Phys 4322 Exam 1
Feb. 10, 2015
You may NOT use the text book or notes to complete this exam. You and may not
receive any aid from anyone other that the instructor. You will have 1.5 hours to nish. DO
YOUR OWN WORK. Express your answers clearly and concise
SET 3
7.26)
The circuit equation is developed from Vc = Q/C and Vl = L dI . Here Q is the charge
dt
on the capacitor, and I is the current in the circuit. As the charge, Q, decreases, the current
increases. Then;
2
I/C + L d I = 0
dt2
The solution has the
Phys 4322 Exam 2 - SOLUTION
Mar. 10, 2015
You may NOT use the text book or notes to complete this exam. You and may not
receive any aid from anyone other that the instructor. You will have 1.5 hours to nish. DO
YOUR OWN WORK. Express your answers clearly
SET 2
7.1)
This problem was essentially worked in class. Solve 2 V = 0 in spherical coordinates,
then surfaces of constant potential are spherical shells of area 4r 2 . The resistance between
two shells a distance dr apart is;
dR =
dr
4r 2
Integrate over
SET 1
2.26)
Figure 1 shows the geometry for the integration. The dierential of charge is;
dq = d dr
The potential is obtained with da the dierential area as;
V =
da
|r r |
R
b
r
z
Figure 1: The geometry of the problem
Then da = d dz with z = rho sin() so
SET 6
12.33)
The pion decays into 2 gamma rays which must conserve momentum and energy. Assume
that the photons are emitted along the direction of motion on the pion.
Energy Conservation
Epi = E1 + E2
Momentum Conservation
p = p1 p2
The relatavistic relat
SET 10
9.20)
a) The energy density is; W = (1/2)[E 2 + b2 /]. This is integrated over time to get
the time average.
E = E0 ex ei[kxt+]
B = B0 ex ei[kxt+ ]
In the above, is the imaginary component and k is the real component of the wave
vector. The phase d
SET 9
9.2)
The function to test is f (z, t) = A sin(kz) cos(kct)
It is easily shown that this is a solution to the wave equation;
2
2f
2 f
2 (1/c)
z
t2
by substitution. The equation can be experssed as a standing wave by superimposing
waves moving in oppo
SET 6
8.1)
The power is obained from the Poynting vector. The E eld is obtained from Gauss
law by placing a uniform charge per unit lenght on the inner conductor. Direction is
determined by symmetry to be radial.
E da E(2rL) = Q/
E = 2 r r
0
The magneti
SET 5
12.11)
Divide the circumference of the loop into a large number of straight lengths. Put each
length in an inertial frame which is instantaneously at rest, so that the length moves with a
1 with = r. The radius
speed, r. Each length then appears con
SET 8
8.7)
The magnetic eld is B = 0 nI z The torque is N = r F and F = I d l B where I
is the leakage current through the spoke. We neglect any induced eld because the discharge
is slow. The force on spoke is;
dF = 0 I nId
The torque on the cylinders is