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PHYS 212 – MT3
Spring 2013
Sample 1 Solutions
Question 1
A rigid circular loop of conducting wire with a current
flowing in it is placed in a non-uniform magnetic field as
shown in the figure. The current is flowing clockwise as
viewed from above. Ignoring other fields such as gravity,
what motion will the loop execute?
A.
B.
C.
D.
E.
Generally upward.
Generally downward.
Rotational, in the plane of the loop.
Rotational, out of the plane of the loop.
More than one of the above
3 ways to do this: (1) Dipole points down, B field is from
magnet pointing up, they repel; (2) Dipole points down,
opposite field, so seeks weaker field (up); (3) IL x B (part of
B that matters points outward)
Question 2
Two identical currents flow along the x and y axes as shown. At which point or points is the
net magnetic field from these two wires zero?
in
out
out
in
in
in
A.
B.
C.
D.
E.
out
out
At point A only
At point B only
At point C only
At points A & C
None of the above
Page 1 of 11
PHYS 212 – MT3
Spring 2013
Sample 1 Solutions
Question 3
A student tries to analyze the circuit shown in the figure. She chooses current directions as
shown. Which of the following equations is correct with the currents as drawn?
A.
B.
C.
D.
E.
+1 – I3R3 – I1R1 = 0
+1 – I3R3 + I1R1 = 0 Walk CW around left loop, starting in lower left corner
+2 – I2R2 + I1R1 = 0
+1 + I3R3 – I1R1 = 0
+2 + I3R3 – I1R1 = 0
Question 4
An electron moves in the negative x direction, through a uniform magnetic field in the
negative y direction. The magnetic force on the electron is:
A.
B.
C.
D.
E.
in the negative x direction
in the positive y direction
in the negative y direction
in the positive z direction
in the negative z direction
   
 
F  qv  B     ˆi  ˆj     ˆi  ˆj  kˆ
Page 2 of 11
PHYS 212 – MT3
Spring 2013
Sample 1 Solutions
Question 5
The figure shows a bar magnet moving vertically upward toward a horizontal coil. The poles of
the bar magnets are labeled X and Y. As the bar magnet approaches the coil it induces an electric
current in the direction indicated on the figure (counter-clockwise as viewed from above). What
are the correct polarities of the magnet?
A.
B.
C.
D.
E.
X is a south pole, Y is a north pole.
X is a north pole, Y is a south pole.
Both X and Y are north poles.
Both X and Y are south poles.
The polarities of the magnet cannot be determined from the information given.
Induced current is trying to make a field up (by RHR) in the center, meaning that there must be a
field down that is increasing as the bar magnet approaches. So the South pole of the bar magnet
is up
Question 6
The diagram shows a straight wire carrying a current out of the page. The wire is between
the poles of the permanent magnet. The direction of the magnetic force exerted on the wire
is:
B
A. Up
B.
C.
D.
E.
F  IL  B   out    right   up by RHR
Left
Right
Out of the page
Into the page
Page 3 of 11
PHYS 212 – MT3
Spring 2013
Sample 1 Solutions
Question 7
A conducting ring moves downward in the magnetic field of a permanent magnet whose south
pole is on top (see sketch). At the instant shown, the induced current in the coil and the magnetic
force on the ring are:
A. Current as seen from above clockwise and magnetic force down.
B. Current as seen from above clockwise and magnetic force up.
C. Current as seen from above counterclockwise and magnetic force down.
D. Current as seen from above counterclockwise and magnetic force up.
Field down increasing, so Lenz says: (1) make B field up (CCW current) and (2) fight the change
of v down to make F up.
Question 8
In a Physics 212 laboratory exercise, a student drops a magnet through a
conducting loop of wire which is connected to an ammeter (not pictured).
The magnet is held as shown with the north pole of the magnet oriented
downward. As viewed from above, describe what happens in the loop of
wire from the time that the student lets go of the magnet until the magnet
is just barely above the loop.
A.
B.
C.
D.
E.
Nothing happens.
A constant counter-clockwise current is induced.
A constant clockwise current is induced.
A time-varying counter-clockwise current is induced.
A time-varying clockwise current is induced.
B down increasing, so loop tries to make B up  CCW current. Changes
with time since d/dt changes with time.
Page 4 of 11
PHYS 212 – MT3
Spring 2013
Sample 1 Solutions
Question 9
A long straight hollow cylindrical tube has an inner radius Ri and an outer radius Ro. It carries a
current i, uniformly distributed over its cross section. A wire which runs along the tube axis
carries a current of the same magnitude but in the opposite direction.
The magnetic field created by these currents is:
A. zero everywhere
B. zero inside the tube (r < Ri), but non-zero elsewhere.
C. zero outside the tube (r > Ro), but non-zero everywhere inside the tube and in the tube
wall (r < Ro). Ampere’s law: where the enclosed current is 0 the field is 0, else not
D. zero outside the tube and in the tube wall (r > Ri), but non-zero inside the tube (r < Ri).
E. non-zero everywhere.
Question 10
A segment of a wire of length dl carries a current I (see sketch below). The unit vector zˆ
points out of the page.
Cross-product:
1
1
dlˆi  aˆi  bˆj
3 dl  r  3
r
r

1
r3

 

bkˆ
The magnetic field dB due to this wire segment at the location P shown in the sketch is
A. 
C. 
o I
a dl
zˆ
2
4  a  b 2  3/ 2


o I
b dl
zˆ
4  a 2  b 2  3/ 2


B.
D.
o I
a dl
zˆ
2
4  a  b 2  3/ 2


o I
b dl
zˆ
4  a 2  b 2  3/ 2


Page 5 of 11
PHYS 212 – MT3
Question 11
A
Spring 2013
Sample 1 Solutions
B
Both of the above pictures show an iron filings representation of the magnetic fields generated
by a loop of current held above a permanent magnet. In both pictures the coil has a current
which is clockwise when viewed from above. The magnet, however, has been flipped over
between A & B. What is true about the direction of the magnet and the force it exerts on the
coil?
A.
B.
C.
D.
In A its North pole is up and it attracts the coil.
In B its North pole is up and it attracts the coil.
In A its North pole is up and it repels the coil.
In B its North pole is up and it repels the coil.
CW current means N pole down. In A the coil is repelled, so the magnet must be
antiparallel (N pole up); In B the coil is attracted so parallel magnet (N pole down)
Question 12
Just after the switch in the circuit shown below is closed, the
current i3 (through the battery) is given by
A.
B.
C.
D.
E.
 / 3R
 / 2R
3 / 2R
2 / 3R
None of the above
Initially C looks like wire, so two R in parallel (R/2) in series
with R (Requivalent = 3R/2). I = /Req.
Page 6 of 11
PHYS 212 – MT3
Spring 2013
Sample 1 Solutions
Question 13
The circuit shown at right consists of an ideal battery, a resistor,
capacitor, light bulb and a switch. Initially the switch is open and
the capacitor uncharged. Which one of the following graphs best
describes the brightness B of the bulb as a function of time t after
closing the switch?
Initially the capacitor is
uncharged, so it shorts out the
bulb. Over time it will “fill up,”
meaning come to a potential, so
the bulb will have the same
potential across it, and will be
bright.
Question 14
The picture shows the field lines outside a permanent magnet. The field line at the center of the
magnet points:
A. up, by magnetic Gauss’s law
B. down, by magnetic Gauss’s law
C. up, by Ampere’s law
D. down, by Ampere’s law
E. nowhere – the field inside the magnet is zero by
symmetry
Magnetic gauss’s law says the net magnetic flux through a closed surface is ALWAYS zero.
So wrap a surface through the center of the magnet and up over the north pole. B goes out
the top so must come in the bottom.
Page 7 of 11
PHYS 212 – MT3
Spring 2013
Sample 1 Solutions
Question 15
A Bainbridge mass spectrometer is shown in the below figure. A charged particle with mass m
and charge q enters from the bottom of the figure and traces out the trajectory shown. The only
electric field is in the region where the trajectory of the charge is a straight line and there has
magnitude E. The magnetic field strength changes from magnitude B in that region to B0 in the
upper region. Field directions are as indicated.
A. r 
mE
qBB0
B. r 
mB0
qBE
C. r 
mB
qB0 E
qBB0
mE
qEB0
E. r 
mB
D. r 
You could solve this by units (only A has the correct units). But the velocity selector tells us
that qvB  qE  v  E B , then you can calculate the circular motion:
qvB0  mv 2 r  r  mv qB0  mE qBB0
Page 8 of 11
PHYS 212 – MT3
Spring 2013
Sample 1 Solutions
Question 16
A loop of wire moves at a constant speed through regions of varying magnetic field as shown.
Which of the graphs correctly depicts the current in the loop? (A clockwise current is to be taken
as positive, and a counter-clockwise current is negative.)
A.
B.
C.
D.
E.
A
B
C
D
E
As the loop enters region B the external B field is out of the page, so there will be an
induced CW (positive) current to oppose that change. Now as it exits region B and enters
region C not only is it losing an external field out but it is gaining an external field into the
page, so the rate of change of the external flux is TWICE as high as it was before, so the
current will be in the opposite direction but twice as large.
Page 9 of 11
PHYS 212 – MT3
Spring 2013
Sample 1 Solutions
Question 17
Consider two identical circular loops of radius a, separated by a distance a, with the second
(right hand) coil rotated slightly clockwise relative to the first when looking from above:
Top view
Side view
A large current is suddenly injected into the left hand loop. What happens to the right hand
loop?
A.
B.
C.
D.
E.
F.
G.
H.
I.
J.
Nothing
Force to the left, no torque
Force to the right, no torque
Force to the left, torque rotates clockwise (in top view)
Force to the left, torque rotates counterclockwise (in top view)
Force to the right, torque rotates clockwise (in top view)
Force to the right, torque rotates counterclockwise (in top view)
No force, torque rotates clockwise (in top view)
No force, torque rotates counterclockwise (in top view)
Can’t tell without knowing which direction current is injected into left loop
You can do this purely by Lenz’s law – when a current appears in the left loop making a B
field, the right loop wants to get away from that field and the flux it creates. It can do that
by moving away (to the right) and by rotating away (CW) so that the field lines won’t
penetrate it.
Page 10 of 11
PHYS 212 – MT3
Spring 2013
Sample 1 Solutions
Question 18
Ions having equal charges but masses of M and 2M are accelerated through the same potential
difference and then enter a uniform magnetic field perpendicular to their path. If the heavier ions
follow a circular arc of radius R, what is the radius of the arc followed by the lighter?
A. 4R
The acceleration: qV  12 mv 2
B. 3R
C.
The circular motion:
2R
qvB  mv 2 R  R  mv qB  2mqV qB  2mV qB 2
D. R 2
E. R/2
If you halve the mass then you decrease the radius by
2
The following description pertains to the next two questions:
Consider an infinite slab of current parallel to the xz plane (between y = 0 and y = a m, where a
is a positive, dimensionless constant). The slab carries a uniform current density J parallel to the
z-axis in the positive z- direction.
Question 19
What is the direction of the magnetic field at the origin (x,y,z) = (0,0,0)?
A.
B.
ˆi
ˆj
C. iˆ
D. jˆ
E. None of the above
You can draw a picture of the axes and see this or do a
cross product with r from the center of the slab:
B  Id s  rˆ  Ikˆ  ˆj  Iˆi
 
Question 20
What is the magnitude of the magnetic field at the origin?
A. B  12 0 Ja
B. B  0 Ja
C. B  20 Ja
D. B 
1
2
0 Ja
E. None of the above
Ampere’s Law:
We did this in class, so you can go look at the class notes for
more detail, but in short, put the Amperean loop with one side at
y = a/2 (middle of the slab where the field is zero) and one side at
y = 0 (where we want to calculate the field).
 B  ds  Bl   Jl  a   B 
0
1
2
1
2
0 Ja
Page 11 of 11