Classical Electrodynamics
Gabriel Barello
6.6
a.
Consider a circular toroidal current distribution with mean radius a and N turns, with a small uniform cross sectional area
A and curent I flowing around it with a charge Q at it’s center. By the elementary application of amperian loops, and
neglecting the spatial extent of the interior relative to the radius of the torus, the magnetic field inside the torus is seen to be
~ x) =
B(~
 IN µ
 ± 2πa0 φ̂

0
IxN
inside the torus
A
else
a
Where the ± depends on the direction of current flow. Recall that the momentum in the fields is given by the equation
Z
Pf ield = µ0 0
E × H d3 x
(1)
V
The E field is given by
Q
r̂
4π0 r2
E(~x) =
(2)
We can express the radial unit vector and the azimuthal unit vector by
r̂ = (cos(φ) sin(θ), sin(φ) sin(θ), cos(θ))
(3)
φ̂ = (− sin(φ), cos(φ), 0)
(4)
so that
E×H=±
Q
IN
Q
IN
(− cos(θ) cos(φ), cos(θ) sin(φ), sin(θ)) ∼
(θ, θ, 1)
=±
2
2πa 4π0 a
2πa 4π0 a2
(5)
When we multiply by µ0 0 and integrate over the volume, which is approximately 2πaA, we get the result for the z
component,
IN QAµ0
(6)
4πa2
And the other components approximately zero, since all positive values of θ are compensated for with negative values of
θ when an integral across the entire cross-sectional area is performed. The approximation given is
(P)z = ±
1
E(0) × m
c2
The total magnetic moment is zero, but if we consider the individual contributions of each loop, and sum them, we get
Pf ield =
Pf ield = ±N
1
Q
IN QAµ0
IA = ±
c2 4π0 a2
4πa2
Beauty.
1
b.
Let Q = 10−6 C ∼
= 6 × 1012 electronic charges, I = 1.0A, N = 2000, A = 10−4 m2 and a = .1m. Then, at the toroid, we have
10−6 C
= 898774 V
4π(8.854 × 10−12 m−3 kg −1 s4 A2 )(.01m2 )
1A × 2000 × (1.256 × 10−6 )
|B|(a) =
= 4 × 10−3 T
2π(.1m)
200 × (10−6 C) × (10−4 m2 ) × (1.25 × 10−6 )
= 2 × 10−12 N · s
(P)z =
4π(.1m)2
|E|(a) =
(7)
(8)
(9)
For comparison, a 10µg insect flying at .1m/s has a momentum of p = 10−7 N · s. So, it is a few orders of magnitude
smaller, even than this.
6.9
Consider a uniform, isotropic medium described by permittivity and µ. In general, for a non-dispersive, linear medium the
following equations apply
1
1
(E · D + B · H) = ((E)2 + µ(B)2 )
2
2
S=E×H
1
g = 2 (E × H) = µE × H
c
1
1
1
1
Tij = [Ei Ej + Bi Bj − (E2 + B2 )δij ] = [Ei Ej + µHi Hj − (E2 + µH2 )δij ]
µ
2
µ
2
u=
eqn (6.106)
eqn (6.109)
eqn (6.118)
eqn (6.120)
Where we have modified the equations in the book by substituting (µ) for 0 (µ0 ). These are all locally defined quantities,
so the only change that needs to be made is to make and µ position-dependent quantities and not pull them out of D and
H. So the functional form of the above stays the same, if we absord and µ into E and B to make them a function of D and
H. However it is interesting to note that since translational invariance is broken, momentum is no longer conversed, which
manifests itself as the divergence of Tµν (in relativistic notation) no longer vanishing since now the derivatives act on E and
B but also and µ.
6.11
A transverse plane wave is incident normally in vacuum on a perfectly abosrbing flat screen. This means that all the
momentum of the plane wave is absorbed by the screen.
a.
The momentum density of an electromagnetic field is given by equation (6.118) as g =
wave travels at speed c, the rate of momentum absorbed by the screen, per area, is cg
newton’s second law, this is the force per unit area, the pressure.
For a plane wave electromagnetic wave, this gives
r
cg =
0 A2
= 0 A2
µ0 c
Whereas the total energy density is given by (6.106) as
1
1 A2
u=
0 A2 +
= 0 A2
2
µ0 c2
So indeed they are equal.
2
1
an electromagnetic
c2 (E × H). Since
q
1
= c (E × H) = µ00 (E × B) and by
b.
Consider an energy flux of F = 1.4W/m2 = 1.4kg/s3 (such as that from the sun at the earth’s mean radius) and a solar sail
of mass of 1g/m2 = .01kg/m2 of area and negligeable other weight. The field energy per unit volume (and thus the pressure)
is then F/c and the maximum acceleration of the sail (that is, normal incidence with no losses) will be
a=
m ∼
1.4
= 4.67 × 10−7 m/s2
8
.01(3 × 10 ) s2
(10)
As for the solar wind, the average pressure of the solar wind is approximately1 2 nP a = 2 × 10−9 kg/ms2 . Thus, the
acceleration due to the solar wind is
2 × 10−9
m/s2 = 2 × 10−7 m/s2
.01
So, these two accelerations are comparable, about a factor of 2 different.
(11)
6.16
a.
The minimum magnetic charge of a dirac monopole is g =
2π~
e .
The generalized lorentz force law (from problem 6.17) is
F = qe E + qm B/µ0 + qe v × B − qm v × 0 E
(12)
Finally, the magnetic field in the median plane of a magnetic dipole (directly above the dipole) of dipole moment e~/2mp ẑ
is.
B(ρ) =
e~µ0
ẑ
6mp πρ3
(13)
So that the force on the magnetic monopole (at rest) is
F=
(1.055 × 10−34 m2 kgs−1 )2
~2
ẑ =
→ −3.529 × 10−33 N
3
3mp ρ
3(1.673 × 10−27 kg)π(.5 × 10−10 m)
(14)
In the ẑ direction. Whoa, thats tiny. Now, the work done in moving this thing in from infinity is this, integrated to
infnity which gives
V = 8.82 × 10−24 J ∼
= 10−5 ev
Which is much less than the binding energy of an electron to a hydrogen nucleus, for example.
Also, it turns out that the median plane is actually the other plane but fortunately due to the form of the magnetic dipole
field, the“real” answer is just this one, divided by −2, or
F = 1.7645 × 10−33 N
(15)
b.
If we use the binding energy above (an ok order of magnitude estimate despite the fact that it as in the wrong position), the
energy of this interaction is on the order of the hyperfine splitting2 ( 4.5 × 10−5 ev). It is also waay smaller than the spin
orbit splitting in a typical atom (.0021 ev in sodium)
6.20
Consider the dipole source (of unit strength and moment in the z direction) described by
ρ(x, t) = δ(x)δ(y)δ 0 (z)δ(t)
0
Jz (x, t) = −δ(x)δ(y)δ(z)δ (t)
(16)
(17)
1 At least is has been for the last fifty years. See http://wattsupwiththat.com/2009/06/09/solar-wind-flow-pressure-another-indication-of-solardowntrend/
2 http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/sodzee.html#c1
3
a.
Using equation (6.23) we see that the instantaneous coloumb potential is
Z
ρ(x0 , t) 3 0
1
d x
Φ(x, t) =
4π0
|x − x0 |
Z
1
δ 0 (z)
=
dz 0
δ(t) p
4π0
x2 + y 2 + (z − z 0 )2
Z
1
δ(z)(z − z 0 )
=−
dz 0
δ(t)
2
4π0
(x + y 2 + (z − z 0 )2 )3/2
1
z
=−
δ(t) 3
4π0
r
(18)
(19)
(20)
(21)
(22)
b.
Recall that the transverse and lognitudinal currents are defined by the equation
J = Jl + Jt , with ∇ × Jl = 0 and ∇ · Jt = 0
(23)
The easiest way to find the transverse current, it seems, will be to calculate the lngitudinal current from equation (6.29)
and subtract it from the total current. This gives
Jl = 0 ∇
1
z
∂Φ
= − δ 0 (t)∇ 3 .
∂t
4π
r
(24)
So that the transverse current is
1
z
Jt = −δ 0 (t) ẑ δ(x) −
∇ 3
4π r
1
zx
zy 1
zz
1
(−3 4 , −3 4 , 3 − 3 4 )
= −δ 0 (t) ẑδ(x) − δ(x)ẑ −
3
4π
r
r r
r
2
ẑ
zr̂
= −δ 0 (t)
ẑδ(x) −
+3
3
4πr3
4πr3
(25)
(26)
(27)
(28)
c.
We will proceed by first putting Jt in the form
1
∂ 1
Jt = −δ (t) ẑδ(x) +
∇
4π ∂z r
0
(29)
In the coloumb gauge, the vector potential satisfies the wave equation
1 ∂2A
= −µ0 Jt
(30)
c2 ∂t2
Since the vector potential is in the form of a wave equation, which is causal, and the E and B fields follow directly from
the vector potential by equation (6.31) in Jackson, the fields are indeed Causal.
∇2 A −
We’re gonna do this by fourier transforming the vector potential and transverse current in the time coordinate in order
to get rid of those pesky delta functions. This gives us
∇2 Ã +
ω2
à = −µ0 J̃t
c2
1
∂ 1
ω
ẑδ(x) +
∇
J̃t = −i
2π
4π ∂z r
4
(31)
(32)
This is the helmholtz equation with k ≡ ω/c and solution3
Ã(x, y, z, ω) =
∞ Z
X
n=0
d3 x0
φn (r)φn (r0 )
J̃t
k 2 − kn2
(33)
where the φ are the solutions to the homogenious holmholtz equation
φn (r, θ, φ) =
∞
X
(al jl (nr) + bl yl (nr))Pl (θ)
(34)
l=0
and we have chosen only the m = 0 solutions since the source is azimuthally symmetric and the j and y are the sherical
bessel functions. Now we ought to expand the current in spherical bessel functions. The delta function is easy, since the
bessel functions are complete. I suppose alternatively we could note that the greens function of the helmholtz equation gives
us a solution
µ0
Ã(x, ω) =
4π
Z
0
d3 x J̃t
eik|x−x |
|x − x0 |
(35)
So that we can write
Ã(x, ω) =
µ0
4π
Z
0
d3 x J̃t
iωµ0
=−
4π
= −iω
µ0
4π
= −iω
µ0
4π
= −iω
µ0
4π
= −iω
µ0
4π
eik|x−x |
|x − x0 |
(36)
0 eik|x−x | 2
1
∂ 1
d x
ẑδ(x) +
∇
|x − x0 | 3
4π ∂z r
!
Z
ik|x−x0 |
eikr 2
e
1
∂
1
ẑ + d3 x
∇
r 3
|x − x0 | 4π ∂z r
ikr
Z
0
1
∂ 1
e 2
ẑ − d3 x eik|x−x | r̂ · ∇(|x − x0 |)
∇
r 3
4π ∂z r
ikr
Z
e 2
1
∂ 1
3
ik|x−x0 |
ẑ − d x e
(2 + ∇(r̂ψ))
∇
r 3
4π ∂z r
ikr
Z
1
∂ 1
e 2
3
ik|x−x0 |
ẑ − d x e
(2 + ∇(r̂ψ))
∇
r 3
4π ∂z r
Z
3
(37)
(38)
(39)
(40)
(41)
(42)
Lets focus on the second term in integral
Z
0
d3 x eik|x−x | ∇(r̂|x − x0 |)
1
∂
∇
4π ∂z
Z
0
1
1
∂ 1
= d3 x ∇(r̂|x − x0 |)∇eik|x−x |
∇
r
4π ∂z r
0
∂
= 0 (r̂|x − x0 |)eik|x−x | |x0 =0
Z∂z
0
1
∂ 1
+ d3 x (r̂|x − x0 |)∇(eik|x−x | )
∇
4π ∂z r
(43)
(44)
(45)
(46)
Ok. I give up... for now.
3 http://mathworld.wolfram.com/GreensFunctionHelmholtzDifferentialEquation.html
5