Spring 2010
PHY 2054 Exam II
Solutions
Problem 1
Solution:
Due to the Doppler effect, the difference in the frequency of the radar signal (electromagnetic wave) f as measured at the
receiver (SUV) and at the source (officer K in the police car) will depend on their relative velocity
Δ
v
f
v
f
c
Δ
Δ
=
,
where c is the speed of light.
Note that the frequency f and wavelength
λ
of an electromagnetic wave are related as c=f
.
So the above equation can be rewritten as
f
v
Δ
⋅=
Δ
.
Given the numbers in the problem,
11
94
0.0285
= 2.679
=
2.679
3600
6
1600
mm
s
m
i
l
e
vf
m
m
s
ss
h
r
h
r
mile
Δ=Δ⋅ =
⋅
⋅
⋅
=
.
Since the frequency as measured at SUV is lower, it must be moving away from the police car.
Hence, it velocity is 70+6=76 mile/hr.
Problem 2
Solution:
Note that the speed of light is defined by
μ
0
and
ε
0
as
00
1
c
με
=
from where it is clear that the units of the product
0
0
must be s
2
/m
2
.
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View Full DocumentProblem 3
Solution:
After closing the switch, the current in the circuit will be I=
E
(
1e
t/
τ
).
At time t=0, the current is zero.
Going over the loop, starting from the left bottom corner and moving clockwise, the account of potential differences is as
follows:
+
Δ
V
L
+
Δ
V
R
= 0.
The latter is zero due to I=0.
Hence
, 
Δ
V
L
 =
= 6
V
.
Problem 4
Solution:
The current in the loop will be directed in such a way so that as attempting to prevent the change of the magnetic flux in
the loop.
In the first case as the dropping magnet comes closer to the loop (see the figure above), the flux is directed downward and
is increasing
—the induced current in the loop should be counterclockwise with its field pointing upward in the loop.
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 Spring '08
 Avery
 Physics, Magnetic Field, Wire, Δf Δv

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