Physics 54!
Summer 2012
Exam II
Solutions
Part A. Multiple choice questions. Check the best answer. Each question carries a value of 4 points.
1.!
Concerning a static magnetic field, which of these is wrong?
!
Its field lines always form closed curves.
!
It never does work on point charges.
!
It exerts no force on a charge at rest.
√ !
2.!
Shown are two parallel straight wires carrying equal currents into
and out of the page. Consider the total B-field of these wires at
points in the plane of the page. Which of the following is wrong?
In
Out
!
There is no point at a finite distance from the wires where the field is zero.
!
At points between the wires the field is directed downward.
√ !
!
3.!
One of the above is not true.
At points to the left of both wires the field is directed downward. [Upward
on both sides.]
At points to the right of both wires the field is directed upward.
A small loop of wire is at rest in a non-uniform B-field, with the plane of the loop
perpendicular to the field. The field strength is increasing with time.
!
The induced magnetic moment of the loop is parallel to the given field.
!
The force on the loop is toward the region of stronger field.
√ !
!
The loop moves in such a way as to decrease the potential energy.
None of the above is true.
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Physics 54!
4.!
Summer 2012
Shown in cross-section is a long solenoid with B-field inside
its windings as shown, and negligible B-field outside. A
positive charge Q is at rest at the point indicated. If the
solenoid current is decreased:!
!
B
⊗
Q
The charge Q will move up the page.
√ !
The charge Q will move down the page.
!
The charge Q will move into the page.
!
The charge Q will not move.
Part B. True-false questions. Check T or F depending on whether the statement is true or false. Each
question carries a value of 3 points.
5.!
In the circuit shown, after the switch is closed bulb A glows less
brightly than before.
√
6.!
F [Voltage across the bulb is lower.]
In the classroom demonstration of a cylindrical magnet dropped
down metal tubes, the magnet moved slower in the copper tube
than in the aluminum tube because copper has lower resistivity
so the induced currents were larger.
√
7.!
T
T
A
B
F
If the current through an inductor is decreasing, the potential falls as one crosses
the inductor in the direction of the current. (Neglect resistance of the inductor.)
T
√
F [Rises.]
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Physics 54!
Summer 2012
Part C. Problems. Work problem in space provided, using extra sheets if needed. Explain your method
clearly. Problems carry the point values shown.
1.!
!
A source is recharging a 12 V automobile battery. The source and battery each
have internal resistance 0.1 Ω , and the emf of the source is 13 V.
a.!
Draw the circuit, representing the source by a battery and denoting with
small circles the terminals of the battery.
b.!
The battery’s capacity is rated as 100 A-hr, meaning it can provide 100 A
current for one hour (3600 s). Find the total charge this represents, and the
amount of energy stored in the battery when it is fully charged.
c.!
How long will it take the source to charge the battery fully if it started out
totally depleted? [Find the current in the circuit.]
d.!
How much energy enters the battery through its terminals in this time?
[What is the terminal voltage during charging?] Why is this different from
the total energy stored?
[20 points]
a.!
Drawing below.
b.!
Qtot = (100)(3600) = 3.6 × 105 C . Energy stored:
Qtot ⋅ E = (3.6 × 105 )(12) = 4.32 × 106 J .
c.!
Loop rule: 13 − 12 − I(0.1 + 0.1) = 0 . This gives I = 5 A . The time to recharge is
T = Qtot /I = 7.2 × 10 4 s (or 20 hr). [This is called a “trickle” charge.]
d.!
The terminal voltage is E + IR = 12 + (5)(0.1) = 12.5 V . Recharging sends Qtot
through this potential difference, so the total energy input from the source is
(12.5)(3.6 × 105 ) = 4.5 × 106 J . The difference is the energy lost to Joule heating in
the resistance.
0.1 Ω 13 V
0.1 Ω 12 V
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Physics 54!
2.!
Summer 2012
In the van Allen belts the earth’s magnetic field runs roughly from south to
north. It intensifies toward the north and south magnetic poles, but at any given
latitude is approximately uniform. A proton emitted by the sun enters this field
above the equator with a velocity that has a component v⊥ perpendicular to the
field and also a component v parallel to the field, toward the north.
a.!
Draw the circle described by the proton in a plane perpendicular to the
field and looking north, if v = 0 . Indicate the direction the proton moves.
b.!
Treating this as a current loop, what is the direction of its magnetic
moment?
c.!
Because of v this loop moves to the north toward more intense B-field.
What happens to the magnitude of v ? Why?
d.!
Answer (c) if v is to the south instead of the north.
[20 points]
a.!
Drawing below. The B-field is to the north, and the proton describes a counterclockwise circle.
b.!
Out of the page (to the south).
c.!
The “loop” is repelled from stronger field, and this force reduces v . (Eventually it
stops and reverses direction.)
d.!
Since the south magnetic pole is also a region of stronger field, the “loop” is
repelled from it also, reducing the value of v . (The proton moves back and forth
between north and south until it loses energy, probably by collision with air
molecules.)
B⊗
!
!
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Physics 54!
3.!
Summer 2012
The diagram shows a rotating bar generator. The circle is a
conducting band on which slides the end of a bar of length 
rotating at constant angular speed ω . The pivot of the bar is
connected to the band by the resistor. You are to find the emf
produced in this system by using Faraday’s law.
!
B
⊗ 
A2
θ A
1
R
Consider the circuit bounding the area called A1 , which
increases as the bar rotates ( θ = ω t ). The magnetic flux
through A1 thus changes, inducing the emf in the bar.
a.!
Write A1 in terms of θ and  . [What fraction of the total area of the circle
is A1 ? Remember that θ is in radians.]
b.!
Find the flux through area A1 and find the emf induced by its change.
c.!
Use Lenz’s law to determine the direction of the current in the bar.
d.!
If we use instead the circuit bounding area A2 , we will find the same emf.
Use Lenz’s law for that circuit to show that the direction of the current in
the bar is the same as in (c).
[20 points]
θ
⋅ π  2 = 12 θ  2 .
2π
a.!
A1 =
b.!
Φ = BA1 = 12 B 2 ⋅ θ . The magnitude of the emf is E =
c!
The flux into the page is increasing, so the induced current runs counter-clockwise
to create flux out of the page. In the bar, the current runs from the tip to the pivot.
d.!
Now the flux into the page is decreasing because A2 is decreasing. The current in
dΦ 1 2 dθ 1 2
= B
= B ω .
dt 2
dt 2
that circuit thus runs clockwise. But the current in the bar is the same as before.
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Physics 54!
4.!
!
a.!
Summer 2012
Two coaxial cables have equal length and both
have a thin cylindrical outer conductor of radius
b carrying current I into the page. Cable A has a
thin hollow cylindrical inner conductor of radius
a, carrying current I out of the page. Cable B has
a solid cylindrical inner conductor, also of radius
a and carrying total current I out of the page, but
distributed uniformly over its cross- section.
A
B
a.!
Use Ampere’s law to calculate the B-field magnitudes in each of the two
cables at a distance r from the symmetry axis, where a < r < b .
b.!
Do the same for r < a .
c.!
Which cable has the larger self-inductance? How do you know?!
[15 points]
Use a circular path following a field line. By the symmetry
∫ B ⋅ dr = 2π r ⋅ B . For
a < r < b the current linked is that in the inner conductor,which is I. Ampere’s law
gives 2π r ⋅ B = µ0 I , or I =
b.!
µ0 I
. This applies to both cables.
2π r
For cable A the current linked for r < a is zero, so the B-field is zero in that region.
For cable B we have for the linked current j ⋅ A =
gives 2π r ⋅ B = µ0 I
c.!
r2
a2
or B =
µ0 I
2π a2
I
π a2
⋅πr2 = I ⋅
r2
a2
. Ampere’s law
⋅r .
The stored energy in the gap ( a < r < b ) is the same for both, but cable B has stored
energy for r < a , while cable A has zero field and zero energy in that region. Thus
cable B has greater total stored energy for given current, and a larger value of L.
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Physics 54!
Summer 2012
Median = 64
SD = 17
Median = 63
SD = 13
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