Back
Lesson
Print
NAME ______________________________________ DATE _______________ CLASS ____________________
Electromagnetic Induction
Problem A
INDUCED emf AND CURRENT
PROBLEM
In 1994, a unicycle with a wheel diameter of 2.5 cm was ridden 3.6 m in Las
Vegas, Nevada. Suppose the wheel has 200 turns of thin wire wrapped
around its rim, creating loops with the same diameter as the wheel. An emf
of 9.6 mV is induced when the wheel is perpendicular to a magnetic field
that steadily decreases from 0.68 T to 0.24 T. For how long is the emf
induced?
SOLUTION
1. DEFINE
2. PLAN
Given:
N = 200 turns
D = 2.5 cm = 2.5 × 10−2 m
Bi = 0.68 T
Bf = 0.24 T
emf = 9.6 mV = 9.6 × 10−3 V
q = 0.0°
Unknown:
∆t = ?
Choose the equation(s) or situation: Use Faraday’s law of magnetic induction to
find the induced emf in the coil. Only the magnetic field strength changes with time.
∆B
∆q m
N∆[AB cos q]
emf = −N = − = −NA cos q 
∆t
∆t
∆t
Use the equation for the area of a circle to calculate the area (A).
D2
A = pr 2 = p 
2
Copyright © Holt, Rinehart and Winston. All rights reserved.
Rearrange the equation(s) to isolate the unknown(s):
∆B
∆t = −NA cos q 
emf
3. CALCULATE
Substitute values into the equation(s) and solve:
2.5 × 10−2 m 2
A = p  = 4.9 × 10−4 m2
2
(0.24 T − 0.68 T )
∆t = −(200)(4.9 × 10−4 m2)[cos 0.0°] 
(9.6 × 10−3 V)
∆t = 4.5 s
4. EVALUATE
The induced emf is directed through the coiled wire so that the magnetic field
produced opposes the decrease in the applied magnetic field. The rate of this
change is indicated by the positive value of time.
Problem A
167
Back
Lesson
Print
NAME ______________________________________ DATE _______________ CLASS ____________________
ADDITIONAL PRACTICE
1. The Pentagon covers an area of 6.04 × 105 m2, making it one of the
world’s largest office buildings. Suppose a huge loop of wire is placed on
the ground so that it covers the same area as the Pentagon. If the loop is
pulled at opposite ends so that its area decreases, an emf will be induced
because of the component of Earth’s magnetic field that is perpendicular
to the ground. If the field component has a strength of 6.0 × 10−5 T and
the average induced emf in the loop is 0.80 V, how much time passes before the loop’s area is reduced by half?
2. A Japanese-built Ferris wheel has a diameter of 100.0 m and can carry
almost 500 people at a time. Suppose a huge magnet is used to create
a field with an average strength of 0.800 T perpendicular to the wheel.
The magnet is then pulled away so that the field steadily decreases to
zero over time. If the wheel is a single conducting circular loop and the
induced emf is 46.7 V, find the time needed for the magnetic field to
decrease to zero.
4. The world’s largest retractable roof is on the SkyDome in Toronto,
Canada. Its area is 3.2 × 104 m2, and it takes 20 min for the roof to fully
retract. If you have a coil with exactly 300 turns of wire that changes its
area from 0.0 m2 to 3.2 × 104 m2 in 20.0 min, what will be the emf induced in the coil? Assume that a uniform perpendicular magnetic field
with a strength of 4.0 × 10−2 T passes through the coil.
5. At Massachusetts Institute of Technology, there is a specially shielded
room in which extremely weak magnetic fields can be measured. As of
1994, the smallest field measured had a magnitude of 8.0 × 10−15 T.
Suppose a loop having an unknown number of turns and an area equal
to 1.00 m2 is placed perpendicular to this field and the magnitude of the
field strength is increased tenfold in 3.0 × 10−2 s. If the emf induced is
equal to −1.92 × 10−11 V, how many turns are in the loop?
6. An electromagnet that has a mass of almost 8.0 × 106 kg was built at the
CERN particle physics research facility in Switzerland. As part of the detector in one of the world’s largest particle accelerators, this magnet creates a fairly large magnetic field with a magnitude of 0.50 T. Consider a
coil of wire that has 880 equal turns. Suppose this loop is placed perpendicular to the magnetic field, which is gradually decreased to zero in 12 s.
If an emf of 147 V is induced, what is the area of the coil?
168
Holt Physics Problem Workbook
Copyright © Holt, Rinehart and Winston. All rights reserved.
3. In 1979, a potential difference of about 32.0 MV was measured in a lab in
Tennessee. The maximum magnetic fields, obtained for very brief periods
of time, had magnitudes around 1.00 × 103 T. Suppose a coil with exactly
50 turns of wire and a cross-sectional area of 4.00 × 10−2 m2 is placed perpendicular to one of these extremely large magnetic fields, which quickly
drops to zero. If the induced emf is 32.0 MV, in how much time does the
magnetic field strength decrease from 1.00 × 103 T to 0.0 T?
Back
Print
Lesson
Problem Workbook Solutions
Electromagnetic
Induction
Additional Practice A
Givens
Solutions
1. Ai = 6.04 × 105 m2
Af =
1
 (6.04
2
5
2
× 10 m )
B = 6.0 × 10−5 T
emf = 0.80 V
∆ΦM −N∆[AB cos q]
emf = −N 
= 
∆t
∆t
−NB cos q
∆t =  ∆A
emf
−NB cos q
∆t =  (Af − Ai )
emf
N = 1 turn
q = 0.0°
−(1)(6.0 × 10−5 T)(cos 0.0°)
1
∆t =  (6.04 × 105 m2)2 − 1
(0.80 V)
II
∆t = 23 s
100.0 m
2. r =  = 50.0 m
2
Bi = 0.800 T
2
∆B −N(pr )(cos q)(Bf − Bi )
∆t = −NA cos q  = 
emf
emf
Bf = 0.000 T
−(1)(p)(50.0 m)2(cos 0.0°)(0.000 T − 0.800 T)
∆t = 
(46.7 V)
q = 0.00°
emf = 46.7 V
N = 1 turn
∆t = 135 s
3. emf = 32.0 × 106 V
Copyright © Holt, Rinehart and Winston. All rights reserved.
∆ΦM −N∆[AB cos q]
emf = −N 
= 
∆t
∆t
3
Bi = 1.00 × 10 T
Bf = 0.00 T
A = 4.00 × 10−2 m2
N = 50 turns
q = 0.00°
∆ΦM −N∆[AB cos q]
emf = −N 
= 
∆t
∆t
∆B −NAcos q(Bf − Bi )
∆t = −NA cos q  = 
emf
emf
−(50)(4.00 × 10−2 m2)(cos 0.00°)[(0.00 T) − (1.00 × 103 T)]
∆t = 
(32.0 × 106 V)
∆t = 6.3 × 10−5 s
4. Af = 3.2 × 104 m2
2
Ai = 0.0 m
∆t = 20.0 min
B = 4.0 × 10−2 T
N = 300 turns
q = 0.0°
∆A −NB cos q
∆ΦM −N∆[AB cos q]
emf = −N 
=  = −NB cos q  =  (Af − Ai )
∆t
∆t
∆t
∆t
−(300)(4.0 × 10−2 T)(cos 0.0°)
emf =  [(3.2 × 104 m2) − (0.0 m2)]
60 s
(20.0 min) 
1 min
emf = −3.2 × 102 V
Section Two—Problem Workbook Solutions
II Ch. 20–1
Back
Print
Lesson
Givens
Solutions
5. Bi = 8.0 × 10−15 T
Bf = 10 Bi = 8.0 × 10
∆t = 3.0 × 10
−2
−14
T
s
A = 1.00 m2
emf = −1.92 × 10−11 V
q = 0.0°
∆ΦM −N∆[AB cos q]
emf = −N 
= 
∆t
∆t
−(emf)(∆t)
−(emf)(∆t)
N =  = 
A cos q ∆B A cos q(Bf − Bi)
−(−1.92 × 10−11 V)(3.0 × 10−2 s)
N = 
(1.00 m2)(cos 0.0°)[(8.0 × 10−14 T) − (8.0 × 10−15 T)]
N = 8 turns
∆B
ΦM −N∆[AB cos q]
emf = −N 
=  = −NA cos q 
∆t
∆t
∆t
6. Bi = 0.50 T
Bf = 0.00 T
−(emf)(∆t)
−(emf)(∆t)
A =  = 
N cos q ∆B N cos q (Bf − Bi)
N = 880 turns
∆t = 12 s
−(147 V)(12 s)
A = 
(880)(cos 0.0°)(0.00 T − 0.50 T)
emf = 147 V
q = 0.0°
II
A = 4.0 m2
Additional Practice B
1. ∆Vrms = 120 V
−2
R = 6.0 × 10
1
 = 0.707
2
Ω
∆Vrms
a. Irms = 
R
(120 V)
Irms = 
(6.0 × 10−2 Ω)
Irms = 2.0 × 103 A
b. Imax = (Irms) 2
Imax = 2.8 × 103 A
c. P = (Irms)(∆Vrms)
P = (2.0 × 103 A)(120 V)
P = 2.4 × 105 W
II Ch. 20–2
Holt Physics Solution Manual
Copyright © Holt, Rinehart and Winston. All rights reserved.
(2.0 × 103 A)
Imax = 
(0.707)