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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
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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
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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
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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
−
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Method 2:
The flux linkage is given by
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•
(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
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= 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
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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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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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258
255
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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Principles of Electromagnetics, 6e
259
256
(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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Principles of Electromagnetics, 6e
260
257
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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Principles of Electromagnetics, 6e
261
258
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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260
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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261
(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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262
(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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263
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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264
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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265
(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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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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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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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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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
292
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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Principles of Electromagnetics, 6e
293
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
294
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
Sadiku & Kulkarni
Principles of Electromagnetics, 6e
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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Sadiku & Kulkarni
Principles of Electromagnetics, 6e
296
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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Sadiku & Kulkarni
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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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
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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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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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Principles of Electromagnetics, 6e
314
(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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Principles of Electromagnetics, 6e
315
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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Principles of Electromagnetics, 6e
316
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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Principles of Electromagnetics, 6e
317
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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Principles of Electromagnetics, 6e
318
(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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y
Sadiku & Kulkarni
Principles of Electromagnetics, 6e
319
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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Sadiku & Kulkarni
Principles of Electromagnetics, 6e
320
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 )
Copyright © 2015 by Oxford University Press
Sadiku & Kulkarni
Principles of Electromagnetics, 6e
321
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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Sadiku & Kulkarni
Principles of Electromagnetics, 6e
322
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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