Sm chapter29

29Magnetic Fields

CHAPTER OUTLINE

29.1 Magnetic Fields and Forces29.2 Motion of a Charged Particle in a

Uniform Magnetic Field29.3 Applications Involving Charged

Particles Moving in a Magnetic Field

29.4 Magnetic Force Acting on a Current-Carrying Conductor

29.5 Torque on a Current Loop in a Uniform Magnetic Field

29.6 The Hall Effect

ANSWERS TO QUESTIONS

*Q29.1 (a) Yes, as described by F E= q .

(b) No, as described by F v B= ×q

(c) Yes. (d) Yes. (e) No. The wire is uncharged. (f ) Yes. (g) Yes. (h) Yes.

*Q29.2 (i) (b). (ii) (a). Electron A has a smaller radius of curvature, as described by qvB = mv2/r.

*Q29.3 (i) (c) (ii) (c) (iii) (c) (iv) (b) (v) (d) (vi) (b) (vii) (b) (viii) (b)

*Q29.4 We consider the quantity | qvB sinq |, in units of e (m/s)(T). For (a) it is 1 × 106 10−3 1 = 103. For (b) it is 1 × 106 10−3 0 = 0. For (c) 2 × 106 10−3 1 = 2 000. For (d) 1 × 106 2 × 10−3 1 = 2 000. For (e) 1 × 106 10−3 1 = 103. For (f ) 1 × 106 10−3 0.707 = 707. The ranking is then c = d > a = e > f > b.

*Q29.5 ˆ ( ˆ ) î k j× − = . Answer (c).

*Q29.6 Answer (c). It is not necessarily zero. If the magnetic fi eld is parallel or antiparallel to the velocity of the charged particle, then the particle will experience no magnetic force.

Q29.7 If they are projected in the same direction into the same magnetic fi eld, the charges are of opposite sign.

Q29.8 Send the particle through the uniform fi eld and look at its path. If the path of the particle is parabolic, then the fi eld must be electric, as the electric fi eld exerts a constant force on a charged particle, independent of its velocity. If you shoot a proton through an electric fi eld, it will feel a constant force in the same direction as the electric fi eld—it’s similar to throwing a ball through a gravitational fi eld.

If the path of the particle is helical or circular, then the fi eld is magnetic. If the path of the particle is straight, then observe the speed of the particle. If the particle

accelerates, then the fi eld is electric, as a constant force on a proton with or against its motion will make its speed change. If the speed remains constant, then the fi eld is magnetic.

*Q29.9 Answer (d). The electrons will feel a constant electric force and a magnetic force that will change in direction and in magnitude as their speed changes.

Q29.10 Yes. If the magnetic fi eld is perpendicular to the plane of the loop, then it exerts no torque on the loop.

151

ISMV2_5104_29.indd 151ISMV2_5104_29.indd 151 6/20/07 6:15:40 PM6/20/07 6:15:40 PM

152 Chapter 29

Q29.11 If you can hook a spring balance to the particle and measure the force on it in a known electric

fi eld, then qF

E= will tell you its charge. You cannot hook a spring balance to an electron.

Measuring the acceleration of small particles by observing their defl ection in known electric and magnetic fi elds can tell you the charge-to-mass ratio, but not separately the charge or mass. Both an acceleration produced by an electric fi eld and an acceleration caused by a magnetic

fi eld depend on the properties of the particle only by being proportional to the ratio q

m.

Q29.12 If the current loop feels a torque, it must be caused by a magnetic fi eld. If the current loop feels no torque, try a different orientation—the torque is zero if the fi eld is along the axis of the loop.

Q29.13 The Earth’s magnetic fi eld exerts force on a charged incoming cosmic ray, tending to make it spiral around a magnetic fi eld line. If the particle energy is low enough, the spiral will be tight enough that the particle will fi rst hit some matter as it follows a fi eld line down into the atmosphere or to the surface at a high geographic latitude.

Q29.14 No. Changing the velocity of a particle requires an accelerating force. The magnetic force is proportional to the speed of the particle. If the particle is not moving, there can be no magnetic force on it.

FIG. Q29.13

SOLUTIONS TO PROBLEMS

Section 29.1 Magnetic Fields and Forces

P29.1 (a) up

(b) out of the page, since the charge is negative.

(c) no defl ection

(d) into the page

FIG. P29.1


Magnetic Fields 153

P29.2 At the equator, the Earth’s magnetic fi eld is horizontally north. Because an electron has negative charge,

F v B= ×q is opposite in direction

to v B× . Figures are drawn looking down.

(a) Down × North = East, so the force is directed West .

(b) North × North = ° =sin :0 0 Zero deflection .

(c) West × North = Down, so the force is directed Up .

(d) Southeast × North = Up, so the force is Down .

P29.3 F ma= = ×( ) ×( ) = ×− −1 67 10 2 00 10 3 34 1027 13. . . kg m s2 114

14

19

90

3 34 10

1 60 10

N

N

= °

= = ××

−

−

q B

BF

q

v

v

sin

.

. C m s T( ) ×( ) = × −

1 00 102 09 10

7

2

..

The right-hand rule shows that B must be in the −y direction to yield a force in the +x direction when v is in the z direction.

P29.4 (a) F q BB = = ×( ) ×( ) ×−v sin . . .θ 1 60 10 3 00 10 3 00 119 6 C m s 00 37 01−( ) ° T sin .

FB = × −8 67 10 14. N

(b) aF

m= = ×

×= ×

−

−

8 67 10

1 67 105 19 10

14

2713.

..

N

kgm s2

P29.5 F q BB = v sinθ so 8 20 10 1 60 10 4 00 10 1 713 19 6. . . .× = ×( ) ×( )− − N C m s 00 T( ) sinθ

sin .θ = 0 754 and θ = ( ) = ° °−sin . .1 0 754 48 9 or 131

P29.6 First fi nd the speed of the electron.

∆ ∆ ∆K m e V U= = =1

22v :

v = =

×( ) ( )×

−

−

2 2 1 60 10 2 400

9 11 10

19

31

e V

m

∆ .

.

C J C

kkgm s( ) = ×2 90 107.

(a) F q BB, . . . max C m s = = ×( ) ×( )−v 1 60 10 2 90 10 1 7019 7 TT N( ) = × −7 90 10 12.

(b) FB, min = 0 occurs whenv is either parallel to or anti-parallel to

B.

P29.7 F v BB q= ×

v B

i j k

i j× = + − ++ + −

= −( ) + +( ) +

ˆ ˆ ˆ

ˆ ˆ2 4 1

1 2 3

12 2 1 6 4 ++( ) = + +

× = + + = ⋅

4 10 7 8

10 7 8 14 62 2 2

ˆ ˆ ˆ ˆ

.

k i j k

v B

T mm s

C T m s F v BB q= × = ×( ) ⋅( ) =−1 60 10 14 6 2 319. . . 44 10 18× − N

FIG. P29.2

FIG. P29.3


154 Chapter 29

P29.8 qE k= − ×( )( ) = − ×(− −1 60 10 20 0 3 20 1019 18. . ˆ .C N C N)) k

F E v B a

k

∑ = + × =

− ×( ) − ×− −

q q m

3 20 10 1 60 1018. ˆ .N 119 31 129 11 10 2 00 10C 1.20 10 m s4×( ) × = ×( ) ×−ˆ . .i B

mm s

N C m s

2( )− ×( ) − × ⋅( )− −

ˆ

. ˆ . ˆ

k

k i3 20 10 1 92 1018 15 ×× = ×( )× ⋅( ) × = −

−

−

B k

i B

1 82 10

1 92 10

18

15

. ˆ

. ˆ

N

C m s 55 02 10 18. ˆ×( )− N k

The magnetic fi eld may have any -componentx . Bz = 0 and By = −2 62. .mT

Section 29.2 Motion of a Charged Particle in a Uniform Magnetic Field

P29.9 (a) B = × −50 0 10 6. T ; v = ×6 20 106. m s

Direction is given by the right-hand-rule: southward

F q B

F

B

B

=

= ×( ) ×( ) ×−

v sin

. . .

θ

1 60 10 6 20 10 50 019 6C m s 110 90 0

4 96 10

6

17

−

−

( ) °

= ×

T

N

sin .

.

(b) Fm

r= v2

so

rm

F= =

×( ) ×( )×

−

−

v2 27 6 2

1

1 67 10 6 20 10

4 96 10

. .

.

kg m s77

1 29N

km= .

*P29.10 (a) The horizontal velocity component of the electrons is given by (1/2)mv

x2 = | q |V.

vx

q V

m= = ×

×

−

−

2 2 1 6 10 19( . C) 2 500 J/C

9.11 10 kg31== ×2 96 107. m/s

Its time of fl ight is t = x/vx = 0.35 m/2.96 × 107 m/s = 1.18 × 10−8 s.

Its vertical defl ection is y = (1/2) gt 2 = (1/2) 9.8 m/s2 (1.18 × 10−8 s)2 = 6.84 × 10−16 m down , which is unobservably

small.

(b) The magnetic force is in the direction –north × down = –west = east. The beam is defl ected into a circular path with radius

rm

q B= = × ×

×

−

−

v 9 11 10

1 6 10

31

19

.

.

kg 2.96 10 m/s

C 2

7

00 10 N s/C mm.–6× ⋅ ⋅

= 8 44.

Their path to the screen subtends at the center of curvature an angle given by

sin q = 0.35 m/8.44 m = 2.38°. Their defl ection is 8.44 m(1 – cos 2.38°) = 7.26 mm east .

It does not move as a projectile, but its northward velocity component stays nearly constant, changing from 2.96 × 107 m/s cos 0° to 2.96 × 107 m/s cos 2.38°. That is, it is constant within 0.09%. It is a good approximation to think of it as moving on a parabola as it really moves on a circle.

FIG. P29.9

r

B

q

W

N

E

S

FIG. P29.10


Magnetic Fields 155

P29.11 q V m∆( ) = 1

22v or v = ( )2q V

m

∆

Also, q Bm

rv

v=2

so rm

qB

m

qB

q V

m

m V

qB= = ( )

= ( )v 2 22

∆ ∆

Therefore, rm V

eBpp2

2

2=

( )∆

rm V

q B

m V

eB

m V

eBdd

d

p p22 2 2

2 2 22

2= ( )

=( )( )

=( )⎛

⎝∆ ∆ ∆

⎜⎜⎞⎠⎟

= 2 2rp

and rm V

q B

m V

e B

m V

eBp p

αα

α

22 2

2 2 4

22

2= ( )

=( )( )

( ) =( )∆ ∆ ∆

2222

⎛⎝⎜

⎞⎠⎟

= rp

The conclusion is: r r rd pα = = 2

P29.12 For each electron, q Bm

rv

vsin .90 0

2

° = and v = eBr

m.

The electrons have no internal structure to absorb energy, so the collision must be perfectly elastic:

K m m m

K me B R

m

i f f= + = +

=⎛

⎝

1

20

1

2

1

2

1

2

12

12

22

2 212

2

v v v

⎜⎜⎞

⎠⎟ +

⎛

⎝⎜

⎞

⎠⎟ = +( )

=

1

2 2

2 222

2

2 2

12

22m

e B R

m

e B

mR R

Ke 11 60 10 0 044 0

2 9 11 10

19 2

31

. .

.

×( ) ⋅ ⋅( )×(

−

−

C N s C m

kg)) ( ) + ( )⎡⎣

⎤⎦ =0 010 0 0 024 0 115

2 2. . m m keV

P29.13 (a) We begin with q Bm

Rv

v=2

or qRB m= v

But L m R qR B= =v 2

Therefore, RL

qB= = × ⋅

×( ) ×(−

− −

4 00 10

1 00 10

25

19 3

.

.

J s

1.60 10 C T)) = =0 050 5 00. .0 m cm

(b) Thus, v = = × ⋅×( )( )

−

−

L

mR

4 00 10

10 0 050 0

25

31

.

.

J s

9.11 kg m== ×8 78 106. m s

P29.14 1

22m q Vv = ( )∆ so v = ( )2q V

m

∆

rm

qB= v

so rm q V m

qB=

( )2 ∆

rm

q

V

B2

2

2= ⋅ ( )∆ and r

m

q

V

B′ ′

′( ) = ⋅ ( )2

2

2 ∆

mqB r

V=

( )2 2

2 ∆ and m

q B r

V′

′ ′( ) = ( ) ( )( )

2 2

2 ∆ so

m

m

q

q

r

r

e

e

R

R

′ ′ ′= ⋅ ( ) = ⎛

⎝⎜

⎞⎠⎟⎛⎝⎜

⎞⎠⎟ =

2

2

22 28


156 Chapter 29

P29.15 E m e V= =1

22v ∆

and e Bm

Rv

vsin 90

2

° =

Bm

eR

m

eR

e V

m R

m V

e

B

= = =

=×

×

v 2 1 2

1

5 80 10

2 1 6710

∆ ∆

.

.

m

110 10 0 10

1 60 107 88 10

27 6

19

−

−

( ) ×( )×

= × kg V

C

.

.. −−12 T

P29.16 (a) The boundary between a region of strong magnetic fi eld and a region of zero fi eld cannot be perfectly sharp, but we ignore the thickness of the transition zone. In the fi eld the electron moves on an arc of a circle:

F ma∑ = :

q Bm

r

r

q B

m

vv

v

sin

.

90

1 60 10 10

2

19 3

° =

= = =×( ) ⋅− −

ωC N s CC m

kgrad s

⋅( )×( ) = ×−9 11 10

1 76 1031

8

..

The time for one half revolution is,

from ∆ ∆θ ω= t

∆ ∆t = =

×= × −θ

ωπ rad

rad ss

1 76 101 79 108

8

..

(b) The maximum depth of penetration is the radius of the path.

Then v = = ×( )( ) = ×−ω r 1 76 10 0 02 3 51 108 1 6. . .s m m s

and

K m= = ×( ) ×( ) = ×−1

2

1

29 11 10 3 51 10 5 622 31 6 2

v . . .kg m s 1105 62 10

1 60 10

35 1

1818

19−

−

−= × ⋅×

=

JJ e

C

eV

.

.

.

Section 29.3 Applications Involving Charged Particles Moving in a Magnetic Field

P29.17 F FB e=

so q B qEv =

where v = 2K

m and K is kinetic energy of the electron.

E BK

mB= = =

( ) ×( )×

−

−v2 2 750 1 60 10

9 11 100 015

19

31

.

.. 00 244( ) = kV m

FIG. P29.16(a)


Magnetic Fields 157

P29.18 K m q V= = ( )1

22v ∆ so v = ( )2q V

m

∆

F v BB q

m

r= × = v2

rm

qB

m

q

q V m

B B

m V

q= = ( )

= ( )v 2 1 2∆ ∆

(a) r238

27

19

2 238 1 66 10 2 000

1 60 10

1

1 20=

× ×( )×

⎛⎝

−

−

.

. .⎜⎜

⎞⎠⎟ = × =−8 28 10 8 282. .m cm

(b) r235 8 23= . cm

r

r

m

m238

235

238

235

238 05

235 041 006 4= = =.

..

The ratios of the orbit radius for different ions are independent of ∆V and B.

P29.19 In the velocity selector: v = = = ×E

B

2 500

0 035 07 14 104V m

Ts

.. m

In the defl ection chamber: rm

qB= =

×( ) ×( )×

−

−

v 2 18 10 7 14 10

1 60 10

26 4

1

. .

.

kg m s99 0 035 0

0 278C T

m( )( ) =.

.

P29.20 Note that the “cyclotron frequency” is an angular speed. The motion of the proton is described by

F ma∑ = :

q B

m

r

q B mr

m

vv

v

sin 902

° =

= = ω

(a) ω = =×( ) ⋅ ⋅( )

×(−

−

q B

m

1 60 10 0 8

1 67 10

19

27

. .

.

C N s C m

kg))⋅⋅

⎛⎝⎜

⎞⎠⎟ = ×kg m

N srad s

27 66 107.

(b) v = = ×( )( )⎛⎝⎜

⎞⎠⎟ =ω r 7 66 10 0 350

1

127. . .rad s m

rad668 107× m s

(c) K m= = ×( ) ×( )−1

2

1

21 67 10 2 68 10

12 27 7 2v . .kg m s

eV

1.66 10 JeV

×⎛⎝⎜

⎞⎠⎟ = ×−19

63 76 10.

(d) The proton gains 600 eV twice during each revolution, so the number of revolutions is

3 76 103 13 10

63.

.×

( ) = ×eV

2 600 eVrevolutions

(e) θ ω= t t = = ××

⎛⎝⎜

⎞θω

π3 13 10 23. rev

7.66 10 rad s

rad

1 rev7 ⎠⎠⎟ = × −2 57 10 4. s

P29.21 (a) F q Bm

RB = =vv2

ω = = = =×( )( )

×

−vR

qBR

mR

qB

m

1 60 10 0 450

1 67 1

19. .

.

C T

004 31 10

277

− = ×kg

rad s.

(b) v = =×( )( )( )

×

−qBR

m

1 60 10 0 450 1 20

1 67 1

19. . .

.

C T m

005 17 10

277

− = ×kg

m s.


158 Chapter 29

*P29.22 (a) The path radius is r = mv/qB, which we can put in terms of energy E by (1/2)mv2 = E.

v = (2E/m)1/2 so r = (m/qB) (2E/m)1/2 = (2m)1/2(qB)−1E1/2

Then dr/dt = = (2m)1/2(qB)−1(1/2) E−1/2 dE/dt = 2

2

2 12m

qB

m

qBr

q B V

m r

V

B

∆ ∆π π

=

(b) The dashed red line should spiral around many times, with its turns relatively far apart on the inside and closer together on the outside.

(c) dr

dt r

V

B= =1 600

0 35

∆π π

V C m N m

m. 00.8 N s V Cm/s= 682

(d) ∆ ∆ ∆r

dr

dtT

r

V

B

m

qB

Vm

rqB= = = = × ×1 2 2 2

2ππ 600 V 1.67 110 kg C m N m

m 1.6 10 C 0.8 N

27 2 2

19 2 2

−

−×0 35. s V Cm

2= 55 9. µ

P29.23 θ =⎛⎝⎜

⎞⎠⎟ = °−tan

.

..1 25 0

10 068 2 and R =

°=1 00

68 21 08

.

sin ..

cm cm

Ignoring relativistic correction, the kinetic energy of the electrons is

1

22m q Vv = ∆ so v = = ×2

1 33 108q V

m

∆. m s

From Newton’s second law, m

Rq B

vv

2

= , we fi nd the magnetic fi eld

Bm

q R= =

×( ) ×( )×

−

−

v 9 11 10 1 33 10

1 60 10

31 8

19

. .

.

kg m s

CC mmT( ) ×( ) =−1 08 10

70 12..

*P29.24 (a) Yes: The constituent of the beam is present in all kinds of atoms.

(b) Yes: Everything in the beam has a single charge-to-mass ratio.

(c) In a charged macroscopic object most of the atoms are uncharged. A molecule never has all of its atoms ionized. Any atom other than hydrogen contains neutrons and so has more mass per charge if it is ionized than hydrogen does. The greatest charge-to-mass ratio Thomson could expect was then for ionized hydrogen, 1.6 × 10−19 C/1.67 × 10−27 kg, smaller than the value e/m he measured, 1.6 × 10−19 C/9.11 × 10−31 kg, by 1 836 times. The particles in his beam could not be whole atoms, but rather must be much smaller in mass.

(d) With kinetic energy 100 eV, an electron has speed given by (1/2)mv2 = 100 eV

v = ⋅ ××

= ×−

−

200 1 6 10

9 11 105 93 10

19

316.

..

C 1 J/C

kgmm/s. The time to travel 40 cm is

0.4 m/(5.93 × 106 m/s) = 6.75 × 10−8 s. If it is fi red horizontally it will fall vertically by (1/2)gt 2 = (1/2)(9.8 m/s2)( 6.75 × 10−8 s)2 = 2.23 × 10−14 m, an immeasurably small amount. An electron with higher energy falls by a smaller amount.

FIG. P29.23


Magnetic Fields 159

Section 29.4 Magnetic Force Acting on a Current-Carrying Conductor

P29.25 F ILBB = sinθ with F F mgB g= =

mg ILB= sinθ so m

Lg IB= sinθ

I = 2 00. A and m

L= ( )⎛

⎝⎜⎞⎠⎟

= ×0 500100

1 0005 00 10. .g cm

cm m

g kg−−2 kg m

Thus 5 00 10 9 80 2 00 90 02. . . sin .×( )( ) = ( ) °− B

B = 0 245. Tesla with the direction given by right-hand rule: eastward .

P29.26

F B i kB I= × = ( )( ) × ( ) = −2 40 0 750 1 60. . ˆ . Â m T 22 88. j( ) N

P29.27 (a) F ILBB = = ( )( )( )sin . . . sin .θ 5 00 2 80 0 390 60 0 A m T °° = 4 73. N

(b) FB = ( )( )( ) ° =5 00 2 80 0 390 90 0 5 46. . . sin . . A m T N

(c) FB = ( )( )( ) ° =5 00 2 80 0 390 120 4 73. . . sin . A m T N

*P29.28 The magnetic force should counterbalance the gravitational force on each section of wire:

IB sin 90° = mg Im g

B= = ×

×=−

−2 4 10

9 88403.

.kg

m

m/s C m

28 10 N s

2

6A

The current should be east so that the magnetic force will be east × north = up.

P29.29 The rod feels force F d B k j iB I Id B IdB= ×( ) = ( ) × −( ) = ( )ˆ ˆ ˆ .

The work-energy theorem is K K E K Ki ftrasn rot trans rot+( ) + = +( )∆

0 01

2

1

22 2+ + = +F m Is cosθ ωv

IdBL m mRR

cos º01

2

1

2

1

22 2

2

= + ⎛⎝

⎞⎠

⎛⎝

⎞⎠v

v

and IdBL m= 3

42v

v = = ( )( )( )4

3

4 48 0 0 120 0 240 0 450IdBL

m

. . . .A m T mm

kgm s

( )( ) =

3 0 7201 07

..

FIG. P29.25

dI

L

B

y

xz

FIG. P29.29


160 Chapter 29

P29.30 The rod feels force F d B k j iB I Id B IdB= ×( ) = ( ) × −( ) = ( )ˆ ˆ ˆ .

The work-energy theorem is K K E K Ki ftrans rot trans rot+( ) + = +( )∆

0 01

2

1

22 2+ + = +Fs m Icosθ ωv

IdBL m mRR

cos º01

2

1

2

1

22 2

2

= + ⎛⎝

⎞⎠

⎛⎝

⎞⎠v

v

and v = 4

3

IdBL

m

P29.31 The magnetic force on each bit of ring is Id IdsB

s B× = radially inward and upward,

at angle q above the radial line. The radi-ally inward components tend to squeeze the ring but all cancel out as forces. The upward components IdsBsinθ all add to

I rB2π θsin . up

*P29.32 (a) For each segment, I = 5 00. A and B j= ⋅0 020 0. ˆ.N A m

Segment

. m

0.400

F B

j

B I

ab

bc

= ×( )−0 400 0ˆ

mm mN

. m . m

ˆ . ˆ

ˆ ˆ

k i

i j

40 0

0 400 0 400

( ) −( )− +cd 440 0

0 400 0 400 40 0

. ˆ

ˆ ˆ .

mN

. m . m

( ) −( )−

k

i kda mmN( ) +( )ˆ ˆk i

(b) The forces on the four segments must add to zero, so the force on the fourth segment must be the negative of the resultant of the forces on the other three.

FIG. P29.31

FIG. P29.32


Magnetic Fields 161

P29.33 Take the x-axis east, the y-axis up, and the z-axis south. The fi eld is

B k= ( ) ° −( )+ ( ) °52 0 60 0 52 0 60 0. cos . ˆ . sin . T Tµ µ −−( )j

The current then has equivalent length: ′ = −( ) + ( )L k j1 40 0 850. ˆ . ˆm m

F L B j kB I= ′ × = ( ) −( ) × −0 035 0 0 850 1 40 4. . ˆ . Â m 55 0 26 0 10

3 50 10 22 1 6

6

8

. ˆ . ˆ

. . ˆ

j k

F i

−( )= × − −

−

−

T

N

B 33 0 2 98 10 2 986. ˆ . ˆ .i i( ) = × −( ) =− N N westµ

Section 29.5 Torque on a Current Loop in a Uniform Magnetic Field

P29.34 (a) 2 2 00πr = . m so r = 0 318. m

µ π= = ×( ) ( )⎡⎣ ⎤⎦ = ⋅−IA 17 0 10 0 318 5 413 2. . .A m mA m2 2

(b) ττ µµ= × B so τ = × ⋅( )( ) = ⋅−5 41 10 0 800 4 333. . .A m T mN m2

P29.35 τ φ

τ

=

= ( ) ×( )NBAI sin

. . . .100 0 800 0 400 0 300 1 2T m2 00 60

9 98

A

N m

( ) °

= ⋅

sin

.τ

Note that f is the angle between the magnetic moment and the

B fi eld. The loop will rotate

so as to align the magnetic moment with the B fi eld.

Looking down along the y-axis, the loop will

rotate in a clockwise direction.

P29.36 Choose U = 0 when the dipole moment is at θ = °90 0. to the fi eld. The fi eld exerts torque of mag-nitude mBsinq on the dipole, tending to turn the dipole moment in the direction of decreasing q. According to the mechanical equations ∆U = – ∫dW and dW = t dq , the potential energy of the dipole-fi eld system is given by

U B d B B− = = −( ) = −∫090 0

90 0µ θ θ µ θ µ θ

θθ

sin cos cos. º

. º++ 0 or U = − ⋅

µµ B .

P29.37 (a) The fi eld exerts torque on the needle tending to align it with the fi eld, so the minimum energy orientation of the needle is:

pointing north at 48.0 below the horizonta° ll

where its energy is U Bmin cos . .= − ° = − × ⋅( ) ×( ) =− −µ 0 9 70 10 55 0 103 6A m T2 −− × −5 34 10 7. J.

It has maximum energy when pointing in the opposite direction,

south at 48.0 above the horizontal°

where its energy is U Bmax cos . .= − ° = + × ⋅( ) ×(− −µ 180 9 70 10 55 0 103 6A m T2 )) = + × −5 34 10 7. J.

(b) U W Umin max+ = : W U U= − = + × − − ×( ) =− −max min . . .5 34 10 5 34 10 1 07 7 J J 77 Jµ

FIG. P29.33

FIG. P29.35


162 Chapter 29

P29.38 (a) τ = ×µµ B , so τ µ θ θ= × = =

µµ B B NIABsin sin

τ πmax sin . . .= ° = ( ) ( )⎡⎣

⎤⎦NIAB 90 0 1 5 00 0 050 0

2 A m 33 00 10 1183. ×( ) = ⋅− T N mµ

(b) U = − ⋅ µ B, so − ≤ ≤ +µ µB U B

Since µ πB NIA B= ( ) = ( ) ( )⎡⎣ ⎤⎦ × −1 5 00 0 050 0 3 00 102

. . .A m 33 118T J,( ) = µ

the range of the potential energy is: − ≤ ≤ +118 118J Jµ µU .

*P29.39 For a single-turn circle, r = 1.5 m/2p. The magnetic moment is

m = N I A = 1 (30 × 10–3 A) p (1.5 m/2p )2 = 5.37 × 10–3 A ⋅ m2

For a single-turn square, = 1.5 m/4. The magnetic moment is

m = N I A = 1 (30 × 10–3 A) (1.5 m/4)2 = 4.22 × 10–3 A ⋅ m2

For a single-turn triangle, = 1.5 m/3 = 0.5 m. The magnetic moment is

m = N I A = 1 (30 × 10–3 A) (1/2) 0.5 m (0.52 – 0.252)1/2 m = 3.25 × 10–3 A ⋅ m2

For a fl at compact circular coil with N turns, r = 1.5 m/2pN. The magnetic moment is

m = N I A = N (30 × 10–3 A) p (1.5 m/2pN)2 = 5.37 × 10–3 A ⋅ m2/N

The magnetic moment cannot go to infi nity. Its maximum value is 5.37 mA⋅m2 for a single-turn circle. Smaller by 21% and by 40% are the magnetic moments for the single-turn square and triangle. Circular coils with several turns have magnetic moments inversely proportional to the number of turns, approaching zero as the number of turns goes to infi nity.

P29.40 (a) ττ µµ= × =B NIABsinθ

τ max . . . sin= ( ) ⋅( ) ⋅( )−80 10 0 025 0 04 0 82 A m m N A m 990 6 40 10 4° = × ⋅−. N m

(b) Pmax max .= = × ⋅ ( )−τ ω π6 40 10

24 N m 3 600 rev minrad

1 rrev

min

60 sW⎛

⎝⎞⎠

⎛⎝

⎞⎠ =1

0 241.

(c) In one half revolution the work is

W U U B B B

NIAB

= − = − °− − °( ) =

=max cos cosmin µ µ µ180 0 2

2 == × ⋅( ) = ×− −2 6 40 10 1 28 104 3. .N m J

In one full revolution,W = ×( ) = ×− −2 1 28 10 2 56 103 3. . .J J

(d) Pavg

J

/60 sW= = ×

( ) =−W

t∆2 56 10

10 154

3..

The peak power in (b) is greater by the factor π2

.


Magnetic Fields 163

Section 29.6 The Hall Effect

P29.41 Bnqt V

I= ( ) =

×( ) ×( )− −∆ Hm C8 46 10 1 60 10 5 0028 3 19. . . ××( ) ×( )− −10 5 10 10

8 00

3 12m V

A

.

.

B = × =−4 31 10 43 15. . T Tµ

P29.42 (a) ∆VIB

nqtH = so nqt

I

B

V= =

×= ×−∆ H

T

0.700 VT V

0 080 0

101 14 106

5..

Then, the unknown fi eld is Bnqt

IV= ⎛

⎝⎞⎠ ( )∆ H

B = ×( ) ×( ) = =−1 14 10 0 330 10 0 037 7 37 75 6. . . .T V V T mTT

(b) nqt

I= ×1 14 105. T V so n

I

qt= ×( )1 14 105. T V

n = ×( )×( ) ×− −1 14 10

0 120

10 2 00 105

19 3..

.T V

A

1.60 C mmm( ) = × −4 29 1025 3.

Additional Problems

*P29.43 (a) Defi ne vector h to have the downward direction of the

current, and vector L to be along the pipe into the page as

shown. The electric current experiences a magnetic forcce .

I h B×( ) in the direction of

L

(b) The sodium, consisting of ions and electrons, fl ows along the pipe transporting no net charge. But inside the section of length L, electrons drift upward to constitute downward electric current J J Lw× ( ) =area .

The current then feels a magnetic force I JLwhB h B× = °sin .90

This force along the pipe axis will make the fl uid move, exerting pressure

F JLwhB

hwJLB

area= =

(c) Charge moves within the fl uid inside the length L, but charge does not accumulate: the fl uid is not charged after it leaves the pump. It is not current-carrying and it is not magnetized.

FIG. P29.43

J


164 Chapter 29

P29.44 (a) At the moment shown in Figure 29.10, the particle must be moving upward in order for the magnetic force on it to be

into the page, toward the center of this turn of its

spiral path. Throughout its motion it circulates clockwise.

(b) After the particle has passed the middle of the bottle and moves into the region of increasing magnetic fi eld, the magnetic force on it has a component to the left (as well as a radially inward component) as shown. This force in the –x direction slows and reverses the particle’s motion along the axis.

(c) The magnetic force is perpendicular to the velocity and does no work on the particle. The particle keeps constant kinetic energy. As its axial velocity component decreases, its tangential velocity component increases.

(d) The orbiting particle constitutes a loop of current in the yz

plane and therefore a magnetic dipole moment IAq

TA=

in the –x direction. It is like a little bar magnet with its N pole on the left.

(e) Problem 31 showed that a nonuniform magnetic fi eld exerts a net force on a magnetic dipole. When the dipole is aligned opposite to the external fi eld, the force pushes it out of the region of stronger fi eld. Here it is to the left, a force of repulsion of one magnetic south pole on another south pole.

*P29.45 The particle moves in an arc of a circle with radius

rm

qB= = × ×

×

−

−

v 1 67 10

1 6 10

27.

.

kg 3 10 m/s C m7

11912 5

C 25 10 N s km

6×=

–. . It will not hit the Earth, but perform

a hairpin turn and go back parallel to its original direction.

P29.46 Fy∑ = 0: + − =n mg 0

Fx∑ = 0: − + ° =µkn IBd sin .90 0 0

B = µk mg

Id=

( )( )( )

0 100 0 200 9 80

10 0 0 5

. . .

. .

kg ms

A

2

00039 2

m mT

( )= .

FB

v

FIG. P29.44(b)

B

N S S

FIG. P29.44(e)

v

B+

FIG. P29.44(a)

FIG. P29.44(d)

+ N S


Magnetic Fields 165

FIG. P29.47

P29.47 The magnetic force on each proton, F v BB q q B= × = °v sin 90 downward

perpendicular to velocity, causes centripetal acceleration, guiding it into a circular path of radius r, with

q Bm

rv

v=2

and rm

qB= v

We compute this radius by fi rst fi nding the proton’s speed:

K m

K

m

=

= =×( ) ×( )−

1

2

2 2 5 00 10 1 60 10

1

2

6 19

v

v. .

.

eV J eV

667 103 10 1027

7

×= ×− kg

m s.

Now, rm

qB= =

×( ) ×( )×

−

−

v 1 67 10 3 10 10

1 60 10

27 7. .

.

kg m s119 0 050 0

6 46C N s C m

m( ) ⋅ ⋅( ) =.

.

(b) We can most conveniently do part (b) fi rst. From the fi gure observe that

sin.

.

α

α

= =

= °

1 00 1

8 90

m m

6.46 mr

(a) The magnitude of the proton momentum stays constant, and its fi nal y component is

− ×( ) ×( ) ° = −−1 67 10 3 10 10 8 90 8 0027 7. . sin . . kg m s ×× ⋅−10 21 kg m s

*P29.48 (a) If B i j k= + +B B Bx y z

ˆ ˆ ˆ , F v B i i j kB i x y z i yq e B B B e B= × = ( ) × + +( ) = +v vˆ ˆ ˆ ˆ 0 ˆ ˆk j− e Bi zv .

Since the force actually experienced is F jB iF= ˆ , observe that

Bx could have any value , By = 0 , and BF

ezi

i

= −v

.

(b) If v i= −vi

ˆ, then F v B i i j kB i x

i

i

q e BF

e= × = −( ) × + −

⎛⎝⎜

⎞⎠⎟

=vv

ˆ ˆ ˆ ˆ0 −−Fi j

(c) If q e= − and v i= −vi

ˆ, then F v B i i j kB i x

i

i

q e BF

e= × = − −( ) × + −

⎛⎝⎜

⎞⎠⎟

vv

ˆ ˆ ˆ ˆ0 == +Fi j

Reversing either the velocity or the sign of the charge reverses the force.

P29.49 (a) The net force is the Lorentz force given by

F E v B E v B

F i

= + × = + ×( )= ×( )−

q q q

3 20 10 419. ˆ −− −( ) + + −( ) × + +( )⎡⎣

⎤⎦1 2 2 3 1 2 4 1ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆj k i j k i j k N

Carrying out the indicated operations, we fi nd:

F i j= −( ) × −3 52 1 60 10 18. ˆ . ˆ .N

(b) θ =⎛⎝⎜

⎞⎠⎟ =

( ) + ( )

⎛

⎝

− −cos cos.

. .

1 1

2 2

3 52

3 52 1 60

F

Fx ⎜⎜

⎜

⎞

⎠

⎟⎟ = °24 4.


166 Chapter 29

*P29.50 (a) The fi eld should be in the +z direction, perpendicular to the fi nal as well as to the initial

velocity, and with ˆ ˆ î k j× = − as the direction of the initial force.

(b) rm

qB= =

×( ) ×( )×

−

−

v 1 67 10 20 10

1 60 10

27 6

19

.

.

kg m s

C N s C mm( ) ⋅ ⋅( ) =

0 30 696

..

(c) The path is a quarter circle, of length (p /2)0.696 m = 1.09 m

(d) ∆ t = 1.09 m/20 × 106 m/s = 54.7 ns

P29.51 A key to solving this problem is that reducing the normal force will reduce

the friction force: F BILB = or BF

ILB= .

When the wire is just able to move, F n F mgy B∑ = + − =cosθ 0

so n mg FB= − cosθ

and f mg FB= −( )µ θcos

Also, F F fx B∑ = − =sinθ 0

so F fB sinθ = : F mg FB Bsin cosθ µ θ= −( ) and F

mgB =

+µ

θ µ θsin cos

We minimize B by minimizing FB: dF

dmgB

θµ θ µ θ

θ µ θµ θ= ( ) −

+( ) = ⇒ =cos sin

sin cossin2 0 ccosθ

Thus, θµ

=⎛

⎝⎜

⎞

⎠⎟ = ( ) = °− −tan tan . .1 11

5 00 78 7 for the smallest fi eld, and

BF

IL

g

I

m L

B

B= =⎛⎝⎜

⎞⎠⎟

( )+

=( )

µθ µ θsin cos

.min

0 200 9..

.

.

sin .

80

1 50

0 100

78 7 0

m s

A

kg m2( )⎡

⎣⎢⎢

⎤

⎦⎥⎥ °+ .. cos .

.

.min

200 78 70 128

0 128

( ) °=

=

T

T pointinB gg north at an angle of 78.7 below the hor° iizontal

P29.52 Let vi represent the original speed of the alpha particle. Let vα and vp represent the particles’ speeds after the collision. We have conservation of momentum 4 4m m mp i p p pv v v= +α and the relative velocity equation v v vi p− = −0 α. Eliminating vi ,

4 4 4v v v vp p− = +α α 3 8v vp = α v vα = 3

8 p

For the proton’s motion in the magnetic fi eld,

F ma∑ = e Bm

Rpp pvv

sin902

° =

eBR

mpp= v

For the alpha particle,

2 904 2

e Bm

rpvv

αα

α

sin ° = rm

eBp

αα=

2 v

r

m

eB

m

eB

eBR

mRp

pp

pα = = =

2 3

8

2 3

8

3

4v

FIG. P29.51

I


Magnetic Fields 167

P29.53 Let ∆x1 be the elongation due to the weight of the wire and let ∆x2 be the additional elongation of the springs when the magnetic fi eld is turned on. Then F k xmagnetic = 2 2∆ where k is the force constant of the spring and can be determined from

kmg

x=

2 1∆. (The factor 2 is included in the two previous

equations since there are 2 springs in parallel.) Combining these two equations, we fi nd

F

mg

xx

mg x

xmagnetic =⎛⎝⎜

⎞⎠⎟

=22 1

22

1∆∆ ∆

∆; but

F L BB I ILB= × =

Therefore, where I = =24 02 00

..

V

12.0A,

Ω B

mg x

IL x= =

( )( ) ×( )( )

−∆∆

2

1

30 100 9 80 3 00 10

2 00

. . .

. 00 050 0 5 00 100 588

3. .. .

( ) ×( ) =− T

P29.54 Suppose the input power is

120 120W V= ( ) I : I ~ 1 100A A=

Suppose ω π= ⎛⎝⎜ ⎞

⎠⎟ ⎛

⎝⎜ ⎞2 000

1 2rev min

min

60 s

rad

1 rev ⎠⎠⎟ ~ 200 rad s

and the output power is 20 200W rad s= = ( )τω τ τ ~ 10 1− ⋅N m

Suppose the area is about 3 4cm cm( ) × ( ), or A ~ 10 3− m2

Suppose that the fi eld is B ~ 10 1− T

Then, the number of turns in the coil may be found from τ ≅ NIAB :

0 1 1 10 103 1. ~N m C s m N s C m2⋅ ( )( ) ⋅ ⋅( )− −N

giving N ~ 103

P29.55 The sphere is in translational equilibrium; thus

f Mgs − =sinθ 0 (1)

The sphere is in rotational equilibrium. If torques are taken about the center of the sphere, the magnetic fi eld produces a clockwise torque of magnitude µ θBsin , and the frictional force a counterclockwise torque of magnitude f Rs , where R is the radius of the sphere. Thus:

f R Bs − =µ θsin 0 (2)

From (1): f Mgs = sin .θ Substituting this in (2) and canceling out sin ,θ one obtains

µB MgR= . (3)

Now µ π= NI R2 . Thus (3) gives IMg

NBR= =

( )( )( )( )π π0 08 9 80

5 0 350 0

. .

. .

kg m s

T

2

220 713

mA

( ) = . .

The current must be counterclockwise as seen from above.

FIG. P29.53

I

fs

B

Mg

θ

θ

µ

FIG. P29.55


168 Chapter 29

P29.56 Call the length of the rod L and the tension in each wire alone T

2. Then, at equilibrium:

F T ILBx∑ = − ° =sin sin .θ 90 0 0 or T ILBsinθ =

F T mgy∑ = − =cos ,θ 0 or T mgcosθ =

tanθ = = ( )ILB

mg

IB

m L g or B

m L g

I

g

I=

( ) =tan tanθ λ θ

P29.57 F ma∑ = or q B

m

rv

vsin .90 0

2

° =

∴ the angular frequency for each ion is vr

qB

mf= = =ω π2 and

∆ω ω ω= − = −

⎛

⎝⎜

⎞

⎠⎟ = ( )−

12 1412 14

191 1 1 60 10 2

qBm m

. × C ..

. . .

40

1 66 10

1

12 0

1

14 027

T

kg u u u

( )( ) −

⎛⎝⎜

⎞⎠−×⎟⎟

= =−∆ω 2 75 10 2 756 1. .× s Mrad/s

P29.58 Let vx and v⊥ be the components of the velocity of the positron parallel to and perpendicular to the direction of the magnetic fi eld.

(a) The pitch of trajectory is the distance moved along x by the positron during each period, T (determined by the cyclotron frequency):

p Tm

Bq

p

x= = °( )⎛

⎝⎜

⎞

⎠⎟

= ( )

v v cos .

. cos

85 02

5 00 106

π

× 885 0 2 9 11 10

0 150 1 60 101

31

19

. .

. .

°( )( )( )( ) =

−

−

π ×

×..04 10 4× − m

(b) The equation about circular motion in a magnetic fi eld still applies to the radius of the

spiral: rm

Bq

m

Bq= = °⊥v v sin .85 0

r = ( )( ) °( )( )

−9 11 10 5 00 10 85 0

0 150 1

31 6. . sin .

. .

× ×

660 101 89 10

19

4

××−

−

( ) = . m

P29.59 τ = IAB where the effective current due to the orbiting electrons is Iq

t

q

T= =∆

∆

and the period of the motion is TR= 2π

v

The electron’s speed in its orbit is found by requiring k q

R

m

Re

2

2

2

= v or v = q

k

mRe

Substituting this expression for v into the equation for T, we fi nd TmR

q ke

= 23

2π

T =( )( )( )

− −

−2

9 11 10 5 29 10

1 60 10 8

31 11 3

19 2π. .

.

× ×

× ...

99 101 52 10

9

16

××

( ) = − s

Therefore, τ π= ⎛⎝⎜

⎞⎠⎟ =

−

−−q

TAB

1 60 10

1 52 105 29 10

19

161.

..

××

× 11 2 240 400 3 70 10( ) ( ) = ⋅−. . .× N m

FIG. P29.58


Magnetic Fields 169

P29.60 (a) K m= = = ( ) −1

26 00 6 00 10 1 60 102 6 19v . . .MeV eV J e× × VV( )

K =

= ( )

−

−

−

9 60 10

2 9 60 10

1 67 10

13

13

27

.

.

.

×

××

J

J

kv

ggm s= 3 39 107. ×

F q Bm

RB = =vv2

so

Rm

qB= = ( )( )−

−

v 1 67 10 3 39 10

1 60 10

27 7

19

. .

.

× ××

kg m s

CC Tm( )( )

=1 00

0 354.

.

Then, from the diagram, x R= ° = ( ) ° =2 45 0 2 0 354 45 0 0 501sin . . sin . . m m .

(b) From the diagram, observe that ′ = °θ 45 0. .

P29.61 (a) The magnetic force acting on ions in the blood stream will defl ect positive charges toward point A and negative charges toward point B. This separation of charges produces an electric fi eld directed from A toward B. At equilibrium, the electric force caused by this fi eld must balance the magnetic force, so

q B qE qV

dv = = ⎛

⎝⎜

⎞⎠⎟∆

or v = = ( )( )( ) =

−

−

∆V

Bd

160 10

0 040 0 3 00 101

6

3

××V

T m. ..333 m s

(b) No . Negative ions moving in the direction of v would be defl ected toward point B,

giving A a higher potential than B. Positive ions moving in the direction of v would be

defl ected toward A, again giving A a higher potential than B. Therefore, the sign of the potential difference does not depend on whether the ions in the blood are positively or negatively charged.

P29.62 (a) See graph to the right. The Hall voltage is directly proportional to the magnetic fi eld. A least-square fi t to the data gives the equation of the best fi tting line as:

∆V BH V T= ( )−1 00 10 4. ×

(b) Comparing the equation of the line which fi ts the data best to

∆Vnqt

BH =⎛⎝⎜

⎞⎠⎟

1

xxxxx

xxxxx

xxxxx

xxxxx

xxxxx

xxxxx

xxxxx45.0°

45°45°

'

x

v

R

Bin = 1.00 T

FIG. P29.60

FIG. P29.61

0

20

40

60

80

100

120

0 0.2 0.4 0.6 0.8 1.0 1.2

B (T)

∆VH (mV)

FIG. P29.62

continued on next page


170 Chapter 29

observe that: I

nqt=1 00 10 4. ,× V T or t

I

nq= ( )1 00 10 4.

.× V T

Then, if I = 0 200. A, q =1 60 10 19. × C, and n =1 00 1026 3. ,× m the thickness of the sample is

t = ( )( )0 200

1 60 10 1 00 103 19

.

. .

A

1.00 10 m C26× × × 441 25 10 0 125

V Tm mm( ) = × =−. .

P29.63 When in the fi eld, the particles follow a circular path

according to q Bm

rv

v=2

, so the radius of the path is

rm

qB= v

.

(a) When r hm

qB= = v

, that is, when v = qBh

m, the

particle will cross the band of fi eld. It will move in a full semicircle of radius h, leaving the fi eld at

2 0 0h, ,( ) with velocityv jf = −vˆ .

(b) When v < qBh

m, the particle will move in a smaller semicircle of radius r

m

qBh= <v.

It will leave the fi eld at 2 0 0r, ,( ) with velocity v jf = −vˆ .

(c) When v > qBh

m, the particle moves in a circular arc of radius r

m

qBh= >v

, centered at

r, , .0 0( ) The arc subtends an angle given by θ = ⎛⎝

⎞⎠

−sin .1 h

r It will leave the fi eld at the

point with coordinates r h1 0−( )[ ]cos , ,θ with velocity v i jf = +v vsin ˆ cos ˆ .θ θ

P29.64 (a) Ie

r= v

2π µ

ππ= =

⎛⎝⎜

⎞⎠⎟ = ⋅−IA

e

rr

v2

9 27 102 24. × A m2

The Bohr model predicts the correct magnetic moment. However, the “planetary model” is seriously defi cient in other regards.

(b) Because the electron is (–), its [conventional] current is clockwise, as seen

from above, and µ points downward .

*P29.65 (a) The torque on the dipole τ µ= × B has magnitude mB sin q ≈ mBq, proportional to the

angular displacement if the angle is small. It is a restoring torque, tending to turn the dipole toward its equilibrium orientation. Then the statement that its motion is simple harmonic is true for small angular displacements.

(b) t = Ia becomes –mBq = I d2q /dt2 d2q /dt2 = –(mB/I)q = –w 2q

where w = (mB/I )1/2 is the angular frequency and f = w /2p = 1

2πµB

I is the frequency

in hertz.

(c) The equilibrium orientation of the needle shows the direction of the fi eld. In a stronger fi eld, the frequency is higher. The frequency is easy to measure precisely over a wide range of values. The equation in part (c) gives

0.680 Hz = (1/2p) (m/I)1/2 (39.2 mT)1/2 and 4.90 Hz = (1/2p) (m/I)1/2 (B2)1/2. We square

and divide to fi nd B2/39.2 mT = (4.90 Hz/0.680 Hz)2 so B

2 = 51.9(39.2 mT) = 2.04 mT .

FIG. P29.63

FIG. P29.64


Magnetic Fields 171

P29.2 (a) west (b) no defl ection (c) up (d) down

P29.4 (a) 86 7. fN (b) 51 9. Tm s2

P29.6 (a) 7 90. pN (b) 0

P29.8 By = −2 62. mT. Bz = 0. Bx may have any value.

P29.10 (a) 6.84 × 10−16 m down (b) 7.24 mm east. The beam moves on an arc of a circle rather than on a parabola, but its northward velocity component stays constant within 0.09%, so it is a good approximation to treat it as constant.

P29.12 115 keV

P29.14 ′ =m

m8

P29.16 (a) 17.9 ns (b) 35.1 eV

P29.18 (a) 8.28 cm (b) 8.23 cm; ratio is independent of both ∆V and B

P29.20 (a) 7.66 × 107 rad/s (b) 26.8 Mm/s (c) 3.76 MeV (d) 3.13 × 103 rev (e) 257 ms

P29.22 (b) The dashed red line should spiral around many times, with its turns relatively far apart on the inside and closer together on the outside. (c) 682 m/s (d) 55.9 mm

P29.24 (a) Yes: The constituent of the beam is present in all kinds of atoms. (b) Yes: Everything in the beam has a single charge-to-mass ratio. (c) Thomson pointed out that ionized hydrogen had the largest charge-to-mass ratio previously known, and that the particles in his beam had a charge-to-mass ratio about 2 000 times larger. The particles in his beam could not be whole atoms, but rather must be much smaller in mass. (d) No. The particles move with speed on the order of ten million meters per second, to fall by an immeasurably small amount over a distance of less than a meter.

P29.26 −( )2 88. j N

P29.28 840 A east

P29.30 4

3

1 2IdBL

m⎛⎝

⎞⎠

P29.32 (a) Fab = 0;

F ibc = −( )40 0. ˆ ;mN

F kcd = −( )40 0. ˆ ;mN

F i kda = ( ) +( )40 0. ˆ ˆmN (b) The forces

on the four segments must add to zero, so the force on the fourth segment must be the negative of the resultant of the forces on the other three.

P29.34 (a) 5 41. mA m2⋅ (b) 4 33. mN m⋅

P29.36 See the solution.

P29.38 (a) 118 mN · m (b) −118 mJ ≤ U ≤ 118 mJ

P29.40 (a) 640 N mm ⋅ (b) 241 mW (c) 2.56 mJ (d) 154 mW

ANSWERS TO EVEN PROBLEMS


172 Chapter 29

P29.42 (a) 37 7. mT (b) 4 29 1025 3. × m

P29.44 See the solution.

P29.46 39 2. mT

P29.48 (a) Bx is indeterminate. By = 0. BF

ezi

i

= −v

(b) −Fi j (c) +Fi j

P29.50 (a) the +z direction (b) 0.696 m (c) 1.09 m (d) 54.7 ns

P29.52 3

4

R

P29.54 B ~ 10 1− T; τ ~ 10 1− ⋅N m; I ~ 1 A; A ~ ;10 3− m2 N ~ 103

P29.56 λ θg

I

tan

P29.58 (a) 0 104. mm; (b) 0 189. mm

P29.60 (a) 0.501 m (b) 45.0°

P29.62 (a) See the solution. Empirically, ∆V BH V T= ( )100 µ (b) 0 125. mm

P28.64 (a) 9.27 × 10−24 A · m2 (b) down


Software

Sm chapter29