
where N is the number of divisions on each arm of the conductor.
The MATLAB codes is as follows:
aa=0.001;
L=2.0;
N=10; %no.of divisions on each arm
NT=N*2;
delta=L/(NT);
x=zeros(NT,1);
y=zeros(NT,1);
%Second calculate the elements of the coefficient matrix
for i=1:N-1
y(i)=0;
x(i)=delta*(i-0.5)
end
for i=N+1:NT
x(i)=0;
y(i)=delta*(i-N-0.5);
end
for i=1:NT
for j=1:NT
if (i ~=j)
R=sqrt((x(i)-x(j))^2 + (y(i)-y(j))^2)
A(i,j)=-delta*R;
else
A(i,j)=-delta*(log(delta)-1.5);
end
end
end
%Determine the matrix of constant vector B and find rho
B=2*pi*eo*vo*ones(NT,1);
rho=inv(A)*B;
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The result is presented below.
Segment
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
x
0.9500
0.8500
0.7500
0.6500
0.5500
0.4500
0.3500
0.2500
0.1500
0.0500
0
0
0
0
0
0
0
0
0
0
y
0
0
0
0
0
0
0
0
0
0
0.0500
0.1500
0.2500
0.3500
0.4500
0.5500
0.6500
0.7500
0.8500
ρ in pC/m
89.6711
80.7171
77.3794
75.4209
74.0605
73.0192
72.1641
71.4150
70.6816
69.6949
69.6949
70.6816
71.4150
72.1641
73.0192
74.0605
75.4209
77.3794
80.7171
89.6711
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Prob. 14.20(a) Exact solution yields
C = 2πε / ln(Δ / a) = 8.02607 × 10−11 F/m and Z o = 41.559Ω
where a = 1cm and Δ = 2cm. The numerical solution is shown below.
N
C (pF/m)
10
20
40
100
82.386
80.966
80.438
80.025
Z o (Ω)
40.486
41.197
41.467
41.562
(b)For this case, the numerical solution is shown below.
N
C (pF/m)
10
20
40
100
109.51
108.71
108.27
107.93
Z o (Ω )
30.458
30.681
30.807
30.905
Prob. 14.21 We modify the MATLAB code in Fig. 14.27 (for Example 14.7) by
changing the input data and matrices [A] and [B]. We let
xi = h + Δ (i-1/2), i = 1,2,… N,
Δ = L/N
yi = h /2, j = 1,2,… N, zk = t/2, k = 1,2,… N
and calculate
Rij = ( xi − x j ) 2 + ( yi − y j )2 + ( zi − z j )2
We obtain matrices [A] and [B]. Inverting [A] gives
N
-1
[q] = [A] [B], [ ρv ] = [q]/(ht Δ ), C =
q
i =1
i
10
The computed values of [ ρv ] and C are shown below.
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Co 2 =
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2h 2
[ PP
1 2 + Q1Q2 ] =
2h 2
[−h × 0 + h × (−h)] = −1
Similarly, C03 = -1 = C04. Thus
Principles of Electromagnetics, 6e
Vo = (V1 + V2 + V3 + V4)/4
475
which is the same result obtained using FDM.
Prob. 14.22
The MATLAB code is similar to the one in Fig.14.34. When the program is run, it gives
Z o = 40.587 Ω .
Prob. 14.23
Sadiku & Kulkarni
Principles of Electromagnetics, 6e
489
On the interface,
ε1
1
ε2
3
= ,
=
2(ε1 + ε 2 ) 8
2(ε1 + ε 2 ) 8
V 3V
12.5 © 2015 by Oxford University Press
V1 = 2 + 3 +Copyright
4
8
3V V
V2 = 12.5 + 4 + 1
8
4
1
V3 = (V1 + V4 )
4
1
V4 = (V2 + V3 )
4
Applying this iteratively, we obtain the results shown in the table below.
No. of iterations
V1
V2
V3
V4
Prob. 14.24
1
0
1
2
3
4
5…
100
0
0
0
0
12.5
15.62
3.125
4.688
17.57
18.65
5.566
6.055
19.25
19.58
6.33
6.477
19.77
19.87
6.56
6.608
19.93
19.96
6.6634
6.649
20
20
6.667
6.667
Copyright © 2015 by Oxford University Press
V
1
4
8
3V V
V2 = 12.5 + 4 + 1
Sadiku & Kulkarni
Principles of Electromagnetics, 6e
8
4
1
V3 = (V1 + V4 )
476
4
1
V4 = (V2 + V3 )
4
Applying this iteratively, we obtain the results shown in the table below.
No. of iterations
V1
V2
V3
V4
0
1
2
3
4
5…
100
0
0
0
0
12.5
15.62
3.125
4.688
17.57
18.65
5.566
6.055
19.25
19.58
6.33
6.477
19.77
19.87
6.56
6.608
19.93
19.96
6.6634
6.649
20
20
6.667
6.667
Prob. 14.24
V
1
V1 = (0 + 0 + 100 + V2 ) = 25 + 2
4
4
V +V
1
V2 = (0 + 100 + V1 + V3 ) = 25 + 1 3
4
4
V
1
V3 = (0 + 0 + 100 + V2 ) = 25 + 2
4
4
V
1
V4 = (0 + 0 + 100 + V5 ) = 25 + 5
4
4
1
(V4 + V6 )
Sadiku & Kulkarni V5 = (0 + 0 + 100 + V4 + V6 ) = 25 +
4
4
V5
1
V6 = (0 + 0 + 100 + V5 ) = 25 +
4
4
490
Principles of Electromagnetics, 6e
We initially set V1 = V2 = V3 = V4 =V5 = V6 = 0 and then apply above formulas
iteratively. The solutions are presented in the table below.
iteration
V1
V2
V3
V4
V5
V6
1st
25
31.25
32.81
25
31.25
32.81
V1 = V4 = 35.71 V,
2nd
32.81
41.41
35.35
23.81
41.41
35.35
3rd
35.35
42.68
35.67
35.35
42.68
35.67
4th
35.67
42.83
35.71
35.67
42.83
35.71
5th
35.71
42.85
35.71
35.71
42.85
35.71
Copyright © 2015 by Oxford University Press
V2 = V5 = 42.85V,
V3 = V6 = 35.71 V
Alternatively, if we take advantage of the symmetry, V1 = V3 = V4 = V6 and V2 = V5 . We
need to find solve two equations, namely,
V1 = 25 + V2 / 4
V2 = 25 + V1 / 2
Solving these gives
V1 = 35.714
V2 = 42.857
Other node voltages follow.
Copyright © 2015 by Oxford University Press
Prob. 14.25
32.81
25
31.25
32.81
V3
V4
Sadiku & Kulkarni
V5
V6
V1 = V4 = 35.71 V,
35.35
23.81
41.41
35.35
V2 = V5 = 42.85V,
35.67
35.35
42.68
35.67
35.71
35.67
42.83
35.71
35.71
35.71
Principles of Electromagnetics, 6e
42.85
35.71
477
V3 = V6 = 35.71 V
Alternatively, if we take advantage of the symmetry, V1 = V3 = V4 = V6 and V2 = V5 . We
need to find solve two equations, namely,
V1 = 25 + V2 / 4
V2 = 25 + V1 / 2
Solving these gives
V1 = 35.714
V2 = 42.857
Other node voltages follow.
Prob. 14.25
The finite difference solution is obtained by following the same steps as in Example 14.10.
We obtain Z o = 43 Ω
Prob.14.26
1
1
V1 = (V2 + 100 + 100 + 100) = V2 + 75
4
4
1
V2 = (V1 + V4 + 2V3 )
4
1
1
V3 = (V2 + V5 + 200) = (V2 + V5 ) + 50
4
4
1
V4 = (V2 + V7 + 2V5 )
4
491
1
V5 = (V3 + V4 + V6 + V8 )
4
1
1
V6 = (V5 + V9 + 200) = (V5 + V9 ) + 50
4
4
Sadiku & Kulkarni
Principles of Electromagnetics, 6e
1
1
V7 = (V4 + 2V8 + 0) = (V4 + 2V8 )
4
4
1
V8 = (V5 + V7 + V9 )
Copyright © 2015 by Oxford University Press
4
1
1
V9 = (V6 + V8 + 100 + 0) = (V6 + V8 ) + 25
4
4
Using these equations, we apply iterative method and obtain the results shown below.
1st
V1
V2
V3
V4
V5
V6
V7
V8
V9
75
18.75
54.69
4.687
14.687
53.71
1.172
4.003
39.43
2nd
3rd
79.687
87.11
48.437
59.64
65.82
73.87
19.824
34.57
35.14
49.45
68.82
74.2
6.958
Copyright
© 2015 by Oxford18.92
University Press
20.557
28.93
47.34
50.78
4th
5th
89.91
68.06
79.38
46.47
57.24
77.01
26.08
33.53
52.63
92.01
74.31
82.89
53.72
61.78
78.6
30.194
36.153
53.69
1
1
V7 = (V4 + 2V8 + 0) = (V4 + 2V8 )
4
4
Sadiku & Kulkarni
Principles of Electromagnetics, 6e
1
V8 = (V5 + V7 + V9 )
4
478
1
1
V9 = (V6 + V8 + 100 + 0) = (V6 + V8 ) + 25
4
4
Using these equations, we apply iterative method and obtain the results shown below.
1st
V1
V2
V3
V4
V5
V6
V7
V8
V9
75
18.75
54.69
4.687
14.687
53.71
1.172
4.003
39.43
2nd
3rd
4th
5th
79.687
48.437
65.82
19.824
35.14
68.82
6.958
20.557
47.34
87.11
59.64
73.87
34.57
49.45
74.2
18.92
28.93
50.78
89.91
68.06
79.38
46.47
57.24
77.01
26.08
33.53
52.63
92.01
74.31
82.89
53.72
61.78
78.6
30.194
36.153
53.69
Prob. 14.27
Sadiku & Kulkarni
Applying the difference method,
V V
V1 = 3 + 2 + 25
4 2
1
V2 = (V1 + V4 ) + 50
4
1
V3 = (V1 + 2V4 )
4
1
V4 = (V2 + V3 + V5 )
4
V
V5 = 4 + 50
4
Principles of Electromagnetics, 6e
492
Applying these equations iteratively, we obtain the results below.
Iterations
0
1
2
3
4
5…
100
V1
V2
V3
V4
V5
0
0
0
0
0
25.0
56.25
6.25
15.63
53.91
54.68
67.58
17.58
35.74
58.74
64.16
74.96
33.91
41.96
60.49
70.97
78.23
38.72
44.86
61.09
73.79
79.54
40.63
45.31
61.37
74.68
80.41
41.89
45.95
51.49
Copyright © 2015 by Oxford University Press
Copyright © 2015 by Oxford University Press
Sadiku & Kulkarni
Principles of Electromagnetics, 6e
479
CHAPTER 15
P. E. 15.1 From Table 15.1, the functional for the two-dimensional diffusion equation
is
Sadiku & Kulkarni
The function of the functional is 457

Principles of Electromagnetics, 6e
  
(b)
Using
theinsame
program with nx
= 12+1=13 and ny = 24+1=25, the potential at the
Putting
this
the Euler-Lagrange
equation,
center is
 

 


 
  
V(7,13) = 23.883 V
    ­ €   in Figs.
 14.23 and 14.27, and dividing each wire into
P. E. 14.7 By combining the 
ideas
­ €  Hence, minimizing the above functional leads to the solution of the diffusion equation.
N segments, the results listed in Table 14.2 is obtained.
P. E.
E. 15.2
15.4
P.
Using
Eq. (15.22):
(a)
Energy stored =
3
Š†‡  Š ‡  Š
‚ „ † ‡ˆ ‰ Š ‡
Š‹  ŠŒŽ‘ ‚ ƒ ‚ ‰†‘ ‚ ‹Š‹†2‚ ’ ŠŒ “”•
‡2
 Š  ŠŒŽ
Š Ž
4
1
P. E. 15.5
œ
–—
‰–— ‘ ™ š ›ž ¡
–—˜
˜ ˜ Ÿ 1
–— „ ˆ
  ˜  ˜  
For element 1, local
1-2-3
with
global
so that 
A1 
= 0.35,
˜  corresponds
 
˜ 1-3-4
 –—
˜ ˜ 





™˜  ˜   ˜ ˜   š ‰–— ‘
P1 = 0.8, P2 = 0.6,
P = -1.4, Q1 = -0.5, Q2 = 0.5, Q3 = 0
–—˜
˜3  
  ˜ 
  
‚
ƒ
˜
– ¢ £¤ ¥  –—¤
 0.6357 0.1643 −0.8  —
¤¦
C (1) =  0.1643 0.4357 −0.6 
„˜  ˜  ˜  ˜ 
£
 −
0.8 −0.6
1.4 
˜  ˜  
    ˜ 
˜corresponds
  ˆ with global 1-2-3 so that A2 = 0.7,
For element 2, local 1-2-3
    ˜ 
P1 = 0.1, P2 = 1.4, P3 = -1.5, Q1 = -1, Q2 = 0, Q3 = 1
C
(2)
 0.3607 0.05 −0.4107 
=  0.05
−0.75 
0.7
Copyright © 2015 by Oxford University Press
 −0.4107 −0.75 1.1607 
P1 = 0.8, P2 = 0.6, P3 = -1.4, Q1 = -0.5, Q2 = 0.5, Q3 = 0
Sadiku & Kulkarni
Principles of Electromagnetics, 6e
C
(1)
 0.6357 0.1643 −0.8 
=  0.1643 0.4357 −0.6 
 −0.8
−0.6
1.4 
480
For element 2, local 1-2-3 corresponds with global 1-2-3 so that A2 = 0.7,
P1 = 0.1, P2 = 1.4, P3 = -1.5, Q1 = -1, Q2 = 0, Q3 = 1
Sadiku & Kulkarni C
(2)
 0.3607 0.05 −0.4107 
=  0.05
−0.75 
0.7
 −0.4107 −0.75 1.1607 
The global coefficient matrix is given by
C (1)11 + C11(2)

C21(2)

C=
 C21(1) + C31(2)

C31(1)

C12 (2)
C22 (2)
C32 (2)
0
C12 (1) + C13(2)
C23(2)
C22 (1) + C33(2)
C32 (2)
Principles of Electromagnetics, 6e
458
C13(1) 

0 
C23(1) 

C33(1) 
0.2464 © 2015
−0.8by Oxford University Press
 0.9964 0.05 −Copyright
 0.05
0.7
0 
−0.75

=
 −0.2464 −0.75 1.596
−0.6 


0
1.4 
−0.75
 −0.8
(b)
3
2
4
2
1
1
For element 1, local 1-2-3 corresponds with global 1-2-4 .
P1 = 0.9000 ; P2 = 0.6000 ; P3 = -1.5000
Q1 = -1.5000 ; Q2 = 0.5000; Q3 = 1;
A1 = 0.6750;
C
(1)
=
 1.1333 -0.0778 -1.0556 
 -0.0778 0.2259 -0.1481 


 -1.0556 -0.1481 1.2037 
For element 2, local numbering 1-2-3 corresponds with global numbering 2-3-4.
Copyright ;© P
2015
University
Press
P1 = 0.8000; P2 = -0.9000
0.1000
;
3 =by Oxford
Q1 = -0.5000 ; Q2 = 1.5000 ; Q3 = -1 ;
1
Sadiku & Kulkarni For
element 1, local 1-2-3 corresponds with global 1-2-4 .
P1 = 0.9000 ; P2 = 0.6000 ; P3 = -1.5000
Q1 = -1.5000 ; Q2 = 0.5000; Q3 = 1;
481
A1 = 0.6750;
C
Sadiku & Kulkarni
(1)
=
Principles of Electromagnetics, 6e
 1.1333 -0.0778 -1.0556 
 -0.0778 0.2259 -0.1481 


 -1.0556 -0.1481 1.2037 
For element 2, local numbering 1-2-3 corresponds with global numbering
2-3-4.
Principles
of Electromagnetics, 6e
P1 = 0.8000; P2 = -0.9000 ; P3 = 0.1000 ;
Q1 = -0.5000 ; Q2 = 1.5000 ; Q3 = -1 ;
459
A2 = 0.3750 ;
C
(2)
=
 0.5933 -0.9800 0.3867 
 -0.9800 2.0400 -1.0600 


 0.3867 -1.0600 0.6733 
The global coefficient matrix is
(1)
University
Press
C (1)11

C12 (1) Copyright
0 © 2015 byCOxford
13
 (1)

C
C22 (1) + C11(2) C12 (2) C23(1) + C13(2) 
C =  21
 0

C12 (2)
C22 (2)
C23(2)
 (1)

C32 (1) + C31(2) C32 (2) C33(1) + C33(2) 
 C31
C
0
 1.1333 -0.0778
 -0.0778 0.8193 -0.9800
= 
 0
-0.9800 2.0400

 -1.0556 0.2385 -1.0600
-1.0556 
0.2385 
-1.0600 

1.8770 
P.E. 15.3 We use the MATLAB program in Fig. 15.8. The input data for the region in
Fig. 15.9 is as follows:
NE = 32; ND = 26; NP = 18;
NL = [ 1 2 4
2 5 4
2 3 5
3 6 5
4 5 9
5 10 9
5 6 10
6 11 10
7 8 12
8 13 12
8 9 13
9 14 13
9 10 14
10 15 14
Copyright © 2015 by Oxford University Press
10 11 15
11 16 15
Sadiku & Kulkarni
Sadiku & Kulkarni
NE = 32; ND = 26; NP = 18;
NL = [ 1 2 4
2 5 4
Principles of Electromagnetics, 6e
2 3 5
3 6 5
4 5 9
482
5 10 9
5 6 10
6 11 10
7 8 12
8 13 12
8 9 13
9 14 13
9 10 14
10 15 14
Principles of Electromagnetics, 6e
10 11 15
11 16 15
12 13 17
460
13 18 17
13 14 18
14 19 18
14 15 19
15 20 19
15 16 20
16 21 20
17 18 22
Copyright © 2015 by Oxford University Press
18 23 22
18 19 23
19 24 23
19 20 24
20 25 24
20 21 25
21 26 25];
X = [ 1.0 1.5 2.0 1.0 1.5 2.0 0.0 0.5 1.0 1.5 2.0 0.0 0.5 1.0 1.5 2.0 0.0 0.5 1.0 1.5
2 0.0 0.5 1.0 1.5 2.0];
Y = [ 0.0 0.0 0.0 0.5 0.5 0.5 1.0 1.0 1.0 1.0 1.0 1.5 1.5 1.5 1.5 1.5 2.0 2.0 2.0
2.0 2.0 2.5 2.5 2.5 2.5 2.5 ];
NDP = [ 1 2 3 6 11 16 21 26 25 24 23 22 17 12 7 8 9 4];
VAL = [0.0 0.0 15.0 30.0 30.0 30.0 30.0 25.0 20.0 20.0 20.0 10.0 0.0 0.0 0.0
0.0 0.0 0.0];
With this data, the finite element (FEM) solution is compared with the finite
difference (FD) solution as shown in the table below.
Node #
5
10
13
14
15
18
19
20
X
1.5
1.5
0.5
1.0
1.0
0.5
1.0
1.5
Y
0.5
1.0
1.5
1.5
1.5
2.0
2.0
2.0
FEM
11.265
15.06
4.958
9.788
18.97
10.04
15.32
21.05
Copyright © 2015 by Oxford University Press
FD
11.25
15.02
4.705
9.545
18.84
9.659
15.85
20.87
Putting this in the Euler-Lagrange equation,

 
 


  
 
Sadiku & Kulkarni
 
    ­ €
­ € 483
 Principles of Electromagnetics, 6e
Hence, minimizing the above functional leads to the solution of the diffusion equation.
P. E. 15.4
Using Eq. (15.22):
Energy stored =
‚
ƒ
Š†‡  Š ‡  Š
‚ „ † ‡ˆ ‰ Š ‡
Š‹  ŠŒŽ‘ ‚  Š  ŠŒŽ
Š Ž
P. E. 15.5
ƒ
‚ ‰†‘ ‚ ‹Š‹† ‚ ‡
œ
–—
‰–— ‘ ™ š ›ž ¡
–—˜
˜ ˜ Ÿ
–— „ ˆ
  ˜  ˜  
  ˜ ˜ ˜  ˜ 

™˜  ˜   ˜  ˜ 
˜  
  ˜ 
˜
–— ¢ £¤ ¥  –—¤
’
ŠŒ “”•
   –—


  š ‰–— ‘
–—˜
  
¤¦
„  ˜  ˜  ˜ 
£   ˜  ˜   ˜
    ˜ 
˜   ˆ     ˜ 
Therefore, for an node of an element having nodes, the shape function is
P. E. 15.6
Using eq. (15.38),
      
Using eq. (15.11), we obtain:
     
    
Copyright © 2015 by Oxford
 University Press



    
Sadiku & Kulkarni
Therefore, for an P. E. 15.6
Principles of Electromagnetics, 6e
node of an element having nodes, the shape function is
484
Using eq. (15.38),
      
Using eq. (15.11), we obtain:
    
    
 


    


     
     
 ­
€ ‚ƒ ­
€ ‚„


  †‡ˆ‰Š‹†‡ˆ‰‹†‰†ŒŽ
‰†‡ˆ‰†ŒŽ
‘ ‚ƒ †‹†‡ˆ‰Š‹†‡’‰‰†ŒŽ
 “‡”‚ƒ “‡•‚„
‰†‡ˆ‰†ŒŽ
‘ ‚„
P. E. 15.7
Area of element 1 “‡” ‰ “Š’ –
Area of element 2 “‡” ‰ “Š’ –
Using equation (15.53),
 “‡” ‰ “Š’  — — ˜  ™ ˜  ™





Assembling of elemental matrices — into global matrix — is similar to assembling
the element coefficient matrix given in Section 15.3c.
Therefore global matrix is
2 1 1 0
1 4 2 1 0.5 106
T
1 2 4 1
12
0 1 1 2
P. E. 15.8
For element 1 the local node numbers are assigned as 1-2-3 for the global node numbers
1-2-4.
Copyright © 2015 by Oxford University Press
From eq. (15.11b) shape function of node 2 in element 1 is
2
1
T 1
0
485
Sadiku & Kulkarni
1
4
2
1
1
2
4
1
0
1 0.5 106
1
12
2
Principles of Electromagnetics, 6e
P. E. 15.8
For element 1 the local node numbers are assigned as 1-2-3 for the global node numbers
1-2-4.
From eq. (15.11b) shape function of node 2 in element 1 is
From eq. (15.12), the area of element 1:

 ­€ ‚ ‚ ƒ„ †
ƒ
Similarly for element 2 the local nodes are assigned as 1-2-3 for the global node
numbers 2-3-4
From eq. (15.12), the area of element 2 is    
­€ ƒ „ 
Line equation of segment 4-2:
  ‡

 ˆ
The expression of the two shape functions of node 2 can be rewritten by eliminating form eq. (A) and eq. (B) using eq. (C).
The expression of the shape function in element 1:
The expressionof the shape function in element 2:
Thus, we can conclude that the expressionof the shape function is same on the segment
4-2 for both elements.
P. E. 15.9
% The geometry is the same as that in Example 14.4
% The geometry is drawn in the pde toolbox and p e t matrices are imported into
% the MATLAB code using procedure described in Section 15.7
clc;
clear all;
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load meshdata;%% loading
p, e,©t2015
matrices
[A3 n_elements]=size(t);% A3=4
The expressionof the shape function in element 2:
Principles
Sadiku & Kulkarni
of Electromagnetics, 6e
Thus, we can conclude that the expressionof
486the shape function is same on the segment
4-2 for both elements.
P. E. 15.9
% The geometry is the same as that in Example 14.4
% The geometry is drawn in the pde toolbox and p e t matrices are imported into
% the MATLAB code using procedure described in Section 15.7
clc;
clear all;
load meshdata;%% loading p, e, t matrices
[A3 n_elements]=size(t);% A3=4
[A2 n_edges]=size(e);% A2=7
[A1 n_nodes]=size(p);% A1=2
C=zeros(n_nodes,n_nodes);% Initialization of the global coefficient matrix
K=zeros(n_nodes,n_nodes);
V1=40;% Potential of the top plate
V2=100;% Potential of the plate on the right hand side
V3=0;% Potential of the plate on the left hand side
V4=10;% Potential of the bottom plate
V=zeros(n_nodes,1);% Initialization of matrix of unknown potentials
B=zeros(n_nodes,1);% Initialization of the right hand side matrix
%contributed by boundary conditions
%Formation of coefficient matrices for all elements
for element=1:n_elements
nodes=t(1:3,element);% 'nodes' is a 3 by 1 matrix containing global
% node numbers of element under consideration
Xc=p(1,nodes'); % X coordinates of nodes
Yc=p(2,nodes'); % Y coordinates of nodes
P=zeros(3,1);
Q=zeros(3,1);
P(1)=Yc(2)-Yc(3);
P(2)=Yc(3)-Yc(1);
P(3)=Yc(1)-Yc(2);
Q(1)=Xc(3)-Xc(2);
Q(2)=Xc(1)-Xc(3);
Q(3)=Xc(2)-Xc(1);
% eq. (15.21c)
delta= 0.5*abs((P(2)*Q(3))-(P(3)*Q(2)));%Absolute value is taken since
% the three nodes may not have been numbered in the anticlockwise direction
for i=1:3
for j=1:3
% eq. (15.21b)
c(element,i,j)=((P(i)*P(j))+(Q(i)*Q(j)))/(4*delta);
end
end
end
for element=1:n_elements
nodes=t(1:3,element);
% Formation of the global coefficient matrix
for i=1:3
for j=1:3
C(nodes(i),nodes(j))=c(element,i,j)+C(nodes(i),nodes(j));
end
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end
end
Sadiku & Kulkarni
for i=1:3
for j=1:3
% eq. (15.21b)
c(element,i,j)=((P(i)*P(j))+(Q(i)*Q(j)))/(4*delta);
end
487
end
end
for element=1:n_elements
nodes=t(1:3,element);
% Formation of the global coefficient matrix
for i=1:3
for j=1:3
C(nodes(i),nodes(j))=c(element,i,j)+C(nodes(i),nodes(j));
end
end
end
% Imposing boundary conditions
K=C;
for edge=1:n_edges
if e(5,edge)==1
node1=e(1,edge);
node2=e(2,edge);
B(node1)=V1;
B(node2)=V1;
K(node1,:)=zeros(1,n_nodes);
K(node1,node1)=1;
K(node2,:)=zeros(1,n_nodes);
K(node2,node2)=1;
end
if e(5,edge)==2
node1=e(1,edge);
node2=e(2,edge);
B(node1)=V2;
B(node2)=V2;
K(node1,:)=zeros(1,n_nodes);
K(node1,node1)=1;
K(node2,:)=zeros(1,n_nodes);
K(node2,node2)=1;
end
if e(5,edge)==3
node1=e(1,edge);
node2=e(2,edge);
B(node1)=V4;
B(node2)=V4;
K(node1,:)=zeros(1,n_nodes);
K(node1,node1)=1;
K(node2,:)=zeros(1,n_nodes);
K(node2,node2)=1;
end
if e(5,edge)==4
node1=e(1,edge);
node2=e(2,edge);
B(node1)=V3;
B(node2)=V3;
K(node1,:)=zeros(1,n_nodes);
K(node1,node1)=1;
K(node2,:)=zeros(1,n_nodes);
K(node2,node2)=1;
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Sadiku & Kulkarni
B(node2)=V4;
K(node1,:)=zeros(1,n_nodes);
K(node1,node1)=1;
K(node2,:)=zeros(1,n_nodes);
K(node2,node2)=1;
end
if e(5,edge)==4
node1=e(1,edge);
node2=e(2,edge);
B(node1)=V3;
B(node2)=V3;
K(node1,:)=zeros(1,n_nodes);
K(node1,node1)=1;
K(node2,:)=zeros(1,n_nodes);
K(node2,node2)=1;
end
Principles of Electromagnetics, 6e
488
end
% When the goemetry of Fig. 14.14 is drawn using the "rectangle"
% option in the pdetoolbox, the four corner nodes get numbered as 1,2,3 and 4
% These nodes need to be made floating to represent gaps between the
% different plates.
for i=1:n_nodes
if (i==1||i==2||i==3||i==4)
K(i,:)=C(i,:);
B(i)=0;
end
end
%solving for unknown potentials
V=K\B;
figure(1);
pdeplot(p,e,t,'xydata',V,'mesh','off','colormap','jet','colorbar','on',
'contour','on','levels',20);
[Ex,Ey]=pdegrad(p,t,V);% Calculation of E field if required
E=sqrt(Ex.^2+Ey.^2);
figure(2)
pdeplot(p,e,t,'xydata',E,'colorbar','off')
disp('potential at the point (0.5, 0.5)')
x=0.5;
y=0.5;
for i=1:n_nodes
xc=p(1,i);
yc=p(2,i);
dist(i)=((x-xc)^2+(y-yc)^2)^0.5;
end
[d,node]=min(dist);
V(node,1)
P. E. 15.10
The code remains essentially the same as in Example 15.10 except for boundary
conditions which need to be explicitly defined as explained in the code for TM modes
and at the end of Section 14.4 above Practice Exercise 14.6.
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I
Sadiku & Kulkarni
ρvi (×10−6 )C / m3Principles of Electromagnetics, 6e
1, 20
0.5104
2, 19
0.4524
489
3, 18
0.4324
4, 17
0.4215
5, 16
0.4144
P. E. 15.10
6,
15
0.409615.10 except for boundary
The code remains essentially the same as in Example
7, 14
0.4063
conditions which need to be explicitly defined as explained in the code for TM modes
8, 13
0.4041
and at the end of Section 14.4
above
Practice
Exercise
14.6.
9, 12
0.4027
10,11
0.4020
V(node,1)
C = 17.02 pF
Prob. 15.1
From the given figure, we obtain
1 x
1
A1
α1 = =
1 x2
A 2A
1 x3
y
1
y2 =
[( x2 y3 − x3 y2 ) + ( y2 − y3 ) x + ( x3 − x2 ) y ]
2A
y3
as expected. The same applies for α 2 and α 3 .
Prob. 15.2
(a)
(b)
P1 = 1.5, P2 = 0.5, P3 = −2, Q1 = −1, Q2 = 1.5, Q3 = −0.5
1
A = ( P2Q3 − PQ
3 2 ) = 1.375
2
1
Cij =
[ PP
i j + Qi Q j ]
4A
 0.5909 −0.1364 −0.4545 
C =  −0.1364 0.4545 −0.3182 
 −0.4545 −0.3182 0.7727 
P1 = −4, P2 = 4, P3 = 0, Q1 = 0, Q2 = −3, Q3 = 3
1
A = ( P2Q3 − P3Q2 ) = 6
2
0 
 0.6667 −0.6667

C =  −0.6667 1.042
−0.375
 0
−0.375 0.375 
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Prob. 15.3 (a)
1 1/ 2 1/ 2
2A = 1
1
3
2
1 / 2 = 15/4
2
α1 =
4
1
4
[(6 − 1) + (−1 ) x + ( −1) y ] = (5 − 1.5 x − y )
15
2
15
α2 =
4
3
3
4
[(1 − 1) + x − y ] = (1.5 x − 1.5 y )
15
2
2
15
α3 =
4
5
4
[(1 / 4 − 3 / 2) + 0 x + y ] = (−1.25 + 2.5 y )
15
2
15
V = α1V1 + α 2V2 + α 3V3
Substituting V=80, V1 = 100, V2 = 50, V3 = 30, α1 , α 2 , and α 3 leads to
20 = 7.5x + 10y + 3.75
Along side 12, y=1/2 so that
20 = 15x/2 + 5 + 15/4
x=3/2, i.e (1.5, 0.5)
Along side 13, x =y
20 = 15x/2 + 10x + 15/4
Along side 23, y = -3x/2 + 5
x=13/4, i.e. (13/14, 13/14)
20 = 15x/2 – 15 + 50 + 15/4
x=-5/2 (not possible)
Hence intersection occurs at
(1.5, 0.5) along 12 and (0.9286, 0.9286) along 13
(b) At (2,1),
α1 =
4
6
5
, α 2 = , α3 =
15
15
15
V (2,1) = α1V1 + α 2V2 + α 3V3 = (400 + 300 + 150)/15 = 56.67 V
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Prob. 15.4
1 0
0
2A = 1 2 −1 = 9
1 1 4
1
9
1
9
1
9
1
9
1
9
1
9
α1 = [(0 − 0) + (4 − 0) x + (0 − 1) y ] = (4 x − y )
α 2 = [(0 − 0) + (0 + 1) x + (2 − 0) y ] = ( x + 2 y )
α 3 = [(8 + 1) + (−1 − 4) x + (1 − 2) y ] = (9 − 5 x − y )
Ve = α1Ve1 + α 2Ve 2 + α 31Ve3
V(1,2) = 8(4-2)/9 + 12(1+4)/9 + 10(9-5-1)/9 = 96/9 = 10.667 V
At the center α1 = α 2 = α 3 = 1/3 so that
V(center) = (8 + 12 + 10)/3 = 10
Or at the center, (x, y) = (0 + 1 + 2, 0 + 4 –1)/3 = (1,1)
V(1,1) = 8(3)/9 + 12(3)/9 + 10(3)/9 = 10 V
Prob. 15.5
(3,12)
(3,12)
(8,12)
2
1
(8,0)
(8,0)
(0,0)
For element 1, local numbering 1-2-3 corresponds to global numbering 4-2-1.
P1 = 12, P2 = 0, P3 = -12, Q1 = -3, Q2 = 8, Q3 = -5,
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A = (0 + 12 x 8)/2 = 48
Cij =
C
(1)
1
[ Pj Pi + Q j Q j ]
4 x 48
 0.7969 −0.125 −0.6719 
=  −0.125 0.3333 −0.2083 
 −0.6719 −0.2083 0.8802 
For element 2, local numbering 1-2-3 corresponds to global numbering 2-4-3.
P1 = -12, P2 = 0, P3 = 12, Q1 = 0, Q2 = -5, Q3 = 5,
A = (0 + 60)/2 = 30
Cij =
C
(2)
1
[ Pj Pi + Q j Q j ]
4 x 48
0
−1.2 
 1.2

0.208 −0.208
= 0
 −1.2 −0.208 1.408 
C (1) 33
 (1)
C
C =  23
 0
 (1)
 C13
C23(1)
(1)
(2)
11
C22 + C
(2)
31
C
C21(1) + C21(2)
C13


C
+ C12 

C32 (2)
(2)
(1) 
C22 + C11 
C31(1)
0
(2)
C33(2)
C23(2)
(1)
21
(2)
0
−0.6719 
 0.8802 −0.2083
 −0.2083 1.533
−1.2
−0.125 
=
 0
−1.2
1.4083 −0.2083


 −0.6719 −0.125 −0.2083 1.0052 
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Prob. 15.6
4
(0,2) 3
For element 1,
1
(0,0)
1
1

1
2
3
3
(2,2)
2

2
2
(4,0)
P1 = 0, P2 = 2, P3 = −2, Q1 = −2, Q2 = 0, Q3 = 2
1
A = (4 − 0) = 2, 4 A = 8
2
0
−0.5
 4 0 −4   0.5
1


0.5 −0.5
C =  0 4 −4  =  0
8
 −4 −4 8   −0.5 −0.5
1 
For element 2,
(1)
P1 = −2, P2 = 2, P3 = 0, Q1 = −2, Q2 = −2, Q3 = 4
1
A = (8 − 0) = 4, 4 A = 16
2
0
−0.5
 8 0 −8  0.5
1 


(2)
0.5 −0.5
C =  0 8 −8 =  0
16
 −8 −8 16   −0.5 −0.5
1 
The global coefficient matrix is
 C11
C
C =  21
C31

C41
C12
C22
C32
C42
C13 C14  C11(1) + C11(2)

C23 C24   C21(2)
=
C33 C34  C21(1) + C31(2)
 
C43 C44   C31(1)
0
−0.5 −0.5
 1
 0
0.5 −0.5
0 

=
 −0.5 −0.5 1.5 −0.5


0
1 
−0.5
 −0.5
C12(2)
C22(2)
C12(1) + C13(2)
(2)
C23
C32(2)
0
(1)
+ C33(2)
C22
C32(1)
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C13(1) 

0 
(1) 
C23

C33(1) 
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Prob. 15.7
4
4
(2,2)
3
(0,1)
1
1
(2,2)
1
1
2
2
2
(1,0)
2
2
(1,0)
3
3
(3,0)
For element 1, local numbering 1-2-3 corresponds to global numbering 1-2-4.
P1 = -2, P2 = 1, P3 = 1, Q1 = 1, Q2 = -2, Q3 = 1,
A = (P2 Q3 - P3 Q2)/2 = 3/2, i.e. 4A = 6
Cij =
C
(1)
1
[ Pj Pi + Q j Q j ]
4A
 5 −4 −1
1
=  −4 5 −1
6
 −1 −1 2 
For element 2, local numbering 1-2-3 corresponds to global numbering 4-2-3.
P1 = 0, P2 = -2, P3 = 2, Q1 = 2, Q2 = -1, Q3 = -1,
A = 2, 4A = 8
C
(2)
 4 −2 −2 
1
=  −2 5 −3
8
 −2 −3 5 
The global coefficient matrix is
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C (1)11
 (1)
C
C =  12
 0
 (1)
 C13
C12 (1)
(1)
C22 + C22
C23
(2)
(2)
C23(1) + C21(2)
C23


C23 + C 

C31(2)

C33(1) + C11(2) 
C13(1)
0
(2)
C33(2)
C31(2)
(1)
(2)
21
0
−0.1667 
 0.8333 −0.667
 −0.6667 1.4583 −0.375 −0.4167 

=
 0
−0.375 0.625
−0.25 


0.833 
 −0.1667 −0.4167 −0.25
Prob. 15.8 We can do it by hand as in Example 15.2. However, it is easier to prepare
an input file and use the program in Fig. 15.8. The MATLAB input data is
NE = 2;
ND = 4;
NP = 2;
NL = [1 2 4
4 2 3];
X = [ 0.0 1.0 3.0 2.0];
Y = [ 1.0 0.0 0.0 2.0];
NDP= [ 1 3 ];
VAL = [ 10.0 30.0]
10 
18 
 
The result is V = 30 
 
 20 
 
From this,
V2 = 18 V, V4 = 20 V
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Prob. 15.9
2
4
1
2
3
5
2
4
1
5
3
1
2
The local numbering 1-2-3 in element 3 corresponds with the global numbering 5-41, while the local number 1-2-3 in element 4 corresponds with the global numbering
5-1-2.
C5,5 = C11(2) + C11(3) + C11(4) + C11(5) , A = 2,
C11(2) = (2 x 2 + 2 x 2)/8 = 1 = C11(5)
C11(3) = (2 x 2 + 0)/8 = ½ = C11(4)
C5,5 = 1 + 1 + ½ + ½ = 3
C5,1 = C31(3) + C21(4)
1
(P3P1 + Q3 Q1) = 0 since P3 = 0 = Q3
8
1
C21(4) = (P2P1 + Q2 Q1) = 0 since P3 = 0 = Q3
8
But C31(3) =
C5,1 = 0
Prob. 15.10 As in P. E. 15.3, we use the program in Fig. 15.8. The input data based
on Fig. 15.19 is as follows.
NE =50; ND= 36; NP= 20;
NL = [1
8
7
1
2
8
2
9
8
2
3
9
3
10
9
3
4
10
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4
11
10
4
5
11
5
12
11
5
6
12
7
14
13
7
8
14
8
15
14
8
9
15
9
16
15
9
10
16
10
17
16
10
11
17
11
18
17
11
12
18
13
20
19
13
14
20
14
21
20
14
15
21
15
22
21
15
16
22
16
23
22
16
17
23
17
24
23
17
18
24
19
26
25
19
20
26
20
27
26
20
21
27
21
28
27
21
22
28
22
29
28
22
23
29
23
30
29
23
24
30
25
32
31
25
26
32
26
33
32
26
27
33
27
34
33
27
28
34
28
35
34
28
29
35
29
36
35
29
30
36];
X = [0.0 0.2 0.4 0.6 0.8 1.0 0.0 0.2 0.4 0.6 0.8 1.0 0.0 0.2 0.4 0.6 0.8 1.0
0.0 0.2 0.4 0.6 0.8 1.0 0.0 0.2 0.4 0.6 0.8 1.0 0.2 0.4 0.6 0.8 1.0];
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Y = [0.0 0.0 0.0 0.0 0.0 0.0 0.2 0.2 0.2 0.2 0.2 0.2 0.4 0.4 0.4 0.4 0.4
0.4 0.6 0.6 0.6 0.6 0.6 0.6 0.8 0.8 0.8 0.8 0.8 0.8 1.0 1.0 1.0 1.0 1.0 1.0];
NDP = [ 1 2 3 4 5 6 12 18 24 30 36 35 34 33 32 31 25 19 13 7];
VAL = [ 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 50.0 100.0 100.0 100.0
100.0 50.0 0.0 0.0 0.0 0.0];
With this data, the potentials at the free nodes are compared with the exact values as
shown below.
Node no.
8
9
10
11
14
15
16
17
20
21
22
23
26
27
28
29
FEM Solution
4.546
7.197
7.197
4.546
10.98
17.05
17.05
10.98
22.35
32.95
32.95
22.35
45.45
59.49
59.49
45.45
Exact Solution
4.366
7.017
7.017
4.366
10.60
16.84
16.84
10.60
21.78
33.16
33.16
21.78
45.63
60.60
60.60
45.63
Prob. 15.11 We use exactly the same input data as in the previous problem except that
the last few lines are replaced by the following lines.
VAL = [0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 29.4 58.8 95.1 95.1
58.8 29.4 0.0 0.0 0.0 0.0];
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Principlesof
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Principles
499
487
The potential at the free nodes obtained with the input data are compared with the exact
solution as shown below.
Node no.
8
9
10
11
14
15
16
17
20
21
22
23
26
27
28
29
FEM Solution
3.635
5.882
5.882
3.635
8.659
14.01
14.01
8.659
16.99
27.49
27.49
16.99
31.81
51.47
51.47
31.81
Exact Solution
3.412
5.521
5.521
3.412
8.217
13.30
13.30
8.217
16.37
26.49
26.49
16.37
31.21
50.5
50.5
31.21
Prob. 15.12
For element 1, the local numbering 1-2-3 corresponds with nodes with V1 , V2 , and V3.
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& Kulkarni
Kulkarni
Principlesof
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Electromagnetics,6e
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Principles
500
488
Vo = −
1 4
ViCio
Coo i =1
4
Coo =  Coj ( e ) =
j =1
1
2
4h / 2
(hh + hh) × 2 +
1
2
4h / 2
(hh + 0) × 4 = 4
Co1 =
2 ×1
2
[ P3 P1 + Q3Q1 ] = 2 [−hh − 0] = −1
2
2h
2h
Co 2 =
2 ×1
2
[ PP
[−h × 0 + h × (−h)] = −1
1 2 + Q1Q2 ] =
2
2h
2h 2
Similarly, C03 = -1 = C04. Thus
Vo = (V1 + V2 + V3 + V4)/4
which is the same result obtained using FDM.
Prob. 14.22
The MATLAB code is similar to the one in Fig.14.34. When the program is run, it gives
Z o = 40.587 Ω .
Prob. 14.23
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