29

CHAPTER OUTLINE

29.1 Magnetic Fields and Forces29.2 Magnetic Force Acting on a Current-Carrying Conductor29.3 Torque on a Current Loop in a Uniform Magnetic Field29.4 Motion of a Charged Particle in a Uniform Magnetic Field29.5 Applications Involving Charged Particles Moving in a Magnetic Field29.6 The Hall Effect

Magnetic Fields

ANSWERS TO QUESTIONS

Q29.1 The force is in the +y direction. No, the proton will notcontinue with constant velocity, but will move in a circularpath in the x-y plane. The magnetic force will always beperpendicular to the magnetic field and also to the velocity ofthe proton. As the velocity changes direction, the magneticforce on the proton does too.

Q29.2 If they are projected in the same direction into the samemagnetic field, the charges are of opposite sign.

Q29.3 Not necessarily. If the magnetic field is parallel or antiparallelto the velocity of the charged particle, then the particle willexperience no magnetic force.

Q29.4 One particle veers in a circular path clockwise in the page,while the other veers in a counterclockwise circular path. If themagnetic field is into the page, the electron goes clockwise andthe proton counterclockwise.

Q29.5 Send the particle through the uniform field and look at its path. If the path of the particle isparabolic, then the field must be electric, as the electric field exerts a constant force on a chargedparticle. If you shoot a proton through an electric field, it will feel a constant force in the samedirection as the electric field—it’s similar to throwing a ball through a gravitational field. If the pathof the particle is helical or circular, then the field is magnetic—see Question 29.1. If the path of theparticle is straight, then observe the speed of the particle. If the particle accelerates, then the field iselectric, as a constant force on a proton with or against its motion will make its speed change. If thespeed remains constant, then the field is magnetic—see Question 29.3.

Q29.6 Similarities: Both can alter the velocity of a charged particle moving through the field. Both exertforces directly proportional to the charge of the particle feeling the force. Positive and negativecharges feel forces in opposite directions. Differences: The direction of the electric force is parallel orantiparallel to the direction of the electric field, but the direction of the magnetic force isperpendicular to the magnetic field and to the velocity of the charged particle. Electric forces canaccelerate a charged particle from rest or stop a moving particle, but magnetic forces cannot.

161

162 Magnetic Fields

Q29.7 Since F v BB q= ×a f , then the acceleration produced by a magnetic field on a particle of mass m is

a v BBqm

= ×a f . For the acceleration to change the speed, a component of the acceleration must be in

the direction of the velocity. The cross product tells us that the acceleration must be perpendicular tothe velocity, and thus can only change the direction of the velocity.

Q29.8 The magnetic field in a cyclotron essentially keeps the charged particle in the electric field for alonger period of time, and thus experiencing a larger change in speed from the electric field, byforcing it in a spiral path. Without the magnetic field, the particle would have to move in a straightline through an electric field over a distance that is very large compared to the size of the cyclotron.

Q29.9 (a) The qv B× force on each electron is down. Since electrons are negative, v B× must be up.With v to the right, B must be into the page, away from you.

(b) Reversing the current in the coils would reverse the direction of B, making it toward you.Then v B× is in the direction right × toward you = down, and qv B× will make the electronbeam curve up.

Q29.10 If the current is in a direction parallel or antiparallel to the magnetic field, then there is no force.

Q29.11 Yes. If the magnetic field is perpendicular to the plane of the loop, then it exerts no torque on the loop.

Q29.12 If you can hook a spring balance to the particle and measure the force on it in a known electric field,

then qFE

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

acceleration of small particles by observing their deflection in known electric and magnetic fields cantell you the charge-to-mass ratio, but not separately the charge or mass. Both an accelerationproduced by an electric field and an acceleration caused by a magnetic field depend on the

properties of the particle only by being proportional to the ratio qm

.

Q29.13 If the current loop feels a torque, it must be caused by a magnetic field. If the current loop feels notorque, try a different orientation—the torque is zero if the field is along the axis of the loop.

Q29.14 The Earth’s magnetic field exerts force on a charged incoming cosmic ray,tending to make it spiral around a magnetic field line. If the particle energy islow enough, the spiral will be tight enough that the particle will first hit somematter as it follows a field line down into the atmosphere or to the surface at ahigh geographic latitude.

FIG. Q29.14

Q29.15 The net force is zero, but not the net torque.

Q29.16 Only a non-uniform field can exert a non-zero force on a magnetic dipole. If the dipole is alignedwith the field, the direction of the resultant force is in the direction of increasing field strength.

Chapter 29 163

Q29.17 The proton will veer upward when it enters the field and move in a counter-clockwise semicirculararc. An electron would turn downward and move in a clockwise semicircular arc of smaller radiusthan that of the proton, due to its smaller mass.

Q29.18 Particles of higher speeds will travel in semicircular paths of proportionately larger radius. They willtake just the same time to travel farther with their higher speeds. As shown in Equation 29.15, thetime it takes to follow the path is independent of particle’s speed.

Q29.19 The spiral tracks are left by charged particles gradually losing kinetic energy. A straight path mightbe left by an uncharged particle that managed to leave a trail of bubbles, or it might be theimperceptibly curving track of a very fast charged particle.

Q29.20 No. Changing the velocity of a particle requires an accelerating force. The magnetic force is proportionalto the speed of the particle. If the particle is not moving, there can be no magnetic force on it.

Q29.21 Increase the current in the probe. If the material is a semiconductor, raising its temperature mayincrease the density of mobile charge carriers in it.

SOLUTIONS TO PROBLEMS

Section 29.1 Magnetic Fields and Forces

P29.1 (a) up

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

(c) no deflection

(d) into the page

FIG. P29.1

P29.2 At the equator, the Earth’s magnetic field ishorizontally north. Because an electron hasnegative charge, F v B= ×q is opposite in directionto v B× . Figures are drawn looking down.

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

(a) (c) (d)

FIG. P29.2

(b) North × North = °=sin0 0 : Zero deflection .

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

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

164 Magnetic Fields

P29.3 F v BB q= × ; F j v i BB e− = − ×e jTherefore, B = −B ke j which indicates the negative directionz .

FIG. P29.3

P29.4 (a) F qvBB = = × × × °− −sin . . . sin .θ 1 60 10 3 00 10 3 00 10 37 019 6 1 C m s Te je je jFB = × −8 67 10 14. N

(b) aFm

= =××

= ×−

−8 67 101 67 10

5 19 1014

2713.

..

N kg

m s2

P29.5 F ma qvB

BFqv

= = × × = × = °

= =×

× ×= ×

− −

−

−−

1 67 10 2 00 10 3 34 10 90

3 34 10

1 60 10 1 00 102 09 10

27 13 14

14

19 72

. . . sin

.

. ..

kg m s N

N

C m s T

2e je j

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

FIG. P29.5

P29.6 First find the speed of the electron.

∆ ∆ ∆K mv e V U= = =12

2 : ve Vm

= =×

×= ×

−

−

2 2 1 60 10 2 400

9 11 102 90 10

19

317∆ .

..

C J C

kg m s

e jb ge j

(a) F qvBB, . . . . max C m s T N= = × × = ×− −1 60 10 2 90 10 1 70 7 90 1019 7 12e je ja f

(b) FB, min = 0 occurs when v is either parallel to or anti-parallel to B.

P29.7 F qvBB = sinθ so 8 20 10 1 60 10 4 00 10 1 7013 19 6. . . . sin× = × ×− − N C m s Te je ja f θ

sin .θ = 0 754 and θ = = ° °−sin . .1 0 754 48 9a f or 131 .

P29.8 Gravitational force: F mgg = = × = ×− −9 11 10 9 80 8 93 1031 30. . . kg m s N down2e je j .

Electric force: F qEe = = − × = ×− −1 60 10 100 1 60 1019 17. . C N C down N upe jb g .

Magnetic force: F v B E NB q= × = − × × × × ⋅ ⋅− −1 60 10 6 00 10 50 0 1019 6 6. . . C m s N s C m e je j e j .

FB = − × = ×− −4 80 10 4 80 1017 17. . N up N down .

Chapter 29 165

P29.9 F v BB q= ×

v Bi j k

i j k i j k

v B

F v B

× = + − ++ + −

= − + + + + = + +

× = + + = ⋅

= × = × ⋅ = ×− −

.

. . .

2 4 11 2 3

12 2 1 6 4 4 10 7 8

10 7 8 14 6

1 60 10 14 6 2 34 10

2 2 2

19 18

a f a f a f

e jb g T m s

C T m s NB q

P29.10 qE k k= − × = − ×− −1 60 10 20 0 3 20 1019 18. . . C N C Ne jb g e jF E v B a

k i B k

k i B k

i B k

∑ = + × =

− × − × × × = × ×

− × − × ⋅ × = ×

× ⋅ × = − ×

− − −

− − −

− −

q q m

3 20 10 1 60 10 9 11 10 2 00 10

3 20 10 1 92 10 1 82 10

1 92 10 5 02 10

18 19 31 12

18 15 18

15 18

. . . .

. . .

. .

N C 1.20 10 m s m s

N C m s N

C m s N

4 2e j e j e je je j e j e je j e jThe magnetic field may have any -componentx . Bz = 0 and By = −2 62. mT .

Section 29.2 Magnetic Force Acting on a Current-Carrying Conductor

P29.11 F ILBB = sinθ with F F mgB g= =

mg ILB= sinθ somL

g IB= sinθ

I = 2 00. A andmL=

FHG

IKJ = × −0 500

1001 000

5 00 10 2. . g cm cm m g kg

kg mb g .

Thus 5 00 10 9 80 2 00 90 02. . . sin .× = °−e ja f a fB

FIG. P29.11

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

P29.12 F B i k jB I= × = × = −2 40 0 750 1 60 2 88. . . . A m T Na fa f a f e j

P29.13 (a) F ILBB = = °=sin . . . sin . .θ 5 00 2 80 0 390 60 0 4 73 A m T Na fa fa f

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

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

166 Magnetic Fields

P29.14F BB mg I

ImgB

= =×

= = =0 040 0 9 80

3 600 109

. .

..

kg m m s

T A

2b ge j

The direction of I in the bar is to the right . Bin

F

FIG. P29.14

P29.15 The rod feels force F d B k j iB I Id B IdB= × = × − =a f e j e j e j .The work-energy theorem is K K E K K

i ftrasn rot trans rot+ + = +b g b g∆

0 012

12

2 2+ + = +F mv Is cosθ ω

IdBL mv mRvR

cos012

12

12

2 22

°= + FHGIKJFHGIKJ and IdBL mv=

34

2

vIdBL

m= = =

43

4 48 0 0 120 0 240 0 4503 0 720

1 07. . . .

..

A m T m kg

m sa fa fa fa f

b g .

dI

L

B

y

xz

FIG. P29.15

P29.16 The rod feels force F d B k j iB I Id B IdB= × = × − =a f e j e j e j .The work-energy theorem is K K E K K

i ftrans rot trans rot+ + = +b g b g∆

0 012

12

2 2+ + = +Fs mv Icosθ ω

IdBL mv mRvR

cos012

12

12

2 22

°= + FHGIKJFHGIKJ and v

IdBLm

=4

3.

P29.17 The magnetic force on each bit of ring isId IdsBs B× = radially inward and upward, atangle θ above the radial line. The radiallyinward components tend to squeeze the ringbut all cancel out as forces. The upwardcomponents IdsBsinθ all add to

I rB2π θsin up .

FIG. P29.17

Chapter 29 167

P29.18 For each segment, I = 5 00. A and B j= ⋅0 020 0. N A m .

Segment. m

0.400 m mN

. m . m mN

. m . m mN

F Bj

k i

i j k

i k k i

B Iab

bc

cd

da

= ×−

−

− + −

− +

a f

a fe j

a fe j

a fe j

0 400 0

40 0

0 400 0 400 40 0

0 400 0 400 40 0

.

.

.FIG. P29.18

P29.19 Take the x-axis east, the y-axis up, and the z-axis south. The field is

B k j= ° − + ° −52 0 60 0 52 0 60 0. cos . . sin . T Tµ µb g e j b g e j.

The current then has equivalent length: ′ = − +L k j1 40 0 850. . m me j e j

F L B j k j k

F i i i

B

B

I= ′ × = − × − −

= × − − = × − =

−

− −

0 035 0 0 850 1 40 45 0 26 0 10

3 50 10 22 1 63 0 2 98 10 2 98

6

8 6

. . . . .

. . . . .

A m T

N N N west

b ge j e je j e j µ

.

FIG. P29.19

Section 29.3 Torque on a Current Loop in a Uniform Magnetic Field

P29.20 (a) 2 2 00π r = . m

so r = 0 318. m

µ π= = × = ⋅−IA 17 0 10 0 318 5 413 2. . . A m mA m2 2e j a f

(b) ττττ µµµµ= ×B

so τ = × ⋅ = ⋅−5 41 10 0 800 4 333. . . A m T mN m2e ja f

P29.21 τ µ θ= Bsin so 4 60 10 0 250 90 03. . sin .× ⋅ = °− N m µa fµ = × ⋅ = ⋅−1 84 10 18 42. . A m mA m2 2

168 Magnetic Fields

P29.22 (a) Let θ represent the unknown angle; L, the total length of the wire; and d, the length of oneside of the square coil. Then, using the definition of magnetic moment and the right-handrule in Figure 29.15, we find

µ = NAI : µµµµ = FHGIKJ

Ld

d I4

2 at angle θ with the horizontal.

At equilibrium, ττττ µµµµ∑ = × − × =B r gb g b gm 0

ILBd mgd4

90 02

0FHGIKJ °− − FHG

IKJ =sin . sinθ θa f

andmgd ILBd

2 4FHGIKJ = FHG

IKJsin cosθ θ

θ =FHGIKJ =

FHGG

IKJJ = °− −tan tan

. . .

. ..1 1

2

3 40 4 00 0 010 0

2 0 100 9 803 97

ILBmg

A m T

kg m s2

a fa fb gb ge j

.

(b) τ θmILBd

= FHGIKJ = °= ⋅

414

3 40 4 00 0 010 0 0 100 3 97 3 39cos . . . . cos . . A m T m mN ma fa fb ga f

P29.23 τ φ

τ

τ

=

= × °

= ⋅

NBAI sin

. . . . sin

.

100 0 800 0 400 0 300 1 20 60

9 98

T m A

N m

2a fe ja f

Note that φ is the angle between the magneticmoment and the B field. The loop will rotate so asto align the magnetic moment with the B field.Looking down along the y-axis, the loop will rotatein a clockwise direction.

FIG. P29.23

P29.24 From τ = × = ×µµµµ B A BI , the magnitude of the torque is IABsin .90 0° .

(a) Each side of the triangle is 40 0. cm

3.

Its altitude is 13 3 6 67 11 52 2. . .− = cm cm and its area is

A = = × −12

11 5 13 3 7 70 10 3. . . cm cm m2a fa f .

Then τ = × ⋅ ⋅ = ⋅−20 0 7 70 10 0 520 80 13. . . . A m N s C m mN m2a fe jb g .

(b) Each side of the square is 10.0 cm and its area is 100 10 2 cm m2 2= − .

τ = = ⋅−20 0 10 0 520 0 1042. . . A m T N m2a fe ja f

(c) r = =0 400

0 063 7.

. m

2 m

πA r= = ×

= × = ⋅

−

−

π

τ

2 2

2

1 27 10

20 0 1 27 10 0 520 0 132

.

. . . .

m

A m N m

2

2a fe ja f

(d) The circular loop experiences the largest torque.

Chapter 29 169

P29.25 Choose U = 0 when the dipole moment is at θ = °90 0. to the field. The field exerts torque of magnitudeµ θBsin on the dipole, tending to turn the dipole moment in the direction of decreasing θ. Accordingto Equations 8.16 and 10.22, the potential energy of the dipole-field system is given by

U B d B B− = = − = − +°

°z0 090 0

90 0µ θ θ µ θ µ θ

θθ

sin cos cos.

.a f or U = − ⋅µµµµ B .

P29.26 (a) The field exerts torque on the needle tending to align it with the field, so the minimumenergy orientation of the needle is:

pointing north at 48.0 below the horizontal°

where its energy is U Bmin cos . . .= − °= − × ⋅ × = − ×− − −µ 0 9 70 10 55 0 10 5 34 103 6 7 A m T J2e je 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 10 5 34 103 6 7 A m T J2e je j .

(b) U W Umin max+ = : W U U= − = + × − − × =− −max min . . .5 34 10 5 34 10 1 077 7 J J Je j µ

P29.27 (a) τ = ×µµµµ B, so τ µ θ θ= × = =µµµµ B B NIABsin sin

τ π µmax sin . . . .= °= × = ⋅−NIAB 90 0 1 5 00 0 050 0 3 00 10 1182 3 A m T N ma f b g e j

(b) U = − ⋅µ B , so − ≤ ≤ +µ µB U B

Since µ π µB NIA B= = × =−a f a f b g e j1 5 00 0 050 0 3 00 10 1182 3. . . A m T J ,

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

*P29.28 (a) ττττ µµµµ= × =B NIABsinθ

τmax . . . sin .= ⋅ ⋅ °= × ⋅− −80 10 0 025 0 04 0 8 90 6 40 102 4 A m m N A m N me ja fb g

(b) Pmax max . .= = × ⋅ FHG

IKJFHGIKJ =

−τ ωπ

6 40 102 1

0 2414 N m 3 600 rev min rad

1 rev min60 s

Wb g

(c) In one half revolution the work is

W U U B B B

NIAB

= − = − °− − ° =

= = × ⋅ = ×− −

max cos cos

. .

min

N m J

µ µ µ180 0 2

2 2 6 40 10 1 28 104 3

b ge j

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

(d) Pavg J

1 60 s W= =

×=

−Wt∆

2 56 100 154

3..b g

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

.

170 Magnetic Fields

Section 29.4 Motion of a Charged Particle in a Uniform Magnetic Field

P29.29 (a) B = × −50 0 10 6. T; v = ×6 20 106. m s

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

F qvB

F

B

B

=

= × × × °

= ×

− −

−

sin

. . . sin .

.

θ

1 60 10 6 20 10 50 0 10 90 0

4 96 10

19 6 6

17

C m s T

N

e je je j

(b) Fmv

r=

2

so rmv

F= =

× ×

×=

−

−

2 27 6 2

17

1 67 10 6 20 10

4 96 101 29

. .

..

kg m s

N km

e je j

FIG. P29.29

P29.3012

2mv q V= ∆a f 12

3 20 10 1 60 10 83326 2 19. .× = ×− − kg C Ve j e ja fv v = 91 3. km s

The magnetic force provides the centripetal force: qvBmv

rsinθ =

2

rmv

qB=

°=

× ×

× ⋅ ⋅=

−

−sin .

. .

. ..

90 0

3 20 10 9 13 10

1 60 10 0 9201 98

26 4

19

kg m s

C N s C m cm

e je je jb g .

P29.31 For each electron, q vBmv

rsin .90 0

2

°= and veBrm

= .

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

K mv mv mv

K me B R

mm

e B Rm

e Bm

R R

Ke

i f f= + = +

=FHG

IKJ +FHG

IKJ = +

=× ⋅ ⋅

×+ =

−

−

12

012

12

12

12 2

1 60 10 0 044 0

2 9 11 100 010 0 0 024 0 115

12

12

22

2 212

2

2 222

2

2 2

12

22

19 2

31

2 2

e j

e jb ge j

b g b g. .

.. .

C N s C m

kg m m keV

P29.32 We begin with qvBmv

R=

2

, so vqRBm

= .

The time to complete one revolution is TR

vR

qRB mm

qB= = =

2 2 2π π π.

Solving for B, Bm

qT= = × −2

6 56 10 2π. T .

Chapter 29 171

P29.33 q V mv∆a f = 12

2 or vq V

m=

2 ∆a f.

Also, qvBmv

r=

2

so rmvqB

mqB

q Vm

m V

qB= = =

2 22

∆ ∆a f a f.

Therefore, rm V

eBpp2

2

2=

∆a f

rm V

q B

m V

eB

m V

eBrd

d

d

p pp

22 2 2

22 2 22

22= = =

FHG

IKJ =

∆ ∆ ∆a f e ja f a f

and rm V

q B

m V

e B

m V

eBr

p ppα

α

α

22 2 2

22 2 4

22

22= = =

FHG

IKJ =

∆ ∆ ∆a f e ja fa f

a f.

The conclusion is: r r rd pα = = 2 .

P29.34 (a) We begin with qvBmv

R=

2

or qRB mv= .

But L mvR qR B= = 2 .

Therefore, RLqB

= =× ⋅

× ×= =

−

− −

4 00 10

1 00 100 050 5 00

25

19 3

.

.. .

J s

1.60 10 C T0 m cm

e je j.

(b) Thus, vL

mR= =

× ⋅

×= ×

−

−

4 00 10

10 0 050 08 78 10

25

316.

..

J s

9.11 kg m m s

e jb g .

P29.35 ω = =×

×= ×

−

−

qBm

1 60 10 5 20

1 67 104 98 10

19

278

. .

..

C T

kg rad s

e ja f

P29.3612

2mv q V= ∆a f so vq V

m=

2 ∆a f

rmvqB

= so rm q V m

qB=

2 ∆a f

rmq

V

B2

2

2= ⋅

∆a fand ′ =

′′⋅r

mq

V

Ba f a f2

2

2 ∆

mqB r

V=

2 2

2 ∆a f and ′ =′ ′

mq B r

Va f b g a fa f

2 2

2 ∆so

′=

′⋅

′= FHGIKJFHGIKJ =

mm

qq

r

re

eR

Ra f2

2

22 28

172 Magnetic Fields

P29.37 E mv e V= =12

2 ∆

and evBmv

Rsin90

2

°=

BmveR

meR

e Vm R

m Ve

B

= = =

=×

× ×

×= ×

−

−−

2 1 2

15 80 10

2 1 67 10 10 0 10

1 60 107 88 1010

27 6

1912

∆ ∆

.

. .

..

m

kg V

C T

e je j

*P29.38 (a) At the moment shown in Figure 29.21, the particle must bemoving 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.

v

B+

FIG. P29.38(a)

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

FB

v

FIG. P29.38(b)

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

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

plane and therefore a magnetic dipole moment IqT

A A=

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

+ N S

FIG. P29.38(d)

(e) Problem 17 showed that a nonuniform magnetic fieldexerts a net force on a magnetic dipole. When the dipole isaligned opposite to the external field, the force pushes itout of the region of stronger field. Here it is to the left, aforce of repulsion of one magnetic south pole on anothersouth pole.

B

N S S

FIG. P29.38(e)

Chapter 29 173

P29.39 rmvqB

= so mrqBv

= =× ×

×

− −7 94 10 1 60 10 1 80

4 60 10

3 19

5

. . .

.

m C T

m s

e je ja f

m = ××

FHG

IKJ =

−−4 97 10 2 9927

27. . kg1 u

1.66 10 kg u

The particle is singly ionized: either a tritium ion, 13 H+ , or a helium ion, 2

3 He+ .

Section 29.5 Applications Involving Charged Particles Moving in a Magnetic Field

P29.40 F FB e=

so qvB qE=

where vK

m=

2 and K is kinetic energy of the electron.

E vBK

mB= = =

×

×=

−

−2 2 750 1 60 10

9 11 100 015 0 244

19

31

a fe j b g.

.. kV m

P29.41 K mv q V= =12

2 ∆a f so vq V

m=

2 ∆a f

F v BB qmv

r= × =

2

rmvqB

mq

q V mB B

m Vq

= = =2 1 2∆ ∆a f a f

(a) r238

27

192

2 238 1 66 10 2 000

1 60 101

1 208 28 10 8 28=

× ×

×FHGIKJ = × =

−

−−

.

. .. .

e j m cm

(b) r235 8 23= . cm

rr

mm

238

235

238

235

238 05235 04

1 006 4= = =..

.

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

P29.42 In the velocity selector: vEB

= = = ×2 5000 035 0

7 14 104 V m T

m s.

. .

In the deflection chamber: rmvqB

= =× ×

×=

−

−

2 18 10 7 14 10

1 60 10 0 035 00 278

26 4

19

. .

. ..

kg m s

C T m

e je je jb g .

174 Magnetic Fields

P29.43 (a) F qvBmv

RB = =2

ω = = = =×

×= ×

−

−vR

qBRmR

qBm

1 60 10 0 450

1 67 104 31 10

19

277

. .

..

C T

kg rad s

e ja f

(b) vqBRm

= =×

×= ×

−

−

1 60 10 0 450 1 20

1 67 105 17 10

19

277

. . .

..

C T m

kg m s

e ja fa f

P29.44 K mv=12

2 : 34 0 10 1 60 1012

1 67 106 19 27 2. . .× × = ×− − eV J eV kge je j e jv

v = ×8 07 107. m s rmvqB

= =× ×

×=

−

−

1 67 10 8 07 10

1 60 10 5 200 162

27 7

19

. .

. ..

kg m s

C T m

e je je ja f

*P29.45 Note that the “cyclotron frequency” is an angular speed. The motion of the proton is described by

F ma∑ = :

q vBmv

r

q B mvr

m

sin902

°=

= = ω

(a) ω = =× ⋅ ⋅

×

⋅

⋅FHGIKJ = ×

−

−

q B

m

1 60 10 0 8

1 67 107 66 10

19

277

. .

..

C N s C m

kg

kg mN s

rad s2

e jb ge j

(b) v r= = × FHGIKJ = ×ω 7 66 10 0 350

11

2 68 107 7. . . rad s m rad

m se ja f

(c) K mv= = × ××

FHG

IKJ = ×−

−12

12

1 67 10 2 68 101

3 76 102 27 7 2

196. . . kg m s

eV1.6 10 J

eVe je j

(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 eV revolutionsa f .

(e) θ ω= t t = =××

FHG

IKJ = × −θ

ωπ3 13 10 2

2 57 103

4..

rev7.66 10 rad s

rad1 rev

s7

P29.46 F qvBmv

rB = =2

Bmvqr

= =× ⋅

×=

−

−

4 80 10

1 60 10 1 0003 00

16

19

.

..

kg m s

C m T

e jb g

Chapter 29 175

P29.47 θ = FHGIKJ = °−tan

.

..1 25 0

10 068 2 and R =

°=

1 0068 2

1 08.

sin ..

cm cm.

Ignoring relativistic correction, the kinetic energy of the electrons is

12

2mv q V= ∆ so vq Vm

= = ×2

1 33 108∆. m s .

From Newton’s second law mv

RqvB

2

= , we find the magnetic field

Bmvq R

= =× ×

× ×=

−

− −

9 11 10 1 33 10

1 60 10 1 08 1070 1

31 8

19 2

. .

. ..

kg m s

C m mT

e je je je j

.

FIG. P29.47

Section 29.6 The Hall Effect

P29.48 (a) RnqH ≡1

so nqR

= =× ×

= ×− −

−1 1

1 60 10 0 840 107 44 10

19 1028 3

H3 C m C

m. .

.e je j

(b) ∆VIBnqtH =

Bnqt V

I= =

× × × ×=

− − − −∆ H

m C m V

A T

b g e je je je j7 44 10 1 60 10 0 200 10 15 0 10

20 01 79

28 3 19 3 6. . . .

..

P29.49 Since ∆VIBnqtH = , and given that I = 50 0. A , B = 1 30. T , and t = 0 330. mm, the number of charge

carriers per unit volume is

nIB

e V t= = × −

∆ H mb g 1 28 1029 3.

The number density of atoms we compute from the density:

n0

23 6288 92 1 6 02 10 10

18 46 10=

FHG

IKJ

×FHG

IKJFHG

IKJ = ×

. ..

gcm

mole63.5 g

atomsmole

cm m

atom m3

3

33

So the number of conduction electrons per atom is

nn0

29

281 28 108 46 10

1 52=××

=..

.

176 Magnetic Fields

P29.50 (a) ∆VIBnqtH = so

nqtI

BV

= =×

= ×−∆ H

T0.700 V

T V0 080 0

101 14 106

5.. .

Then, the unknown field is Bnqt

IV= FHGIKJ ∆ Hb g

B = × × = =−1 14 10 0 330 10 0 037 7 37 75 6. . . . T V V T mTe je j .

(b)nqt

I= ×1 14 105. T V so n

Iqt

= ×1 14 105. T Ve j

n = ×× ×

= ×− −

−1 14 100 120

10 2 00 104 29 105

19 325 3.

.

.. T V

A

1.60 C m me j e je j

.

P29.51 Bnqt V

I= =

× × × ×− − − −∆ H

m C m V

Ab g e je je je j8 49 10 1 60 10 5 00 10 5 10 10

8 00

28 3 19 3 12. . . .

.

B = × =−4 33 10 43 35. . T Tµ

Additional Problems

P29.52 (a) The boundary between a region of strong magnetic field and aregion of zero field cannot be perfectly sharp, but we ignore thethickness of the transition zone. In the field the electron moves onan arc of a circle:

F ma∑ = :

q vBmv

r

vr

q B

m

sin

.

..

90

1 60 10 10

9 11 101 76 10

2

19 3

318

°=

= = =× ⋅ ⋅

×= ×

− −

−ω

C N s C m

kg rad s

e je je j

FIG. P29.52(a)

The time for one half revolution is,

from ∆ ∆θ ω= t

∆∆

t = =×

= × −θω

π rad rad s

s1 76 10

1 79 1088

.. .

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

Then v r= = × = ×−ω 1 76 10 0 02 3 51 108 1 6. . . s m m se ja fand

K mv= = × × = × =× ⋅×

=

− −−

−12

12

9 11 10 3 51 10 5 62 105 62 101 60 10

35 1

2 31 6 2 1818

19. . ...

. .

kg m s J J e C

eV

e je j

Chapter 29 177

P29.53 (a) Define 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 force .

I h B×a f in the direction of L.

(b) The sodium, consisting of ions and electrons, flows along thepipe transporting no net charge. But inside the section oflength L, electrons drift upward to constitute downwardelectric current J J× =areaa f Lw.

The current then feels a magnetic force I JLwhBh B× = °sin90 .

FIG. P29.53

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

F JLwhBhw

JLBarea

= = .

P29.54 Fy∑ = 0 : + − =n mg 0

Fx∑ = 0 : − + °=µ kn IB sin .90 0 0

Bmg

Idk= = =

µ 0 100 0 200 9 80

10 0 0 50039 2

. . .

. ..

kg m s

A m mT

2b ge ja fa f

P29.55 The magnetic force on each proton, F v BB q qvB= × = °sin90 downwardperpendicular to velocity, causes centripetal acceleration, guiding it into acircular path of radius r, with

qvBmv

r=

2

and rmvqB

= .

We compute this radius by first finding the proton’s speed:

K mv

vK

m

=

= =× ×

×= ×

−

−

12

2 2 5 00 10 1 60 10

1 67 103 10 10

2

6 19

277

. .

.. .

eV J eV

kg m s

e je j

Now, rmvqB

= =× ×

× ⋅ ⋅=

−

−

1 67 10 3 10 10

1 60 10 0 050 06 46

27 7

19

. .

. ..

kg m s

C N s C m m

e je je jb g .

FIG. P29.55

(b) From the figure, observe that

sin.

.

α

α

= =

= °

1 00 1

8 90

m m6.46 mr

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

− × × °= − × ⋅− −1 67 10 3 10 10 8 90 8 00 1027 7 21. . sin . . kg m s kg m se je j .

178 Magnetic Fields

P29.56 (a) If B i j k= + +B B Bx y z , F v B i i j k k jB i x y z i y i zq e v B B B ev B ev B= × = × + + = + −e j e j 0 .

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

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

i

i= − .

(b) If v i= −vi , then F v B i i j k jB i xi

iiq e v B

Fev

F= × = − × + −FHG

IKJ = −e j 0 .

(c) If q e= − and v i= vi , then F v B i i j k jB i xi

iiq e v B

Fev

F= × = − × + −FHG

IKJ = −e j 0 .

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

P29.57 (a) The net force is the Lorentz force given by

F E v B E v B

F i j k i j k i j k

= + × = + ×

= × − − + + − × + +−

q q qa fe j e j e j e j3 20 10 4 1 2 2 3 1 2 4 119. N

Carrying out the indicated operations, we find:

F i j= − × −3 52 1 60 10 18. .e j N .

(b) θ = FHGIKJ = +

F

HGG

I

KJJ = °− −cos cos

.

. ..1 1

2 2

3 52

3 52 1 6024 4

FFx

a f a f

P29.58 A key to solving this problem is that reducing the normal force will reduce

the friction force: F BILB = or BFIL

B= .

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 θcosb g .Also, F F fx B∑ = − =sinθ 0 FIG. P29.58

so F fB sinθ = : F mg FB Bsin cosθ µ θ= −b g and Fmg

B =+µ

θ µ θsin cos.

We minimize B by minimizing FB :dFd

mgB

θµ

θ µ θ

θ µ θµ θ θ=

−

+= ⇒ =b g b g

cos sin

sin cossin cos2 0 .

Thus, θµ

=FHGIKJ = = °− −tan tan . .1 11

5 00 78 7a f for the smallest field, and

BFIL

gI

m L

B

B

B= = FHGIKJ +

=L

NMM

O

QPP °+ °

=

= °

µθ µ θb g

a fe ja f

sin cos

. .

..

sin . . cos ..

.

min

min

0 200 9 80

1 500 100

78 7 0 200 78 70 128

0 128

m s

A kg m

T

T pointing north at an angle of 78.7 below the horizontal

2

Chapter 29 179

*P29.59 The electrons are all fired from the electron gun with the same speed v in

U Ki f= qV mv=12

2 − − =e V m vea fa f∆12

2 ve Vme

=2 ∆

For φ small, cosφ is nearly equal to 1. The time T of passage of each electron in the chamber is givenby

d vT= T dme V

e= FHGIKJ2

1 2

∆

Each electron moves in a different helix, around a different axis. If each completes just onerevolution within the chamber, it will be in the right place to pass through the exit port. Itstransverse velocity component v v⊥ = sinφ swings around according to F ma⊥ ⊥=

qv Bmv

r⊥⊥°=sin902

eBm v

rm m

Te

e e= = =⊥ ωπ2

Tm

eBd

me V

e e= = FHGIKJ

22

1 2π∆

Then 2

2

1 2

1 2πB

me

d

VeFHGIKJ =

∆a f Bd

m Vee= F

HGIKJ

2 2 1 2π ∆.

*P29.60 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 v m v m vp i p p p= +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 vpα =38

.

For the proton’s motion in the magnetic field,

F ma∑ = ev Bm v

Rpp psin90

2

°=eBRm

vp

p= .

For the alpha particle,

2 904 2

ev Bm v

rp

αα

αsin °= r

m v

eBp

αα

=2

rm

eBv

m

eBeBRm

Rpp

p

pα = = =

2 38

2 38

34

.

P29.61 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 fieldis 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 inparallel.) Combining these two equations, we find

Fmg

xx

mg xxmagnetic =

FHGIKJ =2

2 12

2

1∆∆

∆∆

; but F L BB I ILB= × = . FIG. P29.61

Therefore, where I = =24 0

2 00.

. V

12.0 A

Ω, B

mg xIL x

= =×

×=

−

−

∆∆

2

1

3

3

0 100 9 80 3 00 10

2 00 0 050 0 5 00 100 588

. . .

. . ..

a fa fe ja fb ge j

T .

180 Magnetic Fields

P29.62 Suppose the input power is

120 120 W V= a fI : I ~1 100 A A= .

Suppose ωπ

= FHGIKJFHG

IKJ2 000

1 2200 rev min

min60 s

rad1 rev

rad s~

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

Suppose the area is about 3 4 cm cma f a f× , or A ~10 3− m2 .

Suppose that the field 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⋅ ⋅ ⋅− −Nb ge je jgiving N ~103 .

*P29.63 The sphere is in translational equilibrium, thus

f Mgs − =sinθ 0 . (1)

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

f R Bs − =µ θsin 0 . (2)

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

µB MgR= . (3)

Ifs

B

Mgθ

θµµµµ

FIG. P29.63

Now µ π= NI R2 . Thus (3) gives IMgNBR

= = =π π

0 08 9 80

5 0 350 0 20 713

. .

. ..

kg m s

T m A

2b ge ja fa fa f . The current must be

counterclockwise as seen from above.

P29.64 Call the length of the rod L and the tension in each wire alone T2

. Then, at equilibrium:

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

F T mgy∑ = − =cosθ 0 , or T mgcosθ =

tanθ = =ILBmg

IBm L gb g or B

m L g

Ig

I= =b g

tan tanθλ

θ

P29.65 F ma∑ = or qvBmv

rsin .90 0

2

°=

∴ the angular frequency for each ion is vr

qBm

f= = =ω π2 and

∆

∆

f f fqB

m m

f f f

= − = −FHG

IKJ =

×

×−F

HGIKJ

= − = × =

−

−

−

12 1412 14

19

27

12 145 1

21 1 1 60 10 2 40

2 1 66 10

112 0

114 0

4 38 10 438

π π

. .

. . .

.

C T

kg u u u

s kHz

e ja fe j

Chapter 29 181

P29.66 Let vx and v⊥ be the components of the velocity of the positron parallelto and perpendicular to the direction of the magnetic field.

(a) The pitch of trajectory is the distance moved along x by thepositron during each period, T (see Equation 29.15)

p v T vm

Bq

p

x= = °FHGIKJ

=× ° ×

×= ×

−

−−

cos .

. cos . .

. ..

85 02

5 00 10 85 0 2 9 11 10

0 150 1 60 101 04 10

6 31

194

a f

e ja fa fe je j

π

π m

FIG. P29.66

(b) From Equation 29.13, rmvBq

mvBq

= =°⊥ sin .85 0

r =× × °

×= ×

−

−−

9 11 10 5 00 10 85 0

0 150 1 60 101 89 10

31 6

194

. . sin .

. ..

e je ja fa fe j

m

P29.67 τ = IAB where the effective current due to the orbiting electrons is Iqt

qT

= =∆∆

and the period of the motion is TR

v=

2π.

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

mvR

e2

2

2

= or v qk

mRe= .

Substituting this expression for v into the equation for T, we find TmRq ke

= 23

2π

T =× ×

× ×= ×

− −

−

−29 11 10 5 29 10

1 60 10 8 99 101 52 10

31 11 3

19 2 9

16π. .

. ..

e je je j e j

s .

Therefore, τ π= FHGIKJ =

××

× = × ⋅−

−− −q

TAB

1 60 101 52 10

5 29 10 0 400 3 70 1019

1611 2 24.

.. . .e j a f N m .

P29.68 Use the equation for cyclotron frequency ω =qBm

or mqB qB

f= =ω π2

m =× ×

×= ×

− −

−−

1 60 10 5 00 10

2 5 00 103 82 10

19 2

325

. .

..

C T

rev 1.50 s kg

e je ja fe jπ

.

182 Magnetic Fields

P29.69 (a) K mv= = = × × −12

6 00 6 00 10 1 60 102 6 19. . . MeV eV J eVe je jK

v

= ×

=×

×= ×

−

−

−

9 60 10

2 9 60 10

1 67 103 39 10

13

13

277

.

.

..

J

J

kg m s

e j

F qvBmv

RB = =2

so

RmvqB

= =× ×

×=

−

−

1 67 10 3 39 10

1 60 10 1 000 354

27 7

19

. .

. ..

kg m s

C T m

e je je ja f

xxxxx

xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx45.0°

45°45°

θ '

x

v

R

Bin T= 1 00.

FIG. P29.69

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

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

P29.70 (a) See graph to the right. TheHall voltage is directlyproportional to the magneticfield. A least-square fit to thedata gives the equation of thebest fitting line as:

∆V BH V T= × −1 00 10 4.e j .

(b) Comparing the equation ofthe line which fits the databest to

∆Vnqt

BH =FHGIKJ

1

0

20

40

60

80

100

120

0 0.2 0.4 0.6 0.8 1.0 1.2

B (T)

∆VH (µV)

FIG. P29.70

observe that: I

nqt= × −1 00 10 4. V T, or t

I

nq=

× −1 00 10 4. V Te j.

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 00 101 25 10 0 125

3 19 44.

. .. .

A

1.00 10 m C V T m mm

26e je je j.

Chapter 29 183

P29.71 (a) The magnetic force acting on ions in the blood stream willdeflect positive charges toward point A and negativecharges toward point B. This separation of chargesproduces an electric field directed from A toward B. Atequilibrium, the electric force caused by this field mustbalance the magnetic force, so

qvB qE qVd

= = FHGIKJ

∆

or vV

Bd= =

×

×=

−

−

∆ 160 10

0 040 0 3 00 101 33

6

3

V

T m m s

e jb ge j. .

. .

FIG. P29.71

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

A a higher potential than B. Positive ions moving in the direction of v would be deflectedtoward A, again giving A a higher potential than B. Therefore, the sign of the potentialdifference does not depend on whether the ions in the blood are positively or negativelycharged.

P29.72 When in the field, the particles follow a circular path

according to qvBmv

r=

2

, so the radius of the path is: rmvqB

=

(a) When r hmvqB

= = , that is, when vqBhm

= , the

particle will cross the band of field. It will move ina full semicircle of radius h, leaving the field at

2 0 0h, ,b g with velocity v jf v= − .

v ji v=

FIG. P29.72

(b) When vqBhm

< , the particle will move in a smaller semicircle of radius rmvqB

h= < . It will

leave the field at 2 0 0r , ,b g with velocity v jf v= − .

(c) When vqBhm

> , the particle moves in a circular arc of radius rmvqB

h= > , centered at

r , ,0 0b g . The arc subtends an angle given by θ = FHGIKJ

−sin 1 hr

. It will leave the field at the point

with coordinates r h1 0− cos , ,θa f with velocity v i jf v v= +sin cosθ θ .

ANSWERS TO EVEN PROBLEMS

P29.2 (a) west; (b) no deflection; (c) up; P29.8 Gravitational force: 8 93 10 30. × − N down;(d) down Electric force: 16 0. aN up;

Magnetic force: 48 0. aN downP29.4 (a) 86 7. fN ; (b) 51 9. Tm s2

P29.10 By = −2 62. mT; Bz = 0; Bx may have any

valueP29.6 (a) 7 90. pN; (b) 0

184 Magnetic Fields

P29.12 −2 88. je j N P29.50 (a) 37 7. mT ; (b) 4 29 1025 3. × m

P29.52 (a) 17.9 ns; (b) 35.1 eVP29.14 109 mA to the right

P29.54 39 2. mTP29.16

43

1 2IdBLm

FHG

IKJ

P29.56 (a) Bx is indeterminate. By = 0 ; BF

evzi

i=−

;

P29.18 Fab = 0; F ibc = −40 0. mNe j ;F kcd = −40 0. mNe j ; F i kda = +40 0. mNa fe j

(b) −Fi j ; (c) −Fi j

P29.58 128 mT north at an angle of 78.7° belowthe horizontal

P29.20 (a) 5 41. mA m2⋅ ; (b) 4 33. mN m⋅

P29.6034R

P29.22 (a)3 97. ° ; (b) 3 39. mN m⋅

P29.24 (a) 80 1. mN m⋅ ; (b) 104 mN m⋅ ; P29.62 B ~10 1− T; τ ~10 1− ⋅ N m; I ~1 A ;A ~10 3− m2 ; N ~103(c) 132 mN m⋅ ;

(d) The torque on the circle.

P29.64λ θg

Itan

P29.26 (a) minimum: pointing north at 48.0°below the horizontal; maximum: pointingsouth at 48.0° above the horizontal; P29.66 (a) 0 104. mm; (b) 0 189. mm(b) 1 07. Jµ

P29.68 3 82 10 25. × − kgP29.28 (a) 640 N mµ ⋅ ; (b) 241 mW; (c) 2.56 mJ;

(d) 154 mW P29.70 (a) see the solution;empirically, ∆V BH V T= 100 µb g ;

P29.30 1 98. cm(b) 0 125. mm

P29.32 65 6. mTP29.72 (a) v

qBhm

= ; The particle moves in a

semicircle of radius h and leaves the fieldwith velocity −vj;

P29.34 (a) 5 00. cm; (b) 8 78. Mm s

P29.36′=

mm

8 (b) The particle moves in a smaller

semicircle of radius mvqB

, attaining final

velocity −vj;P29.38 see the solution

P29.40 244 kV m (c) The particle moves in a circular arc of

radius rmvqB

= , leaving the field with

velocity v vsin cosθ θi j+ where

θ = FHGIKJ

−sin 1 hr

P29.42 278 mm

P29.44 162 mm

P29.46 3 00. T

P29.48 (a) 7 44 1028 3. × m ; (b) 1 79. T