PHYS 3123
Georgia Tech, Spring 2017
Instructor: David Ballantyne
(Thomas) Forrest Kieffer
Homework 8 Solutions
1. (10.15) A particle of charge q moves in a circle of radius a at constant angular velocity ω. (Assume that the circle lies in the
xy plane, centered at the origin, and at time t = 0 the charge is at (a, 0), on the positive x axis.) Find the Liénard-Wiechert
potentials for points on the z-axis.
Solution: In this situation, the trajectory of the particle is w(t) = ha cos (ωt), a sin (ωt), 0i. Therefore, v = ∂t w =
aωh− sin (ωt), cos (ωt), 0i. Unlike example 10.3, we do not need to compute the retarded time tr because (as we will
see) neither r or r · v will depend on tr . Indeed, taking r = h0, 0, zi, we find
p
r = |h−a cos (ωt), −a sin (ωt), zi| = a2 + z 2
and
r · v = r · v − w(tr ) · v
= h0, 0, zi · ha cos (ωt), a sin (ωt), 0i − ha cos (ωt), a sin (ωt), 0i · h−aω sin (ωt), aω cos (ωt), 0i
=0−0
=0
Therefore, Eq. 10.46 now reads
V (z, t) =
1
q
√
4π0 a2 + z 2
A(z, t) =
µ0
q
√
v
2
4π a + z 2
and Eq. 10.47 reads
2. (10.20) Suppose a point charge q is constrained to move along the x axis. Show that the fields at points on the axis to the
right of the charge are given by
q 1 c+v
E=
x̂,
B = 0.
4π0 r2 c − v
(Do not assume v is constant!) What are the fields on the axis to the left of the charge?
Solution: In this situation v = v x̂, a = ax, and r̂ = x̂ for points to the right. The problem is entirely 1-dimensional,
so we can automatically note that any cross product terms will be zero. In particular, B = c−1 r̂ × E = 0. Moreover,
u = (c − v)x̂, u × a = 0, and r · u = (c − v) r. Therefore, the expression for E:
E=
r (c2 − v2 )u +
q
4π0 ( r · u)3
r × (u × a)
becomes
q
E=
4π0
For points to the left of the charge,
q 1
r3 (c − v)3 (c − v )(c − v)x̂ = 4π0 r2
2
2
c+v
c−v
x̂
r̂ = −x̂ and u = −(c + v)x̂, so r · u = r(c + v), and
q
E=−
4π0
and, as before, B = 0.
r
r
q 1
(c − v )(c + v)x̂ = −
3
3
r (c + v)
4π0 r2
2
2
c−v
c+v
x̂
PHYS 3123
Georgia Tech, Spring 2017
Homework 8 Solutions
Instructor: David Ballantyne
(Thomas) Forrest Kieffer
3. (10.23) For the configuration in Prob. 10.15, find the electric and magnetic fields at the center. From your formula for B,
determine the magnetic field at the center of a circular loop carrying a steady current I, and compare your answer with the
result of Ex. 5.6.
Solution: To avoid confusion with the acceleration a, we relabel the radius of the circle in Prob. 10.15 by R instead
of a. Unfortunately, we aren’t able to use the potentials we calculated in Prob. 10.15 since we only calculated those
potentials for points along the z-axis.
Fortunately, r = 0 in this problem and hence r = −w(tr ) = −R cos (ωt)x̂ − R sin (ωtr )ŷ, r = |w(tr )| = R, and
r̂ = − cos (ωtr )x̂ − sin (ωtr )ŷ. Moreover, v(t) = Rω(− sin (ωt)x̂ + cos (ωt)ŷ) and a(t) = −Rω2 (cos (ωt)x̂ +
sin (ωt)ŷ) = −ω 2 w(t). Before calculating E we first need u, r · u and r × (u × a).
u = c r̂ − v(tr ) = − ([c cos (ωtr ) − ωR sin (ωtr )]x̂ + [c sin (ωtr ) + ωR cos (ωtr )]ŷ)
and, using a vector calculus identity for a × (b × c),
r × (u × a) = ( r · a)u − ( r · u)a = ω2 R2 u − Rca
Plugging everything into Eq. 10.72 we find
E=
R 2
q
q cu − Ra
(c − ω 2 R2 )u + ω 2 R2 u − cRa =
3
4π0 (Rc)
4π0 (Rc)2
For B, we compute
r̂ × u = − r̂ × v(tr ) = Rω [cos (ωtr )x̂ + sin (ωtr )ŷ] × [− sin (ωtr )x̂ + cos (ωtr )ŷ] = Rωẑ
and since a ∝ r̂,
r̂ × a = 0. Thus,
B=
1
r̂ × E = q ω 2 ẑ = µ0 q ω ẑ
c
4π0 Rc
4π R
To obtain the field at the center of a circular ring of charge, let q 7→ λ(2πR); for this ring to carry current I, we need
I = λv = λωR, so λ = I/ωR, and hence q 7→ (I/ωR)(2πR) = 2πI/ω. Thus,
B=
2πµ0 I ω
µ0 I
ẑ =
ẑ
4πω R
2R
which is the same as Eq. 5.41, in the case z = 0.
4. (11.1) Check that the retarded potentials of an oscillating dipole (Eqs. 11.12 and 11.17) satisfy the Lorenz gauge condition.
Do not use approximation 3.
Solution: Eq. 11.12 and Eq. 11.17 read
V (r, θ, t) =
p0 cos θ
4π0 r
−
ω
1
sin [ω(t − r/c)] + cos [ω(t − r/c)]
c
r
A(r, θ, t) = −
µ0 p0 ω
sin [ω(t − r/c)]ẑ,
4πr
Page 2
PHYS 3123
Georgia Tech, Spring 2017
Instructor: David Ballantyne
(Thomas) Forrest Kieffer
Homework 8 Solutions
respectively. The Lorenz gauge condition is ∇ · A = −µ0 0 ∂t V . So we compute
µ0 po ω 1
1
2 1
21
∇·A=−
∂
r
sin
[ω(t
−
r/c)]
cos
θ
+
∂
−
sin
θ
sin
[ω(t
−
r/c)]
r
θ
4π
r2
r
r sin θ
r
ωr
µ0 p0 ω 1
2 sin θ cos θ
sin
[ω(t
−
r/c)]
−
=−
cos
[ω(t
−
r/c)]
cos
θ
−
sin
[ω(t
−
r/c)]
4π
r2
c
r2 sin θ
ω
p0 ω
1
sin [ω(t − r/c)] +
cos [ω(t − r/c)] cos θ
= µ0 0
4π0 r2
rc
and
p0 cos θ
ω2
ω
∂t V =
−
cos [ω(t − r/c)] − sin [ω(t − r/c)]
4π0 r
c
r
ω
p0 ω
1
sin [ω(t − r/c)] +
=−
cos [ω(t − r/c)] cos θ
4π0 r2
rc
Hence, the Lorenz gauge condition is satisfied.
5. (11.3) Find the radiation resistance of the wire joining the two ends of the dipole. (This is the resistance that would
give the same average power loss-to heat-as the oscillating dipole in fact puts out in the form of radiation.) Show that
R = 790(d/λ)2 Ω, where λ is the wavelength of the radiation. For the wires in an ordinary radio (say, d = 5 cm), should
you worry about the radiative contribution to the total resistance.
Solution: We start with P = I 2 R and use Eq. 11.15 to find P = q02 ω 2 sin2 (ωt)R, which gives an average power
radiated: hPi = q02 ω 2 R/2. Equating this to Eq. 11.22 and solving for R we find
2
µ0 d2 ω 2 ω=(2πc)/λ
µ0 d2 4π 2 c2
= πµ0 c
R=
−−−−−−−→ R =
6πc
6πc λ2
3
2
d
λ
Plugging in c = 3 × 108 [m] and µ0 = 4π × 10−7 [H/m] we find
R = 790
2
d
[Ω]
λ
For wires with d = 5 × 10−2 [m] and λ = 103 [m], we find R = 2 × 10−6 [Ω], which is negligible.
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