Chapter 1
1-1 The expression in (Ex 1.3.4) may be approximated when R z as
Qk̂
z
Qk̂
R2
Qk̂
1− √
≈
1− 1− 2
=
2
2
2
2
2πε0 R2
2πε
R
2z
4πε
R +z
0
0z
1-2 In cylindrical coordinates, the electric field integral becomes:
E(z)
=
1
4πε0
=
λ0
4πε0
λ0 (1 + sin ϕ )(z k̂ − r )δ(r − a)δ(z )r dr dϕ dz |z k̂ − r |3
2π
0
(1 + sin ϕ )(z k̂ − a cos ϕ ı̂ − a sin ϕ ĵ)adϕ
(z 2 + a2 )3/2
The integrals over ϕ are easily performed to give:
2
ĵ
λ
πa
2πaz
k̂
0
E(z)
=
− 2
2πε0 (z 2 + a2 )3/2
(z + a2 )3/2
λ0 a
1
z
k̂
−
=
aĵ
2
2ε0 (a2 + z 2 )3/2
1-3
After performing the z integration, the electric field integral reduces to:
E(z)
=
1
4πε0
br2 (z k̂ − r r̂)dϕ r dr
(z 2 + r2 )3/2
br3 z k̂ dr
(z 2 + r2 )3/2
where we have used the fact that r̂dϕ = 0. The remaining integral yields:
=
2π
4πε0
a
r3 dr
bz k̂ 2
z2
2 + √
=
z
+
r
2
2 3/2
2ε0
z 2 + r2 0
0 (z + r )
bz k̂ 2z 2 + a2
√
=
− 2|z|
2ε0
z 2 + a2
bz k̂
E(z)
=
2ε0
a
It is easily verified that at large z (compared to a) the field has a 1/z 2 form.
1-4 The field of the cylinders is most easily computed using Gauss’ law. Inside
the inner cylinder the enclosed charge is zero implying that the electric field
must vanish. Between the two cylinders, the charge enclosed by a gaussian
cylinder of radius r and length L is 2πaLσa leading to a field
Er (a < r < b) =
aσa
2πaLσa
=
ε0 2πrL
ε0 r
—1—
2
Classical Electromagnetic Theory
Outside the outer cylinder the charge enclosed by a concentric gaussian cylinder is 2πL(aσa + bσb ) leading to a field
Er (r > b) =
aσa + bσb
ε0 r
1-5 The charge contained in a gaussian sphere of radius r is
r
Qr =
ρ0 e−kr 4πr2 dr
0
r2
2r
2
+ 2 + 3
k
k
k
2
2
r
2r
+ 2+
= 4π 3 − e−kr
k
k
k
= −4πe−kr
r
0
2
k3
The field we deduce is then
Er (r) =
1 2 − e−kr (k 2 r2 + 2kr + 2)
2
3
ε0 r k
1-6 The charge contained in a gaussian sphere of radius r < a is
r r2
2r
+ 2 4πr2 dr
ρ0 1 −
Q(r < a) =
a
a
0
= 4πρ0
r3
2r4
r5
−
+ 2
3
4a
5a
while for radii larger than a, the enclosed charge is 4πρ0 a3 /30. The field in
each case may then be deduced as
⎧
ρ0 r 1
r
1 r2
⎪
⎪
−
+
for r < a
⎪
⎨ ε
3 2a 5 a2
0
Er =
ρ0 a3
⎪
⎪
for r > a
⎪
⎩ 30ε r2
0
1-7 The electric field at either plate is σε0 . It would be tempting to conclude that
= σ 2 ε0 . This is
the force per unit area on the second plate is therefore σ E
wrong however! It is worth noting that the electric field between the plates is
produced by both plates; it is then reasonable to assume that only half of the
electric field is effective at producing a force on either plate. It is probably
more convincing to calculate the force from elementary considerations. Consider an element of charge dq = σdA on one plate and add up all the forces
arising from all the elements of charge dq = σdA on the other plate. The
force is then
σdA
−σ 2 dA
σ (r − r )dA
(z k̂ − x ı̂ − yĵ)dx dy dFq =
=
3
4πε0
|r − r |
4πε0
(z 2 + x2 + y 2 )3/2
Chapter One Solutions
3
−σ 2 dA
=
2ε0
∞
0
z k̂ r dr
−σ 2 k̂dA
=
2ε0
(z 2 + r2 )3/2
1-8 The calculations to this apparently simple problem are surprisingly cumbersome, giving strong motivation to try the dipole approximation of the next
chapter. The net charge on the line is clearly zero while the total electric field
is given by:
a xı̂ + yĵ + (z − z )k̂ z dz = b
E
4πε0 −a [x2 + y 2 + (z − z )2 ]3/2
a
a
(xı̂ + yĵ)(z − z)dz (xı̂ + yĵ)dz b
bz
=
+
4πε0 −a [x2 + y 2 + (z − z)2 ]3/2
4πε0 −a [x2 + y 2 + (z − z)2 ]3/2
a
a
(z − z)2 dz (z − z)dz bk̂
bz k̂
−
−
3/2
4πε0 −a [x2 + y 2 + (z − z)2 ]
4πε0 −a [x2 + y 2 + (z − z)2 ]3/2
b
=
4πε0
+k̂ a
a
z(z − z)(xı̂ + ĵ)
+
2
2
2
2
2
2
2
2
x + y + (z − z) −a (x + y ) x + y + (z − z) −a
−(xı̂ + yĵ)
z
x2 + y 2 + (z − z)2
a 2
2
2
− ln (z − z) + x + y + (z − z)
−a
1-9 Any remaining field tangential to the surface would cause further movement
of the free charges in the conductor.
1-10 It would clearly have been more efficient to do this before calculating the field
and then to obtain the field by differentiating the potential.
a
ρ d3 r bz dz V (z) =
=
2
4πε0 |r − r |
x + y 2 + (z − z )2
−a 4πε0
a
b 2
=
x + y 2 + (z − z )2 4πε0
−a
a
bz
+
ln (z − z ) + x2 + y 2 + (z − z )2 4πε0
−a
1-11 The potential due to the ring along the center line is
2π
λ0 (1 + sin ϕ)adϕ
1
√
V (z) =
4πε0 0
z 2 + a2
=
2π
λ0 a
λ a
√
√0
=
4πε0 z 2 + a2
2ε0 z 2 + a2
4
Classical Electromagnetic Theory
1-12 The potential above the center of the plate is given by
1
2πb
br2 r dϕ dr
r3 dr
√
√
V (z) =
=
4πε0
4πε0
z 2 + r2
z 2 + r2
a
b 1 2
(z + r2 )3/2 − z 2 z 2 + r2 =
2ε0 3
0
b 2
=
(a − 2z 2 ) z 2 + a2 + 2|z|3
6ε0
· J = −∂ρ/∂t, over a sphere surround1-13 Integrating the continuity equation, ∇
ing the point source but excluding the wire, we have 4πR2 J = −I, where I
is the current leading from the point. We conclude then that
J=
I
4πR2
with I the current delivered by the wire.
1-14 The magnetic induction field may be found using the Biot-Savart law as follows:
J × (r − r )d3 r
= μ0
B
4π
|r − r |3
The current density J = ρv = σδ(z )r ω ϕ̂ for r < a. Inserting this into the
integral we have
μ0
B(z)
=
4π
=
μ0 ρω
4π
ρr ω ϕ̂ × (z k̂ − r r̂) r dr dϕ
(z 2 + r2 )3/2
r2 zr̂ + r3 k̂ dϕ dr
(z 2 + r2 )3/2
The integrations over ϕ eliminates the radial component leaving
μ0 σω k̂
B(z)
=
2
r3
dr
(z 2 + r2 )3/2
a
μ0 σω k̂ 2
z2
2
=
z +r + √
2
2
2
z +r
0
2
2
μ0 σω k̂ 2z + a
√
=
− 2 |z|
2
z 2 + a2
Alternatively we could have integrated the magnetic scalar potential of a ring
of radius r carrying current dI = σωr dr over the whole disk to obtain:
zdI
σωz a r dr
√
√
Vm = −
=−
2 0
2 z 2 + r2
z 2 + r2
Chapter One Solutions
5
σωz 2
z + a2 − |z|
2
The z component of the magnetic induction field may then be found by differentiating
=−
∂Vm
μ0 σω 2
z2
2
=
z +a + √
− 2 |z|
Bz = −μ0
∂z
2
z 2 + a2
μ0 σω 2z 2 + a2
√
=
− 2 |z|
2
z 2 + a2
While there appears to be no clear advantage to using the scalar potential in
this problem, the next two problems use it more advantageously.
1-15 The magnetic scalar potential of the hollow sphere will be calculated as a
sum of current loops.
A loop of width dz subtending angle dθ at height z √
2
has radius a = R − z 2 = R sin θ and carries current dI = aσωdz / sin θ
= Rσωdz . The contribution to the magnetic scalar potential from one such
loop is then
(z − z ) dz −(z − z )dI
σRω
=−
dVm = 2
2 (z − z )2 + a2
(z − z )2 + (R2 − z 2 )
=−
(z − z ) dz Rωσ
√
2
z 2 + R2 − 2zz Summing loops from −R to R we find
dz z dz σRωz R
σRω R
√
√
+
Vm (z) = −
2
2
z 2 + R2 − 2zz z 2 + R2 − 2zz −R
−R
R
σRω 2
2
=
z + R − 2zz 2
−R
R
σRω(2z 2 + 2R2 + 2zz ) 2
2 − 2zz −
z
+
R
2
3(2z)
−R
σR4 ω
3z 2
It is now simple to obtain the z -component of the magnetic induction field
Bz :
2 μ0 ωσR4
∂Vm
=
Bz = −μ0
∂z
3
z3
The more complete problem of finding the field anywhere is solved as example
(5–10) of the text.
= −σR2 ω +
1-16 Although we could integrate the field (or potential) of disks such as problem
1-13, it is in fact far simpler to sum spherical shells to fill the sphere. Using
6
Classical Electromagnetic Theory
the scalar potential from the problem above and dropping the inconsequential
constant term, we replace R by r and σ by ρdr and integrate.
R 4
ρr ω ρR5 ω
Vm (z) =
dr
=
3z 2
15z 2
0
Differentiating to find the magnetic induction field we find
Bz (z) =
2μ0 ρR5 ω
15z 3
1-17 The field at z from a circular loop of radius a at ±a is
Bz (z) =
a2
μ0 I
2 (z ∓ 1 a)2 + a2 3/2
2
At z = 0 both coils give the same contribution to yield
a2
8μ0 I
Bz (0) = μ0 I = 3/2
3/2
5 a
( 12 a)2 + a2
1-18 The magnetic field along the axis of a radius a single turn loop located at z is
Iμ0
a2 k̂
B(z)
=
2 [(z − z )2 + a2 ]3/2
The number of turns per length dz of solenoid is given by (ndz )/L, so that
the total field B is
dz nIμ0 a2 k̂ L/2
B(z) =
3/2
2L
−L/2 [(z − z )2 + a2 ]
L/2
z − z
μ0 nIa2 k̂
2L a2 (z − z )2 + a2 −L/2
⎡
⎤
1
1
L
−
z
L
+
z
μ0 nI k̂ ⎣
⎦
2
=
+ 2
2L
1
2
2
(z − L) + a
(z + 1 L)2 + a2
=−
2
2
1-19 Using Faraday’s law,
for r < a
for a < r < b
for r > b
μ0 Iπr2
πa2
⇒
Bϕ =
μ0 Ir
2πa2
2πrBϕ = μ0 I
⇒
Bϕ =
μ0 I
2πr
2πrBϕ = 0
⇒
Bϕ = 0
2πrBϕ = μ0
JdS =
Chapter One Solutions
7
where the thickness of the outer conductor has been neglected.
1-20 The vector potential of a filamentary current may be deduced from (1–52) as
r ) = μ0
A(
4π
r )
J(
d
μ0
3 d
r
=
|r − r |
4π
|r − r |
Alternatively, we recall the magnetic induction field of a filamentary current
loop (1–35)
r ) = μ0
B(
4π
=
μ0 I
4π
μ0
1
Id × (r − r )
=
−
Id × ∇
|r − r |3
4π
|r − r |
×
∇
d d × μ0 I
=
∇
|r − r |
4π
|r − r |
which allows us to identify the vector potential with the integral. We proceed
to find the divergence.
d d μ0 I
μ0 I
∇·
=
∇·A=∇·
4π
|r − r |
4π
|r − r |
1
1
μ0 I
μ0 I
×∇
∇
d · ∇
=
−
=−
dS = 0
4π
|r − r |
4π S |r − r |
where we used Stokes’ theorem (18) to effect the last step and the fact that a
gradient has zero curl. The point of this exercise is to show that the simple
in the Coulomb gauge.
expression derived from (1–52) expresses A
1-21 The Laplacian of the vector potential may be written using (1–52)
r) =
∇2 A(
μ0 2
∇
4π
r )
J(
1
μ0
3 r )∇2
d
J(
d3 r r
=
|r − r |
4π
|r − r |
∇2 (1/|r − r |) may be rewritten with the aid of (26) as −4πδ(r − r ) so that
the above becomes
2
r )δ(r − r )d3 r = −μ0 J(
r)
∇ A(r ) = −μ0 J(
1-22 For simplicity we place the z axis along one of the wires. From this wire alone,
the nonzero component of the field is given by
Bϕ (r < a) =
μ0 Ir
∂Az
=−
2πa2
∂r
from which we conclude that
Az (r < a) = −
μ0 Ir2
+C
4πa2
8
Classical Electromagnetic Theory
and similar reasoning gives the vector potential due to this wire outside the
wire as
Bϕ (r > a) =
μ0 I
μ0 I
⇒ Az (r > a) =
ln r + D
2πr
2π
At a point r outside the two wires, the vector potential is then given by
r
μ0 I
ln
Az (r > a) = −
+F
2π
r2
where r2 is the distance of the point from the center of the second wire’s
center and F is an arbitrary constant. At a point inside the wire containing
the z-axis, the vector potential is
Az (r < a) = −
μ0 Ir2
μ0 I
ln r2 + E
−
4πa2
2π
where E is an arbitrary constant. In either case the distance r2 of the field
point from the second wire may be expressed in terms of the point’s polar
angle using
r2 = h2 + r2 − 2hr cos θ
1-23 In line with the considerations of section 1.2.2, The transverse force
on par
−1
−1
=
γ
F
.
The
inverse
Lorentz
factor
γ
=
1 − β2 =
ticles
must
obey
F
√
2
1 − .99 = 0.141 The mutual repulsive force is therefore decreased to 14.1%
of its electrostatic (rest frame) value.
1-24 The magnetic scalar potential along the axis is given by (p. 29)
Vm (z) = −
N I (z + 12 L)2 + a2 − (z − 12 L)2 + a2
2L
Differentiating to obtain the magnetic induction field,
1
L−z
z + 12 L
I
k̂
μ
0
m=
= −μ0 ∇V
B
+ 2
2L
(z + 1 L)2 + a2
(z − 1 L)2 + a2
2
2
Figure 1.1: Geometry of solenoid in question 1.24.
With reference to the diagram, it is evident that the two fractions are respectively the cosines of the half angles subtended by the ends of the solenoid:
Bz =
μ0 I
(cos θ1 + cos θ2 )
2L
Chapter One Solutions
9
1-25 In general a charged particle in an electromagnetic field experiences a Lorentz
force
+ v × B)
F = q(E
We must chose v so that no further acceleration or deceleration occurs, in
other words
= −v × B
E
to obtain
To isolate v , we cross multiply both sides with B
×B
= −(v × B)
×B
= (v · B)
B
− B 2v
E
cannot give rise to forces, we find
Since the component of v along B
v = −
×B
E
2
B
Despite my best efforts a few typos have crept into the text, they will be
reported at the end of each chapter.
The denominator of the first term of (1–36) is missing an r2 to become
∞
z
μ0 I2 (k̂ × r̂)
√
=
2
2
2
4π
r r + z −∞
Chapter 2
2-1 The dipole moment of the ring is
p = r λδ(r − a)δ(z) cos ϕ rdrdϕdz
= a(ı̂ cos ϕ + ĵ sin ϕ)a cos ϕdϕ
= λπa2 ı̂
2-2 The dipole moment of the rod is
a
2λa3 k̂
p =
z k̂λzdz =
3
−a
of a dipole as given in (2–23):
2-3 We use (10) to expand the curl of A
r
μ0
· r − (m
× A(
r) = ∇
× μ0 (m
∇
∇
×
r
)
=
·
∇)
m
4πr3
4π
r3
r3
· (r/r3 ) occurring in the first term is readily evaluated using (7)
The ∇
· r = ∇
· r = 0
1 · r + 1 ∇
∇
r3
r3
r3
and the second term may be evaluated from
r
∂ r
∂ r
∂ r
(m
· ∇) 3 = mx
+ my
+ my
3
3
r
∂x r
∂y r
∂y r3
focussing on the first of these three terms,
ı̂
∂ r
3xr
mx
=
m
−
x
∂x r3
r3
r5
and similar expressions obtain for the other two terms. Adding the three
terms gives
3(m
· r )r
× A(
r ) = −m
∇
+
r3
r5
2-4 Rather than computing the dipole moment about the origin let us calculate a
modified dipole moment about an arbitrary point a.
pa = (r − a )ρ(r)d3 r
3
= rρ(r)d r − a
ρ(r)d3 r
= p0 − aQ
— 10—
Chapter Two Solutions
11
We conclude that whenever the total charge Q vanishes, the dipole moment
about a is identical to that about the origin.
2-5 The zz component is easily found:
Qzz = ρ(3z 2 − r2 )d3 r = q(b2 − a2 )
and Qxx = q(3x2a −a2 )−q(3x2b −b2 ) = 12 q(a2 −b2 ) = Qyy . The off-diagonal
elements vanish.
2-6 The representation of this quadrupole will depend on the relative orientation
of the square and the coordinate system. Let us place the corners at ± 12 a
with a positive charges along the y ± 12 a edges while negative charges reside
on the x = ± 12 a edges. Then
Qzz = (3z 2 − r2 )ρ(r )d3 r = 0
Qyy = (3y 2 − r2 )ρ(r )d3 r
1
1
2q 2 a
2q 2 a 3 2
( 4 a − x2 − 14 a2 )dx −
(3y 2 − 14 a2 − y 2 )dy
=
a − 12 a
a − 12 a
1 a
1 a
a2 y 2
2q a2 x x3 2
2q 2y 3
−
−
=
−
a
2
3 1
a
3
4 1
= qa2
−2a
−2a
and Qxx = −qa2 . The remaining components such as Qxz and Qxy all vanish.
2-7 For an object spinning about the z-axis, let us denote by r the distance of a
mass or volume element from the axis and let ρm denote the mass density.
is given by
Then the angular momentum L
2
L = I
ω = k̂ω r dm = k̂ω r2 ρm d3 r
The magnetic moment is given by
3r =
m
= 12 r × Jd
= 12 r2 ω k̂ρd3 r
1
2
r × (
ω × r )ρd3 r
Comparing the two expressions, when the functional form of ρ and ρm is the
= 1 ρ/ρm = 1 q/m.
same, the ratio of (m/
L)
2
2
2-8 The moment of inertia of the solid sphere is 25 mR2 while the magnetic moment
of the hollow shell may be found as a sum of plane loops:
R
dz
2
m
= k̂ πr dI = k̂
π(R2 − z 2 )R sin θωσ
sin
θ
−R
4
2R
= πσω k̂ 2R4 −
= 13 4πωσR4 k̂ = 13 QR2 ω k̂
3
12
Classical Electromagnetic Theory
The gyromagnetic ratio is then 56 Q/m.
2-9 The zz component of the quadrupole moment is given by
Qzz =
=
1
2L
− 12 L
5
η(z 2 −
L2
)(3z 2 − z 2 )dz
12
L η
90
while the xx and yy components are each −L5 η/180. The off-diagonal elements vanish as would be expected.
2-10 The potential due to the quadrupole when Qxx = Qyy = − 12 Qzz is
" 2
#
1
1 ! xi xj Qij
z Qzz + x2 Qxx + y 2 Qyy
=
5
5
4πε0
2r
8πε0 r
#
Qzz " 2
r cos2 θ − 12 r2 sin2 θ cos2 ϕ − 12 r2 sin2 θ sin2 ϕ
=
5
8πε0 r
#
Qzz
Qzz "
2 cos2 θ − sin2 θ =
(3 cos2 θ − 1)
=
3
16πε0 r
16πε0 r3
V (r ) =
2-11 The charge on a radius R sphere with charge density ρ = ρ0 z 2 = ρ0 r2 cos2 θ
is
3
Q = ρd r = ρ0 r2 cos2 θr2 dr sin θdθdϕ
4πρ0 R5
2πρ0 R5 π
sin θ cos2 θdθ =
=
5
15
0
2-12 The zz component of the quadrupole moment of the sphere in problem 2-11
is
Qzz = (3z 2 − r2 )ρd3 r
= ρ0 3z 4 d3 r − ρ0 r2 z 2 d3 r
= 2πρ0
=
3r cos θ sin θdθdr − 2πρ0
6
4
r6 cos2 θ sin θdrdθ
12πρ0 a7
4πρ0 a7
16πρ0 a7
−
=
5·7
3·7
105
2-13 Expanding the given expression,
1 !
1
qir (i) · ∇
4πε0
r
r
p · r
1 !
qir (i) · 3 ≡
=
4πε0
r
4πε0 r3
V2 = −
and we recover (2–4).
Chapter Two Solutions
13
2-14 We expand the form given:
(i) 1 ! (i) −xj xj
V3 =
qir · ∇
8πε0 i,j
r3
xj
−1 ! (i) (i)
=
qi xk xj ∂k 3
8πε0
r
i,j,k
δjk
1 ! (i) (i) 3xj xk
qi xk xj
−
=
8πε0
r5
r3
i,j,k
!
xj xk Qjk
δjk xj xk
1 ! (i) (i) 3xj xk
qi xk xj
−
=
=
8πε0
r5
r5
8πε0 r5
i,j,k
j,k
and we recover the point charge form of (2–16).
2-15 Substituting the charge density ρ = ρ0 z for r on the z axis z ∈ (−a, a) and
zero elsewhere into (2–32), we note that γ = θ, the field point polar angle so
that we may remove it from the integral to obtain form (2–33). Then
∞
1 ! Pn (cos θ) a
ρ0 z z n dz V (r ) =
4πε0 n=0 rn+1
−a
a
ρ0 ! Pn (cos θ) z n+2 ρ0 ! P2n+1 (cos θ)a2n+3
=
=
4πε0
rn+1 n + 2 −a 2πε0
(2n + 3)r2n+2
2-16 When r is less than a, we must split the integral into two portions, the first
running from −r to r where r = r< and the second running from r to a
where r = r> . Thus
r
1
ρ0 !
Pn (cos θ) n+1
z n z dz V (r < a, θ) =
4πε0
r
−r
a −r z dz
z dz
n
+r
+
n+1
n+1
r z
−a z
$
−r %
a
r
z n+2
rn
1 ρ0 !
1 Pn (cos θ)
−
+
=
4πε0
(n + 2)rn+1 −r (n − 1) z n−1 r z n−1 −a
2rn+2
2
rn
2
ρ0 !
Pn (cos θ)
−
−
=
4πε0
(n + 2)rn+1
(n − 1) an−1
rn−1
n odd
r
ρ0 !
r
rn−1
=
+
Pn (cos θ)
1 − n−1
2πε0
(n + 2) (n − 1)
a
n odd
2-17 We take the electrons to be at x = ± a at time t = 0 and to rotate in the x -y
plane. In (a), the electrons have coordinates:
x1 = a cos ωt
y1 = a sin ωt
x2 = −a cos ωt
y2 = −a sin ωt
14
Classical Electromagnetic Theory
The dipole moment is then easily seen to be zero while the quadrupole moment
has nonzero components:
Qxx
Qyx = Qxy = −6ea2 cos ωt sin ωt
"
"
#
#
= −2ea2 3 cos2 ωt − 1
Qyy = −2ea2 3 sin2 ωt − 1
and Qzz = 2ea2 . The counter-rotating electrons of (b) have coordinates
x1 = a cos ωt
y1 = a sin ωt
x2 = −a cos ωt
y2 = a sin ωt
The dipole moment is then p = −2eaĵ sin ωt while the quadrupole moment
has components (about the nucleus since the result is no longer unique) Qxy =
Qzx = Qyz = 0 while the diagonal components are the same as those for (a).
2-18 For simplicity we place one dipole at the origin so that the second is located
at r. The force on the dipole may be found from
E
F = (
p2 · ∇)
is the field at r due to the first dipole p1 . With the dipole field found
where E
in Example 2.1, and abbreviating ∂/∂x ≡ ∂x , etc.
−1
1
3(
p1 · r )r
F =
(p2x ∂x + p2y ∂y + p2z ∂z ) 3 p1 −
2
4πε0
r
r
3 (
p2 · r )
p1 + (
p1 · r )
p2 + (
p1 · p2 )r
p2 · r )r
5(
p1 · r )(
1
−
=
4πε0
r5
r7
The more general result of arbitrarily located dipoles is obtained by replacing
r by (r2 − r1 ).
2-19 The potential at r due to a quadrupole at the origin is (assuming summation
over repeated indices)
V (r ) =
1 xi xj Qij
4πε0 2r5
The potential energy of a dipole in this potential is
i j
= − 1 p ∂ x x Qij
W = −
p·E
4πε0
2r5
The force on the dipole may be found as minus the gradient of its potential
energy. The k component of the force F is then
i j
x x Qij
1 Fk =
p ∂ ∂ k
4πε0
2r5
Chapter Two Solutions
15
ik j
δ x Qij + δ jk xi Qij
1 5xk xi xj Qij
p ∂
−
4πε0
2r5
2r7
j k
x Qj + xi Qki
1 5xk xi xj Qij
=
p ∂
−
4πε0
2r5
2r7
k
Q +Qk
p
δk xi xj Qij +xk xj Qj + xk xi Qi 35x xk xi xj Qij
=
−5
+
4πε0
2r5
2r7
2r9
=
1
=
4πε0
p Qk
pk (xi xj Qij ) + 2xk (xj p Qj ) 35xk (p x )(xi xj Qij )
−
5
+
r5
2r7
2r9
Thus
F =
1
4πε0
↔
p · Q
p (xi xj Qij ) + 2r (xj x Qj )
r (
p · r )(xi xj Qij )
−
5
+
35
r5
2r7
2r6
2-20 The potential from the hypothetical monopoles would be
1
qm
1
−
Vm =
4π |r − 12 a| |r + 12 a|
−1/2 −1/2 a · r
a2
a · r
a2
qm
1− 2 + 2
− 1+ 2 + 2
=
4πr
r
4r
r
4r
qm
=
a · r + O (a/r)3
4πr3
Defining the magnetic moment as m
= qma, we find in the limit as a → 0
· r/4πr3 so that we reproduce the magnetic scalar potential of
that Vm → m
the dipole.
and
2-21 Generally, the equation of motion is τ = L/dt.
Substituting τ = m
×B
−1
L = am
with a = 2.79 e/me we write
= a dm
m
×B
dt
to be directed along the z axis so that the equation
For simplicity we take B
of motion, one component at a time, may be written
my Bz = a
dmx
dt
mx Bz = −a
dmy
dt
0=
dmz
dt
We differentiate the the first of these once more with respect to time and
substitute for dmy /dt using the second
d 2 mx
Bz2
Bz dmy
=
−
=
mx
dt2
a dt
a2
16
Classical Electromagnetic Theory
The solution for mx is easily found to be mx = m cos(ω0 t+δ) with ω0 = B/a =
2.79 eB/me . Reverting to the first order equations we have ω0 my = dmx /dt,
or my = −m sin(ω0 t + δ). The dipole moment has a constant z component
and the x and y components rotate about the z axis at angular frequency ω0 .
2-22 The potential can immediately be written as a summation of the individual
dipole fields:
D · (r − r ) 2 1
V (r ) =
d r
4πε0 S |r − r |
m·
= ∇(m
2-23 The force on the dipole takes the form F = −∇(−
B)
x Bx +my By +
mz Bz ) . Assuming that B has only a variation with z we have
∂Bx
∂By
∂Bz
F = mx
ı̂ + my
ĵ + mz
k̂
∂z
∂z
∂z
= αm cos θk̂
2-24 The residence time of the quadrupole in the field is 10−3 m/(100 m/s) = 10−5 s.
The force required to impart the required impulse is then 10−21 N. In general
the force may be obtained from the potential energy of the quadrupole:
∂2V
1
F = −∇W = − 6 ∇ Qm
∂x ∂xm
zz ∂ − ∂V = 1 ∇
Qzz ∂Ez
= 16 ∇Q
6
∂z
∂z
∂z
The k component of the force is then
Fk = 16 Qzz
∂ 2 Ez
∂z∂xk
leading to a requirement of (∂ 2Ez )/(∂z∂xk ) of order 6×1018 V/m3 . Assuming
this field is to be established at the tip of a charged needle, it is interesting
to consider the size of tip required. Let us assume the tip is charged to
V0 = 1000 V; the third derivative of the potential around a spherical tip just
becomes 6V0 /R3 . Equating this to the required inhomogeneity above leads
to R = 10−5 m. While a 10 μm radius tip is not unreasonable, it is clearly
impossible to maintain the required field gradient over distance the order of
a millimeter.
Chapter 3
3-1
From the definition of the magnetic flux and with the aid of Stokes’ theorem
(18),
· dS
= (∇
× A)
· dS
= A
· d
Φ= B
3-2
The current I circulating in the loop may be found from the induced EMF
(disregarding signs),
πa2 dB
E
=
I=
R
R dt
The resulting field at the center of the loop is
Bind =
μ0 πa2 I
μ0 πa dB
=
2π a3
2R dt
The direction of the induced field must be chosen, according to Lenz’ law to
oppose the increase or decrease of the background field. There is a difficulty
with this result: the induced field becomes arbitrarily large as the resistance
is decreased. The solution is of course that the induced field’s dB/dt must
be included in the flux change, when the resistance vanishes the the induced
current’s flux keeps the total flux in the loop constant.
3-3
To have a stable orbit the magnetic force on the electron must provide the
centripetal acceleration so that, disregarding signs,
me v 2
= qvB
Fcent =
r
⇒
|v| =
qrB(r )
me
The tangential electric field that accelerates the electron may be obtained
(again disregarding signs) from
r ) · d = −
E(
r)
dB(
· dS
dt
dB̄
r dB̄
⇒
Eϕ (r ) =
dt
2 dt
(where B̄ is the magnetic field averaged over the area included by the electron’s orbit) and substituted into Newton’s second law to obtain the acceleration:
qr dB̄
d|v|
= qE =
me
dt
2 dt
in other words,
qr dB̄
d|v|
=
dt
2me dt
2πrEϕ = πr2
Integrating both sides, we find |v| =
qrB̄
.
2me
— 17—
18
Classical Electromagnetic Theory
Comparing the two expressions we find that we require B̄ = 12 B(r). The
space average field inside the orbit must be half the field at the orbit.
3-4
The magnetic induction field in the torus with center line in the x-y plane
at distance x from the center, [ x ∈ (a − b, a + b)] is easily seen to be B =
μ0 N I/2πx. The flux Φ may be found by integrating this field over a crosssection in the x -z plane.
a+b √b2 −(x−a)2
μ0 N I
dzdx
Φ = B · dS =
√2
a−b
− b −(x−a)2 2πx
μ0 N I a+b 2 b2 − (x − a)2
=
dx
2π
x
a−b
√
√
#
#
"
μ0 N I "
=
π a − a2 − b2 = μ0 N I a − a2 − b2
π
The integral is far from trivial, however, it may be verified that as a b, the
flux reduces to that of a long solenoid.
3-5
The electric field between the plates is given in terms of the surface charge
= σ k̂ε0 , so that
density by E
×B
= μ0 ε0 ∂ E = μ0 k̂ dσ = μ0 k̂ I
∇
∂t
dt
A
We draw a circle of radius r between the plates, centered on the center of
symmetry and integrate both sides over the area of this circle. The left hand
side may be replaced by an integral along the boundary so that, assuming a
uniform electric field, we obtain
I dS
Bϕ (r) · d = μ0 k̂ ·
A
giving
3-6
μ0 πr2 I
A
⇒
Bϕ =
μ0 rI
2A
At large distances r the electric field included within the loop is that corresponding to the total charge on the plates. Repeating the reasoning of
problem 3-5,
· d = μ0 I
B
giving
3-7
2πrBϕ (r) =
Bϕ =
μ0 I
2πr
The power dissipated by the pendulum written in mechanical terms may be
equated to that written in electrical terms to give
ωτ = EI =
E2
R
Chapter Three Solutions
19
and the EMF generated is
E =A
dB⊥ dθ
dB⊥
dB⊥
=A
=
Aω
dt
dθ dt
dθ
so that the dissipative torque on the loop due to power generation is
τ=
E2
=
Rω
A2
dB⊥
dθ
R
2
ω
The moment of inertia of the loop about the point of suspension is m(2 +a2 ),
so the equation of motion becomes
" 2 #2 2
πa
dB⊥ dθ
d2 θ
+ mg sin θ = 0
m( + a ) 2 +
dt
R
dθ
dt
2
3-8
2
The charge delivered by the coil may be written as the time integral of the
current in the coil which itself may be calculated from the EMF induced in
the coil as the coil flips.
I=
E
1 d(BA)
BA d
=
=
(cos θ)
R
R dt
R dt
Identifying the time before the coil is flipped as t0 and the time after as tπ ,
we find the charge may be found as
tπ
tπ
Idt =
Q=
t0
t0
BA d
(cos θ)dt
R dt
π
BA
2BA
=
cos θ =
R
R
0
It is evident that the charge generated is independent of the speed of flipping
the coil.
3-9
Writing out the Lagrangian in terms of the cartesian components of position
and velocity we have
L = 12 m(ẋ2 + ẏ 2 + ż 2 ) − qV + q(ẋAx + ẏAy + żAz )
∂L
∂L
∂L
= mẋ + qAx ,
= mẏ + qAy and
= mż + qAz . The canonical
∂ ẋ
∂ ẏ
∂ ż
in agreement with (3–38)
momentum is then given by p + q A
Then
0 inde =E
0 ei(k·r−ωt) , with E
3-10 We take the electric field to have the form E
·E
is given by
pendent of the coordinates. Then ∇
20
Classical Electromagnetic Theory
·E
= ∂j E j = E j ∂j ei(k·r−ωt)
∇
0
"
#
= iE0j ei(k·r−ωt) ∂j ki xi − ωt
= iE0j δji ki = ikj E j = ik · E
×E
is given by
In similar fashion, the j component of ∇
× E)
j = jk ∂k E = ijk E ∂k (km rm − ωt)
(∇
= ijk δkm km E
j
= ijk kk E = i(k × E)
we take the curl of
3-11 To obtain a wave equation for B
×B
= μ0 J + μ0 ε0 ∂ E
∇
∂t
× (∇
× B)
= ∇(
∇
· B)
− ∇2 B to get
and use (13), ∇
∇
· B)
− ∇2 B
= μ0 ∇
× J + μ0 ε0 ∂ (∇
× E)
∇(
∂t
is replaced by −∂ B/∂t,
·B
and J both vanish to leave
The curl of E
and ∇
= −μ0 ε0
−∇2 B
∂2B
∂t2
3-12 To obtain a reasonable ionization rate, we need eV = eE/d. The electric field
must therefore be at least
E=
10 V
= 1010 V/m
10−9 m
The corresponding irradiance I is found from
×B
E
I = μ0
|E|2
=
= 2.65 × 1017 W/m2
μ0 c
A (modest) 10 MW pulse, focussed to a (10 μm)2 spot would provide I =
1017 W/m2 so that we see that air sparks should be easily produced.
and ξ = r × ∇ψ
solve the vector wave equation
3-13 We wish to show that ξ = ∇ψ
∇2 ξ =
1 ∂ 2 ξ
c2 ∂t2
Chapter Three Solutions
21
whenever ψ solves the scalar wave equation. We recall the definition of the
vector Laplacian (13),
−∇
∇
· ξ)
× (∇
× ξ)
∇2 ξ = ∇(
and substitute each of the forms ξ and ξ in turn. Thus for ξ = ∇ψ
× [∇
× (∇ψ)]
∇
· ∇ψ)
2 ψ)
∇2 (∇ψ)
= −∇
+ ∇(
= ∇(∇
since the gradient of any function is curl free. Substituting for ∇2 ψ from the
scalar wave equation, we find
2
2
1 ∂ ψ = 1 ∂ (∇ψ)
=∇
∇2 (∇ψ)
2
2
2
c ∂t
c ∂t2
which proves the required result for ∇ψ.
The equivalent result for ψ is a
little more laborious.
∇
· (r × ∇ψ)]
× [∇
× (r × ∇ψ)]
∇2 (r × ∇ψ)
= ∇[
−∇
∇
· [−∇
× (rψ)]} − ∇
× [∇
× (r × ∇ψ)]
= ∇{
The first term on the right side vanishes because the curl of any function
has no divergence. In the remaining term, we expand the expression within
square brackets
× (r × ∇ψ)
· ∇ψ)
· r ) + (∇ψ
· ∇)
r − (r · ∇)
∇ψ
∇
= r (∇
− (∇ψ)(
∇
+ ∇ψ
− (r · ∇)
∇ψ
= r ∇2 ψ − 3∇ψ
· ∇)
r − ∇ψ
The last term is most easily evaluated by adding the null term (∇ψ
to it. Thus
∇ψ+
· ∇)
r − ∇ψ
(r · ∇)
(∇ψ
r · ∇ψ)
× ∇ψ)
× (∇
× r) − ∇ψ
= ∇(
− r × (∇
−∇
have zero curl, the two middle terms vanish. FurtherBecause both r and ∇ψ
and ∇(
r · ∇ψ)
more, taking the curl once more (as we must) eliminates ∇ψ
as well so that we find
2
× [∇
× (r × ∇ψ)]
× (r ∇2 ψ) = ∇
× r 1 ∂ ψ
∇
=∇
c2 ∂t2
2 2 1 ∂
× rψ = − 1 ∂
r
×
∇ψ
= 2 2 ∇
c ∂t
c2 ∂t2
Finally,
∇2 (r × ∇ψ)
=
1 ∂2
(r × ∇ψ)
c2 ∂t2
22
Classical Electromagnetic Theory
The relative difficulty of the foregoing is largely circumvented if tensor
notation and the Levi-Cevita symbol are used instead, as will be demonstrated
below.
"
#
k
= ∂i ∂ i km x ∂m ψ
[∇2 (r × ∇ψ)]
= km ∂i ∂ i (x ∂m ψ)
= km (∂i δi ∂m ψ + x ∂i ∂ i ∂m ψ)
= km ∂ (∂m ψ) + km x ∂m (∂i ∂ i ψ)
k
× (∇ψ)]
+ km x ∂m (∇2 ψ)
= [∇
k
1 ∂2ψ
2
k
= [r × ∇(∇ ψ)] = r × ∇ 2 2
c ∂t
1 ∂2
k
= 2 2 (r × ∇ψ)
c ∂t
where the first term in the sum was eliminated because a gradient has no curl.
r + δr, t) may be expanded to first order as
3-14 Generally, the function A(
r + δr, t) = A(
r, t) + ∂ A(r, t) · δx + ∂ A(r, t) · δy + ∂ A(r, t) · δz
A(
∂x
∂y
∂z
Substituting δx = vx dt, δy = vy dt, and δz = vz dt, we find
r + v dt, t) = A(
r, t) + (v · ∇)
Adt
A(
·A
= 0. Replacing A
by A
= A
+ ∇Λ,
3-15 The Coulomb gauge poses ∇
we find
·A
= ∇
·A
+ ∇2 Λ
∇
If ∇2 Λ = 0, the Coulomb gauge is clearly preserved.
and V satisfy
3-16 In the Lorenz gauge, A
·A
+ 1 ∂V = 0
∇
c2 ∂t
by A
= A
+ ∇Λ
and V by V = V − ∂Λ/dt, the gauge condition
Replacing A
becomes
2
·A
+ ∇2 Λ + 1 ∂V − 1 ∂ Λ
·A
+ 1 ∂V = ∇
∇
c2 ∂t
c2 ∂t
c2 ∂t2
It is evident that in order to preserve the Lorentz gauge, the gauge function
Λ must satisfy
1 ∂2Λ
∇2 Λ − 2 2 = 0
c ∂t
Chapter Three Solutions
23
Neglecting the vector
3-17 Electrons in the spinning disk experience a force ev × B.
character and writing only the magnitude this is rωB. Integrating the force
per unit charge from r = 0 to r = a on the rim, we obtain the EMF
a
r2 ωB
rωBdr =
E=
2
0
3-18 The “paradox” arises because the current in the loop has been neglected in
determining the potential measured by the voltmeter. Suppose, for simplicity
that the loop has the same resistance R in the bottom half and in the top
half of the loop. The current of magnitude
I=
1 dΦ
2R dt
flows in a clockwise direction around the loop in response to the changing
magnetic field indicated. The potential measured in circuit b is then IR =
1
1
2 dΦ/dt and the potential measured in c is dΦ/dt − IR = 2 dΦ/dt. This
argument works equally well if we define unequal resistances for the top and
bottom half of the loop.
Chapter 4
A number of readers have complained that the integral in (Ex 4.4.12) is not
obvious. Admittedly I used tables to to find
π
2π ln a a ≥ b ≥ 0
2
2
ln(a − 2ab cos θ + b )dθ =
2π ln b b ≥ a ≥ 0
0
Indeed, there seems to be no elementary method for obtaining this result.
However, computing the potential of a charged cylinder of radius a via two
different methods closely reproduces the required result. We begin by Using Gauss’ law to find the electric field and consequent potential around the
cylinder. Assuming charge density λ per unit length,
r
−λ
−λ
= λr̂
⇒ V (r > a) =
ln r + A =
ln
E
2πε0 r
2πε0
2πε0
b
where we have imposed that V vanish at a distant point b. Alternatively,
we consider each the cylindrical shell to be composed of filaments running
parallel to the axis, each subtending angle dϕ with respect to the axis and
carrying line charge density (λ/2π)dϕ. Summing the potentials from each of
these filaments
&
−λdϕ 1
a2 + r2 − 2ar cos ϕ
|r − r |
−λdϕ 1
=
ln
ln
dV (r ) =
2π 2πε0
b
2π 2πε0
b2
which may be integrated to give
−λ 1
V (r ) =
2πε0 2π
&
2π
ln
0
a2 + r2 − 2ar cos ϕ dϕ
b2
Comparing the two expressions for V (r > a),
2π & 2
r
a + r2 − 2ar cos ϕ ln
dϕ
=
2π
ln
2
b
b
0
For a not so distant point b, the correct form to integrate would be
dV (r ) =
−λdϕ
|r − r |
ln
2
(2π) ε0 |b − r |
leading to
−λ 1
V (r ) =
2π0 2π
2π
ln
a2 + r2 − 2ar cos ϕ dϕ
0
−
2π
ln
0
— 24—
a2 + b2 − 2ab cos ϕ dϕ
Chapter Four Solutions
25
Denoting
the first
integral as F (r), the second is F (b) and we have ln(r/b) =
1
F
(r)
−
F
(b)
allowing us to conclude that except for a constant multiplier,
2π
F (r) = 2π ln r. Coupled with the foregoing discussion that multiplier should
be 1.
4-1 The potential energy of all the charges is given by
W =
8
8
1 ! ! qi qj
8πε0 i=1 j=1 ri,j
j=i
The inner sum is easily evaluated with the aid of√
a diagram. Any charge
has
√
three neighbors at distance a, three at distance 2a, and one at 3a. The
inner sum is therefore
8
!
qi qj
j=1
j=i
ri,j
1
q2
3
=
3+ √ + √
a
2
3
Every other charge contributes exactly the same to the potential energy sum,
therefore
1
8q 2
3
5.7 q 2
W =
3+ √ + √
=
8πε0 a
πε0 a
2
3
4-2 The inductance of the solenoid may be found by equating the energy of the
enclosed field to 12 LI 2 or by differentiating the flux at any turn with respect to I and summing over the turns. Since the magnetic induction field is
nearly constant through the volume of the solenoid and nearly zero outside
the solenoid, either method ought to work. For either method we need the
field of a solenoid of length and N turns: B = μ0 N I.
Using energy:
B2 3
d r
2μ0
W = 12 LI 2 =
=
μ0 N I
2
from which we deduce
L=
πR2 πμ0 I 2 N 2 R2
=
2μ0
2
πμ0 N 2 R2
From the flux:
∂Φ
∂
L=N
=N
∂I
∂I
μ0 πR2 N 2
μ0 N I
dS =
4-3 For the centerline of the torus in the x-y plane, the magnetic induction field
over a cross-section in the x-z plane is given by (see also problem 3-4, note
26
Classical Electromagnetic Theory
however, that a and b are reversed) By = μ0 N I/(2πx). A volume element in
the neighborhood of the x-z plane is d3 r = xϕdS so that we may write
2 2 2
B2 3
μ0 N I
W =
d r=
xdϕdS
2μ0
8μ0 π 2 x2
√a2 −(x−b)2
μ0 N 2 I 2 b+a
1
=
dzdx
√ 2
4π
x
2
− a −(x−b)
b−a
Exactly the same integral was evaluated in problem 3-4 to give
μ0 N 2 I 2 W =
b − b2 − a2
2
leading us to deduce that the inductance of the torus is
L = μ0 N 2 b − b2 − a2
We could of course have differentiated the flux Φ found in problem 3-4 with
respect to the current to obtain the same result.
4-4 In order to have the same cross-section, the outer conductor must have an
outer radius c that satisfies π(c2 − b2 ) = πa2 implying c2 = a2 + b2 . The
field in the various regions is easily obtained; in particular, because a loop
encircling both inner and outer conductor has zero net current included, the
field outside the wire must vanish. This fact makes it practical to calculate
the total energy content of the fields established by the currents.
The field inside the inner wire is obtained from
2
= μ0 r I ⇒ B(r
< a) = μ0 Ir
B · d = μ0 J · dS
a2
2πa2
ϕ = μ0 I ⇒ B = μ0 I . Finally, in the outer conductor
Between the wires, 2πrB
2πr
r
r
= μ0 I 1 − 1
J · dS
2πr dr
2πrBϕ = μ0 I −
2
πa
b
b
r 2 − b2
= μ0 I 1 −
a2
The energy for a unit length of the field is then
$
b
a 3
μ0 I 2
r
1
W =
dr
dr +
4
4π
0 a
a r
√
2
a2 +b2 1
2r2
b2
+
1+ 2 − 2 1+
r
a
a
b
&
2
b
b2
μ0 I 2 1
a2
+ ln + 1 + 2 ln 1 + 2
=
4π 4
a
a
b
2
2
2
(a + b ) − b
b2
−
1
+
+
a2
a2
b2
a2
"
%
r4
+ 2 dr
a
#2
a2 + b2 − b4
4a4
Chapter Four Solutions
27
2 1
μ0 I 2 1
a2
b
b2
b2
=
− 1 − 2 + ln +
1 + 2 ln 1 + 2
4π 2
a
a 2
a
b
The inductance is now easily deduced to be
2 μ0
b2
a2
b2
b2
L=
ln 2 + 1 + 2 ln 1 + 2 − 1 + 2
4π
a
a
b
a
4-5 We place one of the wires along the z axis and the other parallel to the first
in the x-z pane at distance h from the first. The magnetic induction field in
the x-z plane between the wires of the first wire is then,
r ) = μ0 I ĵ
B(
2πx
and that from the second wire is
r) =
B(
μ0 I ĵ
2π(h − x)
The flux through a loop of width (h − 2a) between the wires is
1
1
1
μ0 I
μ0 I h−a 1
+
+
dx dz =
dx
2π
x h−x
2π a
x h−x
h−a
μ0 I
ln
=
π
a
The inductance may be found as dΦ/dI
h
μ0 dΦ
=
ln
−1
dI
π
a
When h a, this result is similar to (Ex 4.4.13).
4-6 The electric field inside the uniformly charged sphere is obtained from Gauss’
law
ρ 3
· dS
=
E
d r
ε0
4πr2 E =
or
Qr3
ε0 a3
Outside the sphere, the field is E =
ε0
W =
2
⇒
E=
Qr
4πε0 a3
Q
. The energy is then
4πε0 r2
E 2 d3 r
Q2 r2
Q2
ε0 ∞
2
4πr
dr
+
4πr2 dr
2 6
2 a (4πε0 )2 r4
0 (4πε0 ) a
a 4
∞
r
1
3Q2
Q2
dr
+
dr
=
=
6
2
8πε0
r
20πε0 a
0 a
a
=
ε0
2
a
28
Classical Electromagnetic Theory
4-7 Using Gauss’ law to find the field inside and outside the charge
1
E(r < a) =
4πε0 r2
r
ρ0
0
E(r ≥ a) =
and
r3
ρ0 r
r 2
2 − 2
1 − 2 4πr dr =
a
ε0 3 5a
Q
4πε0 r2
with
Q = ρ0
8πa3
15
The energy is then
2
a 2 2
Q2 4πr2 dr
ρ0 r 1
r2
2
−
+
4πr dr
(4πε0 )2 r4
ε20
3 5a2
a
0
Q2
2πρ20 a5
2a7
a9
=
+
−
+
8πε0 a
ε0
45 105a2
225a4
ε0
W =
2
=
∞
16πρ20 a5
5Q2
=
315ε0
28πε0 a
4-8 The magnitude of the Poynting vector S is
S=
E2
E2
=
−7
2μ0 c
2 × 4π × 10 × 3 × 108 m/s
If we set S = 5W/.5 × 10−6 m2 = 107 W/m2 , we find E = 8.68 × 104 V/m.
is parallel to z and B
at sufficiently large distance is azimuthal, A
×B
4-9 Since A
is radial and the surface integral over the end faces of our cylinder make no
at large r is the difference between the
contribution. The vector potential A
exterior field terms of the two wires, ie. Az (r h) = μ0 I/(2π)(ln r2 /a −
ln r1 /a). We rewrite this as
2
r1 + h2 − 2hr1 cos ϕ
h2
μ0 I
2h cos ϕ
μ0 I
Az (r > h) =
ln
ln 1 + 2 −
=
4π
r12
4π
r
r
When r h we may approximate the logarithm using ln(1 + ) ≈ to write
the vector potential as
μ0 I h2
2h
cos ϕ
−
Az (r h) ≈
4π r2
r
The magnetic induction field of two opposing currents may be written as
1
μ0 I 1
Bϕ =
−
2π r
r2 + h2 − 2rh cos ϕ
μ0 I h2
h
μ0 I
1
≈
− cos ϕ
=
1− 2πr
2πr 2r2
r
1 + (h2 /r2 ) − 2(h/r) cos ϕ
Chapter Four Solutions
29
in other words it vanishes at least as fast as r−2 . The surface integral of
(4–43) then becomes
μ0 I 2
1 C
(A × B) · dS < lim
2πrdrdz → 0
r→∞ (4π)2
2μ0 r3
4-10 As a first approximation, we might be tempted to use the central field B =
μ0 I/ 2a of the coil as an estimate of the average field through the second coil.
The flux would then be
πaμ0 I
Φ=
2
1
and M12 = dΦ2 /dI1 = 2 μ0 πa. Unfortunately there is no good reason to
assume that the magnetic induction field is uniform across the second loop.
In particular it seems likely that that the field near the edges, close to the
primary loop might well be much larger. Indeed this is so. We now pursue a
more rigorous approach to an approximate
solution. We begin by writing the
r ) · d2 , where the integration is carried
flux threading loop 2 as Φ2 = A(
by its integral form, we find
out along loop 2. Replacing A
d1
μ0 I
r on loop 1
Φ2 =
d2 ·
r on loop 2
4π loop 2
r − r |
loop 1 |
The mutual inductance M12 is then
μ0
d2
M12 =
d1 ·
4π
b2 + (2a sin 1 ϕ)2
2
where ϕ is the projection of the angle between r and r on the plane of one
of the coils as shown in figure 4.1. Replacing the scalar product of the unit
tangent vectors d1 · d2 by cos ϕd1 d2 we get
μ0
cos ϕd2
M12 =
d1
4π
b2 + (2a sin 12 ϕ)2
Figure 1.2: The current loops of problem 4-8
30
Classical Electromagnetic Theory
For any fixed point r , we compute the second integral as follows.
Except when r and r are very close to each other, say within 2a sin 12 ϕ0 =
±Δ b, we make no great mistake neglecting b. For the region inside Δ, we
ignore the curvature of the loops and set cos ϕ = 1. Thus
ϕ0
2π−ϕ0
cos ϕd1
d
μ0
1
d2
M12 =
+
1
4π loop 2
|2a
sin
ϕ|
b2 + 21
ϕ0
−ϕ0
2
μ0 2πa
=
4π
2
π
ϕ0
cos ϕd1
+
|2a sin 12 ϕ|
Δ
−Δ
d1
b2 + 21
where Δ ≈ aϕ0 .
We will now recast the second integral in parentheses in a form that will
allow us to rewrite it as an integral with the same form as the first, but
eliminating the arbitrary Δ (or ϕ0 ). We integrate the second integral to get
Δ
−Δ
d1
b2 +
21
≈ 2 sinh−1
Δ
2Δ
≈ 2 ln
=2
b
b
Δ
1
2b
dx
≈2
x
ϕ0
1
2 b/a
adϕ
2a sin 12 ϕ
The second integral may now be included within the first by merely changing
the lower limit of the first
π
a cos ϕdϕ
M12 = μ0 a
1
2a sin 12 ϕ
2 b/a
'
(
= μ0 a − ln tan[ 14 (b/2a)] − 2 cos[ 12 (b/2a)]
b
8a
−2
= μ0 a − ln
− 2 = μ0 a ln
8a
b
The self-inductance of a single loop is found in very similar manner with the
radius of the wire replacing b. There will, however, be another term (+ 14 μ0 a),
the contribution of the local current as opposed to that of the remainder of
the loop.
4-11 Using the magnetic field portion of the Maxwell stress tensor (in Cartesian
coordinates) we find the x and y components of force on an element dSr of
the curved wall
dFx =
B2
dSx
2μ0
and
dFy =
B2
dSy
2μ0
leading to
dFr = dFx cos ϕ + dFy sin ϕ
=
B2
B2
(dSx cos ϕ + dSy sin ϕ) =
dSr
2μ0
2μ0
Chapter Four Solutions
31
The pressure ℘ on the coil is the found as
℘=
dFr
B2
μ0 N 2 I 2
=
=
dSr
2μ0
2L2
This pressure must be balanced by the inward resultant of the tension on the
wires. There are N/L wires in a unit width sharing the outward directed force
from the pressure on those wires. A simple diagram shows that the inward
force of a wire under tension T is T/R where R is the radius of curvature.
Therefore N/L such wires produce an inward force NT/LR per unit area.
Equating this to the pressure, we find
T =
μ0 RN I 2
2L
Alternatively, the average field at the wire is μ0 N I 2 /2L, giving an outward
force μ0 N I 2 /2L on the wire. The tensile force required to sustain this force
is given by equating it to T/R.
T
μ0 N I 2
=
R
2L
⇒
T =
μ0 N I 2 R
2L
Although this seems an easier approach, it is by no means self evident that
one should use half the field in the solenoid as the field at the wire.
4-12 The magnetic force between two hypothetical magnetic monopoles qm , is
Fm =
2
μ0 qm
μ0
=
4π r2
4πr2
2πε0 c2 h̄
e
2
=
πh̄2
μ0 e2 r2
The force between two electric charges e separated by the same distance is
e2
4πε0 r2
Fe
μ0 e4
The ratio of the two forces is
=
= 2.13 × 10−4 .
Fm
4π 2 ε0 h̄2
4-13 We transform each equation in turn:
Fe =
·E
= ∇
·E
cos θ + c∇
·B
sin θ
∇
ρe cos θ
=
+ cμ0 ρm sin θ
ε0
# ε0 μ0 c
ρm
ρ
1" sin θ +
ρm −
ρm sin θ = e
=
ε0
c
ε0
ε0
·B
= ∇
·B
cos θ − ∇ · E sin θ
∇
c
ρe
sin θ
= μ0 ρm cos θ −
ε0 c
ρe
sin θ = μ0 ρm
= μ0 (ρm + cρe sin θ) −
ε0 c
32
Classical Electromagnetic Theory
× E)
cos θ + c(∇
× B)
sin θ
×E
= (∇
∇
1 ∂E
∂B
cos θ +
sin θ − μ0 Jm cos θ + μ0 cJe sin θ
=−
∂t
c ∂t
∂B
− μ0 Jm
=−
∂t
E
×B
cos θ
∇ × B = ∇ × − sin θ + ∇
c
1 ∂B
μ0 1 ∂E
+
Jm sin θ + 2
cos θ + μ0 Je cos θ
=
c ∂t
c
c ∂t
#
1 ∂ "
sin θ + μ0 Jm sin θ + μ0 Je cos θ
E cos θ + cB
2
c ∂t
c
1 ∂E
+ μ0 Je
= 2
c ∂t
=
4-14 The capacitance of each sphere is
C=
Q
Q
=
= 4πε0 R
V
Q/4πε0 R
(= 111pF)
The approximate mutual capacitance of the pair of spheres, using the results
from example 4.2 is
C12 (4πε0 R)2
4πε0 × 1 m2
−C1 C2
=−
=−
= −11.1 pF
4πε0 r
4πε0 r
10 m
4-15 The charge on the concentric cylinders must lie entirely in the region where
the two cylinders overlap (and the fringing region), because any charge on an
exposed long cylinder leads to an infinite (logarithmically divergent) potential
on the cylinder. The electric field between the cylinders is easily found to be
E=
V
r ln(b/a)
whence the energy of a length z of concentric cylinders is
z0 +z b
ε0
V2
1
πε0 V 2
W =
z
2πrdrdz =
2
2
2 [ln(b/a)] z0
[ln(b/a)]
a r
The energy of the field in the fringing region is irrelevant since it will not
change as one cylinder moves with respect to the other. The force pulling one
cylinder into the other is
∂W
πε0 V 2
=
∂z
ln(b/a)
Chapter Four Solutions
33
4-16 The electric field between the overlapping plates is E = 2V/d and when the
plates don’t overlap the field vanishes. The energy of the 14 gaps between the
plates overlapping by angle θ is
4V 2 3
14ε0 V 2 R2 θ
ε0
d
r
=
W = 14 ×
2
d2
d
The torque is given by
14ε0 V 2 R2
∂W
=
∂θ
d
Substitution of reasonable values for the radius and the spacing d, say 10
cm and 1 mm, respectively, gives a torque τ = 1.24 × 10−9 V 2 Nm/V2 =
1.24 V 2 × 10−2 dyne·cm/V2 . It would appear to be entirely feasible to build
an electrostatic voltmeter on this principle.
4-17 The flux through the loop is Φ = I0 e−iωt /(2μ0 r)dA giving rise to an EMF
−iωI0 e−iωt /(2μ0 ) ln(b/d)Δz. The potential on the plate (working into an infinite impedance) is V0 e−iωt d/ ln(b/a) where d is the distance of the plate from
the central conductor and a and b are respectively the inner and outer radii
of the coaxial wire. The impedance (1/iωC) of the source (plate) decreases
linearly with the area of the plate. If the power meter impedance is small
compared to 1/ωC, we conclude that the inductive and capacitive current
each increase linearly with ω resulting in a torque proportional to ω 2 .
4-18 The electrostatic energy of the charged bubble is
W =
ε0
2
∞
R
Q
4πε0 r2
2
4πr2 dr =
∞
Q2 1 Q2
=
8πε0 −r R
8πε0 R
The outward pressure from the electrical repulsion is then
℘=
dW dR
Q2
dW
=
=
dτ
dR dτ
32πε0 R4
Equating this to the pressure 4T/R (there is both an inside and an outside
surface) required to balance surface tension gives
R=
Q2
128π 2 ε0 T
1/3
If the radius were to increase beyond the equilibrium point, the electrical (expansive) pressure decreases as R−4 whereas the (contractile) surface-tension
derived force only decreases as R−1 and if the radius were to decrease, the
electrical force would grow faster. The bubble would tend to return to its
equilibrium radius in either case.
34
Classical Electromagnetic Theory
The center term of (4–6) has an erroneous subscript. The expression
should read:
W4 =
1
8πε0
q1 q4 + q4 q1
q2 q4 + q4 q2
q3 q4 + q4 q3
+
+
r14
r24
r34
(4–6)
Chapter 5
5-1 Although there are elegant ways to prove this we use the brute force approach
as it requires little thought. The general expansion for the potential in spherical polar coordinates is
!
B
V (r ) =
A r + +1 Ym (θ, ϕ)
r
Each of the constants B must vanish since otherwise the potential would
be infinite at the origin. To evaluate the constants A we multiply both
(θ, ϕ) and integrate over the surface of
sides of the expression above by Y∗m
the enclosing sphere. Using the orthogonality of the spherical harmonics we
obtain
A a = V (R)Y∗m (θ, ϕ)dΩ
At the center of the sphere, r = 0, only the first term does not vanish and we
need evaluate only this term. Thus
V (R, θ, ϕ)
1
V 4π
√
dΩ = √
A0 =
V dΩ = √
4π
4π
4π
4π
4π
The potential at r = 0 then becomes
V (0) = A0 r0 Y00 (θ, ϕ) =
5-2
V 4π
= V 4π
We use plane polar coordinates to describe the geometry. Because the boundary conditions do not depend on r we expect V to be independent of r.
Laplace’s equation then becomes
1 ∂2V
=0
r2 ∂ϕ2
The solution is easily found to have the form V = Aϕ + B. Evaluating A and
B from the boundary conditions we obtain
V (ϕ) =
V0 ϕ
α
5-3 The potential between the cones is most easily solved for in spherical polar
coordinates since the boundary conditions are independent of r and ϕ. It
remains to solve
d
1
dV
sin θ
=0
dθ
r2 sin2 θ dθ
Excluding r = 0, we have
sin θ
dV
=a
dθ
and integrating once more,
— 35—
36
Classical Electromagnetic Theory
V = a ln(tan 12 θ) + b
Fitting the boundary conditions at θ = 12 α1 and θ = 12 α2 we find
a=
V2 − V1
tan( 14 α2 )
ln
tan( 14 α1 )
b = V1 − a ln tan( 14 α1 )
and
Substituting these constants into the solution we find
tan( 12 θ)
ln
tan( 14 α1 )
V = V1 + (V2 − V1 ) tan( 14 α2 )
ln
tan( 14 α1 )
5-4 The general solution of Laplace’s equation compatible with the boundary conditions is
!
V =
Aλ sin λy sinh λx
λ
=
!
An sin
nπx
nπy
sinh
b
b
An sinh
nπy
nπa
sin
= V0
b
b
n
At x = a,
V (a, y) =
!
then
An sinh
nπa b
· =
b
2
b
V0 sin
0
=−
nπy
dy
b
b
(cos nπ − cos 0)
nπ
⎧
⎨ 2V0 b
nπ
=
⎩
0
for n odd
for n even
We evaluate the first four nonzero coefficients
A1 =
A3 =
A5 =
4V0
π sinh
πa = 0.05019V0
b
4V0
3πa
3π sinh
b
4V0
5πa
5π sinh
b
= 6.492 × 10−6 V0
= 1.51 × 10−9 V0
Chapter Five Solutions
37
A7 =
4V0
7πa
7π sinh
b
= 4.2 × 10−13 V0
We therefore write the potential to the required precision as
πx
3πx
πy
3πy
sinh
+ 6.5 × 10−6 V0 sin
sinh
b
b
b
b
5πx
7πx
5πy
7πy
+1.51 × 10−9 V0 sin
sinh
+ 4.2 × 10−13 V0 sin
sinh
b
b
b
b
V = 0.0502V0 sin
which may be numerically evaluated at the required points to give
V (a/10, b/2) = 0.0502 sinh 18 π − 6.5 × 10−6 sinh 38 π = .0202V0
V (a/2, b/2) = .0502 sinh 58 π − 6.5 × 10−6 sinh 15
8 π = 0.1753V0
and
V (9a/10, b/2) = 0.0502 sinh 98 π − 6.5 × 10−6 sinh 27
8 π
−13
sinh 63
+ 1.51 × 10−6 sinh 45
8 π − 4.2 × 10
8 π = .758V0
5-5 In cylindrical polars, the potential in the region including r = 0 is
V (r, ϕ) =
!
A
r
sin ϕ
a
At r = a this becomes
V (a) =
!
A sin ϕ = ±
A =
leading to
V0
2
2V0
π
when is odd and A = 0 when is even. Therefore
V (r < a) =
! 2V0 r sin ϕ
π a
odd
Outside the pipe, the general expansion is
a
V (r > a) = B
sin ϕ
r
The boundary conditions lead to the same coefficients so that
V (r > a) =
! 2V0 a sin ϕ
π r
odd
38
Classical Electromagnetic Theory
5-6 The general solution in spherical polar coordinates may be written
+1 !
r
a
+ B
P (cos θ)
V =
A
a
r
For the interior solutions all the B coefficients must vanish while the outside
solutions must have A = 0. In either case, at the surface of the sphere, the
coefficients must satisfy
! A P (cos θ) = ±V0
B
leading to
12 π
1
2A
= 2V0
P (cos θ) sin θdθ = 2V0
P (x)dx
2 + 1
0
0
for odd, and 0 when is even and an identical expression for B . The
integral may be evaluated using (F–30), (2 + 1)P (x) = P+1 (x) − P−1 (x).
1
(−1)(+3)/2 ( − 2)!!
1 P−1 (0) − P+1 (0) =
P (x)dx =
2 + 1
( + 1)!!
0
The coefficients A for the interior solution and B for the exterior solution
follow immediately.
5-7 In Example 1.16 we found the scalar potential along the axis of a solenoid to
be
N I (z + )2 + R2 − (z − )2 + R2
Vm (z) =
4
where = L/2 is the half length of the solenoid. We factor (R2 + 2 )1/2 from
the radicals
1/2
z 2 ± 2z
2
2
2
2
2
R + + z ± 2z = R + 1 + 2
R + 2
and expand the term in parentheses using the binomial theorem
1/2
1 z 2 ± 2z
1 (z 2 ± 2z)2
1 (z 2 ± 2z)3
z 2 ± 2
=
1
+
−
+
− ···
1+ 2
R + 2
2 R2 + 2
8 (R2 + 2 )2
16 (R2 + 2 )3
yielding
Vm
√
N I R2 + 2
z 3
3 z 3
2z
=
+ 2
− 2
+ ···
− 2
4
R + 2
(R + 2 )2
(R + 2 )3
NI
2
1
3
= √
− 2
−2z + + 2
z + ···
R + 2
(R + 2 )2
4 R2 + 2
The general form of the scalar potential in a region that includes the origin
is (in spherical polars)
!
Vm (r, θ) =
Al rl Pl (cos θ)
Chapter Five Solutions
39
which for cos θ = 0 specializes to
Vm (z) =
!
Al z l
along the z axis. Comparing the coefficients of the two series we find
NI
,
A1 = − √
2 R2 + 2
A2 = 0,
A3 =
N IR2
4(R2 + 2 )5/2
so that the general solution in the neighborhood of r = 0 is
Vm = A1 r cos θ + A3 r3 P3 (cos θ) + · · ·
3
z
z
z
1
−3
= A1 r
+ A3 r3 5
+ ···
r
2
r
r
= A1 z + 52 A3 z 3 − 32 A3 r2 z + · · ·
We may usefully write this in terms of the cylindrical coordinates ρ and z so
that r2 = ρ2 + z 2 and
V (ρ, z) = A1 z + A3 z 3 − 32 A3 ρ2 z + · · ·
The resulting second order axial and first order radial magnetic field components are now easily calculated.
5-8 We express f in terms of its real and imaginary part u and v as
2
a + x + iy
a − x2 − y 2 + 2iay
u + iv = ln
= ln
a − x − iy
a2 − 2ax + x2 + y 2
= ln Reiα = ln R + iα
which allows us to identify v as α. But we can also find α from
tan α =
2ay
Im(f )
= 2
Re(f )
a − x2 − y 2
On the circle x2 + y 2 = a2− we find when y > 0 that
tan α = +∞
⇒
v=α=
π
2
and when y < 0
tan α = −∞
⇒
v=α=−
π
2
In other words, the the upper half circle x2 + y 2 = a2− is mapped from the
v = π/2 line while the lower half of the interior circle is mapped from the v
= −π/2 line. It may be verified using l’Hôpital’s rule that as x varies from
−a to a, u varies from −∞ to ∞.
40
Classical Electromagnetic Theory
The potential V = V0 v/π between the v = ±π/2 lines becomes in terms
of x and y
2ay
V0
tan−1
V =
π
a2 − x2 − y 2
At y = 0.2a and x = 0, this becomes
0.4
2V0
1
2V0
−1
V =
× tan
× (0.1973956)
=
π
2
0.96
π
Using the series solution from problem 5-5 to find the potential at this point,
we have
! 2V0 r sin ϕ
V =
π a
odd
2V0 1(.2) − 13 (.2)3 + 15 (.2)5 − 17 (.2)7 + 19 (.2)9 + · · ·
π
2V0
× (0.1973956)
=
π
=
5-9 The required mapping must change the polar angle of f from π to α while
leaving a polar angle of zero unaffected. The mapping
#α/π
"
= Rα/π eiθα/π
z = f α/π = Reiθ
clearly has this property. When v = 0, R = u and
z = x + iy = Rα/π
⇒
0 ≤ x < ∞ and y = 0
When v = π, R = |u|, then
z = x + iy = Rα/π eiα
= Rα/π (cos α + i sin α)
Thus, as R increases, a line with x = Rα/π cosα and y = Rα/π sin α is
swept out. Despite the fact that the mapping is obvious it is a good exercise
to use the Schwarz Christoffel transformations to derive it. Thus
dz
= A(f − 0)α/π−1
df
which is readily integrated to yield
z=
Af α/π
+k
α/π
The denominator may be absorbed into the constant A and the choice of
k = 0 fixes the inflection point to the origin.
Chapter Five Solutions
41
5-10 As found on page 118, the potential in the vicinity of the right-angled plate
has the form V = V0 − 2axy. (It is trivial to verify that this satisfies ∇2 V = 0
and satisfies the boundary conditions.) The electric field is then
= −∇V
= 2ayı̂ + 2axĵ
E
= 2axĵ. We equate the normal
Along the x axis (y = 0) this becomes E
component of the electric field to σ/ε0 to obtain σ(x, 0) = 2aε0 x. Similarly,
along the y axis σ(0, y) = 2aε0 y.
5-11 When v = 0, (f = u + iv), cosh f = 12 (eu + e−u ). Therefore, as u varies from
−∞ to +∞, cosh f varies from ∞ to 1 to ∞, meaning that z ∝ ln(cosh f )
ranges from ∞ to 0 to ∞ and is pure real. In other words, the image of the
u axis is the positive x axis.
When v = − 12 π,
eu+iπ/2 + e−u−iπ/2
2
ieu − ie−u
= i sinh u = (sinh u)eiπ/2
=
2
cosh f =
Then x + iy = (2a/π)[ln(sinh u) + 12 iπ]. We conclude then that when v =
π/2, y = a and as u varies from 0 to ∞, sinh u varies from 0 to ∞ and x =
(2a/π) ln(sinh u) varies from −∞ to +∞. In the same fashion, the line with v
= −π/2 maps to a line at y = −a running from −∞ to ∞. The capacitance
of a unit length and width w (large compared to a) including the end of the
plate) capacitor made by the central plate of one polarity and the outside
plate of opposite polarity is easily obtained.
We begin by finding the values of u that correspond to the ends of the
segment x = -∞ to x = w a on the upper plate. We have
πx
= ln(sinh u)
2a
⇒
eπx/2a = sinh u =
eu − e−u
2
The value of u corresponding to x large and negative is clearly u = 0. When
x a, u 1 implying that to an excellent approximation
sinh u = 12 eu
We find then
eu = 2eπw/2a
or
u = ln 2 +
πw
2a
Using the results of p 126,
ln 2 + πw/2a
u2 − u1
C = −ε0
= ε0
v2 − v 1
π/2
2 ln 2 w
+
= ε0
π
a
42
Classical Electromagnetic Theory
The term ε0 w/a is just what we would have expected in the absence of fringing
fields at the x = 0 edge. The bottom plate makes an identical contribution
to the capacitance so that we have for such a capacitor
2 ln 2 w
+
C = 2ε0
π
a
5-12 Expression (5–74) gives the capacitance per unit length of a single sided strip
of parallel plate capacitor. Specializing this result to the circular capacitor we
note that the capacitance should become the sum of the area term ε0 πr2 /a
and a contribution of the perimeter due to a strip of width r and length 2πr
so that
2
πr
2πr
+ r ln
C = ε0
a
a
W have arbitrarily taken r as the effective width of the strip. However, writing
the result as
2πr
ε0 πr2
a
ln
C=
1+
a
πr
a
we note that fractional contribution diminishes as r/a increases.
5-13 The mapping
z=
a f
b
b
a u + iv
+
+
=
2 b
f
2
b
u + iv
may be rationalized as
a u + iv b(u − iv)
+ 2
x + iy =
2
b
u + v2
bu
bv
a u
ai v
+ 2
−
=
+
2 b
u + v2
2 b
u2 + v 2
so that
bu
bv
a u
a v
+ 2
− 2
x=
and y =
2 b
u + v2
2 b
u + v2
a u
bu
b
a 1
For v = 1,
x=
+ 2
− 2
and y =
2 b
u +1
2 b u +1
b
1
b
au 1
+ 2
− 2
and for v = 2, x =
and y = a
2 b u +4
b u +4
We plot these curves parametrically for b = 1 in figure 5.1.
5-14 The mapping indicated may be expanded for f = u + iv as
z=
i−f
i − u − iv
u + (v − 1)i
=
=−
i+f
i + u + iv
u + (v + 1)i
Chapter Five Solutions
43
Figure 5.1: Images of the v = 1 and v = 2 lines in the x-y plane
Along the real axis of the f plane, v = 0, reducing z to
z=−
u2 − 1 − 2ui
u−i
=−
u+i
u2 + 1
The magnitude of z when v = 0 is
u − i
≡1
|z| = u + i
Moreover, u = 0 → z = 1, u = 1 → z = i, u = −1 → z = −i, u = ∞ → z =
−1. More generally, when v = 0 we can write z = e−2iα with α = tan−1 (1/u).
Evidently the real axis of f maps into the unit circle in the z plane.
When v = 0, we can again find |z|2 ,
2
2
u + (v − 1)i 2
= [u + (v − 1)i][u − (v − 1)i] = u + (v − 1)
|z|2 = 2
u + (v + 1)i
[u + (v + 1)i][u − (v + 1)i]
u + (v + 1)2
=
u2 + (v + 1)2 − 4v
4v
=1− 2
2
2
u + (v + 1)
u + (v + 1)2
We see then that for v < 0, |z| increases to give points outside the unit circle.
In similar fashion the second mapping may be investigated.
z=
i + u + iv
u + (v + 1)i
i+f
=
=−
i−f
i − u − iv
u + (v − 1)i
For f on the real axis,
u + i
≡1
|z| = u − i
Moreover, u = 0 → z = −1, u = 1 → z = i, u = −1 → z = −i, u = ∞ →
z = −1. Again, the real axis is mapped to a unit circle. When v = 0,
2
2
u + (v + 1)i 2
2
= [u + (v + 1)i][u − (v + 1)i] = u + (v + 1)
|z| = u + (v − 1)i [u + (v − 1)i][u − (v − 1)i]
u2 + (v − 1)2
=
u2 + (v − 1)2 + 4v
4v
=1+ 2
u2 + (v − 1)2
u + (v − 1)2
5-15 We investigate first
√ the behavior of the mapping for f along the real axis.
When u is real, f is real. Thus we readily find that u = 0 → z = −1, u =
44
Classical Electromagnetic Theory
1 → z = 0, u = ∞ → z = +1. In other words, the positive real axis maps to
x ∈ (−1, 1). When u is negative, the square root is imaginary and we write
it as ri. Then
ri − 1 ≡1
|z| = ri + 1 In particular u = −1 maps to i and points ∈ (0, −1) map to the left quadrant
quarter circle whereas points more negative than −1 map to the right quarter
circle. The whole axis then is mapped into a closed half circle in the upper
half plane.
The second mapping is easily deduced from the first. Taking the n’th
root of any point in the complex plane reduces its magnitude to the n’th root
and divides the argument by n. Thus the points at |z| = 1 remain at that
distance but have their polar angles reduced by a factor n. In particular, the
image of 0 at x = −1 gets moved to polar angle π/n. The result is that the
semicircle of the previous map now gets compressed like a closing hand-held
fan to become a wedge with interior angle π/n.
5-16 To find the image of the real axis set f = u and note that when u is positive,
the mapping is straightforward, u = 0 → x = −∞, u = 1 → x = 0, u = ∞ →
x = ∞. In other word, the positive u axis √
maps to the entire (−∞, ∞) x axis.
When u < 0, we write the square root as u = ir. Then
1
1
= id r +
z = d ir −
ir
r
Thus the negative u axis maps to a vertical line along the y axis that terminates at height 2d above the x axis (the image point of u = −1).
5-17 This mapping presents several points where the image of the real line will
abruptly change directions. We note that if u is large and positive, the ln f
and −2 ln[(f + 1)1/2 + 1] essentially cancel leaving only the first term which
clearly increases to ∞ as f → ∞. As u → 0, the ln f will dominate and image
0 to −∞ The first singularity occurs when f = 0 at that point the ln f term
has to be replaced by ln(|u|eiπ ) = ln |u| + iπ. To sum up to this point then,
u ∈ (0, ∞) → x ∈ (−∞, ∞). At u = 0, the image is displaced upwards a
distance π while still at x = −∞. As u becomes more negative, v remains at
π until u reaches −1 when the argument of the two square root terms both
become negative. Just before the arguments turn negative, we abbreviate
1 + u = . The expression for z then becomes
z = 21/2 − 2 ln(1 + 1/2 ) + iπ + ln |u| ≈ 21/2 − 21/2 + iπ + ln(1) = 0 + iπ
In other words, u = (0, −1) maps to x = (−∞, 0), y = π After u passes −1,
set (1 + u)1/2 = |1 + u|1/2 eiπ/2 = riπ/2 . We again express z in this domain.
z = 2ir − 2 ln(1 + reiπ/2 ) + iπ + ln(1 + r2 )
= 2ir − 2 ln[eiπ/2 (e−iπ/2 + r] + iπ + ln(1 + r2 )
Chapter Five Solutions
45
= 2ir − iπ − 2 ln(r − i) + iπ + ln(1 + r2 )
= 2ir − ln[(1 + r2 )e−2iα ] + ln(1 + r2 ) = 2ir + 2iα
with α = tan−1 (1/r).
We see that z becomes pure imaginary with y = 1 at r = 0 increasing to
infinity as r → ∞. The entire image is illustrated in Figure 5.2.
Figure 5.2: Image of the f plane real axis in the z plane. The corresponding
values of u are given at the vertices.
5-18 The mapping z = a f 2 − 1 is readily investigated. The real axis has three
regions of interest. When
√ u > 1, z is also real and increases with u. When
|u| < 1 we write z = ai 1 − u2 and finally when u < −1 we revert to the
original. we find the, u ∈ (1, ∞) → x ∈ (0, ∞), y = 0; u ∈ (−1, 0, 1) → x =
0, y ∈ (0, a, 0) and u ∈ (−1, −∞) → x ∈ (0, ∞), y = 0. If f is in the first
quadrant then f 2 is in the positive half plane as is f 2 − 1. Taking the square
root maps all such points into the first quadrant. When f is in the second
quadrant, f 2 will be in the third or fourth, and f 2 − 1 will lie in the third
or the fourth quadrant. The square root therefore maps the second quadrant
f ’s into the second quadrant.
5-19 For a cylinder of length L and when the potential has no ϕ dependence,
!
Aλ sinh λzJ0 (λr)
V (r, z) =
λ
From V (r = a) = 0 we deduce that λa is a root of J0 say ρ0i ⇒ λ = ρ0i /a.
Thus
!
ρ0i z
ρ0i r
Ai sinh
V (r, z) =
J0
a
a
i
At z = L this specializes to
V (r, L) =
!
i
Ai sinh
ρ0i L
ρ0i r
J0
a
a
46
Classical Electromagnetic Theory
We abbreviate the constant terms by Ci and write
!
ρ0i r
r2
V (r, L) =
Ci J0
= V0 1 − 2
a
a
i
The coefficients Ci may be evaluated by multiplying both sides of the equation
by J0 (ρ0j r/a) and integrating over the surface.
a ρ0i r
r2
1
2 2
V 0 1 − 2 J0
rdr
2 Ci a J1 (ρ0i ) =
a
a
0
Changing variables to x = ρ0i r/a, the integral on the right becomes
a2 ρ0i
a2 ρ0i 3
xJ
(x)dx
−
x J0 (x)dx
0
ρ20i 0
ρ40i 0
We evaluate each term independently.
ρ0i
xJ0 (x)dx = ρ0i J1 (ρ0i )
0
ρ0i
3
d
(xJ1 )dx
x (xJ0 ) dx = x2
dx
ρ0i = x3 J1 − 2x (xJ1 )dx
0
ρ0i
= ρ30i J1 (ρ0i ) − 2x2 J2 x J0 (x)dx =
0
2
0
= ρ30i J1 (ρ0i ) − 2ρ20i J2 (ρ0i )
Gathering the terms, we may solve for Ci to get
2V0
2
4V0 J2 (ρ0i )
(ρ
)
−
J
(ρ
)
+
J
(ρ
)
= 2 2
Ci =
J
1 0i
1 0i
2 0i
ρ0i J21 (ρ0i )
ρ0i
ρ0i J1 (ρ0i )
5-20 The boundary conditions are obtained from Maxwell’s equations:
·E
= ρ
∇
ε0
⇒
Erext − Erint =
×E
=0
∇
⇒
Eθext − Eθint = 0
σ
σ0 cos θ
=
ε0
ε0
Using these boundary conditions for the general solution of ∇2 V = 0 in
spherical polar coordinates
!
B
V (r, θ) =
A r + +1 P (cos θ)
r
we have from the radial equation
∂V
∂V −
∂r a−
∂r
=
a+
σ0 cos θ
ε0
Chapter Five Solutions
47
or
∞
!
A a−1 P (cos θ) −
=1
∞
!
−( + 1)B
=0
which means for = 1
a+2
P (cos θ) =
σ0 cos θ
ε0
2B1
σ0
+ A1 =
a3
ε0
and for = 1
( + 1)B
+ A a−1 = 0
a+2
Similarly, for the θ component of the electric field we find
1 ∂V 1 ∂V =
a ∂θ a+
a ∂θ a−
giving
B
= A a−1
a+2
The θ and r equations may be solved simultaneously to give
B1 =
σ0 a3
,
3ε0
A1 =
σ0
,
3ε0
and
A = B = 0 for = 1
The potentials inside and outside the sphere and the associated fields are then
σ0 r cos θ
3ε0
σ0 a3 cos θ
V (r > a) =
3ε0 r2
V (r < a) =
⇒
⇒
= − σ0 k̂ = σ0 −r̂ cos θ + θ̂ sin θ
E
3ε0
3ε0
3 a
σ
0
=
2r̂
cos
θ
+
θ̂
sin
θ
E
3ε0 r3
The continuity of Eθ across the surface is evident, as is the discontinuity of
Er by σ/ε0 .
5-21 As there is both a ϕ and θ dependence in this problem we use the general
spherical solution to Laplace’s equation.
!
r
V (r < a) =
Am
Ym (θ, ϕ)
a
,m
and
V (r > a) =
!
,m
+1
a
Bm
Ym (θ, ϕ)
r
The boundary condition V (a) = sin 2θ cos ϕ may be written in terms of spherical harmonics using (F–44) as
& 8π −1
sin 2θ sin ϕ =
Y2 (θ, ϕ) − Y21 (θ, ϕ)
15
48
Classical Electromagnetic Theory
Equating the inside
and outside solutions at to the potential at a, we find
A2,−1 = A2,1 = 8π/15 to get
&
r2
r2 8π −1
Y2 − Y21 = 2 sin 2θ cos ϕ
V (r < a) = 2
a
15
a
Similarly when r > a, we get
&
a3
a3 8π −1
Y2 − Y21 = 3 sin 2θ cos ϕ
V (r > a) = 3
r
15
r
5-22 We again take the general solution for the potential as and denote the radius
of the sphere by R.
+1 ∞ !
!
r
R
Vm (r, θ, ϕ) =
+ Bm
Am
Ym (θ, ϕ)
R
r
=0 m=−
Inside the spherical shell, the solution takes the form
!
r
V (r < R) =
Am
Ym (θ, ϕ)
R
,m
whereas outside the shell V is given by
+1
!
R
Bm
Ym (θ, ϕ)
V (r > R) =
r
,m
V is continuous across the boundary since otherwise its gradient would diverge
leading to infinite electric field. The linear independence of the spherical
harmonics require Am = Bm The boundary condition on Er is
∂V ∂V σ
σ0 sin θ sin ϕ
−
=
=
∂r R−
∂r R+
ε0
ε0
The trigonometric terms may be written in terms of spherical harmonics using
(F–44)
&
&
3
3
−1
1
iϕ
−iϕ
= −i
sin θ e − e
sin θ sin ϕ
Y1 (θ, ϕ) + Y1 (θ, ϕ) = −
8π
2π
so that the condition above becomes
&
2π σ0 1
1 !
m
Y1 + Y1−1
Am + ( + 1)Bm Y = i
R
3 ε0
and the continuity of V at R gives Am = Bm . Thus for (, m) = (1, ±1),
Am = Bm = 0 and
&
2π σ0
R
A1,1 = B1,1 = A1,−1 = B1,−1 = i
3
3 ε0
Chapter Five Solutions
49
Inserting these values into the general solutions above, we obtain
r
r
V (r < R) = A1,1
Y11 + A1,−1
Y1−1
R
R
&
2π σ0 1
r
σ0
= i
r sin θ sin ϕ
Y1 + Y1−1 =
3
3 ε0
3ε0
and
2
R2
R
1
+
B
Y
Y1−1
1,−1
1
2
r
r2
&
2π σ0 1
R3
σ0 R 3
= 2i
sin θ sin ϕ
Y1 + Y1−1 =
3r
3 ε0
3ε0 r2
V (r > R) = B1,1
5-23 Using the methods of Apendix B, the nonzero elements of the metric tensor
for oblate ellipsoidal coordinates are
gρρ =
a2 (ρ2 − cos2 α)
ρ2 − 1
gαα = a2 (ρ2 − cos2 α)
gϕϕ = a2 (ρ2 − 1) sin2 α
Since on a conducting prolate ellipsoid (the coordinate surface for ρ held
constant), the potential is independent of α and ϕ we anticipate the potential
everywhere will be independent of α and ϕ. Therefore eliminating all terms
in ∂V /∂α and ∂V /∂ϕ from the Laplacian we obtain as the only remaining
term
# ∂V
∂ " 2
ρ −1
=0
∂ρ
∂ρ
which is easily integrated to give
k
∂V
= 2
∂ρ
ρ −1
with k an arbitrary constant of integration. This result may be integrated
once more to give
ρ−1
1
V = 2 k ln
+C
ρ+1
As ρ goes to infinity we anticipate that V would tend to zero, implying that
C = 0. For the ellipsoid defined by ρ = ρ0 ,
k=
2V0
ρ0 − 1
ln
ρ0 + 1
Unfortunately, this result is not usable for the infinitely thin needle (ρ0 = 1)
as k would would be indeterminate. Instead, we relate k to the charge on
50
Classical Electromagnetic Theory
the surface. The electric field is the negative gradient of V so that in these
coordinates,
2
k
= − √ρ̂ ∂V = − ρ̂ ρ − 1
E
2−1
2
2
gρρ ∂ρ
ρ
a ρ − cos α
We equate this field to the surface charge density ε0 and integrate over the
surface to obtain
Q
· dS
= Eρ √gαα gϕϕ dαdϕ
= E
ε0
a2 ρ2 − 1
= −k
(ρ2 − cos2 α)(ρ2 − 1) sin2 α dαdϕ
2
2
2
a(ρ − 1) ρ − cos α
2π π
sin αdαdϕ = −4πka
= −ka
0
0
We conclude then that k = −Q/(4πε0 a).
Instead of integrating over α, we can change variables to integrate over z.
From the definition, dz = −aρ0 sin α dα, leading to
Q
2πk −a
4πka
=
dz = −
ε0
ρ0 a
ρ0
The charge per unit length λ is evidently −2πkε0 ρ0 . Substituting the value
of k we get the constant linear charge density λ = Q/(2aρ0 ). For the needle
of length 2a with ρ0 = 1 this becomes λ = 12 Q/a.
5-24 The solution to this problem is rather lengthy, closely following the discussion
of page 364 to 367. Completing exercise (B-12) should be a prerequisite.
Taking the form of the Laplace equation from (B-12),
1
∂V
(cosh ρ − cos α)3
∂
2
∇ V =
sin α
a3 sin α
∂ρ cosh ρ − cos α ∂ρ
sin α
∂V
∂
(cosh ρ − cos α)2 ∂ 2 V
+
+
∂α cosh ρ − cos α ∂α
∂ϕ2
a2 sin2 α
We place the origin midway between the two spheres so that their centers lie
at z = a coth ρ = ±5 m and the radius a/ sinh
√ρ = 1 m. Thus the surfaces lie
at cosh ρ = ±5 which implies a = sinh ρ = 24 m. It should be clear that
the solution will be independent of ϕ and we expect the solution to depend
only on ρ. Unfortunately, setting ∂V /∂α = 0 does not make the remaining
equation α independent. The equation remaining from ∇2 V = 0 is
∂
1
∂V
sin α
∂V
1 ∂
+
=0
∂ρ cosh ρ − cos α ∂ρ
sin α ∂α cosh ρ − cos α ∂α
√
Substituting V = cosh ρ − cos α U as suggested, changes the equation to
2
∂ U
1
∂U
1
∂2U
√
+
cot
α
−
U
=0
+
∂2α
∂α
4
cosh ρ − cos α ∂ρ2
Chapter Five Solutions
51
which may be separated. Trying U = R(ρ)A(α) we obtain
d2 R
d2 A
dA
+ cot α
2
2
dρ
dα − 1 = 0
+ dα
R
A
4
resulting in
d2 R
− ( 14 + λ)R = 0 and
dρ2
d2 A cos α dA
+ λA = 0
+
dα2
sin α dα
The equation for A is just the Legendre equation, and insisting on reasonably
behaved solutions we require λ = ( + 1) meaning A(α) = P (cos α). The
remaining equation for R is now easily solved.
d2 R
d2 R
1
1
−
[
+
(
+
1)]R
=
− ( + 12 )2 R = 0 ⇒ R(ρ) = ±e±(+ 2 )ρ
4
2
2
dρ
dρ
We can now write the general solution as
!
1
1
C e(+ 2 )ρ + C− e−(+ 2 )ρ P (cos α)
V (ρ, α) = cosh ρ − cos α
−1
on the upper sphere, ρ = cosh (5) = ρ0 and v = Va is independent of α. We
substitute these values into the general solution to get
!
√
B P (cos α)
Va = 5 − cos α
1
1
where we have abbreviated B = C e(+ 2 )ρ0 + C− e−(+ 2 )ρ0 . Evidently, the
B are
√ just the expansion coefficients in the Legendre polynomial expansion of
Va / 5 − cos α (see example D.3). At the lower sphere ρ = −ρ0 and defining
1
1
D = C e−(+ 2 )ρ0 + C− e(+ 2 )ρ0 we similarly get
!
√
Vb = 5 − cos α
D P (cos α)
After solving for B
their
definitions simultaneously
#
" and 1D we may solve
1
to obtain C = B e(+ 2 )ρ0 − D e−(+ 2 )ρ0 /[sinh(2 + 1)ρ0 ] and C− =
"
#
1
1
D e(+ 2 )ρ0 − B e−(+ 2 )ρ0 /[sinh(2 + 1)ρ0 ].
For
√ the special case of Va = const. a Legendre polynomial expansion of
Va / cosh ρ − cos α may be constructed from the generating function for Legendre polynomials:
!
1
√
t P (cos α)
=
2
1 + t − 2t cos α
with |t| < 1. If we substitute t = e−ρ the radical becomes:
1
1
√
=√
1 ρ
1 + e−2ρ − 2e−ρ cos α
−ρ/2
−ρ ) − cos α
2e
2 (e + e
52
Classical Electromagnetic Theory
=√
or
√
!
1
=
e−ρ P (cos α)
√
2e−ρ/2 cosh ρ − cos α
√ ! −(+ 1 )|ρ|
1
2
e
P (cos α)
= 2
cosh ρ − cos α
The boundary condition at ρ = ρa = cosh−1 5 may then be written
√
√ ! −(+ 1 )|ρ |
Va
a
2
= Va 2
e
P (cos α)
cosh ρa − cos α
=
!
1
1
C e(+ 2 )ρa + C− e−(+ 2 )ρa P (cos α)
Equating coefficients gives
√
1
1
1
Va 2e−(+ 2 )|ρa | = C e(+ 2 )ρa + C− e−(+ 2 )ρa
similarly at ρb = −ρa we have
√
1
1
1
Vb 2e−(+ 2 )|ρa | = C e−(+ 2 )ρa + C− e(+ 2 )ρa
with solutions
C =
√ −(+ 1 )|ρ | Va e(+ 12 )ρa − Vb e−(+ 12 )ρa
a
2
2e
2 sinh((2 + 1)ρa )
and
C− =
√ −(+ 1 )|ρ | Vb e(+ 12 )ρa − Va e−(+ 12 )ρa
a
2
2e
2 sinh((2 + 1)ρa )
We could of course accommodate spheres of any size simply by moving the
x-y plane up or down the z axis. If we pick our zero of potential so that
Vb = −Va we get
√
1
√
cosh( + 12 )ρa
1
2Va e−(+ 2 )|ρa |
C = −C− = 2Va e−(+ 2 )|ρa |
=
sinh(2 + 1)ρa
2 sinh( + 12 )ρa
Finally the compete solution is
V (ρ, α) =
√
2Va
1
∞
!
e−(+ 2 )|ρa | sinh( + 12 ρ)
cosh ρ − sinh α
P (cos α)
sinh( + 12 )ρa
=0
which, incidentally, also constitutes the solution for a charged sphere in the
neighborhood of a conducting, neutral plane.
Errata: Ex 5.4.3 and 5.4.4 should each have s a summation symbol preceding the left hand side. Just above Ex 5.4.4, the phrase should have read:
Multiplying by sin mϕ and integrating, we have
Chapter 6
6-1 The charge q located at distance h from the center of the isolated conducting
sphere has an image charge q = −(R/h)q located at h − R2 /h from the
charge and neutrality requires us to place a neutralizing charge (+R/h)q at
the center of the sphere. The force on q is therefore
−1
1
q2 R
+ 2
F =
4πε0 h (h − b)2
h
with b = R2 /h.
6-2 If we differentiate the given potential
−
d
q
q
∂V
=−
=
∂z
dz 8πε0 z
8πε0 z 2
Comparing this to the field we expect at z from the charge at −z,
q
q
E=
=
4πε0 (2z)2
16πε0 z 2
The reason for the discrepancy is that when we take the gradient of V in the
neighborhood of z, we are determining how it varies when the source is held
constant. By expressing the source coordinate in terms of the field coordinate
we are allowing it to vary as well in the differentiation.
6-3 We assume the sphere of radius R centered on the origin. For a charge q
located at r, its image q = −q(R/r) is located at r = (R2 /r2 )r. The
potential at r due to its image is
V (r ) =
−q(R/r)
−qR
q
=
=
4πε0 |r − r |
4πε0 |r − (R2 /r2 )r |
4πε0 r2 (1 − R2 /r2 )
Then
−2qrR
qR
d
=
2
2
dr 4πε0 (r − R )
4πε0 (r2 − R2 )2
The r component of the electric field we compute directly is
)r =
−(∇V
Er (r ) =
q
−qR
1
1
=
4πε0 (r − r )2
4πε0 (r2 − R2 )2
We see again that the gradient overcalculates the field by precisely a factor
of two for the same reason as in the previous problem.
6-4 The dipole, which we take to lie along the z axis at distance z from the plane,
p + 2pz k̂ located at z = −z. The electric field at
has an image p = −
displacement r from the dipole p is in general given by
p · r
1
E = −∇V = −∇
4πε0
r3
3r(r · p ) p 1
=
−
4πε0
r5
r3
— 53—
54
Classical Electromagnetic Theory
In this case r = z + z . The potential energy of dipole p in this field is
−1 3(r · p )(r · p ) p · p W = −
p·E =
− 3
4πε0
r5
r
3pz pz
−1
p · p −(p2 + p2z )
=
−
=
4πε0 (z + z )3
(z + z )3
4πε0 (z + z )3
The force on the dipole p is then given by (note that had we written z + z as 2z, then dW = −2F dz)
Fz = −
−3(p2 + p2z )
∂W
−3(p2 + p2z )
=
=
4
∂z
4πε0 (z + z )
64πε0 z 4
If we express the potential energy of the dipole in terms of the angle it makes
with the z axis we easily obtain the torque (we have combined the z and z coordinates in the energy expression and recognize that changing θ changes
θ whereas the virtual work requires that only θ change):
τθ = −
1 ∂ p2 (1 + cos2 θ)
1 ∂W
−p2 sin(2θ)
=
=
2 ∂θ
2 ∂θ 32πε0 z 3
64πε0 z 3
where the θ axis is normal to the plane containing the dipole moment and the
z axis.
6-5 We place the right plate along the x-y plane meaning that the charge is located
at z = −a. The various images formed are shown in figure 6.1. It is evident
that the forces from all the positive images cancel. Writing the remaining
terms, we have
1
q2
1
1
1
+
+
+
+
·
·
·
Fz =
4πε0 (2a)2
(2D + 2a)2
(4D + 2a)2
(6D + 2a)2
q2
4πε0
1
1
1
+
+
+ ···
2
2
(2D − 2a)
(4D − 2a)
(6D − 2a)2
1
q2
4aD
8aD
16aD
=
−
−
−
−
·
·
·
16πε0 a2
(D2 − a2 )2
(4D2 − a2 )2
(9D2 − a2 )2
−
Figure 6.1: Images of the charge located at z = −a.
Chapter Six Solutions
55
6-6 Since the sphere is conducting, the potential on it must be constant, and
outside it must be spherically symmetric. The total charge enclosed is zero so
that we conclude the potential everywhere outside the sphere is zero. To find
the potential and hence the field inside the sphere, we pick the z axis along
the dipole axis and construct an extended dipole p at the center of the sphere
by placing a positive charge p/a at z = 12 a and a negative charge at − 12 a.
The point dipole will be recovered by letting a tend to 0. The charges at ± 12 a
have images of magnitude ∓2R/a located at h = ±2R/ a. The potential along
the z axis is then
1
2R/a
1
2R/a
p 1
+
−
−
V (z) =
a 4πε0 z − 12 a z + 12 a 2R2 /a − z
2R2 /a + z
1
1
1
1
1
1
p
−
−
−
=
4 πε0 a z 1 − 12 a/z
R 1 − 12 az/R2
1 + 12 a/z
1 + 12 az/R2
p
a
az
=
− 3 + O(a2 )
2
4πε0 a z
R
taking the limit as a tends to 0, we find
1
p
z
V (z) =
−
4πε0 z 2
R3
Comparing this to the general form of the potential in spherical polar coordinates, we conclude immediately that V inside the sphere is
1
p
r
V (r) =
− 3 cos θ
4πε0 r2
R
and the electric field inside the sphere is
2
1
p
1
1
p
E = −∇V =
+ 3 r̂ cos θ −
− 3 θ̂ sin θ
4πε0 r3
R
4πε0 r3
R
Outside the sphere the electric field vanishes.
6-7 The ring bearing charge Q has an image ring with charge Q = (−b/a)Qand
radius a = −b2 a. The potential along the z axis is then
1
b/a
Q
√
−
V =
4πε0
z 2 + a2
z 2 + b4 /a2
To generalize this result, we expand each of the terms using the binomial
theorem. The first term may be expanded for z ∈ (a, b) as
−1/2
1
1
a2
√
=
1+ 2
z
z
z 2 + a2
2
3
1 a2
1
1 3 1 a2
1 3 5 1 a2
1−
=
+
−
+ ···
z
2 z2
2 2 2! z 2
2 2 2 3! z 2
∞
1 ! (−1)n (2n − 1)!! a2n
=
z n=0
2n n!
z 2n
56
Classical Electromagnetic Theory
The second term in V,
−1/2
b/a
1
1
z 2 a2
=
=
1+ 4
b
b
z 2 + b4 /a2
z 2 a2 /b2 + b2
is expanded as
−1/2
∞
1
1 ! (−1)n (2n − 1)!! z 2n a2n
z 2 a2
=
1+ 4
b
b
b n=0
2n n!
b4n
The potential between a and b along the z axis may therefore be expressed
as
1
Q ! (−1)n (2n − 1)!! 2n
z 2n
V (z) =
a
−
4πε0
2n n!
z 2n + 1 b4n+1
Comparing this to the general spherical polar solution
!
Bn
An rn + n+1 Pn (cos θ)
V =
r
we find the general solution in the region (a, b) to be
1
r2n
Q ! (−1)n (2n − 1)!! 2n
a
− 4n+1 P2n (cos θ)
V (a < r < b) =
4πε0
2n n!
r2n+1
b
6-8 According to example 6.3 on page 148, a dipole p at r with respect to the
center of a grounded conducting sphere has an image
R3
3(
p · r)r
p = 3 −
p+
r
r2
located at r = (R2 /r2 )r. We can use the expression for the potential energy
of two dipoles from problem 6-4, noting that the vector running from r to r
is r = (1 − R2 /r2 )r.
3(
p · r )(
p · r )
−1
W =
−
(
p
·
p
)
4πε0 r3
r2
3
−R
3(r · p )2
2
=
p +
4πε0 r6 (1 − R2 /r2 )3
r2
For r R, this gives a 1/r6 potential for the dipole–induced dipole interaction.
6-9 The force and the torque on the dipole of the preceding problem may be found
by differentiating the potential energy. If we take the dipole to be inclined at
angle θ with respect to its position vector, we may write (r · p )/r = p cos θ
and
−R3 p2 (1 + 3 cos2 θ)
−R3 p2 (1 + 3 cos2 θ)
=
W =
3
3
4πε0 r6 (1 − R2 /r2 )
4πε0 (r2 − R2 )
Chapter Six Solutions
57
so that
Fr = −
−3R3 p2 r(1 + 3 cos2 θ)
1 ∂W
=
4
2 ∂r
4πε0 (r2 − R2 )
and
τθ = −
3R3 p2 sin θ cos θ
1 ∂W
=−
3
2 ∂θ
4πε0 (r2 − R2 )
6-10 Let the distance of closest approach of the wire to the center of the sphere
be b and choose the x axis parallel to the wire lying in the x y plane, with x
= 0 at the point of closest approach. A point x on the line charge is imaged
at r = R2 /r where r2 = x2 + b2 . Calling the angle between r and the y axis
θ, we note that the image falls at the same angle θ. In terms of θ, we may
find r as
R4
R4
R4 cos2 θ
r2 = 2
=
=
2
b + x2
b2
b2 (1 + tan θ)
the equation of a circle with radius R2 /2b centered at y = R2 /2b as is shown
below.
R2
R2 cos2 θ
y = r cos θ =
=
(1 + cos 2θ)
b
2b 2
2
cos
θ
sin
θ
R
R
x = r sin θ =
=
sin 2θ
b
2b
leading to
2 2 2
R
R2
2
x + y −
=
2b
2b
The charge dq = λ dθ on a segment subtending dθ about θ of the ring is the
image of the charge dq = λbdθ/ cos2 θ. The segment on the ring lies a distance
r = R2 cos θ/b from the origin so that
dq = −
−R cos θ
R cos θ
λR
R
dq =
dq = −
λdx = −
dθ
b/ cos θ
b
b
cos θ
so that, with the help of d = R2 /b dθ
dq λR
dq dq dθ
λR b
λb
=−
, and
=
=−
= − cos θ
dθ
cos θ
d
dθ d
cos θ R2
R
6-11 In order to establish an electric field E0 at the origin, we place a charge Q =
−2πε0 L2 E0 at +L and Q = +2πε0 L2 E0 at −L. These charges would have
images Q = −(R/L)Q at +R2 /L and −R2 /L, respectively. The potential
along the axis arising from the four charges is
2πε0 L2
2πε0 L2
2πε0 RL
2πε0 RL
E0
−
+ 2
−
V (z) = −
4πε0
L−z
L−z
R /L + z
z − R2 /L
1
1
1
1
−E0
RL
−
−
=
L
+
2
1 − z/L 1 + z/L
z 1 + R2 /Lz
1 − R2 /Lz
58
Classical Electromagnetic Theory
We expand this expression to first order to obtain
z
R2
1
−E0
z
LR
R2
−1−
V (z) =
L 1+ −1+
+
1−
+O
2
L
L
z
Lz
Lz
L
E0 2R3
−
2z
=
2
z2
Comparing this to the general solution in spherical polar coordinates we generalize immediately to obtain
3
2R
V (r > R) = E0
−
r
cos θ
r2
6-12 The images produced by a 60◦ plate vertex are shown in figure 6.2. It is readily
ascertained that the y component of force vanishes whereas the x component
may be calculated with the aid of figure 6.2 as
Figure 6.2: The location of the images when the vertex angle is α = 60◦
q ! qi (x − xi )
4πε0
|r − ri |3
2( 3 )b
2( 1 )b
q2
2b
=
−
− 23 + √ 2
4πε0
b
(2b)3
( 3b)3
q2
1
1
1
q2
=
−
− 2+√
= −0.6726
2
4πε0
b
(2b)
4πε0 b2
3b2
Fx =
6-13 Although the diagram above suggests a 60◦ angle between the plates, the
generalization to α = π/n is immediate. When the angle α of the bend is not
commensurate with π, a solution may be interpolated from the π/n solutions.
We take the x axis to contain the vertex and the charge. Consider the force
due to the first images located at distance b along a line at ±α to the x axis.
Chapter Six Solutions
59
The distance between the charge and each of these images is 2b sin 12 α so that
we deduce a force
q2
F =
4πε0 (2b)2 sin2 12 α
from each of these. The y components exactly cancel, and the x components
are given by
−q 2
Fx = F sin 12 α =
4πε0 (2b)2 sin 12 α
The second pair of images lie at 2α and each contribute
Fx =
+q 2
4πε0 (2b)2 sin α
to the x component of the force, while the each charge of the third pair
contributes
−q 2
Fx =
4πε0 (2b)2 sin 32 α
and so forth until the successive images reach the angle π at which point a
single charge of sign (−1)π/α contributes
Fx =
(−1)π/α+1 q 2
4πε0 (2b)2
The net force from all these images is
Fx =
π/α−1
! 2(−1)(π/α+1)
−q 2
+ (−1)π/α
2
4πε0 (2b) i=1
sin( 12 iα)
It is not immediately obvious how to extend this result to
values of π/α.
Because we anticipate that the the force will
vary smoothly with the angle of the vertex it
is useful to examine how the sum varies as a n
α
function of integral π/α. Calculating the first
1
180◦
few sums, we find for n = π/α = 1, 2, 3, . . .and
2
90◦
so forth.
3
60◦
One is struck by the near constancy of the
4
45◦
increase as we go to successive n. In fact,
5
36◦
to excellent precision, the value of the sum
6
30◦
is given by (try it for other values)
7 25.71◦
8
22.5◦
0.13
Sn = 0.8825425n +
9
20◦
n
10
18◦
This result must surely work for the non20
9◦
integral values of n as well. In all cases the
100
1.8◦
force is one of attraction toward the vertex of 1000
0.18◦
the bend.
non- integral
Sum
1
1.82842706
2.69059886
3.56260903
4.43874411
5.31698487
6.19644235
7.07666496
7.95739990
8.83849498
17.6573902
88.2555449
8882.54251
60
Classical Electromagnetic Theory
6-14 The images must produce a zero potential line along the plane interface, which
can be accomplished by any symmetric arrangement of charges. Therefore,
we place one image an equal distance below the plane. The original and
this image produce images in the sphere extrapolated from the hemisphere.
These three image charges complete the set required to satisfy the boundary
condition.
6-15 We pick the axes so that the line charge lies at x = b and y = 0 and runs
parallel to the z axis. Then an image line charge −λ located at h = R2 /b will
ensure a constant potential on the cylinder. We write the potential as
−λ
(ln r1 − ln r2 ) + C
4πε0 2
r12
r + b2 − 2rb cos θ
−λ
−λ
ln
ln
=
+C =
+C
4πε0
r22
4πε0
r2 + h2 − 2rh cos θ
V (r, θ) =
At r = R the ratio r12 /r22 = b/h, so that in order to obtain zero potential
on the cylinder we must set
b
λ
ln
C=
4πε0
h
The potential inside the cylinder is therefore
h(r2 + b2 − 2rb cos θ)
−λ
V (r, θ) =
ln
4πε0
b(r2 + h2 − 2rh cos θ)
with h = R 2 /b.
6-16 We know that two parallel line charges ±λ produce a set of nonconcentric
cylindrical equipotentials. Our task will be to pick the locations of these line
charges so that their equipotentials coincide with the two cylinders. Let h
denote displacement of the nearer line charge from axis of cylinder a while
h denotes the distance of the farther line charge from the axis of cylinder a.
The distances of these line charges from the axis of cylinder b are then h + D
and h + D. The geometry is illustrated in Figure 6.3. In order that the two
line charges produce the equipotential desired, they must satisfy
hh = a2
(D + h)(D + h ) = b2
Expanding the second of these and substituting hh = a2 we obtain
D2 + (h + h )D + a2 = b2
Substituting a2 /h for h we obtain the quadratic
Dh2 + (D2 + a2 − b2 )h + a2 D = 0
Chapter Six Solutions
61
Figure 6.3: The nested conducting cylinders of problem 6-16.
which may be solved to yield
h=
(b2 − a2 − D2 ) ±
(b2 − a2 − D2 )2 − 4a2 D2
2D
The negative sign gives h while the + sign gives h .
The potentials on the two cylinders are given by
a
b
−λ
−λ
ln
ln
Va =
Vb =
2πε0
h
2πε0
h+D
so that the capacitance per unit length, λ/ΔV , becomes
C
=
ln
2πε0
a h+D
·
h
b
with h given above.
6-17 The zero potential plane lies halfway between the wires of example 6.4. The
potential difference between ground and one of the wires is just half that
between the wires so that we deduce the capacitance per unit length is twice
that of the pair of wires (D = 2d ).
2πε0
C
=
cosh−1 (d/R)
6-18 We again use the equipotentials around two line charges to do this problem.
Denoting the distance of the hypothetical line charge λ from the center of
cylinder a by ha and that of −λ from the axis of b as hb we write, referring
to Figure 6.4
62
Classical Electromagnetic Theory
Figure 6.4: The equipotential cylinders of problem 6-18.
ha (D − hb ) = a2
hb (D − ha ) = b2
Using the second equation to eliminate hb from the first equation, we find
Dh2a − (D2 − b2 + a2 )ha + a2 D = 0
which we solve to obtain
ha =
D2 − b2 + a2 ±
(D2 − b2 + a2 )2 − 4a2 D2
2D
The negative root must be chosen to give ha as illustrated. hb is found in the
same manner to give
D2 − a2 + b2 ± (D2 − a2 + b2 )2 − 4b2 D2
hb =
2D
Again the positive root must be chosen. The potentials on cylinder the cylinders are
a
b
−λ
λ
Va =
ln
ln
Vb =
2πε0
ha
2πε0
hb
from which we conclude the capacitance per unit length is
C
=
ln
2πε0
a
b
·
ha hb
6-19 When there is no inner sphere, and the charge distribution is independent of
ϕ, the Green’s function (6-48) reduces to
∞
!
r>
+1
1
P (cos θ )P (cos θ)r<
−
G(r, r ) =
b2+1
r>
=0
Chapter Six Solutions
63
With the charge density given,
Q
[δ(cos θ + 1) + δ(cos θ − 1)]
2b2 πr2
the potential becomes
b ∞
r>
1
Q 1 !
V (r ) =
[P (1) + P (−1)] P (cos θ)
r<
−
dr
+1
2+1
2b 4πε0
b
r
0
>
=1
ρ=
The integral must be split into a region where r is r< and a second region
where r is r> .
b
r
b
(. . .)dr =
(r = r< )dr +
(r = r> )dr
0
0
rr
b
1
1
r
r
=
− 2+1
r dr + r
− 2+1 dr
r+1
b
r+1
b
0
r
b
−1
1
r
r2+1
=
−
1 − 2+1 + r
+1
b
r
( + 1)b2+1 r
(2 + 1)
r
=
1− ( + 1)
b
The = 0 term is best obtained by direct integration
b
r
b
1 1
1 1
1
1
1
−
−
dr +
−
=
r
− (b − r)
dr
+
ln
r
r
b
r
b
r
b
b
0
r
r
=1−
r r
ln b
ln b
+ −1+
=
b
b
r
r
The Legendre polynomial P (−1) = (−1) so that P (1) + P (−1) = 2 for
even and vanishes for odd . Then
2 ∞
r
4 + 1
Q
b !
P2 (cos θ)
V (r ) =
ln +
1−
4πε0 b
r
2(2 + 1)
b
=1
6-20 The charge density may be written ρ(r ) = qδ[r − 12 (a + b)k̂] = qδ[r − 12 (a +
b)]δ(cos θ − 1)/2πr2 . Both surfaces are grounded; hence the surface integral
of (6–34) vanishes. We conclude, therefore, that
1
ρ(r )G(r, r )d3 r
V (r ) =
ε0 τ
with G )
given by (6–74). As the solution should have no ϕ dependence we
replace m Ym (θ, ϕ)Y∗m (θ , ϕ ) by [(2 + 1)/4π]P (cos θ), leading to
1 ! P (cos θ)
V (r ) =
×
4πε0
1 − (a/b)2+1
r>
1
a2+1
− +1
−
r<
δ[r − 12 (a + b)]dr
+1
b2+1
r>
r>
τ
64
Classical Electromagnetic Theory
When r < 12 (a + b) it is r< so that performing the trivial integration, the
potential reduces to
2+1
q ! P (cos θ)
(a + b)
a2+1
V (r < r ) =
−
r
−
4πε0
1 − (a/b)2+1
r+1
(a + b)+1
2 b2+1
When r > 12 (a + b), r = r> so that we may write
1
(a + b)
q ! P (cos θ)
r
2+1 a2+1
V (r > r ) =
− 2+1
−
4πε0
1 − (a/b)2+1 r+1
b
2
(a + b)+1
6-21 We develop the Dirichlet Green’s function for the interior of a cylindrical box
of radius a and length L. The equation we need to solve is
1 ∂
∂G(r, r )
1 ∂ 2 G(r, r ) ∂ 2 G(r, r )
+
= −δ(r − r )
r
+ 2
r ∂r
∂r
r
∂ϕ2
∂z 2
We consider G(r, r ) as a function of r and expand it in terms products of of
Bessel functions that vanish at r = a and complex exponentials as
ρ r
!
mj
G(r, r ) =
eimϕ
Amj (θ , ϕ )F(z, z )Jm
a
,m
Inspection of Bessel’s equation, (E–1) gives
1 d dJm m2
r
=
−
1
Jm
r dr
dr
r2
so that substituting the expansion of G into ∇2 G = δ(r − r ) we simplify
ρ r
! d2 F(z, z ) ρ2mj
mj
eimϕ = δ(r − r )
−
F(z,
z
)
A
(θ
,
ϕ
)J
mj
m
2
2
dz
a
a
,m
Next we must expand the δ function in terms of Bessel functions and
complex exponentials. Using Example D.4 we can write
δ(ϕ − ϕ ) =
∞
1 ! im(ϕ−ϕ )
e
2π m=−∞
and using (D–15) we write
δ(r − r ) =
ρ r ρ r ! r
mj
mj
Jm
Jm
ν
a
a
mj
m,j
δ(z − z )δ(r − r )δ(ϕ − ϕ )
may be written
r
ρ r ρ r eim(ϕ−ϕ )
! 1
mj
mj
δ(r − r ) = δ(z − z )
Jm
Jm
ν
a
a
2π
mj
m,j
so that δ(r − r ) =
Chapter Six Solutions
65
with
νmj =
ρ
a
J
0
ρ r
a2
mj
J
r dr = [Jm+1 (ρmj )]2
a
a
2
mj r
Substituting the δ function expansion into the equation for F(z, z ) above, we
immediately identify the expansion coefficients as
ρ r mj
e−imϕ
Jmj
a
Ajm (r , ϕ ) = 2
a π[Jm+1 (ρmj )]2
leaving only the equation in (z, z ) to be solved.
d2 F(z, z ) ρ2mj
− 2 F(z, z ) = −δ(z − z )
dz 2
a
Recognizing the discontinuity provided by the δ function we solve the
homogeneous equation when z < z and when z > z ρ z mj
F(z < z ) = Cmj (r ) sinh
a
ρ (L − z) mj
F(z > z ) = Dmj (r ) sinh
a
We impose the symmetry between z and z and the requirement of continuity
to write the solution as
ρ (L − z ) ρ z mj
mj
sinh
F(z < z ) = Cmj
sinh
a
a
and
F(z > z ) = Cmj
sinh
ρ
mj z
sinh
ρ
mj (L
− z) a
a
Finally, we determine the constant C by integrating the differential equation over a small region containing z .
z + 2
z +
z + 2
ρmj
d F
dz
−
Fdz
=
−
δ(z − z )dz = −1
2
a2
z − dz
z −
z −
dF dz or
sinh
= −1
z −
ρ (L − z ) ρmj
mj
cosh
a
a
a
ρ z ρ (L − z ) ρ
ρ L
ρmj
1
mj
mj
mj
mj
+
cosh
sinh
=
sinh
= a
a
a
a
a
Cmj
ρ
mj z
z +
dF −
dz We conclude F(z, z ) may be written
F(z, z ) =
ρ z ρ (L − z ) a
mj <
mj
>
sinh
sinh
ρmj sinh(ρmj L/a)
a
a
66
Classical Electromagnetic Theory
We can now reassemble the Green’s function to get
G(r, r ) =
1
aπ
m=−∞ j=1
ρ
− z> ) a
a
ρmj sinh(ρmj L/a)J2m+1 (ρmj )
∞ sinh
∞
!
!
mj z<
sinh
ρ
× eim(ϕ−ϕ ) Jm
mj (L
ρ
mj r
Jm
ρ
mj r
a
a
The Green’s function for locations surrounding a cylinder or for regions
between two cylinders may be constructed in similar (laborious) fashion.
Errata: The line above Ex 6.4.5 should read: The capacitance per unit length
is now obtained as λ/ΔV .
The discussion following (6–33) has the wrong sign for the surface charge and
the dipole layer.
Chapter 7
7-1 This problem is essentially solved in the example 7.2. It suffices to replace the
0.9 by t/d and 0.1 by 1 − t/d so that
Ed =
V
κ(d − t) + t
and Eair =
κV
κ(d − t) + t
7-1 Since the needle like cavity is parallel to the polarization, the bound surface
charge has little effect on the field inside the cavity; instead, we must use
to determine the field in the
the continuity of the parallel component of E
diel. =
cavity. Taking the dielectric to be linear and isotropic, we have E
P /ε0 χ. Therefore, in the cavity
cav = P
E
ε0 χ
=P
cav = ε0 E
and D
χ
7-3 This time the exposed ends of the dipoles contribute an field P/ε0 to the field
perpendicular
in the cavity. We use the continuity of the component of D
to the interface to determine the fields in the cavity. In the dielectric, the
displacement field is
+ P = P + P
= ε0 E
D
χ
⊥ is continuous, we find
Since D
1
Dcav
P
P
P
=
+
=
+ Ediel.
Dcav = P 1 +
and Ecav =
χ
ε0
ε0
χε0
ε0
7-4 The electric field outside the dielectric cylinder may be found as the field
created by the exposed ends of the dipoles on the two end faces. Thus,
1
(P · n̂)(z k̂ − r ) (P · n̂)(z k̂ − r ) E(z) =
dS
+
dS
4πε0 z=0
|r − r |3
|r − r |3
z=L
=
1
4πε0
0
a
0
2π
−P (z k̂ − r r̂)r dr dϕ
(z 2 + r2 )3/2
a 2π
P [(z − L)k̂ − r r̂]r dr dϕ
1
+
4πε0 0 0
[(z − L)2 + r2 ]3/2
Inspection shows that these integrals are indistinguishable from those of
(Ex 7.1.1). The electric field when z > L is then:
z
P k̂
z−L
E(z)
=
−
2ε0 (z 2 + a2 )1/2
[(z − L)2 + a2 ]1/2
— 67—
68
Classical Electromagnetic Theory
is the same as that of
7-5 When the slot is parallel, the internal field strength H
the magnetized material. In other words, for Hmat = M/χ, Hslot = M/χ and
B = μ0 Hslot = μ0 M/χ. When the slot is perpendicular to the magnetization,
B⊥ is continuous. Therefore Bdisk = μ0 (H + M ) = μ0 M (1/χ + 1) and
Hdisk = M (1/χ + 1).
7-6
·D
= ρ and Gauss’ law to conclude
To find the force on charge 2, we use ∇
D=
q1
2
4πr12
The force on charge 2 is
q2 D
q2
=
2
ε1
4πε1 r12
q2 E =
7-7
If q2 lies in the material with permittivity ε2 , the field at this point is that
due to the screened charge q1 and that due to the image q2
q1 =
2ε2
q1
ε1 + ε2
and
q2 = −
ε 1 − ε2
q2
ε1 + ε2
Hence the apparent charge at position q2 is
3ε2 − ε1
q
ε1 + ε2
and the forces on the charges neglecting the signs are
F2 =
(3ε2 − ε1 )q 2
4πε2 (ε1 + ε2 )d2
and
F1 =
(3ε1 − ε2 )q 2
4πε1 (ε1 + ε2 )d2
where d is the distance between the charges. The difference reflects the force
that the discontinuity of polarization at the interface exerts on the charges.
7-8 The potential V (R) at the surface of the sphere is Q/(4πε0 R) so that the
potential at the center of the sphere may be found from
V (0) − V (R) = −
0
E(r)
· dr =
R
1
4πε1
0
R
Qr
dr
R3
Q
=
8πε1 R
where we have used Gauss’ law to find E(r)
inside the sphere. The potential
at the center is therefore
1
1
Q
+
V (0) =
4πR ε0
2ε1
Chapter Seven Solutions
7-9
69
Using expression (7–114), we find for a uniform magnetization
= − μ0
B
4π
μ0 (M
· n̂)(r − r )dS M · dS
∇
=
|r − r |
4π
|r − r |3
When r is 0, the integral over the inside and outside hemisphere precisely
· n̂)rdS =
cancel leaving only the integral over the disk. On this surface, (M
M r̂r dr dϕ . Clearly this also vanishes since 2π r̂dϕ ≡ 0.
7-10 Noting that the image dipole p is given by
p =
ε0 − ε
(
p − 2pz k̂)
ε0 + ε
we adopt the expression for the energy of the dipoles found in problem 6-4:
1
(3pz pz − p · p )
32πε0 z 3
ε1 − ε0 2
1
p (1 + cos2 θ)
=
32πε0 z 3 ε0 + ε1
W =
where the z axis was chosen to lie along the line joining the dipole and its
image. The force on the dipole and the torque are now easily obtained.
Fz = −
ε0 − ε 3p2 (1 + cos2 θ)
1 ∂W
=
2 ∂z
ε0 + ε
64πε0 z 4
and
τθ = −
ε0 − ε p2 sin(2θ)
1 ∂W
=
2 ∂θ
ε0 + ε 64πε0 z 3
7-11 In the region of the capacitor where the dielectric slab lies between the plates,
we may use the results of problem 7-1 to write the electric field
Eair =
2εV
(ε0 + ε)d
and Ediel =
2ε0 V
(ε0 + ε)d
and the field in the remaining area is E = V /d. The potential energy of the
capacitor when charged to voltage V is then
W = 12 εE 2 dr3
=
ε0 ΔxΔy 12 d (2εV )2
εΔxΔy 12 d (2ε0 V )2
ε0 (ab − ΔxΔy)V 2
+
+
2d
2
(ε0 + ε)2 d2
2
(ε0 + ε)2 d2
The force in the x direction (when V is held constant) is
−ε0
ε0 ε 2
εε20
∂W
ε0 (ε − ε0 )
=
+
+
ΔyV 2
Fx =
ΔyV 2 =
∂x
2d
(ε0 + ε)2 d (ε0 + ε)2 d
2(ε0 + ε)d
70
Classical Electromagnetic Theory
and the y component is
Fy =
ε0 (ε − ε0 )
∂W
=
ΔxV 2
∂y
2(ε0 + ε)d
7-12 The electric field between the cylinders takes the form
ΔV
Va − Vb
≡
r ln(a/b)
r ln(b/a)
E=
so that, assuming that oil fills the capacitor of length to height z, the potential energy may be written
W =
π(ΔV )2
[εz + ε0 ( − z)]
ln(b/a)
The vertical electrical force on the oil may be equated to the gravitational
force
(ε − ε0 )π(ΔV )2
= ρgzπ(b2 − a2 )
Fz =
ln(b/a)
to obtain the height to which the oil rises:
z=
(ε − ε0 )(ΔV )2
ρg(b2 − a2 ) ln(b/a)
7-13 We start this problem by recognizing that any line charge λ at distance d from
the plane dielectric interface has an image λ = (ε0 − ε1 )/(ε0 + ε1 )λ ≡ Kλ
a distance d from the interface inside the dielectric. Assuming |K| is small
a rapidly diminishing series of image line charges will establish a constant
potential on the cylinder surface. Let us assume the cylinder center is at
distance D from the interface. As a first approximation we place a line charge
λ (not the total charge per length on the cylinder which will be the sum of all
the line charges in the cylinder) at the center which we denote by x = 0. An
image charge λ = Kλ at x1 = 2D is required. This image must be balanced
by an image −Kλ placed at x1 = a2 /(2D). This charge in turn produces an
image −K 2 λ in the plane at x2 = 2D − 2a2 /2D. This image in turn yields an
image K 2 λ inside the cylinder located at x2 = 2a2 D/(4D2 − a2 ). We iterate
once more to get an image K 3 λ in the dielectric at x3 = 2D − 2a2 D/(4D2 −
a2 ) whose image −K 3 λ inside the cylinder lies at x3 = a2 /x3 = a2 (4D2 −
a2 )/[2D(4D2 − 3a2 )]. This process could be continued ad infinitum although
the location of images hardly changes after the first few iterations. The charge
density on he cylinder is λ(1 − K + K 2 − K 3 + · · ·) = λ/(1 + K) = λ0 meaning
the total image charge is λ0 K. We would make little error by approximating
the charges inside the cylinder by a central line charge density λ and the
remaining −Kλ0 at x1 = a2 /2D. The image line charge in the dielectric may
Chapter Seven Solutions
71
be approximated at Kλ0 located at x1 = 2D. The potential just outside the
surface of the cylinder is then
−λ0 "
ln r − K ln r2 + x21 − 2rx1 cos ϕ
V (r ) =
2πε0
+K ln r2 + (2D)2 − 4rD cos ϕ
whence
K(a − x1 cos ϕ)
−λ0 1
K(a − 2D cos ϕ)
−
Er (a) =
+
2πε0 a
a2 + (2D)2 − 4Da cos ϕ
a2 + x21 − 2ax1 cos ϕ
σ
=
ε0
If higher accuracy is desired, it suffices to carry more terms of the series as
outlined. The term for the central charge in relation to the total charge is
correct to all orders
7-14 The field strength between the faces of the magnetron magnet is
1
H=−
HP M d
L PM
where L is the length of the gap. Assuming that inserting the screwdriver
into the gap does not significantly perturb the field internal to the magnet,
we find the energy of the field increased by
2
(μ − μ0 ) PM HP M d A r
W =
L
when the screwdriver is inserted to a distance r. The radial force is then
2
A(μ − μ0 ) PM HP M d
∂W
=
Fr = −
∂r
L
In reality , of course, the field in the magnet will be affected by the change of
reluctance as the screwdriver is inserted.
7-15 Several different methods may be used to solve this problem. We will explore
three taking the z axis along the magnetization. For the first method we use
equation (7–110) and proceed much like the example 7.10.
Vm = −
1 ∇·
4π
τ
M0 k̂d3 r
|r − r |
To effect the integration, we expand |r − r |−1 as a Legendre series:
1
1 ! r<
=
P (cos θ)
|r − r |
r>
r>
72
Classical Electromagnetic Theory
All the = 0 terms vanish when integrated over the solid angle, leaving only
2 r dr dΩ
M0 ∂
Vm = −
4π ∂z
r>
2 r dr
∂
= −M0
∂z
r>
r must always be less than a, the radius of the sphere. Likewise, when r lies
outside the sphere it is always r> , leading to
a 2 ∂
r dr
∂ a3
= −M0
Vm (r > a) = −M0
∂z 0
r
∂z 3r
3
M0 a z
=
3r3
Differentiating, we obtain
r
k̂
3 z
B(r > a) = −μ0 ∇Vm = μ0 M0 a
− 3
r5
3r
When r lies inside the sphere, some care is required. r is r< from 0 until
it reaches r whereupon it becomes r> for the rest of the integration. Thus
r 2 a 2 ∂
r dr
r dr
+
Vm (r < a) = −M0
∂z
r
r
0
r
2
2
2
r
a
∂ r
= −M0
−
+
= 13 M0 z
∂z 3
2
2
The induction field this time is given by
= μ0 (H
+M
) = μ0 (−∇V
m+M
)
B
= μ0 (− 13 M0 k̂ + M0 k̂) = 23 μ0 M0 k̂
An alternative approach is to replace the magnetization by a body current
×M
and a surface current j = M
× n̂. The body current vanishes
Jm = ∇
leaving only the surface current
× n̂ = M k̂ × r̂ = M sin θ ϕ̂
M
Comparing this current to that on the rotating charged sphere (Example 5.10)
× n̂ with σaω and adapt the solution immediately.
we identify M
We could also have treated this problem as a boundary condition problem.
The scalar potential takes the form
!
A r P (cos θ)
Vm (r < a) =
! B
Vm (r > a) =
P (cos θ)
r+1
Chapter Seven Solutions
73
From this we find
= −∇V
m
H
and
⎧ !
⎨
− r−1 A P (cos θ)
! ( + 1)B
Hr =
⎩
P (cos θ)
r+2
⇒
⎧ !
∂P (cos θ)
⎪
⎨
−A r−1
∂θ
Hθ =
! B ∂P (cos
θ)
⎪
⎩
− +2
r
∂θ
(r < a)
(r > a)
(r < a)
(r > a)
= μ0 (H
+M
),
The magnetic induction field may generally be found as B
whence we obtain the radial (perpendicular) component of B
⎧
!
⎨ μ0
−A r−1 P (cos θ) + M cos θ
(r < a)
! B ( + 1)
Br =
⎩ μ0
P (cos θ)
(r > a)
r+2
We equate the interior and exterior forms of Br at the boundary
−A1 + M =
2B1
a3
and
A a−1 =
( + 1)B
a+2
when = 1
The requirement of continuity of Hθ at the boundary gives
A1 =
B1
a3
and
A a−1 =
B
a+2
when = 1
Solving the = 1 equation gives us A1 = 13 M while the = 1 equations give
+1
B
B
= 2+1
A =
2+1
a
a
having no nonzero solutions. The scalar magnetic potential may now be
written
Vm (r < a) = 13 M r cos θ = 13 M z
Vm (r > a) =
a3
M cos θ
3r2
is performed as in the first solution.
The evaluation of B
7-16 We write the general solution for the scalar potential in spherical polar coor 0 at large distances.
dinates, including explicitly the term to give H
!
Vm (r ≤ R) =
A r P (cos θ)
! B
Vm (r > R) = −H0 r cos θ +
P (cos θ)
r+1
74
Classical Electromagnetic Theory
The boundary conditions to be applied to this solution are
Br (R+ ) = Br (R− ) and Hθ (R+ ) = Hθ (R− )
The first of these yields
!
A r−1 P (cos θ) =
! −( + 1)B
r+2
P (cos θ) − H0 cos θ − M cos θ
while the second yields
! B
!
P (cos θ)
A r−1 P (cos θ) = −H0 P1 (cos θ) +
r+2 The = 1 terms yield
A1 = −H0 +
B1
R3
andA1 = −H0 − M −
2B1
R3
which we solve to obtain B1 = − 13 M R3 and A1 = −H0 − 13 M . The = 1
terms must vanish to give the following interior and exterior potentials
Vm (r ≤ R) = −(H0 + 13 M )r cos θ = −(H0 + 13 M )z
M R3
Vm (r > R) = − H0 +
z
3r3
The magnetic induction field is now easily obtained:
≤ R) = μ0 (H0 + 1 M )k̂
B(r
3
M R3
μ0 M R3 zr
B(r > R) = μ0 H0 + M +
k̂
−
3r3
r5
7-17 Adapting the results for the magnetized cylinder to this problem by letting
a → ∞, we find Vm = M z inside the slab where we have taken M to lie along
. Hence B
= μ0 (H
+M
)=0
= −∇V
m = −M
the z axis. We deduce that H
inside the slab. That B = 0 outside the slab is easily demonstrated by
calculating the field of an infinite sheet of dipoles as follows:
∞ M · (r − r ) M · (z k̂ − ρr̂)
= −μ0 ∇
B
∇
dS
=
−μ
2πρ dρ = 0
0
2 + ρ2 )3/2
4π|r − r |3
4π(z
0
7-18 The contribution from the face centered at 12 b to the scalar potential at z
assuming that M is z directed is
M · dS
M a
2πr dr
1
=
−
Vm (z) =
4π
|r − r |
4π 0
r2 + (z − 12 b)2
a
M
M 2
r2 + (z − 12 b)2 = −
a + (z − 12 b)2 − (z − 12 b)2
=−
2
2
0
Similarly the face at − 12 b contributes
Chapter Seven Solutions
75
Vm (z) =
M 2
a + (z + 12 b)2 − (z + 12 b)
2
to the scalar potential. Adding the two terms, we obtain
M 2
Vm (z) =
a + (z + 12 b)2 − a2 + (z − 12 b)2 − 2z
2
0 , implying that
7-19 At large distances the magnetic field intensity must tend to H
Vm → −H0 r cos ϕ. Thus
Vm (r > a) = −H0 r cos ϕ +
and
Vm (r < a) =
!
! Bn
rn
cos nϕ
An rn cos nϕ
The boundary conditions require the continuity of Br = −μ∂Vm /∂r and
Hϕ = −(1/r)∂Vm /∂ϕ at r = a. Applying the first of these,
! nBn
!
n−1
μ0 H0 cos ϕ +
cos
nϕ
=
μ
nA
r
cos
nϕ
n
n+1
r
r=a
r=a
which gives
B1
−μ0 H0 + 2 = μA1
a
and
−
Bn
μ
=
An
a2n
μ0
(n = 1)
The continuity of Hϕ gives
−H0 +
B1
= A1
a2
and
Bn
= An
a2n
(n = 1)
The n = 1 equations admit only the trivial solution when μ = μ0 while the n
= 1 equations give
(μ − μ0 )A1 = −
or
B1 =
2μB1
a2
and
μ − μ0
H0 a2
μ + μ0
(μ − μ0 )H0 = (μ + μ0 )
and A1 =
B1
a2
−2μ0
H0
μ + μ0
0
Thus, for an x -directed field B
V (r < a) = −
2μ0
2B0 x
H0 r cos ϕ = −
μ + μ0
μ + μ0
and
V (r > a) = −
B0
B0
μ − μ0 a2
μ − μ0 a2
cos ϕ = −
x
r cos ϕ −
x−
μ0
μ + μ0 r
μ0
μ + μ0 r2
76
Classical Electromagnetic Theory
The fields are now easily computed. Interior to the cylinder,
< a) = −μ∇V
m=
B(r
and
> a) = B
0 − μ − μ0
B(r
μ + μ0
2μ B0
μ + μ0
0
0 · r)r a2 B
2a2 (B
−
r2
r4
d3 r must lie along the symmetry axis which
7-20 The magnetic dipole moment M
we take to be the z axis. The magnetic dipole moment is therefore simply
M τ , where τ is the volume of the magnet.
7-21 The magnet at perpendicular distance z from the interface induces an image
− 2Mz k̂) inside the iron. Following the method
dipole −(1 − 2 × 10−4 )V (M
of problem 6-4, we find the force on the dipole to be
Fz =
−3μ0 (m2 + m2z )
64πz 4
We specialize this result to the dipole being parallel to the plate, mz = 0, Fz =
−3μ0 (M V )2 /(64πz 4 ), or perpendicular, mz = m, Fz = −3μ0 (M V )2 /(32πz 2 ).
The force is directed towards the iron surface.
7-22 We consider the magnetic circuit constituted by the yoke, the airspace and
the magnet. For the entire circuit
· d = 0 ⇒ HP M P M = −
· d
H
H
rest
· d = Φ, with
H
Further,
rest
=
d
20 cm
1 cm
25, 020 cm−1
=
+
=
μA
25, 000 μ0 cm2
μ0 cm2
25, 000 μ0
so that the flux may be found in terms of the PM field,
Φ=
−HP M × 10cm
= −9.992μ0 cm2 HP M
(25, 020/25, 000)cm
Φ is assumed constant around the circuit so that we may relate it the magnetic induction field in the magnet to obtain a relation between BP M and
HP M : BP M = ΦA = −9.992μ0 HP M . Solving this simultaneously with the
hysteresis curve relation between B and H, we obtain BP M ∼ 0.54T and
HP M ∼ −0.054T/4π ×10−7 = −4.3×104 A/m. Because the cross section does
not vary, B is constant around the loop, so that Hair = B/μ0 = 4.3×105 A/m
and Hyoke = B/(25, 000μ0 ) = 17.2A/m.
Chapter Seven Solutions
77
= ε0 E
+ P . Adding the polariza7-23 The displacement field is generally given D
tion to ε0 E as given in (Ex 7.7.1) gives
P
z
L−z
D(z)
=
+
2 (z 2 + a2 )1/2
[(L − z)2 + a2 ]1/2
We translate the origin to z = 12 L so that z, the distance to the bottom face
is replaced by z + 12 L and (L − z) is replaced by 12 L − z to obtain
1
1
P
2L + z
2L − z
D(z) =
+ 1
2 [( 12 L + z)2 + a2 ]3/2
[( 2 L − z)2 + a2 ]3/2
Comparison to the result of Exercise 7.4 shows that replacing P by μ0 M
makes the results identical.
7-24 This problem is very similar to Example 7.10 (or 7.12) and we proceed in the
same way. Thus,
M0 x ı̂ 3 1 Vm (r ) = − ∇ ·
d r
4π
r − r |
τ |
x
M0 ∂
d3 r =−
4π ∂x τ |r − r |
where the volume τ includes the boundary. We again resort to spherical
harmonics
to effect the integration. x = r sin θ cos ϕ may be written as
−1 1 x = 2r 8π
r − r |
3 Y1 (θ , ϕ ) − Y1 (θ , ϕ ) and we use (F–47) to expand 1/|
resulting in
&
8π 4π
2M0 ∂ ! m
Vm (r ) = −
Y (θ, ϕ)
4π ∂x
3 2 + 1
,m
×
r<
Y1−1 (Ω ) − Y11 (Ω ) Y∗m (Ω )r r2 dr dΩ
+1
r>
The integration over Ω eliminates all but the = 1, m = ±1 terms each of
which integrates to unity.
M0 ∂ x a r< 3 Vm (r ) = −
2 r dr
3 ∂x r 0 r>
When r > a, the scalar magnetic potential is
M0 ∂ x a r 3 M0 a5 ∂ x
M0 a5 1
3x2 Vm (r ) = −
r
dr
=
−
=
−
−
3 ∂x r 0 r2
15 ∂x r3
15
r3
r5
When r < a we need to break the integral into two intervals.
r a
r 3 r 3 M0 ∂ x
Vm (r ) = −
r
dr
+
r
dr
2
3 ∂x r 0 r2
r r
78
Classical Electromagnetic Theory
xa2
xr2 M0 2
M0 ∂ xr2
+
−
=−
5a − 3(3x2 + y 2 + z 2 )
3 ∂x 5
2
2
30
m
= −μ0 ∇V
The magnetic induction field may be found as B
=−
5
2 > a) = μ0 M0 a − r − 3xı̂ + 15x r
B(r
15
r5
r5
r7
Similarly, the interior induction field is
< a) = − μ0 M0 3xı̂ + yĵ + z k̂
B(r
5
Alternatively, we might use the methods of Example 7.12. The divergence
of the magnetization ∇·M
0 xı̂ = M0 which is easily integrated over the volume
to give the first of the two integrals in (7–114).
a
M0
1 2 1
2 r
dr
dΩ
=
−M
r dr
−
0
4π τ |r − r |
0 r>
where the volume τ does not include the boundary.
Vm (r > a) = −
M0 a3
3r
Vm (r < a) = −M0
a2
2
−
r2 6
= sin θ cos ϕ a2 dΩ
The remaining surface integral may be evaluated using ı̂·dS
and on the surface x = a sin θ cos ϕ so that
1
4π
· dS
M
M0
a sin2 θ cos2 ϕ a2 dΩ
=
|r − r |
4π
|r − r |
! Ym (θ, ϕ) Y∗m (θ , ϕ ) sin2 θ cos2 ϕ r dΩ
<
= M0 a
+1
2 + 1
r>
3
,m
sin θ cos ϕ may be written as 14 sin2 θ(eiϕ +e−iϕ )2 = 14 sin2 θ(e2iϕ +e−2iϕ +2)
1
4π 0 4π 0
π −2
π −2
2
2
2
Y
+
Y
−
Y
sin
θ
=
−
Y
=
2
2 −
2
2
30
2
30
45 Y2 +
9 Y0 which
allows us to eliminate the terms other than = 2, m = 0, ±2 and = 0, m = 0
from the sum. the sum above, integrated over Ω then becomes
√
√
√
& √ −2
2
π
3Y2 (θ, ϕ) − 3Y22 (θ, ϕ) − 8Y20 r<
40Y00 (θ, ϕ)
3
M0 a
3 +
90
5
r>
r>
2
2
=
2
r2 M0 a3 r<
M0 a3 1
sin2 θ cos2 ϕ +
− <3
3
5 r>
3
r>
5r>
When r > a, meaning r< = a, the scalar magnetic potential, using both the
volume and surface integral becomes
Vm (r > a) =
M0 a5
(3x2 − r2 )
15r5
Chapter Seven Solutions
79
The interior (r> = a) magnetic scalar potential becomes
a2
r2 M0 x2
M0 2 r 2 −
+
Vm (r < a) = −M0
+
a −
2
6
5
3
5
M0
(9x2 + 3y 2 + 3z 2 − 5a2 )
=
30
7-25 Following the hint, we compute
*
"
#+
= (ε + δε) E
+ δE
D
+ δεδ E
= εE
vanish. To a zeroth approximation, the permittivity of the
as δε and δ E
mixture is just the average of the permittivities. To obtain the first order
Consider ∇
·D
= 0,
correction we must evaluate δεδ E.
"
#
+ δ E)
=∇
· (εδ E
+ δεE
· ε + δε)(E
∇
"
#
· δ E)
+ E
·∇
δε = 0
= ε(∇
In other words,
ε∂k δE k = −E k ∂k δε
We differentiate once more with respect to xj to get
ε∂ j ∂k δE k = −E k ∂ j ∂k δε
· E)
= ∇2 E j so that
If the medium is isotropic, ∂ j ∂k ε = 13 δkj ∇2 ε while ∂ j (∇
we rewrite the equation as
E j 2
∇ δε
3ε
E j δε
δE j = −
3ε
∇2 δE j = −
⇒
Multiplying both sides by δε and averaging, we get the required correction to
Thus
D.
E j (δε)2 δE j δε = −
3ε
we have
Substitute this into the expression for D
(δε)2 D = ε −
E 3ε
From this expression we read immediately
εmix = ε −
(δε)2
3ε
To show that this is the desired result, we expand
80
Classical Electromagnetic Theory
"
ε + δε
#1/3
= ε1/3 + 13 ε−2/3 (δε) −
1
−5/3
(δε)2
9 ε
+ ···
We take the average of both sides to get
*
+
1/3
(ε + δε)
= ε1/3 − 19 ε−5/3 (δε)2 + · · ·
(δε)2
+
·
·
·
= ε1/3 1 −
9ε2
Finally we cube both sides to obtain
*
+3
(δε)2
1/3
(ε + δε)
= ε 1 −
+ ···
3ε2
completing the demonstration.
7-26 Equation (7–54) readily reduces to the diffusion equation
∂B
∂t
when the conductor is stationary. This problem is better described as an
initial value problem than a boundary condition problem. Let us look at the
were a scalar (it is not of course) the
form of the solutions to this equation. If B
separable solutions of the form R(r)f (t) would have spherical Bessel functions
R(r) = j0 (kr) = (sin kr)/(kr). B would then satisfy ∇2 B = −k 2 B so that
the temporal function f (t) satisfies
= gμ
∇2 B
df
k2
=− f
dt
gμ
⇒
f = e−(k
2
/gμ)t
The characteristic decay time for a disturbance of length k −1 is gμ/k 2 . Before
it is objected that B is a vector field and all this is irrelevant, we recall that the
of the scalar solution solves the vector equation so
gradient (as well as r × ∇)
that conclusions about the temporal behavior are valid. Continuing therefore
with the scalar solution, we can synthesize an arbitrary B in the sphere by
superimposing solutions with increasing k. The lowest k (slowest decay) is
k = π/R. Thus, τ = gμR2 /π 2 is the decay time for global fields. For a
copper sphere of 1 m radius this implies a decay time of
5.9 × 107 × 4π × 10−7
= 7.5s
π2
For the hypothetical earth (R = 3 × 106 m for the core),
τ=
106 × 4π × 10−7 × 9 × 1012
≈ 1.15 × 1012 s ≈ 3.6 × 104 y
π2
These considerations would suggest that it would be very difficult to produce
large-scale changes in the earth’s magnetic field in times less than 104 years.
τ=
Errata: Ex 7.8.3 on page 187 has a misplaced superscript ( + 2) which should
be in the denominator R of the last term.
On page 188, following Ex 7.8.6 the phrase should read: which, when solved
for A1 , give A1 = −E0 − 13 P/ε0 .
Chapter 8
8-1
of a plane wave satisfies (8–123)
Generally, in a conductor, the wave vector K
ig
K 2 = μεω 2 1 +
εω
A good conductor is characterized by g εω, so that to a good approximation
√
2
2 ig
K = μω
⇒ K = μgωeiπ/4
ω
in terms of E
using (8–122), H
0 = (K
× E)/(ωμ)
Writing H
it is evident that
◦
H and E oscillate 45 out of phase.
8-2
The skin depth δ is given by
,
2 βε
with
δ=
g 2μ
,
β =1+
1+
g
εω
2
Putting the explicit values in we find
4.3
g
=
= 9.7 × 106 1
εω
80 × 8.85 × 10−12 × 2π × 100
so that
&
δ=
2
=
gωμ0
&
2
= 24.3 m
4.3 × 4π × 10−7 × 2π × 102
Such ELF waves would be useful only if the submarine was within a few
meters of the water surface.
8-3
A spherical membrane with surface tension σ and radius of curvature R exerts
inward pressure ℘ = σ/2R per surface. The outward pressure from blackbody
radiation is 13 U = 43 σS-B T 4 /c.
Putting in numeric values as far as possible, the interior pressure supported
by two surfaces is
⇒
4 × 5.67 × 10−8 × (300)4 N
σ
=
= 0.51 × 10−6 Pascal
R
3 × 3 × 108
m2
σ
R =
= 1.96 × 106 σ
0.51 × 10−6
× B,
whereas the
8-4 According to (8–39) the momentum density is given by D
Poynting vector S = E × H according to (8–36). In the dielectric,
×B
= εE
× B = ε S
D
μ
μ
— 81—
82
Classical Electromagnetic Theory
8-5 Forces arise from two different sources—from the change of momentum of the
reflected light and from the change of momentum of the transmitted light.
The reflected power is just [(1 − n)/(1 + n)]2 dW/dt, with attendant change
in momentum flux
2
2 1 − n dW
Fr =
c 1+n
dt
The portion of the beam that will be transmitted has energy flux (4n)/(1 +
n)2 dW/dt and carries (in air) momentum flux
4n(dW/dt)
c(1 + n)2
toward the interface while it carries momentum at a rate
4n(dW/dt)
v(1 + n)2
away from the interface. The net change in momentum of the light beam per
unit time is therefore
4n 4n
2
(1 − n)2 +
−
c
c
v dW
Fr =
2
(1 + n)
dt
=
2 − 4n + 2n2 + 4n − 4n2 dW
(1 + n)2 c
dt
=
2(1 − n2 ) dW
(1 + n)2 c dt
The corresponding force on the dielectric is just minus this result.
It is instructive to obtain this result from the Maxwell stress tensor using
dFz = −Tzz dSz with
·D
+B
· H)
= 1 (E
·D
+B
· H)
Tzz = −Ez Dz − Bz Hz + 12 (E
2
for a z directed transverse wave. In accordance with the discussion on page
84, the force from a parallel field is directed toward the wall.
On the incident side,
1−n
2
Ei
Etot = Ei + Er = 1 +
Ei =
1+n
n+1
and
&
Htot =
ε0
(Ei − Er ) =
μ0
&
&
ε0
ε0 2n
1−n
Ei
1−
=
μ0
1+n
μ0 1 + n
leading to
Tzz
·D
+B
· H)
= 1 ε0
= 12 (E
2
2
n+1
2
ε0
+ μ0
μ0
2n
n+1
2 Ei2 Chapter Eight Solutions
=
83
2(1 + n2 )
2(1 + n2 ) d2 W
ε0 cEi2 =
2
c(n + 1)
c(n, +1)2 dAdt
In the same fashion, on the dielectric side of the interface
&
ε 2
2
Ei
Ei
Ht =
Et =
n+1
μ0 n + 1
whence
1 2 E
4ε
4ε
+
Ei2 (n + 1)2
(n + 1)2
d2 W
4n2
4n2
ε0 cEi2 =
=
2
2
c(n + 1)
c(n + 1) dAdt
+B
· H
=
·D
1
2
The force on the dielectric is then
F =
2
2(n2 − 1) dW
2
2 dW
=
[2n
−
(1
+
n
)]
c(n + 1)2
dt
c(n + 1)2 dt
8-6 The decay constant is 1/α, with
&
sin2 θi
2π
2
2 =
α = kt
−
1
=
k
sin
θ
−
n
sin2 60◦ − (1/1.7)2
i
i
n2
λ
3.993
=
λ
Thus the decay distance is nearly 14 λ.
8-7 Using the formulas for the half angles, the phase angles are easily computed.
sin2 θi − n2
tan 12 ϕs =
= 1.27
⇒ ϕs = 103.6◦
cos θi
tan 12 ϕp = (1.7)2 × 1.27 = 3.67
⇒
ϕp = 149.52◦
8-8 An EM plane wave propagating through a conducting medium satisfies
ig
−k 2 + μεω 2 1 +
=0
ωε
Taking the view that the plasma is a conducting rather than a polarizable
medium, we set ε = ε0 and using
mv˙ = q E
⇒
v =
iq E
mω
⇒
inq 2 J = nqv =
E
mω
we get the conductivity g = inq 2 /mω. We substitute this into the dispersion
relation above to find
nq 2
k 2 = με0 ω 2 1 −
mω 2 ε0
84
Classical Electromagnetic Theory
With the identification ωp2 = nq 2 /(mε0 ) this becomes
ωp2
ω2
k = 2 1− 2
c
ω
2
8-9 Newton’s equation of motion for a charge e in an axial field B and an electric
(in the x-y plane) is
field E
−iωt + eẏBı̂ − eẋBĵ
mr̈ = eEe
or
eEx −iωt
eEx −iωt
eB
ẏ +
e
e
≡ ωc ẏ +
m
m
m
eEy −iωt
eB
eEy −iωt
ẋ +
e
e
ÿ = −
≡ −ωc ẋ +
m
m
m
We untangle these equations by taking (complex) linear combinations of x
and y.
e(Ex + iEy ) −iωt
e
ẍ + iÿ = ωc (ẏ − iẋ) +
m
e(Ex − iEy ) −iωt
e
ẍ − iÿ = ωc (ẏ + iẋ) +
m
Now assuming a harmonic time dependence with angular frequency ω for each
of the components, we write
ẍ =
e(Ex + iEy )
eE+
= −ωωc (x + iy) +
m
m
e(Ex − iEy )
eE−
2
= ωωc (x − iy) +
−ω (x − iy) = ωc (−iω)(y + ix) +
m
m
− ω 2 (x + iy) = ωc (−iω)(y − ix) +
Grouping terms we find
−(ω 2 ± ωωc )(x ± iy) =
eE±
m
⇒
x ± iy = −
eE±
m(ω 2 ± ωωc )
The corresponding components of the polarization of the medium are
P± ≡ Px ± iPy = nex ± iy =
from which we deduce that
ε± = ε 0
−ωp2 ε0 E±
−ne2 E±
=
m(ω 2 ± ωωc )
(ω 2 ± ωωc )
ωp2
1− 2
(ω ± ωωc )
The dispersion relation k 2 = ω 2 μ0 ε becomes
ωp2
ω2
2
k±
= 2 1− 2
c
ω ± ωωc
Chapter Eight Solutions
85
A right hand circularly polarized wave propagating in the +z direction has
electric field components E− = E0 and E+ = 0 and therefore propagates with
k− while a left hand circularly polarized wave propagates with k+ .
8-10 The total current consists of both the real current and the bound current arising from the changing polarization of the medium. Thus the current traversing
a cross sectional area A is
∂ P 0 (g cos ωt − ωχε0 sin ωt) · A
I = J +
·A=E
∂t
#
VA "
g cos ωt − (ε − ε0 )ω sin ωt
=
d
The amplitude of the current is
1/2
VA 2
g + (ε − ε0 )2
d
1/2
VA = −2 4.32 + (3.14 × 109 × 79 × 8.85 × 10−12 )2
10 m
= 482.8 × VA (A/V m)
|I| =
8-11 The permittivities may be computed from the principal refractive indices,
εxx = 1.69ε0 , εyy = 2.25ε0 and εzz = 2.89ε0 . The electric displacement vector
associated with E
is then D
= ε0 (0.976ı̂, 1.26ĵ, 1.674k̂). The Poynting
D
vector is now easily computed, except for a numerical constant
×H
∝ (ı̂, ĵ, −2k̂)
S = E
The direction of the wave vector k may similarly be found
k = D
×B
∝ (ı̂, ĵ, −1.36k̂)
The non-colinearity of the Poynting and wave vector is evident.
8-12 The total power incident on the interface where it has footprint S is
i × H
i) · S
= S(E
i × H
i ) cos θi ∝ ni S cos θi E 2
Pi = (E
i
On the other side of the interface, the footprint is the same, but the cross
section of the beam is different,
t × H
t) · S
∝ nt S cos θt Et2
Pt = (E
The ratio of transmitted to incident power is therefore
nt cos θt 2
Pt
=
t
Pi
ni cos θi
It is readily verified that r2 + Pt /Pi = 1.
86
Classical Electromagnetic Theory
8-13 If the expansion occurred adiabatically the expanded volume must have the
same energy contained within it as did the unexpanded. The radiation does
work as it is expanding and we use the adiabatic expansion law P V γ = constant with γ = 43 and noting that V ∝ L3 and P ∝ T 4 , we have T 4 L4 =
constant. Thus
Lf
Ti
=
= 366.3
Li
Tf
8-14 The ratio of reflected to incident power is given (for a good conductor)
&
E0,r 2
= 1 − 2 2ωε0 = 0.9898
E0,i g
The transmitted and reflected field “amplitudes” may be found from
E0,t =
2
E0,i
1+η
and E0,r =
1−η
E0,i
1+η
with η given by
&
η=
g
(1 + i) = 196 + 196i
2ωε0
Substituting these values we find
E0,t = 0.0072e−0.7828i E0,i
and
E0,r = −(0.9949 + .005i)E0,i
yielding phase shifts of 44.85◦ for the transmitted wave and 179.72◦ for the
reflected wave. The reflection coefficient found earlier is easily verified from
the amplitude of the reflected wave.
Chapter 9
9-1 TM modes are characterized by Bz = 0. It remains, therefore, to solve for Ez
from
∇2t Ez + (μεω 2 − k 2 )Ez = 0
The solutions are of the form
cos αx
cos βy
Ez =
sin αx
sin βy
Applying the boundary conditions: Ez = 0 at x = 0, x = a, y = 0 and y =
b the solution reduces to
mπy
nπx
sin
Ez = A sin
a
b
If this is substituted into the wave equation (above) we find immediately the
relevant dispersion relation
2 2
mπ
nπ
+
+ k 2 − μεω 2 = 0
a
b
The transverse components of the magnetic field intensity Ht are found from
(9–23)
t = iεω k̂ × ∇t Ez
H
μεω 2 − k 2
nπ
nπx
mπy mπ
nπx
mπy
iεωA
ı̂ cos
sin
+
ĵ sin
cos
=
k̂ ×
μεω 2 − k 2
a
a
b
b
a
b
nπ
nπx
mπy mπ
nπx
mπy
iεωA
ĵ cos
sin
−
ı̂ sin
cos
=
μεω 2 − k 2 a
a
b
b
a
b
9-2 TE modes satisfy
(∇2 + γ 2 )H0,z = 0
with boundary conditions H0,z (0) = H0,z (c) = 0 and ∂H0,z /∂n = 0 at the
side walls, leading to solutions of the form
H0,z = H0 sin
nπx
mπy
πz
cos
cos
c
a
b
whence we find the transverse fields (page 254)
0,t = − πH0 cos πz ı̂ nπ sin nπx cos mπy + ĵ mπ cos nπx sin mπy
H
cγ 2
c
a
a
b
b
a
b
and
0,t = iμωH0 sin πz ĵ nπ nπx cos mπy − ı̂ mπ cos nπx sin mπy
E
γ2
c
a sin a
b
b
a
b
— 87—
88
Classical Electromagnetic Theory
where
γ = μεω −
2
2
π
c
2
=
nπ
a
2
+
mπ
b
2
2
The second equality may be rewritten to give
2
μεω =
π
c
2
or
+
,
1
ω
= √
f=
2π
2 με
nπ
a
n 2
a
2
+
+
mπ
b
m 2
b
2
+
c
The = 0 term yields only the trivial solution so that we must have ≥ 1.
Some of the lower modes have then the mode numbers (n, m, ) = (0, 0, 1),
(1, 0, 1), (0, 1, 1), (0, 0, 2), (2, 0, 1), (0, 2, 1) and (1, 1, 1). It is not possible,
without further information about the dimensions a, b, and c to judge which
of the second to sixth term will be the lowest.
9-3 The fundamental modes of the air-spaced parallel plates are TEM modes.
(Note that the argument forbidding these fails when the enclosure is not
complete.) If we ignore the fringing field, (more accurately we should use the
field distribution illustrated in figure 5.12 to account for the fringing field)
the TEM electric field may be written
r, t) = E0 ı̂ei(kz−ωt)
E(
where we have taken the x axis perpendicular to the plates. The corresponding
magnetic field intensity is
= ∇ × E = E0 ĵ ei(kz−ωt)
H
iμ0 ω
μ0 c
As pointed out these waves are non-dispersive.
The parallel plates also support are also TE and TM modes. The TM
modes satisfy
[∇2t + (μεω 2 − k 2 )]E0 z = 0
with solutions
cos αx
sin βy
E0,z = A
with α2 + β 2 = μεω 2 − k 2
sin αx
cos βy
The boundary condition at the plates, x = 0 and x = a is that Ez must
vanish, leading to
nπx cos βy
E0,z = A sin
sin βy
a
Chapter Nine Solutions
89
At the open sides, y = 0 and y = b, the most obvious boundary condition is
that the surface current must vanish. Writing the surface current in terms of
H we have
= ±ĵ × (H0,x ı̂ + H0,y ĵ) = H0,x k̂
j = n̂ × H
which, in turn, implies that ∂Ez /∂y = 0. The form of the z component of
the electric field is therefore
TM
= A sin
E0,z
mπy
nπx
cos
a
b
TE modes similarly have
H0,z = α cos
nπx
a
sin βy
cos βy
z /∂x, and the fact that this vanishes at y = 0 and at y = b,
Using Hx ∝ ∂ H
we obtain
mπy
nπx
TE
sin
= A cos
H0,z
a
b
9-4 We first calculate kz and then find λ as 2π/kz . To this end we evaluate
ω
2π × 12 × 109 s−1
=
= 80π m−1
c
3 × 108 m/s
so that
kz = π
802 − (0.0228)−2 = 66.9π m−1
From this we conclude that λ = 2π/kz = 2.99 cm.
9-5 Using the results of example 9.3, we have
2
2δ
m2
dP/dz
4
2 2 n
= 2
+
(a + b)γ + π k
P
γ kab
a
b
For the TE1,0 mode, m = 0 and n = 1, meaning γ = π/a and ωc = (cπ/a)
which we use to rewrite the expression above as
2δ (a + b)π 2
dP/dz
2
=
+k
P
kb
a3
where
ω2
π2
− 2
2
c
a
We simplify by assuming that b = 2a, (the usual shape for rectangular waveguides) and compute
√
3π
π
and k(1.05ωc ) = 0.32
k(2ωc ) =
a
a
k2 =
90
Classical Electromagnetic Theory
as well as
and
δ(2ωc ) = 4.89 × 10−6 a1/2 m1/2
δ(1.05ωc ) = 9.314 × 10−6 a1/2 m1/2
Substituting these values we get
dP/dz
= 5.32 × 10−5 a−3/2 m1/2
P
dP/dz
= 2.84 × 10−4 a−3/2 m1/2
P
ω = 2ωc
ω = 1.05ωc
For a typical value of a = 1 cm, these attenuation coefficients reduce to
0.0532/m and 0.283/m respectively.
9-6 TE modes have Ez = 0. For simplicity we restrict ourselves to TE0,m modes
in a rectangle of width 2a so that
$
A sin αx + B cos βx
−a < x < a
H0,z =
|x| > a
Ce−|γx|
We further subdivide the modes into symmetric and antisymmetric modes.
Fixing our attention on the antisymmetric modes we have
E0,y =
−iωμα cos αx
α2
inside the guide
iωμγe−γ|x|
outside the guide
γ2
with the + sign pertaining to positive x. (E0,x = 0 in either case.) Matching
E
and H
at the boundary (x = ±a), we find
and
E0,y = ±C
A sin αa = Ce−γa
e−γa
−A cos αa
=
α
γ
A sin(−αa) = Ceγa
−A cos(−αa)
−eγa
=
α
γ
In either case, αa must be a root of
−
e−|γa|
tan αa
=
αa
γa
where α and γ are related by α2 + γ 2 = (n2 − 1)ω 2 /c2 . The roots may be
determined graphically or numerically.
The symmetric modes are found in the same fashion to obey
−
e−|γa|
cot βa
=
βa
γa
9-7 The general solution the the wave equation in spherical polar coordinates is
given on page 66. For TE modes we have
Br =
1 ( + 1)
j (kr)Ym (θ, ϕ)
c
r
Chapter Nine Solutions
91
Br , Eϕ , and Eθ must each vanish at the conducting wall, which leads us to
conclude that ka must be a root of j . The = 0 solution would have nonzero divergence for B and is therefore eliminated. The first few zeros are r11
= 4.4934, r21 = 5.76349, r31 = 6.987932, r12 = 7.72525, .... The radial B
component is therefore
A ri m
Br = j
r Y (θ, ϕ)
r
a
For TM modes we find Er =
( + 1)
j (kr)Ym (θ, ϕ) and setting Eθ = Eϕ = 0
r
d
(rj (kr))
=0
dr
r=a
⇒
j (ka) + akj (ka) = 0
Again = 0 does not give valid solution. For = 0, solutions must be found
numerically. In particular, when = 1, the first two solutions are ka =
2.74371, and ka = 6.11676.
9-8 For the TM1,0,1 mode,
E0,z = A sin
πz
πx
cos
a
c
and from (9–50, 51)
$
0,t
E
H
0,t
%
⎧
ı̂π 2
πz
πx
⎪
⎨ −
sin
cos
A
ac
c
a
=
εμω 2 − k 2 ⎪
⎩ ĵiωεπ cos πz cos πx
a
c
a
⎫
⎪
⎬
⎪
⎭
The Q of the cavity may be defined as the ω0 × energy stored divided by the
energy lost per second. The resonant frequency is easily found to be
, 2
2
π
π
1
+
ω0 ≡ ω1,0,1 = √
με
a
c
and the energy stored may be computed from
#
"
ε|E|2 + μ|H|2 d3 r
W = 14
(as E 2 = 12 |E 2 | and H 2 = 12 |H 2 |) or
π4
2 πz
2 πx
1 2
sin
W = 4A
ε 4 2 2 cos
γ a c
a
c
2 2 2
ω ε π
2 πz
2 πx
cos
cos
+μ
d3 r
γ 4 a2
c
a
The integration is straightforward and gives
92
Classical Electromagnetic Theory
W = 14 A2 ε
π4
μεω 2 π 2
+
4
2
2
γ a c
γ 4 a2
abc
4
The rate of energy loss by surface currents may be expressed as
μωδ
P=
|H
|2 dS
4 μωδ
=
Hy2 dydz + Hy2 dydz +
4
x=0
x=a
+ Hy2 dxdy Hy2 dxdy
z=0 z=c
μωδ A2 ω 2 ε2 π 2
2 πx
2 πz
dxdy + cos
dydz
=
cos
2 γ 4 a2
a
c
μωδ A2 ω 2 ε2 π 2 ab bc
=
+
4 γ 4 a2
2
2
The Q of the cavity is then
π 2 /c2 + μεω 2 abc
ω0 W
=
=
P
2μεω 2 δ
ab + bc
2a2 + c2
a2 + c2
abc
2δ(a + c)b
π2
π2
where we have used μεω 2 = 2 + 2 . We see that the Q of the cavity
a
c
increases as the volume to surface area.
9-9 A TM mode has Hz = 0 and in a circular waveguide Ez must take the form
⎧
⎪
with γ 2 = με1 ω 2 − k 2 ;
r≤a
⎨ J0 (γr)
E0,z =
⎪
⎩ AK0 (βr) with β 2 = k 2 − με0 ω 2 ;
r≥a
The corresponding transverse fields are
⎫
ik
ik
⎪
⎬
E0,r = − AK0 (βr)
E0,r = J0 (γr)
γ
β
r≤a
iωε1 iωε0
⎪
J (γr) ⎭
AK0 (βr)
H0,ϕ =
H0,ϕ = −
γ 0
β
⎫
⎪
⎬
⎪
⎭
r≥a
At the fiber-air interface, we require that εEr , H ϕ , and Ez be continuous. In
other words,
J0 (γa) = AK0 (βa)
iε1 k iε0 k
J0 (γa) = −
AK0 (βa)
γ
β
iωε0
iωε1 J (γa) = −
AK0 (βa)
γ 0
β
The second and third of these equations are equivalent. Dividing the third
equation by the first, we obtain the characteristic equation (apart from a
common factor a) (9–101)
ε1 J0 (γa) ε0 K0 (βa)
+
=0
γJ0 (γa)
βK0 (βa)
Chapter Nine Solutions
93
9-10 TEM nodes have Ez = 0 and Bz = 0. The wave equation (9–15) then reduces
to
t
t
t
# E
" 2
∂2
E
E
2
2
2
2
∇ + μεω
= ∇t + 2 + εμω
= ∇t
=0
t
t
t
∂z
H
H
H
t and H
t avoid
Just the equation governing the static fields. In order that E
t and H
t have the simple solution
vanishing, we have set εμω 2 = k 2 . E
t = A r̂
E
r
t = εω A ϕ̂
and H
k r
There are also TE and TM modes of propagation with ψ =
Bm Nm (γr)]e±imϕ .
)
[Am Jm (γr) +
9-11 The boundary conditions to these waves require that Eϕ and Eθ vanish at
r = a and r = b. TE modes, having only these components to the field must
have at least half a wave between a and b leading to ω ∼ c/(a − b). For TM
modes, on the other hand, it is possible to have Eϕ and Eθ vanish while Er
varies only slightly over the interval (a, b). In this case we expect ω ∼ c/a.
In terms of the expressions for the tangential electric fields the boundary
conditions are,
1 ∂(rf ) 1
= [f (ka) + ak f (ka)] = 0
r ∂r a
a
and
1 ∂(rf ) 1
= [f (kb) + bk f (kb)] = 0
r ∂r b
b
where f ≡ Aj + Bn .
Writing the expressions out in full and rearranging the terms, we obtain
A j (ka) − Aka j (ka) = −B n (ka) + Bka n (ka)
A j (kb) − Akb j (kb) = −B n (kb) + Bkb n (kb)
Dividing one equation by the other eliminate the expansion constants A and
B to produce the characteristic equations for TM modes
n (ka) − ak n (ka)
j (ka) − ak j (ka)
=−
j (kb) − bk j (kb)
n (kb) − bk n (kb)
The lowest (non-trivial) frequency mode will correspond to = 1; numerical
means must be employed to find the roots of the characteristic equation.
Jackson, on page 376 of Classical Electrodynamics, 3rd ed., gives an alternative
solution by solving the wave equation for Bϕ when this is the only nonvanishing component of B obtaining
c
ω ( + 1)
a
94
Classical Electromagnetic Theory
9-12 The characteristic impedance of the coaxial line carrying a TEM mode is most
easily determined from Z = V /I as given by (Ex 9.1.8):
&
2πV
V
1
μ
I=
⇒ Z=
=
ln(b/a)
vμ ln(b/a)
I
2π ε
Inserting numerical values gives Z = 74 Ω.
9-13 In terms of the non-vanishing longitudinal field Hz , the transverse fields are
given by
t = −iω k̂ × ∇t Hz and H
t = ik ∇t Hz
E
γ2
γ2
t in terms of E
t , we find
Re-expressing H
= |E
× H|
= k |E|2
|S|
μω
We conclude that the impedance Z is given by
&
&
μ2πv/λ0
1 λ
μ λ
μω
=
=μ
=
Z=
k
2π/λ
με λ0
ε λ0
9-14 We begin with wave equation valid for either Hz or Ez either of which may
be denoted by ψe−iωt
∇2t ψ +
∂2ψ
+ μ0 ε0 ω 2 ψ = 0
∂z 2
we separate variables by setting ψ(r, ϕ, z) = R(r)Φ(ϕ)Z(z). Substituting this
into the differential equation we obtain after dividing by ψ
∇2t (RΦ)
RΦ
d2 Z
2
+ μ0 ε0 ω 2 = − dz = λ2
Z
where we have chosen a positive separation constant to ensure that either Z
or its derivative will vanish at the ends. Moreover as the sine or cosine needs
roots at 0 and L we can conclude that λ = nπ/L with n an integer. The
remaining equation in will be separated again this time in polar coordinates.
1 ∂
∂
1 ∂2
2
+ γ RΦ = 0
r
+ 2
r ∂r ∂r
r ∂ϕ2
Where we have abbreviated γ 2 = μ0 ε0 ω 2 −n2 π 2 /L2 . Anticipating the solution
we try Φ = e±imϕ which reduces the equation to Bessel’s equation.
r2
d2 R
dR
+ (γ 2 r2 − m2 )R = 0
+r
2
dr
dr
Chapter Nine Solutions
95
with general solution Rm (r) = Am Jm (γr) + Bm N(γr) we conclude therefore
that ψ has the form
cos(nπz/L)
Jm (γr)
e±imϕ
ψ=
sin(nπz/L)
Nm (γr)
It remains to pick out the solutions that fit the boundary conditions.
TE modes: TE modes must have Hz = 0 at the ends, meaning that if the
ends lie at z = 0 and z = L. The solution must be of the form
nπz ±imϕ
Jm (γr)
e
sin
Hz =
Nm (γr)
L
The magnetic field intensity satisfies dHz /dn = 0 at the side walls meaning
that the (n, m) mode must satisfy
Anm Jm (γR1 ) + Bnm Nm (γR1 ) = 0
and
Anm Jm (γR2 ) + Bnm Nm (γR2 ) = 0
In order that these equation have a solution, we must have
Jm (γR1 )Nm (γR2 ) = Jm (γR2 )Nm (γR1 )
Plotting each product as a function of γ we find roots γm,i at
the intersections.
2
The resonant frequency is then given by ωn,m,i
= (με0 )−1
2 + (nπ/L)2 .
γm,i
TM modes:For TM modes, Ez = ψ. The normal derivative of Ez must
vanish at the end walls meaning that Ez simplifies to
nπz ±imϕ
Jm (γr)
e
Ez =
cos
Nm (γr)
L
At the side walls, Ez = E
= 0, meaning
Anm Jm (γR1 ) + Bnm Nm (γR1 ) = 0
and
Anm Jm (γR2 ) + Bnm Nm (γR2 ) = 0
In order that these equation have a solution, we must have
Jm (γR1 )Nm (γR2 ) = Jm (γR2 )Nm (γR1 )
The roots of this equation which we denote ρm,i give the resonant frequencies
2
ωn,m,i
= (με0 )−1
ρ2m,i + (nπ/L)2 .
9-15 According to the discussion on page 262, the cut-off angular frequency ωc =
√
2.405/(a εclad μclad − εcore μcore ). We replace με by 1/(n2 c2 ) to get
2.405c
=
ωc = 2
a 1/nclad − 1/n2core
2.405 × 3 × 108
≈ 4.80 × 1015 s−1
1
−5 × .03009
2 × 10
The lowest frequency radiation that the fiber can transmit in the TE1,0 mode
is fc = ωc /2π = 7.63 × 1014 Hz.
Chapter 10
10-1 The first step is to compute the quadrupole moment neglecting the time variation.
) i
Qxx =
q (3xi 2 − ri 2 )
2
2
2
2
b
b
b
b
−3
+3
−3
± 14 (a2 + b2 ) = 0
=q 3
2
2
2
2
Similarly Qyy = Qzz = 0. The off-diagonal elements are
!
Qxy =
3q i xi y i = 3qab
and Qxz = Qyz = 0. The terms of the angular distribution (10–77) of power
to be calculated are
xβ Qαβ Qαγ xγ = (xQyx + yQyx )2 = (x2 + y 2 + 2xy)Q2xy
= 9(qab)2 (x2 + y 2 + 2xy)
= 9(qab)2 r2 sin2 θ(1 + 2 cos ϕ sin ϕ)
and
(xβ Qαβ xα )2 = (xQxy y + yQyx x)2
= 36(qab)2 x2 y 2 = 36(qab)2 r4 sin4 θ sin2 ϕ cos2 ϕ
= 9(qab)2 r4 sin4 θ sin2 2ϕ
so that
μ0 ω 6
dP
=
(qab)2 sin2 θ(1 + sin 2ϕ − sin2 θ sin2 2ϕ)
dΩ
128π 2 c3
The total power output is given by (10–81) as
P=
μ0 ω 6
μ0 ω 6 (qab)2
2
2
(Q
+
Q
)
=
xy
yx
1440πc3
80πc3
10-2 Starting with the expression (10–59) on page 280, the power emitted by an
oscillating electric dipole may be written
P =
1 |p̈|2
μ0 ε0 ω 4 p20
=
12πε0 c
4πε0 3c3
The expressions (10–67) and (10–68) may be used to obtain for the magnetic
dipole,
1
0
dPM
μ0 ω 4 m2 sin2 θ
1 k 2 μ0 kωμ0
|m sin θ|2 =
=
dΩ
2μ0 4π 4π
4π
8πc3
— 96—
Chapter Ten Solutions
97
We can integrate this over the complete solid angle (or simply compare to the
electric dipole result) to obtain
PM =
μ0 ω 4 m20
μ0 |m̈|2
=
4π 3c3
4π 3c3
It is worth noting that this result closely parallels that for the electric dipole;
replacing 1/4πε0 by μ0 /4π and p by m converts the electric dipole result to
that for the magnetic dipole.
The electric quadrupole result may be written
¨ αβ Q̈
¨ αβ
ω 6 Qαβ Qαβ
1 Q̈
PQ =
=
4πε0 360c5
4πε0 360c5
A reasonable guess for the equivalent gravitational result is obtained replacing
1/4πε0 by G and Qαβ by the mass quadrupole moment. For cylindrically
symmetric quadrupoles, Qxx = Qyy = − 21 Qzz in the principal axis system.
In this case the single number Qxx suffices. The term Qαβ Qαβ may be written
as Q2xx + Q2yy + Q2zz = 6Q2xx . The formula above then reduces to
P=
¨ 2xx
GQ̈
60c5
¨ 2 /45c5 for the equivalent gravitational result.
Landau gives GQ̈
10-3 The rotating magnet may be thought of as two linear oscillators oscillating
τ
in quadrature. The magnetic dipole moment of the magnet is just m
=M
where τ is the volume. The power emitted is
P = 2 ×
μ0 ω 4 m2
4π 3c3
10-4 The equation of motion for a nucleus with spin angular momentum I and
0 is
magnetic moment m
= γ I in a z -directed magnetic induction field B
dIx
= my B0 = γIy B0
dt
dIy
= −mx Bz = −γIx , Bz
dt
dIz
=0
dt
with solution
mx = m0 cos(ωt + ϕ),
my = m0 sin(ωt + ϕ),
mz = m0
with ω = γB0 . A rotating dipole can always be viewed as two orthogonal
linear oscillating dipoles in quadrature, meaning that the energy is radiated
at a rate
2
μ0 γ 6 B04 ( 12 h̄)2
μ0 ω 4 (γ I)
=
P = 2
4π 3c3
6πc3
For a typical field of 2T and γ = 2.68 × 108 (SI units) this amounts to
2 × 10−50 W. In a realistic situation, one would have not one spin, but the
98
Classical Electromagnetic Theory
appropriate Boltzman fraction of about 1022 spins. In the same 2T field at
room temperature, W/kT = 7 × 10−4 so that the total magnetic moment
would be that of 7 × 1018 nuclei. The power output would be about 10−12 W.
The total magnetic energy of the spins is 2 × 10−7 J leading us to estimate
a relaxation time of order 105 seconds due to radiation. In fact relaxation
due to the induction field would be considerably larger than this, and local
phenomena dominate the relaxation.
10-5 (a) The angular frequency ω of the postulated dipole is 2πc/λ = 3.77 ×
1015 s−1 . Taking the dipole moment as e×(10−10 m) we find the power emitted
by such a rotating dipole to be
P=
μ0 ω 4 p2
= 5.74×, 10−12 W ≈ 3 × 107 eV/s
6πc
The lifetime of such an atom with total energy ∼ 2 eV would be about
2eV/(3 × 107 eV/s) ∼ 10−7 s. (b) A quadrupole of the same dimensions would
typically have Qαβ ∼ 3e × (10−20 m2 ) to give emitted power of order
P
μ0 ω 6
10−7 ω 6 e2 × 10−40
αβ
Q
Q
∼
= 6.8 × 10−17 W
αβ
1440πc3
40c3
approximately a factor of 105 smaller than the power emitted by the dipole.
(c) The magnetic dipole radiates according to
P=
μ0 ω 4 m2
12πc3
Taking the magnetic moment as m ∼ ( 12 e/m)L and taking L = h̄, we find
P ∼ 2 × 10−17 W
about the same as that emitted by the quadrupole.
When the radiation has a wavelength of 50 nm, the frequency will be
ten times as large giving 104 times the power output for the dipoles and 106
times the output for the quadrupole. Since the energy of the emitting state
must initially have been 10 times as large as that required to produce 500
nm radiation, the lifetimes would be decreased by 103 for the dipole and
106 for the quadrupole. It should be clear that for very short wavelength
x-rays, the various multipoles lead to comparable lifetimes. Of course, once
the wavelength becomes of order 0.1 nm, the multipole approximation will
fail.
10-6 The magnetic moment of a single neutron is −1.913 × 5.501 × 10−27 J/T =
−9.657 × 10−27 J/T and the number of neutrons in the star is
N=
2 × 1030 kg
= 1.194 × 1057
1.675 × 10−27 kg
Chapter Ten Solutions
99
The magnetic moment of the star would then be m
= 1.153 × 1031 J/T. The
power emitted by the rotating dipole would be
μ0 ω 4 |m|2
= 3.28 × 1029 ω 4 J s3
6πc3
P=
The energy of rotation of the star is W = 12 Iω 2 = 15 M R2 ω 2 = 4 × 1037 ω 2 Js2
which we abbreviate as aω 2 . Writing P = b ω 4 , we have
dW
bW 2
= −b ω 4 = − 2
dt
a
or
W0 /10
W0
dW
=
W2
t9/10
−
0
b
dt
a2
This is easily integrated to give
9
b
= 2 t9/10
W0
a
⇒
t9/10 =
9a2
bW0
Substituting the numerical values of a, b, and W0 , we obtain
t9/10 =
9 × (4 × 1037 )2
= 1.2 × 104 s
3.3 × 1029 × 3.6 × 1042
This is clearly much faster than we would expect the rotation of the neutron
star to decay. The difficulty lies in the assumption that all the neutrons point
in the same direction. This alignment would impose a huge degeneracy energy
to the star and is clearly unrealistic. If we assume a polarization of 0.1%, the
radiation rate decreases by a factor of 106 and the lifetime increases by a
corresponding factor of 106 , still somewhat short but probably in the right
ballpark.
10-7 To find the radiation resistance, we equate the power P radiated to
For a half wave antenna, kd = π, whence
R=
1 2
2 I0 R.
μ0 c
2P
× 2.438 = 10−7 × 3 × 108 × 2.438 = 73Ω
=
I2
4π
The radiation resistance of the full-wave antenna is similarly found to be
200Ω.
10-8 (a) When kd 1, the supply current I0 = I sin 12 kd ∼ 12 Ikd. The charge on
each side of the wire may be found from
I0 =
dQ
= −iωQ
dt
⇒
Q=
Ikd −iωt
e
−2iω
which leads to a charge per unit length of Q/(2d). The dipole moment of the
two segments assumed to lie along the z axis is then
d/2
0
1
−2Qzdz
2Qzdz
Ikd2 k̂ −iωt
+ k̂
= k̂Qd =
e
p = k̂
d
d
2
−4iω
d/2
0
100
Classical Electromagnetic Theory
This result could have been more easily obtained using (10–44), where the
the dipole moment is related to the current.
(b) The power per solid angle radiated by such an oscillating dipole is
μ0 ω 4 |p|2 sin2 θ
μ0 ω 2 k 2 I 2 d4 sin2 θ
dP
=
=
2
dΩ
32π c
29 π 2 c
(c) We approximate the various terms as follows:
cos( 12 kd cos θ) 1 −
1 1
( kd cos θ)2
2! 2
and
cos 12 kd 1 −
1 1
( kd)2
2! 2
to write
cos( 12 kd cos θ) − cos 12 kd = 12 ( 12 kd)2 (1 − cos2 θ) = 18 k 2 d2 sin2 θ
Equation (10–125) then becomes
2
dP
2μ0 2 18 k 2 d2 sin2 θ
=
I c
dΩ
(4π)2
sin θ
=
μ0 k 4 d4 I 2 c sin2 θ
μ0 k 2 ω 2 I 2 d4 sin2 θ
=
9
2
2 π
29 π 2 c
10-9 The large conducting sheet (the z-y plane) may be maintained at zero potential by an image half wave antenna placed λ/4 behind the conducting sheet
that has, at any instant, the opposite polarity of the real antenna. The current
distribution is therefore given by
r, t) = k̂I sin( 1 kd − k|z|)[δ(x − 1 λ) − δ(x + 1 λ)]δ(y)e−iωt
J(
2
4
4
Following (10–121), we approximate eikR /R in the radiation zone by
eik|r−r |
eikr −ik(r·r )/r
eikr −i(kz cos θ± 1 π sin θ cos ϕ)
2
·e
·e
=
|r − r |
r
r
With this approximation, the vector potential may be written
ikr 0 (r ) = μ0 e k̂ e 12 π sin θ cos ϕ − e− 12 π sin θ cos ϕ
A
4π r
d/2
−d/2
I sin( 12 kd − k|z|)e−ikz
The angular distribution of power becomes
cos2 ( 12 π cos θ)
μ0 I 2 c
dP
2 1
=
sin
(
π
sin
θ
cos
ϕ)
2
dΩ
2π 2
sin2 θ
cos θ
dz Chapter Ten Solutions
101
To obtain the total power emitted we must integrate this expression over the
half sphere above the plane. By numerical integration, we find P = 42.8 I 2 .
We deduce that the radiation resistance is 85.7Ω.
10-10 The angular velocity of the electron about the nucleus may be computed by
setting
1 e2
e2
2
⇒
ω
=
mω 2 r =
4πε0 r2
4πε0 mr3
and the dipole moment is p = er. The power emitted by the two orthogonal
dipoles constituted by the orbiting electron is then
P=
2e6
2 ω4 2
p =
3
4πε0 3c
(4πε0 )3 3c3 m2 r4
If the energy is changed by only a small fraction during each revolution, the
orbit will remain roughly circular and we can equate the rate of energy loss
to the power emitted.
e2 d 1
dW
=
P=−
dt
8πε0 dt r
so that
−
e4
1 dr
=
3
2
2
r dt
12π ε0 c3 m2 r4
or
− r2
e4
dr
=
2
dt
12π ε30 c3 m2
Integrating both sides over the trajectory, from r = r0 to 0, we have
tf
0
e4
−r2 dr =
dt
12π 2 ε20 c3 m2 0
r0
0
−r3 e4 tf
=
3 r0
12π 2 ε20 c3 m2
or
tf =
4π 2 ε20 c3 m2 r03
= 1.5 × 10−11 s
e4
where we have set r0 = 0.52 × 10−10 m.
10-11 The two counter-rotating electrons form an oscillating dipole of magnitude
2er. The power radiated by such a dipole is
P =
ω 4 (2er)2
4πε0 3c3
10-12 The power radiated by a nonrelativistic accelerated particle is according to
(10–149)
q 2 a2
P=
6πε0 c3
102
Classical Electromagnetic Theory
Applying this to the classical hydrogen atom we have
a=
e2
4πε0 me r2
which we substitute into (10-149) to obtain
P=
e6
96π 2 ε30 c3 m2e r4
10-13 Rather than dealing with a varying acceleration for an electron in an elliptical
orbit, we consider the motion as the superposition of orthogonal two dipoles of
amplitude A and B respectively, oscillating in quadrature. The power emitted
two such by harmonically oscillating dipoles is
P=
ω 4 q 2 A2
ω4 q2 B 2
ω 4 q 2 (A2 + B 2 )
+
=
12πε0 c3
12πε0 c3
12πε0 c3
10-14 At maximum,
d
dθ
dP
dΩ
= 0 or
sin2 θ
d
=0
dθ (1 − β cos θ)5
Expanding the derivative, we obtain
2 sin θ cos θ
5β sin2 θ sin θ
−
=0
(1 − β cos θ)5
(1 − β cos θ)6
which may be reduced to
⇒
3β cos2 θ + 2 cos θ − 5β = 0
1 + 15β 2 − 1
cos θ =
3β
When β ≈ 1, we set β = 1 − , leading to cos θ 1 − 14 . But cos θ 1 − 12 θ2 .
We conclude that θ2 12 . Now, using 1 + β 2, we have 1 − β 2 = 1/γ 2 2
so that we find θ2 1/(4γ 2 ) ⇒ θ 1/(2γ). At the angle of maximum power,
θmax = 1/(2γ) the power per solid angle is readily calculated.
−5
dP sin2 (1/2γ)
q 2 β̇ 2
1
−2
∝
(2γ)
1
−
β
1
−
dΩ θmax
(4π)2 ε0 c [1 − β cos(1/2γ)]5
8γ 2
For β ≈ 1, the square bracketed term becomes
−5 −5
5
1
1
1
=
−
1− 2
1− 1− 2
2γ
8γ
8γ 2
16γ 4
Chapter Ten Solutions
103
We retain only the leading term to replace the square bracketed term by
−5
5
−5
[· · · · · ·] =
8γ 2
Gathering terms, we find
5
dP
q 2 β̇ 2
8 1 8
=
γ
dΩ
(4π)2 ε0 c 5 4
10-15 The angular distribution of power in (10–162) goes to zero when the term in
square brackets vanishes. Taking for simplicity the angle ϕ = 0, the angle θ
may be found from
sin2 θ = γ 2 (1 − β cos θ)2 =
(1 − β cos θ)2
1 − β2
we rationalize this and write
(1 − β 2 )(1 − cos2 θ) = (1 − β cos θ)2
1 − cos2 θ − β 2 + β 2 cos2 θ = 1 − 2β cos θ + β 2 cos θ2
which reduces to
cos2 θ − 2β cos θ + β 2 = 0
⇒
cos θ = β
Substituting β for cos θ in first equation, we have
sin2 θ = 1 − β 2 =
1
γ2
⇒
θ=
1
γ
A direction parallel to the velocity has θ = 0 and ϕ = 0, so that the
angular power density becomes
dP q 2 β̇ 2
=
dΩ θ=0
(4π)2 ε0 c(1 − β)3
Substituting 2γ 2 for (1 − β)−1 this becomes
dP q 2 β̇ 2 γ 6
=
dΩ θ=0
2π 2 ε0 c
in agreement with (10–164) except for the numerical factor which would result
from the integration over the beam width.
Chapter 11
11-1 The phase of a wave may be written as
k · r − ωt = (ω/c, k ) · (−ct, r ) = K μ Xμ
The quotient theorem now guarantees that because the phase is a zero order
tensor and Xμ is a first rank tensor, K μ must also be a first rank tensor.
we take K 1 = k and when it is
When k is parallel to the frame velocity V,
1
perpendicular to V we take K = 0. The transformation law for the frequency
is then obtained from that of the 0-component of K μ :
" #
ω
= K 0 = ΓK 0 − βΓK 1
c
this leads to ω = Γω. When k is parallel to V
we
For k perpendicular to V
find
,
ω V
Γω
ω
V
ω 1 − V /c
=Γ
− k =
1−
=
c
c
c
c
c
c 1 + V /c
11-2 Differentiating the scalar Vμ V μ = c2 with respect to the proper time τ , we
find
d(Vμ V μ )
= Aμ V μ + Vμ Aμ = 2Aμ V μ
0=
dτ
11-3 We evaluate the components of f j one-at-a time
f 1 = F 10 J0 + F 11 J1 + F 12 J2 + F 13 J3
Ex
ρc + 0 − Bz Jy + By Jz
=
c
x
= ρEx + (J × B)
In the same fashion
f 2 = F 20J0 + F 21J1 + F 22J2 + F 23J3
Ey
ρc + Bz Jx + 0 − Bx Jz
=
c
y
= ρEy + (J × B)
f 3 = F 30J0 + F 31J1 + F 32J2 + F 33J3
Ez
ρc − By Jx + Bx Jy + 0
=
c
z
= ρEz + (J × B)
11-4 Expression (11-12), ∂μ F μν = μ0 J ν leads to
· E = μ0 J 0 = μ0 ρc = ρ
∂μ F μ,0 = ∂0 F 00 + ∂1 F 10 + ∂2 F 20 + ∂3 F 30 = ∇
c
ε0 c
— 104—
Chapter Eleven Solutions
105
μ0 J 1 = ∂0 F 01 + +∂2 F 21 + ∂3 F 31
∂Bz
∂By
1 ∂Ex
−
+
=− 2
c ∂t
∂y
∂z
∂E
1
x
× B)
x−
= (∇
c2 ∂t
μ0 J 2 = ∂0 F 0,2 + +∂1 F 1,2 + ∂3 F 3,2
∂Bz
∂Bx
1 ∂Ey
+
−
=− 2
c ∂t
∂x
∂z
∂E
1
y
× B)
y−
= (∇
c2 ∂t
The fourth component is found in the same fashion. We similarly interpret
(11-14) to obtain
∂0 F12 + ∂1 F20 + ∂2 F01 = −
∂Ey
∂Ex
1 ∂B
∂Bz
×E
=0
−
+
=−
+∇
c∂t
c∂x
c∂y
c ∂t
z
∂0 F23 + ∂2 F30 + ∂3 F02 = −
∂Ez
∂Ey
1 ∂B
∂Bx
×E
−
+
=−
+∇
=0
c∂t
c∂y
c∂z
c ∂t
x
∂0 F13 + ∂1 F30 + ∂3 F01 =
∂Ez
∂Ex
1 ∂B
∂By
×E
=0
−
+
=
+∇
c∂t
c∂x
c∂z
c ∂t
y
∂1 F23 + ∂2 F31 + ∂3 F12 = −
∂By
∂Bz
∂Bx
·B
=0
−
−
= −∇
∂x
∂y
∂z
Explicitly evaluating the expression implicit in (11–23) and noting that
H μν is just Hμν with rows and columns interchanged, we have
J 0 = ∂0 H 00 + ∂1 H 10 + ∂2 H 20 + ∂3 H 30 = ∂x cDx + ∂y cDy + ∂z cDz
·D
ρc = c∇
J 1 = ∂0 H 01 + ∂1 H 11 + ∂2 H 21 + ∂3 H 31 = −∂t Dx + 0 − ∂y Hz + ∂z Hy
× H)
x
Jx = −∂t Dx + (∇
J 2 = ∂0 H 02 + ∂1 H 12 + ∂2 H 22 + ∂3 H 32 = −∂t Dy + ∂x Hz + 0 + ∂z Hx
× H)
y
Jy = −∂t Dy + (∇
J 3 = ∂0 H 03 + ∂1 H 13 + ∂2 H 23 + ∂3 H 33 = −∂t Dz − ∂x Hy + ∂y Hx + 0
× H)
z
Jz = −∂t Dx + (∇
·D
= ρ and
In summary, the four equations resulting from (11-23) yield ∇
×H
= J + ∂ D/∂t.
∇
106
Classical Electromagnetic Theory
11-5 Using the definition of the dual tensor Gμν we calculate
G01 =
1
2
G02 =
1
2
G03 =
1
2
G23
#
0123 F23 + 0132 F32 =
1
2
(F23 − F32 ) = −Bx = −G10
"
#
0213 F13 + 0231 F31 =
1
2
(−F13 + F31 ) = −By = −G20
"
#
0312 F12 + 0321 F21 =
1
2
(F12 − F21 ) = −Bz = −G30
"
#
1230 F30 + 1203 F03 =
1
2
Ez
= −G21
c
"
#
Ey
= −G31
= 12 1320 F20 + 1302 F02 = 12 (F20 − F02 ) = −
c
"
#
Ex
= −G32
= 12 2310 F10 + 2301 F01 = 12 (−F10 + F01 ) =
c
G12 =
G13
"
1
2
(−F30 + F03 ) =
The entire array may be written in matrix form as
⎛
0 −Bx −By
−Bz
⎜
Ey
Ez
⎜ Bx
−
0
⎜
c
c
⎜
Gμν = ⎜
Ex
Ez
⎜
0
⎜ By −
⎜
c
c
⎝
Ex
Ey
−
0
Bz
c
c
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠
I will be noted that Gμν may be generated from F μν by replacing the terms
Ei /c → Bi and Bi → −Ei /c. The requisite divergence is now easily calculated:
·B
∂μ Gμ0 = ∇
∂Ez
∂Ey
1
∂Bx
1 ∂B
μ1
× E)
+ −
+
+ (∇
∂μ G =
−
=−
c
dt
∂y
∂z
c ∂t
x
∂Ez
∂Ex
1
∂By
1 ∂B
× E)
+
−
+ (∇
∂μ Gμ2 =
−
=−
c
dt
∂x
∂z
c ∂t
y
∂Ey
∂Ex
1
∂Bz
1 ∂B
μ3
× E)
+ −
+
+ (∇
∂μ G =
−
=−
c
dt
∂x
∂y
c ∂t
z
Equating each of these terms to zero gives the required result.
#
"
= Γ B
⊥ −v × E
⊥ /c2 . Taking,
11-6 We wish to demonstrate that B
= B
and B
⊥
as usual, the velocity of the frame along the x axis
#
"
Bx = F 32 = αμ3 αν2 F μν = δμ3 δν2 F μν = F 32 = Bx
#
"
By = F 13 = αμ1 αν3 F μν = αμ1 F μ3 = α01 F 03 + α11 F 13
Ez
= −βΓF 03 + ΓF 13 = Γ By + β
c
Chapter Eleven Solutions
107
#
"
Bz = F 21 = αμ2 αν1 F μν = αμ1 F 2ν = α01 F 20 + α11 F 21
Ey
20
21
= −βΓF + ΓF = Γ Bz − β
c
Combining the last two expression as
vx E y
vx Ez
v
B⊥ = By ĵ + Bz k̂ = Γ B⊥ +
ĵ −
k̂ = Γ B⊥ − 2 × E⊥
c c
c c
c
we obtain the requisite relation.
11-7 To obtain the new potential we merely apply the Lorentz transformation to
to obtain
the four vector Φμ = (V /c, A)
Φ0 = ΓΦ0 − Γβx Φ1
V
, Ay = Ay , and Az = Az .
or V = Γ(V −vx ·Ax ). Similarly, Ax = Γ Ax −β
c
11-8 Consider a charged particle moving with velocity v along the x axis passing the
observer at distance b. For simplicity we take the rest frame of the observer
and that of the particle to coincide at t = t = 0. In the frame of the particle
at time t , the observer is located at x = −vt and y = b. The non vanishing
components of the field at the observer in the particle’s frame are therefore
Ex =
−qvt
4πε0 r3
Ey =
qb
4πε0 r3
Expressing r and t in terms of r and t we have
t = γ t and r = b2 + γ 2 v 2 t2
so that
Ex =
−qγvt
4πε0 (b2 + γ 2 v 2 t2 )3/2
and Ey =
qb
4πε0 (b2 + γ 2 v 2 t2 )3/2
We transform these fields to those measured in Σ using (11–19) to obtain
Ex = Ex =
−qγvt
4πε0 (b2 + γ 2 v 2 t2 )3/2
and Ey = γEy =
qγb
4πε0 (b2 + γ 2 v 2 t2 )3/2
We can relate this result to the geometry by noting that r2 = b2 + (vt)2 ,
whence
b2 + γ 2 v 2 t2 = γ 2 r2 + (1 − γ 2 )b2
1 − γ 2 b2
2 2
=γ r 1+
γ 2 r2
b2
= γ 2 r2 1 − β 2 2
r
= γ 2 r2 (1 − β 2 sin2 θ)
108
Classical Electromagnetic Theory
where θ is the angle between r and v . Substituting this result, we find
=
E
qr
4πε0 γ 2 r3 (1 − β 2 sin2 θ)3/2
where r is the location of the field point with respect to the current (not
)r.
retarded) position of the particle. As was shown in figure 10.9, r = (n̂ − β
11-9 Let the rectangle have length a along the x direction taken to be the direction
of motion, and width b in the y direction. The loop will have a magnetic
moment m
= Iab in the ±z direction. Taking the current to run in the +x
direction along the top of the loop (so that m points in the −z direction) we
use the Lorentz transformation to find J μ in the rest frame:
ρc = 0 + βΓJx
Integrating this over the cross section of the wire we obtain a line charge
density λ
V
λ = Γ 2 I
c
The total charge on the top of the loop is found by summing this over the
contracted length a/Γ to give
Q=
aV I
c2
Similarly along the bottom, Q = −aIV /c2 , while the sides perpendicular to
v remain charge free. The dipole moment of the moving loop becomes
p = Qb =
m
×V
abIV
=
−
c2
c2
The direction is easily verified with the aid of a diagram.
11-10 In the moving dielectric with J = 0 and ρ = 0, the absence of magnetic
·B
= 0 and
monopoles implies ∇
×E
= − ∂B
∇
∂t
remains unchanged as it involves only the fields, independent of any media.
The remaining two Maxwell equations do involve the media and therefore will
be the polarization
have to take account of the movement. Letting P and M
and magnetization of the medium as measured by a stationary observer, we
have
· (ε0 E
+ P ) = 0
∇
and
×
∇
B
= ∂D
−M
μ0
∂t
Chapter Eleven Solutions
109
where, for small V /c,
×M
V
P = P +
2
c
and
=M
−V
× P
M
+ P = ε0 E
+ P + V × M so that ∇
= ε0 E
·D
= −V × M
leading to D
2
2
c
c
and
× B −M
× V ×P
= ∂P − ∇
∇
2
μ0
∂t
c
or
×H
= ∂D + ∂ V × M − ∇
× (V
× P )
∇
∂t
∂t c2
and a polarThe movement imparts to the medium a magnetization P × V
2
ization (V × M )/c .
11-11 The moving magnetization appears in the lab frame to have polarization
ωρϕ̂ × M
ωM (xı̂ + yĵ)
v × M
=
=
P =
2
2
c
c
c2
· P = −2ωM/c2 distributed throughout
The bound charge density is then −∇
the sphere and bound surface charge P · n̂ = aωM sin2 θ/c2 on the surface. It
is readily verified that the “net charge” resulting from these bound charges is
zero. We find the potential of these distributions as
1
V (r ) =
4πε0
=
τ
−2ωM
4πε0 c2
· P (r )
−∇
1
r2 dΩ dr +
|r − r |
4πε0
τ
a3 ωM
r2 dΩ dr
+
|r − r |
4πε0 c2
S
S
P · n̂ 2 a dΩ
|r − r |
sin2 θdΩ
|r − r |
To perform the integration, we expand
! 4π r
1
<
=
Ym (θ, ϕ)Ym∗ (θ , ϕ )
+1 |r − r |
2 + 1 r>
,m
√
and write 3 sin2 θ = (3 − 3 cos2 θ) = 2 + (1 − 3 cos2 θ) = 2 4πY00 − 16π/5Y20 .
Thus in the first integral only the Y00 term and in the second, only the Y00
and Y02 terms of the sum survives the integration over Ω .
When r > a, r< = r and r> = r so that
−2ωM 4π a 2 a3 ωM 2 4π
r
dr
+
|Y0 (θ , ϕ )|2 dΩ
V (r > a) =
4πε0 c2 r 0
4πε0 c2 3 r 4π 0
&
16π 0
ωM a5
Y
−
(θ,
ϕ)
|Y20 (θ , ϕ )|2 dΩ
15ε2 c2 r3
5 2
4π
110
Classical Electromagnetic Theory
The first two terms cancel precisely leaving only
V (r > a) =
ωM a5 (1 − 3 cos2 θ)
15ε0 c2 r3
Evidently, the external potential is that of an electric quadrupole. The
interior potential is calculated in much the same fashion. For the surface
integral r = r> and only a single evaluation is required. For the volume
integral, however, for values of r up to r, r = r< while for the remaining
volume it is r> .
−2ωM r r2 2ωM a dr
−
r dr
V (r < a) =
ε0 c2
r
ε0 c2 r
0
2 3
a ωM
+3
4πε0 ac2
r2 ωM a3
− 3
a 15ε0 c2
&
16π 0
Y (θ, ϕ)
5 2
4π
|Y20 (θ , ϕ )|2 dΩ
ωM 2
5(a − r2 ) + r2 (3 cos2 θ − 1)
=−
2
15ε0 c
The electric field may be now be found by differentiating the potentials.
It is instructive to solve this problem using an alternative approach. If we
· P /R over a region of space that includes the
write V as the integral of ∇
boundary, then there is no need to include an integral over the boundary
· P has a singularity at the discontinuity.
separately. However, as we noted, ∇
We can sidestep this discontinuity in the following fashion.
V (r ) = −
1
4πε0
1
=−
4πε0
∇ · P (r ) 3 d r
r − r |
τ |
P (r )
1
1
∇ ·
P (r ) · ∇
+
d3 r |r − r |
4πε0 τ
|r − r |
τ
The first of the two integrals may be converted to a vanishing surface integral
over a surface outside the sphere (where P vanishes). In the second integral,
(1/|r − r |) with −∇(1/|
we replace ∇
r − r |) to obtain
1
V (r ) = −
4πε0
P (r ) · ∇
τ
Mω =−
∇·
4πε0 c2
τ
1
|r − r |
3 P (r )d r
1 ∇·
d3 r = −
4πε0
r − r |
τ |
xı̂ + yĵ 2 r dr dΩ
|r − r |
To perform the integration, we expand
! 4π r
1
<
=
Ym (θ, ϕ)Ym∗ (θ , ϕ )
+1 |r − r |
2 + 1 r>
,m
Chapter Eleven Solutions
111
&
and write
2π 1 r Y1 (θ , ϕ ) − Y1−1 (θ , ϕ )
3
&
2π 1 r Y1 (θ , ϕ ) + Y1−1 (θ , ϕ )
y = i
3
It is evident that only the = 1, m = ±1 terms will survive the integration
over Ω . Performing the integration over Ω , we are left with
xı̂ + yĵ a r< 2 ωM V (r ) = −
∇
·
r
dr
2
3ε0 c2
r
0 r>
x =−
When r > a, r = r> and r = r< , so that
ωM a5
V (r > a) = −
∇
·
(xı̂
+
yĵ)
3ε0 c2
5r5
ωM a5 2
3(x2 + y 2 )
=−
−
15ε0 c2 r3
r5
=−
ωM a5 (1 − 3 cos2 θ)
15ε0 c2 r3
The internal potential is obtained as
r
r4 dr
+
r2
rr3 dr
r2
0
r
5
2
2
r
r
a
ωM ∇ · (xı̂ + yĵ)
−
+
=−
2
3
3ε0 c
5r
2
2
ωM 2
(a − r2 ) − r2 (3 cos2 −1)
=−
15ε0 c2
ωM xı̂ + yĵ
∇·
V (r < a) = −
3ε0 c2
r
a
11-12 As in section 11.4 we start with the four force density K μ = F μν = F μν Jν
generalization of the Lorentz force and wish to express K μ as K μ = −∂λ η λμ
where η μν depends only on the fields. Replacing Jν by the field derivatives,
we get
Kμ = Fμν J ν = Fνμ ∂λ H λν = ∂λ (Fμν H λν ) − H λν ∂λ Fμν
The first term is in the required form, the second can be expanded to give
H λν ∂λ Fμν = 12 H λν ∂λ Fμν + 12 H νλ ∂ν Fμλ
= 12 H λν ∂λ Fμν + 12 H λν ∂ν Fλμ
= 12 H λν (∂λ Fμν + ∂ν Fλμ )
The source-free field equation ∂λ Fμν + ∂μ Fνλ + ∂ν Fλμ = 0 allows us to replace
the two terms in parentheses by −∂μ Fνλ , leading to
112
Classical Electromagnetic Theory
H λν ∂λ Fμν = − 12 H λν ∂μ Fνλ = 12 H λν ∂μ Fλν
If the medium is linear, we can replace the latter term by 14 ∂μ (H λν Fλν ) and
write
Kμ = ∂λ (Fμν H λν ) − 14 ∂μ (H λν Fλν )
"
#
= ∂λ Fμν H λν − 14 δμλ Fλν H λν
We can restate the equation in contravariant form by multiplying by g μα to
obtain
"
#
K α = g αμ Kμ = ∂λ g αμ Fμν H λν − 14 g αμ δμλ Fλν H λν
"
#
= ∂λ g αμ Fμν H λν − 14 g αλ Fλν H λν
The resulting Stress-Energy-Momentum tensor is
η λα = −(g αμ Fμν H λν − 14 g αλ Fλν H λν
#
11-13 We start with the form of the Lorentz transformation when the motion is
along one axis, labelling the component of r along this axis r
:
· r )
(X 0 ) = Γ(X 0 − β
r
= Γ(r
− βX 0 )
r⊥ = r⊥
In general, the parallel component may be found as r
=
(r · β)
so that r β
may be written
β
Γ(r · β)
0
− ΓβX
2
β
β
(Γ − 1)(r · β)
0
= r⊥ + r
+
− ΓβX
β2
r = r
+ r⊥ = r⊥ +
This transformation may be written in matrix form
⎛
Γ
−Γβx
−Γβy
⎜
2
⎜
(Γ − 1)βx βy
(Γ − 1)βx
⎜ −Γβ
x 1+
⎜
2
β
β2
=⎜
L(β)
⎜
(Γ
− 1)βy2
(Γ
−
1)β
β
x y
⎜ −Γβ
1
+
⎜
y
β2
β2
⎜
⎝
(Γ − 1)βx βz
(Γ − 1)βy βz
−Γβz
β2
β2
as
−Γβz
(Γ − 1)βx βz
β2
(Γ − 1)βy βz
β2
(Γ − 1)βz2
1+
β2
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠
· B/c
is 0 in any frame then it must be
11-14 If the invariant μνρσ F μν F ρσ = −E
zero in all frames, proving the required result.
Chapter Eleven Solutions
113
and
11-15 Take the direction of the moving frame to be perpendicular to both E
B. Then, according to (11–19) when |E| < c |B| it is possible to pick v so
= −E
, making E
vanish. When |E| > c|B| we can pick v so
that v × B
2
that B c = v × E , making B vanish. When |E| = c|B| the required frame
velocity would be c.
11-16 In the moving medium, the polarization is produced by the effective field,
+V
× B.
Thus,
E
+V
× B)
P = ε0 (E
, and limiting ourselves to terms linear in V
, we
Neglecting the term in M
have for the curl of B,
∂E
(κ − 1) ∂E
∂B
+ ε0 (κ − 1)
+ ε0
+ ∇ × (E × V )
∇ × B = μ0 ε0
−V ×
∂t
∂t
κ
∂t
κ ∂E
κ − 1 × E)
+∇
× (E
×V
)
= 2
+
− V × (∇
c
∂t
κ
κ − 1 κ ∂E
+
∇(V · E) − (V · ∇)E − (V · ∇)E
= 2
c
∂t
κ
× (E
×V
) and ∇(
V
· E)
and assumed
We have used the vector identities for ∇
V constant. Taking the curl once more we get
κ − 1 ∂B
κ ∂2B
2
+2
∇ B= 2
(V · ∇)
c ∂t2
κ
∂t
For a plane wave, travelling with wave vector k we have
· ∇)
B
= (V
· k)B
= − V · k ∂ B = V · n̂ ∂ B
(V
ω ∂t
v0 ∂t
so that the wave equation becomes
2
= κ ∂ B 1 − 2 κ − 1 V · n̂
∇2 B
c2 ∂t2
κ
v0
from which we deduce the speed v of the wave through the moving medium:
· n̂ −1/2
κ−1 κ−1V
V · n̂
v = v0 1 − 2
v0 +
κ
v0
κ
11-17 Using (11–83), we find that the total emitted power is
2
dp dW 2
q2 c
q2 c
P =
=
6πε0 (m0 c2 )2 dt 6πε0 (m0 c2 )2 dS =
6 × 109 × (1.6 × 10−19 )2 × 3 × 108
× (1MeV/m)2
0.511MeV)2
= 1.76 × 10−19 W ∼ 1.1eV/s
114
Classical Electromagnetic Theory
Clearly this is not a major energy loss mechanism.
11-18 Using (11–84), the power may be written
q2 γ 2
P=
6πε0 m20 c3
d
p
dt
2
q 2 γ 4 a2
q2 γ 4
=
=
6πε0 c3
6πε0 c3
v2
r
2
When v c, this becomes
4
q2 c γ 4
9
−19 2
8γ
=
6
×
10
×
(1.6
×
10
)
×
3
×
10
6πε0 r2
r2
4
γ
= 4.6 × 10−20 2 Wm2
r
P =
A 10-GeV electron has γ = 104 /0.511 = 19569 ⇒ γ 4 1.5 × 1017 , so that
for R = 20 m, P = 1.69 × 10−5 W 108 MeV/s. A revolution takes 2πr/c =
4.19 × 10−7 seconds; hence the energy loss per revolution is about 44 MeV. As
it is difficult to increase the electron’s energy by this much in one revolution
the radiation loss represents a major loss. Much larger radii are needed to
get the losses down to a manageable amount.
Chapter 12
12-1 Making the substitution suggested, we have y = (t − t)/τ , t = t + yτ and dt
= τ dy. The integral (12–41) then becomes
∞
F (t + τ y)e−y dy
v˙ =
m0
0
12-2 Using the expression above, the acceleration of the particle may be written
∞
˙v =
−ω02 x(t + τ y)e−y dy
0
and trying a solution of the form x = x0 e−αt we find on substitution
∞
2
−αt
=
−ω02 x0 e−α(t+τ y) e−y dy
α x0 e
0
or
α2 =
0
∞
−ω02 e−(ατ +1)y dy
In order that the integral converge, it is required that Re(ατ + 1) > 0. Under
this condition the integral is easily evaluated to give
∞
ω 2 e−(ατ +1)y ω02
α2 = 0
=−
−(ατ + 1) 0
ατ + 1
Upon rationalization and multiplication by τ 2 we obtain the characteristic
equation
α3 τ 3 + α2 τ 2 + ω02 τ 2 = 0
which may in principle be solved for α. When ω0 τ is very small, an approximate solution may be found. It is probably easiest to use an iterative
approach. We abbreviate ατ = z and ω0 τ = r and note that when r is
small z must also be small. Therefore a zeroth order solution may be found
from z 2 + r2 = 0, or z0 = ±ir. We now use Newton-Raphson iteration
[zi+1 = zi − f (zi )/f (zi )] to obtain successively better estimates of the root.
Thus, taking z0 = ir
(ir)3 − r2 + r2
= ir + 12 r2
2ir
where we have neglected terms in r3 in the denominator. We iterate once
more, carrying terms to order r5 to obtain
z1 = ir −
(ir + 12 r2 )3 + (ir + 12 r2 )2 + r2
2ir
= 12 r2 + i(r − 58 r3 )
z2 = ir + 12 r2 −
The solution correct through third order in ωτ is then
— 115—
116
Classical Electromagnetic Theory
α = 12 ω02 τ ± i(ω0 − 58 ω03 τ 2 )
The decay constant is 12 ω02 τ and the damping decreases the resonant angular
frequency by 58 ω03 τ 2 .
12-3 The differential equation of motion for the charged particle is
d2 x
d3 x
+ ω02 x − τ 3 = 0
2
dt
dt
Again trying a solution of the form x = x0 e−αt , we find
α2 x + ω02 x + α3 τ x = 0
which gives the characteristic equation
α3 τ 3 + α2 τ 2 + ω02 τ 2 = 0
identical to the characteristic equation found in 12-2.
12-4 (a) Neglecting radiation damping, the relativistic equation of motion is
m0 c
dβ μ
μ
= Fext
dτ
f).
We note that in terms of the three-vectors, β μ = γ(1, v /c) and F μ = γ(f·β,
Taking f to be in the x direction, we rewrite
m0 c
dβ 1
d(γvx )
= m0 γ
= F1
dτ
dt
⇒
fx
d(γvx )
=
dt
m0
This equation is easily integrated to give
γvx =
fx t
m0
Equating the expression for γv to the numerical value fx t/m0 = c, we solve
for v to obtain v = .707c.
(b) In terms of the four-momentum, the covariant equation of motion (12–46)
may be written
2 μ
d P
dP μ
q2
P μ dP ν dPν
−
+ 2 2
= Fμ
dτ
6πε0 m0 c3 dτ 2
m0 c dτ dτ
Writing
d W
d 2
dP ν
=
, p =
p + m20 c2 , p
dτ
dτ c
dτ
pdp/dt
d
p
= ,
p2 + m20 c2 dt
Chapter Twelve Solutions
117
we find
dP ν dP ν
=
dτ dτ
Further,
2
2
p2
dp
dp
−m20 c2
−1
= 2
2
2
2
2
2
p + m0 c
dτ
p + m0 c dτ
,
f · v p2
f · v ,f = 1 + 2 2
,f
F =γ
c
m0 c
c
μ
Taking all motion and forces to be directed along x we extract the μ = 1
component of the resulting equation to obtain
2 ,
dpx
dpx
d 2 px
px
p2
−a
=
−
1
+
fx
2
dτ
dτ 2
p2 + m0 c2 dτ
m20 c2
where we have relabelled q 2 /(6πε0 m0 c3 ) as a to avoid confusion with the
proper time τ . The equation may be considerably simplified with the substitution px = m0 c sinh s. With this substitution, the equation of motion
becomes
2
2 ds
ds
d2 s
ds
fx
− a cosh s 2 − sinh s
+ sinh s
cosh s
= cosh s
dτ
dτ
dτ
dτ
m0 c
or
d2 s
ds
fx
=a 2 =
dτ
dτ
m0 c
The solution to this equation is easily seen to be
ds
fx
= Aeτ /a +
dτ
m0 c
We might try to set A = 0 in order to avoid the runaway solution and integrate
once more to obtain
fx τ
s=
m0 c
whence,
p = m0 c sinh s = m0 c sinh
fx τ
m0 c
It remains to relate t to τ . We note that
t=
τ
γ(τ )dτ =
0
,
τ
dτ
0
,
where
fx τ
p = sinh
,
m0 c
and
1+
1+
p2
m20 c2
p2
fx τ
= cosh
2
2
m0 c
m0 c
118
Classical Electromagnetic Theory
We abbreviate b = fx /m0 c and integrate,
τ
m0 c sinh s
sinh bτ
p
t=
=
cosh bτ dτ =
=
b
fx
fx
0
which leads to p = fx t or γv = c ⇒ v = .707c, precisely the same result
as obtained in (a). Clearly setting A = 0 eliminates all effects of radiation
damping.
(c) We may convert the differential equation in s and τ to an integral equation
using exactly the same stratagem as in section 12.4 to obtain
ds
1 ∞ fx (τ ) −(τ −τ )/a =
e
dτ
dτ
a τ
m0 c
If the force continues constantly for all time, then we find as before
ds
fx
=
dτ
m0 c
and radiation damping appears to have no effect. Only when the force terminates at some time τ0 do we get a different effect. Now
1 τ0 fx e−(τ −τ )/a ds
=
dτ
dτ
a τ
m0 c
#
fx eτ /a τ0 −τ /a fx "
1 − e−τ0 /a
=
e
dτ =
am0 c τ
m0 c
As we have no information about the continuation of the force, we abandon
the calculation at this point. It should be clear that it would be considerably
easier to estimate the energy loss due to radiation assuming the acceleration
is f /γm0 and calculate the final energy from
dW
af 2
= f · v − 2
dt
γ m0
Appendix B
B-1 The result is most easily obtained by using raising a lowering operators.
Bj ∂i Aj = gjk B k ∂i g j A = gjk g j B k ∂i A = δk B k ∂i A = B ∂i A
B-2 To demonstrate the equivalence of the second and third line of (Ex B.8.18) it
suffices to show that
sinh2 α − 2 cosh α(cosh α − cos β) + sin2 β
= −1
(cosh α − cos β)2
To this end we expand the numerator
sinh2 α − 2 cosh2 α + 2 cosh α cos β + sin2 β
= −1 − cosh2 α + 2 cosh α cos β + sin2 β
= − cos2 β + 2 cosh α cos β − cosh2 α
by inspection, this is just minus the denominator.
B-3 A particle’s four momentum is
P μ = m0 V μ = γm0 (c, v )
where m0 is the particle’s proper or rest mass. The components of the 3momentum have the same transformation law as those of the three-velocity.
p
=
p
− m0 βc
1 − βvx /c
and p⊥ =
p⊥
Γ(1 − βvx /c)
B-4 The standard expansion of a determinant in terms of its cofactors is precisely
what this expression describes.
B-5 The square interval (dS)2 is generally given by
(dS)2 = dqi dq i = gij dq j dq i
For spherical polar coordinates this becomes
grr (dr)2 + gθθ (dθ)2 + gϕϕ (dϕ)2 = (dr)2 + r2 (dθ)2 + r2 sin2 θ(dϕ)2
B-6 Again this should be almost self-evident from the definition of gik = ei · ek
We transform to the primed system
(gik ) = ei · ek = αij ej · αk e = αij αk gj
Comparing this with (B–30) we find this to be exactly the transformation law
for a second rank fully covariant tensor.
— 119—
120
Classical Electromagnetic Theory
· ej although
B-7 The simplest method is probably to use the definition Aj = A
i
the alternative Aj = gij A could also be used.
A1 = (5ı̂ + 6ĵ + 7k̂) · (2ı̂ + 3ĵ + k̂) = 10 + 18 + 7 = 35
A2 = (5ı̂ + 6ĵ + 7k̂) · (ı̂ − ĵ + k̂) = 5 − 6 + 7 = 6
A3 = (5ı̂ + 6ĵ + 7k̂) · (ĵ + k̂) = 6 + 7 = 13
We quickly verify that Ai Ai is independent of which coordinate system is
chosen. Indeed, Calculate in the Cartesian frame, |A|2 = 25 + 36 + 49 = 110
and in the oblique frame Ai Ai = 35 × 32 + 6 × 2 + 13 × 72 = 110.
As an illustration of using the alternate method we compute A1 = g1j Aj .
We first must find g1j .
g11 = e1 · e1 = 14
g12 = e1 · e2 = 0 g13 = e1 · e3 = 4
leading to
A1 = g11 A1 + g12 A2 + g13 A3 = 14 × 1.5 + 0 × 2 + 4 × 3.5 = 35
The other components are computed in the same manner.
B-8 We continue to find the remaining components of the metric tensor in addition
to the g1j already obtained above.
g21 = 0 g22 = 3
g31 = 4 g32 = 0
g23 = 0
g33 = 2
Applying these elements to find |A|2 we have
|A|2 = g11 A1 A1 + g13 A1 A3 + +g22 A2 A2 + g31 A1 A3 + g33 A3 A3
= 14 × (1.5)2 + 4 × (1.5)(3.5) + 3 × (2)(2) + 4 × (3.5)(1.5) + 2 × (3.5)2
= 110
B-9 The angular momentum of the rigid system is
!
=
L
m(j)r(j) × (
ω × r(j) )
j
Dropping the summation over particles (we will restore it at the end of the
exercise) to avoid the rather cumbersome notation, we write
Li = mijk xj km ω xm = mkij km xj ω xm
j
i
= m(δi δm
− δj δm
)xj ω xm
= m(ω i xm xm − xj ω j xi )
= ω j m(δji xm xm − xj xi ) = ω j Iji
Appendix B Solutions
121
We can generally exchange the upper and lower indices of an inner product
(Aj Bj = Aj B j ) so that we can write the latter expression as
Li = ωj I ij
with
I ij = m(δ ij xm xm − xi xj )
Restoring the sum over particles we have
!
2
I ik =
m(j) δ ik r(j)
− xi(j) xk(j)
j
In this sum, r2 is a scalar and we know each of δ ik and xi xk to be a second
rank contravariant tensor, which allows us to conclude that I ik is a second
rank tensor.
B-10 It is simple to construct the mixed tensor
Uij = 12 (∂i U j + ∂j U i )
The fully covariant tensor is obtained by multiplying by g ik to get
U jk = g ik Uij = 12 (∂ k U j + g ik ∂j U i )
Now, ∂j U i = ∂ j Ui so that the second term in the sum may be written ∂ j U k
to complete the problem.
B-11 The hyperbolic coordinates are defined by
√
√
x= ρ+v
y = ρ−v
z=z
√
with ρ = u2 + v 2 . The basis may be constructed as
1
1
dx
dy
u
dr
2 ı̂
2 ĵ
√
√
=
ı̂ +
ĵ =
+
eu =
du
du
du
ρ
ρ+v
ρ−v
1
1
(v/ρ + 1)ı̂
(v/ρ − 1)ĵ
dx
dy
dr
=
ı̂ +
ĵ = 2 √
+ 2 √
ev =
dv
dv
dv
ρ+v
ρ−v
and, of course ez = k̂. The nonzero metric tensor elements are gzz = 1,
1
1
1
u2
1
4
4
guu = eu · eu = 2
+
= √
=
2
ρ ρ+v ρ−v
2ρ
2 u + v2
ρ − v 2
v + ρ 2
1
1
= √
2
2ρ
2 u + v2
The remaining elements all vanish. The Laplacian may be constructed following the prescription √of (B–69).√ The determinant of the metric tensor,
G = 1/[4(u2 + v 2 )] and Gg uu = Gg vv = 1 so that ∇2 V becomes
2
∂ V
∂2V
d2 V
∇2 V = 2 u2 + v 2
+
+
2
2
∂u
∂v
dz 2
gvv = ev · ev =
2ρ
v+ρ
+
2ρ
ρ−v
=
122
Classical Electromagnetic Theory
B-12 We begin by defining r =
r=
a sin θ
cosh η − cos θ
x2 + y 2 to write
and
z=
a sinh η
cosh η − cos θ
Referring to Example B.8, we note that r and z have been reversed, but
otherwise the problems are identical. The toroidal system has a circle in the
centered on and perpendicular to the x-y plane that, rotated about the z axis,
sweeps out a torus, whereas the bispherical coordinates have a circle centered
on the z axis which rotated about the z axis sweeps out a sphere. Adapting
the results of Example B.8 simply exchanging z and r we have from (ExB.8.8)
r2 + (z − coth η)2 =
a2
sinh2 η
the equation of a circle of radius a/ sinh η centered at z = coth η. Positive
η lead to spheres centered on positive z whereas negative η yield spheres at
negative z.
To obtain the metric tensor we first construct the local basis.
eφ =
eη =
a cosh η
a sinh2 η
−a sin θ sinh η(ı̂ cos φ + ĵ sin φ)
−
+
k̂
(cosh η − cos θ)2
cosh η − cos θ (cosh η − cos θ)2
eθ =
d
a sin θ
dr
=
(xı̂ + yĵ + z k̂) =
(−ı̂ sin φ + ĵ cos φ)
dφ
dφ
cosh η − cos θ
a cos θ
a sin2 θ
a sinh η sin θk̂
−
(ı̂ cos φ + ĵ sin φ) −
cosh η − cos θ (cosh η − cos θ)2
(cosh η − cos θ)2
The metric tensor elements may now be computed,
gφφ = eφ · eφ =
a2 sin2 θ
(cosh η − cos θ)2
2
a cos θ
a sin2 θ
a sinh2 η sin2 θ
−
+
2
cosh η − cos θ (cosh η − cos θ)
(cosh η − cos θ)4
a2
=
cos2 θ + sin2 θ
(cosh η − cos θ)2
2
sin θ − 2 cos θ(cosh η − cos θ) + sinh2 η
×
(cosh η − cos θ)2
2
a
=
(cosh η − cos θ)2
gθθ =
Appendix B Solutions
123
and
gηη
2
a cosh η
a sinh2 η
a2 sin2 θ sinh2 η
−
=
+
(cosh η − cos θ)4
cosh η − cos θ (cosh η − cos θ)2
a2
=
cosh2 η + sinh2 η
(cosh η − cos θ)2
2
sin θ − 2 cosh η(cosh η − cos θ) + sinh2 η
×
(cosh η − cos θ)2
2
a
=
(cosh η − cos θ)2
The off-diagonal elements vanish.
The Laplacian is constructed according to the prescription of (B–69):
1
∂V
(cosh η − cos θ)3
∂
2
∇ V =
sin θ
a3 sin θ
∂η cosh η − cos θ ∂η
sin θ
∂V
∂
(cosh η − cos θ)2 ∂ 2 V
+
+
∂θ cosh η − cos θ ∂θ
∂φ2
a2 sin2 θ
c
Jack
Vanderlinde, Reproduction without express permission of the author
is strictly prohibited.