Solutions to Final Exam for GP II, RE, Spring 2011.
1. 20%
The LW rectangular loop in the figure has a mass per unit
length of  and is pivoted about the side ab on a frictionless
axis. The current in the loop is I in the direction shown.
Find the magnitude and direction of the magnetic field
parallel to the y-axis that will cause the loop to swing up until
its plane makes an angle of  < 90 with the yz-plane.
Solution
In equilibrium,
τ  0 .
m  2 W  L 
 I A IW L
B  B yˆ
With respect to point A, the z-components of ’s are
 mg  m g
W
sin    W W  L  g sin 
2
    B cos   I W L B cos 
 Wg
B  1  
tan 
L I

For  < 90, B is positive, so B is parallel to ŷ .
2. 15%
The wire in the figure carries a current I in the direction shown. The wire
consists of two long, straight sections connected by a quarter circle of radius R.
What are the magnitude and direction of the net magnetic
field at the center of curvature of the quarter-circle section
(point P) ?
Solution
0 I d l  rˆ
4
r2
For the straightline sections, d B  0 since d l  rˆ  0 .
For the quarter-circle section,
Biot-Savart law:
dB 
d l  rˆ  d l kˆ  R d kˆ ,
where k̂ points out of the paper and  is the polar angle around P. Hence,
 I  R 0 I ˆ
B  kˆ 0 2

k
4 R 2
8R
3. 25%
The infinitely long, straight wire in the figure carries a constant current I.
The metal bar of length L beneath it moves with a constant velocity v.
(a) Calcuate the emf in the bar.
(b) Which point, a or b, is at a higher potential?
(c) If the bar is replaced by a square wire loop of
sides L, resistance R, and lying in the plane of
the paper, what is the magnitude of the induced
current?
Solution (a)
Using Ampere’s law, the magnetic field of wire at the metal bar
is easily found to be
 I
B   0 kˆ ,
2 r
where k̂ points out of the paper, and r is the perpendicular
distance from the wire.
Motional emf in dr is
d E  v  B  d l  v
0 I
dr
2 r
Since there is no current, the potential difference is equal to .
b
 I b1
 I r
 I  L
Vb  Va   d E  v 0  d r  v 0 ln b   v 0 ln  1  
a
2 a r
2
ra
2
 d
Solution (b)
Thus, potential is higher at a.
Solution (c)
Since there’s no change in magnetic flux through the loop during a
d
 0.
uniform motion, E  
dt
Hence, I  0 .
This result can also be obtained by calculating the motional emf’s
as in (a), which gives
Vb  Va  E  Vc  Vd
Vb  Vc  0  Va  Vd
so that E  0 going around the loop.
4. 15%
A resistor, inductor, and capacitor are connected in parallel to voltage V  V p cos  t .
(a) Find the impedance of the circuit.
(b) Draw the phasors for IR , IC , and IL for a given V.
Solution (a)
1 1
1
 i C 
Z R
i L
1

1 
Z    i  C 

 L  

R
Z 
Note:
1 
1 
  C 
2
R 
 L 
2
either Z or |Z| can be called the impedance.
Solution (b)
1
5. 25%
Find the reading in each ammeter and voltmeter
(a) just after switch S is closed.
(b) after S is closed a long time.
Solution (a)
Just after switch S is closed, i  0 (open circuit) for each inductor and V  0 (short
circuit) for each capacitor. Thus
A2  A3  0
V5  0
and the circuit simplifies to
or
Hence,
IT 
40.0 V
 0.800 A
50.0 
so that
A1  A4  0.800 A
A2  A3  0
V1  40.0 V
No current
V2  V4


V3  V4  0
V2  0
V5  0
Solution (b)
After S is closed a long time, V  0 (short circuit) for each inductor and i  0 (open
circuit) for each capacitor. Thus
V3  V4  V5
V2  0
A4  0
The circuit simplifies to
or
Hence,
IT 
40.0 V
 0.480 A
50.0  33.33 
so that
A1  0.480 A
V1   0.480 A50.0   24.0 V
V2  0
V3  V4  V5   40.0  24.0V  16.0 V
A2 
16.0 V
 0.160 A
100.0 
A3 
16.0 V
 0.320 A
50.0 
A4  0