chapter 20 induced voltages and inductance · chapter 20 induced voltages and inductance problem...

275

Chapter 20

Induced Voltages and Inductance

Problem Solutions

20.1 The angle between the d irection of the constant field and the normal to the plane of the

loop is 0 , so

2 2 3 2cos 0.50 T 8.0 10 m 12 10 m cos0 4.8 10 T m 4.8 mWbB BA

20.2 The magnetic flux through the loop is given by cosB BA where B is the magnitude

of the magnetic field , A is the area enclosed by the loop, and is the angle the magnetic

field makes with the normal to the plane of the loop. Thus,

2

-25 2 7 210 m

cos 5.00 10 T 20.0 cm cos 1.00 10 T m cos1 cm

B BA

(a) When B is perpendicular to the plane of the loop, 0 and

7 21.00 10 T mB

(b) If 7 2 8 230.0 , then 1.00 10 T m cos30.0 8.66 10 T mB

(c) If 7 290.0 , then 1.00 10 T m cos90.0 0B

20.3 2cos ( )cosB BA B r where is the angle between the d irection of the field and

the normal to the plane of the loop.

(a) If the field is perpendicular to the plane of the loop, 0 , and

3 2

22

8.00 10 T m0.177 T

cos 0.12 m cos0

BBr

276 CHAPTER 20

(b) If the field is d irected parallel to the plane of the loop, 90 , and

cos cos90 0B BA BA

20.4 The magnetic field lines are tangent to the surface of the cylinder, so that no magnetic

field lines penetrate the cylindrical surface. The total flux through the cylinder is zero

20.5 (a) Every field line that comes up through the area A on one side of the wire goes back

down through area A on the other side of the wire. Thus, the net flux through the

coil is zero

(b) The magnetic field is parallel to the plane of the coil , so 90.0 . Therefore,

cos cos90.0 0B BA BA

20.6 (a) The magnitude of the field inside the solenoid is

7 3

0 0

4004 10 T m A 5.00 A 6.98 10 T 6.98 mT

0.360 m

NB nI I

(b) The field inside a solenoid is d irected perpendicular to the cross-sectional area, so

0 and the flux through a loop of the solenoid is

2

3 2 5 2

cos cos

6.98 10 T 3.00 10 m cos0 1.97 10 T m

B BA B r

20.7 (a) The magnetic flux through an area A may be written as

cos

component of perpendicular to

B B A

B A A

Thus, the flux through the shaded side of the cube is

2

2 3 25.0 T 2.5 10 m 3.1 10 T mB xB A

(b) Unlike electric field lines, magnetic field lines always form closed loops, without

beginning or end . Therefore, no magnetic field lines originate or terminate within

the cube and any line entering the cube at one point must emerge from the cube at

some other point. The net flux through the cube, and indeed through any closed

surface, is zero .

Induced Voltages and Inductance 277

20.8

23

4

3

1.5 T 0 1.6 10 m cos0cos1.0 10 V 0.10 mV

120 10 s

B B A

t t

20.9 With the constant field d irected perpendicular to the plane of the coil, the flux through

the coil is cos0B BA BA . As the enclosed area increases, the magnitude of the

induced emf in the coil is

3 2 30.30 T 5.0 10 m s 1.5 10 V 1.5 mVB A

Bt t

20.10 cos

BB A

t t

2

20.15 T 0.12 m 0 cos0

3.4 10 V 34 mV0.20 s

20.11 The magnitude of the induced emf is cos

BB A

t t

If the normal to the plane of the loop is considered to point in the original d irection of

the magnetic field , then 0 and 180i f . Thus, we find

2

20.20 T cos180 0.30 T cos0 0.30 m

9.4 10 V 94 mV1.5 s

20.12 With the field d irected perpendicular to the plane of the coil, the flux through the coil is

cos0B BA BA . As the magnitude of the field increases, the magnitude of the

induced emf in the coil is

2 30.050 0 T s 0.120 m 2.26 10 V 2.26 mV

B BA

t t

278 CHAPTER 20

20.13 The required induced emf is 0.10 A 8.0 0.80 VIR .

From cosB BNA

t t

0.80 V2.7 T s

cos 75 0.050 m 0.080 m cos0

B

t NA

20.14 The initial magnetic field inside the solenoid is

7 3

0

1004 10 T m A 3.00 A 1.88 10 T

0.200 mB nI

(a) 2

3 2

7 2

cos 1.88 10 T 1.00 10 m cos0

1.88 10 T m

B BA

(b) When the current is zero, the flux through the loop is 0B and the average

induced emf has been

7 2

80 1.88 10 T m

6.28 10 V3.00 s

B

t

20.15 (a) The initial field inside the solenoid is

7 3

0

3004 10 T m A 2.00 A 3.77 10 T

0.200 mi iB nI

(b) The final field inside the solenoid is

7 3

0

3004 10 T m A 5.00 A 9.42 10 T

0.200 mf fB nI

(c) The 4-turn coil encloses an area 2

2 2 4 21.50 10 m 7.07 10 mA r

(d) The change in flux through each turn of the 4-turn coil during the 0.900-s period is

3 3 4 2 69.42 10 T 3.77 10 T 7.07 10 m 3.99 10 WbB B A


(e) The average induced emf in the 4-turn coil is

6

5

2

3.99 10 Wb4 1.77 10 V

0.900 s

BNt

Since the current increases at a constant rate during this time interval, the induced

emf at any instant during the interval is the same as the average value given above.

(f) The induced emf is small, so the current in the 4-turn coil will also be very small.

This means that the magnetic field generated by this current will be negligibly small

in comparison to the field generated by the solenoid.

20.16 The magnitude of the average emf is

4 2

cos

200 1.1 T 100 10 m cos0 cos18044 V

0.10 s

BNBAN

t t

Therefore, the average induced current is 44 V

8.8 A5.0

IR

20.17 If the magnetic field makes an angle of 28.0° with the plane of the coil, the angle it

makes with the normal to the plane of the coil is 62.0 . Thus,

-6 2 2 4 2

5

cos

200 50.0 10 T 39.0 cm 1 m 10 cm cos62.01.02 10 V 10.2 V

1.80 s

BN NB A

t t

20.18 With the magnetic field perpendicular to the plane of the coil, the flux through each turn

of the coil is 2

B BA B r . Since the area remains constant, the change in flux due

to the changing magnitude of the magnetic field is 2

B B r .

(a) The induced emf is: 2 2

0 00

0

B r NB rN N

t t t

280 CHAPTER 20

(b) When looking down on the coil from a location on the positive z-axis, the magnetic

field (in the positive z-d irection) is d irected up toward you and increasing in

magnitude. This means the change in the flux through the coil is d irected upw ard .

In order to oppose this change in flux, the current must produce a flux d irected

downward through the area enclosed by the coil. Thus, the current must flow

clockwise as seen from your viewing location.

(c) Since the turns of the coil are connected in series, the total resistance of the coil is

eqR NR . Thus, the magnitude of the induced current is

2 2

0 0

eq

NB r t B rI

R NR tR

20.19 The vertical component of the Earth’ s magnetic field is perpendicular to the horizontal

velocity of the metallic truck body. Thus, the motional emf induced across the width of

the truck is

6 31 m35 10 T 79.8 in 37 m s 2.6 10 V 2.6 mV

39.37 inB v

20.20 The vertical component of the Earth’ s magnetic field is perpendicular to the horizontal

velocity of the wire. Thus, the magnitude of the motional emf induced in the wire is

6 340.0 10 T 2.00 m 15.0 m s 1.20 10 V 1.20 mVB v

Imagine hold ing your right hand horizontal with the fingers pointing north (the

direction of the wire’ s velocity), such that when you close your hand the fingers curl

downward (in the d irection of B). Your thumb will then be pointing westward . By

right-hand rule 1, the magnetic force on charges in the wire would tend to move positive

charges westward . Thus, the west end of the wire will be positive relative to the east end .

20.21 (a) Observe that only the horizontal component, hB , of Earth’ s magnetic field is

effective in exerting a vertical force on charged particles in the antenna. For the

magnetic force, sinm hF qvB , on positive charges in the antenna to be d irected

upward and have maximum magnitude (when =90°), the car should move

toward the east through the northward horizontal component of the magnetic

field .


(b) hB v , where

hB is the horizontal component of the magnetic field .

6

4

km 0.278 m s50.0 10 T cos65.0 1.20 m 65.0

h 1 km h

4.58 10 V

20.22 (a) Since B v , the magnitude of the vertical component of the Earth’ s magnetic

field at this location is

6

vertical 3

0.45 V6.0 10 T 6.0 T

25 m 3.0 10 m sB B

v

(b) Yes. The magnitude and d irection of the Earth’ s field varies from one location to

the other, so the induced voltage in the wire changes. Further, the voltage will

change if the tether cord changes its orientation relative to the Earth’ s field .

20.23 B v , where B is the component of the magnetic field perpendicular to the velocity

v . Thus,

650.0 10 T sin58.0 60.0 m 300 m s 0.763 V

20.24 From B v , the required speed is

0.500 A 6.00 1.00 m s

2.50 T 1.20 m

IRv

B B

20.25 (a) To oppose the motion of the magnet, the magnetic field generated by the induced

current should be d irected to the right along the axis of the coil. The current must

then be left to right through the resistor.

(b) The magnetic field produced by the current should be d irected to the left along the

axis of the coil, so the current must be right to left through the resistor.

20.26 When the switch is closed , the magnetic field due to the current from the battery will be

d irected to the left along the axis of the cylinder. To oppose this increasing leftward flux,

the induced current in the other loop must produce a field d irected to the right through

the area it encloses. Thus, the induced current is left to right through the resistor.

282 CHAPTER 20

20.27 Since the magnetic force, sinmF qvB , on a positive charge is d irected toward the top

of the bar when the velocity is to the right, the right hand rule says that the magnetic

field is d irected into the page .

20.28 When the switch is closed , the current from the battery produces a magnetic field

d irected toward the right along the axis of both coils.

(a) As the battery current is growing in magnitude, the induced current in the

rightmost coil opposes the increasing rightward d irected field by generating a field

toward to the left along the axis. Thus, the induced current must be left to right

through the resistor.

(b) Once the battery current, and the field it produces, have stabilized , the flux throug h

the rightmost coil is constant and there is no induced current .

(c) As the switch is opened, the battery current and the field it produces rapid ly

decrease in magnitude. To oppose this decrease in the rightward d irected field , the

induced current must produce a field toward the right along the axis, so the

induced current is right to left through the resistor.

20.29 When the switch is closed , the current from the battery produces a magnetic field

d irected toward the left along the axis of both coils.

(a) As the current from the battery, and the leftward field it produces, increase in

magnitude, the induced current in the leftmost coil opposes the increased leftward

field by flowing right to left through R and producing a field d irected toward the

right along the axis.

(b) As the variable resistance is decreased , the battery current and the leftward field

generated by it increase in magnitude. To oppose this, the induced current is

right to left through R, producing a field d irected toward the right along the axis.

(c) Moving the circuit containing R to the left decreases the leftward field (due to the

battery current) along its axis. To oppose this decrease, the induced current is

left to right through R, producing an additional field d irected toward the left

along the axis.

(d) As the switch is opened, the battery current and the leftward field it produces

decrease rapid ly in magnitude. To oppose this decrease, the induced current is

left to right through R, generating additional magnetic field d irected toward the

left along the axis.


20.30 (a) As the bottom conductor of the loop falls, it cuts across

the magnetic field lines coming out of the page. This

induces an emf of magnitude Bwv in this conductor,

with the left end at the higher potential. As a result, an

induced current of magnitude

Bwv

IR R

flows clockwise around the loop. The field then exerts

an upward force of magnitude

2 2

m

Bwv B w vF BIw B w

R R

on this current-carrying conductor forming the bottom of the loop. If the loop is

falling at terminal speed , the magnitude of this force must equal the downward

gravitational force acting on the loop. That is, when Tv v , we must have

2 2

tB w vMg

R or

2 2T

MgRv

B w

(b) A larger resistance would make the current smaller, so the loop must reach higher

speed before the magnitude of the magnetic force will equal the g ravitational force.

(c) The magnetic force is proportional to the product of the field and the current, while

the current itself is proportional to the field . If B is cut in half, the speed must

become four times larger to compensate and yield a magnetic force with magnitude

equal to the that of the gravitational force.

284 CHAPTER 20

20.31 (a) After the right end of the coil has

entered the field , but the left end

has not, the flux through the area

enclosed by the coil is d irected into

the page and is increasing in

magnitude. This increasing flux

induces an emf of magnitude

B

NB ANBwv

t t

in the loop. Note that in the above equation, A is the area enclosed by the coil that

enters the field in time t . This emf produces a counterclockwise current in the

loop to oppose the increasing inward flux. The magnitude of this current is

I R NBwv R . The right end of the loop is now a conductor, of length Nw,

carrying a current toward the top of the page through a field d irected into the page.

The field exerts a magnetic force of magnitude

2 2 2

directed toward the leftNBwv N B w v

F BI Nw B NwR R

on this conductor, and hence, on the loop.

(b) When the loop is entirely within the magnetic field , the flux through the area

enclosed by the loop is constant. Hence, there is no induced emf or current in the

loop, and the field exerts zero force on the loop.

(c) After the right end of the loop emerges from the field , and before the left end

emerges, the flux through the loop is d irected into the page and decreasing. This

decreasing flux induces an emf of magnitude NBwv in the loop, which

produces an induced current d irected clockwise around the loop so as to oppose the

decreasing flux. The current has magnitude I R NBwv R . This current

flowing upward , through conductors of total length Nw, in the left end of the loop,

experiences a magnetic force given by

2 2 2

directed toward the leftNBwv N B w v

F BI Nw B NwR R


20.32 (a) The motional emf induced in the bar must be IR , where I is the current in this

series circuit. Since B v , the speed of the moving bar must be

38.5 10 A 9.0 0.73 m s

0.30 T 0.35 m

IRv

B B

The flux through the closed loop formed by the rails, the bar, and the resistor is

d irected into the page and is increasing in magnitude. To oppose this change in

flux, the current must flow in a manner so as to produce flux out of the page

through the area enclosed by the loop. This means the current will flow

countercloclockwise .

(b) The rate at which energy is delivered to the resistor is

3.00 s1.30 s

ln 0.100

(c) An external force d irected to the right acts on the bar to balance the magnetic force

to the left. Hence, work is being done by the external force , which is transformed into

the resistor’ s thermal energy.

20.33 The emf induced in a rotating coil is d irectly proportional to the angular frequency of

the coil. Thus,

2 2

1 1

or 22 1

1

500 rev min24.0 V 13.3 V

900 rev min

20.34 25

max horizontal

2

rev 2 rad 1 min100 2.0 10 T 0.20 m 1500

min 1 rev 60 s

1.3 10 V 13 mV

NB A

286 CHAPTER 20

20.35 Note the similarity between the situation in this problem and a generator. In a generator,

one normally has a loop rotating in a constant magnetic field so the flux through the

loop varies sinusoidally in time. In this problem, we have a s tationary loop in an

oscillating magnetic field , and the flux through the loop varies sinusoidally in time. In

both cases, a sinusoidal emf max sin t where

max NBA is induced in the loop.

The loop in this case consists of a single band 1N around the perimeter of a red

blood cell with d iameter 68.0 10 md . The angular frequency of the oscillating flux

through the area of this loop is 2 2 60 Hz 120 rad sf . The maximum

induced emf is then

23 6 -12

11

max

1.0 10 T 8.0 10 m 120 s1.9 10 V

4 4

dNBA B

20.36 (a) Using max NBA ,

2 3

max

rev 2 rad1000 0.20 T 0.10 m 60 7.5 10 7.5 kV

s 1 rev

(b) max occurs when the flux through the loop is changing the most rapid ly. This is

when the plane of the loop is parallel to the magnetic field .

20.37 rev 1 min 2 rad rad

120 4 min 60 s 1 rev s

and the period is 2

0.50 sT

(a) max 500 0.60 T 0.080 m 0.20 m 4 rad s 60 VNBA

(b) max sin 60 V sin 4 rad s s 57 V32

t

(c) The emf is first maximum at 0.50 s

0.13 s4 4

Tt


20.38 (a) Immediately after the switch is closed , the motor coils are still stationary and the

back emf is zero. Thus, 240 V

8.0 A30

IR

(b) At maximum speed, 145 Vback and

240 V 145 V

3.2 A30

backIR

(c) 240 V 6.0 A 30 60 Vback IR

20.39 (a) When a coil having N turns and enclosing area A rotates at angular frequency in

a constant magnetic field , the emf induced in the coil is

max maxsin where t NB A

Here, B is the magnitude of the magnetic field perpendicular to the rotation axis

of the coil. In the given case, 55.0 TB ; A ab where 10.0 cm 2a and

4.00 cm 2b ; and

rev 1 min

2 2 100 10.5 rad smin 60.0 s

f

Thus, 6

max 10.0 55.0 10 T 0.100 m 0.040 0 m 10.5 rad s4

or 5

max 1.81 10 V 18.1 V

(b) When the rotation axis is parallel to the field , then 0B giving max 0

It is easily understood that the induced emf is always zero in this case if you

recognize that the magnetic field lines are always parallel to the plane of the coil,

and the flux through the coil has a constant value of zero.

20.40 (a) In terms of its cross-sectional area (A ), length ( ), and number of turns (N ), the self

inductance of a solenoid is given as 2

0L N A . Thus, for the given solenoid ,

222 2 7 2

0 34 4 10 T m A 580 8.0 10 m

5.9 10 H 5.9 mH4 0.36 m

N dL

288 CHAPTER 20

(b) 3 35.9 10 H 4.0 A s 24 10 V 24 mVI

Lt

20.41 From L I t , we have

3

312 10 V 0.50 s

4.0 10 H 4.0 mH2.0 A 3.5 A

tL

I t I

20.42 The units of BN

I

are

2T m

A

From the force on a moving charged particle, F qvB , the magnetic field is F

Bqv

and

we find that

N N s1 T 1 1

C m s C m

Thus, 2 2N m sN s J

T m m s V sC m C C

and 2T m V s

A A

which is the same as the units of

I t

20.43 (a)

227 22

0

3

4 10 T m A 400 2.5 10 m

0.20 m

2.0 10 H 2.0 mH

N AL

(b) From L I t , 3

3

75 10 V38 A s

2.0 10 H

I

t L


20.44 From L I t , the self-inductance is

3

324.0 10 V2.40 10 H

10.0 A sL

I t

Then, from BL N I , the magnetic flux through each turn is

3

5 22.40 10 H 4.00 A

1.92 10 T m500

B

L I

N

20.45 (a) 3

312 10 H4.0 10 s 4.0 ms

3.0

L

R

(b) maxIR

, so 3 1

max 150 10 A 3.0 4.5 10 V 0.45 VI R

20.46 (a) The time constant of the RL circuit is L R , and that of the RC circuit is RC . If

the two time constants have the same value, then

L

RCR

, or 3

-6

3.00 H1.00 10 1.00 k

3.00 10 F

LR

C

(b) The common value of the two time constants is

-3

3

3.00 H3.00 10 s 3.00 ms

1.00 10

L

R

20.47 (a) maxIR

, so max 8.0 A 0.30 2.4 VI R

(b) The time constant is L

R , giving

20.25 s 0.30 7.5 10 H 75 mHL R

(c) The current as a function of time is max 1 tI I e , so at t ,

1

max max1 0.632 0.632 8.0 A 5.1 AI I e I

290 CHAPTER 20

(d) At t , 5.1 AI and the voltage drop across the resistor is

5.1 A 0.30 1.5 VRV IR

Applying Kirchhoff’ s loop rule to the circuit shown in Figure 20.27 gives

0R LV V . Thus, at t , we have

2.4 V 1.5 V 0.90 VL RV v

20.48 The current in the RL circuit at time t is 1 tI eR

. The potential d ifference across

the resistor is 1 t

RV RI e , and from Kirchhoff’ s loop rule, the potential

difference across the inductor is

1 1 t t

L RV V e e

(a) At 0t , 01 1 1 0RV e

(b) At t , 11 6.0 V 1 0.368 3.8 VRV e

(c) At 0t , 0 6.0 VLV e

(d) At t , 1 6.0 V 0.368 2.2 VLV e

20.49 From max 1 tI I e , max

1t Ie

I

If max

0.900I

I at 3.00 st , then

3.00 s 0.100e or

3.00 s1.30 s

ln 0.100

Since the time constant of an RL circuit is L R , the resistance is

2.50 H

1.92 1.30 s

LR


20.50 (a) 8.00 mH

2.00 ms4.00

L

R

(b) 6 -3250 10 s 2.00 10 s6.00 V

1 1 0.176 A4.00

tI e eR

(c) max

6.00 V1.50 A

4.00 I

R

(d) max 1 tI I e yields max1te I I ,

and maxln 1 2.00 ms ln 1 0.800 3.22 mst I I

20.51 (a) The energy stored by an inductor is 212LPE LI , so the self inductance is

3

4

2 2

2 0.300 10 J22.08 10 H 0.208 mH

1.70 A

LPEL

I

(b) If 3.0 AI , the stored energy will be

22 4 41 1

2.08 10 H 3.0 A 9.36 10 J 0.936 mJ2 2

LPE LI

20.52 (a) The inductance of a solenoid is given by 2

0L N A , where N is the number of

turns on the solenoid , A is its cross-sectional area, and is its length. For the given

solenoid ,

222 2 7 2

0 34 10 T m A 300 5.00 10 m

4.44 10 H0.200 m

N rL

(b) When the solenoid described above carries a current of 0.500 AI , the stored

energy is

22 3 41 1

4.44 10 H 0.500 A 5.55 10 J2 2

LPE LI

20.53 The current in the circuit at time t is 1 tI eR

, and the energy stored in the

inductor is 21

2LPE LI

292 CHAPTER 20

(a) As t , max

24 V3.0 A

8.0 I I

R

, and

22

max

1 14.0 H 3.0 A 18 J

2 2LPE LI

(b) At t , 1

max 1 3.0 A 1 0.368 1.9 AI I e

and 21

4.0 H 1.9 A 7.2 J2

LPE

20.54 (a) Use Table 17.1 to obtain the resistivity of the copper wire and find

8

Cu Cuwire 2 2

3wire wire

1.7 10 m 60.0 m1.3

0.50 10 m

L LR

A r

(b)

2

2solenoid

60.0 m4.8 10 turns

Circumference of a loop 2 2 2.0 10 m

L LN

r

(c) The length of the solenoid is

3

wirediameter of wire 2 480 2 0.50 10 m 0.48 mN N r

(d)

227 22 2 2

0 solenoid 0 solenoid4 10 T m A 480 2.0 10 m

0.48 m

N A N rL

giving 47.6 10 H 0.76 mHL

(e) 4

4

total wire internal

7.6 10 H4.6 10 s 0.46 ms

1.3 0.350

L L

R R r

(f) max

total

6.0 V3.6 A

1.3 0.350 I

R

(g) max 1 tI I e , so when max0.999I I , we have 1 0.999te and

1 0.999 0.001te . Thus, ln 0.001t

or ln 0.001t

giving 0.46 ms ln 0.001 3.2 mst


(h) 22 4 3

maxmax

1 17.6 10 H 3.6 A 4.9 10 J 4.9 mJ

2 2LPE LI

20.55 According to Lenz’ s law, a current will be induced in the coil to oppose the change in

magnetic flux due to the magnet. Therefore, current must be d irected from b to a

through the resistor, and a bV V will be negative .

20.56 cos

BNBA

t t

, so

cos

tB

NA

or

3

5

2

0.166 V 2.77 10 s5.20 10 T 52.0 T

500 0.150 m 4 cos0 cos90B

20.57 (a) The current in the solenoid reaches max0.632I I in a time of t L R , where

27 4 22

0

-2

4 10 T m A 12 500 1.00 10 m0.280 H

7.00 10 m

N AL

Thus, 20.280 H2.00 10 s 20.0 ms

14.0 t

(b) The change in the solenoid current during this time is

max

60.0 V0.632 0 0.632 0.632 2.71 A

14.0

VI I

R

so the average back emf is

back -2

2.71 A0.280 H 37.9 V

2.00 10 s

IL

t

(c)

102 0

7 4 2

3

-2 -2

4 10 T m A 12 500 2.71 A 1.00 10 m1.52 10 V

2 7.00 10 m 2.00 10 s

2B

n I AB A N I A

t t t t

(d) 3

coil coil

coil coil

820 1.52 10 V0.0519 A 51.9 mA

24.0

BN tI

R R

294 CHAPTER 20

20.58 (a) The gravitational force exerted on the ship by the pulsar supplies the centripetal

acceleration needed to hold the ship in orbit. Thus,

2

pulsar ship ship

2

orbitorbit

g

GM m m vF

rr ,

giving

11 2 30

pulsar 6

7

orbit

6.67 10 N m kg 2.0 10 kg2.1 10 m s

3.0 10 m

GMv

r

(b) The magnetic force acting on charged particles moving through a magnetic field is

perpendicular to both the magnetic field and the velocity of the particles (and

therefore perpendicular to the ship’ s length). Thus, the charged particles in the

materials making up the spacecraft experience magnetic forces d irected from one

side of the ship to the other, meaning that the induced emf is d irected

from side to side within the ship.

(c) B v , where ship 0.080 km 80 m2r is the side to side d imension of the ship.

This yields

2 6 101.0 10 T 80 m 2.1 10 m s 1.7 10 V

(d) The very large induced emf would lead to powerful spontaneous electric

d ischarges. The strong electric and magnetic fields would d isrupt the flow of ions in

their bodies.

20.59 (a) To move the bar at uniform speed, the magnitude of the applied force must equal

that of the magnetic force retard ing the motion of the bar. Therefore, appF BI .

The magnitude of the induced current is

B t B A t B v

IR R R R

so the field strength is IR

Bv

, giving 2

appF I R v

Thus, the current is

app 1.00 N 2.00 m s

0.500 A8.00

F vI

R

(b) 22 0.500 A 8.00 2.00 WI R P

(c) input app 1.00 N 2.00 m s 2.00 WF v P


20.60 (a) When the motor is first turned on, the coil is not rotating so the back emf is zero.

Thus, the current is a maximum with only the resistance of the windings limiting its

value. This gives maxI R , or

max

120 V11

11 AR

I

(b) When the motor has reached maximum speed, the steady state value of the current

is backI R , giving the back emf as

back 120 V 4.0 A 11 76 VIR

20.61 If d is the d istance from the lightning bolt to the center of the coil, then

0 0

av

27 6

6

5

2

2

100 4 10 T m A 6.02 10 A 0 0.800 m

2 200 m 10.5 10 s

1.15 10 V= 115 kV

BN I d AN N B A N I A

t t t d t

20.62 When A and B are 3.00 m apart, the area enclosed

by the loop consists of four triangular sections,

each having hypotenuse of 3.00 m, altitude of

1.50 m, and base of 2 2

3.00 m 1.50 m 2.60 m

The decrease in the enclosed area has been

2 21

3.00 m 4 1.50 m 2.60 m 1.21 m2

i fA A A

The average induced current has been

2

av

av

0.100 T 1.21 m 0.100 s0.121 A

10.0

B t B A tI

R R R

As the enclosed area decreases, the flux (d irected into the page) through this area also

decreases. Thus, the induced current will be d irected clockwise around the loop to

create additional flux d irected into the page through the enclosed area.

296 CHAPTER 20

20.63 (a)

2

av

2-2

3

4 0

25.0 mT 2.00 10 m0.157 mV

4 50.0 10 s

BB dB A

t t t

As the inward d irected flux through the loop decreases, the induced current goes

clockwise around the loop in an attempt to create additional inward flux through

the enclosed area. With positive charges accumulating at B,

point is at a higher potential than B A

(b)

2-2

av 3

100 25.0 mT 2.00 10 m5.89 mV

4 4.00 10 s

BB A

t t

As the inward d irected flux through the enclosed area increases, the induced

current goes counterclockwise around the loop in an attempt to create flux d irected

outward through the enclosed area.

With positive charges now accumulating at A ,

point is at a higher potential than A B

20.64 The induced emf in the ring is

solenoid solenoidsolenoid solenoidav 0 solenoid

27 2 4

2 1

2

14 10 T m A 1000 270 A s 3.00 10 m 4.80 10 V

2

BB AB A I

n At t t t

Thus, the induced current in the ring is

4

av

ring -4

4.80 10 V1.60 A

3.00 10 I

R


20.65 (a) As the rolling axle (of length 1.50 m ) moves perpendicularly to the uniform

magnetic field , an induced emf of magnitude B v will exist between its ends.

The current produced in the closed -loop circuit by this induced emf has magnitude

av 0.800 T 1.50 m 3.00 m s

9.00 A0.400

B t B A t B vI

R R R R

(b) The induced current through the axle will cause the magnetic field to exert a

retard ing force of magnitude rF BI on the axle. The d irection of this force will be

opposite to that of the velocity v so as to oppose the motion of the axle. If the axle

is to continue moving at constant speed , an applied force in the d irection of v and

having magnitude app rF F must be exerted on the axle.

app 0.800 T 9.00 A 1.50 m 10.8 NF BI

(c) Using the right-hand rule, observe that positive charges within the moving axle

experience a magnetic force toward the rail containing point b, and negative charges

experience a force d irected toward the rail containing point a. Thus, the rail

containing b will be positive relative to the other rail. Point

is then at a higher potential than b a , and the current goes from b to a through the

resistor R.

(d) No . Both the velocity v of the rolling axle and the magnetic field B are

unchanged. Thus, the polarity of the induced emf in the moving axle is unchanged ,

and the current continues to be d irected from b to a through the resistor R.

20.66 (a) The time required for the coil to move d istance and exit the field is t v , where

v is the constant speed of the coil. Since the speed of the coil is constant, the flux

through the area enclosed by the coil decreases at a constant rate. Thus, the

instantaneous induced emf is the same as the average emf over the interval t

seconds in duration, or

2 20

0

BA B NBN N N NB v

t t t v

(b) The current induced in the coil is NB v

IR R

(c) The power delivered to the coil is given by 2I RP , or

2 2 2 2 2 2 2 2

2

N B v N B vR

RR

P

298 CHAPTER 20

(d) The rate that the applied force does work must equal the power delivered to the

coil, so appF v P or

2 2 2 2 2 2 2

app

N B v R N B vF

v v R P

(e) As the coil is emerging from the field , the flux through the area it encloses is

d irected into the page and decreasing in magnitude. Thus, the change in the flux

through the coil is d irected out of the page. The induced current must then flow

around the coil in such a d irection as to produce flux into the page through the

enclosed area, opposing the change that is occurring. This means that the current

must flow clockwise around the coil.

As the coil is emerging from the field , the left side of the coil is carrying an induced

current d irected toward the top of the page through a magnetic field that is d irected

into the page. Right-hand rule 1, then shows that this side of the coil will experience

a magnetic force directed to the left , opposing the motion of the coil.

chapter 20 induced voltages and inductance · chapter 20 induced voltages and inductance problem...

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