Solucionario Capitulo 31 - Paul E. Tippens

Physics, 6th Edition

Chapter 31. Electromagnetic Induction

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? A=

π D 2 π (0.08 m) 2 = = 5.03 x 10-3 m 2 ; 4 4

E = −N

∆φ NA ∆B =− ; ∆t ∆t

E= −

∆B = 1.8 T – 0.6 T = 1.20 T

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

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? E = −N

31-3

∆φ NB ∆A =− ; ∆t ∆t

E= −

(100)(0.004 T)(0.044 m 2 − 0) ; 0.3 s

E = -58.7 mV

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? E = −N

∆φ 0.23 x 10-3 Wb ; = −(300) ∆t 0.002 s

E = –34.5 V.

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? E = −N

∆φ 5 x 10-3 Wb − 2 x 10-3 Wb ; = −(1) ∆t 0.100 s

173

E = –30.0 mV

Physics, 6th Edition

Chapter 31. Electromagnetic Induction

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 ] E = −N

∆φ NA ∆B − NA ∆B −(120)(6.36 x 10-3 m 2 )(0.06 T) =− ; ∆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 =− = ∆t ∆t 0.006 s

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? 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 = Blv sin θ ;

v=

v=

E Bl sin θ

E 60 x 10-3V = Bl sin θ (0.7 T)(0.2 m)sin 280 v = 79.9 m/s

174

Physics, 6th Edition

Chapter 31. Electromagnetic Induction

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? v=

E 1.00 V = ; 0 Bl sin 90 (0.7 T)(0.5 m)(1)

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? A=

π D 2 π (0.06 m) 2 = = 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 ] f =

Emax 7.00 V = ; 2π NBA 2π (300)(0.006 T)(0.314 m 2 )

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? f = 3600

rev  1 min    = 60 rev/s; min  60 s 

175

f = 60.0 Hz

Physics, 6th Edition

Chapter 31. Electromagnetic Induction

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 ; f =

Emax 2.00 V = ; 2π NBA 2π (100)(0.06 T)(7.85 x 10-3 m 2 ) f = 6.75

rev  60 s   ; s  min 

f = 6.75 rev/s

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?

π D 2 π (0.05 m) 2 A= = = 1.96 x 10-3 m 2 ; 4 4 f =

Emax 110 V = ; 2π NBA 2π (70)(0.8 T)(1.96 x 10-3m 2 )

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

176

Physics, 6th Edition

Chapter 31. Electromagnetic Induction

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 ] A=

π D 2 π (0.3 m) 2 = = 0.0707 m 2 ; 4 4

Einst = 2π fNBA sin(2π ft )

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); I inst =

1.47 V ; 300 Ω

Einst = 1.47 V

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 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 I0 =

120 V ; 8Ω

I0 = 15.0 A

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? I0 =

12.0 V = 240 A; 0.05 Ω

177

I0 = 240 A

Physics, 6th Edition

Chapter 31. Electromagnetic Induction I=

V − Eb 12 V − 6 V = ; R 0.05 Ω

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 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); I0 =

220 V ; 0.4 Ω

Eb = 216 V

I0 = 550 A

*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? (a) RT = 90 Ω + 10 Ω = (c) I =

(b) I 0 =

1 00 Ω ;

120 V - 80 V ; 100 Ω

120 V ; 100 Ω

I0 = 1.20 A

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. E=

Pout Pin − Pout = ; Pin Pin

Pout = I 2 R = (0.4 A) 2 (100 Ω);

Pin = IV = (0.4 A)(120 V) = 48.0 W;

E=

Pout = 16.0 W

48 W- 16 W ; 48 W

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?

178

Physics, 6th Edition

Chapter 31. Electromagnetic Induction Ep

=

Es

Np Ns

;

Ep N s

Es =

Np

=

(120 V)(400) ; (100)

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? Np

=

Ns Pout = Pin ;

Ep Es

=

500V 1 = ; 10,000 V 20

I p Ep = I s Es ;

Is =

Ratio = 1:20

(1 A)(10,000 V) ; (500 V)

Is = 20 A

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 Es E=

=

Np Ns

;

Es =

Ep N s

Pout Es I s = = 0.95; Pin Ep I p

Np

=

Is =

(120 V)(720) ; (80)

Es = 1080 V

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

Is = 2.11 A

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. Pout =

2 Vout ; R

Vout = Pout R = (25 W)(8 Ω); N s Es 14.1 V = = ; N p Ep 120 V

179

Vout = 14.1 V

Ns = 0.118 Np

Physics, 6th Edition

Chapter 31. Electromagnetic Induction

Challenge Problems 31-27. A 70-turn coil of wire has an area of 0.06 m2 and is placed perpendicular to a constant 8mT magnetic field. Calculate the induced emf if the coil flips 900 in 0.02 s. E = −N

∆φ NB ∆A =− ; ∆t ∆t

E= −

(70)(0.008 T)(0.06 m 2 − 0) ; 0.02 s

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

2V

as the coil is flipped parallel to the field in 0.5 s? E = −N

∆φ NB ∆A =− ; ∆t ∆t

B=−

(−2 V)(0.5 s) ; (80)(0.2 m 2 − 0)

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 Ω. What is the average induced emf and what average current is delivered to the light filament? E = −N

∆φ 200(0.06 Wb - 0.025 Wb =− ; ∆t 0.5 s I=

14.0 V ; 2Ω

E = -14.0 V

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 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);

180

E = 0.252 V

Physics, 6th Edition

Chapter 31. Electromagnetic Induction I=

0.252 V ; 0.005 Ω

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 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 = 2π fNBA;

B=

Emax 170 V = ; 2π fNA 2π (60 Hz)(500)(0.004 m 2 )

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) E2 f 2 = ; E1 f1

E2 =

(1800 rpm)(24 V) ; 600 rpm

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: ΣE = ΣIR : E – Eb = IARA 117 V – 112 V = (10 A)RA ; RA =

5V = 0.500 Ω 10 A

181

117 V

RA

RF

Physics, 6th Edition

Chapter 31. Electromagnetic Induction

*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? RT =

RA RF (10 Ω)(200 Ω) = ; RA + RF 10 Ω + 200 Ω

At start Eb = 0:

I0 =

110 V ; 9.524 Ω

Loop rule for first loop ΣE = ΣIR: Loop rule for outside loop:

RT = 9.524 Ω

RF

RA

110 V

I0 = 11.6 A 110 V - 90 V = IA(10 Ω);

110 V = IFRF ;

IF =

IA = 2.00 A

110 V = 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? RT =

RA RF (1.0 Ω)(160 Ω) = ; RA + RF 1.0 Ω + 160 Ω

At start Eb = 0:

Added Rs:

I0 =

120 V ; 0.994 Ω

RT = 0.994 Ω 120 V

I0 = 120.75 A

120 V I0 = = 30 A; R s + 0.994 Ω

RA

Rs Rs=? Added

Rs = 3.01 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? RT =

RA RF (2.0 Ω)(400 Ω) = ; RA + RF 2.0 Ω + 400 Ω

182

RT = 1.99 Ω

RF

Physics, 6th Edition

Chapter 31. Electromagnetic Induction

E – Ir = 120 V;

*31-37. (Cont.)

P0 = (120 V)I;

E = 120 V – (33.3 A)(1.99 Ω);

I=

P0 4000 W = = 33.3 A; V 120 V

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? E = −N

∆φ NA ∆B =− ; ∆t ∆t

A=

π D2 ; 4

D=

A=

−E N ∆B

(

∆t

)

=

−( −0.004 V) ; (10)(2.5 x 10-3 wb/s)

4A 4(0.160 m 2 ) = ; π π

A = 0.160 m 2

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)

183

Physics, 6th Edition

Chapter 31. Electromagnetic Induction

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 Motion

speed? First lets reason the direction from Lenz’s Law: (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. E = −N

∆φ = −(1)(0.002 Wb/s) ; E = -2.00 mV; ∆t

I=

−0.002 V ; 0.004 Ω

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

184

I Loop

R Varying current I

Physics, 6th Edition

Chapter 31. Electromagnetic Induction

*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 counterclockwise)? 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

N

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

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 = Emax sin(2π ft ) ;

sin(2π ft ) =

2π ft = 0.5294 rad;

t=

Einst 2V = ; Emax 3.96 V 0.5294 rad ; 2π (60 Hz)

2π ft = sin −1 (0.505);

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 120V power source, what is the back emf generated while the motor is running? At start:

I0 =

120 V = 40.0 A; RA

RA =

120 V = 3.0 Ω ; 40 A

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

185

Eb = 84.0 V