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Engineering Electromagnetics 9th solution
Engineering Electromagnetics (Chungnam National University)
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CHAPTER 11 – 9th Edition
11.1. Show that 𝐸𝑥𝑠 = 𝐴𝑒𝑗𝑘0 𝑧+𝜙 is a solution to the vector Helmholtz equation, Sec. 11.1, Eq. (30), for
√
𝑘0 = 𝜔 𝜇0 𝜖0 and any 𝜙 and 𝐴: We take
𝑑2
𝐴𝑒𝑗𝑘0 𝑧+𝜙 = (𝑗𝑘0 )2 𝐴𝑒𝑗𝑘0 𝑧+𝜙 = −𝑘20 𝐸𝑥𝑠
𝑑𝑧2
11.2. A 20-GHz uniform plane wave propagates in the forward 𝑧 direction in a lossless medium for which
𝜖𝑟 = 𝜇𝑟 = 3. Assume a real electric field amplitude, 𝐸0 , and take the wave as polarized in the 𝑥
direction. Find:
√
√
a) 𝑣𝑝 : 𝑣𝑝 = 1∕ 𝜇𝜖 = 𝑐∕ 𝜇𝑟 𝜖𝑟 = 𝑐∕3 = 1.0 × 108 m∕s.
b) 𝛽: In this lossless medium, 𝑘 = 𝛽 = 𝜔∕𝑣𝑝 = (4𝜋 × 1010 )∕(1.0 × 108 ) = 1.26 × 103 m−1 .
c) 𝜆: 𝜆 = 2𝜋∕𝛽 = 5.0 × 10−3 m = 5.0 mm.
d) 𝐄𝑠 : With real amplitude 𝐸0 , forward 𝑧 travel, and 𝑥 polarization, we write
𝐄𝑠 = 𝐸0 exp(−𝑗𝛽𝑧)𝐚𝑥 = 𝐸0 exp(−𝑗1.26 × 103 𝑧) 𝐚𝑥 V∕m.
√
√
e) 𝐇𝑠 : First, the intrinsic impedance of the medium is 𝜂 = 𝜇∕𝜖 = 𝜂0 𝜇𝑟 ∕𝜖𝑟 = 𝜂0 = 377 ohms.
Then 𝐇𝑠 = (𝐸0 ∕𝜂0 ) exp(−𝑗𝛽𝑧) 𝐚𝑦 = (𝐸0 ∕377) exp(−𝑗1.26 × 103 𝑧) 𝐚𝑦 A∕m.
{
}
f) < 𝐒 >= (1∕2)𝑒 𝐄𝑠 × 𝐇∗𝑠 = (𝐸02 ∕754) 𝐚𝑧 W∕m2
11.3. An 𝐇 field in free space is given as (𝑥, 𝑡) = 10 cos(108 𝑡 − 𝛽𝑥)𝐚𝑦 A/m. Find
a) 𝛽: Since we have a uniform plane wave, 𝛽 = 𝜔∕𝑐, where we identify 𝜔 = 108 sec−1 . Thus
𝛽 = 108 ∕(3 × 108 ) = 0.33 rad∕m.
b) 𝜆: We know 𝜆 = 2𝜋∕𝛽 = 18.9 m.
c) (𝑥, 𝑡) at 𝑃 (0.1, 0.2, 0.3) at 𝑡 = 1 ns: Use 𝐸(𝑥, 𝑡) = −𝜂0 𝐻(𝑥, 𝑡) = −(377)(10) cos(108 𝑡 −
𝛽𝑥) = −3.77 × 103 cos(108 𝑡 − 𝛽𝑥). The vector direction of 𝐄 will be −𝐚𝑧 , since we require that
𝐒 = 𝐄 × 𝐇, where 𝐒 is 𝑥-directed. At the given point, the relevant coordinate is 𝑥 = 0.1. Using
this, along with 𝑡 = 10−9 sec, we finally obtain
𝐄(𝑥, 𝑡) = −3.77 × 103 cos[(108 )(10−9 ) − (0.33)(0.1)]𝐚𝑧 = −3.77 × 103 cos(6.7 × 10−2 )𝐚𝑧
= −3.76 × 103 𝐚𝑧 V∕m
11.4. Small antennas have low efficiencies (as will be seen in Chapter 14) and the efficiency increases
with size up to the point at which a critical dimension of the antenna is an appreciable fraction of a
wavelength, say 𝜆∕8.
a) An antenna is that is 12cm long is operated in air at 1 MHz. What fraction of a wavelength long
is it? The free space wavelength will be
𝜆𝑎𝑖𝑟 =
3.0 × 108 m∕s
𝑐
0.12
= 300 m, so that the f raction =
=
= 4.0 × 10−4
𝑓
300
106 s−1
b) The same antenna is embedded in a ferrite material for which 𝜖𝑟 = 20 and 𝜇𝑟 = 2, 000. What
fraction of a wavelength is it now?
𝜆
0.12
300
𝜆𝑓 𝑒𝑟𝑟𝑖𝑡𝑒 = √ 𝑎𝑖𝑟 = √
= 0.08
= 1.5m ⇒ f raction =
1.5
𝜇𝑟 𝜖𝑟
(20)(2000)
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11.5. Consider two 𝑥-polarized light waves that counter-propagate along the 𝑧 axis. The wave traveling in
the forward 𝑧 direction is of frequency 𝜔2 ; the backward 𝑧-directed wave is of frequency 𝜔1 , where
𝜔1 < 𝜔2 . Both frequencies are very slightly detuned on either side of their mean frequency 𝜔0 , such
that 𝜔0 − 𝜔1 = 𝜔2 − 𝜔0 << 𝜔0 . Using the complex field forms, construct the expression for the total
electric field, and from this find the light intensity distribution (proportional to 𝐸𝐸 ∗ ). Your answer
should be in the form of a “standing wave” that in fact moves along the 𝑧 axis. Find an expression
for the velocity of the standing wave pattern in terms of given or known parameters. Does the pattern
move in the forward or backward 𝑧 direction?
The fact that the frequencies differ means that the wavenumbers will differ as well. If we assume that
the medium is free space, we would have 𝑘2 = 𝜔2 ∕𝑐 and 𝑘1 = 𝜔1 ∕𝑐. Assuming equal amplitudes,
𝐸0 , the total complex field, representing the sum of the forward and backward waves is
[
]
𝐄𝑐𝑇 (𝑧) = 𝐸0 exp(−𝑗𝑘2 𝑧) exp(𝑗𝜔2 𝑡) + exp(+𝑗𝑘1 𝑧) exp(𝑗𝜔1 𝑡) 𝐚𝑥
Now define the mean frequency and wavenumber:
𝜔0 =
𝜔2 + 𝜔1
2
and
𝑘𝑚 =
𝑘2 + 𝑘1
2
then define Δ𝑘 = 𝑘2 − 𝑘1 and Δ𝜔 = 𝜔2 − 𝜔1 , from which
𝑘2 = 𝑘𝑚 +
Δ𝑘
,
2
𝑘1 = 𝑘𝑚 −
Δ𝑘
,
2
𝜔2 = 𝜔0 +
Δ𝜔
,
2
and
𝜔1 = 𝜔0 −
Δ𝜔
2
The total complex field can now be expressed as
[
]
𝐄𝑐𝑇 (𝑧) = 𝐸0 𝑒−𝑗Δ𝑘𝑧∕2 𝑒𝑗𝜔0 𝑡 𝑒−𝑗𝑘𝑚 𝑧 𝑒𝑗Δ𝜔𝑡∕2 + 𝑒+𝑗𝑘𝑚 𝑧 𝑒−𝑗Δ𝜔𝑡∕2 𝐚𝑥
)
(
Δ𝜔
𝑡 − 𝑘𝑚 𝑧 𝐚𝑥
= 2𝐸0 𝑒−𝑗Δ𝑘𝑧∕2 𝑒𝑗𝜔0 𝑡 cos
2
The power density (light intensity) is now proportional to
(
)
Δ𝜔
∗
𝐸𝑐𝑇 𝐸𝑐𝑇
= 4𝐸02 cos2
𝑡 − 𝑘𝑚 𝑧
2
The intensity pattern moves in the forward 𝑧 direction at velocity 𝑣 = Δ𝜔∕2𝑘𝑚 = 𝑐Δ𝜔∕2𝜔0 .
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11.6. A uniform plane wave has electric field 𝐄𝑠 = (𝐸𝑦0 𝐚𝑦 − 𝐸𝑧0 𝐚𝑧 ) 𝑒−𝛼𝑥 𝑒−𝑗𝛽𝑥 V∕m. The intrinsic
impedance of the medium is given as 𝜂 = |𝜂| 𝑒𝑗𝜙 , where 𝜙 is a constant phase.
a) Describe the wave polarization and state the direction of propagation: The wave is linearly
polarized in the 𝑦-𝑧 plane, and propagates in the forward 𝑥 direction (from the 𝑒−𝑗𝛽𝑥 factor).
b) Find 𝐇𝑠 : Each component of 𝐄𝑠 , when crossed into its companion component of 𝐇∗𝑠 , must give
a vector in the positive-𝑥 direction of travel. Using this rule, we find
] [
]
[
𝐸𝑦
𝐸𝑦0
𝐸𝑧
𝐸𝑧0
𝐇𝑠 =
𝐚 +
𝐚 =
𝐚 +
𝐚 𝑒−𝛼𝑥 𝑒−𝑗𝜙 𝑒−𝑗𝛽𝑥 A∕m
𝜂 𝑧
𝜂 𝑦
|𝜂| 𝑧 |𝜂| 𝑦
} [
{
]
c) Find (𝑥, 𝑡) and (𝑥, 𝑡): (𝑥, 𝑡) = 𝑒 𝐄𝑠 𝑒𝑗𝜔𝑡 = 𝐸𝑦0 𝐚𝑦 − 𝐸𝑧0 𝐚𝑧 𝑒−𝛼𝑥 cos(𝜔𝑡 − 𝛽𝑥)
]
{
}
1 [
𝐸𝑦0 𝐚𝑧 + 𝐸𝑧0 𝐚𝑦 𝑒−𝛼𝑥 cos(𝜔𝑡 − 𝛽𝑥 − 𝜙)
(𝑥, 𝑡) = 𝑒 𝐇𝑠 𝑒𝑗𝜔𝑡 =
|𝜂|
where all amplitudes are assumed real.
d) Find < 𝐒 > in W∕m2 :
)
(
{
}
1
1
2
2
< 𝐒 >= 𝑒 𝐄𝑠 × 𝐇∗𝑠 =
𝑒−2𝛼𝑥 cos 𝜙 𝐚𝑥 W∕m2
𝐸𝑦0
+ 𝐸𝑧0
2
2|𝜂|
e) Find the time-average power in watts that is intercepted by an antenna of rectangular crosssection, having width 𝑤 and height ℎ, suspended parallel to the 𝑦𝑧 plane, and at a distance 𝑑
from the wave source. This will be
(
)
1
2
2
(𝑤ℎ) 𝐸𝑦0
+ 𝐸𝑧0
𝑒−2𝛼𝑑 cos 𝜙 W
𝑃 =
< 𝐒 > ⋅ 𝑑𝐒 = | < 𝐒 > |𝑥=𝑑 × 𝑎𝑟𝑒𝑎 =
∫ ∫𝑝𝑙𝑎𝑡𝑒
2|𝜂|
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11.7. Express in terms of the penetration depth, 𝛿, the distance into a lossy medium at which the wave
power has attenuated by
a) 1 dB: The decibel loss is given by
Loss (dB) = 8.69𝛼𝐿 =
𝛿
8.69𝐿
⇒ 𝐿1 =
= 0.12𝛿
𝛿
8.69
b) 3 dB: This would be 𝐿3 = 3𝐿1 = 0.35𝛿
c) 30 dB: 𝐿30 = 30𝐿1 = 3.45𝛿.
11.8. An electric field in free space is given in spherical coordinates as 𝐄𝑠 (𝑟) = 𝐸0 (𝑟)𝑒−𝑗𝑘𝑟 𝐚𝜃 V/m.
a) find 𝐇𝑠 (𝑟) assuming uniform plane wave behavior: Knowing that the cross product of 𝐄𝑠 with
the complex conjugate of the phasor 𝐇𝑠 field must give a vector in the direction of propagation,
we obtain,
𝐸 (𝑟)
𝐇𝑠 (𝑟) = 0 𝑒−𝑗𝑘𝑟 𝐚𝜙 A∕m
𝜂0
b) Find < 𝐒 >: This will be
{
} 𝐸 2 (𝑟)
1
< 𝐒 >= 𝑒 𝐄𝑠 × 𝐇∗𝑠 = 0 𝐚𝑟 W∕m2
2
2𝜂0
c) Express the average outward power in watts through a closed spherical shell of radius 𝑟, centered
at the origin: The power will be (in this case) just the product of the power density magnitude
in part 𝑏 with the sphere area, or
𝐸 2 (𝑟)
W
𝑃 = 4𝜋𝑟2 0
2𝜂0
where 𝐸0 (𝑟) is assumed real.
d) Establish the required functional form of 𝐸0 (𝑟) that will enable the power flow in part 𝑐 to be
independent of radius: Evidently this condition is met when 𝐸0 (𝑟) ∝ 1∕𝑟
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11.9. An example of a nonuniform plane wave is a surface or evanescent wave, an example of which in
phasor form is shown here, exhibiting diminishing amplitude with 𝑥 while propagating in the forward
𝑧 direction:
𝐄𝑠 (𝑥, 𝑧) = 𝐸0 𝑒−𝛼𝑥 𝑒−𝑗𝛽𝑧 𝐚𝑦
Such fields form part of the mode structure in dielectric waveguides, as will be explored in
Chapter 13.
a) Assuming free space conditions, use Eq. (23) to find the associated phasor magnetic field,
𝐇𝑠 (𝑥, 𝑧) (note that there will be two components). Use ∇ × 𝐄𝑠 = −𝑗𝜔𝜇0 𝐇𝑠 , which leads to
(
)
𝜕𝐸𝑦
𝜕𝐸𝑦
)
𝑗
𝑗 (
𝑗
𝐇𝑠 =
−𝛼𝐸𝑦 𝐚𝑧 + 𝑗𝛽𝐸𝑦 𝐚𝑥
∇ × 𝐸𝑦 𝐚𝑦 =
𝐚𝑧 −
𝐚𝑥 =
𝜔𝜇0
𝜔𝜇0 𝜕𝑥
𝜕𝑧
𝜔𝜇0
where the subscript 𝑠 has been dropped everywhere, with the understanding that that all fields
are in phasor form. Finally,
𝐇(𝑥, 𝑧) = −
)
𝐸0 (
𝛽𝐚𝑥 + 𝑗𝛼𝐚𝑧 𝑒−𝛼𝑥 𝑒−𝑗𝛽𝑧
𝜔𝜇0
b) Find 𝛼 as a function of 𝛽, 𝜔, 𝜖0 , and 𝜇0 , such that all of Maxwell’s equations are satisfied by the
electric and magnetic fields: With a source-free medium, we first of all have ∇ ⋅ 𝐄 = ∇ ⋅ 𝐇 = 0,
which by inspection are seen to be satisfied by the above fields. The ∇ × 𝐄 = −𝑗𝜔𝜇0 𝐇 equation
is of course automatically satisfied. This leaves ∇ × 𝐇 = 𝑗𝜔𝜖0 𝐄. Substituting the known fields,
we find
)
(
𝜕𝐻𝑥 𝜕𝐻𝑧
∇×𝐇=
−
𝐚𝑦 = 𝑗𝜔𝜖0 𝐸𝑦 𝐚𝑦
𝜕𝑧
𝜕𝑥
or:
)
𝑗𝐸0 ( 2
𝛽 − 𝛼 2 𝑒−𝛼𝑥 𝑒−𝑗𝛽𝑧 = 𝑗𝜔𝜖0 𝐸0 𝑒−𝛼𝑥 𝑒−𝑗𝛽𝑧
𝜔𝜇0
which simplifies to
𝛽 2 − 𝛼 2 = 𝜔2 𝜇0 𝜖0 ⇒ 𝛼 =
√
𝛽 2 − 𝜔2 𝜇0 𝜖0
11.10. In a medium characterized by intrinsic impedance
𝜂 = |𝜂|𝑒)𝑗𝜙 , a linearly-polarized plane wave prop(
agates, with magnetic field given as 𝐇𝑠 = 𝐻0𝑦 𝐚𝑦 + 𝐻0𝑧 𝐚𝑧 𝑒−𝛼𝑥 𝑒−𝑗𝛽𝑥 . Find:
a) 𝐄𝑠 : Requiring orthogonal components of 𝐄𝑠 for each component of 𝐇𝑠 , we find
[
]
𝐄𝑠 = |𝜂| 𝐻0𝑧 𝐚𝑦 − 𝐻0𝑦 𝐚𝑧 𝑒−𝛼𝑥 𝑒−𝑗𝛽𝑥 𝑒𝑗𝜙
[
]
b) (𝑥, 𝑡) = 𝑒 {𝐄𝑠 𝑒𝑗𝜔𝑡 } = |𝜂| 𝐻0𝑧 𝐚𝑦 − 𝐻0𝑦 𝐚𝑧 𝑒−𝛼𝑥 cos(𝜔𝑡 − 𝛽𝑥 + 𝜙).
[
]
c) (𝑥, 𝑡) = 𝑒 {𝐇𝑠 𝑒𝑗𝜔𝑡 } = 𝐻0𝑦 𝐚𝑦 + 𝐻0𝑧 𝐚𝑧 𝑒−𝛼𝑥 cos(𝜔𝑡 − 𝛽𝑥).
]
[
2 + 𝐻 2 𝑒−2𝛼𝑥 cos 𝜙 𝐚 W∕m2
d) < 𝐒 >= 21 𝑒{𝐄𝑠 × 𝐇∗𝑠 } = 12 |𝜂| 𝐻0𝑦
𝑥
0𝑧
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11.11. A uniform plane wave at frequency 𝑓 = 100 MHz propagates in a material having conductivity
𝜎 = 3.0 S/m and dielectric constant 𝜖𝑟′ = 8.00. The wave carries electric field amplitude 𝐸0 = 100
V/m.
a) Calculate the loss tangent and determine whether the medium would qualify as a good dielectric
or a good conductor.
3.0
𝜎
=
= 67.5
′
8
𝜔𝜖
(2𝜋 × 10 )(8.00)(8.85 × 10−12 )
This falls within the good conductor approximation.
b) Calculate 𝛼, 𝛽, and 𝜂. Assuming a good conductor, we have:
√
.
. √
𝛼 = 𝛽 = 𝜋𝑓 𝜇0 𝜎 = 𝜋(108 )(4𝜋 × 10−7 )(3) = 34.4 m−1
Then
. 1+𝑗
(1 + 𝑗)𝛼 (1 + 𝑗)(34.4)
(Eq. (85)) =
=
= 11.5(1 + 𝑗) ohms
𝜂=
𝜎𝛿
𝜎
3.0
c) Assuming forward 𝑧 propagation and 𝑥 polarization, write the electric field in phasor form:
𝐄𝑠 = 100𝑒−𝛼𝑧 𝑒−𝑗𝛽𝑧 𝐚𝑥
where 𝛼 and 𝛽 are as found in part 𝑏.
d) Write the magnetic field in phasor form:
𝐇𝑠 =
100
𝑒−𝛼𝑧 𝑒−𝑗𝛽𝑧 𝐚𝑦 = 6.15 𝑒−𝑗0.79 𝑒−𝛼𝑧 𝑒−𝑗𝛽𝑧 𝐚𝑦
11.5(1 + 𝑗)
e) Write the time-average Poynting vector, < 𝐒 > in W∕m2 .
{
} 1
1
< 𝐒 >= 𝑒 𝐄𝑠 × 𝐇∗𝑠 = (100)(6.15)𝑒−2𝛼𝑧 cos(0.79) 𝐚𝑧 = 216 𝑒−2𝛼𝑧 𝐚𝑧 W∕m2
2
2
f) Find the 6-dB material thickness; i.e., the propagation distance over which the wave power drops
to 25% of its initial value.
𝑧(25%) =
6
6
=
= 2.0 × 10−2 m = 2.0 cm
8.69𝛼 (8.69)(34.4)
g) What dB power drop corresponds to one penetration depth, 𝛿? In general, the decibel power
loss is given by 8.69𝛼𝑧. So, when 𝑧 = 𝛿 = 1∕𝛼, we have 8.69 dB.
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11.12. Repeat Problem 11.11, except that the wave now propagates in fresh water, having conductivity 𝜎 =
10−3 S/m, dielectric constant 𝜖𝑟′ = 80.0, and permeability 𝜇0 :
a) Calculate the loss tangent and determine whether the medium would qualify as a good dielectric
or a good conductor:
10−3
𝜎
=
= 0.002
𝜔𝜖 ′
(2𝜋 × 108 )(80.0)(8.85 × 10−12 )
This falls within the good dielectric approximation.
b) Calculate 𝛼, 𝛽, and 𝜂. Assuming a good dielectric, we have:
√
𝜂0 𝜎
. 𝜎 𝜇0
377(10−3 )
𝛼=
= 0.0211 Np∕m
=
= √
√
2 𝜖0
2 𝜖𝑟′
2 80
√
. √
𝜔
2𝜋 × 108 √
𝛽 = 𝜔 𝜇0 𝜖0 𝜖𝑟′ =
𝜖𝑟′ =
80 = 18.7 rad∕m
𝑐
3 × 108
√
𝜇0
.
377
𝜂=
= √ = 42.2 ohms
′
𝜖𝑟 𝜖0
80
c) Assuming forward 𝑧 propagation and 𝑥 polarization, write the electric field in phasor form:
𝐄𝑠 = 100𝑒−𝛼𝑧 𝑒−𝑗𝛽𝑧 𝐚𝑥
where 𝛼 and 𝛽 are as found in part 𝑏.
d) Write the magnetic field in phasor form:
𝐇𝑠 =
100 −𝛼𝑧 −𝑗𝛽𝑧
𝑒 𝑒
𝐚𝑦 = 2.37𝑒−𝛼𝑧 𝑒−𝑗𝛽𝑧 𝐚𝑦
42.2
e) Write the time-average Poynting vector, < 𝐒 > in W∕m2 .
{
} 1
1
< 𝐒 >= 𝑒 𝐄𝑠 × 𝐇∗𝑠 = (100)(2.37)𝑒−2𝛼𝑧 𝐚𝑧 = 118 𝑒−2𝛼𝑧 𝐚𝑧 W∕m2
2
2
f) Find the 6-dB material thickness; i.e., the propagation distance over which the wave power drops
to 25% of its initial value.
𝑧(25%) =
6
6
=
= 32.7 m
8.69𝛼 (8.69)(0.0211)
g) What dB power drop corresponds to one penetration depth, 𝛿? In general, the decibel power
loss is given by 8.69𝛼𝑧. So, when 𝑧 = 𝛿 = 1∕𝛼, we have 8.69 dB.
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11.13. Let 𝑗𝑘 = 0.2+𝑗1.5 m−1 and 𝜂 = 450+𝑗60 Ω for a uniform plane wave propagating in the 𝐚𝑧 direction.
If 𝜔 = 300 Mrad/s, find 𝜇, 𝜖 ′ , and 𝜖 ′′ : We begin with
√
𝜇
1
= 450 + 𝑗60
𝜂=
√
′
𝜖
1 − 𝑗(𝜖 ′′ ∕𝜖 ′ )
and
√
√
𝑗𝑘 = 𝑗𝜔 𝜇𝜖 ′ 1 − 𝑗(𝜖 ′′ ∕𝜖 ′ ) = 0.2 + 𝑗1.5
Then
𝜂𝜂 ∗ =
𝜇
1
= (450 + 𝑗60)(450 − 𝑗60) = 2.06 × 105
√
′
𝜖
′′
′
2
1 + (𝜖 ∕𝜖 )
and
(𝑗𝑘)(𝑗𝑘)∗ = 𝜔2 𝜇𝜖 ′
√
1 + (𝜖 ′′ ∕𝜖 ′ )2 = (0.2 + 𝑗1.5)(0.2 − 𝑗1.5) = 2.29
(1)
(2)
Taking the ratio of (2) to (1),
(
)
(𝑗𝑘)(𝑗𝑘)∗
2.29
′′ ′ 2
2 ′ 2
=
𝜔
(𝜖
)
1
+
(𝜖
∕𝜖
)
=
= 1.11 × 10−5
∗
𝜂𝜂
2.06 × 105
Then with 𝜔 = 3 × 108 ,
(𝜖 ′ )2 =
1.23 × 10−22
1.11 × 10−5
(
)=(
)
(3 × 108 )2 1 + (𝜖 ′′ ∕𝜖 ′ )2
1 + (𝜖 ′′ ∕𝜖 ′ )2
(3)
Now, we use Eqs. (35) and (36). Squaring these and taking their ratio gives
√
1 + (𝜖 ′′ ∕𝜖 ′ )2
(0.2)2
𝛼2
=
=
√
𝛽2
(1.5)2
1 + (𝜖 ′′ ∕𝜖 ′ )2
We solve this to find 𝜖 ′′ ∕𝜖 ′ = 0.271. Substituting this result into (3) gives 𝜖 ′ = 1.07 × 10−11 F/m.
Since 𝜖 ′′ ∕𝜖 ′ = 0.271, we then find 𝜖 ′′ = 2.90 × 10−12 F/m. Finally, using these results in either (1)
or (2) we find 𝜇 = 2.28 × 10−6 H/m. Summary: 𝜇 = 2.28 × 10−6 H∕m,
𝜖 ′ = 1.07 × 10−11 F∕m, and 𝜖 ′′ = 2.90 × 10−12 F∕m.
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11.14. A certain nonmagnetic material has the material constants 𝜖𝑟′ = 2 and 𝜖 ′′ ∕𝜖 ′ = 4 × 10−4 at 𝜔 = 1.5
Grad/s. Find the distance a uniform plane wave can propagate through the material before:
a) it is attenuated by 1 Np: First, 𝜖 ′′ = (4 × 104 )(2)(8.854 × 10−12 ) = 7.1 × 10−15 F/m. Then, since
𝜖 ′′ ∕𝜖 ′ << 1, we use the approximate form for 𝛼, given by Eq. (51) (written in terms of 𝜖 ′′ ):
√
. 𝜔𝜖 ′′ 𝜇
(1.5 × 109 )(7.1 × 10−15 ) 377
−3
=
𝛼=
√ = 1.42 × 10 Np∕m
2
𝜖′
2
2
The required distance is now 𝑧1 = (1.42 × 10−3 )−1 = 706 m
b) the power level is reduced by one-half: The governing relation is 𝑒−2𝛼𝑧1∕2 = 1∕2, or 𝑧1∕2 =
ln 2∕2𝛼 = ln 2∕2(1.42 × 10− 3) = 244 m.
c) the phase shifts 360◦ : This distance is defined as
√ one wavelength, where 𝜆 = 2𝜋∕𝛽
√
= (2𝜋𝑐)∕(𝜔 𝜖𝑟′ ) = [2𝜋(3 × 108 )]∕[(1.5 × 109 ) 2] = 0.89 m.
11.15. A 10 GHz radar signal may be represented as a uniform plane wave in a sufficiently small region.
Calculate the wavelength in centimeters and the attenuation in nepers per meter if the wave is propagating in a non-magnetic material for which
a) 𝜖𝑟′ = 1 and 𝜖𝑟′′ = 0: In a non-magnetic material, we would have:
√
1∕2
√
( ′′ )2
⎤
𝜇0 𝜖0 𝜖𝑟′ ⎡
𝜖𝑟
⎢ 1+
− 1⎥
𝛼=𝜔
⎥
2 ⎢
𝜖𝑟′
⎣
⎦
and
√
1∕2
√
( ′′ )2
⎤
𝜖𝑟
𝜇0 𝜖0 𝜖𝑟′ ⎡
⎢ 1+
𝛽=𝜔
+ 1⎥
⎥
2 ⎢
𝜖𝑟′
⎣
⎦
√
′
′′
With the given values of 𝜖𝑟 and 𝜖𝑟 , it is clear that 𝛽 = 𝜔 𝜇0 𝜖0 = 𝜔∕𝑐, and so
𝜆 = 2𝜋∕𝛽 = 2𝜋𝑐∕𝜔 = 3 × 1010 ∕1010 = 3 cm. It is also clear that 𝛼 = 0.
. √
b) 𝜖𝑟′ = 1.04 and 𝜖𝑟′′ = 9.00 × 10−4 : In this case 𝜖𝑟′′ ∕𝜖𝑟′ << 1, and so 𝛽 = 𝜔 𝜖𝑟′ ∕𝑐 = 2.13 cm−1 .
Thus 𝜆 = 2𝜋∕𝛽 = 2.95 cm. Then
√
√
′′
𝜔𝜖𝑟′′ 𝜇0 𝜖0
. 𝜔𝜖 ′′ 𝜇
𝜔 𝜖𝑟
2𝜋 × 1010 (9.00 × 10−4 )
=
=
=
𝛼=
√
√
√
2
𝜖′
2
2𝑐 𝜖 ′
2 × 3 × 108
𝜖𝑟′
1.04
𝑟
−2
= 9.24 × 10 Np∕m
c) 𝜖𝑟′ = 2.5 and 𝜖𝑟′′ = 7.2: Using the above formulas, we obtain
√
1∕2
√
⎤
⎡
( )2
2𝜋 × 1010 2.5 ⎢
7.2
+ 1⎥ = 4.71 cm−1
1+
𝛽=
√
⎥
⎢
2.5
10
(3 × 10 ) 2 ⎣
⎦
and so 𝜆 = 2𝜋∕𝛽 = 1.33 cm. Then
√
1∕2
√
⎤
⎡
( )2
2𝜋 ×
2.5 ⎢
7.2
− 1⎥ = 335 Np∕m
1+
𝛼=
√
⎥
⎢
2.5
8
(3 × 10 ) 2 ⎣
⎦
1010
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11.16. Consider the power dissipation term, ∫ 𝐄⋅𝐉𝑑𝑣 in Poynting’s theorem (Eq.(70)). This gives the power
lost to heat within a volume into which electromagnetic waves enter. The term 𝑝𝑑 = 𝐄 ⋅ 𝐉 is thus the
power dissipation per unit volume in W∕m3 . Following the same reasoning
in Eq.(77),
{ that resulted
}
the time-average power dissipation per volume will be < 𝑝𝑑 >= (1∕2)𝑒 𝐄𝑠 ⋅ 𝐉∗𝑠 .
a) Show that in a conducting medium, through which a uniform plane wave of amplitude 𝐸0 propagates in the forward 𝑧 direction, < 𝑝𝑑 >= (𝜎∕2)|𝐸0 |2 𝑒−2𝛼𝑧 : Begin with the phasor expression
for the electric field, assuming complex amplitude 𝐸0 , and 𝑥-polarization:
𝐄𝑠 = 𝐸0 𝑒−𝛼𝑧 𝑒−𝑗𝛽𝑧 𝐚𝑥 V∕m2
Then
𝐉𝑠 = 𝜎𝐄𝑠 = 𝜎𝐸0 𝑒−𝛼𝑧 𝑒−𝑗𝛽𝑧 𝐚𝑥 A∕m2
So that
{
}
< 𝑝𝑑 >= (1∕2)𝑒 𝐸0 𝑒−𝛼𝑧 𝑒−𝑗𝛽𝑧 𝐚𝑥 ⋅ 𝜎𝐸0∗ 𝑒−𝛼𝑧 𝑒+𝑗𝛽𝑧 𝐚𝑥 = (𝜎∕2)|𝐸0 |2 𝑒−2𝛼𝑧
b) Confirm this result for the special case of a good conductor by using the left hand side of
Eq. (70), and consider a very small volume. In a good conductor, the intrinsic impedance
is, from Eq. (85), 𝜂𝑐 = (1 + 𝑗)∕(𝜎𝛿), where the skin depth, 𝛿 = 1∕𝛼. The magnetic field phasor
is then
𝐸
𝜎
𝐇𝑠 = 𝑠 𝐚𝑦 =
𝐸 𝑒−𝛼𝑧 𝑒−𝑗𝛽𝑧 𝐚𝑦 A∕m
𝜂𝑐
(1 + 𝑗)𝛼 0
The time-average Poynting vector is then
{
}
𝜎
1
|𝐸 |2 𝑒−2𝛼𝑧 𝐚𝑧 W∕m2
< 𝐒 >= 𝑒 𝐄𝑠 × 𝐇∗𝑠 =
2
4𝛼 0
Now, consider a rectangular volume of side lengths, Δ𝑥, Δ𝑦, and Δ𝑧, all of which are very
small. As the wave passes through this volume in the forward 𝑧 direction, the power dissipated
will be the difference between the power at entry (at 𝑧 = 0), and the power that exits the volume
.
(at 𝑧 = Δ𝑧). With small 𝑧, we may approximate 𝑒−2𝛼𝑧 = 1 − 2𝛼𝑧, and the dissipated power in
the volume becomes
]
[
]
[
𝜎
𝜎
𝜎
𝑃𝑑 = 𝑃𝑖𝑛 − 𝑃𝑜𝑢𝑡 =
|𝐸0 |2 Δ𝑥Δ𝑦 −
|𝐸0 |2 (1 − 2𝛼Δ𝑧) Δ𝑥Δ𝑦 = |𝐸0 |2 (Δ𝑥Δ𝑦Δ𝑧)
4𝛼
4𝛼
2
This is just the result of part 𝑎, evaluated at 𝑧 = 0 and multiplied by the volume. The relation
becomes exact as Δ𝑧 → 0, in which case < 𝑝𝑑 >→ (𝜎∕2)|𝐸0 |2 .
It is also possible to show the relation by using Eq. (69) (which involves taking the divergence
of < 𝐒 >), or by removing the restriction of a small volume and evaluating the integrals in
Eq. (70) without approximations. Either method is straightforward.
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11.17. Consider a long solid cylindrical wire having uniform conductivity 𝜎. The wire is oriented along the
𝑧 axis, and has radius 𝑎 and length 𝐿. Constant voltage 𝑉0 is applied between the two ends.
a. Write the electric field intensity, 𝐄, in terms of 𝑉0 and 𝐿. Assume a positive 𝑧 directed field:
Would have 𝐄 = 𝑉0 ∕𝐿 𝐚𝑧 .
b. Using Ampere’s circuital law, find the magnetic field intensity, 𝐇, inside the wire as a function
of 𝜌.
∮
𝐇 ⋅ 𝑑𝐋 =
∫ ∫
𝐉 ⋅ 𝐧𝑑𝑎 ⇒ 2𝜋𝜌𝐻𝜙 = 𝜋𝜌2 𝐽 = 𝜋𝜌2 𝜎𝑉0 ∕𝐿
where 𝐉 = 𝜎𝐄 = (𝜎𝑉0 ∕𝐿) 𝐚𝑧 . So
𝜎𝜌𝑉0
𝐚 A∕m (0 < 𝜌 < 𝑎)
2𝐿 𝜙
𝐇=
c. Find the Poynting vector 𝐒 = 𝐄 × 𝐇:
𝐒=
−𝜎𝜌𝑉02
𝑉0
𝜎𝜌𝑉0
𝐚𝜌 W∕m2
𝐚𝑧 ×
𝐚𝜙 =
𝐿
2𝐿
2𝐿2
d. Evaluate the left hand side of Poynting’s theorem by integrating the Poynting vector over the
wire surface to find the power transferred into the volume:
−
∮
𝐒 ⋅ 𝐧 𝑑𝑎 = −
∫0
2𝜋
∫0
𝐿
−𝜎𝜌𝑉02
2𝐿2
𝐚𝜌 ⋅ 𝐚𝜌 𝑎𝑑𝜙𝑑𝑧 =
𝜋𝑎2 𝜎 2
𝑉0 = 𝑉02 ∕𝑅 W
𝐿
where the wire resistance is 𝑅 = 𝐿∕(𝜋𝑎2 𝜎) ohms.
e. Evaluate the right hand side of Poynting’s theorem, and thus verify that the theorem is satisfied
in this situation: In the absence of time variation, all time derivatives are zero, and Poynting’s
theorem is simplified to read:
−
∮𝑠
𝐒 ⋅ 𝐧 𝑑𝑎 =
∫𝑣
𝐄 ⋅ 𝐉 𝑑𝑣
The right hand side evaluates as
∫𝑣
𝐿
𝐄 ⋅ 𝐉 𝑑𝑣 =
∫0 ∫0
2𝜋
∫0
𝑎
𝜎𝑉02
𝐿2
𝜌 𝑑𝜌 𝑑𝜙 𝑑𝑧 =
𝜋𝑎2 𝜎 2
𝑉0 = 𝑉02 ∕𝑅
𝐿
which is the same as the left side evaluation, found in part 𝑑. So the theorem works as it should!
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11.18. Given, a 100MHz uniform plane wave in a medium known to be a good dielectric. The phasor electric
field is 𝐄𝑠 = 4𝑒−0.5𝑧 𝑒−𝑗20𝑧 𝐚𝑥 V/m. Not stated in the problem is the permeability, which we take to be
𝜇0 . Determine:
a) 𝜖 ′ : As a first step, it is useful to see just how much of a good dielectric we have. We use the
good dielectric approximations, Eqs. (60a) and (60b), with 𝜎 = 𝜔𝜖 ′′ . Using these, we take the
ratio, 𝛽∕𝛼, to find
√
]
[
( ′)
( )
𝜔 𝜇𝜖 ′ 1 + (1∕8)(𝜖 ′′ ∕𝜖 ′ )2
𝛽
𝜖
20
1 𝜖 ′′
= 2 ′′ +
=
=
√
𝛼 0.5
𝜖
4 𝜖′
(𝜔𝜖 ′′ ∕2) 𝜇∕𝜖 ′
This becomes the quadratic equation:
(
𝜖 ′′
𝜖′
)2
(
− 160
𝜖 ′′
𝜖′
)
+8=0
(
)
The solution to the quadratic is 𝜖 ′′ ∕𝜖 ′ = 0.05, which means that we can neglect the second
√
. √
term in Eq. (60b), so that 𝛽 = 𝜔 𝜇𝜖 ′ = (𝜔∕𝑐) 𝜖𝑟′ . With the given frequency of 100 MHz,
√
and with 𝜇 = 𝜇0 , we find 𝜖𝑟′ = 20(3∕2𝜋) = 9.55, so that 𝜖𝑟′ = 91.3, and finally 𝜖 ′ = 𝜖𝑟′ 𝜖0 =
8.1 × 10−10 F∕m.
b) 𝜖 ′′ : Using Eq. (60a), the set up is
√
√
2(0.5)
𝜔𝜖 ′′ 𝜇
𝜖′
10−8 √
′′
𝛼 = 0.5 =
91.3 = 4.0 × 10−11 F∕m
⇒
𝜖
=
=
2
𝜖′
2𝜋(377)
2𝜋 × 108 𝜇
c) 𝜂: Using Eq. (62b), we find
√ [
( )]
.
𝜇
1 𝜖 ′′
377
1
+
𝑗
=√
𝜂=
(1 + 𝑗.025) = (39.5 + 𝑗0.99) ohms
𝜖′
2 𝜖′
91.3
d) 𝐇𝑠 : This will be a 𝑦-directed field, and will be
𝐇𝑠 =
𝐸𝑠
4
𝐚𝑦 =
𝑒−0.5𝑧 𝑒−𝑗20𝑧 𝐚𝑦 = 0.101𝑒−0.5𝑧 𝑒−𝑗20𝑧 𝑒−𝑗0.025 𝐚𝑦 A∕m
𝜂
(39.5 + 𝑗0.99)
e) < 𝐒 >: Using the given field and the result of part 𝑑, obtain
(0.101)(4) −2(0.5)𝑧
1
𝑒
cos(0.025) 𝐚𝑧 = 0.202𝑒−𝑧 𝐚𝑧 W∕m2
< 𝐒 >= 𝑒{𝐄𝑠 × 𝐇∗𝑠 } =
2
2
f) the power in watts that is incident on a rectangular surface measuring 20m x 30m at 𝑧 = 10m:
At 10m, the power density is < 𝐒 >= 0.202𝑒−10 = 9.2 × 10−6 W∕m2 . The incident power on
the given area is then 𝑃 = 9.2 × 10−6 × (20)(30) = 5.5 mW.
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11.19. Perfectly-conducting cylinders with radii of 8 mm and 20 mm are coaxial. The region between the
cylinders is filled with a perfect dielectric for which 𝜖 = 10−9 ∕4𝜋 F/m and 𝜇𝑟 = 1. If 𝐄 in this region
is (500∕𝜌) cos(𝜔𝑡 − 4𝑧)𝐚𝜌 V/m, find:
a) 𝜔, with the help of Maxwell’s equations in cylindrical coordinates: We use the two curl equations, beginning with ∇ × 𝐄 = −𝜕𝐁∕𝜕𝑡, where in this case,
∇×𝐄=
So
𝐵𝜙 =
𝜕𝐸𝜌
𝜕𝑧
𝐚𝜙 =
𝜕𝐵𝜙
2000
sin(𝜔𝑡 − 4𝑧)𝐚𝜙 = −
𝐚
𝜌
𝜕𝑡 𝜙
2000
2000
sin(𝜔𝑡 − 4𝑧)𝑑𝑡 =
cos(𝜔𝑡 − 4𝑧) T
∫
𝜌
𝜔𝜌
Then
𝐻𝜙 =
𝐵𝜙
𝜇0
=
2000
cos(𝜔𝑡 − 4𝑧) A∕m
(4𝜋 × 10−7 )𝜔𝜌
We next use ∇ × 𝐇 = 𝜕𝐃∕𝜕𝑡, where in this case
∇×𝐇=−
𝜕𝐻𝜙
𝜕𝑧
𝐚𝜌 +
1 𝜕(𝜌𝐻𝜙 )
𝐚𝑧
𝜌 𝜕𝜌
where the second term on the right hand side becomes zero when substituting our 𝐻𝜙 . So
∇×𝐇=−
𝜕𝐻𝜙
𝜕𝑧
𝐚𝜌 = −
𝜕𝐷𝜌
8000
𝐚
sin(𝜔𝑡
−
4𝑧)𝐚
=
𝜌
𝜕𝑡 𝜌
(4𝜋 × 10−7 )𝜔𝜌
And
𝐷𝜌 =
∫
−
8000
8000
cos(𝜔𝑡 − 4𝑧) C∕m2
sin(𝜔𝑡 − 4𝑧)𝑑𝑡 =
(4𝜋 × 10−7 )𝜔𝜌
(4𝜋 × 10−7 )𝜔2 𝜌
Finally, using the given 𝜖,
𝐷𝜌
𝐸𝜌 =
𝜖
=
8000
cos(𝜔𝑡 − 4𝑧) V∕m
(10−16 )𝜔2 𝜌
This must be the same as the given field, so we require
500
8000
=
⇒ 𝜔 = 4 × 108 rad∕s
−16
2
𝜌
(10 )𝜔 𝜌
b) 𝐇(𝜌, 𝑧, 𝑡): From part 𝑎, we have
𝐇(𝜌, 𝑧, 𝑡) =
4.0
2000
cos(4 × 108 𝑡 − 4𝑧)𝐚𝜙 A∕m
cos(𝜔𝑡 − 4𝑧)𝐚𝜙 =
−7
𝜌
(4𝜋 × 10 )𝜔𝜌
c) 𝐒(𝜌, 𝜙, 𝑧): This will be
𝐒(𝜌, 𝜙, 𝑧) = 𝐄 × 𝐇 =
=
500
4.0
cos(4 × 108 𝑡 − 4𝑧)𝐚𝜌 ×
cos(4 × 108 𝑡 − 4𝑧)𝐚𝜙
𝜌
𝜌
2.0 × 10−3
cos2 (4 × 108 𝑡 − 4𝑧)𝐚𝑧 W∕m2
𝜌2
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11.19
d) the average power passing through every cross-section 8 < 𝜌 < 20 mm, 0 < 𝜙 < 2𝜋. Using
the result of part 𝑐, we find < 𝐒 >= (1.0 × 103 )∕𝜌2 𝐚𝑧 W∕m2 . The power through the given
cross-section is now
P=
∫0
2𝜋
.020
∫.008
( )
20
1.0 × 103
3
= 5.7 kW
𝜌
𝑑𝜌
𝑑𝜙
=
2𝜋
×
10
ln
2
8
𝜌
11.20. Voltage breakdown in air at standard temperature and pressure occurs at an electric field strength
of approximately 3 × 106 V/m. This becomes an issue in some high-power optical experiments,
in which tight focusing of light may be necessary. Estimate the lightwave power in watts that can
be focused into a cylindrical beam of 10𝜇m radius before breakdown occurs. Assume uniform plane
wave behavior (although this assumption will produce an answer that is higher than the actual number
by as much as a factor of 2, depending on the actual beam shape).
The power density in the beam in free space can be found as a special case of Eq. (76) (with
𝜂 = 𝜂0 , 𝜃𝜂 = 𝛼 = 0):
|<𝐒>|=
𝐸02
2𝜂0
=
(3 × 106 )2
= 1.2 × 1010 W∕m2
2(377)
To avoid breakdown, the power in a 10-𝜇m radius cylinder is then bounded by
𝑃 < (1.2 × 1010 )(𝜋 × (10−5 )2 ) = 3.75 W
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11.21. The cylindrical shell, 1 cm < 𝜌 < 1.2 cm, is composed of a conducting material for which 𝜎 = 106
S/m. The external and internal regions are non-conducting. Let 𝐻𝜙 = 2000 A/m at 𝜌 = 1.2 cm.
a) Find 𝐇 everywhere: Use Ampere’s circuital law, which states:
∮
𝐇 ⋅ 𝑑𝐋 = 2𝜋𝜌(2000) = 2𝜋(1.2 × 10−2 )(2000) = 48𝜋 A = 𝐼𝑒𝑛𝑐𝑙
Then in this case
𝐉=
48
𝐼
𝐚 =
𝐚 = 1.09 × 106 𝐚𝑧 A∕m2
𝐴𝑟𝑒𝑎 𝑧 (1.44 − 1.00) × 10−4 𝑧
With this result we again use Ampere’s circuital law to find 𝐇 everywhere within the shell as a
function of 𝜌 (in meters):
𝐻𝜙1 (𝜌) =
1
2𝜋𝜌 ∫0
2𝜋
𝜌
∫.01
1.09 × 106 𝜌 𝑑𝜌 𝑑𝜙 =
54.5 4 2
(10 𝜌 − 1) A∕m (.01 < 𝜌 < .012)
𝜌
Outside the shell, we would have
𝐻𝜙2 (𝜌) =
48𝜋
= 24∕𝜌 A∕m (𝜌 > .012)
2𝜋𝜌
Inside the shell (𝜌 < .01 m), 𝐻𝜙 = 0 since there is no enclosed current.
b) Find 𝐄 everywhere: We use
𝐄=
1.09 × 106
𝐉
=
𝐚𝑧 = 1.09 𝐚𝑧 V∕m
𝜎
106
which is valid, presumeably, outside as well as inside the shell.
c) Find 𝐒 everywhere: Use
𝐏 = 𝐄 × 𝐇 = 1.09 𝐚𝑧 ×
=−
54.5 4 2
(10 𝜌 − 1) 𝐚𝜙
𝜌
59.4 4 2
(10 𝜌 − 1) 𝐚𝜌 W∕m2 (.01 < 𝜌 < .012 m)
𝜌
Outside the shell,
𝐒 = 1.09 𝐚𝑧 ×
26
24
𝐚𝜙 = − 𝐚𝜌 W∕m2 (𝜌 > .012 m)
𝜌
𝜌
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11.22. The inner and outer dimensions of a copper coaxial transmission line are 2 and 7 mm, respectively.
Both conductors have thicknesses much greater than 𝛿. The dielectric is lossless and the operating
frequency is 400 MHz. Calculate the resistance per meter length of the:
a) inner conductor: First
1
1
𝛿=√
=√
= 3.3 × 10−6 m = 3.3𝜇m
𝜋𝑓 𝜇𝜎
𝜋(4 × 108 )(4𝜋 × 10−7 )(5.8 × 107 )
Now, using (90) with a unit length, we find
𝑅𝑖𝑛 =
1
1
=
= 0.42 ohms∕m
−3
2𝜋𝑎𝜎𝛿 2𝜋(2 × 10 )(5.8 × 107 )(3.3 × 10−6 )
b) outer conductor: Again, (90) applies but with a different conductor radius. Thus
2
𝑎
𝑅𝑜𝑢𝑡 = 𝑅𝑖𝑛 = (0.42) = 0.12 ohms∕m
𝑏
7
c) transmission line: Since the two resistances found above are in series, the line resistance is their
sum, or 𝑅 = 𝑅𝑖𝑛 + 𝑅𝑜𝑢𝑡 = 0.54 ohms∕m.
11.23. A hollow tubular conductor is constructed from a type of brass having a conductivity of 1.2 × 107
S/m. The inner and outer radii are 9 mm and 10 mm respectively. Calculate the resistance per meter
length at a frequency of
a) dc: In this case the current density is uniform over the entire tube cross-section. We write:
𝑅(dc) =
1
𝐿
= 1.4 × 10−3 Ω∕m
=
7
𝜎𝐴 (1.2 × 10 )𝜋(.012 − .0092 )
b) 20 MHz: Now the skin effect will limit the effective cross-section. At 20 MHz, the skin depth
is
𝛿(20MHz) = [𝜋𝑓 𝜇0 𝜎]−1∕2 = [𝜋(20 × 106 )(4𝜋 × 10−7 )(1.2 × 107 )]−1∕2 = 3.25 × 10−5 m
This is much less than the outer radius of the tube. Therefore we can approximate the resistance
using the formula:
𝑅(20MHz) =
1
1
𝐿
=
=
= 4.1 × 10−2 Ω∕m
7
𝜎𝐴 2𝜋𝑏𝛿 (1.2 × 10 )(2𝜋(.01))(3.25 × 10−5 )
c) 2 GHz: Using the same formula as in part 𝑏, we find the skin depth at 2 GHz to be 𝛿 = 3.25×10−6
m. The resistance (using the other formula) is 𝑅(2GHz) = 4.1 × 10−1 Ω∕m.
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11.24
a) Most microwave ovens operate at 2.45 GHz. Assume that 𝜎 = 1.2 × 106 S/m and 𝜇𝑟 = 500 for
the stainless steel interior, and find the depth of penetration:
1
1
𝛿=√
=√
= 9.28 × 10−6 m = 9.28𝜇m
9
−7
6
𝜋𝑓 𝜇𝜎
𝜋(2.45 × 10 )(4𝜋 × 10 )(1.2 × 10 )
b) Let 𝐸𝑠 = 50∠ 0◦ V/m at the surface of the conductor, and plot a curve of the amplitude of 𝐸𝑠 vs.
the angle of 𝐸𝑠 as the field propagates
into the stainless steel: Since the conductivity is high, we
.
. √
use (82) to write 𝛼 = 𝛽 = 𝜋𝑓 𝜇𝜎 = 1∕𝛿. So, assuming that the direction into the conductor
is 𝑧, the depth-dependent field is written as
𝐸𝑠 (𝑧) = 50𝑒−𝛼𝑧 𝑒−𝑗𝛽𝑧 = 50𝑒−𝑧∕𝛿 𝑒−𝑗𝑧∕𝛿 = 50 exp(−𝑧∕9.28) exp(−𝑗 𝑧∕9.28 )
⏟⏞⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏞⏟
⏟⏟⏟
amplitude
angle
where 𝑧 is in microns. Therefore, the plot of amplitude versus angle is simply a plot of 𝑒−𝑥 versus
𝑥, where 𝑥 = 𝑧∕9.28; the starting amplitude is 50 and the 1∕𝑒 amplitude (at 𝑧 = 9.28 𝜇m) is
18.4.
11.25. A good conductor is planar in form and carries a uniform plane wave that has a wavelength of 0.3
mm and a velocity of 3 × 105 m/s. Assuming the conductor is non-magnetic, determine the frequency
and the conductivity: First, we use
𝑓=
𝑣
3 × 105
= 109 Hz = 1 GHz
=
𝜆 3 × 10−4
Next, for a good conductor,
𝛿=
𝜆
4𝜋
4𝜋
1
=
= 1.1 × 105 S∕m
⇒ 𝜎= 2
=√
−8 )(109 )(4𝜋 × 10−7 )
2𝜋
𝜆
𝑓
𝜇
(9
×
10
𝜋𝑓 𝜇𝜎
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11.26. The dimensions of a certain coaxial transmission line are 𝑎 = 0.8mm and 𝑏 = 4mm. The outer
conductor thickness is 0.6mm, and all conductors have 𝜎 = 1.6 × 107 S/m.
a) Find 𝑅, the resistance per unit length, at an operating frequency of 2.4 GHz: First
𝛿=√
1
=√
= 2.57 × 10−6 m = 2.57𝜇m
8
−7
7
𝜋𝑓 𝜇𝜎
𝜋(2.4 × 10 )(4𝜋 × 10 )(1.6 × 10 )
1
Then, using (90) with a unit length, we find
𝑅𝑖𝑛 =
1
1
= 4.84 ohms∕m
=
−3
2𝜋𝑎𝜎𝛿 2𝜋(0.8 × 10 )(1.6 × 107 )(2.57 × 10−6 )
The outer conductor resistance is then found from the inner through
𝑎
0.8
𝑅𝑜𝑢𝑡 = 𝑅𝑖𝑛 =
(4.84) = 0.97 ohms∕m
𝑏
4
The net resistance per length is then the sum, 𝑅 = 𝑅𝑖𝑛 + 𝑅𝑜𝑢𝑡 = 5.81 ohms∕m.
b) Use information from Secs. 6.3 and 8.10 to find 𝐶 and 𝐿, the capacitance and inductance per
unit length, respectively. The coax is air-filled. From those sections, we find (in free space)
𝐶=
2𝜋𝜖0
2𝜋(8.854 × 10−12 )
=
= 3.46 × 10−11 F∕m
ln(𝑏∕𝑎)
ln(4∕.8)
𝜇0
4𝜋 × 10−7
ln(𝑏∕𝑎) =
ln(4∕.8) = 3.22 × 10−7 H∕m
2𝜋
2𝜋
√
c) Find 𝛼 and 𝛽 if 𝛼 + 𝑗𝛽 = 𝑗𝜔𝐶(𝑅 + 𝑗𝜔𝐿): Taking real and imaginary parts of the given
expression, we find
𝐿=
]1∕2
[√
√
)
} 𝜔 𝐿𝐶
(
{√
𝑅 2
−1
𝑗𝜔𝐶(𝑅 + 𝑗𝜔𝐿) = √
1+
𝛼 = Re
𝜔𝐿
2
and
𝛽 = Im
{√
[√
]1∕2
√
(
)
𝜔 𝐿𝐶
𝑅 2
𝑗𝜔𝐶(𝑅 + 𝑗𝜔𝐿) = √
1+
+1
𝜔𝐿
2
}
These can be found by writing out
}
{√
√
𝑗𝜔𝐶(𝑅 + 𝑗𝜔𝐿) = (1∕2) 𝑗𝜔𝐶(𝑅 + 𝑗𝜔𝐿) + 𝑐.𝑐.
𝛼 = Re
where 𝑐.𝑐 denotes the complex conjugate. The result is squared, terms collected, and the
square root taken. Now, using the values of 𝑅, 𝐶, and 𝐿 found in parts 𝑎 and 𝑏, we find
𝛼 = 3.0 × 10−2 Np∕m and 𝛽 = 50.3 rad∕m.
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11.27. The planar surface at 𝑧 = 0 is a brass-Teflon interface. Use data available in Appendix C to evaluate
the following ratios for a uniform plane wave having 𝜔 = 4 × 1010 rad/s:
a) 𝛼Tef ∕𝛼brass : From the appendix we find 𝜖 ′′ ∕𝜖 ′ = .0003 for Teflon, making the material a good
dielectric. Also, for Teflon, 𝜖𝑟′ = 2.1. For brass, we find 𝜎 = 1.5×107 S/m, making brass a good
conductor at the stated frequency. For a good dielectric (Teflon) we use the approximations:
√
( ′′ ) ( ) √
( ) √
. 𝜎 𝜇
𝜖
1
1 𝜖 ′′ 𝜔
′
𝜔 𝜇𝜖 =
=
𝛼=
𝜖𝑟′
2 𝜖′
𝜖′
2
2 𝜖′ 𝑐
( )]
[
√
. √
. √ ′
1 𝜖 ′′
𝜔
=
𝜔
𝛽 = 𝜔 𝜇𝜖 1 +
𝜇𝜖′
=
𝜖𝑟′
8 𝜖′
𝑐
For brass (good conductor) we have
√ ( )
.
. √
1
(4 × 1010 )(4𝜋 × 10−7 )(1.5 × 107 ) = 6.14 × 105 m−1
𝛼 = 𝛽 = 𝜋𝑓 𝜇𝜎brass = 𝜋
2𝜋
Now
√
(
)
√
1∕2 𝜖 ′′ ∕𝜖 ′ (𝜔∕𝑐) 𝜖𝑟′
𝛼Tef
(1∕2)(.0003)(4 × 1010 ∕3 × 108 ) 2.1
= 4.7 × 10−8
=
=
√
5
𝛼brass
6.14
×
10
𝜋𝑓 𝜇𝜎brass
b)
√
𝑐 𝜋𝑓 𝜇𝜎brass
(2𝜋∕𝛽Tef )
𝛽brass
𝜆Tef
(3 × 108 )(6.14 × 105 )
=
=
=
=
= 3.2 × 103
√
√
𝜆brass
(2𝜋∕𝛽brass )
𝛽Tef
(4 × 1010 ) 2.1
𝜔 𝜖𝑟′ Tef
c)
𝑣Tef
(𝜔∕𝛽Tef )
𝛽
=
= brass = 3.2 × 103 as before
𝑣brass
(𝜔∕𝛽brass )
𝛽Tef
11.28. A uniform plane wave in free space has electric field given by 𝐄𝑠 = 10𝑒−𝑗𝛽𝑥 𝐚𝑧 + 15𝑒−𝑗𝛽𝑥 𝐚𝑦 V/m.
a) Describe the wave polarization: Since the two components have a fixed phase difference (in
this case zero) with respect to time and position, the wave has linear polarization, with the field
vector in the 𝑦𝑧 plane at angle 𝜙 = tan−1 (10∕15) = 33.7◦ to the 𝑦 axis.
b) Find 𝐇𝑠 : With propagation in forward 𝑥, we would have
𝐇𝑠 =
15 −𝑗𝛽𝑥
−10 −𝑗𝛽𝑥
𝑒
𝐚𝑦 +
𝑒
𝐚𝑧 A∕m = −26.5𝑒−𝑗𝛽𝑥 𝐚𝑦 + 39.8𝑒−𝑗𝛽𝑥 𝐚𝑧 mA∕m
377
377
c) determine the average power density in the wave in W∕m2 : Use
[
]
} 1 (10)2
(15)2
1 {
𝐏𝑎𝑣𝑔 = Re 𝐄𝑠 × 𝐇∗𝑠 =
𝐚𝑥 +
𝐚𝑥 = 0.43𝐚𝑥 W∕m2 or 𝑃𝑎𝑣𝑔 = 0.43 W∕m2
2
2 377
377
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11.29. Consider a left-circularly polarized wave in free space that propagates in the forward 𝑧 direction. The
electric field is given by the appropriate form of Eq. (100).
a) Determine the magnetic field phasor, 𝐇𝑠 :
We begin, using (100), with 𝐄𝑠 = 𝐸0 (𝐚𝑥 + 𝑗𝐚𝑦 )𝑒−𝑗𝛽𝑧 . We find the two components of 𝐇𝑠
separately, using the two components of 𝐄𝑠 . Specifically, the 𝑥 component of 𝐄𝑠 is associated
with a 𝑦 component of 𝐇𝑠 , and the 𝑦 component of 𝐄𝑠 is associated with a negative 𝑥 component
of 𝐇𝑠 . The result is
)
𝐸 (
𝐇𝑠 = 0 𝐚𝑦 − 𝑗𝐚𝑥 𝑒−𝑗𝛽𝑧
𝜂0
b) Determine an expression for the average power density in the wave in W∕m2 by direct application of Eq. (77): We have
)
(
𝐸0
1
1
+𝑗𝛽𝑧
∗
−𝑗𝛽𝑧
(𝐚 − 𝑗𝐚𝑥 )𝑒
×
𝐏𝑧,𝑎𝑣𝑔 = 𝑅𝑒(𝐄𝑠 × 𝐇𝑠 ) = 𝑅𝑒 𝐸0 (𝐚𝑥 + 𝑗𝐚𝑦 )𝑒
2
2
𝜂0 𝑦
=
𝐸02
𝜂0
𝐚𝑧 W∕m2 (assuming 𝐸0 is real)
11.30. In an anisotropic medium, permittivity varies with electric field direction, and is a property seen in
most crystals. Consider a uniform plane wave propagating in the 𝑧 direction in such a medium, and
which enters the material with equal field components along the 𝑥 and 𝑦 axes. The field phasor will
take the form:
𝐄𝑠 (𝑧) = 𝐸0 (𝐚𝑥 + 𝐚𝑦 𝑒𝑗Δ𝛽𝑧 ) 𝑒−𝑗𝛽𝑧
where Δ𝛽 = 𝛽𝑥 − 𝛽𝑦 is the difference in phase constants for waves that are linearly-polarized in the 𝑥
and 𝑦 directions. Find distances into the material (in terms of Δ𝛽) at which the field is:
a) Linearly-polarized: We want the 𝑥 and 𝑦 components to be in phase, so therefore
𝑚𝜋
, (𝑚 = 1, 2, 3...)
Δ𝛽𝑧𝑙𝑖𝑛 = 𝑚𝜋 ⇒ 𝑧𝑙𝑖𝑛 =
Δ𝛽
b) Circularly-polarized: In this case, we want the two field components to be in quadrature phase,
such that the total field is of the form, 𝐄𝑠 = 𝐸0 (𝐚𝑥 ± 𝑗𝐚𝑦 )𝑒−𝑗𝛽𝑧 . Therefore,
Δ𝛽𝑧𝑐𝑖𝑟𝑐 =
(2𝑛 + 1)𝜋
(2𝑛 + 1)𝜋
⇒ 𝑧𝑐𝑖𝑟𝑐 =
, (𝑛 = 0, 1, 2, 3...)
2
2Δ𝛽
c) Assume intrinsic impedance 𝜂 that is approximately constant with field orientation and find 𝐇𝑠
and < 𝐒 >: Magnetic field is found by looking at the individual components:
𝐇𝑠 (𝑧) =
)
𝐸0 (
𝐚𝑦 − 𝐚𝑥 𝑒𝑗Δ𝛽𝑧 𝑒−𝑗𝛽𝑧 and
𝜂
{
} 𝐸2
1
< 𝐒 >= 𝑒 𝐄𝑠 × 𝐇∗𝑠 = 0 𝐚𝑧 W∕m2
2
𝜂
where it is assumed that 𝐸0 is real. 𝜂 is real because the medium is evidently lossless.
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11.31. A linearly-polarized uniform plane wave, propagating in the forward 𝑧 direction, is input to a lossless anisotropic material, in which the dielectric constant encountered by waves polarized along
𝑦 (𝜖𝑟𝑦 ) differs from that seen by waves polarized along 𝑥 (𝜖𝑟𝑥 ). Suppose 𝜖𝑟𝑥 = 2.15, 𝜖𝑟𝑦 = 2.10,
and the wave electric field at input is polarized at 45◦ to the positive 𝑥 and 𝑦 axes. Assume free space
wavelength 𝜆.
a) Determine the shortest length of the material such that the wave as it emerges from the output
end is circularly polarized: With the input field at 45◦ , the 𝑥 and 𝑦 components are of equal
magnitude, and circular polarization will result if the phase difference between the components
is 𝜋∕2. Our requirement over length 𝐿 is thus 𝛽𝑥 𝐿 − 𝛽𝑦 𝐿 = 𝜋∕2, or
𝐿=
𝜋
𝜋𝑐
=
√
√
2(𝛽𝑥 − 𝛽𝑦 ) 2𝜔( 𝜖𝑟𝑥 − 𝜖𝑟𝑦 )
With the given values, we find,
𝐿=
(58.3)𝜋𝑐
𝜆
= 58.3 = 14.6 𝜆
2𝜔
4
b) Will the output wave be right- or left-circularly-polarized? With the dielectric constant greater
for 𝑥-polarized waves, the 𝑥 component will lag the 𝑦 component in time at the output. The
field can thus be written as 𝐄 = 𝐸0 (𝐚𝑦 − 𝑗𝐚𝑥 ), which is lef t circular polarization.
11.32. Suppose that the length of the medium of Problem 11.31 is made to be twice that as determined in
the problem. Describe the polarization of the output wave in this case: With the length doubled, a
phase shift of 𝜋 radians develops between the two components. At the input, we can write the field
as 𝐄𝑠 (0) = 𝐸0 (𝐚𝑥 + 𝐚𝑦 ). After propagating through length 𝐿, we would have,
𝐄𝑠 (𝐿) = 𝐸0 [𝑒−𝑗𝛽𝑥 𝐿 𝐚𝑥 + 𝑒−𝑗𝛽𝑦 𝐿 𝐚𝑦 ] = 𝐸0 𝑒−𝑗𝛽𝑥 𝐿 [𝐚𝑥 + 𝑒−𝑗(𝛽𝑦 −𝛽𝑥 )𝐿 𝐚𝑦 ]
where (𝛽𝑦 − 𝛽𝑥 )𝐿 = −𝜋 (since 𝛽𝑥 > 𝛽𝑦 ), and so 𝐄𝑠 (𝐿) = 𝐸0 𝑒−𝑗𝛽𝑥 𝐿 [𝐚𝑥 − 𝐚𝑦 ]. With the reversal of the
𝑦 component, the wave polarization is rotated by 90◦ , but is still linear polarization.
11.33. Given a wave for which 𝐄𝑠 = 15𝑒−𝑗𝛽𝑧 𝐚𝑥 + 18𝑒−𝑗𝛽𝑧 𝑒𝑗𝜙 𝐚𝑦 V/m, propagating in a medium characterized
by complex intrinsic impedance, 𝜂.
a) Find 𝐇𝑠 : With the wave propagating in the forward 𝑧 direction, we find:
𝐇𝑠 =
]
1[
−18𝑒𝑗𝜙 𝐚𝑥 + 15𝐚𝑦 𝑒−𝑗𝛽𝑧 A∕m
𝜂
b) Determine the average power density in W∕m2 : We find
{
}
{ }
} 1
(15)2 (18)2
1 {
1
𝑃𝑧,𝑎𝑣𝑔 = Re 𝐄𝑠 × 𝐇∗𝑠 = Re
+
=
275
Re
W∕m2
2
2
𝜂∗
𝜂∗
𝜂∗
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11.34. Given the general elliptically-polarized wave as per Eq. (93):
𝐄𝑠 = [𝐸𝑥0 𝐚𝑥 + 𝐸𝑦0 𝑒𝑗𝜙 𝐚𝑦 ]𝑒−𝑗𝛽𝑧
a) Show, using methods similar to those of Example 11.7, that a linearly polarized wave results
when superimposing the given field and a phase-shifted field of the form:
𝐄𝑠 = [𝐸𝑥0 𝐚𝑥 + 𝐸𝑦0 𝑒−𝑗𝜙 𝐚𝑦 ]𝑒−𝑗𝛽𝑧 𝑒𝑗𝛿
where 𝛿 is a constant: Adding the two fields gives
[
(
)
(
) ]
𝐄𝑠,𝑡𝑜𝑡 = 𝐸𝑥0 1 + 𝑒𝑗𝛿 𝐚𝑥 + 𝐸𝑦0 𝑒𝑗𝜙 + 𝑒−𝑗𝜙 𝑒𝑗𝛿 𝐚𝑦 𝑒−𝑗𝛽𝑧
⎤
⎡
⎢
⎥
(
)
(
)
= ⎢𝐸𝑥0 𝑒𝑗𝛿∕2 𝑒−𝑗𝛿∕2 + 𝑒𝑗𝛿∕2 𝐚𝑥 + 𝐸𝑦0 𝑒𝑗𝛿∕2 𝑒−𝑗𝛿∕2 𝑒𝑗𝜙 + 𝑒−𝑗𝜙 𝑒𝑗𝛿∕2 𝐚𝑦 ⎥ 𝑒−𝑗𝛽𝑧
⎢
⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟ ⎥
⏟⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏟
⎢
⎥
2 cos(𝛿∕2)
2 cos(𝜙−𝛿∕2)
⎣
⎦
[
] 𝑗𝛿∕2 −𝑗𝛽𝑧
This simplifies to 𝐄𝑠,𝑡𝑜𝑡 = 2 𝐸𝑥0 cos(𝛿∕2)𝐚𝑥 + 𝐸𝑦0 cos(𝜙 − 𝛿∕2)𝐚𝑦 𝑒 𝑒
, which is
linearly polarized.
b) Find 𝛿 in terms of 𝜙 such that the resultant wave is polarized along 𝑥: By inspecting the part 𝑎
result, we achieve a zero 𝑦 component when 2𝜙 − 𝛿 = 𝜋 (or odd multiples of 𝜋).
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