P HYSICS 210A, W INTER 2015
H OMEWORK 7 ( DUE T HURS . M AR . 5, IN CLASS )
READING FOR WEEK 9: Zangwill Chapter 13 (skip 13.9), 14 (skip 14.13) or Jackson Chapter 5 &
6; We will cover magnetic materials and start electrodynamics
[1.] Jackson Problem 5.26
PHYSICS 210A: HOMEWORK 7 SOLUTIONS
TA: BIJAN POURHAMZEH
1.
Jackson 5.26
We can compute the inductance using equation (5.154) in Jackson. We can write it as
L=
1
I2
d3 xJ (x) A (x)
where the integral is taken over the volume of the loop. From Ampere's law,
PHYSICS 210A: HOMEWORK 3 SOLUTIONS
TA: BIJAN POURHAMZEH
1.
Jackson 4.10
(a) We can solve this problem by considering two separate congurations, one with the
space between the shells lled completely with dielectric material and one with a
vacuum between th
P HYSICS 210A, W INTER 2015
H OMEWORK 6 ( DUE T HURS . F EB . 26,
IN CLASS )
READING FOR WEEK 8: Zangwill Chapter 12, 13 or Jackson Chapter 5
We will cover more of orbits, magnetic energy, dipole interaction, magnetic materials
[1.] Jackson Problem 5.3
[2
P HYSICS 210A, W INTER 2015
H OMEWORK 4 ( DUE T HURS . F EB . 5, IN CLASS )
[1.] Jackson Problem 3.4
[2.] Jackson Problem 3.9
[3.] Finish the in-class example: consider an innitely-long, rectangular conducting duct (length
of each side is L). Each side of
P HYSICS 210A, W INTER 2015
H OMEWORK 3 ( DUE T HURS . J AN . 29, IN CLASS )
[1.] Jackson Problem 4.10
[2.] A thick spherical shell with inner radius a and outer radius b is made of a material with
uniform polarization P = Po z. Find E(r ) everywhere.
[3.
P HYSICS 210A, W INTER 2015
H OMEWORK 2 ( DUE T HURS . J AN . 22, IN CLASS )
[1.] Jackson Problem 4.2
[2.] Jackson Problem 4.7 (a), (b)
[3.] (Zangwill Problem 4.5) Place a point electric dipole p at the origin and release a point charge
q (initially at re
P HYSICS 210A, W INTER 2015
H OMEWORK 1 ( DUE T HURS . J AN . 15, IN CLASS )
Reading: Jackson Chapter 1 through 1.12
[1.] Jackson Problem 1.7
[2.] Jackson Problem 1.9
[3.] Jackson Problem 1.13
[4.] A long cylindrical wire carries current I along the z dir
P HYSICS 210A, W INTER 2015
H OMEWORK 8 ( DUE T HURS . M AR . 12,
IN CLASS )
READING FOR WEEK 10: Zangwill Chapter 14 (skip 14.13) or Jackson Chapter 5.15 & 5.18
[1.] Jackson Problem 5.18
[2.] (Zangwill 13.9) A Hole Drilled through a Permanent Magnet. The
v =15.55 mr' s cfw_35s misl (see s] = s43 mi's
which can also be convened to3l].3 kmi'h.
45. We neglect air resistance, which justies setting a = g = 49.3 mis1 cfw_taking down as
the - direction for the ouration ofthe fall. This is constant acceleration m
PHYSICS 210A: HOMEWORK 4 SOLUTIONS
TA: BIJAN POURHAMZEH
1. Jackson 3.4
(a) In spherical coordinates, the general solution to the potential inside the sphere takes
the form
Alm rl Ylm (, )
(r, , ) =
l=0
where we set the coecients in front of
r = 0,
to zer
PHYSICS 210A: HOMEWORK 2 SOLUTIONS
TA: BIJAN POURHAMZEH
1.
Jackson 4.2
You can check the result by plugging it into the equation for the potential, , and seeing
that it equals the potential for a point dipole. Let's do that.
e (x ) 3
1
dx
4 0
|x x |
p
(x
PHYSICS 210A: HOMEWORK 1 SOLUTIONS
TA: BIJAN POURHAMZEH
1.
Jackson 1.7
We can assume that each cylinder is very long and has a uniform linear charge density
and , respectively on its surface. In reality, if you're trying to maintain a constant voltage
on
PHYSICS 210A: HOMEWORK 6 SOLUTIONS
TA: BIJAN POURHAMZEH
1.
Jackson 5.3
We can divide the solenoid until innitesimal rings with current IN dz . The magnetic eld
from one of these rings is
0 IN dz
a2
3
2
(a2 + z 2 ) 2
dBz =
Adding these up to get the total
PHYSICS 210A: HOMEWORK 8 SOLUTIONS
TA: BIJAN POURHAMZEH
1.
Jackson 5.18
We can cheat a little and use the result from problem 5.17. The idea is to place an image
loop a distance d below the slab interface with current I = r 1 I and compute the
r +1
force
Lecture Notes #14
Physics 210A, Fall 2010
Review
r
r
The E and B field of a moving charge
rr
r rr
r
r
q
r
2
2r
E ( x, t ) =
where u cr v
r r 3 ( c v )u + r ( u a )
ret
4 0 ( r u )
rr
rr
1
B( x, t ) = r E ( x, t )
c
Jefimenkos Equations
Time-dependent Coul
Lecture Notes #13
Physics 210A, Fall 2010
Review
Wave equations.
1 2
2 2 2 =
c t
0
r
r 1 2 A
r
2 A 2 2 = 0 J
c t
The general solutions to the wave equations are
r
r
( x ' , tr )
1
( x, t ) =
r r dV '
4 0 | x x ' |
rr
rr
0 J ( x ' , t r )
A( x , t ) =
r