SOLUTION MANUAL

CHAPTER
Static Magnetic
7 Fields

7.1 Since (a > b), the dominant mode is the TE10 whose cutoff frequency is defined using equation
(7 – 90)
m 2+ n 2
(fc)10 = 1
2√με a b √   
which simplifies to
c
= 3 × 10 = 9 GHz
m 1
(fc)10 = 1 = 1
8
=
2√με a 2√μ0ε0 a 2a 2a
 a = 1.667 × 10–2 m
The cutoff frequency for TE20 is
m cm 2c
(fc)20 = 1 = = = 18 GHz > 14 GHz
2√με a 2a 2a
Therefore the next higher-order mode is TE01, which cutoff frequency is then
(fc)01 = 1 n = 1 1 = c = 3 × 108 = 14 GHz
2√με b 2√μ0ε0 b 2b 2ab
 b = 1.071 × 10 m
–2

7.2 From equation (7 – 90), we have (for both TEmn & TMmn modes)
m 2 n 2
(fc)mn =
1
2√με a √   
+
b
(a) The lowest-order mode is TE10, then equation (7 – 90) simplifies to
1 3 × 108
(fc)10 = 1 m = 1 = c = = 4.101 GHz
2√με a 2√μ0ε0εr a 2√μ0a 2√2.56 × 2.286 × 10–2
Given the f = 10 × 109 Hz > (fc)10 = 4.101 GHz,
We can then apply equation (7 – 39) such that

λg = 2π = λ
β
1 – fc
√  
2

f
(Note: same equation for both TE and TM waves)

Therefore, λ0 (3× 108)

λ √εr (10× 109 × √2.56
λg = = = = 2.056× 10–2 m
4.101 2
1 – fc
√   √1 –  ff  √ 
2

2
c 1–
f 10
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(b) From equation (7 – 50), we have

ZTE = jωμ = η
γ fc
√1 –  f 
2

377
= η √2.56 = 258.36 Ω
1 – fc
√   √  4.101
2 2

f
1–
10 
7.3 Given:
a = 7.214 × 10–2 m; b = 3.404 × 10–2 m; f = 3 × 109 Hz
(a) Since (a > b), the dominant mode is the TE10 whose cutoff frequency is defined using
equation (7 – 90)
m 2+ n 2
(fc)mn = 1
2√μϵ a √   
b
which simplified to
1 m= 1 1 3 × 108
(fc)10 = = c = = 2.079 GHz (1)
2√μϵ a 2√μ0ϵ0 a 2a 2 × 7.214 × 10–2
(1) is the same equation as defined equation (7 – 64).

Alternatively, we have

λc = c = 2a = 2 × 7.214 × 10–2 = 14.428 × 10–2 m

fc
c 3 × 108
fc = = = 2.079 GHz
λc 14.428 × 10–2
(b) Since
c 3 × 108
λ= = = 0.1 m
f 3 × 109
c c
 (λc)10 = = c = 2a = 14.4 × 10–2 m
(fc)10
2a
The above equation can also be found using equation (7 – 91) such that

(λc)mn = 2     (λc)10 = 2 = 2a
n
√   
m
√   
1 0
2 2 2 2
+ –
a b a b
Equation (7 – 39) gives

λg = λ = λ = 0.139 m
√   1 – fc 1– λ
√  
2 2

f 2a
(c) The distance at which the field intensities is attenuated by 70%

e–αz = 03
z = – 1  ln 0.3
α
To determine α for TE10, we would refer to equation (10 – 187), page 557 on “Field and
Wave Electromagnetics, 2nd edition, by David K. Chieng”, that is,
 Static Magnetic Fields
3
2b 2 fc 2
Rs 1 +  a   f   = Np/m   πfμ = 1.429 × 10–2 Ω
(αc)TE =
10
where  Rs = √ σ
√1 –  f 
fc 2
η0b

= 2.26 × 10–2 Np/m

1 1
z = –  ln 0.3 = – (–0.5228) = 0.23 × 102 = 23 m
α 2.26 × 10–2
7.4 From equation (7 – 90), we have (for both TEmn & TMmn modes)
m 2+ n 2
(fc)mn = 1
2√μϵ √   
a b
c 3 × 108
TE10: (fc)10 = 1 m = = = 6.56 GHz
2√μϵ a 2a 2(2.286 × 10–2)
n c 3 × 108
TE01: (fc)01 = 1 = = = 9.84 GHz
2√μϵ b 2b 2(1.524 × 10–2)
TE20: (fc)20 = 13.12 GHz
TE11: (fc)11 = 11.83 GHz
TM11: (fc)11 = 11.83 GHz
TE21: (fc)21 = 16.40 GHz
TM30: (fc)30 = 19.69 GHz
TE02: (fc)02 = 19.69 GHz

7.5 (a) Given:

a = 2.5 × 10–2 m; b = 1.5 × 10–2 m; f = 7 × 109 Hz

u = c 3 × 108
λ= = = 0.04386 = 4.386 × 10–2 m
f 7 × 109
Equation (7 – 39) gives
λ λ λ
λδ = = = = 0.08332 m
f F λ
√   √  
2 2
1– c 1
1–
f 2a
where
f 2 λ 2
√   √   √  = 1 – 4.286 × 10 –2
–2 2
F1 = 1 – c = 1 –
f 2a 2 × 2.5 × 10 
= √1 – (0.8572)2 = 0.515

36π √
 β = ω√μϵ = F1 = 2π × 1010 = 4π × 10–7 × 2.25 × 1 × 10–9 × 0.515

= 161.79 rad/m

ug = u ∙ F1 = c ∙ F1 = 1.545 × 108 m/s

uρ = c = 3 × 10 = 5.825 × 108 m/s

8

F1 0.515
jωμ η
ZTE = = Ω (7 – 50)
v fc 2
1–
f √  
η0
= = 377 = 732 Ω
F1 0.516
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(b) a = 2.5 × 10–2 m; b = 1.5 × 10–2 m; f = 7 × 109 Hz; ϵr = 2 and μr = 1.

From equation (7 – 90), we have
m 2 + n 2
(fc)mn = 1
2√μϵ √   
a b
(7 – 90)

For TE10 waves,

1
√ a1  = 2a√cμϵ = 2a√1μ ϵ = 2ac√2
2
(fc)10′ =
2√μϵ r r

The λ is λ′ such that

0.04286
λ′ = u = λ = m = 3.031 × 10–2 m
f √2 √2
Equation (7 – 39) gives
λ λ′ λ′
λg′ = = = = 0.03811 m
fc 2 F2 λ′
√   √  
2
1– 1–
f 2a
where
f 2 λ′ 2
√   √   √  = 1 – 3.031 × 10 –2
–2 2
F2 = 1 – c = 1 –
f 2a 2 × 2.5 × 10 
= √1 – (0.6062) = 0.7953
2

√
 β′ = ω√μϵ ∙ F2 = 2π × 1010 = 4π × 10–7 × 2.25 ×
1
36π
× 10–9 × 0.7953

= 249.85 rad/m
c
× 0.7953 = 2.121 × 108 m/s
ug′ = u ∙ F2 =
√2
up′ = c = 3 × 10 = 2.667 × 108 m/s
8

√2F2 √2 × 0.7953
jωμ η′
ZTE′ = = Ω
v fc′ 2
1–
f √  
η0 377
= = = 474 Ω
F2 0.7953

7.6 The dominant (the lowest-order mode) in circular waveguide is TE11 mode. From equation
(7 – 134), we have

(fc)TE11 = 0.293
a√μϵ
The cutoff frequency is 0.8 × 10 = 8 GHz.

The waveguide is filled with air, it implies

0.293
8 × 109 = C    a = 0.2939 3 × 108 = 10.46 GHz > 8GHz
a 8 × 10
TM01 is the next higher mode. Using equation (7 – 126), we obtain

(fc)TM = 0.383 = 0.383 c = 0.383 3 × 108 =10.46 GHz   >  8 GHz

01 a√μϵ a a
10.5 GHz is not 20% less than (fc)TM01, so such waveguide cannot exist.

Therefore only one mode, that is, TE11, can propagate through the circular waveguide.
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7.7 The general formula for the cutoff frequency of theTMmn mode is given by

(fc)TEmn = Xmn (1)

2πa√μϵ
Note: Equation (1) is taken from
http://olaschool.chania.teicrete.gr/lectures2008/Cylindrical%20Waveguides%20-%20Dr%20
Vardiambasis.pdf

From Table 7.1, we have

Xmn = 3.833
 (fc)TMmn = Xmn = 3.833 × 3 × 108 = 1.83 × 108 Hz
2πa√μϵ 2πa a

 λ11 = 2π = c = 3 × 108 m
β
√f 2 – ((fc)TMmn)2 (14 × 109)2 – 1.03 × 10 2
8

√ a  
 (ηTM)11 = η0 = √f 2 – ((fc)TMmn)2 = 377 (14 × 109)2 – 1.03 × 10
8

√
2

a   Ω

7.8 For circuit waveguides, the dominant mode is the TE11 mode whose cutoff frequency according
to equation (7 – 134) is

(fc)TE11 = 0.293
a√μϵ
The next higher-order mode is TM01, whose cutoff frequency is according to equation (7 – 126),
is

(fc)TM01 = 0.383
a√μϵ
Then,
∆ f = (fc)TM01 – (fc)TE11 = 0.383 – 0.293 = 0.383 – 0.293 = 0.09
a√μϵ a√μϵ a√μϵ a√μϵ
0.09 0.09c 0.09c 0.009 × 108
 √ϵr = = = = = 1.60714285
a∆f√μ0ϵ0 a∆f a∆f 1.12 × 10–2 (1.5 × 109)
 ϵr = 2.583

7.9 Figure 7.10 is reproduced below for easy reference.

n2
d n1
n2

Figure 7.10  Symmetric dielectric slab waveguide, with slab region (refractive index n1)
surrounded by two dielectrics of index n2 < n1.

(a) The figure below shows the longitudinal cross section of Figure 7.10, with the ray (wave)
path as indicated.
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5°

0 I

(b) Using equation (7 – 135), the critical angle is defined such that
n
sin θc = 2
n1 
The critical angel at the lower boundary is
n 1.46
θc1 = sin–1 2 = sin–1
n1   1.48 
= 80.6° 
The critical angel at the upper boundary is
n 1.0
θc2 = sin–1 0 = sin–1
n2   1.46 
= 43.23° 
Snell’s Law of refraction states that
sin θt k1 ω√μ1ε1 n1
= = =
sin θt k2 ω√μ2ε2 n2

 
θt = sin–1 sin θi
n1
n2 
= sin–1 sin 75°
1.48
1.46  
= 78.282°

Therefore the rays are totally reflected. And there ae no cladding modes for θ > θc1 = 80.6°.
For θt > θc2 = 43.23°.


θ < sin–1 sin θc2
n2
n1 
= sin–1 sin 43.23°
1.48
1.46 
= 42.5° 
7.10 Figure 7.16 is reproduced here for easy reference.
y

ε0 ,µ0

εd ,µd d
2d O z
d

ε0 ,µ0

Figure 7.16  A longitudinal cross-section of a dielectric slab.

From equation (7 – 160), we have (for both TEmn & TMmn modes)

(fc)m = 1 m = 0,2,4, ..., TMm (odd) and TEm (odd)

4d√μdϵdμ0ϵ0 m = 1,3,5, ..., TMm (even) and TEm (even)
Given:
2d = 0.01 m = 1 cm, εr = 5, μr = 1
m = 0: (fc)m = 0; TM0 (odd) and TE0 (odd)
1
m = 1: (fc)1 = = 7.4949 × 109; TM1 (even) and TE1 (even)
4(0.5 × 10–2) √5 – 1