Oberlin College Physics 411, Electrodynamics, Spring 2014
Assignment 4
Friday, 18 April
Reading: Within Griths chapter 10 on Potentials, read sections 10.1 and 10.2.
Problems: Due Friday, 25 April.
Griths 9.12: Time averages
Griths 9.13: Maxwell stress
Group velocity problems
Griths Electrodynamics, fourth edition, problem 9.23: Water waves and quantal waves
(a) Deep water waves
v
=
v
=
=
vg
2
k
2
=
k
k
2k
d
1
2
1
=
= v
dk
2
k
2
=
=
(b) Quantum mechanical waves
h
The form of the wave Aei(pxEt)/ , co
Time averages
Griths, Electrodynamics, fourth edition, problem 9.12
We have
f (r, t) = A cos(k r t + a ) = ecfw_Aei(krt+a ) = ecfw_f ei(krt)
where f = Aeia . And we have
g(r, t) = B cos(k r t + b ) = ecfw_Bei(krt+b ) =
ecfw_ei(krt)
g
where g = Beib .
Maxwell stress tensor for light
Griths, Electrodynamics, fourth edition, problem 9.13
The wave in question is (see Griths equation 9.48)
E(z, t)
=
B(z, t)
=
E0 cos(kz t + ) Eg (z, t)
x
x
1
1
E0 cos(kz t + ) = Eg (z, t)
y
y
c
c
(The subscript g stands for
Oberlin College Physics 411, Electrodynamics, Spring 2014
Assignment 1
Monday, 31 March
Reading: Griths chapter 7 on Electrodynamics. Section 7.1, Electromotive
Force, should be pretty straightforward. Sections 7.2, Electromagnetic Induction, and 7.3 Maxw
Energy in capacitor discharge
Griths, Electrodynamics, fourth edition, problem 7.2
part (a)
I(t)
+Q(t)
Q(t)
part (c)
+Q(t)
I(t)
C
R
Q(t)
V0
C
R
a. Initial charge is Q0 = V0 C.
Integrate
E d around loop, giving
Q(t)
I(t)R = 0.
C
But
I(t) =
so
dQ(t)
dt
dQ
Electrostatics of a circuit
Griths, Electrodynamics, fourth edition, problem 7.42
(a) Find the potential.
The potential everywhere is the solution to
2
V (s, ) =
1
s s
s
V
s
+
1 2V
=0
s2 2
subject to boundary condition
V (a, ) =
V0
.
2
Use separation of
A solution to the Maxwell equations
Griths, Electrodynamics, fourth edition, problem 7.37
Recall that (vt r) = 0 when r vt and that
d(x)
= (x).
dx
The situation described is: A charge of +q sits directly on top of a charge of q. At t = 0, the charge
+q ex
Solenoid
Griths, Electrodynamics, fourth edition, problem 7.15
Part I: Qualitative character of eld. The eld inside has magnitude B(t) = n0 I(t), the eld outside is
zero. (Current shown in red, B shown in green.)
Side View
Top View
If the current increase
Experimental detection of magnetic monopoles
Griths, Electrodynamics, fourth edition, problem 7.39
Do you remember the Faraday law problems you did in Physics 111 involving changing magnetic ux?
Ed =E =
But also
E = L
so
dB
.
dt
dI
dt
dB
dI
=L .
dt
dt
Int
Thomsons dipole
Griths, Electrodynamics, fourth edition, problem 8.19
E
Angular momentum
B
rm
qm
d
r
qe
At the point in question:
Electric eld is
E=
1 qe
r.
4 0 r2
B=
0 qm
r .
2 m
4 rm
Magnetic eld is
Momentum density is
gem =
0E
B
so it points into page
Capacitor elds and energy
Griths, Electrodynamics, fourth edition, problem 7.34: Charging a capacitor
side view
end view
dE/dt
E
I
I
B
The changing E makes B. . . in fact,
view sketch), because
0
E
makes B in exactly the same way that J makes B (see end
t
Apparent oversight in the ux rule
Griths, Electrodynamics, fourth edition, problem 7.9
surface 2
loop
surface 1
(sense of unit normal
for flux integral)
(sense of unit normal
for flux integral)
(direction of line
integration around loop)
Heres the loop. D
Electric induction
Griths, Electrodynamics, fourth edition, problem 7.8
a. For a long straight wire
B=
0 i
2r
(directed out of the page)
so for this square
a+s
B =
B n da =
s
0 ia a + s
0 i
0 ia
0 ia
a+s
a dr =
[ln r]s =
ln
=
ln(1 + a/s).
2r
2
2
s
2
b. Th
Inductance of a hairpin loop
Griths, Electrodynamics, fourth edition, problem 7.25
y
I
d
x
l
If the current is I, then inside the loop
B(x, y) =
0 I
0 I
+
2y 2(d y)
k.
Clearly, the B due to the top wire doesnt equal the B due to the bottom wire, yet the u
Oberlin College Physics 411, Electrodynamics, Spring 2014
Assignment 3
Friday, 11 April
Reading: Griths chapter 9 on Light. Section 9.4.3 on waves in dispersive media
is particularly important.
Problems: Due Friday, 18 April.
Griths 7.2: Energy in capaci
Alfvens theorem
Griths, Electrodynamics, fourth edition, problem 7.63
(a) We have J = (E + v B), but and J is nite, so E + v B 0 or, to good approximation,
E = v B.
Thus Faradays law
E =
B
t
implies
B
=
t
(v B).
(b) The sum of surfaces S plus R plus S co
Magnetic force between two moving charged particles
a. According to the Biot-Savart law, the magnetic eld at particle 2 due to particle 1 is
B=
0 q1
2 v1 r12 .
4 r12
But according to the Lorentz force law, the force on particle 2 from this eld is
Fon 2 by
Oberlin College Physics 411, Electrodynamics, Spring 2014
Assignment 2
Friday, 4 April
Reading: Griths chapter 8 on Conservation Laws. Section 8.2.2 on Maxwells
Stress Tensor is particularly hairy. Ill try to explicate in class.
Problems: Due Friday, 11 A