29
Magnetic Fields
CHAPTER OUTLINE
29.1
29.2
29.3
29.4
29.5
29.6
Magnetic Fields and Forces
Magnetic Force Acting on a
Current-Carrying Conductor
Torque on a Current Loop in
a Uniform Magnetic Field
Motion of a Charged Particle
in a Uniform Magnetic Field
Applications Involving
Charged Particles Moving in
a Magnetic Field
The Hall Effect
ANSWERS TO QUESTIONS
Q29.1
The force is in the +y direction. No, the proton will not
continue with constant velocity, but will move in a circular
path in the x-y plane. The magnetic force will always be
perpendicular to the magnetic field and also to the velocity of
the proton. As the velocity changes direction, the magnetic
force on the proton does too.
Q29.2
If they are projected in the same direction into the same
magnetic field, the charges are of opposite sign.
Q29.3
Not necessarily. If the magnetic field is parallel or antiparallel
to the velocity of the charged particle, then the particle will
experience 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 the
magnetic field is into the page, the electron goes clockwise and
the proton counterclockwise.
Q29.5
Send the particle through the uniform field and look at its path. If the path of the particle is
parabolic, then the field must be electric, as the electric field exerts a constant force on a charged
particle. If you shoot a proton through an electric field, it will feel a constant force in the same
direction as the electric field—it’s similar to throwing a ball through a gravitational field. If the path
of the particle is helical or circular, then the field is magnetic—see Question 29.1. If the path of the
particle is straight, then observe the speed of the particle. If the particle accelerates, then the field 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 field is magnetic—see Question 29.3.
Q29.6
Similarities: Both can alter the velocity of a charged particle moving through the field. Both exert
forces directly proportional to the charge of the particle feeling the force. Positive and negative
charges feel forces in opposite directions. Differences: The direction of the electric force is parallel or
antiparallel to the direction of the electric field, but the direction of the magnetic force is
perpendicular to the magnetic field and to the velocity of the charged particle. Electric forces can
accelerate a charged particle from rest or stop a moving particle, but magnetic forces cannot.
161
162
Q29.7
Magnetic Fields
a
f
Since FB = q v × B , then the acceleration produced by a magnetic field on a particle of mass m is
q
aB =
v × B . For the acceleration to change the speed, a component of the acceleration must be in
m
the direction of the velocity. The cross product tells us that the acceleration must be perpendicular to
the velocity, and thus can only change the direction of the velocity.
a
f
Q29.8
The magnetic field in a cyclotron essentially keeps the charged particle in the electric field for a
longer period of time, and thus experiencing a larger change in speed from the electric field, by
forcing it in a spiral path. Without the magnetic field, the particle would have to move in a straight
line 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 electron
beam 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,
F
then q = will tell you its charge. You cannot hook a spring balance to an electron. Measuring the
E
acceleration of small particles by observing their deflection in known electric and magnetic fields can
tell you the charge-to-mass ratio, but not separately the charge or mass. Both an acceleration
produced by an electric field and an acceleration caused by a magnetic field depend on the
q
properties of the particle only by being proportional to the ratio .
m
Q29.13
If the current loop feels a torque, it must be caused by a magnetic field. If the current loop feels no
torque, 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 is
low enough, the spiral will be tight enough that the particle will first hit some
matter as it follows a field line down into the atmosphere or to the surface at a
high 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 aligned
with 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 semicircular
arc. An electron would turn downward and move in a clockwise semicircular arc of smaller radius
than 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 will
take just the same time to travel farther with their higher speeds. As shown in Equation 29.15, the
time 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 might
be left by an uncharged particle that managed to leave a trail of bubbles, or it might be the
imperceptibly 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 proportional
to 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 may
increase the density of mobile charge carriers in it.
SOLUTIONS TO PROBLEMS
Section 29.1
P29.1
Magnetic Fields and Forces
(a)
up
(b)
out of the page, since the
charge is negative.
(c)
no deflection
(d)
into the page
FIG. P29.1
P29.2
At the equator, the Earth’s magnetic field is
horizontally north. Because an electron has
negative charge, F = qv × B 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 .
(a)
(c)
FIG. P29.2
(d)
164
P29.3
Magnetic Fields
e j
Therefore, B = B e− k j which indicates the
FB = qv × B ; FB − j = − e v i × B
negative z direction .
FIG. P29.3
P29.4
e
je
je
j
FB = qvB sin θ = 1.60 × 10 −19 C 3.00 × 10 6 m s 3.00 × 10 −1 T sin 37.0°
(a)
FB = 8.67 × 10 −14 N
a=
(b)
P29.5
F 8.67 × 10 −14 N
=
= 5.19 × 10 13 m s 2
m 1.67 × 10 −27 kg
e
je
j
F = ma = 1.67 × 10 −27 kg 2.00 × 10 13 m s 2 = 3.34 × 10 −14 N = qvB sin 90°
B=
F
3.34 × 10 −14 N
=
= 2.09 × 10 −2 T
−19
7
qv
1.60 × 10
C 1.00 × 10 m s
e
je
j
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.
P29.6
First find the speed of the electron.
∆K =
P29.7
P29.8
1
mv 2 = e∆V = ∆U :
2
v=
e
2 e∆ V
=
m
e
jb
2 1.60 × 10 −19 C 2 400 J C
e9.11 × 10
ja
je
−31
kg
j
FIG. P29.5
g = 2.90 × 10
7
ms
f
(a)
FB, max = qvB = 1.60 × 10 −19 C 2.90 × 10 7 m s 1.70 T = 7.90 × 10 −12 N
(b)
FB, min = 0 occurs when v is either parallel to or anti-parallel to B.
e
je
ja
f
FB = qvB sin θ
so
8.20 × 10 −13 N = 1.60 × 10 −19 C 4.00 × 10 6 m s 1.70 T sin θ
sin θ = 0.754
and
θ = sin −1 0.754 = 48.9° or 131° .
Gravitational force:
Electric force:
Magnetic force:
a
f
e
je
j
F = qE = e −1.60 × 10
C jb100 N C downg = 1.60 × 10
N up .
.
F = qv × B = e −1.60 × 10
C je6.00 × 10 m s E j × e50.0 × 10 N ⋅ s C ⋅ m N
j
Fg = mg = 9.11 × 10 −31 kg 9.80 m s 2 = 8.93 × 10 −30 N down .
e
B
−19
−17
−19
FB = −4.80 × 10 −17 N up = 4.80 × 10 −17 N down .
6
−6
Chapter 29
P29.9
FB = qv × B
i
j
k
a
f a f a f
v × B = +2 −4 +1 = 12 − 2 i + 1 + 6 j + 4 + 4 k = 10 i + 7 j + 8k
+1 +2 −3
v × B = 10 2 + 7 2 + 8 2 = 14.6 T ⋅ m s
jb
e
g
FB = q v × B = 1.60 × 10 −19 C 14.6 T ⋅ m s = 2.34 × 10 −18 N
P29.10
jb
e
g e
j
qE = −1.60 × 10 −19 C 20.0 N C k = −3.20 × 10 −18 N k
∑ F = qE + qv × B = ma
e−3.20 × 10
−e3.20 × 10
e1.92 × 10
j
e
j e
N jk − e1.92 × 10
C ⋅ m sji × B = e1.82 × 10
N jk
C ⋅ m sji × B = −e5.02 × 10
N jk
−18
−18
−15
je
j
N k − 1.60 × 10 −19 C 1.20 × 10 4 m s i × B = 9.11 × 10 −31 2.00 × 10 12 m s 2 k
−15
−18
−18
The magnetic field may have any x -component . Bz = 0 and By = −2.62 mT .
Section 29.2
P29.11
Magnetic Force Acting on a Current-Carrying Conductor
FB = ILB sin θ
with FB = Fg = mg
mg = ILB sin θ so
m
g = IB sin θ
L
I = 2.00 A
100 cm m
m
= 0.500 g cm
= 5.00 × 10 −2 kg m .
L
1 000 g kg
b
and
e5.00 × 10
Thus
−2
I
gFGH
JK
ja9.80f = a2.00fB sin 90.0°
B = 0.245 Tesla with the direction given by right-hand rule: eastward .
a
fa
f a
f
fa
fa
e−2.88jj N
P29.12
FB = IA × B = 2.40 A 0.750 m i × 1.60 T k =
P29.13
(a)
FB = ILB sin θ = 5.00 A 2.80 m 0.390 T sin 60.0° = 4.73 N
(b)
FB = 5.00 A 2.80 m 0.390 T sin 90.0° = 5.46 N
(c)
FB = 5.00 A 2.80 m 0.390 T sin 120° = 4.73 N
a
a
fa
fa
f
a
fa
fa
f
f
FIG. P29.11
165
166
P29.14
Magnetic Fields
FB
A
I=
=
mg I A × B
=
A
A
b
F
ge
j
0.040 0 kg m 9.80 m s 2
mg
=
= 0.109 A
BA
3.60 T
Bin
The direction of I in the bar is to the right .
FIG. P29.14
P29.15
a
f e j e j ej
+ K g + ∆E = b K
The work-energy theorem is bK
trasn
0 + 0 + Fs cos θ =
IdBL cos 0° =
v=
P29.16
trans
+ K rot
g
d
f
I
1
1
mv 2 + Iω 2
2
2
FG
H
IJ FG v IJ and IdBL = 3 mv
KH RK
4
4a 48.0 A fa0.120 mfa0.240 Tfa0.450 mf
=
3b0.720 kg g
2
1
1 1
mv 2 +
mR 2
2
2 2
4IdBL
=
3m
rot i
L
y
2
x
1.07 m s .
f e j e j ej
+ K g + ∆E = bK
The work-energy theorem is bK
z
FIG. P29.15
a
The rod feels force FB = I d × B = Id k × B − j = IdB i .
trans
0 + 0 + Fs cos θ =
IdBL cos 0° =
P29.17
B
The rod feels force FB = I d × B = Id k × B − j = IdB i .
rot i
trans
+ K rot
g
f
1
1
mv 2 + Iω 2
2
2
FG
H
1
1 1
mv 2 +
mR 2
2
2 2
IJ FG v IJ
KH RK
2
and v =
4IdBL
.
3m
The magnetic force on each bit of ring is
Ids × B = IdsB radially inward and upward, at
angle θ above the radial line. The radially
inward components tend to squeeze the ring
but all cancel out as forces. The upward
components IdsB sin θ all add to
I 2π rB sin θ up .
FIG. P29.17
Chapter 29
P29.18
For each segment, I = 5.00 A and B = 0.020 0 N A ⋅ m j .
Segment
P29.19
a f
FB = I A × B
A
ab
−0.400 m j
bc
0.400 m k
cd
−0.400 m i + 0.400 m j
da
0.400 m i − 0.400 m k
0
a40.0 mNfe− ij
a40.0 mNfe−k j
a40.0 mNfek + ij
FIG. P29.18
Take the x-axis east, the y-axis up, and the z-axis south. The field is
b
g
e j b
g
e j
B = 52.0 µT cos 60.0° − k + 52.0 µT sin 60.0° − j .
e j
ej
The current then has equivalent length: L ′ = 1.40 m − k + 0.850 m j
b
ge
j e
j
FB = IL ′ × B = 0.035 0 A 0.850 j − 1.40k m × −45.0 j − 26.0 k 10 −6 T
e
j
e j
FB = 3.50 × 10 −8 N −22.1i − 63.0 i = 2.98 × 10 −6 N − i = 2.98 µN west
Section 29.3
P29.20
(a)
Torque on a Current Loop in a Uniform Magnetic Field
2π r = 2.00 m
so
r = 0.318 m
j a
f
ja
f
e
µ = IA = 17.0 × 10 −3 A π 0.318
(b)
m 2 = 5.41 mA ⋅ m 2
τ = µ ×B
so
P29.21
2
e
τ = 5.41 × 10 −3 A ⋅ m 2 0.800 T = 4.33 mN ⋅ m
a
f
τ = µB sin θ so 4.60 × 10 −3 N ⋅ m = µ 0.250 sin 90.0°
µ = 1.84 × 10 −2 A ⋅ m 2 = 18.4 mA ⋅ m 2
FIG. P29.19
.
167
168
P29.22
Magnetic Fields
Let θ represent the unknown angle; L, the total length of the wire; and d, the length of one
side of the square coil. Then, using the definition of magnetic moment and the right-hand
rule in Figure 29.15, we find
(a)
FG L IJ d I at angle θ with the horizontal.
H 4d K
∑ τ = bµ × Bg − br × mgg = 0
FG ILBd IJ sina90.0°−θ f − FG mgd IJ sinθ = 0
H 4 K
H 2K
FG mgd IJ sinθ = FG ILBd IJ cos θ
H 4 K
H 2K
F ILB I = tan FG a3.40 Afa4.00 mfb0.010 0 Tg IJ =
θ = tan G
GH 2b0.100 kg ge9.80 m s j JK
H 2mg JK
µ = NAI :
2
µ=
At equilibrium,
and
−1
−1
2
τm =
(b)
P29.23
FG ILBd IJ cosθ = 1 a3.40 Afa4.00 mfb0.010 0 Tga0.100 mf cos 3.97° =
H 4 K
4
τ = NBAI sin φ
a
ja
fe
3.39 mN ⋅ m
f
τ = 100 0.800 T 0.400 × 0.300 m 2 1.20 A sin 60°
τ = 9.98 N ⋅ m
Note that φ is the angle between the magnetic
moment and the B field. The loop will rotate so as
to align the magnetic moment with the B field.
Looking down along the y-axis, the loop will rotate
in a clockwise direction.
P29.24
FIG. P29.23
From τ = µ × B = IA × B , the magnitude of the torque is IAB sin 90.0°.
(a)
Each side of the triangle is
40.0 cm
.
3
Its altitude is 13.3 2 − 6.67 2 cm = 11.5 cm and its area is
1
A = 11.5 cm 13.3 cm = 7.70 × 10 −3 m 2 .
2
Then τ = 20.0 A 7.70 × 10 −3 m 2 0.520 N ⋅ s C ⋅ m = 80.1 mN ⋅ m .
a
fa
a
(b)
f
fe
jb
Each side of the square is 10.0 cm and its area is 100 cm 2 = 10 −2 m 2 .
a
fe
ja
f
τ = 20.0 A 10 −2 m 2 0.520 T = 0.104 N ⋅ m
(c)
0.400 m
= 0.063 7 m
2π
A = π r 2 = 1.27 × 10 −2 m 2
r=
a
fe
ja
f
τ = 20.0 A 1.27 × 10 −2 m 2 0.520 = 0.132 N ⋅ m
(d)
g
The circular loop experiences the largest torque.
3.97° .
Chapter 29
P29.25
Choose U = 0 when the dipole moment is at θ = 90.0° to the field. The field exerts torque of magnitude
µB sin θ on the dipole, tending to turn the dipole moment in the direction of decreasing θ. According
to Equations 8.16 and 10.22, the potential energy of the dipole-field system is given by
U −0 =
z
θ
a
µB sin θ dθ = µB − cos θ
f
90 .0 °
P29.26
169
(a)
θ
90 .0 °
= − µB cos θ + 0
U = −µ ⋅ B .
or
The field exerts torque on the needle tending to align it with the field, so the minimum
energy orientation of the needle is:
pointing north at 48.0° below the horizontal
e
je
j
where its energy is U min = − µB cos 0° = − 9.70 × 10 −3 A ⋅ m 2 55.0 × 10 −6 T = −5.34 × 10 −7 J .
It has maximum energy when pointing in the opposite direction,
south at 48.0° above the horizontal
e
je
j
where its energy is U max = − µB cos 180° = + 9.70 × 10 −3 A ⋅ m 2 55.0 × 10 −6 T = +5.34 × 10 −7 J .
P29.27
e
j
(b)
U min + W = U max : W = U max − U min = +5.34 × 10 −7 J − −5.34 × 10 −7 J = 1.07 µJ
(a)
τ = µ × B,
τ = µ × B = µB sin θ = NIAB sin θ
so
a
f b
τ max = NIAB sin 90.0° = 1 5.00 A π 0.050 0 m
(b)
g e3.00 × 10 Tj =
2
−3
118 µN ⋅ m
U = − µ ⋅ B, so − µB ≤ U ≤ + µB
a f
a
f b
Since µB = NIA B = 1 5.00 A π 0.050 0 m
g e3.00 × 10 Tj = 118 µJ ,
2
−3
the range of the potential energy is: −118 µJ ≤ U ≤ +118 µJ .
*P29.28
(a)
τ = µ × B = NIAB sin θ
e
ja
fb
g
F 2π rad IJ FG 1 min IJ =
N ⋅ mb3 600 rev mingG
H 1 rev K H 60 s K
τ max = 80 10 −2 A 0.025 m ⋅ 0.04 m 0.8 N A ⋅ m sin 90° = 6.40 × 10 −4 N ⋅ m
(b)
Pmax = τ maxω = 6.40 × 10 −4
(c)
In one half revolution the work is
b
g
= 2 NIAB = 2e6.40 × 10 N ⋅ mj = 1.28 × 10 J
In one full revolution, W = 2e1.28 × 10 Jj = 2.56 × 10
W = U max − U min = − µB cos 180°− − µB cos 0° = 2 µB
−4
−3
−3
(d)
Pavg =
W 2.56 × 10 −3 J
=
= 0.154 W
∆t
1 60 s
b g
The peak power in (b) is greater by the factor
π
.
2
−3
J .
0.241 W
170
Magnetic Fields
Section 29.4
P29.29
Motion of a Charged Particle in a Uniform Magnetic Field
B = 50.0 × 10 −6 T; v = 6.20 × 10 6 m s
(a)
Direction is given by the right-hand-rule: southward
FB = qvB sin θ
e
je
je
j
FB = 1.60 × 10 −19 C 6.20 × 10 6 m s 50.0 × 10 −6 T sin 90.0°
FIG. P29.29
= 4.96 × 10 −17 N
e
(b)
P29.30
je
1.67 × 10 −27 kg 6.20 × 10 6 m s
mv 2
mv 2
=
F=
so r =
r
F
4.96 × 10 −17 N
a f
1
mv 2 = q ∆V
2
P29.31
2
= 1.29 km
ja
1
3.20 × 10 −26 kg v 2 = 1.60 × 10 −19 C 833 V
2
e
j
e
The magnetic force provides the centripetal force: qvB sin θ =
r=
j
e
e
je
jb
f
v = 91.3 km s
mv 2
r
j
g
3.20 × 10 −26 kg 9.13 × 10 4 m s
mv
=
= 1.98 cm .
qB sin 90.0° 1.60 × 10 −19 C 0.920 N ⋅ s C ⋅ m
For each electron, q vB sin 90.0° =
mv 2
eBr
.
and v =
r
m
The electrons have no internal structure to absorb energy, so the collision must be perfectly elastic:
K=
1
1
1
mv12i + 0 = mv12 f + mv 22 f
2
2
2
K=
e 2 B 2 R 22
e 2 B 2 R12
e2B2 2
1
1
R1 + R 22
m
+
=
m
2
2
2m
m2
m2
K=
P29.32
F
GH
e
I
JK
F
GH
jb
I
JK
e
g b0.010 0 mg + b0.024 0 mg
e 1.60 × 10 −19 C 0.044 0 N ⋅ s C ⋅ m
e
2 9.11 × 10
We begin with qvB =
−31
kg
j
2
2
2
qRB
mv 2
, so v =
.
R
m
The time to complete one revolution is T =
Solving for B, B =
j
2π m
= 6.56 × 10 −2 T .
qT
2π R
2π R
2π m
=
=
.
v
qRB m
qB
= 115 keV
Chapter 29
P29.33
a f
q ∆V =
1
mv 2
2
Also, qvB =
mv 2
r
or
v=
so
r=
a f
2 q ∆V
.
m
rp2 =
Therefore,
rα2
(a)
(b)
a f
2 m p ∆V
eB 2
2 m d ∆V
p
2
d
p
α
2
α
or
qRB = mv .
But
L = mvR = qR 2 B .
Therefore,
R=
Thus,
v=
ja
e
e1.60 × 10
2
p
−19
je
C 1.00 × 10 −3 T
j
= 0.050 0 m = 5.00 cm .
L
4.00 × 10 −25 J ⋅ s
=
= 8.78 × 10 6 m s .
−31
mR
9.11 × 10
kg 0.050 0 m
jb
e
g
f
P29.36
1
mv 2 = q ∆V
2
so
v=
so
r=
a f
2 q ∆V
m
a f
m 2 q ∆V m
qB
a f
r2 =
m 2 ∆V
⋅
q
B2
and
ar ′ f
m=
qB 2 r 2
2 ∆V
and
am ′f = bq′2gBa∆Varf′f
a f
2
4.00 × 10 −25 J ⋅ s
L
=
qB
P29.35
a f
2
p
mv 2
R
qvB =
mv
qB
2
p
2
−19
C 5.20 T
qB 1.60 × 10
ω=
=
= 4.98 × 10 8 rad s
m
1.67 × 10 −27 kg
r=
p
2
We begin with
a f
qB 2
rα = rd = 2 rp .
The conclusion is:
P29.34
a f.
2 m ∆V
a f = 2e2m ja∆V f = 2FG 2m a∆V f IJ = 2r
q B
eB
H eB K
F 2 m a ∆V f I = 2 r .
2m a ∆V f 2e 4m ja ∆V f
=
=
= 2G
q B
a2efB
H eB JK
rd2 =
and
a f
2 q ∆V
=
m
mv m
=
qB qB
2
=
m ′ 2 ∆V
⋅
q′
B2
2
2
a f FG IJ FG 2R IJ
H KH R K
2
so
m ′ q′ r ′
2e
= ⋅ 2 =
m
q r
e
2
= 8
171
172
P29.37
Magnetic Fields
E=
and
1
mv 2 = e∆V
2
evB sin 90° =
B=
mv m
=
eR eR
mv 2
R
2 e∆ V 1
=
m
R
1
B=
5.80 × 10 10 m
*P29.38
(a)
e
2m∆V
e
je
2 1.67 × 10 −27 kg 10.0 × 10 6 V
1.60 × 10
−19
C
j=
7.88 × 10 −12 T
v
At the moment shown in Figure 29.21, 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)
B
FIG. P29.38(a)
After the particle has passed the middle of the bottle and
moves into the region of increasing magnetic field, 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.
v
B
F
FIG. P29.38(b)
(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
q
plane and therefore a magnetic dipole moment I A = A
T
in the –x direction. It is like a little bar magnet with its N
pole on the left.
(e)
Problem 17 showed that a nonuniform magnetic field
exerts a net force on a magnetic dipole. When the dipole is
aligned opposite to the external field, the force pushes it
out of the region of stronger field. Here it is to the left, a
force of repulsion of one magnetic south pole on another
south pole.
+
N
FIG. P29.38(d)
B
N
S
S
FIG. P29.38(e)
S
Chapter 29
P29.39
e
ja
je
7.94 × 10 −3 m 1.60 × 10 −19 C 1.80 T
rqB
=
m=
v
4.60 × 10 5 m s
mv
so
r=
qB
m = 4.97 × 10 −27 kg
F 1u
GH 1.66 × 10
−27
I=
J
kg K
The particle is singly ionized: either a tritium ion,
Section 29.5
P29.40
2.99 u
+
3
1H
, or a helium ion,
+
3
2 He
.
Applications Involving Charged Particles Moving in a Magnetic Field
FB = Fe
so
qvB = qE
where
v=
2K
and K is kinetic energy of the electron.
m
E = vB =
P29.41
f
K=
2K
B=
m
a f
so
mv 2
r
r=
1
mv 2 = q ∆V
2
FB = qv × B =
9.11 × 10
j
(b)
r235 = 8. 23 cm
1.60 × 10
m 238
=
m 235
−19
j b0.015 0g =
244 kV m
a f
a f
2 q ∆V m 1
=
B
B
mv m
=
qB q
e
r238 =
−31
2 q ∆V
m
v=
2 238 × 1.66 × 10 −27 2 000
(a)
r238
=
r235
a fe
2 750 1.60 × 10 −19
a f
2m ∆V
q
FG 1 IJ = 8.28 × 10
H 1.20 K
−2
m = 8.28 cm
238.05
= 1.006 4
235.04
The ratios of the orbit radius for different ions are independent of ∆V and B.
P29.42
E 2 500 V m
=
= 7.14 × 10 4 m s .
B
0.035 0 T
In the velocity selector:
v=
In the deflection chamber:
2.18 × 10 −26 kg 7.14 × 10 4 m s
mv
r=
=
= 0.278 m .
qB
1.60 × 10 −19 C 0.035 0 T
e
e
je
jb
g
j
173
174
P29.43
Magnetic Fields
(a)
FB = qvB =
mv 2
R
ja
e
f
−19
C 0.450 T
v qBR qB 1.60 × 10
ω= =
=
=
= 4.31 × 10 7 rad s
−27
R mR
m
kg
1.67 × 10
(b)
P29.44
K=
v=
ja
e
fa
f
−19
C 0.450 T 1.20 m
qBR 1.60 × 10
=
= 5.17 × 10 7 m s
m
1.67 × 10 −27 kg
1
mv 2 :
2
j 12 e1.67 × 10 kg jv
kg je8.07 × 10 m sj
mv e1.67 × 10
r=
=
= 0.162 m
qB
e1.60 × 10 Cja5.20 Tf
e34.0 × 10
je
eV 1.60 × 10 −19 J eV =
6
−27
7
v = 8.07 × 10 m s
*P29.45
−27
2
7
−19
Note that the “cyclotron frequency” is an angular speed. The motion of the proton is described by
∑ F = ma :
q vB sin 90° =
qB=m
(a)
(b)
(c)
(d)
mv 2
r
v
= mω
r
e1.60 × 10 Cjb0.8 N ⋅ s C ⋅ mg FG kg ⋅ m IJ = 7.66 × 10 rad s
ω=
=
H N ⋅s K
m
e1.67 × 10 kgj
F 1 IJ = 2.68 × 10 m s
v = ω r = e7.66 × 10 rad sja0.350 mfG
H 1 rad K
F 1 eV IJ = 3.76 × 10
1
1
K = mv = e1.67 × 10
kg je 2.68 × 10 m sj G
2
2
H 1.6 × 10 J K
−19
qB
7
7
−27
2
7
2
−27
7
2
−19
a
P29.46
θ =ωt
FB = qvB =
B=
eV
The proton gains 600 eV twice during each revolution, so the number of revolutions is
3.76 × 10 6 eV
= 3.13 × 10 3 revolutions .
2 600 eV
(e)
6
t=
FG
H
mv 2
r
jb
IJ
K
θ
3.13 × 10 3 rev 2π rad
= 2.57 × 10 −4 s
=
ω 7.66 × 10 7 rad s 1 rev
4.80 × 10 −16 kg ⋅ m s
mv
=
= 3.00 T
qr
1.60 × 10 −19 C 1 000 m
e
f
g
Chapter 29
P29.47
θ = tan −1
FG 25.0 IJ = 68.2°
H 10.0 K
R=
and
1.00 cm
= 1.08 cm .
sin 68.2°
Ignoring relativistic correction, the kinetic energy of the electrons is
1
mv 2 = q∆V
2
From Newton’s second law
e
2 q ∆V
= 1.33 × 10 8 m s .
m
v=
so
mv 2
= qvB , we find the magnetic field
R
je
je
FIG. P29.47
j
9.11 × 10 −31 kg 1.33 × 10 8 m s
mv
B=
=
= 70.1 mT .
qR
1.60 × 10 −19 C 1.08 × 10 −2 m
Section 29.6
P29.48
e
The Hall Effect
1
nq
(a)
RH ≡
(b)
∆VH =
B=
P29.49
j
so
n=
1
1
=
= 7.44 × 10 28 m −3
−
19
qRH
C 0.840 × 10 −10 m3 C
1.60 × 10
e
je
j
IB
nqt
b
nqt ∆VH
I
g = e7.44 × 10
28
je
je
je
m −3 1.60 × 10 −19 C 0.200 × 10 −3 m 15.0 × 10 −6 V
20.0 A
j=
1.79 T
IB
, and given that I = 50.0 A , B = 1.30 T , and t = 0.330 mm, the number of charge
nqt
carriers per unit volume is
Since ∆VH =
n=
IB
= 1.28 × 10 29 m −3
e ∆VH t
b
g
The number density of atoms we compute from the density:
n0 =
F
GH
8.92 g 1 mole
cm 3 63.5 g
I F 6.02 × 10 atoms I F 10 cm I = 8.46 × 10
JK GH mole JK GH 1 m JK
23
3
6
3
So the number of conduction electrons per atom is
n 1.28 × 10 29
=
= 1.52
n 0 8.46 × 10 28
28
atom m3
175
176
P29.50
Magnetic Fields
∆VH =
(a)
IB
nqt
nqt
0.080 0 T
B
=
=
= 1.14 × 10 5 T V .
∆VH 0.700 × 10 −6 V
I
so
B=
Then, the unknown field is
e
je
FG nqt IJ b∆V g
HIK
H
j
B = 1.14 × 10 5 T V 0.330 × 10 −6 V = 0.037 7 T = 37.7 mT .
nqt
= 1.14 × 10 5 T V
I
(b)
e
n = 1.14 × 10 5 T V
P29.51
B=
b
nqt ∆VH
I
g = e8.49 × 10
e
n = 1.14 × 10 5 T V
so
j e1.60 × 10
0.120 A
−19
je
C 2.00 × 10
je
−3
j
m
j qtI
= 4.29 × 10 25 m −3 .
je
je
m −3 1.60 × 10 −19 C 5.00 × 10 −3 m 5.10 × 10 −12 V
28
j
8.00 A
B = 4.33 × 10 −5 T = 43.3 µT
Additional Problems
P29.52
(a)
The boundary between a region of strong magnetic field and a
region of zero field cannot be perfectly sharp, but we ignore the
thickness of the transition zone. In the field the electron moves on
an arc of a circle:
∑ F = ma :
q vB sin 90° =
mv 2
r
e
je
j
−19
C 10 −3 N ⋅ s C ⋅ m
q B 1.60 × 10
v
=ω =
=
= 1.76 × 10 8 rad s
r
m
9.11 × 10 −31 kg
e
j
FIG. P29.52(a)
The time for one half revolution is,
from
∆ θ = ω∆ t
∆t =
(b)
∆θ
ω
=
π rad
= 1.79 × 10 −8 s .
1.76 × 10 8 rad s
The maximum depth of penetration is the radius of the path.
Then
ja
e
f
v = ω r = 1.76 × 10 8 s −1 0.02 m = 3.51 × 10 6 m s
and
K=
1
1
mv 2 = 9.11 × 10 −31 kg 3.51 × 10 6 m s
2
2
e
= 35.1 eV .
je
j
2
= 5.62 × 10 −18 J =
5.62 × 10 −18 J ⋅ e
1.60 × 10 −19 C
Chapter 29
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 .
a
f
I h × B in the direction of L.
(b)
The sodium, consisting of ions and electrons, flows along the
pipe transporting no net charge. But inside the section of
length L, electrons drift upward to constitute downward
electric current J × area = J Lw .
FIG. P29.53
a f
The current then feels a magnetic force I h × B = JLwhB sin 90° .
This force along the pipe axis will make the fluid move, exerting pressure
F
JLwhB
=
= JLB .
area
hw
P29.54
∑ Fy = 0 :
+n − mg = 0
∑ Fx = 0 :
− µ k n + IB sin 90.0° = 0
B=
P29.55
b
a
ge
j
2
µ k mg 0.100 0.200 kg 9.80 m s
=
= 39.2 mT
10.0 A 0.500 m
Id
fa
f
The magnetic force on each proton, FB = qv × B = qvB sin 90° downward
perpendicular to velocity, causes centripetal acceleration, guiding it into a
circular path of radius r, with
mv 2
r
mv
.
r=
qB
qvB =
and
We compute this radius by first finding the proton’s speed:
Now,
(b)
K=
1
mv 2
2
v=
2K
=
m
FIG. P29.55
e
je
2 5.00 × 10 6 eV 1.60 × 10 −19 J eV
1.67 × 10
e
−27
je
kg
j
j = 3.10 × 10
7
m s.
1.67 × 10 −27 kg 3.10 × 10 7 m s
mv
=
= 6.46 m .
r=
qB
1.60 × 10 −19 C 0.050 0 N ⋅ s C ⋅ m
e
jb
g
From the figure, observe that
1.00 m
1m
=
r
6.46 m
α = 8.90°
sin α =
(a)
The magnitude of the proton momentum stays constant, and its final y component is
e
je
j
− 1.67 × 10 −27 kg 3.10 × 10 7 m s sin 8.90° = −8.00 × 10 −21 kg ⋅ m s .
177
178
P29.56
Magnetic Fields
e j e
j
If B = Bx i + By j + Bz k , FB = qv × B = e vi i × Bx i + By j + Bz k = 0 + evi B y k − evi Bz j .
(a)
Since the force actually experienced is FB = Fi j , observe that
B x could have any value , B y = 0 , and B z = −
Fi
.
evi
IJ
e j FGH
K
F
F I
= qv × B = − ee v i j × G B i + 0 j −
kJ =
ev K
H
(b)
If v = −vi i , then
F
FB = qv × B = e − vi i × Bx i + 0 j − i k = − Fi j .
evi
(c)
If q = − e and v = vi i , then
FB
i
x
i
i
− Fi j .
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
a
f
j e4i − 1j − 2k j + e2 i + 3 j − 1k j × e2 i + 4j + 1k j N
F = qE + qv × B = q E + v × B
e
F = 3.20 × 10 −19
Carrying out the indicated operations, we find:
e3.52i − 1.60jj × 10 N .
3.52
FG F IJ = cos FG
H FK
GH a3.52f + a1.60f
−18
F=
θ = cos
(b)
P29.58
−1
−1
x
2
2
I
JJ =
K
24.4°
A key to solving this problem is that reducing the normal force will reduce
F
the friction force: FB = BIL or B = B .
IL
When the wire is just able to move,
∑ Fy = n + FB cos θ − mg = 0
so
n = mg − FB cos θ
and
f = µ mg − FB cos θ .
Also,
∑ Fx = FB sin θ − f = 0
so FB sin θ = f :
FB sin θ = µ mg − FB cos θ and FB =
We minimize B by minimizing FB :
dFB
cos θ − µ sin θ
= µmg
= 0 ⇒ µ sin θ = cos θ .
2
dθ
sin θ + µ cos θ
b
b
b gb
g
FIG. P29.58
g
g
FG 1 IJ = tan a5.00f = 78.7° for the smallest field, and
H µK
F
F µg I bm Lg
=G J
B=
IL H I K sin θ + µ cos θ
L a0.200fe9.80 m s j OP
0.100 kg m
=M
B
MN 1.50 A PQ sin 78.7°+a0.200f cos 78.7° = 0.128 T
Thus, θ = tan −1
−1
B
2
min
Bmin = 0.128 T pointing north at an angle of 78.7° below the horizontal
µmg
.
sin θ + µ cos θ
Chapter 29
*P29.59
179
The electrons are all fired from the electron gun with the same speed v in
Ui = K f
qV =
a− efa−∆V f = 12 m v
1
mv 2
2
e
2
v=
2 e∆V
me
For φ small, cos φ is nearly equal to 1. The time T of passage of each electron in the chamber is given
by
d = vT
T=d
FG m IJ
H 2 e∆ V K
e
12
Each electron moves in a different helix, around a different axis. If each completes just one
revolution within the chamber, it will be in the right place to pass through the exit port. Its
transverse velocity component v ⊥ = v sin φ swings around according to F⊥ = ma ⊥
qv ⊥ B sin 90° =
Then
*P29.60
mv ⊥2
r
FG IJ
H K
2π m e
B
e
eB =
12
=
me v⊥
2π
= m eω = m e
r
T
d
a2 ∆V f
12
B=
FG
H
2π 2m e ∆V
d
e
T=
IJ
K
FG
H
m e 2π
me
=d
2 e∆ V
eB
IJ
K
12
12
.
Let vi represent the original speed of the alpha particle. Let vα and v p represent the particles’
speeds after the collision. We have conservation of momentum 4m p vi = 4m p vα + m p v p and the
relative velocity equation vi − 0 = v p − vα . Eliminating vi ,
4 v p − 4 v α = 4 vα + v p
3 v p = 8 vα
vα =
3
vp .
8
For the proton’s motion in the magnetic field,
∑ F = ma
ev p B sin 90° =
m p v p2
eBR
= vp .
mp
R
For the alpha particle,
2 evα B sin 90° =
P29.61
4m p vα2
rα =
rα
2 m p vα
rα =
eB
2m p 3
2m p 3 eBR
3
vp =
R .
=
eB 8
4
eB 8 m p
Let ∆x 1 be the elongation due to the weight of the wire and let ∆x 2
be the additional elongation of the springs when the magnetic field
is turned on. Then Fmagnetic = 2 k∆x 2 where k is the force constant of
mg
. (The factor 2 is
2 ∆x 1
included in the two previous equations since there are 2 springs in
parallel.) Combining these two equations, we find
the spring and can be determined from k =
Fmagnetic = 2
FG mg IJ ∆x
H 2 ∆x K
1
2
=
mg∆x 2
; but FB = I L × B = ILB .
∆x 1
a
fa fe
a fb ge
FIG. P29.61
j
0.100 9.80 3.00 × 10 −3
mg∆x 2
24.0 V
Therefore, where I =
= 2.00 A , B =
=
= 0.588 T .
IL∆x1
12.0 Ω
2.00 0.050 0 5.00 × 10 −3
j
180
P29.62
Magnetic Fields
Suppose the input power is
a
f
I ~ 1 A = 10 0 A .
120 W = 120 V I :
FG 1 min IJ FG 2π rad IJ ~ 200 rad s
H 60 s K H 1 rev K
20 W = τω = τ b 200 rad sg τ ~ 10 N ⋅ m .
A ~ 10 m .
a3 cmf × a4 cmf, or
ω = 2 000 rev min
Suppose
−1
and the output power is
−3
Suppose the area is about
2
B ~ 10 −1 T .
Suppose that the field is
Then, the number of turns in the coil may be found from τ ≅ NIAB :
b
ge
je
j
0.1 N ⋅ m ~ N 1 C s 10 −3 m 2 10 −1 N ⋅ s C ⋅ m
N ~ 10
giving
*P29.63
3
.
The sphere is in translational equilibrium, thus
fs − Mg sin θ = 0 .
G
µ
(1)
θ
G
B
The sphere is in rotational equilibrium. If torques are taken about the
center of the sphere, the magnetic field produces a clockwise torque of
magnitude µB sin θ , and the frictional force a counterclockwise torque
of magnitude fs R , where R is the radius of the sphere. Thus:
fs R − µB sin θ = 0 .
(2)
From (1): fs = Mg sin θ . Substituting this in (2) and canceling out sin θ ,
one obtains
µB = MgR .
b
ge
j
f
(3)
fs
I
θ
Mg
FIG. P29.63
0.08 kg 9.80 m s 2
Mg
=
= 0.713 A . The current must be
π NBR π 5 0.350 T 0.2 m
counterclockwise as seen from above.
Now µ = NIπ R 2 . Thus (3) gives I =
P29.64
fa
Call the length of the rod L and the tension in each wire alone
∑ Fx = T sin θ − ILB sin 90.0° = 0
∑ Fy = T cos θ − mg = 0 ,
tan θ =
P29.65
a fa
ILB
IB
=
mg
mL g
b g
∑ F = ma or qvB sin 90.0° =
or
T sin θ = ILB
or
T cos θ = mg
or
B=
mv 2
r
∴ the angular frequency for each ion is
∆f = f12 − f14 =
FG
H
IJ e
K e
bm Lgg tanθ =
I
qB
v
=ω =
= 2π f and
r
m
ja
λg
tan θ
I
fF 1 − 1 I
G
J
j H 12.0 u 14.0 u K
1.60 × 10 −19 C 2.40 T
qB 1
1
=
−
2π m12 m14
2π 1.66 × 10 −27 kg u
∆f = f12 − f14 = 4.38 × 10 5 s −1 = 438 kHz
T
. Then, at equilibrium:
2
Chapter 29
P29.66
Let v x and v ⊥ be the components of the velocity of the positron parallel
to and perpendicular to the direction of the magnetic field.
(a)
The pitch of trajectory is the distance moved along x by the
positron during each period, T (see Equation 29.15)
fFGH 2πBqm IJK
e5.00 × 10 jacos 85.0°fa2π fe9.11 × 10 j =
p=
0.150e1.60 × 10 j
a
p = v x T = v cos 85.0°
−19
(b)
FIG. P29.66
−31
6
From Equation 29.13,
1.04 × 10 −4 m
r=
mv ⊥ mv sin 85.0°
=
Bq
Bq
r=
e9.11 × 10 je5.00 × 10 jasin 85.0°f =
a0.150fe1.60 × 10 j
−31
P29.67
6
−19
τ = IAB where the effective current due to the orbiting electrons is
I=
∆q q
=
∆t T
and the period of the motion is
T=
2π R
.
v
The electron’s speed in its orbit is found by requiring
keq2
R
2
=
mv 2
or
R
Substituting this expression for v into the equation for T, we find
e9.11 × 10 je5.29 × 10 j = 1.52 × 10 s .
T = 2π
e1.60 × 10 j e8.99 × 10 j
F q I 1.60 × 10 π e5.29 × 10 j a0.400f =
Therefore, τ = G J AB =
H T K 1.52 × 10
mR 3
q2 ke
9
−19
−11 2
−16
Use the equation for cyclotron frequency ω =
e1.60 × 10 Cje5.00 × 10 Tj =
a2π fe5.00 rev 1.50 × 10 sj
−19
T = 2π
ke
.
mR
−16
−19 2
m=
v=q
−11 3
−31
P29.68
1.89 × 10 −4 m
qB
qB
qB
=
or m =
ω 2π f
m
−2
−3
3.70 × 10 −24 N ⋅ m .
3.82 × 10 −25 kg .
181
182
P29.69
Magnetic Fields
(a)
K=
1
mv 2 = 6.00 MeV = 6.00 × 10 6 eV 1.60 × 10 −19 J eV
2
e
je
j
K = 9.60 × 10 −13 J
e
2 9.60 × 10
v=
1.67 × 10
FB = qvB =
R=
mv
=
qB
−13
−27
J
j = 3.39 × 10
kg
θ' x x x x x
B in = 1.00 T
x x x x x
45° x x x x x
45° x x x x x
R
x x x x x
x x x x x
45.0° x x x x x
x
7
ms
v
2
mv
so
R
1.67 × 10 −27 kg 3.39 × 10 7 m s
e
e1.60 × 10
je
−19
ja
C 1.00 T
f
FIG. P29.69
j = 0.354 m
a
f
Then, from the diagram, x = 2 R sin 45.0° = 2 0.354 m sin 45.0° = 0.501 m
P29.70
(b)
From the diagram, observe that θ ′ = 45.0° .
(a)
See graph to the right. The
Hall voltage is directly
proportional to the magnetic
field. A least-square fit to the
data gives the equation of the
best fitting line as:
120
100
80
∆V H (µV)
60
40
20
e
∆VH = 1.00 × 10
−4
j
V TB .
0
0
0.2
0.4
0.6
0.8
1.0
1.2
B (T)
(b)
Comparing the equation of
the line which fits the data
best to
∆VH =
FIG. P29.70
F 1 IB
GH nqt JK
observe that:
I
I
= 1.00 × 10 −4 V T, or t =
.
nqt
nq 1.00 × 10 −4 V T
e
j
Then, if I = 0.200 A , q = 1.60 × 10 −19 C , and n = 1.00 × 10 26 m −3 , the thickness of the sample is
t=
e1.00 × 10
0.200 A
26
m
−3
je1.60 × 10
−19
je
C 1.00 × 10 −4 V T
j
= 1.25 × 10 −4 m = 0.125 mm .
Chapter 29
P29.71
(a)
183
The magnetic force acting on ions in the blood stream will
deflect positive charges toward point A and negative
charges toward point B. This separation of charges
produces an electric field directed from A toward B. At
equilibrium, the electric force caused by this field must
balance the magnetic force, so
FG ∆V IJ
HdK
e160 × 10 Vj
∆V
=
v=
Bd b0.040 0 Tge3.00 × 10
qvB = qE = q
FIG. P29.71
−6
or
−3
j
m
= 1.33 m s .
No . Negative ions moving in the direction of v would be deflected toward point B, giving
(b)
A a higher potential than B. Positive ions moving in the direction of v would be deflected
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.
P29.72
When in the field, the particles follow a circular path
mv 2
mv
, so the radius of the path is: r =
according to qvB =
r
qB
(a)
qBh
mv
, that is, when v =
, the
qB
m
particle will cross the band of field. It will move in
a full semicircle of radius h, leaving the field at
2 h, 0 , 0 with velocity v = − vj .
When r = h =
b
(b)
g
f
FIG. P29.72
qBh
mv
, the particle will move in a smaller semicircle of radius r =
< h . It will
m
qB
leave the field at 2r, 0 , 0 with velocity v f = − vj .
When v <
b
(c)
v i = vj
g
qBh
mv
, the particle moves in a circular arc of radius r =
> h , centered at
qB
m
h
r, 0 , 0 . The arc subtends an angle given by θ = sin −1
. It will leave the field at the point
r
with coordinates r 1 − cos θ , h, 0 with velocity v = v sin θ i + v cos θ j .
When v >
b
FG IJ
HK
g
a
f
f
ANSWERS TO EVEN PROBLEMS
P29.2
(a) west; (b) no deflection; (c) up;
(d) down
P29.4
(a) 86.7 fN ; (b) 51.9 Tm s 2
P29.6
(a) 7.90 pN ; (b) 0
P29.8
Gravitational force: 8.93 × 10 −30 N down;
Electric force: 16.0 aN up ;
Magnetic force: 48.0 aN down
P29.10
By = −2.62 mT ; Bz = 0; Bx may have any
value
184
Magnetic Fields
P29.12
e−2.88jj N
P29.14
109 mA to the right
P29.16
FG 4IdBL IJ
H 3m K
P29.18
12
e j
= 40.0 mN e − k j ; F = a 40.0 mN fe i + k j
P29.50
(a) 37.7 mT ; (b) 4.29 × 10 25 m3
P29.52
(a) 17.9 ns; (b) 35.1 eV
P29.54
39.2 mT
P29.56
(a) Bx is indeterminate. By = 0 ; Bz =
Fab = 0; Fbc = 40.0 mN − i ;
Fcd
da
(b) −Fi j ; (c) −Fi j
P29.58
128 mT north at an angle of 78.7° below
the horizontal
P29.20
(a) 5.41 mA ⋅ m 2 ; (b) 4.33 mN ⋅ m
P29.22
(a)3.97° ; (b) 3.39 mN ⋅ m
P29.24
(a) 80.1 mN ⋅ m ; (b) 104 mN ⋅ m ;
(c) 132 mN ⋅ m ;
(d) The torque on the circle.
P29.62
P29.26
(a) minimum: pointing north at 48.0°
below the horizontal; maximum: pointing
south at 48.0° above the horizontal;
(b) 1.07 µJ
P29.64
λ g tan θ
I
P29.66
(a) 0.104 mm; (b) 0.189 mm
P29.68
3.82 × 10 −25 kg
P29.70
(a) see the solution;
empirically, ∆VH = 100 µ V T B ;
(b) 0.125 mm
P29.28
(a) 640 µN ⋅ m ; (b) 241 mW; (c) 2.56 mJ;
(d) 154 mW
P29.30
1.98 cm
P29.32
65.6 mT
P29.34
(a) 5.00 cm ; (b) 8.78 Mm s
P29.36
m′
=8
m
P29.38
see the solution
P29.40
244 kV m
P29.42
278 mm
P29.44
162 mm
P29.46
3.00 T
P29.48
(a) 7.44 × 10 28 m3 ; (b) 1.79 T
P29.60
P29.72
− Fi
;
evi
3R
4
B ~ 10 −1 T; τ ~ 10 −1 N ⋅ m; I ~ 1 A ;
A ~ 10 −3 m 2 ; N ~ 10 3
b
g
qBh
; The particle moves in a
m
semicircle of radius h and leaves the field
with velocity −vj;
(b) The particle moves in a smaller
mv
semicircle of radius
, attaining final
qB
velocity −vj;
(a) v =
(c) The particle moves in a circular arc of
mv
, leaving the field with
radius r =
qB
velocity v sin θ i + v cos θ j where
θ = sin −1
FG h IJ
HrK