solucionario capitulo 31 - paul e. tippens

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?

2 2-3 2(0.08 m) 5.03 x 10 m ;

4 4DA

B = 1.8 T – 0.6 T = 1.20 T

-3 2(50)(5.03 x 10 m )(1.20 T); 0.002 s

NA BNt t

E E; E = -151 V

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?

2(100)(0.004 T)(0.044 m 0); 0.3 s

NB ANt t

E E; E = -58.7 mV

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?

-30.23 x 10 Wb(300)0.002 s

Nt

E; E = –34.5 V.

429


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?

-3 -35 x 10 Wb 2 x 10 Wb(1)0.100 s

Nt

E; E = –30.0 mV

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 = R2 = (0.045 m)2 = 6.36 x 10-3 m2 ]

-3 2(120)(6.36 x 10 m )(0.06 T); 6.00 V

NA B NA BN tt t

E E

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?

256(0.3 m )(0.007 T - 0)0.006 s

NA BNt t

E; E = -19.6 V

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?

0sin (0.4 T)(0.15 m)(4 m/s)sin 36Blv E ; E = 141 mV

430


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?

sin ; sin

Blv vBl

EE

-3

0

60 x 10 Vsin (0.7 T)(0.2 m)sin 28

vBl

E

v = 79.9 m/s

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?

0

1.00 Vsin 90 (0.7 T)(0.5 m)(1)

vBl

E

; v = 2.86 m/s

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?

2 2-3 2(0.06 m) 2.83 x 10 m ;

4 4DA

f = 200 rpm = 3.33 rev/s

431


-32 2 (3.33 rev/s)(1)(0.004 T)(2.83 x 10 T)fNBA maxE ; E = 237 V

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 = R2 = (0.10 m)2 = 0.314 m2 ]

max2

7.00 V ;2 2 (300)(0.006 T)(0.314 m )

fNBA

E

f = 19.7 rev/s

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 min3600 60 rev/s;min 60 s

f f = 60.0 Hz

-4 2max 2 2 (60 Hz)(500)(0.002 T)(60 x 10 m )fNBA E ; 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 = 900 – 600 = 300 ]

0max sin 30 (2.26 V)(0.500);inst E E 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?

432


2 2 -3 2(0.05 m) 7.85 x 10 m ;A R

max-3 2

2.00 V ;2 2 (100)(0.06 T)(7.85 x 10 m )

fNBA

E

f = 6.75 rev/s

rev 60 s6.75 ;s min

f f = 405 rpm

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?

2 2-3 2(0.05 m) 1.96 x 10 m ;

4 4DA

max-3 2

110 V ;2 2 (70)(0.8 T)(1.96 x 10 m )

fNBA

E

f = 159 rev/s

*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]

2max 2 2 (10 Hz)(800)(0.003 T)(0.25 m )fNBA E ; Emax = 37.7 V

max sin(2 ) (37.7 V)sin 2 (10 Hz)(0.43 s)inst ft E E ; Einst = 35.9 V

433


*31-17. A 300- 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 ]

2 22(0.3 m) 0.0707 m ;

4 4DA

2 sin(2 )inst fNBA ft E

22 (5 Hz)(200)(0.005 T)(0.0707 m )sin[2 (5 Hz)(0.377 s)]inst E

Einst = (2.22 V) sin (0.118 rad); Einst = 1.47 V

1.47 V ;300 instI

Iinst = 4.90 mA

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 .

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 ); Eb = 96.0 V

0120 V ;

8 I

I0 = 15.0 A

31-19. The armature coil of the starting motor in an automobile has a resistance of 0.05 . 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?

434


012.0 V 240 A;0.05

I I0 = 240 A

12 V 6 V ;0.05

bVIR

E I = 120 A

31-20. A 220-V dc motor draws a current of 10 A in operation and has an armature resistance of

0.4 . What is the back emf when the motor is operating and what is the starting current?

220 V (10A)(0.4 );b V IR E Eb = 216 V

0220 V ;0.4

I I0 = 550 A

*31-21. A 120-V series-wound dc motor has a field resistance of 90 and an armature resistance

of 10 . 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?

(a) RT = 90 + 10 ; (b) 0

120 V ;100

I I0 = 1.20 A

(c)

120 V - 80 V ;100

I I = 0.400 A

*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.

435


2 2; (0.4 ) (100 ); 16.0 Wout in outout out

in in

P P PE P I R A PP P

48 W- 16 W(0.4 A)(120 V) 48.0 W; ;48 WinP IV E

E = 66.7%

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?

(120 V)(400); ;(100)

p p p ss

s s p

N NN N

E EEE Es = 480 V

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?

500V 1 ;10,000 V 20

p p

s s

NN

EE Ratio = 1:20

(1 A)(10,000 V); ; (500 V)out in p p s s sP P I I I E E

; Is = 20 A

436


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?

(120 V)(720); ;(80)

p p p ss

s s p

N NN N

E EEE Es = 1080 V

0.95(20 A)(120 V)0.95; ;1080 V

out s ss

in p p

P IE IP I

EE Is = 2.11 A

31-26. A 25-W light bulb has a resistance of 8.0 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

; (25 W)(8 ); 14.1 Voutout out out out

VP V P R VR

14.1 V ; 0.118120 V

s s s

p p p

N NN N

EE

Challenge Problems

437


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.

2(70)(0.008 T)(0.06 m 0); 0.02 s

NB ANt t

E E; E = -1.68 mV

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?

2

( 2 V)(0.5 s); ;(80)(0.2 m 0)

NB AN Bt t

E B = 62.5 mT

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 . What is the average

induced emf and what average current is delivered to the light filament?

200(0.06 Wb - 0.025 Wb0.5 s

Nt

E; E = -14.0 V

14.0 V ;2

I I = 7.00 A

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 m, what are the magnitude and direction of the induced current?

438


-3sin (80 x 10 T)(0.09 m)(35 m/s)(1);Blv E E = 0.252 V

0.252 V ;0.005

I I = 50.4 A, into paper

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

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?

maxmax 2

170 V2 ; ;2 2 (60 Hz)(500)(0.004 m )

fNBA BfNA

EE

B = 225 mT

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)

2 22

1 1

(1800 rpm)(24 V); ;600 rpm

ff

E EE E2 = 72.0 V

*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: E = IR : E – Eb = IARA

439

RA RF

117 V


117 V – 112 V = (10 A)RA ;

5 V10 AAR

= 0.500

*31-35. A 110-V shunt-wound motor has a field resistance of 200 connected in parallel with an

armature resistance of 10 . 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?

(10 )(200 ) ; 9.524 10 + 200

A FT T

A F

R RR RR R

At start Eb = 0: 0

110 V9.524

I ; I0 = 11.6 A

Loop rule for first loop E = IR: 110 V - 90 V = IA(10 ); IA = 2.00 A

Loop rule for outside loop: 110 V = IFRF ;

110 V 0.550 A200 FI

Current rule: IT = IF + IA = 0.550 + 2.00 ; IT = 2.55 A

*31-36. A 120-V shunt motor has a field resistance of 160 and an armature resistance of 1.00 .

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?

440

RA RF

110 V

Rs=? Added

Rs

RA RF

120 V


(1.0 )(160 ) ; 0.994 1.0 + 160

A FT T

A F

R RR RR R

At start Eb = 0: 0

120 V0.994

I ; I0 = 120.75 A

Added Rs: 0

s

120 V 30 A;R 0.994

I Rs = 3.01

*31-37. A shunt generator has a field resistance of 400 and an armature resistance of 2.00 .

The generator delivers a power of 4000 W to an external line at 120 V. What is the emf

of the generator?

(2.0 )(400 ) ; 1.99 2.0 + 400

A FT T

A F

R RR RR R

*31-37. (Cont.) E – Ir = 120 V; P0 = (120 V)I; 0 4000 W 33.3 A;

120 VPIV

E = 120 V – (33.3 A)(1.99 ); 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?

441


2

-3

( 0.004 V); ; 0.160 m(10)(2.5 x 10 wb/s)

NA BN A ABt t N t

EE

2 24 4(0.160 m ); ;4D AA D

D = 0.451 m = 45.1 cm

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)

31-39. In Fig. 31-18a, the single loop of area 0.024 m2 is connected to a 4 m 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:

442

N S

Motion

RR


(a) The flux lines are increasing to the left, which means that

the induced emf must generate a B field to the right which

opposes the cause that gave rise to it. The current must be

counterclockwise viewed from the magnet side.

(1)(0.002 Wb/s)Nt

E; E = -2.00 mV;

0.002 V0.004

I

; I = 0.500 A

(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

left, so induced current is upward in near loop.

(b) Reverse reasoning for decreasing current shows B

will be down in near loop. Ans. (a) Up; (b) Down

*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

443

IRLoop

Varying current I

B

NS


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

the induced current to be out at that point. Thus, the

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.

2max 2 2 (60 Hz)(50)(0.003 T)(0.07 m )fNBA E ; Emax = 3.96 V

max sin(2 )inst ftE E ;

1

max

2 Vsin(2 ) ; 2 sin (0.505);3.96 V

instft ft EE

0.5294 rad2 0.5294 rad; ;2 (60 Hz)

ft t

t = 1.40 ms

*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?

At start: 0

A

120 V 120 V40.0 A; 3.0 R 40 AAI R

;

Operating: 120 V – Eb = (12 A)(3.0 ); Eb = 84.0 V

444

solucionario capitulo 31 - paul e. tippens

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