1. Chapter 31. Electromagnetic Induction Physics, 6th Edition
Chapter 31. Electromagnetic Induction
Induced Electromotive Force
31-1. A coil of wire 8 cm in diameter has 50 turns and is placed in a B field of 1.8 T. If the B
field in reduced to 0.6 T in 0.002 s, what is the induced emf?
π D 2 π (0.08 m) 2
A= = = 5.03 x 10-3 m 2 ; ∆B = 1.8 T – 0.6 T = 1.20 T
4 4
∆φ NA ∆B (50)(5.03 x 10-3 m 2 )(1.20 T)
E = −N =− ; E=− ; E = -151 V
∆t ∆t 0.002 s
31-2. A square coil of wire having 100 turns of area 0.044 m2 is placed with its plane
perpendicular to a constant B field of 4 mT. The coil is flipped to a position parallel with
the field in a time of 0.3 s. What is the induced emf?
∆φ NB ∆A (100)(0.004 T)(0.044 m 2 − 0)
E = −N =− ; E=− ; E = -58.7 mV
∆t ∆t 0.3 s
31-3 A coil of 300 turns moving perpendicular to the flux in a uniform magnetic field
experiences a flux linkage of 0.23 mWb in 0.002 s. What is the induced emf?
∆φ 0.23 x 10-3 Wb
E = −N = −(300) ; E = –34.5 V.
∆t 0.002 s
31-4. The magnetic flux linking a loop of wire changes from 5 mWb to 2 mWb in 0.1 s. What
is the average induced emf?
∆φ 5 x 10-3 Wb − 2 x 10-3 Wb
E = −N = −(1) ; E = –30.0 mV
∆t 0.100 s
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2. Chapter 31. Electromagnetic Induction Physics, 6th Edition
31-5. A coil of 120 turns is 90 mm in diameter and has its plane perpendicular to a 60-mT
magnetic field produced by a nearby electromagnet. The current in the electromagnet is
cut off, and as the filed collapses, an emf of 6 V is induced in the coil. How long does it
take for the field to disappear? [ A = πR2 = π(0.045 m)2 = 6.36 x 10-3 m2 ]
∆φ NA ∆B − NA ∆B −(120)(6.36 x 10-3 m 2 )(0.06 T)
E = −N =− ; ∆t = =
∆t ∆t E −6.00 V
∆t = 7.63 ms.
31-6. A coil of 56 turns has an area of 0.3 m2. Its plane is perpendicular to a 7-mT magnetic
field. If this field collapses to zero in 6 ms, what is the induced emf?
∆φ NA ∆B −56(0.3 m 2 )(0.007 T - 0)
E = −N =− = ; E = -19.6 V
∆t ∆t 0.006 s
31-7. A wire 0.15 m long moves at a constant velocity of 4 m/s in a direction that is 360 with
respect to a 0.4-T magnetic field. The axis of the wire is perpendicular to the magnetic
flux lines. What is the induced emf?
E = Blv sin θ = (0.4 T)(0.15 m)(4 m/s)sin 360 ; E = 141 mV
31-8. A 0.2-m wire moves at an angle of 280 with an 8-mT magnetic field. The wire length is
perpendicular to the flux. What velocity v is required to induce an emf of 60 mV?
E
E = Blv sin θ ; v=
Bl sin θ
E 60 x 10-3V
v= =
Bl sin θ (0.7 T)(0.2 m)sin 280
v = 79.9 m/s
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3. Chapter 31. Electromagnetic Induction Physics, 6th Edition
Generators
31-9. The magnetic field in the air gap between the magnetic poles and the armature of an
electric generator has a flux density of 0.7 T. The length of the wires on the armature is
0.5 m. How fast must these wires move to generate a maximum emf of 1.00 V in each
armature wire?
E 1.00 V
v= 0
= ; v = 2.86 m/s
Bl sin 90 (0.7 T)(0.5 m)(1)
31-10. A single loop of wire has a diameter of 60 mm and makes 200 rpm in a constant 4-mT
magnetic field. What is the maximum emf generated?
π D 2 π (0.06 m) 2
A= = = 2.83 x 10-3m 2 ; f = 200 rpm = 3.33 rev/s
4 4
Emax = 2π fNBA = 2π (3.33 rev/s)(1)(0.004 T)(2.83 x 10-3T) ; E = 237 µV
31-11. The armature of a simple generator has 300 loops of diameter 20 cm in a constant 6-mT
magnetic field. What must be the frequency of rotation in revolutions per second in order
to induce a maximum emf of 7.00 V? [ A = πR2 = π(0.10 m)2 = 0.314 m2 ]
Emax 7.00 V
f = = ; f = 19.7 rev/s
2π NBA 2π (300)(0.006 T)(0.314 m 2 )
31-12. An armature in an ac generator consists of 500 turns, each of area 60 cm2. The armature
is rotated at a frequency of 3600 rpm in a uniform 2-mT magnetic field. What is the
frequency of the alternating emf? What is the maximum emf generated?
rev  1 min 
f = 3600   = 60 rev/s; f = 60.0 Hz
min  60 s 
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4. Chapter 31. Electromagnetic Induction Physics, 6th Edition
Emax = 2π fNBA = 2π (60 Hz)(500)(0.002 T)(60 x 10-4 m 2 ) ; Emax = 2.26 V
31-13. In Problem 37-12, what is the instantaneous emf at the time when the plane of the coil
makes an angle of 600 with the magnetic flux? [ Note θ = 900 – 600 = 300 ]
Einst = Emax sin 300 = (2.26 V)(0.500); Einst = 1.13 V
31-14. The armature of a simple ac generator has 100 turns of wire, each having a radius of 5.00
cm. The armature turns in a constant 0.06-T magnetic field. What must be the rotational
frequency in rpm to generate a maximum voltage of 2.00 V?
A = π R 2 = π (0.05 m) 2 = 7.85 x 10-3m 2 ;
Emax 2.00 V
f = = ; f = 6.75 rev/s
2π NBA 2π (100)(0.06 T)(7.85 x 10-3 m 2 )
rev  60 s 
f = 6.75  ; f = 405 rpm
s  min 
31-15. A circular coil has 70 turns, each 50 mm in diameter. Assume that the coil rotates about
an axis that is perpendicular to a magnetic filed of 0.8 T. How many revolutions per
second must the coil make to generate a maximum emf of 110 V?
π D 2 π (0.05 m) 2
A= = = 1.96 x 10-3 m 2 ;
4 4
Emax 110 V
f = = ; f = 159 rev/s
2π NBA 2π (70)(0.8 T)(1.96 x 10-3m 2 )
*31-16. The armature of an ac generator has 800 turns, each of area 0.25 m2. The coil rotates at a
constant 600 rpm in a 3-mT field. What is the maximum induced emf? What is the
instantaneous emf 0.43 s after the coil passes a position of zero emf? [600 rpm = 10 rev/s]
Emax = 2π fNBA = 2π (10 Hz)(800)(0.003 T)(0.25 m 2 ) ; Emax = 37.7 V
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5. Chapter 31. Electromagnetic Induction Physics, 6th Edition
Einst = Emax sin(2π ft ) = (37.7 V)sin [ 2π (10 Hz)(0.43 s) ] ; Einst = 35.9 V
*31-17. A 300-Ω resistor is connected in series with an ac generator of negligible internal
resistance. The armature of the generator has 200 turns of wire 30 cm in diameter and it
turns at 300 rpm in a constant 5-mT field. What is the instantaneous current through the
resistor 0.377 s after the coil passes a position of zero emf? [ 300 rpm = 5.00 rev/s ]
π D 2 π (0.3 m) 2 Einst = 2π fNBA sin(2π ft )
A= = = 0.0707 m 2 ;
4 4
Einst = 2π (5 Hz)(200)(0.005 T)(0.0707 m 2 ) sin[2π (5 Hz)(0.377 s)]
Einst = (2.22 V) sin (0.118 rad); Einst = 1.47 V
1.47 V
I inst = ; Iinst = 4.90 mA
300 Ω
DC Motors and Back EMF
31-18. A 120-V dc motor draws a current of 3.00 A in operation and has a resistance of 8.00 Ω.
What is the back emf when the motor is operating and what is the starting current?
V – Eb = IR; Eb = V – IR = 120 V – (3.00 A)(8.00 Ω); Eb = 96.0 V
120 V
I0 = ; I0 = 15.0 A
8Ω
31-19. The armature coil of the starting motor in an automobile has a resistance of 0.05 Ω. The
motor is driven by a 12-V battery, and the back emf at operating speed is 6.00 V. What is
the starting current? What is the current at full speed?
12.0 V
I0 = = 240 A; I0 = 240 A
0.05 Ω
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6. Chapter 31. Electromagnetic Induction Physics, 6th Edition
V − Eb 12 V − 6 V
I= = ; I = 120 A
R 0.05 Ω
31-20. A 220-V dc motor draws a current of 10 A in operation and has an armature resistance of
0.4 Ω. What is the back emf when the motor is operating and what is the starting current?
Eb = V − IR = 220 V − (10A)(0.4 Ω); Eb = 216 V
220 V
I0 = ; I0 = 550 A
0.4 Ω
*31-21. A 120-V series-wound dc motor has a field resistance of 90 Ω and an armature resistance
of 10 Ω. When operating at full speed, a back emf of 80 V is generated. What is the total
resistance of the motor? What is the starting current? What is the operating current?
120 V
(a) RT = 90 Ω + 10 Ω = 100 Ω ; (b) I 0 = ; I0 = 1.20 A
100 Ω
120 V - 80 V
(c) I = ; I = 0.400 A
100 Ω
*31-22. The efficiency of the motor in Problem 37-21 is the ratio of the power output to the
power input. Determine the efficiency based on the known data.
Pout Pin − Pout
E= = ; Pout = I 2 R = (0.4 A) 2 (100 Ω); Pout = 16.0 W
Pin Pin
48 W- 16 W
Pin = IV = (0.4 A)(120 V) = 48.0 W; E= ; E = 66.7%
48 W
Transformers
31-23. A step-up transformer has 400 secondary turns and only 100 primary turns. A 120-V
alternating voltage is connected to the primary coil. What is the output voltage?
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7. Chapter 31. Electromagnetic Induction Physics, 6th Edition
Ep Np Ep N s (120 V)(400)
= ; Es = = ; Es = 480 V
Es Ns Np (100)
31-24. A step-down transformer is used to drop an alternating voltage from 10,000 to 500 V.
What must be the ratio of secondary turns to primary turns? If the input current is 1.00 A
and the transformer is 100 percent efficient, what is the output current?
Np Ep 500V 1
= = = ; Ratio = 1:20
Ns Es 10,000 V 20
(1 A)(10,000 V)
Pout = Pin ; I p Ep = I s Es ; Is = ; Is = 20 A
(500 V)
31-25. A step-up transformer is 95 percent efficient and has 80 primary turns and 720 secondary
turns. If the primary draws a current of 20 A at 120 V, what are the current and voltage
for the secondary?
Ep Np Ep N s (120 V)(720)
= ; Es = = ; Es = 1080 V
Es Ns Np (80)
Pout Es I s 0.95(20 A)(120 V)
E= = = 0.95; Is = ; Is = 2.11 A
Pin Ep I p 1080 V
31-26. A 25-W light bulb has a resistance of 8.0 Ω while burning. The light is powered from the
secondary of a small transformer connected to a 120-V circuit. What must be the ratio of
secondary turns to primary turns in this application? Assume 100 percent efficiency.
2
Vout
Pout = ; Vout = Pout R = (25 W)(8 Ω); Vout = 14.1 V
R
N s Es 14.1 V Ns
= = ; = 0.118
N p Ep 120 V Np
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8. Chapter 31. Electromagnetic Induction Physics, 6th Edition
Challenge Problems
31-27. A 70-turn coil of wire has an area of 0.06 m2 and is placed perpendicular to a constant 8-
mT magnetic field. Calculate the induced emf if the coil flips 900 in 0.02 s.
∆φ NB ∆A (70)(0.008 T)(0.06 m 2 − 0)
E = −N =− ; E=− ; E = -1.68 mV
∆t ∆t 0.02 s
31-28. A coil of area 0.2 m2 has 80 turns of wire and is suspended with its plane perpendicular to
a uniform magnetic field. What must be the flux density to produce an average emf of
2 V as the coil is flipped parallel to the field in 0.5 s?
∆φ NB ∆A (−2 V)(0.5 s)
E = −N =− ; B=− ; B = 62.5 mT
∆t ∆t (80)(0.2 m 2 − 0)
31-29. The flux through a 200-turn coil changes from 0.06 to 0.025 Wb in 0.5 s. The coil is
connected to an electric light, and the combined resistance is 2 Ω. What is the average
induced emf and what average current is delivered to the light filament?
∆φ 200(0.06 Wb - 0.025 Wb
E = −N =− ; E = -14.0 V
∆t 0.5 s
14.0 V
I= ; I = 7.00 A
2Ω
31-30. A 90-mm length of wire moves with an upward velocity of 35 m/s between the poles of a
magnet. The magnetic field is 80 mT directed to the right. If the resistance in the wire is
5.00 mΩ, what are the magnitude and direction of the induced current?
E = Blv sin θ = (80 x 10-3T)(0.09 m)(35 m/s)(1); E = 0.252 V
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9. Chapter 31. Electromagnetic Induction Physics, 6th Edition
0.252 V
I= ; I = 50.4 A, into paper
0.005 Ω
31-31. A generator develops an emf of 120 V and has a terminal potential difference of 115 V
when the armature current is 25.0 A. What is the resistance of the armature?
E – VT = IARA ; 120 V – 115 V = (25 A)RA; RA = 0.200 Ω
31-32. The coil of an ac generator rotates at a frequency of 60 Hz and develops a maximum emf
of 170 V. The coil has 500 turns, each of area 4 x 10-3 m2. What is the magnitude of the
magnetic field in which the coil rotates?
Emax 170 V
Emax = 2π fNBA; B= = ; B = 225 mT
2π fNA 2π (60 Hz)(500)(0.004 m 2 )
31-33. A generator produces a maximum emf of 24 V when the armature rotates at 600 rpm.
Assuming nothing else changes, what is the maximum emf when the armature rotates at
1800 rpm? ( Maximum emf is proportional to the frequency)
E2 f 2 (1800 rpm)(24 V)
= ; E2 = ; E2 = 72.0 V
E1 f1 600 rpm
*31-34. A shunt-wound motor connected across a 117-V line generates a back emf of 112 V
when the armature current is 10 A. What is the armature resistance?
Loop rule for first loop: ΣE = ΣIR : E – Eb = IARA
RA RF
117 V
5V
117 V – 112 V = (10 A)RA ; RA = = 0.500 Ω
10 A
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10. Chapter 31. Electromagnetic Induction Physics, 6th Edition
*31-35. A 110-V shunt-wound motor has a field resistance of 200 Ω connected in parallel with an
armature resistance of 10 Ω. When the motor is operating at full speed, the back emf is
90 V. What is the starting current and what is the operating current?
RA RF (10 Ω)(200 Ω)
RT = = ; RT = 9.524 Ω RF
RA + RF 10 Ω + 200 Ω RA
110 V
110 V
At start Eb = 0: I0 = ; I0 = 11.6 A
9.524 Ω
Loop rule for first loop ΣE = ΣIR: 110 V - 90 V = IA(10 Ω); IA = 2.00 A
110 V
Loop rule for outside loop: 110 V = IFRF ; IF = = 0.550 A
200 Ω
Current rule: IT = IF + IA = 0.550 Ω + 2.00 Ω; IT = 2.55 A
*31-36. A 120-V shunt motor has a field resistance of 160 Ω and an armature resistance of 1.00 Ω.
When the motor is operating at full speed, it draws a current of 8.00 A. Find the starting
current. What series resistance must be added to reduce the starting current to 30 A?
RA RF (1.0 Ω)(160 Ω)
RT = = ; RT = 0.994 Ω RF
RA + RF 1.0 Ω + 160 Ω RA
120 V
120 V
At start Eb = 0: I0 = ; I0 = 120.75 A
0.994 Ω Rs
Rs=? Added
120 V
Added Rs: I0 = = 30 A; Rs = 3.01 Ω
R s + 0.994 Ω
*31-37. A shunt generator has a field resistance of 400 Ω and an armature resistance of 2.00 Ω.
The generator delivers a power of 4000 W to an external line at 120 V. What is the emf
of the generator?
RA RF (2.0 Ω)(400 Ω)
RT = = ; RT = 1.99 Ω
RA + RF 2.0 Ω + 400 Ω
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11. Chapter 31. Electromagnetic Induction Physics, 6th Edition
P0 4000 W
*31-37. (Cont.) E – Ir = 120 V; P0 = (120 V)I; I= = = 33.3 A;
V 120 V
E = 120 V – (33.3 A)(1.99 Ω); E = 187 V
Critical Thinking Problems
31-38. A coil of wire has 10 loops, each of diameter D, placed inside a B field that varies at the
rate of 2.5 mWb/s. If the induced emf is 4 mV, what is the diameter of the coil? What
will the induced emf be if the diameter is doubled? What is the induced emf if the rate
of change in the B field is doubled?
∆φ NA ∆B −E −( −0.004 V)
E = −N =− ; A= = ; A = 0.160 m 2
∆t ∆t (
N ∆B
∆t ) (10)(2.5 x 10-3 wb/s)
π D2 4A 4(0.160 m 2 )
A= ; D= = ; D = 0.451 m = 45.1 cm
4 π π
Doubling D quadruples the area A and thus increases the emf by factor of four:
E2D = 4(-0.004 V); E2D= -16.0 mV (Doubling the diameter)
The emf varies directly with ∆B/∆t, therefore emf is doubled when it doubles:
E2B = 2(-0.004 V); E2B= -8.00 mV (Doubling the diameter)
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12. Chapter 31. Electromagnetic Induction Physics, 6th Edition
31-39. In Fig. 31-18a, the single loop of area 0.024 m2 is connected to a 4 mΩ resistor. The
magnet moves to the left through the center of the loop causing the magnetic flux to
increase at the rate of 2 mWb/s. (a) What are the magnitude and direction of the current
through the resistor? What if the magnet is then pulled back out of the loop with the
same speed? First lets reason the direction from Lenz’s Law: Motion
(a) The flux lines are increasing to the left, which means that
N S
the induced emf must generate a B field to the right which
opposes the cause that gave rise to it. The current must be
R
R
counterclockwise viewed from the magnet side.
∆φ −0.002 V
E = −N = −(1)(0.002 Wb/s) ; E = -2.00 mV; I= ; I = 0.500 A
∆t 0.004 Ω
(b) When magnet moves out to the right, the flux lines are decreasing, so the induced emf
in the coil must be such as to oppose that motion, i.e., to the right. The induced current
must be clockwise viewed from the magnet side. Magnitude is unchanged:
I = 0.500 A
31-40. In Fig. 31-18b a changing B field is produced first by an increasing current through the
loops and then by a decreasing current through the loops. In each case will the induced
current be downward or upward on the near side of the single loop?
(a) Grasping near wires of coil with thumb down
shows field increasing to right. Induced B must be
I
left, so induced current is upward in near loop. R
Loop
(b) Reverse reasoning for decreasing current shows B
Varying current
will be down in near loop. Ans. (a) Up; (b) Down
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13. Chapter 31. Electromagnetic Induction Physics, 6th Edition
*31-41. A coil of 50 turns is rotated clockwise with a frequency of 60 Hz in a constant 3.0-mT
magnetic field. The direction of the B field is along the positive x-axis and the coil of
area 0.070 m2 rotates clockwise in the x-y plane. When the plane of the loop is parallel to
the field, what is the direction of the current as viewed from the top (clockwise or
counter-clockwise)? What is the maximum induced emf? Starting time t = 0 when the
emf is zero, at what later time will its emf first rise to 2.00 V?
The right edge of the loop moves downward causing B
the induced current to be out at that point. Thus, the N
S
current in the coil is clockwise viewed from the top.
Note also from Lenz’s law that as the loop rotates, the “back” field induced inside the
loop is directed left, or opposite to the B field from the magnets.
Emax = 2π fNBA = 2π (60 Hz)(50)(0.003 T)(0.07 m 2 ) ; Emax = 3.96 V
Einst 2V
Einst = Emax sin(2π ft ) ; sin(2π ft ) = = ; 2π ft = sin −1 (0.505);
Emax 3.96 V
0.5294 rad
2π ft = 0.5294 rad; t= ; t = 1.40 ms
2π (60 Hz)
*31-42. When the motor on a heat pump is first turned on, it momentarily draws 40.0 A. The
current then immediately drops to a steady value of 12.0 A. If the motor operates on a
120-V power source, what is the back emf generated while the motor is running?
120 V 120 V
At start: I0 = = 40.0 A; RA = = 3.0 Ω ;
RA 40 A
Operating: 120 V – Eb = (12 A)(3.0 Ω); Eb = 84.0 V
185