Name _________________
Solutions to Test 3
April 11, 2012
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
Spheres
A  4 R 2
V  43  R 3
Faraday’s Law
d
 E  ds   dt B
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. In the simple circuit illustrated at right, at what frequencies will
the most power be transferred to the resistor?
A) Low frequencies
B) High frequencies
C) A broad range of intermediate frequencies
D) A narrow band of intermediate frequencies
E) Insufficient information
2. Electricity is produced in commercial generators by
A) Giant banks of chemical batteries
B) Extracting electrons from atoms and pumping them into wires
C) Rubbing objects together to produce static charges
D) Moving loops of wire near magnetic fields (or vice versa)
E) Accelerating electrons using light waves
0
dE
dt
3. If an electromagnetic wave has frequency f = 2 Hz, what is the angular frequency ?
B) 2 s-1
C) 42 s-1
D) 1/(2) s E) 1/(42) s
A) 1 s-1
4. Which two components work best if you want to create an oscillator?
A) Inductor and capacitor
B) Inductor and resistor
C) Resistor and capacitor
D) Inductor and diode
E) Capacitor and diode
5. If you put power into an RLC circuit, the power gradually drains away. Which
component is actually consuming the power?
A) The resistor (only)
B) The capacitor (only)
C) The inductor (only)
D) The capacitor and the resistor
E) The inductor and the resistor
6. Which of the following types of electromagnetic radiation moves fastest in vacuum?
A) Infrared
B) Radio C) X-rays
D) Gamma rays E) None; it is a tie
7. In an LC circuit, where is most of the energy stored at any given time?
A) In the capacitor
B) In the inductor
C) It is split evenly at all times, with 50% in each component
D) It is constantly being passed back and forth between the two components
E) It is impossible to answer this question without know the capacitance and/or the
inductance
8. Why do we use step-up transformers to increase the voltage very high before we
transport it across the country?
A) It has to go to many customers who each get just a little bit of this voltage
B) Some customers (like industry) require very high voltage
C) Since power is voltage by current, it vastly increases the power being transmitted
D) High voltage is less dangerous, so it is just a safety consideration
E) Since it is at high voltage, there is much less current needed, which decreases
the losses in transmission
9. A resistor with resistance R = 3  is connected in series to a capacitor with
impedance XC = 4 . What is the total impedance?
B) 4 
C) 5 
D) 7 
E) 25 
A) 1 
10. Which of the following is the correct relation between the wavelength  and wave
number k?
B)  k  2 C) 2 k  1 D)   2 k E) k  2
A)  k  1
Part II: Short answer [20 points]
Choose two of the following questions and give a short answer (1-3 sentences).
11. Explain qualitatively Lenz’s Law. Suppose I took a superconducting loop of
wire out of a region of magnetic field. What does Lenz’s Law tell us in this case?
Lenz’s Law says when you try to change the magnetic flux passing through a
conducting loop, a current will arise in the loop to try to oppose that change. In
particular, if you remove a superconducting loop of wire out of a region of magnetic
field, a current will spontaneously appear in the loop of wire which recreates that
magnetic flux. Because a superconductor has no resistance, this current will persist
indefinitely.
12. The graph at right shows the
current (dashed curve) that
results from putting an AC
voltage (solid curve) through
a single component (resistor,
capacitor, inductor).
Identify which one, and
explain how you can tell.
The current (dashed curve) rises and falls one-quarter cycle before the voltage
(solid curve); that is, the current leads the voltage. A resistor has no phase shift, and an
inductor causes the current to lag the voltage, so this must be a capacitor, which indeed
does cause current to lead the voltage.
13. Place the following in order from highest
frequency to lowest: blue, green, orange, red,
violet, yellow. You may include additional colors if
you wish, but you don’t get extra credit.
Listed at right, with highest frequency first, and
extra colors squeezed in on the right.
Violet
Blue
Green
Yellow
Orange
Red
Indigo
Turquoise
Chartreuse
Saffron
Vermillion
Part III: Calculation: [60 points]
Choose three of the following four questions and perform the indicated
calculations (20 points each)
magnetic field
14. In 2187, a spaceship approaches neutron star EC50
which has a magnetic field of 1.53×107 T and rotates
12.0 times per second. The captain then places a
circular loop of wire of radius 500 m with 10 turns in
it in the vicinity of this magnetic field, so at t = 0, the
magnetic field is perpendicular to the loop.
(a) Find the magnetic flux through the loop at t = 0.
Wire
loop
EC50
Magnetic flux is given by  B  B  A , where A is the
area vector perpendicular to the loop. This makes A parallel
to B, so we then have


magnetic field
 B  BA  B R 2   1.53  107 T  500 m   1.20 1013 Wb .
2
The problem is ambiguous; this is the flux through a single loop. If you count the twelve
loops (relevant for part (d)), you would multiply this by 12 to get  B  1.44  1014 Wb .
(b) Half a cycle later, the neutron star has rotated, so that the field is now
reversed. What is the flux now? What time is it now?
Since it makes 12 rotations a second, it will be reversed 241 s later, or 0.0417 s.
The flux will be the same, but reversed. Counting the twelve loops, it is now
 B  1.44 1014 Wb .
(c) Assume the magnetic field changed linearly between part (a) and part (b).
What was the EMF generated in the loop of wire during this half-rotation?
Faraday’s Law says that the EMF generated is the negative of the time derivative
of the magnetic field. If we assume the magnetic field changed linearly (it probably
didn’t), then we would have


1.44 1014 Wb  1.44 1014 Wb
dB
 B
 


 6.92 1015 V .
0.0417 s
dt
t
None of the numbers given in the problem are unrealistic for a real neutron star. Whether
we could ever travel to such a star, and build a conductor that could stand up to this
voltage, however, is more in doubt.
5A
1.50 k 
4.0 H
15. A 4.0 H inductor and a 1.50 k resistor are connected
as shown at right, and 5.0 A of current is then fed
through it until it attains a steady state. Assume the
inductor has no resistance.
(a) In a steady state, how much current flows through
the inductor? The resistor?
5A
In a steady state, the inductor has unchanging current, and hence there is no EMF
across it. This tells us the voltage across it is zero, and thus the current in the resistor is
zero. The inductor carries all the current.
I L  5.0 A , I R  0.0 A .
(b) At t = 0, the external current is suddenly switched off. What is the current
through the inductor and resistor a moment later? What is the voltage
across the resistor at this moment?
Inductors resist changes in current, and hence the inductor will still carry 5.0 A
through it. Because this current has nowhere else to go, it will force its way through the
resistor, and hence this is the current in the resistor as well.
I R  I L  5.0 A .
Since resistors are governed by the equation V  IR , the voltage across the resistor will
now be
V  IR   5.0 A 1.50 k   7.5 kV .
(c) How long will it take until the current in the inductor has dropped to 0.0010
A?
The current is given by the equation I  I 0 et  , where   L R . We therefore
have:
0.001 A   5.0 A  e t  ,
et  
t

5.0 A
 5000 ,
0.001 A
 ln  5000   8.517,
t  8.517  8.517
L 8.517  4.0 H 

 0.0227 s  22.7 ms .
1500 
R
16. A certain device requires AC voltage Vmax  5.00 V , with a maximum current
I max  1.40 A , but all that is available is house current with Vrms  120 V .
(a) What is the rms voltage and current required by the device?
The rms voltage and current can simply be found by dividing the maximum by
2 , so we have
Vrms 
Vmax 5.00 V
I
1.40 A

 3.54 V , I rms  max 
 0.990 A .
2
2
2
2
(b) A transformer with 50 turns on the secondary (output) side will be used to
supply this current. How many turns should there be on the primary (input)
side of the transformer? Assume the transformer is 100% efficient.
For an ideal transformer, we have
V1 V2
,

N1
N2
N1V2  N 2 V1 ,
N1  N 2
V1
120 V
 50
 1700 turns
3.54 V
V2
(c) What will be the rms current drawn from the primary source?
Because we assume our transformer is 100% efficient, the power in will equal the
power out; that is,
I1V1  I 2 V2 ,
where all quantities are rms values. This allows us to compute the current in:
I1  I 2
V2
3.54 V
  0.990 A 
 0.0292 A  29.2 mA .
V1
120.0 V
8.0 m
6.0 m
directly at the spaceship’s pointy end for a duration
of 1.00 hour.
(a) What is the maximum intensity of the magnetic
field and electric field from this wave?
6.0 m
17. A black spaceship in the shape of a cone with radius
6.0 m and height 8.0 m is being impacted by a light
beam with intensity S  5.00 108 W/m 2 beamed
We solve for the magnetic field using the equation
S 
cB02
,
2 0
2 0
2  4 107 T  m/A
T  N s
S 
5.00 108 W/m 2   4.188  106

8
c
3 10 m/s
Cm
2
6
 4.188 10 T ,
B02 
B0  2.05 103 T  2.05 mT.
We then get the electric field strength from
E0  cB0   3.00 108 m/s  2.05 103 T   6.14 105 V/m .
(b) What is the total momentum transferred to the spacecraft by the light beam?
To get this right, we have to think carefully about the concept of cross-section.
The cross-sectional area is the area as viewed by the incoming light ray. For this light
ray, the cone is a circle of radius 6.0 m, so the cross sectional area is
   R 2    6.0 m   113 m 2 .
2
We now multiply this by the pressure to get the force:
8
2
2
S   5.00 10 W/m 113 m 
W s
F  P 

 189
 189 N .
8
c
3.00 10 m/s
m
We then multiply by the time to get the total momentum transferred:
p  FT  189 N  3600 s   6.79 105 kg  m/s .