Name _________________
Solutions to Test 3
April 8, 2015
This test consists of three parts. Please note that in parts II and III,
Constant
you can skip one question of those offered.
0  4 107 T  m/A
RL
Circuits
LC Circuits
Possibly useful formulas:
 L R
AC Circuits
  1 LC
I  I 0 e t 
V  Vmax sin t 
Q  Q0 cos t 
EM waves

I  I max sin t   
E0  cB0
I  1  e t  
R
Impedance
1
S  E  B
X C  1 C , X L  L
RMS values
0
RLC Circuits
2
2
2
V
Z  R   X L  XC 
Vrms  max
S  cB02 20
1
R2
2


  tan 1  X L  X C  R 
LC 4 L2
P S c
I
I rms  max
Q  Q0 e  Rt 2 L cos t 
0  1 LC
2
Inductor
L  0 N 2 A 
U  12 LI 2
Circles
C  2 R
A   R2
Generator
  N  BA sin t 
Triangles
A  12 BH
Faraday’s Law
d
 E  ds   dt B
Spheres
A  4 R 2
V  43  R 3
Cylinders
V   R2 L
Alat  2 RL
Ampere’s Law
 B  ds   I   
0
0
Cones
V   R2 L
1
3
Alat   R L2  R 2
Part I: Multiple Choice [20 points]
For each question, choose the best answer (2 points each)
1. Which of the following is a property that a transformer would be useful for?
A) Increasing or decreasing voltage (only)
B) Increasing or decreasing current (only)
C) Changing alternating current to direct current (only)
D) Both (A) and (B), but not (C)
E) None of the above
2. The unit of inductance is the
A) Volt
B) Ohm
C) Ampere
D) Farad
E) Henry
3. Which of the following is not a type of electromagnetic radiation?
A) Ultraviolet B) Alpha rays C) Gamma rays D) Radio
E) Microwaves
0
dE
dt
4. Which of the following explains the properties of a diode?
A) It blocks low frequencies
B) It blocks high frequencies
C) It tries to keep voltage constant
D) It tries to keep current constant
E) It lets current through one direction, but not the other
5. If we want to induce an EMF in a loop of wire, which of the following methods
would work?
A) Change the size of the loop, so that the magnetic flux through it changes (only)
B) Change the strength of the magnetic field (only)
C) Move the loop to a new location, where the magnetic flux is different (only)
D) All of the above
E) None of the above
6. Which of the following colors has the longest wavelength?
A) Red
B) Blue
C) Yellow
D) Violet
E) Orange
7. Which of the following formulas would give the time averaged power  consumed
by a resistor with resistance R in an AC circuit?
2
2
2
A) RI max
B) I max
R
C) RI rms
2
D) I rms
R
E) None of these
8. Electrical power in the US is generated by
A) Strings of batteries connected in series
B) Rubbing materials of different types together; triboelectric generators
C) Loops of wire mechanically rotated in strong magnetic fields
D) Conductive materials under high pressure pushed through filters that only allow
electrons through
E) Harnessing natural sources of electricity, such as lightning
9. In a perfect LC circuit, power is never lost. This is unrealistic for real circuits
because
A) We want to consume power, so such a circuit is useless
B) Real circuits, especially inductors, tend to have a lot of resistance
C) There must be batteries or other sources of power, which have resistance
D) Real capacitors are also inductors, and inductors are capacitors, so these cancel
out
E) This is only true at one frequency, and you can never match that frequency
exactly
10. To construct an inductor, one would typically use
A) A solenoid; stacked coils of wire
B) A material of known resistivity and known dimensions
C) Two parallel plates a fixed distance apart
D) Doped semi-conductors which allow current through only one way
E) A chemical source of a voltage difference
Part II: Short answer [20 points]
Choose two of the following questions and give a short answer (1-3 sentences).
11. For the simple circuit sketched at right, assume at t = 0, there is
charge in the capacitor but no current flowing. Explain
qualitatively what happens when the switch is closed, giving any
relevant equations.
C
L
When the switch is closed, current will begin to flow through the inductor,
moving the energy to the inductor, which then pushes it back into the capacitor,
alternating back and forth, with the charge on the capacitor oscillating according to the
equation Q  Q0 cos t  , where the frequency is given by the formula   1 LC .
12. Explain qualitatively how the two terms on the right side of Ampere’s Law
 B  ds  0 I   0 0  d  E dt  each allow different ways to create a magnetic
field.
The presence of current causes magnetic field; this is represented by the first
term. Magnetic fields can also be caused by changing magnetic fields, specifically,
changing electric flux, as represented by the second term.
13. State Lenz’s Law. In particular, tell what would happen if a
permanent magnet, with field lines as sketched in, were
suddenly withdrawn from a conducting ring, as sketched at
right.
According to Lenz’s Law, when magnetic flux through a conducting loop
changes, current will flow in the loop which attempts to maintain constant flux through
the loop. In particular, if you pull the magnet to the right, which would eliminate the
magnetic field through the loop, a current will flow to recreate that magnetic field. In
this case, this current would have to flow counter-clockwise as viewed end-on.
14. A loop of copper wire is in the shape of a right triangle of size 30.0
cm  40.0 cm, with a hypotenuse of 50.0 cm, as sketched at right,
with N = 125 turns. It is in the presence of a magnetic field
pointing into the paper, which is changing as a function of time
given by: B  t    0.40 T/s 2  t 2
40 cm
Part III: Calculation: [60 points]
Choose three of the following four questions and perform the indicated
calculations (20 points each)
30 cm
(a) Find the flux as a function of time, for the single loop and for the 125 turns.
Flux is magnetic field times area. A triangle of size 30.0 cm  40.0 cm will have area
A  12 BH 
1
2
 0.300 m  0.400 m   0.06 m 2
The flux from one loop is then just
 B1  AB   0.0600 m 2  0.40 T/s 2  t 2   0.0240 Wb/s 2  t 2 .
For N loops, we simply multiply by N to give
 BN  N  B1  125   0.0240 Wb/s 2  t 2   3.00 Wb/s 2  t 2 .
(b) Calculate the EMF around the 125 loops as a function of time.
The EMF can just be calculated from
 
d
   3.00 Wb/s 2  2t    6.00 V/s  t .
dt
The sign is a little hard to figure out, but it means that the EMF will try to create a current
that opposes the increasing magnetic field. For t > 0, this means the EMF is counterclockwise.
(c) If the entire 125 loops have a resistance of 0.78 , what will be the current at
t = 2.00 s? Which direction does it flow?
The EMF at this time is clearly
   6.00 V/s  2.00 s   12.00 V
The current is therefore
I
V 12.0 V

 15.4 A.
R
0.78 
The current will flow in the direction of the EMF, or counter-clockwise.
125 
5H
15. A circuit consists of a 12.0 V battery, an ideal 5.00 H
inductor, and two resistors of magnitude 50.0  and 125.0
, as sketched at right. Initially, the switch is closed and
has remained closed for a long time.
(a) With the switch closed, in a steady state, which circuit
elements will current flow through? Find the initial
current I0 flowing through the inductor.
12 V
– +
50 
In the steady state, the inductor acts like a wire. Hence both sides of the inductor
will be at the same voltage, and hence there is no voltage across the 125  resistor, and
no current will flow through it. All the current will flow through the 50  resistor and
the inductor. The current flowing will just be
I0 
V 12.0 V

 0.24 A  240 mA .
R
50.0 
(b) At t = 0, the switch is opened. Now which way will the current flow?
Calculate the current as a function of time. Find the time when the current
drops to 1.0 mA.
The current can no longer pass through the battery nor the 50  resistor, but the
inductor wants to keep the current constant, so it will flow through the 160  resistor,
initially still operating at 240 mA. It will decay with time constant

L 5.00 H

 0.0400s .
R 125 
We therefore have
I  t   I 0 et    240 mA  e t  0.0400 s 
We then equate the current of 1.0 mA to I(t) and solve for t, to yield
1 mA   240 mA  e t 
et   240 ,
t   ln  240   5.48 ,
t  5.48    0.0400s   0.219 s
(c) Immediately after the switch is opened, at t = 0, what will be the voltage
across the 125  resistor?
Since the current running through it at this moment is 240 mA, the voltage across
it is just
V  IR  125   0.240 A   30.0 V .
V
I
V (V), I (mA)
16. A circuit consists of an AC source of unknown
frequency and voltage being fed through a
resistor with R = 500  and a capacitor of
unknown capacitance. A graph of the voltage
(solid line, in V) from the source and the
current (dashed line, in mA) through the
circuit is sketched at right.
(a) From the graph, estimate Vmax and Imax.
What is the approximate impedance Z for
the two components?
The maximum current and voltage can be approximately read
from the graph:
t (s)
500 
Vmax  65 V , I max  50 mA .
We then can use the equation Vmax  I max Z to find the impedance,
Z
Vmax
65 V

 1300  .
I max
50 103 A
(b) From the graph, estimate the frequency f and calculate the angular
frequency .
The wave goes through a whole cycle in 0.02 s, so T = 0.02 s. The frequency is
the reciprocal of this, so f = 50 Hz. The angular frequency is then
  2 f  2  50 s 1   314 s 1
(c) Estimate the capactative reactance XC for the capacitor at this frequency?
What is the capacitance of the capacitor?
We first use the formula for the impedance in terms of the resistance R and the
capacitative reactance XC, namely Z  R 2  X C2 . Squaring this and solving for XC, we
have
Z 2  R 2  X C2 ,
X c  Z 2  R2 
1300     500  
2
2
 1, 440, 000  2  1200  .
We then use the equation for capacitative reactance, namely X C  1 C  , which we
rearrange to find the capacitance:
C
1
1

 2.65 106 F  2.65  F .
1
 X C  314 s  1200  
17. The James Webb space telescope has a large sunshield
designed to reflect the light coming from the Sun. For
purposes of this problem, assume the sunshield is a
rectangle 21 m  14 m perpendicular to sunlight. The
intensity of sunlight is approximately S  1361 W/m 2 .
(a) What is the light pressure on the telescope, assuming it is perfectly
reflecting? What is the total force pushing on this telescope?
Since the light is reflecting, the pressure is 2 S c ,* which works out to
P
2 S
c


2 1361 W/m 2
  9.08 10
6
2.998  10 m/s
8
N/m 2 .
Multiplying by the area of the sunshield, the force will be


F  P  9.08 106 N/m 2  21 m 14 m   0.00267 N .
(b) The total mass of the telescope is 6500 kg. Calculate the acceleration of the
telescope due to sunlight. Calculate how long it would take sunlight to cause
it to move 1.00 m, assuming it starts from rest.
We simply use F  ma to find the acceleration, which is
a
F 0.00267 N

 4.107 105 m/s 2
m
6500 kg
For uniform acceleration starting from rest, the displacement is d  v0t  12 at 2  12 at 2 , or
solving for t, we have
t2 
2 1.00 m 
2d

 4.87 106 s 2 ,
a 4.107  107 m/s 2
t  4.87  106 s 2  2207 s  36.8 m .
(c) Suppose that the mirror surface were mistakenly painted black. How long
would it take for the “shield” to absorb one million Joules of energy from the
Sun?
Since the energy is power times time, E   t , we solve for time and get
t
E
E
1.00 106 J


 2.50 s .

S 
1361 W/m 2  21 m 14 m 


* Some people did the calculation as if the pressure is unchanged for reflection, but the
force is doubled. This is equally acceptable.