PHYS 212 – MT3 Spring 2013 Sample 3 Solutions

PHYS 212 – MT3
Spring 2013
Sample 3 Solutions
Question 1
Consider the Bohr model of the hydrogen atom, in which circles the nucleus, in this case in
a counterclockwise direction as shown. At the moment the electron is at the position
indicated, it will produce a magnetic field at the nucleus (“+”) that points:
A.
B.
C.
D.
E.
into the page.
out of the page.
to the left.
to the right.
None of the above
+
A counterclockwise circling electron is a clockwise circling current. So by the RHR this is a
dipole into the page (B field at the center is into the page).
Question 2
Magnetic dipole X is fixed and dipole Y is free to move. Dipole Y will initially:
A.
B.
C.
D.
E.
move away from X and rotate.
move away from X but not rotate.
rotate but not move toward or away from X.
move toward X but not rotate.
move toward X and rotate.
It wants to align with the B field created by X (which is in the direction of X at the center of Y).
But energetically there won’t be any change in the energy by moving because the dipole moment
is perpendicular to the field.
Page 1 of 11
PHYS 212 – MT3
Spring 2013
Sample 3 Solutions
Question 3
A conducting ring moves downward in the magnetic field of a permanent magnet whose north
pole points upward. 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 up increasing, so Lenz says: (1) make B field down (CW current) and (2) fight the change
of v down to make F up.
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PHYS 212 – MT3
Spring 2013
Sample 3 Solutions
Question 4
A long straight wire is in the plane of a rectangular conducting loop. The straight wire carries an
decreasing current in the direction shown. The current in the rectangle is:
i
A.
B.
C.
D.
E.
counterclockwise in the left side and clockwise in the right side.
zero.
counterclockwise.
clockwise in the left side and counterclockwise in the right side.
clockwise.
The current makes a B field into the page at the rectangular loop (RHR). As I decreases so does
B, so the rectangle tries to make more B into the page  CW current
Question 5
An long wire with current i1 is tangent to a circular current loop with current i2, as shown below.
The B field at the center of the loop is measured to be zero.
From this data, you deduce that:
A. i1  i2 
B. i1   i2 
C. i1   i2
D. i1   i2
E. You can’t relate the currents without
knowing the radius
The linear current is making a field into the page of magnitude B  0i1 2 r
The circular current is making a field out of the page of magnitude B  0i2 2r
These are equal and opposite at the center to make the B field there zero, so
0i1 2 r  0i2 2r  i1   i2
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PHYS 212 – MT3
Spring 2013
Sample 3 Solutions
Question 6
An electron moving in the positive x-direction experiences a magnetic deflection in the negative
y-direction. Which could be the direction of the magnetic field?
A. Positive z-direction
B. Negative z-direction

F  qv  B     ˆi   ??   -ˆj  ˆi   ??   ˆj
C. Positive y-direction
D. Negative y-direction
E. Positive x-direction
Question 7
A bar magnet, centered on the origin, lies along the x-axis with its North pole pointing in the +x
direction. A very long straight wire runs parallel to the z-axis at x = 0, y = +1 cm and carries
current in the +z direction. Looking at the very short segment of this wire which penetrates the
xy-plane (in other words, which is located at (x,y,z) = (0, +1cm, 0)), what is the direction of the
force on it due to the magnetic field from the bar magnet?
A. Positive z-direction
B. Negative z-direction
C. Positive y-direction
   
D. Negative y-direction F  IL  B  kˆ  -ˆi  ˆj (B field direction? Draw dipole field!)
E. None of the above
Question 8
A vertical wire carries a current straight down (from the sky to the Earth).
To the east of this wire, the magnetic field points
A.
B.
C.
D.
E.
toward the north.
toward the east.
toward the west.
toward the south.
downward.
The field wraps clockwise around the wire when viewed from above
Page 4 of 11
PHYS 212 – MT3
Spring 2013
Sample 3 Solutions
Question 9
A current-carrying wire that is parallel to the y-axis experiences a force completely in the
z-direction in the presence of an unknown magnetic field. The magnetic field:
A. must have an x component, and cannot have any other components.
B. must have a z component, and cannot have any other components.
C. must have some x component, but could have at least one other component.
D. must have some y component, but could have at least one other component.
E. must have some z component, but could have at least one other component.


F  IL  B  ˆj   ??   kˆ , so must have ˆi but could also have ˆj but not any kˆ

Question 10
The below circuit is constructed with three identical resistors R, a capacitor C and a battery with
EMF . The switch is open for a long time.
What is the current supplied by the battery immediately after the switch is closed?
A. 2
The capacitor is empty, so acts like a wire and shorts out
R
the resistor above it. It is then in series with two resistors
B. 
R
in parallel (equivalent resistance R/2)
C. 2
3R
D. 
E. 
2R
3R
F. 0
G. None of the above
Page 5 of 11
PHYS 212 – MT3
Spring 2013
Sample 3 Solutions
Question 11
A square piece of copper slides on a horizontal frictionless table. There is a square region of
constant uniform magnetic field perpendicular to the table, as shown. Which graph correctly
shows the speed v of the copper as a function of time t as it passes into and out of the region of
field?
Whenever the flux changes (when the loop enters or exits the field) there will be a force trying to
“fight the change!” and hence slow down the loop
Question 12
You are given two long straight pieces of wire and a semicircle of wire and asked to construct a
shape that, when a current is passed through it, creates a large B field at the center of the
semicircle (that is, the center of the circle from which the semi-circle is cut). You construct:
The B field created at the center of the semi-circle in the above circuits is:
A. Largest in II
B. Largest in II & III (they are equal)
C. Largest in I
D. Largest in I & III (they are equal)
E. Largest in III
In {I} the straight lengths don’t contribute to B. In II they subtract but in III they add.
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PHYS 212 – MT3
Spring 2013
Sample 3 Solutions
Question 13
Which of the following equations can be used to show that magnetic field lines form closed
loops?
Qin
A.
 E  dA 
B.
 B  dA  0
C.
 E  ds  
D.
 B  ds  I
o
Magnetic Gauss’s Law (see the CQ “B field inside a magnet”)
dB
dt
o through
E. None of the above
Question 14
A rectangular loop is placed in a uniform magnetic field aligned in the positive z-direction. There
is a current I circulating in the loop. In which of the following positions will the loop rotate
around the z-axis?
A.
B.
C.
D.
E.
a
b
c
d
none of the above
The dipole moments will want to rotate to align with the external B field, that is, with the z-axis.
So the dipoles might rotate around the x- or y- axes, but they will never rotate around the z-axis
since it won’t help them align with the z-axis.
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PHYS 212 – MT3
Spring 2013
Sample 3 Solutions
Question 15
A current I is uniformly distributed over the cross-section of a conducting wire of radius a. For
distances r from the center of the wire with r < a,
A.
B.
C.
D.
E.
The magnitude of the magnetic field does not depend on r.
The magnitude of the magnetic field is proportional to 1/r
The magnitude of the magnetic field is proportional to r
The magnitude of the magnetic field is proportional to 1/r2
The magnitude of the magnetic field is proportional to r2
Ampere’s law. Uniform current means constant current density, call it J. Then:
 B  ds  B2 r  
0
I enc  0 J  r 2  B  r The field grows linearly inside the wire and
decays like 1/r outside of the wire.
Question 16
A square coil rotates in the direction shown (see sketch) in a magnetic field directed to the
right. At the time shown, the current in the square when looking down along its normal nˆ
and the magnetic torque vector on the coil will be:
A. Current counterclockwise and torque out of the page along the rotation axis
B. Current counterclockwise and torque into the page along the rotation axis
C. Current clockwise and torque out of the page along the rotation axis
D. Current clockwise and torque into the page along the rotation axis
The flux to the right is increasing so we want to make a flux to the left, so current runs CW as
seen from above (or from the right). Lenz also says “fight the rotation!” so there will be a torque
out of the page along the rotation axis to try to stop the rotation (use right hand screw rule to see
how this works)
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PHYS 212 – MT3
Spring 2013
Sample 3 Solutions
Question 17
Consider the below circuit consisting of two fixed conducting rails (top and bottom) and
two bars which make good electrical contact to the rails and are free to slide (no friction).
The bars each have some (not necessarily the same) resistance, while the rails are
superconducting (no resistance).
The entire circuit is put in a uniform magnetic field B, pointing out of the page as pictured
(it is everywhere!)
The right hand bar is pulled with a force F such that the bar moves at a constant velocity v.
What, if any, force will the left bar feel because of this?
A. The same force F to the right
B. The same force F but to the left
C. A smaller force (Fleft < F) to the right
D. A smaller force (Fleft < F) to the left
E. None of the above/impossible to determine
Whatever current runs through the right wire also goes through the left, and they are in the
same field so will feel the same force. The net force on the right wire is zero (constant v) so
the magnetic force must be F to the left. By RHR (or by Lenz – fight the change of the
separation between the wires) the force on the left wire is to the right.
Question 18
An electron is accelerated from rest through a potential V and enters a magnetic field, where it
undergoes cyclotron motion (i.e. it moves in a circle of radius Re with period Te). If a proton
were to be accelerated from rest through a potential of the same magnitude and entered the same
magnetic field, what would be true of its radius Rp and period Tp?
A.
B.
C.
D.
E.
Rp > Re and Tp > Te
Rp > Re and Tp < Te
Rp < Re and Tp > Te
Rp < Re and Tp < Te
None of the above
The acceleration: qV  12 mv 2
The circular motion:
qvB  mv 2 R  R  mv qB  2mqV qB  2mV qB 2
T  2 R v  2 mv qBv  2 m qB
Both R and T will grow if you increase the mass
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PHYS 212 – MT3
Spring 2013
Sample 3 Solutions
Question 19
Consider the circuit shown in the figure. Note that two currents are shown, and that while some
of the battery EMFs are given, two (1 and 3) are not. What is the EMF ε1?
A.
B.
C.
D.
E.
12 V
20 V
28 V
44 V
Impossible to determine with the given information
Walk CW around the outer loop, starting in the upper left corner:
 I1  4    2  I 2  6   I1  4   1  0
Solving for 1
1  I1  4    2  I 2  6   I1  4 
  4A  4   20V-  4A  6    4A  4 
 16V  20V-24V  16V  28V  1
Page 10 of 11
PHYS 212 – MT3
Spring 2013
Sample 3 Solutions
Question 20
The circuit at left consists of three identical
resistors each with resistance R, two
identical batteries with emfs , and a
capacitor with capacitance C. The capacitor
is initially uncharged at t = 0. After a very
long time, what is the charge Q (as labeled
in the picture) on the capacitor?
A. Q  23 C
B. Q  C
C. Q  12 C
D. Q  0
E. None of the above (maybe the sign is wrong?)
After a long time the capacitor is “fully charged” meaning that no current flows through the
branch with the capacitor. So the two resistors effectively in series with the right battery
(the top R and central R) act as a voltage divider, each with /2. Since the left branch is in
parallel with the central R that means it also has /2 across it. So, start at the bottom center
and walk up the left branch: Q/C –  = /2
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