Resoluções de Exercícios - Cap. 23 - Princípios de Física Vol. 3

8/2/2019 Resolues de Exerccios - Cap. 23 - Princpios de Fsica Vol. 3

1/14


2/14

Chapter 23

198

23.5 (a)

E= = =d

dtAdB

dt

B

ABe tmax /

(b)E=

( ) ( )=

0 160 0 350

2 00

24 00 2 00. .

.. / .m T

se

3 79. mV

(c) At t = 0, E = 28 0. mV

*23.6 (a)dB = B dA =

0I

2xLdx :

B h

h w IL dx

x= =

+

0

2

0

2

IL h w

hln

+

(b)E= =

+

=

+

d

dt

d

dt

IL h w

h

L h w

h

dI

dtB

0 0

2 2ln ln

E=

( )( ) +

( )

=4 10 1 00

2

1 00 10 0

1 0010 0

7

T m/A mA/s

.ln

. .

..

4 80. V

The long wire produces magnetic flux into the page through therectangle, shown by the first hand in the figure to the right.

As the magnetic flux increases, the rectangle produces its own magnetic

field out of the page, which it does by carrying counterclockwise

current (second hand in the figure).

*23.7

E = =

= +( )

B

tN

dB

dtA N t A0 0100 0 0800. .

At t = 5.00 s,E = ( ) ( )[ ] =30 0 0 410 0 0400 2. . .T m 61 8. mV

23.8

B

nI A= ( )0 solenoid

E= = ( )N

d

dtN n r

dI

dt

B 02solenoid

E= ( ) ( ) ( ) ( ) 15 0 4 10 1 00 10 0 0200 600 1207 3 1 2. . . cos( ) T m/A m m A/s t

E = ( )14 2. cos 120 mVt


3/14

Chapter 23

199

23.9 (a) B iext extB= and Bext decreases; therefore, theinduced field is B i0 0= B (to the right) and the

current in the resistor is directed to the right .

(b)

B iext extB= ( ) increases; therefore, the inducedfield B i0 0= +( )B is to the right, and the current in

the resistor is directed to the right .

(c) B kext extB= ( ) into the paper and Bext decreases;therefore, the induced field is B k0 0= ( )B into thepaper, and the current in the resistor is directed

to the right

(d) By the magnetic force law, F qB = ( )v B .Therefore, a positive charge will move to the top

of the bar if B is into the paper .

23.10

I

R

B v

R= =E l

v = 1 00. m/s

*23.11 (a)F BB I I B= =ll l

When I= E/R

and E = Blv

we get FB v

RB

B v

RB = = = =

ll

l( )

( . ) ( . ) ( . )

.

2 2 2 22 50 1 20 2 00

6 003.00 N

The applied force is 3 00. N to the right

(b)P

l= = =I R

B v

R

22 2 2

6 0. 0 W or P = =Fv 6 00. W


4/14

Chapter 23

200

23.12 FB = IlB and E l= B v

I

R

B v

R= =E l

so

B =IR

lv

(a) F

I R

vB =

2l

l and I

=

FBv

R = 0 500. A

(b) I R2 =

2 00. W

(c) For constant force, P = F v = 1.00 N( ) 2.00 m/ s( ) = 2 00. W

23.13 (a) For maximum induced emf, with positive charge at the top of the antenna,

F v B+ += ( )q , so the auto must move east

(b)

E l= = ( )( )

=B v 5 00 10 1 2065 0 10

65 053

. ..

cos .T mm

3600 s 4 58 10 4. V

23.14 = (2.00 rev/s)(2rad/rev) = 4.00rad/s

E l= =1

22B

2 83. mV

*23.15 (a) Suppose, first, that the central wire is long and straight. The enclosed current of unknownamplitude creates a circular magnetic field around it, with the magnitude of the field given by

Amperes Law.

B s = d I0 :

B

I t

R=

0

2max sin

at the location of the Rogowski coil, which we assume is centered on the wire. This field passesperpendicularly through each turn of the toroid, producing flux

B A =

0

2

I A t

R

max sin

The toroid has 2Rn turns. As the magnetic field varies, the emf induced in it is

E = = = Nd

dt

RnI A

R

d

dt

t I nA toB A 2

20

max maxsin cos

This is an alternating voltage with amplitude Emax max= 0nA I . Measuring the amplitudedetermines the size Imax of the central current. Our assumptions that the central wire is long andstraight and passes perpendicularly through the center of the Rogowski coil are all unnecessary.

(b) If the wire is not centered, the coil will respond to stronger magnetic fields on one side, but tocorrespondingly weaker fields on the opposite side. The emf induced in the coil is proportional tothe line integral of the magnetic field around the circular axis of the toroid. Amperes Law saysthat this line integral depends only on the amount of current the coil encloses. It does not dependon the shape or location of the current within the coil, or on any currents outside the coil.


5/14

Chapter 23

201

*23.16 (a) The force on the side of the coil entering the field (consisting ofNwires) is

F = N(ILB) = N(IwB)

The induced emf in the coil is

E = = ( ) =N

d

dtNd Bwx

dtNBwvB

so the current is

IR

NBwv

R= =

counterclockwise.

The force on the leading side of the coil is then:

F NNBwv

RwB=

=

N B w v

R

2 2 2

to the left

(b) Once the coil is entirely inside the field, B NBA= = constant,

so E= 0, I= 0, and F = 0

(c) As the coil starts to leave the field, the flux decreases at the rate

Bwv, so the magnitude of the current is the same as in part (a),but now the current is clockwise. Thus, the force exerted on thetrailing side of the coil is:

F =

N B w v

R

2 2 2

to the left again

23.17 (a) At terminal speed,

Mg F I wB

RwB

Bwv

RwB

B w v

RB

T T= = =

=

=E

2 2

orv

MgR

B wT = 2 2

(b) The emf is directly proportional to vT , but the current isinversely proportional to R. A large R means a small current ata given speed, so the loop must travel faster to get F mgB = .

(c) At a given speed, the current is directly proportional to themagnetic field. But the force is proportional to the product ofthe current and the field. For a small B, the speed must increaseto compensate for both the small B and also the current, so

v BT 2 .


6/14

Chapter 23

202

*23.18 Observe that the homopolar generator has no commutator andproduces a voltage constant in time: dc with no ripple. In time dt,the disk turns by angle d dt = . The outer brush slides overdistance r d.

The radial line to the outer brush sweeps over area

dA rr d r dt= =12 12 2

The emf generated is E = Nd

dtB A

E = ( ) = ( )1 0 12

2B

dA

dtB rcos

(We could think of this as following from the result of Example 23.3.)

The magnitude of the emf isE = ( )B r12

2= ( ) ( )

0 9 0 4 320021

2

2. .N s/C m mrev

min

rad/rev

60 s/min

E = 24 1. V

A free positive charge q shown, turning with the disk, feels a magnetic force qv B Z radiallyoutward. Thus the outer contact is positive .

23.19 (a) Emax . .= = ( )( )( )( ) =NAB 1000 0 100 0 200 120 7 54. kV

(b) E t NBA t NBA( ) = = sin sin

E is maximal when sin = 1

or

=

2

so the plane of coil is parallel to B

*23.20B = 0nI= 4 10

7 T m A( ) 200 m1( ) 15.0 A( ) = 3.77 103 T

For the small coil,

B N NBA t NB r t= = = ( )B A cos cos 2

Thus,E = =

d

dtNB r t

B 2 sin

E = ( ) ( ) ( ) ( ) ( ) 30 0 3 77 10 0 0800 4 00 4 003 2. . . . sin .T m s 1 t

E =

28 6 4 00. sin .mV( ) ( )t


7/14

Chapter 23

203

23.21

dB

dtt= 0 0600.

E = = =

d

dtr

dB

dtr E

B 12

12

At t = 3.00 s,

E rr

dBdt

=

= ( )( )

12

120 0200

20 0600 3 00 1. . .m T/s s N s

1 T C m2

E = 1 80 10 3 1. N/C perpendicular to and counterclockwiser

23.22 (a)

dB

dtt t= 6 00 8 002. .

E =d

dt

B

At t = 2.00 s,

ER dB dt

r=

( )= ( )

( )

2

2

2

2

8 00 0 0250

2 0 0500

/ . .

.

F = qE = 8 00 10 21. N clockwise for electron

(b) When 6 00 8 00 02. .t t = , t = 1 33. s

23.23

E =

= ( )

= = LI

t3 00 10

1 50 0 200

0 2001 95 103 2.

. .

..H

A A

sV

19 5. mV

*23.24 Treating the telephone cord as a solenoid, we have:

L

N A= =

02

l

4 T m/A 70.0 6.50 m

0.600 m

2 2 ( )( ) ( )[ ]

= 10 107 3

1 36. H

23.25

E Eback = = = ( ) = = ( )( )( )LdI

dtLd

dtI t L I t tmax maxsin cos . . cos 10 0 10 120 5 00

3

Eback = ( ) ( ) =6 00 120. cos t

18 8 377. cosV( ) ( )t

*23.26

E =

= ( )

=LI

t2 00

0 0 500 1.

.H

A

0.0100 s

V s

1 H A 100 V


8/14

Chapter 23

204

23.27 From E =

L

I

t , we have

L

I t=

( )=

=

E

/

.

..

24 0 10

10 02 40 10

33V

A/sH

From LN

I

B=

, we have

BLI

N

= =( )( )

=2 40 10 4 00

500

3. .H A

19 2. T m2

23.28

E = = ( ) ( )L

dI

dt

d

dtt t90 0 10 63 2. V

(a) At t = 1.00 s, E=

360 mV

(b) At t = 4.00 s, E= 180 mV

(c)E= ( ) ( ) =90 0 10 2 6 03. t

when t = 3 00. s

23.29

L

N

I

NBA

I

NA

I

NI

R

B= = =

0

2

0

2

2

N A

R

23.30

I

Re Rt L=

E( )/1 :

0 900 1 3 0 2 5. ( . )/ .

E E

R Re R= [ ] 0 s 0 H

exp

R(3.00 s)

2.50 H

= 0.100

R =

2.50 H

3.00 sln 10.0 =

1 92.


9/14

Chapter 23

205

23.31 At time t,I t

e

R

t

( ) =( )E 1 /

where = =L R/ .0 200 s

After a long time,I

e

R Rmax =

( ) =E E1

At I t I( ) = 0 500. max 0 500

1 0 200

.

/ .

( ) =( )E E

R

e

R

t s

so 0 500 10 200. / .= e t s

Isolating the constants on the right,ln ln ./ .e t( ) = ( )0 200 0 500s

and solving for t,

= t

0 200

0 693

.

.

s

or t = 0 139. s

(b) Similarly, to reach 90% of Imax, 0 900 1./= e t

and t = ( )ln .1 0 900

Thus, t = ( ) ( ) =0 200 0 100. ln .s 0 461. s

23.32 Taking = L/R , I= I0et/:

dI

dt= I0e

t/ 1

IR + L

dI

dt= 0 will be true if I0Re

t/ + L I0 et/( ) 1

= 0

Because = L R/ , we have agreement with 0 = 0

23.33 (a) = = =L R/ .2 00 10 3 s

2 00. ms

(b)I I e e

t= ( ) =

( ) = max / . / ..

16 00

1 0 250 2 00V

4.00 0 176. A

(c)I

Rmax

.= =

=

E 6 00 V

4.001 50. A

(d) 0 800 1 2 00 0 2002 00. . ln ./ .= = ( ) ( ) =e tt ms ms

3 22. ms


10/14

Chapter 23

206

*23.34I I e t= ( )max /1 : 0 980 1 3 00 10

3. . /=

e

0 02003 00 10 3. . /=

e

= =

3 00 100 0200 7 67 10

3

4.ln( . ) . s

= L R/ , so L R= = ( )( ) = 7 67 10 10 04. . 7 67. mH

23.35 (a) VR = IR = (8.00 )(2.00 A) = 16.0 V

and V VL R= = =E 36 0 16 0 20 0. . .V V V

Therefore,

VR

VL =

16.0 V

20.0 V = 0 800.

(b) VR = IR = (4.50 A)(8.00 ) = 36.0 V

V VL R= =E 0

*23.36 Name the currents as shown. By Kirchhoffs laws:

I1 = I2 + I3 (1)

+ =10 0 4 00 4 00 01 2. . .V I I (2)

+ ( ) =10 0 4 00 8 00 1 00 01 3

3. . . .V I IdI

dt(3)

From (1) and (2), +10.0 4.00 I1 4.00 I1 + 4.00 I3 = 0

and I1 = 0.500 I3 + 1.25 A

Then (3) becomes10 0 4 00 0 500 1 25 8 00 1 00 03 3

3. . . . . .V A +( ) ( ) =I IdI

dt

1 00 10 0 5 003 3. / . .H V( )( ) + ( ) =dI dt I

We solve the differential equation using Equations 23.13 and 23.14:

I t e

t

310 0 1 005 00 1( ) =

[ ] = ( ). . / .V10.0H

0 500 1 10. A /s( ) [ ]e t

I1 = 1.25 + 0.500 I3 = 1 50 0 250 10. . /A A s ( ) e t


11/14

Chapter 23

207

*23.37

= = =

L

R

0 140

4 9028 6

.

.. ms

I

Rmax

..= = =

E 6 001 22

V

4.90A

(a) I= Imax 1 et/

( ) so 0.220 = 1.22 1 et/

( )e

t/ = 0.820: t = ln(0.820) = 5 66. ms

(b)

I= Imax 1 e 10.0

0.0286

= (1.22 A) 1 e

350( ) = 1 22. A

(c) I= Imaxet/ and 0 160 1 22. .

/= e t r

so t = ( ) =ln .0 131 58 1. ms

23.38 (a)I

R= =

=

E 12 0. V

12.01 00. A

(b) Initial current is 1.00 A: = ( ) ( ) =V12 1 00 12 00. .A 12 0. V

= ( ) ( ) =V1200 1 00 1200. A 1 20. kV

=VL 1 21. kV

(c) I I eRt L= max / :

dIdt

I RLe Rt L= max /

and = = LdI

dtV I ReL

Rt Lmax

/

Solving 12 0 12121212 2 00. / .V V= ( ) e t

so 9 90 103 606. = e t

Thus,t = 7 62. ms

*23.39

L

N A= =

[ ]=

0

2

0

2 2 268 0 0 600 10

0 08008 21

l

( . ) ( . )

.. H

U LI= = ( )( ) =12

2 1

26 28 21 10 0 770. .H A

2 44. J


12/14

Chapter 23

208

23.40 (a) The magnetic energy density is given by

uB

= = ( )

=

2

062

4 50

2 1 26 10

( .

.

T)

T m/A

2

8 06 106. J/m3

(b) The magnetic energy stored in the field equals u times the volume of the solenoid (the volume inwhich B is non-zero).

U uV= = ( )8 06 106. J/m3 (0.260 m)(0.0310 m)

2[ ] = 6 32. kJ

23.41

u

E= =e0

2

2 44 2. nJ/m3

u

B= =

2

02 995 J/m3

23.42

(a) U LI= = ( )( )

1

2

2 1

2

2

4 00 0 500. .H AU

= 0 500. J

(b) When the current is 1.00 A,

Kirchhoffs loop rule reads+ ( )( ) =22 0 1 00 5 00 0. . .V A VL

Then VL= 17.0 V

The power being stored in the inductor is I VL = ( )( ) =1 00 17 0. .A V 17 0. W

(c) P = = ( )( )I V 0 500 22 0. .A V P = 11 0. W

*23.43 The induced emf in the leading edge of the loop is

E l= = ( )( )

= ( )B v B B0 2 4001000

22 2. . /m km/hm

km

h

3600 sm s2

The induced current isI

R

B= =

( )

E 22 2

25

. m

s

2

The magnetic force on the lower section of one loop is F BB I= l :

F upB B B B=

( ) =

( )

22 20 1 90 0 0889 2

.. sin .

m A

25 V sm m C /J s

23 2 2J

We require =Fy 0 : 100 0F mgB =

B2 48 89 5 10 9 80. / .m C N s N2 2 2 ( ) = ( )

B = =235 N s/C m 2 102 T


13/14

Chapter 23

217

ANSWERS TO EVEN NUMBERED PROBLEMS

2.9.82 mV

4. (a)

0 22

2

n r

R

I

t

(b)

02

22

14

n r

r R

I

t

(c) upward

6. (a)

0 2 1IL w h/ ln /( ) +( ) (b) 4.80 V; current is counterclockwise

8. (14.2 mV) cos (120 t)

10.1.00 m/s

12. (a) 0.500 A (b) 2.00 W (c) 2.00 W

14. 2.83 mV

16. (a) F N B w v R=2 2 2 / to the left (b) 0

(c) F N B w v R=2 2 2 / to the left

18. 24.1 V with the outer contact positive

20. (28.6 mV) sin (4.00t)

22. (a) 8 00 1021. N, clockwise (b) 1.33 s

24. 1.36 H

26. 100 V

28. (a) 360 mV (b) 180 mV (c) 3.00 s

30. 1.92

32. See the solution

34. 7.67 mH


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Resoluções de Exercícios - Cap. 23 - Princípios de Física Vol. 3

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