Solutions to Problems in Jackson,
Classical Electrodynamics, Third Edition
Homer Reid
May 24, 2002
Chapter 8: Problems 1-8
1
Homer Reid’s Solutions to Jackson Problems: Chapter 8
2
Problem 8.2
A transmission line consisting of two concentric cylinders of metal with conductivity
σ and skin depth δ, as shown, is filled with a uniform lossless dielectric (µ, ). A
TEM mode is propagated along this line.
(a) Show that the time-averaged power flow along the line is
r
b
µ 2
P =
πa |H0 |2 ln
a
where H0 is the peak value of the azimuthal magnetic field at the surface of
the inner conductor.
(b) Show that the transmitted power is attenuated along the line as
P (z) = P0 e−2γz
where
1
γ=
2σδ
r
a1 + 1b
.
µ ln ab
(c) The characteristic impedance Z0 of the line is defined as the ratio of the voltage
between the cylinders to the axial current flowing in one of them at any
position z. Show that for this line
r
1
µ
b
Z0 =
ln
.
2π a
(d) Show that the series resistance and inductance per unit length of the line are
1
1 1
R=
+
2πσδ a b
µ
µc δ 1 1
b
L=
+
.
ln
+
2π
a
4π a b
(a) For the TEM mode, the electric field in the waveguide may be written
E(x, y, z, t) = Et (x, y)e−ikz e−iωt
where Et has only x and y components and may be derived from a scalar
potential, i.e. Et = −∇t Φ. Since Φ satisfies the Laplace equation, we may
write its general form immediately (neglecting an arbitrary constant):
Φ(ρ, θ) = β0 ln ρ +
∞
X
l=1
(αl ρl + βl ρ−l ) sin(lθ + αl ).
Homer Reid’s Solutions to Jackson Problems: Chapter 8
3
In this part of the problem we’ll neglect dissipation in the waveguide walls,
so the boundary condition on Et is that its components transverse to the walls
vanish, i.e.
∂Φ ∂Φ =
= 0.
∂θ r=b
∂θ r=a
This yields no condition on β0 , since the θ derivative of that term vanishes
anyway, but on the terms in the summation we obtain the conditions
αl al + βl a−l = αl bl + βl b−l = 0
which can only be satisfied if αl = βl = 0 for l 6= 0. Hence we have
Φ(ρ) = β0 ln ρ
−→
1
E = −β0 ρ̂.
ρ
The magnetic field is found from Jackson’s (8.28):
r
1
H=− B=
(z × E)
µ
µ
r
β0
=
θ̂.
µ ρ
(1)
(2)
The time-averaged Poynting vector is
1
1
S = (E × H∗ ) =
2
2
r
|β0 |2
µ
2
1
ẑ
ρ
Integrating over the cross section of the waveguide, we obtain the power transfer:
P =
Z
b
a
Z
2π
0
S · dA
r
Z b
2πρ dρ
1 |β0 |2
2 µ
ρ2
a
r
b
=
|β0 |2 · π ln
µ
a
r
µ
|β0 |2
b
=
·
(πa2 ) ln
a
µ a2
=
(3)
Referring back to (2) to rewrite the term in brackets, we obtain
P =
r
µ
(πa2 ) ln
b
|H(a)|2
a
(4)
(b) Without going back and completely re-solving for the fields in the waveguide
for the case of finite conductivity, we can calculate the power loss per unit length
Homer Reid’s Solutions to Jackson Problems: Chapter 8
4
approximately using Jackson’s equation (8.58):
I
dP
1
|ρ̂ × H|2 dl
−
=
dz
2σδ c
2 !
2
1
β0
β0
=
+ 2πa ·
2πb ·
2σδ µ
b
a
2
1 1
πβ0 =
.
+
σδ µ
b a
Dividing by (3), we obtain
1 dP
1
γ=−
=
2P dz
2σδ
r
(c) The fields inside the waveguide are
a1 + 1b
.
µ ln ab
β0
E(ρ, z, t) = − ei(kz−ωt) ρ̂
ρ
r
β0 i(kz−ωt)
H(ρ, z, t) =
e
θ̂
µ ρ
From the E field we can compute the voltage difference between the cylinders:
V (z, t) = −β0 ei(kz−ωt)
Z
b
a
b
dρ
= −β0 ei(kz−ωt) ln
ρ
a
(5)
while from the H field we can compute the axial current flowing in, say, the
outer cylinder:
r
I = 2πb|Kb | = 2πb|ρ̂ × H(ρ = b)| = 2π
β0 ei(kz−ωt) .
(6)
µ
Dividing (5) by (6), we have
Z=
V
1
=
I
2π
r
µ b
ln .
a
Homer Reid’s Solutions to Jackson Problems: Chapter 8
5
Problem 8.4
Transverse electric and magnetic waves are propagated along a hollow, right circular
cylinder of brass with inner radius R.
(a) Find the cutoff frequencies of the various TE and TM modes. Determine numerically the lowest cutoff frequency (the dominant mode) in terms of the
tube radius and the ratio of cutoff frequencies of the next four higher modes
to that of the dominant mode. For this part assume that hte conductivity of
brass is infinite.
(b) Calculate the attenuation constant of the waveguide as a function of frequency
for the lowest two modes and plot it as a function of frequency.
(a) The equation we have to solve is
(∇2t + γ 2 )Ψ(ρ, θ) = 0,
i.e. the Helmholtz equation. Ψ is Ez for the TM case and Hz for the TE case.
The boundary conditions are Ψ(ρ = R) = 0 for the TM case, and (∂Ψ/∂ρ)(ρ =
R) = 0 for the TE case.
The general solution of Helmholtz in 2D is
Ψ(ρ, θ) =
∞
X
JL (γρ)(AL eiLθ + BL e−iLθ ) + NL (γρ)(CL eiLθ + DL eiLθ ).
L=0
Since this solution must be valid everywhere in the interior of the waveguide,
including at ρ = 0, the part of the solution involving NL must vanish. Also,
for a physical solution we must have L an integer. But otherwise I don’t think
there are any constraints on AL and BL . I guess these guys are determined by
the field configuration one forces into the waveguide at one of its ends.
The allowed values of γ are determined by the boundary conditions. These
are
TM case :
TE case :
Ψ|ρ=R = 0
∂Ψ =0
∂ρ =⇒
JL (γR) = 0
(7)
=⇒
JL0 (γR) = 0
(8)
ρ=R
(9)
Hence the allowable eigenvalues are given by
γi =
xi
R
Homer Reid’s Solutions to Jackson Problems: Chapter 8
6
where the xi are the roots of JL (x) = 0 and JL0 (x) = 0. Referring to Jackson’s
tables on pages 114 and 370, we can write down the five lowest-lying eigenvalues:
γ1 =
γ2 =
γ3 =
γ4a =
γ4b =
1.841
,
R
2.405
,
R
3.054
,
R
3.832
,
R
3.832
,
R
TE, L = 1
TM, L = 0
TE, L = 2
TE, L = 1
TM, L = 0.
The last two eigenvalues are degenerate.
The lowest cutoff frequency is
γ1
1.841
ωc = √ = √ .
µ
R µ
(b) The lowest-lying mode is the TE mode with L = 1. For this mode we have
Hz (ρ, θ, z, t) = H0 J1 (γ1 ρ)eiθ ei(kz−ωt)
(10)
with k 2 = µω 2 − γ12 . The tangential component of the field, from Jackson
(8.33), is
k
(11)
Hθ (ρ, θ, z, t) = − 2 Hz
ργ
Using (10) and (11), we can find the current induced in the wall of the conductor
at ρ = R:
k
Keff = ρ̂ × H(ρ = R) = −H0 J1 (γ1 R)eiθ ei(kz−ωt ) θ̂ +
ẑ
.
Rγ12
Then Jackson (8.58) is
"
2 #
dP
1
k
2 2
−
=
H J (γ1 R) · 2πR 1 +
dz
2σδ 0 1
Rγ12
On the other hand, the transmitted power is given by Jackson (8.51):
2 ω 2 1/2 Z
µ ω
λ
1−
Hz∗ Hz dA
ωλ
ω
A
r 2 ω 2 1/2 Z R
µ
ω
λ
ρJ12 (γ1 ρ) dρ
1−
= πH02
ωλ
ω
0
P =
1
2
r
(12)
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Homer Reid’s Solutions to Jackson Problems: Chapter 8
The integral can be evaluated with Jackson (3.95):
= πH02
r
µ
ω
ωλ
2 1−
ω 2 1/2 R2
λ
ω
2
[J2 (γ1 R)]2
(13)
Dividing (12) by (13), we calculate the attenuation coefficient:
2 "
1/2
2 # r J1 (1.841)
k
ω2
2
1 dP
ωλ 2
1+
=
β=
2P dz
σδR µ J2 (1.841)
Rγ12
ω
ω 2 − ωλ2
2 "
1/2
2 # r 1 dP
J1 (1.841)
Rk
ω2
2
ωλ 2
=
1+
=
2P dz
σδR µ J2 (1.841)
(1.841)2
ω
ω 2 − ωλ2
2 1/2
r 1 dP
2
ω2
J1 (1.841)
µR2 ω 2 ωλ 2
=
=
.
2P dz
σδR µ J2 (1.841)
(1.841)2
ω
ω 2 − ωλ2
Problem 8.5
A waveguide is constructed so that
√ the cross section of the guide forms a right
triangle with sides of length a, a, 2a, as shown. The medium inside has µr = r =
1.
(a) Assuming infinite conductivity for the walls, determine the possible modes of
propagation and their cutoff frequencies.
(b) For the lowest mode of each type calculate the attenuation constant, assuming
that the walls have large, but finite, conductivity. Compare the result with
that for a square guide of side a made from the same material.
(a) We’ll take the origin of coordinates at the lower left corner of the triangle.
Then the boundary surfaces are defined by x = 0, y = a, and x = y. The
task is to solve (∇2t + γ 2 )Ψ = 0 subject to the vanishing of Ψ, or its normal
derivative, at the walls. In the text, Jackson finds the form of the solutions
for a rectangular waveguide. A little bit of staring at the triangular waveguide
reveals that appropriate solutions for this geometry can be assembled from linear
combinations of the solutions for the rectangular case. For example, a term like
sin kx x sin ky y, for suitable choices of kx and ky , already vanishes on the two
legs of the triangle. To get it to vanish on the third boundary surface, i.e. the
hypotenuse (x = y), we can simply subtract the same term with kx and ky
swapped. In other words, we take
h mπx nπy nπx mπy i
sin
− sin
sin
Amn sin
a a a a i
h
mπx
nπy
nπx
mπy
X
Hz (x, y) =
Bmn cos
cos
+ cos
cos
a
a
a
a
Ez (x, y) =
X
(TM)
(TE)
Homer Reid’s Solutions to Jackson Problems: Chapter 8
8
The TE case involves the plus sign because in the normal derivative on the
diagonal boundary surface the x derivative comes in with the opposite sign as
the y derivative.
2
These satisfy (∇2t + γmn
)Ψ = 0, where
π 2
2
(n2x + n2y ).
=
γmn
a
In contrast to the rectangular case, TM modes with m = n vanish identically.
For both TM and TE modes, mode (m, n) is the same mode as (n, m).
As in the case of the rectangular waveguide, the smallest value of γ is to be
had for the TE1 , 0 mode, in which case
π
γ10 =
a
√
and the cutoff frequency is ωc(1,0) = π/(a µ). For the TM case the lowest
√
propagating mode is (m, n) = (2, 1), for which γ21 = 5π/a and ωc(2,1) =
√
5ωc(1,0) .
(b) The lowest-frequency TE mode has
πy i
h
πx + cos
Hz = H0 cos
a
a
h
ikπ
πx
πy i
Ht =
î + sin
ĵ .
H0 sin
2
aγ10
a
a
The power loss is
I
1
dP
=
|n × H|2 dl
(14)
−
dz
2σδ
On the lower surface (y = 0) we have
h
i2
πx k2 π2
2 πx
2
2
2
2
+ 1 + 2 4 sin
|n × H| = |Hy + Hz | = H0
cos
a
a γ10
a
The contribution of the lower surface to the integral in (14) is thus
Z
k2 π2
3
+ 2 4
= aH02
2 a γ10
lower
(15)
The contribution of the right (vertical) boundary surface is the same. On the
diagonal boundary surface, we have
1
n = √ (−î + ĵ)
2
=⇒
1
n × H = √ [Hz î + Hz ĵ − (Hy + Hx )k̂]
2
with magnitude
1 2
[Hx + Hy2 + Hz2 + 2Hx Hy ]
2 (
)
2
kπ
H02
2 πγ
2 πγ
4 cos
+4
sin
=
2
a
aγ10
a
|n × H|2 =
9
Homer Reid’s Solutions to Jackson Problems: Chapter 8
where γ = x = y is the common coordinate
as we move from (0, 0) to (a, a). In
√
the integral in (14) we can put dl = 2dγ and integrate over γ from 0 to a to
obtain
Z
√
k2 π2
2
= 2aH0 1 + 2 4 .
a γ10
diagonal
Adding this to two times (15) and inserting into (14), we have
√
√ k2 π2
aH02
dP
=
−
3 + 2 + (2 + 2) 2 4 .
dz
2σδ
a γ10
(16)
On the other hand, from Jackson (8.51) we have
H2
P = 0
2
r
1/2
2 ω
ωλ2
1− 2
ωλ
ω
Z aZ yh
πx πy πx πy i
cos2
×
+ cos2
+ 2 cos
cos
dx dy
a
a
a
a
0
0
µ
By symmetry, the integral is just half of what we would get from integrating
the integrand over a square of side a, which, by inspection, is a2 . Hence
a2 H02
P =
4
r
µ
ω
ωλ
2 ω2
1 − λ2
ω
.1/2
We could at this point proceed to write out the explicit form of the attenuation
constant, but what’s the point?
Problem 8.6
A resonant cavity of copper consists of a hollow, right circular cylinder of inner
radius R and length L, with flat end faces.
(a) Determine the resonant frequencies of the cavity for all types of waves. With
√
(1/ µR) as a unit of frequency, plot the lowest four resonant frequencies of
each type as a function of R/L for 0 < R/L < 2. Does the same mode have
the lowest frequency for all R/L ?
(b) If R=2 cm, L=3 cm, and the cavity is made of pure copper, what is the
numerical value of Q for the lowest resonant mode?
(a) Taking the origin at the center of the cavity, the ρ and φ components of the
fields must vanish at z = ±L/2. Since the z dependence of all field components
is e±ikz , the allowed values of k are k = nπ/L, with E ∝ sin kz for k even and
E ∝ cos kz for k odd.
Homer Reid’s Solutions to Jackson Problems: Chapter 8
10
The equation characterizing T M modes is
(∇2t + γ 2 )Ez = 0,
Ez |∂S = 0.
Expanding this in cylindrical coordinates, we obtain
∂ 2 Ez
1 ∂Ez
1 ∂ 2 Ez
+
+
+ γ 2 Ez = 0
∂ρ2
ρ ∂ρ
ρ2 ∂φ2
We put Ez (ρ, φ) = R(ρ)P (φ) to obtain
∂2P
+ νP = 0
∂φ2
ν2
∂ 2 R 1 ∂R
2
+
+ γ − 2 = 0.
∂ρ2
ρ ∂ρ
ρ
The solutions are
P (φ) = e±iνφ
R(ρ) = Jν (γρ).
For single-valuedness we require ν ∈ , and to ensure Ez (ρ = R) = 0 we require
γ = xνm /R where xνm is the mth root of Jν (x) = 0. Hence
(
keven
ρ ±iνφ −iωt sin kz,
(TM modes).
e
e
Ez = AJν xνm
R
cos kz,
kodd
For TE modes the relevant equation is
(∇2t
2
+ γ )Bz = 0,
∂Bz = 0.
∂n ∂S
The general solution to the differential equation is the same as above, but now
the boundary condition requires Jν0 (γR) = 0, so γ = yνm /R where yνm is the
mth root of Jν0 (y) = 0. Then the solutions look like
(
ρ ±iνφ −iωt sin kz,
keven
Bz = AJν yνm
(TE modes).
e
e
R
cos kz,
kodd
As we saw above, the allowed wavevectors are k = nπ/L. The frequency is
related to the wavenumber according to
p
2 + k2
ωνmn = cm γνm
n
s

2

cm
R

2 + π2

x
n2 ,
(TM)

νm
 R
L
=
s
2



c
R
m

2
2

yνm + π
n2 ,
(TE)
R
L
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Homer Reid’s Solutions to Jackson Problems: Chapter 8
PSfrag replacements
9
TE modes
TM modes
ω (units of c/R)
8
7
6
5
4
3
2
1
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
R/L
Figure 1: TM and TE mode frequencies for the resonant cavity of Problem 8.6.
2
Homer Reid’s Solutions to Jackson Problems: Chapter 8
(The m subscript on cm is not related to the m subscripts on ω and ν).
The lowest four TM and TE mode frequencies are shown in Figure 1.
(b) The lowest resonant mode is the TE1,1,1 mode.info
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