Name ___________________________________ Student ID _______________
last
first
II.
[25 pts] This questions consists of two unrelated parts.
Part 1. In the circuits below, bulbs 1-5 are identical, and the batteries are identical and ideal. Boxes X, Y,
and Z contain unknown arrangements of linear resistors. However, it has been determined that RX = RY > RZ.
1
2
X
4
Y
Z
3
5
A student uses a voltmeter to measure the electric potential difference across the three boxes.
For the following solutions: circuit I is the circuit containing bulb 1 and box X, circuit II is the circuit
containing bulb 2 and the parallel network of bulb 3 and box Y, and circuit III is the circuit containing bulb
4 and the parallel network of bulb 5 and box Z.
17. [5 pts] The absolute value of the measured electric potential difference across box X is:
A. Greater than that across box Y.
B. Less than that across box Y.
C. Equal to that across box Y.
D. There is not enough information provided to compare these quantities.
Circuit I has more resistance and therefore less current through the battery than circuit II (since adding
a parallel branch decreases the total resistance). Since the current through bulb 1 is less than the
current through bulb 2 and the brightness of a bulb is an indicator of current through the bulb, bulb 1 is
dimmer than bulb 2. Since brightness is also an indicator of the potential difference across a bulb, the
potential difference across bulb 1 is less than the potential difference across bulb 2, or ΔV1 < ΔV2 .
From Kirchhoff’s loop rule, Vbat = ΔV1 +ΔVX and Vbat = ΔV2 +ΔVY , so it follows that ΔVX >ΔVY .
18. [5 pts] Bulb 1 is:
A. Brighter than bulb 2.
B. Dimmer than bulb 2.
C. As bright as bulb 2.
D. There is not enough information provided to compare the bulbs according to brightness.
For the reasoning described in question 17, bulb 1 is dimmer than bulb 2.
19. [5 pts] The absolute value of the measured electric potential difference across box Y is:
A. Greater than that across box Z.
B. Less than that across box Z.
C. Equal to that across box Z.
D. There is not enough information provided to compare these quantities.
Circuit II has a greater resistance than circuit III since RY > RZ. Using the same reasoning as in
question 23, ΔV2 <ΔV4. Vbat = ΔV2 +ΔVY and Vbat = ΔV4 +ΔVZ, so therefore ΔVY >ΔVZ.
Physics 122B, Autumn 2009
Final Exam
EM-UWA122B094L-EF(EC1,EC2)mc_soln.doc
Name ___________________________________ Student ID _______________
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20. [5 pts] Bulb 3 is:
A. Brighter than bulb 5.
B. Dimmer than bulb 5.
C. As bright as bulb 5.
D. There is not enough information provided to compare the bulbs according to brightness.
The potential difference across bulb 3 is the same as the potential difference across box Y, orΔV3 =ΔVY.
Similarly, ΔV5 =ΔVZ. Since ΔVY >ΔVZ , ΔV3 >ΔV5 and bulb 3 is brighter than bulb 5 (since bulb
brightness is an indicator of the potential difference across a bulb).
Part 2. Shown at right is a physical arrangement of circuit elements, including
one battery, four wires, and three bulbs.
21. [5 pts] Which of the circuit diagrams below, if any, represent the physical
arrangement of elements shown at right?
A.
B.
X
C.
Y
Z
Z
Z
X
X
Y
D.
Y
Y
X
Z
E.
Y
X
None of the circuits shown represent
the given physical arrangement.
E. None of the circuits shown represent the given physical arrangement. Current from the positive
terminal of the battery can flow (1) through the first wire, (2) through bulb Z [i.e., into the threaded side
of the base of bulb Z, through its filament, and out from the tip of the base], (3) through the short wire
connecting bulbs Z and Y, (4) through bulb Y, and (5) back into the negative terminal of the battery,
thereby making a complete loop. Alternatively, current may flow (1) through the first wire, (2) through
the threaded base of bulb Z [simply using it as a conductor], (3) through the wire connecting bulbs Z
and X, (4) through bulb X, (5) through the short wire connecting bulbs X and
Y, (6) through bulb Y, and (7) back into the negative terminal of the battery,
again making a complete loop. In both cases, current must pass through bulb
X
Z
Y after passing through either bulb X or bulb Z. Therefore, bulbs X and Z
must be connected in parallel with one another and bulb Y must be connected
Y
in series with the XZ network, as shown at right.
Physics 122B, Autumn 2009
Final Exam
EM-UWA122B094L-EF(EC1,EC2)mc_soln.doc
Name ___________________________________ Student ID _______________ Score ___________
last
first
III. Two infinite insulating sheets, each with the same positive charge density
+o, intersect at a right angle as shown. Points A and B are both a distance
s to the right of the vertical sheet. Point B is also a distance s above the
horizontal sheet. The angle between path I and path III is 45˚.
23. [5 pts] Which, if any of the three paths (I, II and III) are equipotentials?
C. III only
A
s
+σo
45˚
I
A path will be an equipotential if the electric field is perpendicular
to the path at every point. Using superposition, we find that the
electric field in this case points up and to the right, at a 45˚angle
from the horizontal sheet. The only path shown that is
perpendicular to this field is path III.
III
C
II
s B
45˚
+σo
24. [5 pts] How does the electric potential at point C compare to the electric potential at point B?
B. The electric potential at point C is less than the electric potential at point B.
Consider a positive point charge that is moved along path II, from point B to point C: the path is
parallel to the magnetic field at every point, which means the field does positive work on the
charge. Thus the electric potential difference in going from point B to point C is negative, which
implies that the electric potential at B is greater than that at C.
25. [5 pts] Suppose a third identical infinite sheet with charge density
+o is added such that it forms a 45˚ angle with the vertical sheet.
How does |VBC|, the magnitude of the electric potential difference
in going from point B to point C, change with the addition of the
third sheet?
45˚
s
B. |VBC| decreases with the addition of the third sheet.
The electric field from the third sheet is antiparallel to and is of
lesser magnitude than the field from the two original sheets.
Using superposition, we find that the resulting field still points
up and to the right at a 45˚angle from the horizontal sheet, but is
weaker than before. Because the field has smaller magnitude,
the work done on a charge moving from B to C will be less,
which means the magnitude of the potential difference between
the two points will be less.
Physics 122B, Autumn 2009
Final Exam
+σo
A
I
+σo
III
C
II
s B
+σo
EM-UWA122B094T-EF(EPD,CHG,LNZ)mc_sol.doc
Name ___________________________________ Student ID _______________ Score ___________
last
first
An initially neutral conducting sphere (sphere A) is on an insulating
stand. A positively charged rod is slowly brought near the conducting
sphere, as shown in the top view diagram at right. The rod and sphere
A never touch, and no spark jumps between the rod and the sphere.
Positively
charged rod
26. [5 pts] Which of the following best describes the net charge on
sphere A?
C. The net charge on sphere A is zero
A
Conducting sphere
on insulating stand
Top view
Since the sphere is isolated, and no charge was transferred
from or to the sphere (no touch, no spark), the net charge on sphere A remains zero.
Case I: Two identical initially neutral conducting spheres on insulating
stands are placed such that all three spheres touch as shown.
Case I
Positively
charged rod
Case II: A person touches sphere A as shown below right.
A
In both cases, the charged rod is held stationary, and no spark jumps between
the rod and sphere A.
27. [5 pts] Choose the statement below that best describes the net charge on
sphere A in each case.
A. The sign of the net charge on sphere A is the same in both cases, and
is not zero.
The positively charged rod repels the positive charges in sphere A.
In both cases there is a conducting body onto which the positive
charges can move (the spheres in case I and the hand in case II).
Thus the net charge on sphere A is negative in both cases.
Top view
Case II
Positively
charged rod
A
Top view
Physics 122B, Autumn 2009
Final Exam
EM-UWA122B094T-EF(EPD,CHG,LNZ)mc_sol.doc
Name ___________________________________ Student ID _______________ Score ___________
last
first
A conducting rod moves to the right with
constant speed vo while touching a stationary
rectangular loop of wire, as shown. An external uniform, static magnetic field Bo
points perpendicular to the plane of the loop,
but its direction is not given. An induced
current, Ileft, is observed to flow down in the
left portion of the loop formed by the wire and
rod, as shown. Note: no batteries are present.
?
?
?
?
?
?
Wire ?
?
?
?
?
?
?
?
vo
?
? Bo ?
Direction
unknown
?
?
28. [5 pts] Which of the following statements
is most correct regarding thedirection of
the external magnetic field Bo ?
?
?
?
?
Ileft
?
?
?
Conducting rod
?
?
A. It points into the page.
One can apply the right hand rule to the given direction of Ileft to obtain the direction of the
induced magnetic field at points inside the left loop---it points out of the page. For definiteness,
one needs to choose an area vector for the left loop. Suppose that it is chosen to point into the
page. Therefore, the induced magnetic flux is negative. From Lenz’ law, one knows that when a
current is induced, it flows in such a way that the corresponding induced magnetic flux opposes
the change in flux due to the external field. As the rod moves, the left loop’s area increases. If the
external field also points into the page then the change in external flux over a short time interval
will be positive. This sign is opposite to that of the induced flux as Lenz’ law requires.
29. [5 pts] What is the direction of the magnetic force on the rod?
D. To the left.
From question 28 one knows that the external field points
into the page. The current in the rod is
up. [To see this, one may apply the force law F = qv o Bo to a positive charge carrier, which is
initially stationary relative to the rod, to get the direction of charge flow. Also, note that current
up the rod generates a magnetic field that points out of the page at points inside the plane of the
left loop, just
does.] The magnetic force on the current-carrying rod is
as the
field
due to Ileft
given by F = Irod L Bo , where L points up the rod. The cross product yields a force vector that
points to the left.
30. [5 pts] As the rod is moving, there will be a nonzero induced current Irod through the rod and an
induced current Iright through the right portion of the loop. How do their magnitudes compare?
B. |Irod| is always greater than |Iright|.
From question 28 one knows that the external field points into the page. If the area vector for the
right loop is taken as pointing into the page, then the change in external flux is negative over a
short time interval. From Lenz’ law, one knows that the induced current Iright flows clockwise in
the right loop since this contributes a positive flux. The current in the rod flows up along the rod.
Thus, at the top junction where the rod and wire touch, Irod = Ileft + Iright. By construction, these
currents represent positive charge flow. Since Ileft is non-zero, it means that Irod is always greater
than Iright.
Physics 122B, Autumn 2009
Final Exam
EM-UWA122B094T-EF(EPD,CHG,LNZ)mc_sol.doc