1
Two identical current loops have currents I flowing in opposite directions, as shown.
1
The net force between the two loops is: attractive or repulsive ?
Solution: Two different principles are at play here: we need to know the direction of the magnetic field created by one of the wires, and then what direction the force on the other wire due to that field is. For the first part, we need to use the
Biot-Savart law:
B =
µ
0
4 π
I d l × ˆ r 2
(1)
Let’s think about the bottom wire. Using the right hand rule and considering a point on the wire , we see that the vector in the direction of the current (dl) crossed into the vector in the direction of ˆ gives a magnetic field out of the page.
If you do that all the way around the circle, you can see that the magnetic field points radially outward from that upper loop. Now we need to see what the force on the top loop is due to this field. Using the Lorentz force law,
B = Id l × B (2)

we see that the direction is given through the right hand rule:
2
So, the force is up. By symmetry, the force on the bottom loop will be down. Thus, the net force between the loops is repulsive .

2
For the circuit shown in the figure below use the values:
3
E
0
= 150V
R
1
= 317Ω
R
2
= 951Ω
C = 7 .
15 µ F
What is the initial battery current immediately after the switch S is closed?
Solution: Right after the switch is closed, the capacitor is uncharged. Thus, there is essentially no voltage drop across it, so it behaves like a wire - current can pass through it without any resistance. Thus, resistor battery current is simply:
R
2 is short-circuited and so the
I = V /R = E
0
/R
1
= 150V / 317Ω = 473 mA (3)

3
What is the battery current a long time after the switch S is closed?
Solution: A long time after the current has been running through the circuit, the capacitor will be fully charged and so no current can pass through it. Thus, the current is routed through both R
1 and the capacitor on it. The current is then:
R
2 and will completely ignore the wire with
I = V /R = E
0
/ ( R
1
+ R
2
) = 150V / (317 + 951)Ω = 118 mA (4)
4

4
4) A solenoid is required to produce a magnetic field at its center of magnitude B =
15T. It is to have length L = 2m, bore radius a = 10cm, and can be approximated as an infinitely long object. It is to be constructed of 10,000 turns of superconducting wire. What current must the wire carry to produce the desired field?
5
Solution: Ampere’s law (or your crib sheet!) tells us that the magnetic field in the center of this solenoid is:
B = µ
0 nI.
(5) where n is the turn density. We can then solve for I :
B
I =
µ
0 n
=
(15T)(2m)
4 π × 10 − 7 (V · s/(A · m))(10000)
≈ 2 .
4 × 10
3
A (6)

5
Two wires are made from the same material. Which of the following is true?
A) If one is twice as long as the other, it needs to be twice as thick in order to have the same resistance.
B) As long as their diameter is the same, they have equal resistance, independent from their length.
C) If one is 9 times longer than the other, it needs to be 3 times thicker in order to have the same resistance.
D) As long as their length is the same, they have equal resistance, independent of their diameter.
Solution: We know that the resistance of a material is proportional to its length and inversely proportional to its area. Since area is proportional to diameter squared
( A = πr
2
= π ( d/ 2)
2
), then we know that A cannot be correct since twice as thick means 4 times the area, so the resistance would decrease by a factor of 2. B also cannot be correct, since that ignores the length dependence. C) could be correct, since if the length is 9 times bigger, the area needs to be 9 times bigger as well, which, since it depends on diameter squared, requires that it be 3 times thicker. Finally, D) ignores the area dependence of resistance. The answer then is C .
6

6
Two capacitors half-filled with a dielectric are shown in the figure below:
7
If C a is the capacitance of the configuration on the left and C the configuration on the right, which of the following is true?
b is the capacitance of
A) C a
< C b
B) C a
= C b
C) Need to know the value of the dielectric constant
D) C a
> C b
A key thing to recognize with this problem is that the setup on the left is two capacitors in parallel, while the setup on the right is two capacitors in series. Recognizing that, let’s find out what the equivalent capacitance for each layout is. Starting with the capacitance of a parallel plate capacitor and remembering that in free space,
κ = 1, we have:
C =
κ
0
A
.
d
(7)

8
Start with the left configuration. Here we have two capacitors, each with separation distance d but only half the total area:
C a
=
=
κ
0
A
0
A
+
2 d 2 d
0
A
(1 + κ ) .
2 d
Now let’s consider the right configuration, where we have two capacitors in series with area A but now only half the separation distance:
1
C b
=
=
= ⇒ C b
= d
+
2 κ
0
A d ( κ + 1)
2
2 κ
0
A
0
A κ d
2
κ + 1 d
0
A
Comparing these two results, we see that the value of each depends on the value of
κ . If κ is small (take it to be close to one), then we have:
C a
C b
≈
≈
0
A
2
2 d
0
A
(1 + 1) =
1
= d 1 + 1
= ⇒ C a
= C b
, κ → 1
0
A d
0
A d
However, notice that this only ever occurs if κ = 1. Since κ is always slightly greater than 1, C a will actually be slightly greater than large. In this case,
C b in this limit. Now take κ to be
= ⇒
C a
≈
C
C b a
≈
2
0
A
κ
2 d
0
A κ
= d κ
C b
, κ 1
2
0 d
A
So we see that in both cases C a function of k justifies this further:
> C b
. A numerical plot of the ratio C a
/C b as a

In[22]:=
Ca = a

2 *
H
1 + k
L
;
Cb = 2 * a *
H k
 H k + 1
LL
;
Plot
@
Ca

Cb

. a ® 1.,
8 k , 0 , 2
<D
1.5
Out[24]=
1.4
1.3
1.2
1.7
1.6
1.1
0.5
1.0
1.5
2.0
9

10
7
7) The next three questions pertain to the situation described below: An electron of mass m (9 .
1 × 10
− 31 kg) and charge q ( − 1 .
6 × 10
− 19 C) is accelerated to the right (in the plane of the page) from rest through a potential difference V = 1 , 500V. The electron then enters a region, defined by x > 0, containing a uniform magnetic field
B = 0 .
3T out of the page.
When the electron enters the region with the magnetic field the force on it is directed
A) Toward the top of the page
B) Into the page
C) Toward the bottom of the page.
Solution: The force is given by the Lorentz force equation:
F = q v × B (8)
Using the right hand rule, fingers point to the right (direction of v ), curl up out of the page (towards B ), so that the thumb points down. However, since it is an electron, q is negative and so the force is opposite what the RH rule gives. So the answer is A .

11
8
When the electron is in the magnetic field region, its speed
A) Remains constant
B) Decreases
C) Increases
Solution: The Lorentz force law shows us that the force acts perpendicularly to the velocity. Thus, it does no work, since this force acts perpendicular to the displacement of the particle. Therefore there is no energy loss, so its kinetic energy (
1
2 mv 2 ) remains the same, so v is constant.

12
9
How much time T does the electron spend in the magnetic field region?
Solution: The force on the electron in the field region is always perpendicular to the velocity; thus, it follows a circular path through the magnetic field. We can find the radius of this path through Newton’s second law, setting the Lorentz force equal to its mass times the (centripetal) acceleration: qvB = mv 2 r mv
= ⇒ r = qB
Thus, the electron travels a semi-circular path of circumference will take time:
πmv qB
, so at speed v it
= ⇒ vT = mvπ
T = qB mπ
.
qB
Notice that this is independent of how fast the electron was moving! That means that we don’t need to worry about using conservation of energy and the potential difference that the electron was accelerated through to calculate its initial velocity.
We can simply calculate
T = π (9 .
1 × 10
− 31 kg) / (1 .
6 × 10
− 19
C) / (0 .
3T) ≈ 6 × 10
− 11 s (9)

13
10
Doubling the potential difference across a capacitor:
A) Quadruples the charge stored on the capacitor.
B) Doubles its capacitance
C) Doubles the charge stored on the capacitor
D) Halves its capacitance
E) Produces none of the other results.
Solution: Let’s step through each possibility. We know that the capacitance is the proportionality constant between the stored charge and the voltage:
C = Q/V.
(10)
If we double the potential difference, this relationship tells us that either the charge must double or the capacitance must get a factor of 2 smaller. This rules out options
A, B, and E. At face value from this equation, it seems like either C or D could be true
- but keep in mind that the capacitance of an object is a purely geometric/physical quantity. It only depends on what you could actually build (separation distance, surface area, radius, etc...) and does not depend on the charge stored on it or voltage put across it. So this eliminates option D, so we must choose C

14
11
The figure below applies to the next two problems. Consider the two cases shown below. In each case a conductor carries the same total current I = 2 amps into the page, and in each case the current is uniformly distributed over the cross-section of the conductor. In Case 1 the conductor is a cylindrical shell of outer radius R
0 and inner radius R
1
= 10cm
= 7 .
5cm. In Case 2 the conductor is a solid cylinder having the same outer radius R
0
= 10cm.
Compare B
1
( a ), the magnitude of the magnetic field at point a ( r = 5cm) in Case
1 to B
2
( a ), the magnitude of the magnetic field at point a ( r = 5cm) in Case 2.
A) B
1
( a ) = B
2
( a ).
B) B
1
( a ) < B
2
( a ).
C) B
1
( a ) > B
2
( a ).
Solution: This problem is an application of Ampere’s law. By drawing a circular
Amperian loop with radius 5 cm inside each wire, we see that in case A there is no enclosed current, whereas in case B there is some enclosed current (which we could calculate by comparing the ratio of enclosed area / total area vs enclosed current / total current, but don’t need to here). Thus the magnetic field at a must be greater in case 2, so we choose B .

15
12
What is the y component of the magnetic field at point b ( r = 15cm) in Case 2?
Solution: We can solve for the magnitude of the magnetic field using Ampere’s law, this time with a circular amperian loop of radius 15cm:
I
B · dl = µ
0
I enc
= ⇒ B 2 πr = µ
0
I enc
µ
0
(2 A )
= ⇒ B =
2 π 15 cm
By using the right hand rule we can see that, with the thumb pointing into the page, the magnetic field at this point points only in the -y direction. Thus, we get:
~
= 2 .
67 × 10
− 6
T ( − ˆ ) = − 2 .
67 × 10
− 6
T in the y direction (11)

16
13
An inductance L and resistance R are connected in series with a battery as in the figure below. A long time after switch S
1 is closed, the current is 2 battery is switched out of the circuit by opening switch S
1 drops to 1.5 A in 44 ms.
and closing
.
1 A. When the
S
2
, the current
What is the time constant for this circuit?
A) 14 ms
B) 29 ms
C) 130 ms
D) 261 ms
E) 42 ms
Solution: If you didn’t remember the correct equation here, you could still have come up with its general form by thinking about what will happen physically to the inductor. For a long time switch S
1 is closed, meaning that current will flow from the battery through both resistors and then through the inductor. This steady-state current is given to be 2 .
1 A. Then, we know if we close S
2 and open S
1
, the stored energy in the inductor will produce a current that will then drop asymptotically to zero. Rembering that decay of this form always takes an exponential form, we see that an equation that will satisfy both of these conditions is:
I ( t ) = I
0 e
− t/τ
(12)

17 where I
0
= 2 .
1A and τ is the time constant. We can now solve for the time constant by using the other information:
I (44 ms ) = 1 .
5 A = 2 .
1 Ae
− 44 ms/τ
=
=
⇒ −
⇒ ln(
1 .
5
2 .
1
1 .
5
) = 44 ms/τ
2 .
1
= e
− 44 ms/τ
44 ms
= ⇒ τ =
− ln(1 .
5 / 2 .
1)
= 130 ms

18
14
14) The next three questions pertain to the situation described below: Two fixed conductors are connected by a resistor R = 20Ω. The two fixed conductors are separated by L = 2 .
5m and lie horizontally. A moving conductor of mass m slides on them at a constant speed v , producing a current of 3.75 amps. A magnetic field
(shown by the black dots in the figure) with magnitude 5 T points out of the page.
In what direction does the current flow through the moving conductor when the bar is sliding in the direction shown?
A) To the left
B) To the right
Solution: Lenz’s law states that the induced current will attempt to preserve the status quo. Since the effect of the rod falling is to increase the area that the magnetic field lines pass through, the flux is increasing (with the magnetic field pointing out of the page). Thus, a current will be induced such that a magnetic field into the page is increases. The right hand rule gives this as a clockwise current, so it will flow the the left through the moving conductor.

19
15
15) At what speed is the bar moving?
A) 6 m/s
B) 3 m/s
C) 5 m/s
D) 1 m/s
E) 9 m/s
Solution: We know the induced current, so we can use that to find out what the change in flux is using Faraday’s law:
E = − d Φ
B dt d
= ⇒ (3 .
75 A )(20Ω) = − dt
( B ( Lh ))
= − (5 T )(2 .
5 m ) dh dt
= ⇒
(3 .
75 A )(20Ω)
(5 T )(2 .
5 m )
= − dh dt
= v
= ⇒ v = 6 m/s

20
16
16) What is the magnetic force on the bar?
A) 28 N
B) 47 N
C) 51 N
D) 37 N
E) 19 N
Solution: Even though the bar is moving, it is essentially neutral and so there is no magnetic force on it due to its velocity and the charges that it is made out of.
However, since there is a current flowing through it, there is a magnetic force due to this net current:
F = ILB = (3 .
75 A )(2 .
5 m )(5 T ) ≈ 47 N (13)