Name _________________
Solutions to Test 2
March 4, 2015
This test consists of three parts. Please note that in parts II and III, you can skip one
question of those offered.
Solenoid
Loops
Force
0 NI
τ  IA  B
Between
B
Possibly useful formulas:
U   IA  B
Wires
L
Cyclotron
Constants
0 I1 I 2
F
Motion

19
e  1.602 10 C
Field from Straight Wire
L
2 d
mv  qRB
7
0  4 10 T  m/A
I
qB
B  0  cos 1  cos  2 

Hall Effect
4 a
m
Resistance
IB
0 I

V


B
H
  0 1   T  T0  
tnq
RC Circuits
2 a
t 
Q  Q0 e
R  R0 1   T  T0  
Biot-Savart
Ampere’s Law
Q  C 1  e t  
L
0 I ds  rˆ
R
B
A
 B  ds  0 I
  RC
4  r 2
Circles
C  2 R
A   R2
Triangles
A  12 BH
Spheres
A  4 R 2
V  43  R 3
Cylinders
V   R2 L
Alat  2 RL
Metric Prefixes
k = 103  = 10–6
c = 10–2 n = 10–9
m = 10–3 p = 10 –12
Part I: Multiple Choice [20 points]
For each question, choose the best answer (2 points each)
1. In the Hall effect, a wire is placed in a magnetic field. An electric potential is
generated that is __________ to the direction of current flow and _________ to the
magnetic field.
A) Perpendicular, perpendicular
B) Perpendicular, parallel
C) Parallel, perpendicular
D) Parallel, parallel
E) None of the above
2. Which of the following units is equivalent to an ohm ()?
A) V/A
B) A/V
C) A·V
D) W/A
E) W/V
3. The cross product of the unit vectors ˆi  kˆ is
A) î
B) iˆ
C) ĵ
D) jˆ
E) k̂
4. The loop at right contains currents as marked. What is the equation
derived from Kirchoff’s second law for this loop?
+
–
5
3
7
I2
I1
A)
B)
C)
D)
E)
5 + 3I1 + 7I2 = 0
5 – 3I1 + 7I2 = 0
5 + 3I1 – 7I2 = 0
5 – 3I1 – 7I2 = 0
None of the above
5. If 10 A of current is running through a wire for 20 s, how much total charge passes
through?
E) 200 C
A) 1/200 C B) ½ C
C) 1 C
D) 2 C
I1
6. Three currents are joined at a point as sketched at right. What is the
I3
equation derived from Kirchoff’s first law from this vertex?
A) I1 + I2 = I3 B) I1 + I3 = I2 C) I2 + I3 = I1 D) I1 + I2 + I3 = 0 E) None of these
7. When a capacitor is discharged through a resistor, the capacitor initially has energy in
it, but at the end it does not. Where does the energy go?
A) It is stored as electric energy in the resistor
B) It is stored as electric energy in the wires
C) It is converted to heat in the capacitor
D) It is converted to heat in the wires
E) It is converted to heat in the resistor
8. The formula for the magnetic field given for a long solenoid, B  0 NI L , is
A) Correct everywhere
B) Correct inside the solenoid; outside it is approximately zero
C) Correct outside the solenoid; inside it is approximately zero
D) Correct only at the exact center of the solenoid
E) Correct only on the surface of the solenoid
9. Qualitatively, two parallel wires with current running through them feel a force
because
A) Current from one creates a magnetic field, which the other feels
B) Current from one creates an electric field, which the other feels
C) The Hall effect causes charges to accumulate on one side of each of the wires,
which then attract each other
D) One wire creates an electric potential which the other one feels
E) None of the above
10. Which of the following is a correct relation for the current density J in terms of the
current I, the cross sectional area A, and the length L?
A) J = IL B) J = I/L C) J = I/A D) J = IA E) J = I/(LA)
Part II: Short answer [20 points]
I2
Choose two of the following questions and give a short answer (1-3 sentences) or
brief sketch (10 points each).
11. A light bulb is essentially a resistor. Explain qualitatively how a light bulb
works. It turns out that the resistance of a light bulb that is sitting on a shelf is
rather different from the resistance of a light bulb in operation. Explain why,
giving one relevant formula.
A light bulb is a very thin wire of tungsten, which has resistance and, therefore,
when current passes through it, it consumes power. This makes the wire so hot it glows.
The resistance of a resistor tends to be temperature dependent, according to the equation
R  R0 1   T  T0   , where  is the temperature coefficient of resistivity. If it isn’t
zero (and it rarely is), then the resistance of an ultra-hot tungsten filament will tend to be
very different from the room-temperature value.
12. A student is trying to measure the current and the voltage
for the resistor at right. Sketch where an ammeter and
a voltmeter should be added to make this work. You
may draw them directly on the diagram provided, if you V
wish.
–
+

R
C
A
The dashed lines show what you need to add.
13. Define magnetic flux, giving any relevant formula. Then give a simple way to
compute the total magnetic flux out of an enclosed region.
Magnetic flux is the dot product of magnetic field over area,  B  B  nˆ A , or for
more complicated situations, where the surface is curved or the magnetic field varies,
 B   B  nˆ dA . The total magnetic flux out of an enclosed region is always zero, so
 B   B  nˆ dA  0 . You can’t get much simpler than that!
Part III: Calculation: [60 points]
`
Choose three of the following four questions and perform the indicated
calculations (20 points each)
14. A gallon of gasoline can produce approximately 1.20  108 J of energy. Suppose
an electric car has a battery with the equivalent of five gallons of gasoline, or
6.00  108 J.
(a) Suppose it is being charged by a 500 V source that is pumping 16.0 A into the
car (larger voltages or currents can be dangerous). How many hours will it
take to recharge the car?
The power is given by   I V  16.0 A  500 V   8000 W . Power is energy
over time,   E t , so
t
E 6.00  108 J
1m 1h

  75, 000 s  

 20.83 h .

8000 W
60 s 60 m
(b) You don’t want the cables to consume too much power. If the cables
charging the car (and therefore carrying the current) are consuming 320 W
of power, what must their resistance be?
We start with the equation   I V  I  IR   I 2 R , so
R
320 W


 1.25  .
2
2
I
16.0 A 
(c) Assume the cables are made of copper (   1.68 108   m ), have a circular
cross-section, and have a total length of l  25.0 m . What is the radius r of
cable required to obtain the resistance found in part (b)?
We use the given formula for resistance, R   l A , which we rearrange to have
A
l
R
1.68 10

8
  m   25.0 m 
1.25 
 3.36 107 m 2 .
We know that the cross-sectional area of a circle is A   r 2 , which we rearrange to give
r 2  A  , or taking the square root, we have
r
A


3.36 107 m 2
 3.27 104 m  0.327 mm .
3.1416
The 10 k and 15 k resistor are connected in parallel,
because they are joined at both ends. Hence their equivalent
resistance is
15.0 k
10.0 V 25.0 F
– +
10.0 k
15. A 10.0 V battery is going to charge a 25.0 F capacitor
through the combination of resistors as sketched at right
(a) What is the equivalent resistance of the three
resistors?
14.0 k
1
1
1
23
1




R 10.0 k 15.0 k 30.0 k 6.00 k
Hence they have a resistance of 6 k. This is then in series with the 14.0 k resistor for
a total resistance of
R  6.0 k  14.0 k  20.0 k .
(b) What is the time constant  for this charging process?
The time constant is given by
  RC   20.0 103   25.0 106 F   0.500 s .
(c) The switch is closed at t = 0. At what time t will there be 225 C of charge
on the capacitor?
We are charging a capacitor, so we use the appropriate equation, which is
Q  C 1  et   . We are trying to solve for the time t, so we start rearranging it to have
Q
 1  e t  ,
C
Q
225 106 C
e t   1 
 1
 1  0.9  0.1 ,
C
 25.0 106 F 10.0 V 
 t   ln  0.1  2.303 ,
t  2.303  2.303  0.500 s   1.15 s .
16. A certain mass spectrometer contains a magnetic field of magnitude B = 0.124
T. Then Na+ ions with charge q = + 1.60210-19 C are injected. The velocity
selector only allows ions with a velocity of v = 8.64104 m/s to pass through.
After passing through the straight section, the sodium ions curve, as sketched.
(a) What direction is the magnetic field pointing?
The magnetic field must be perpendicular both to the direction the charge is
moving (up) and the direction it accelerates after it leaves the straight part (to the left).
Therefore it is into or out of the paper. By making it into the paper, we can convince
ourselves that v  B will be to the left, and since the charge is positive, the force
F  qv  B is to the left, and this is the direction it will accelerate.
(b) What magnitude of electric field is required in the straight section to let only
this velocity through? What direction is the electric field pointing?
The total force in the straight section is given by F  q  E  v  B  to cancel
v  B , which points to the left, E must be as large as it and point to the right. The
magnitude will be
E  v B   8.64 104 m/s   0.124 T   1.07 104 V/m
(c) A typical Na+ ion has a mass of 3.8210-26 kg. Find the radius in the curved
portion of its track.
This is straightforward from the formula mv  qRB , which can be rearranged to
yield
R
26
4
mv  3.82 10 kg  8.64 10 m/s 

 0.166 m  16.6 cm .
qB
1.602 1019 C   0.124 T 
17. A long tri-coaxial cable has a thin center wire (current 1.00 A out) surrounded
by a thin cylindrical layer with radius r = 2.00 mm
1 A out
(current 5.00 A in), surrounded by another thin
cylindrical layer with radius r = 4.00 mm (current
4 mm
4.00 A out). Calculate the magnitude of the magnetic
5 A in
field, as well as its direction (where appropriate) at
2 mm
an arbitrary point the following distances from the
center:
4 A out
(a) r = 1.00 mm
(b) r = 3.00 mm
(c) r = 5.00 mm
Each of these is going to be computed the same way, using Ampere’s Law.
According to Ampere’s Law, the integral of the magnetic field around a loop is
proportional to the current running through that loop,  B  ds  0 I . The magnetic fields
will make circles around the center. We will calculate those loops counter-clockwise in
every case, which means we will compute the current coming out of the paper towards
us. In each case, the magnetic field will be parallel to the direction of integration, so the
dot product just yields a factor of 1, and we have  B  ds  B  ds  2 rB . We therefore
have, for all three parts,
B
0 I
.
2 r
In each case, we have to include only the current smaller than the given radius. Our three
answers will therefore be
 4 10 T  m/A  1.00 A   2.00 10 T  200  T ,

2 1.00  10 m 
 4 10 T  m/A  1.00 A  5.00 A   2.67 10 T  267  T ,

2  3.00 10 m 
 4 10 T  m/A  1.00 A  5.00 A  4.00 A   0 .

2  5.00  10 m 
7
Ba
4
3
7
Bb
4
3
7
Bc
3
Because the second one came out negative, that means it is actually going clockwise, so
our final answers are (a) 200 T counter-clockwise, (b) 267 T clockwise, and (c) 0 T
(no direction)