SET 10
1.)
The radiated power due to acceleration for non-relativisitc motion (eqn 14.22 - Larmor
equation in Gaussian units);
Prad = (2/3)(e2 /c) 2
The energy loss is (0 = v0 /c);
0
Eloss =
2
dt Prad = (2/3)(e /c)
dt| |2
0
d
| |2 =
dt
For constant ;
El
Phys 6321 Midterm - Solution
March 18, 2013
You must do all problems in the time allocated. You may NOT use any notes or
books other than those provided by the instructor do you own work. You may
NOT receive aid from anyone other that the instructor. Show
SET 1
1.)
z
(0,y,z)
r
a
a
y
dl
I
I
x
Figure 1: The coordinate geometry for the problem
From the gure above r 2 = x2 + (y a)2 + z 2 . Then we nd the vector potential for each wire.
The currents are in opposite directions which accounts for the dierence in
SET 2
1.)
We assume that is 1. We will need an expansion in powers of .
[1 + P2 ] = 1 + P2 +
( 1)( P2 )2 ( 1)( 2)( P2 )3
+
+
2!
3!
Then evaluate the multipole moment expression;
qlm =
d3 x r l Ylm
The density is constant, 0 , and the surface is given by
SET 3
1.)
In a frame where the B eld is at rest, B = B0 z . We determine A where B = A .
In a cylindrical coordinate system we choose A = B0 (/2). Note that A = 0. Now
assume a quasi-static situation so that is assumed small and constant. In a frame movin
SET 4
1.)
2
y
d/2
1
d/2
x
Figure 1: The coordinate geometry for the problem
Assume the wires have negligible radius. If not, the problem requires a solution in toroidal
coordinates, or a solution where the wire surface is kept at constant potential. Recal
SET 6
1.)
The 4-velocity is u = (c, u):
The length is then;
u u = 2 c2 2 u 2
2 =
1.
1 2
If c = 1 (relativistic units);
u u = 1.
2.)
For 2 reference frames observing the same event, we assume that one frame (aircraft rest
frame) travels faster than c towar
SET 7
1.)
Charge transforms as a scalar and volume transforms as a length. Use Maxwells eqns;
E = /
E = B
t
B =0
B = J + E
t
The Lorentz transformation is chosen without loss of generality to have velocity along the
x1 axis.
x0
x1
x2
x3
= (x0 x1 )
= (x
SET 8
1.)
E d1z
E d2z
E d1z
E d2z
Figure 1: The gure shows the geometry of the problem
The eld due to one end-cap is:
40
(, z ) =
0
(, z ) =
2
a
d
d
0
0
a
1
2
[| | + z 2 ]1/2
dkJ0 (k)J0 (k ) ekz
d
0
0
The eld is (z 0);
Ez =
0
a
dk k J0 (k)J0 (k ) ekz
d
SET 9
1.)
The development is shown in the text.
2.)
Suppose the Lagrangian;
L = (m + g (x) 1 v 2
Substitute into the Euler-Lagrange equations;
(m
L = v + g ) .
v
1 v2
d = x0 + x1 + x2 + x3 .
d
x0
x1
x2
x3
d = u .
d
x
Thus;
and;
d ( L ) = gV
v.
2 x
d
SET 10
1.)
R = R0 [1 + P2 (x)] ;
x = cos()
(t) = 0 eit
The volume is then obtained from;
R
V0 =
d =
d
0
drr 2[1 + P2 (x)]2 (4/3)R3
0
To rst order;
=
Q0
4 R3 /3
0
In the long-wavelength approximation;
l+1
ck l+2
l
i(2l + 1)!
l+2
l+1
aE (l, m) = ck
l
i(2l
Phys 6321 Final Exam - Solutions
May 3, 2013
You may NOT use any book or notes other than that supplied with this test. You
will have 3 hours to nish. DO YOUR OWN WORK. Express your answers clearly and
concisely so that appropriate credit can be assigned