Name _________________
Solutions to Test 3
April 9, 2014
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
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 an AC circuit with an ideal resistor, capacitor, and inductor, which component(s)
actually consume power
A) Resistor (only)
B) Capacitor (only)
C) Inductor (only)
D) Resistor and capacitor
E) Capacitor and inductor
2. Magnetic fields can be made by electric current, as well as by
A) Constant electric fields
B) Changing electric fields
C) Constant magnetic fields
D) Changing magnetic fields
E) Loops of wire in motion
dE
dt
3. According to Lenz’s law, if you move a loop of wire in a region with a magnetic
field, currents will appear in the wire that try to keep the __________ constant
through the wire
A) Magnetic field
B) Magnetic flux
C) Electric field
D) Electric flux
E) Current
4. The electronic component that resists changes in current is a
D) Inductor
A) Resistor
B) Capacitor C) Diode
E) Switch
5. Suppose you have 240 V AC going into a transformer, and you want to get 120 V AC
out. If there are 200 turns on the primary winding, how many should there be on the
secondary?
E) 100
A) 800
B) 400
C) 200
D) 143
6. Suppose I had an air-core inductor, and I decided I wanted to increase the inductance.
Which of the following would be a bad way to do this?
A) Increase the number of turns
B) Increase the cross-sectional area
C) Increase the length (while leaving the number of turns the same)
D) Add an iron core
E) Actually, all of these would be a good idea
7. In a steady state, an ideal inductor acts like
A) A wire, no resistance
B) An open circuit, no current
C) A capacitor
D) A switch
E) None of the above
8. Which of the following describes the physical meaning of wavelength ?
A) How many times it repeats in a meter
B) How many times it repeats in a second
C) The distance between one peak of a wave and the next
D) The time between one peak of a wave and the next
E) The velocity of the wave
9. What type of signals tend to be blocked by a capacitor?
A) High voltage
B) Low voltage
C) High current
D) High frequency
E) Low frequency
10. If Vmax = 100 V for an AC source, what is Vrms?
A) 200 V
B) 141 V
C) 100 V
D) 71 V
E) 50 V
Part II: Short answer [20 points]
Choose two of the following questions and give a short answer (1-3 sentences).
11. An ideal transformer can, in principle, increase the voltage by a large factor.
Since   I  V  , doesn’t this mean that the power increases a lot? Doesn’t this
mean that energy is being created for free?
No, energy is not created for free. Though the voltage out is greater than the
voltage in, the current decreases by the same factor, so that I1  V1   I 2  V2  , and there
is no increase in power for an ideal transformer. For a realistic transformer, there will in
fact be a drop in voltage or current on the output, so it is even worse.
12. Explain what it means when the current lags or leads the voltage in an AC
circuit. When you have a capacitor, which of these does it do?
The current lags (or leads) the voltage when the current has its peaks and valleys
before (after) the voltage has its peaks and valleys. When you put an AC source through
a capacitor, the current will lead the voltage.
13. Place the following in order from longest wavelength to shortest: gamms-rays,
infrared, microwaves, radio, ultraviolet, visible, X-rays.
Radio, microwaves, infrared, visible, ultraviolet, X-rays, gamma rays
Part III: Calculation: [60 points]
Choose three of the following four questions and perform the indicated
calculations (20 points each)
14. A loop of wire or radius R is in the shape of a circle with N =
500 loops in a region with a magnetic field of B = 1.30 T,
pointing perpendicular to the loop, as sketched at right. The
circle is expanding its radius at a velocity of 2.00 m/s, so that
at time t, the radius is R = 2t.
(a) Calculate the magnetic flux through a single loop, and through all 500 loops,
as a function of time.
Magnetic flux is magnetic field times area. Therefore, the magnetic flux is
 B  BA  B R 2  1.3   2t   16.3t 2 Wb .
2
This is for one loop. We now multiply this by the number of loops, so we have


 B  NBA  500 16.3t 2 Wb  8170t 2 Wb
(b) Calculate the EMF around the 500 loops as a function of time.
We calculate this using the formula
 
dB
 8170  2t  V  16,340t V .
dt
The minus sign means it is opposite to the right-hand rule. So basically, if we point our
thumb in the direction of the magnetic field, our fingers curl clockwise, so the current
goes counter-clockwise.
This is consistent with Lenz’s Law, which says that as you increase the size of the
circle, the magnetic flux increases, so to compensate, a current will flow counterclockwise to cancel out some of the magnetic field.
(c) At what time is the EMF equal to 150 V? At this time, which direction is the
EMF generated, clockwise or counter-clockwise?
Since it didn’t specify which direction in the question, we can simply set the emf
to the voltage we found, and not worry about sign.
150 V  16,340t V ,
t
150 V
 0.00918 s  9.18 ms .
16,340 V
The direction, as already described, will be counter-clockwise.
15. An inductor with L = 0.120 H is connected via a switch to a
capacitor with C = 32.00 F, as shown at right. Initially,
L C
there is a charge Q0 on the right hand side of the capacitor.
At t = 0, the switch is closed (ignore the dashed figure for
now)
(a) Find the angular frequency , the frequency f, and the period T over which
this oscillator will oscillate. Also find a formula for the charge Q as a
function of time.
The angular frequency can be found using the formula

1

LC
1
 0.120 H   32.0 10
6
F

 510.3 s 1 .
We can then use some simple formulas to find the frequency and period.
f 

 81.2 Hz ,
2
T
1
1

 0.0123 s  12.3 ms .
f 81.2 s 1
The formula for the charge as a function of time is given by
Q  Q0 cos t   Q0 cos  510.3t  .
(b) What is the first positive time t when the capacitor has no charge on it? At
this moment, the inductor is measured to have a current I = 0.200 A running
through it. Find the energy in the inductor at this moment. Is there any
energy in the capacitor at this moment?
The capacitor has no charge on it when cos t   0 . The first positive time when
this happens is when t  12  , so
t



 0.00308 s  3.08 ms .
2 2  510.3 s 1 
This is, in fact, exactly one-fourth of the period T. At this point, the energy in the
inductor is
U  12 LI 2 
1
2
 0.120 H  0.200 A 
2
 0.00240 J  2.40 mJ .
Since the capacitor has no charge on it, it has no energy, so this is in fact the energy of
the whole system.
16. A certain circuit consists of a Vrms  120.0 V at
A
f  60.0 Hz source fed into a heater with a resistance of
R  20.0  . Initially, boxes A and B contain simple wires,
as shown.
(a) Calculate the current Irms through the resistor. This is
the amount of current our heroes need to keep from freezing.
B
The current is given by
I rms 
Vrms Vrms 120.0 V


 6.00 A .
Z
R
20.0 
(b) Horrible Henry sneaks in and replaces box A with an inductor with L = 0.500
H. Find the impedance Z of the resulting circuit, and the current now
passing through the resistor.
We first need to find the reactance of the inductor, which requires that we find the
angular frequency. This is simply
  2 f  2  60.0 s 1   377.0 s 1
The reactance is then
X L  L   377.0 s 1   0.500 H   188.5  .
The impedance then increases to
Z  R 2  X L2 
 20.0    188.5  
2
2
 189.6  .
The current is now
I rms 
Vrms 120.0 V

 0.633 A .
189.6 
Z
(c) Friendly Fred then replaces box B with a capacitor that exactly cancels out
the effects of the inductor, thereby restoring the current to the value found in
part (a). What capacitance did he use?
The impedance of the inductor can be cancelled by an equal impedance of the
capacitor, so that X L  X C . Substituting the corresponding formulas in, we find
1
,
C
1
1
C

 1.407 105 C  14.07  C .
2
2
1
L
 0.500 H   377.0 s 
L 
17. An electromagnetic wave has a wavelength of 457 nm and an electric field given
by
E   324 V/m  ˆi sin  kz  t 
where the î denotes the direction of the electric field.
(a) What is the value of k and  for this wavelength?
We can use the formulas k   2 and   ck to find these two quantities:
2
 1.375 107 m 1 ,
 457 109 m
  ck   2.998 108 m/s 1.375 107 m 1   4.122 1015 s 1 .
k
2

(b) What is the magnitude and direction of the magnetic field B for this wave?
First note that because our wave is like sin  kz  t  , the wave is moving in the zor k̂ direction. The magnetic field must be perpendicular to both the electric field and
the direction of motion, so it must be (up to sign) in the ĵ -direction. In fact, since
S  E  B , this would imply S would be in the ˆi  ˆj  kˆ direction, so this is correct.
The magnitude is related by E0  cB0 , so
B0 
E0
324 V

 1.081106 T  1.081  T .
c 2.998 108 m/s
This, in fact, tells us everything we need to know about the magnetic field, which is then
B  1.081  T  ˆj sin  kz  t  .
(c) What is the time-averaged intensity S ?
We use the formula
6
8
cB02  2.998 10 m/s 1.0807 10 T 
S 

 139.3 W/m 2 .
7
2 0
2  4 10 T  m/A 
2