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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
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