CHAPTER 20
ELECTROMAGNETIC INDUCTION AND WAVES
Multiple Choice Questions:
I.
(a) and (b). Magnetic flux is the product of the magnetic field strength (B) and the area (A) over which the field acts,
so Tm
, which is defined as the weber (Wb), and the weber are both proper units of magnetic flux.
2
2.
(a). (b), (c), and (d) are all correct choices. Magnetic flux is equal to BA cos 0. so any change in B, A, or 0 will
change the flux.
3.
(d).
4.
(b), because the change in magnetic flux is greater in loop A.
5.
(d).
6.
(b). The induced magnetic field due to the induced clockwise current in the coil points upward through the coil.
Since the external flux through the coil due to the magnet is increasing as the magnet approach the coil, the induced
current is flowing to oppose this increase, which means that the induced field through the coil due to the induced
current must be opposite to the field due to the magnet. Hence the field due to the magnet points downward, which
means that the pole nearest the coil must be a south magnetic pole.
7.
(a). As you view the coil, the magnetic field from the north pole of the magnet is pointing away from you but is
decreasing. Therefore the induced current will cause an induced field in the same direction to oppose this decrease.
Thus a clockwise current will flow in the coil.
8.
(c). The induced emf is
9.
(a). The magnetic flux is
sin ax
10.
=
I (when ax
=
NBA osin ax, Thus an increase in the area A increases the amplitude of the ernf,
=
NBA cos ax but the induced emf is
r/2), but at that instant cos ax
cos ir/2
=
=
NBAwsin ax. The maximum emf occurs when
0.
(a). The induced emf is equal to the rate of change of the magnetic flux with respect to time. Therefore when the rate
of change of the flux is a maximum, the emf output is also a maximum.
11.
(c).
12.
(a).
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355
Chapter 20
Electromagnetic Induction and Waves
13.
(b).
14.
(b). This transformer steps up the voltage from the power plant, so we need more turns in the secondary (outside the
plant) than in the primary (inside the plant).
15.
(a). The voltage is stepped up but the current is stepped down to reduce power losses in the lines.
16.
(b).
17.
(d).
18.
(e). All electromagnetic waves travel through vacuum at the speed of light.
19.
(f). The wavelength of orange light is about 620 nm. If we halve its frequency, we double its wavelength to about
1240 nm, which puts it into the IR (infrared) region.
Conceptual Questions:
1.
(a) When the bar magnet enters the coil, the needle deflects to one side, and when it leaves the coil, the needle
reverses direction.
(b)
,
because of induced currents according to Lenz’s law, it is repelled as it moves toward the loop and
attracted as it leaves the loop.
2.
The direction is [counterclockwise (head-on view).
3.
Move the coil with the
4.
The one from the Iplast_] will emerge first. When the one falls into the copper tube, a current will be induced
velccity as the bar magnet
,
so there is no change in magnetic flux through the coil.
by the changing magnetic flux in the copper tube, This current will generate a magnetic field opposite to the one of
the magnet. This will exert a force on the magnet that will slow it as it falls. There is no such current in the plastic
tube, because it is an insulator.
5.
Faraday’s law is
i/c
=
=
—N—— The units of are
.
At
N’m/C
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
356
College Physics Seventh Edition: Instructor Solutions Manual
The units of N— are
At
Tmis
N
rn-/s
Cm/s
Nsm
2
C-rn-s
,,
=
—
=
N-rn/C
Hence the units are the same.
6.
Since neither the area nor the field have changed, the only thing that could have changed is the orientation of the
loop with respect to the magnetic field Therefore the loop must be ing in the field
7.
To prevent induced current, the magnetic flux must remain constant. Since the magnitude of the magnetic field has
increased, the area must decrease, so the diameter must also sj1.
8.
(a) If the rod moves down, the outward magnetic flux would decrease. To counter this change, the induced current
must flow [terclockwise (from 2 to 1)j.
(b) The induced magnetic field is in the same directiqpj as the external magnetic field to oppose the decrease in the
magnetic flux.
9.
ax.
The
sin 0
0,
(a) The magnetic flux through the coil is proportional to cos at while the induced emf is proportional to sin
max emf occurs when sin
ax
=
1. which is when
ir/2. But at this time, cos
at
(b) As in part (a), the flux is a maximum when cos
ax
1, which is when ax
01
=
cos ir/2
0.
0. But at this time, sin
ax
=
so the emf is zero. The basic answer to parts (a) and (b) is that the flux and emf are 90° out of phase. When one is at
is maximum magnitude, the other is zero.
10.
The magnet moving through the coil produces a current. As the magnet moves up and down in the coil it will induce
a current in the coil that will light the bulb. However, the magnet produces the current (and energy to light the bulb)
according to Faraday’s law of induction at the expense of its kinetic energy and potential energy. The magnet’s
motion will therefore damp out as its total mechanical energy dwindles.
11.
If the armature is jammed or tums very slowly, there is no back emf and thus there is a large current.
12.
The induced emf is
=
NBA co sin
at.
Since we cannot change the frequency, all we can change are the number of
loops (N) or the magnitude of the external magnetic field (B). To keep the emf the same, we would have to
increase B or NI to make up for the decrease in the area A.
13.
Power is transmitted at high voltage and thus low current to reduce the rate of Joule heatingj which depends on the
square
of the current (P
=
).
1
R
2
Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist, No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.