Download Physics 1402 Homework Solutions - Walker, Chapter 23 Conceptual

Physics 1402
Homework Solutions - Walker, Chapter 23
Conceptual Questions
CQ5. Before the switch is closed, there is no current in the coil, and therefore no magnetic flux through the metal
ring. When the switch is closed there begins to be a magnetic flux through the ring. Let us assume, for example,
that this flux is downward through the ring. By Lenz’s law, this change in magnetic flux through the ring will give
rise to a induced B-field that will be upward through the ring. Therefore, it is as though this induced B-field were
produced by a bar magnet with its south pole at the bottom of the ring and the north pole at the top of the ring. The
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south pole of this Bind repels the south pole of the B due to the current in the coil. This repulsive force makes the
ring fly off.
CQ9. Anytime a conductor is moved in a B-field, there is an emf induced in the conductor. The wire is pulled
through the B-field of the Earth, and this generates an emf in the wire. If this wire were then connected to an
external circuit, the wire would act like a battery, pushing charge through the external circuit.
Problems
2. Let us refer to the bottom surface of the box as Side 1 and the lateral surfaces (the sides) as Sides 2, 3, 4, and 5.
(It will not matter which is 2, which is 3, and so on.) The magnetic flux through any surface is defined by
Φ mag = BA cos θ ,
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in which θ is the angle between B and the normal to the surface you’re talking about. Looking at the figure, we
see that everywhere on Sides 2, 3, 4, and 5, the normal points horizontally. The B-field is vertically upward.
Therefore the angle θ is 90°, and the flux through these sides is zero:
( Φ mag )2 = ( Φ mag )3 = ( Φ mag )4 = ( Φ mag )5 = BA cos 90D = 0 .
For Side 1, we have:
( Φ mag )1 = BA cos θ .
For Side 1, the normal is either vertically upward (giving θ = 0D ) or vertically downward (giving θ = 180D ). Since
I’m interested in the magnitude of the flux, I’ll avoid the minus sign by taking the normal to point upward . Then:
( Φ mag )1 = BA cos 0D = BA
The area A is the area of Side 1. From the figure, this is LW. So we have:
( Φ mag )1 = B ( LW ) = ( 0.0250 T )( 0.325 m )( 0.120 m ) = 9.75 ×10−4 Wb
9.
The problem should say, “what is the magnitude of the average induced emf?” The absolute value of the
average induced emf is given by Faraday’s law:
(ε ind )av
=N
∆Φ mag
∆t
.
∆Φ mag = ∆ ( BA cos θ )
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Physics 1402
Homework Solutions - Walker, Chapter 23
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The angle θ is the angle between B and the normal to the cross-sectional area of the loop. This angle is either 0°
or 180°. I will take it to be 0° to avoid a minus sign, since I’m interested in the absolute value of the induced emf.
The area of the loop is not changing, so
∆Φ mag = A∆B
and Faraday’s law becomes:
(ε ind )av
= NA
∆B
2 −0.25 T
= ( 53) ⎡π ( 0.15 m ) ⎤
= 7.8 V ,
⎣
⎦ 0.12 s
∆t
in which I’ve used A = π r 2 for the area of the circular loop.
11.
(ε ind )av = − N
∆Φ mag
∆t
=−
∆Φ mag
∆t
,
since the number of turns is 1. (It’s a “single-loop” coil, the problem says.)
∆Φ mag
represents on a graph of Φ mag vs t: it represents the slope of the graph.
∆t
(a.) At t = 0.050 s, we are on a segment of the graph that has slope
The trick here is to realize what
∆Φ mag
∆t
=
10 Wb − 0 Wb
= 100 V .
0.1 s − 0 s
(I used the two endpoints of this segment to calculate the slope, since I know exactly where those endpoints are.)
Therefore, at t = 0.050 s,
(ε ind )av = −100 V .
(b.) At t = 0.15 s, we’re on a segment of the graph that has a slope of zero. (This segment is horizontal.)
Therefore ( ε ind )av = 0 V . (The flux is not changing during this segment, therefore there is no induced emf.)
(c.) At t = 0.50 s, we’re on a segment of the graph that has slope
∆Φ mag
∆t
Therefore,
18.
=
−5 Wb − 10 Wb
= −38 V
0.6 s − 0.2 s
(ε ind )av = − ( −38 V ) = +38 V
The B-field due to the magnet is downward and getting larger as the magnet falls toward the ring. Therefore,
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by Lenz’s law, this will cause an induced B-field, Bind , that is upward in the interior of the ring, to partially
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cancel some of the increase in the downward flux due to the magnet. By the RHR for the sense in which B
circulates around a current-carrying wire, if you imagine grasping the ring anywhere in such a way that your
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Physics 1402
Homework Solutions - Walker, Chapter 23
fingers curl around and point upward in the interior of the ring, your thumb will have to point in the
counterclockwise sense. Therefore, I ind must flow counterclockwise, as viewed from above.
27.
For Ring A:
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By the RHR for the sense in which B circulates around a current-carrying wire, the B-field due to the wire is
out of the page through Ring A. Furthermore, if the current in the wire is increasing with time, this B-field is
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getting larger. Therefore, by Lenz’s law, Bind will point into the page through Ring A. By the RHR for the
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sense in which B circulates around a current-carrying wire, this means that I ind will have to flow clockwise
around Ring A.
I will come back to Ring B in just a moment. But first, let’s take care of Ring C.
For Ring C:
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The B-field due to the wire is into the page and increasing. Therefore, by Lenz’s law, Bind will be out of the
page through Ring C. This means that I ind will have to flow counterclockwise around Ring C.
For Ring B:
Think of Ring B as consisting of two halves, one above the wire and the other below the wire.
For the half of Ring B that’s above the wire, the flux due to the wire is out of the page. For the half of Ring B
that’s below the wire, the flux due to the wire is into the page. Furthermore, because the area of the top half of
Ring B is the same as the area of the bottom half, Φ mag through the top half is equal in magnitude to Φ mag
through the bottom half. This means that the total flux through Ring B is zero. And it’s zero at all times! As
the current in the wire increases, the B-field through the top half gets bigger, so Φ mag through the top half gets
bigger. But Φ mag through the bottom half gets bigger by the same amount, so the total flux through the ring
remains zero. By Faraday’s law,
(ε ind )av = − N
∆Φ mag
∆t
,
if Φ mag is not changing with time, there will be no induced emf in Ring B. Therefore, no induced current
flows in Ring B.
30. (a.)
I ind =
v=
ε ind
R
=
BLv
R
I ind R ( 0.125 A )(12.5 Ω )
=
= 4.6 m/s
BL
( 0.750 T )( 0.45 m )
(b.) If the bar were moving to the left, the polarity of the induced emf would be swapped. With the bar
moving to the right, the bottom of the bar is at higher potential than the top of the bar. If the bar moved to the
left, the top of the bar would be at higher potential than the bottom. But it would still be true that ε ind = BLv .
Therefore, we would still get
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Physics 1402
Homework Solutions - Walker, Chapter 23
I ind =
BLv
R
and this would still lead to
v=
I ind R
BL
as before. Therefore, there would be no change in the speed required.
31. (a.) Once some current I ind begins to flow, the rod becomes a current-carrying wire in a B-field. Therefore,
there’s a magnetic force exerted on the rod given by
Fmag = I ind LB sin θ ,
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in which θ means the angle between I ind and B . The current I ind flows downward through the rod; B is
out of the page. Therefore, θ = 90D and we have
Fmag = I ind LB
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By the RHR for the direction of Fmag , this force points to the left if the rod is moving to the right. Therefore,
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Fmag opposes the motion and would cause the rod to slow down, were it not for some other force – an
externally applied force – continuing to pull the bar to the right. Let us call this externally applied force
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Fapplied . And now we come to the bit about I ind staying constant. If I ind is to stay constant, the bar must be
pulled with constant speed. (See the expression for I ind in Problem 30 if this isn’t immediately clear.) Well,
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if the rod is to be pulled with constant speed, the force Fapplied must equal Fmag in magnitude! Therefore,
Fapplied = I ind LB
Fapplied = ( 0.125 A )( 0.45 m )( 0.750 T ) = 0.042 N ,
to the right.
(b.) The rate of energy dissipation in the resistor is the power dissipated in the resistor. This is given by
P = IV = I ind ε ind
So:
B 2 L2 v 2
⎛ BLv ⎞
P=⎜
⎟ ( BLv ) =
R
⎝ R ⎠
P=
( 0.750 T )2 ( 0.45 m )2 ( 4.6 m/s )2
12.5 Ω
= 0.20 W
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Physics 1402
Homework Solutions - Walker, Chapter 23
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(c.) If the rod is not to slow down, then the externally applied force Fapplied must deliver an amount of power
equal to the power being dissipated in the resistor. So the mechanical power delivered to the rod must be
0.20 W.
39.
Faraday’s law, written in terms of the inductance L, says
(ε ind )av
(ε ind )av
42.
=L
∆I
∆t
⎛ 515 × 10−3 A ⎞
= 45.0 × 10−3 H ⎜⎜
⎟⎟ = 1.4 V
−3
⎝ 16.5 × 10 s ⎠
(
)
Assuming the solenoid to be an ideal solenoid (and they should have said this), the inductance L is given by
⎛ N2
L = µ0 ⎜⎜
⎝ A
⎞
⎟⎟ A
⎠
2⎤
⎡
2
⎛
−7 T ⋅ m ⎞ ( 640 )
⎢
⎥ ⎡π ( 0.043 m ) ⎤ = 0.012 H ,
=
×
L ⎜ 4π 10
⎟
⎣
⎦
A
0.25
m
⎝
⎠ ⎢⎣
⎥⎦
in which I’ve used the fact that the area of a circle is π r 2 .
61. We’re told that N P = 25 and N S = 750 . We want VS = 4800 V and I S = 0.012 A . What do I P and VP
need to be? Well,
IP NS
=
IS NP
⎛N
IP = IS ⎜ S
⎝ NP
⎞
⎛ 750 ⎞
⎟ = ( 0.012 A ) ⎜
⎟ = 0.36 A
⎝ 25 ⎠
⎠
And
VP N P
=
VS N S
⎛N ⎞
⎛ 25 ⎞
VP = VS ⎜ P ⎟ = ( 4800 V ) ⎜
⎟ = 160 V
⎝ 750 ⎠
⎝ NS ⎠
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