Solutions to Final Exam for GP II,
SNME,
Spring 2012.
1. 10%
A network of resistors of resistance R1
and R2 extends to infinity towards the
right (see figure). Let the total
resistance of the network be RT . If a
potential difference V is applied across
points a and b, what is the potential difference across points c and d.
Hint:
What is the resistance of the network to the right of points c and d ?
Solution:
Since the network is infinite, the resistance of the network to the right of points c
and d is also RT . The total resistance between points c and d is therefore
R R
Rdc  T 2
RT  R2
Applying the voltage divider rule,
R
R2
Vdc  cd V 
V
RT
RT  R2
2. ( 5% + 5% + 10% )
(a) What is Vab  Vb  Va when switch S is
open?
(b) What is the final Vb when S is closed?
(c) How much does the charge on each
capacitor change when S is closed?
Solution:
(a) There’s no current through the capacitors.
Va  18 V
Vb  0 V
Vab  18 V
(b) There’s only a current I through the resistors.
18 V
 I
 2A
 6  3 
Vb  2 A  3   6 V
(c) When switch S is open,
Q6  F  C V  6  F  18 V  108 C
Q3 F  C V  3  F  18 V  54 C
When switch S is closed,
Q6 F  C V  6  F  18  6V  72 C
Q3F  C V  3  F  18 V  2 A  6   18 C

Q6 F   7 2 1 0 
8 C 
3 6C
Q3F  18  54 C  36 C
3.
20%
A circular loop has radius R and carries
current I in a clockwise direction (see
figure). The center of the loop is a
distance D above a long, straight wire.
What are the magnitude and direction of
the current I1 in the wire if the magnetic
field at the center of the loop is zero?
Solution:
Biot-Savart law:
B
0
4

I d l  rˆ
r2
Let the z –axis points out of the paper.
B
Field due to the loop current is
0 I 2
 I
 2 R zˆ   0 2 zˆ
2
4 R
2R
Ampere’s law

C
B  d r  0 I . Let I1 be positive when flowing to the right.
Choose C as a circle of radius D, centered on the axis, and passing through the
center of the loop.
B  2 D  0 I 1
B
At the center of the loop,
0 I 1 0 I 2

0
2 D
2R
Total field vanaishes:

I1 
0 I1
zˆ .
2 D
 D I2
R
( flows to the right )
4. ( 10% + 10% + 10% )
A bar of length L = 0.8 m is free to
slide without friction on horizontal rails
(see figure). A uniform magnetic field
B = 1.5 T is directed into the plane of
the figure. At one end of the rails is a
battery with emf  = 12 V and a switch.
The bar has mass 0.90 kg and resistance
5.0 , and all other resistance in the circuit can be ignored. The switch is closed
at t = 0.
(a) Sketch the speed of the bar as a function of time.
(b) Just after the switch is closed, what is the acceleration of the bar?
(c) What is the terminal speed of the bar?
Solution:
Let the x-axis be pointing to the right.
Let the bar be moving with a velocity v to the right. Choose surface as
pointing out of the paper. Then positive induced emf means induced current
flowing upwards in the bar.
d
Ei n d  B L x  B L v (opposes the battery emf  )
dt
The current on bar is therefore
I
E  BLv
R
( positive if flowing downward )
Force on the bar is
F  I L  B  I L B xˆ

m a
E BLv
L B
R
dv
L2 B 2
E
L2 B 2 
E 

v
LB  
v


dt
mR
mR
mR 
LB 
Let w  v 
E
LB
dw
L2 B 2

w
dt
mR


Thus,
w  w0 e  L B
2 2
v
t/R

2 2
E
1  e  L B t / mR
LB

with v0  0 ,
v 
E
LB
(a)
(b)
Just after the switch is closed (t = 0), v = 0. Hence,
a
(c)
 0.8 m 1.5 T  12 V  3.2 m / s 2
L2 B 2 
E  LB
E 
 
v 

mR 
LB  mR
 0.9 kg  5.0  
Terminal speed of the bar is
v 
E
12 V

 10 m / s
LB  0.8 m 1.5 T 
5. 20%
Vs in the figure is a signal source.
Derive an expression for the
amplitude ratio Vout / Vs as a function of the angular frequency  of the source.
Find the limits for
0
and
.
Solution:
Vs 
Q
dI
L
 IR  0
C
dt
For a sinusoidal signal of frequency ,
I
Vs 
 i  LI  I R  0
iC
Voltage divider:
Vout 
i LR
Vs
 1

R i
 L
 C

Amplitude ratio:
Vout

Vs

 L 
i LR

 1

R i
 L
 C


2
LC    C R 
 C R 
2
2
0,
Vout
 0.
Vs
For
,
Vout
 1.
Vs
 R2
 1

R  
 L
 C

2
2
  1   2 LC 
For
2
2
2