serway1 (1)

29

Magnetic Fields

CHAPTER OUTLINE ANSWERS TO QUESTIONS

29.1 Magnetic Fields and Forces *Q29.1 (a) Yes, as described by F = qE.
29.2 Motion of a Charged Particle in a
(b) No, as described by F = qv × B
Uniform Magnetic Field (c) Yes. (d) Yes. (e) No. The wire is uncharged.
29.3 Applications Involving Charged (f) Yes. (g) Yes. (h) Yes.

Particles Moving in a Magnetic *Q29.2 (i) (b). (ii) (a). Electron A has a smaller radius of
Field curvature, as described by qvB = mv2/r.
29.4 Magnetic Force Acting on a
Current-Carrying Conductor
29.5 Torque on a Current Loop in a
Uniform Magnetic Field
29.6 The Hall Effect

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

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

c = d > a = e > f > b.

*Q29.5 ˆi × (−kˆ ) = ˆj . Answer (c).

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

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

Q29.8 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, independent of its velocity. 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.
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.

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

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

151

152 Chapter 29

Q29.11 If you can hook a spring balance to the particle and measure the force on it in a known electric
Q29.12 field, then q = F will tell you its charge. You cannot hook a spring balance to an electron.

E
Measuring the 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 properties of the particle only by being proportional to the ratio q .

m

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.

Q29.13 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. Q29.13

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

SOLUTIONS TO PROBLEMS

Section 29.1 Magnetic Fields and Forces

P29.1 (a) up

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

(c) no deflection

(d) into the page

FIG. P29.1

Magnetic Fields 153

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 FIG. P29.2
directed West .

(b) North × North = sin 0° = 0: Zero deflection .
(c) West × North = Down, so the force is directed Up .

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

P29.3 ( ) ( )F = ma = 1.67 × 10−27 kg 2.00 × 1013 m s2 = 3.34 × 10−14 N = qvB sin 90°

3.34 × 10−14 N
( )( )B = F = = 2.09 × 10−2 T
qv 1.60 × 10−19 C 1.00 × 107 m s

The right-hand rule shows that B must be in the −y direction to yield a force FIG. P29.3
in the +x direction when v is in the z direction.

P29.4 ( ) ( ) ( )(a) FB = qvB sinθ = 1.60 ×10−19 C 3.00 ×106 m s 3.00 ×10−1 T sin 37.0°
P29.5
FB = 8.67 × 10−14 N

(b) a = F = 8.67 × 10−14 N = 5.19 × 1013 m s2
m 1.67 × 10−27 kg

FB = qvB sinθ ( ) ( )so 8.20 ×10−13 N = 1.60 ×10−19 C 4.00 ×106 m s (1.70 T) sinθ

sinθ = 0.754 and θ = sin−1 (0.754) = 48.9° or 131°

P29.6 First find the speed of the electron.

( ( ) )2
∆K = 1 mv2 = e∆V = ∆U: v= 2e∆V = 1.60 × 10−19 C (2 400 J C) = 2.90 × 107 m s
2 m
9.11× 10−31 kg

( ) ( )(a) FB, max = qvB = 1.60 × 10−19 C 2.90 × 107 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.

P29.7 FB = qv × B

ˆi ˆj kˆ

v × B = +2 −4 +1 = (12 − 2) ˆi + (1+ 6) ˆj + (4 + 4)kˆ = 10ˆi + 7ˆj + 8kˆ

+1 +2 −3

v × B = 102 + 72 + 82 = 14.6 T ⋅ m s

( )FB = q v × B = 1.60 × 10−19 C (14.6 T ⋅ m s) = 2.34 × 10−18 N

154 Chapter 29

P29.8 ( ) ( )qE = −1.60 ×10−19 C (20.0 N C) kˆ = −3.20 ×10−18 N kˆ
∑ F = qE + qv × B = ma

( ) ( ) ( )( )−3.20 × 10−18 N kˆ − 1.60 × 10−19 C 1.20 × 104 m s ˆi × B = 9.11 × 10−31 2.00 × 1012 m s2 kˆ

( ) ( ) ( )− 3.20 × 10−18 N kˆ − 1.92 × 10−15 C ⋅ m s ˆi × B = 1.82 × 10−18 N kˆ

( ) ( )1.92 × 10−15 C ⋅ m s ˆi × B = − 5.02 × 10−18 N kˆ

The magnetic field may have any x-component . Bz = 0 and By = −2.62 mT .

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

P29.9 (a) B = 50.0 × 10−6 T ; v = 6.20 × 106 m s

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

FB = qvB sinθ

( ) ( ) ( )FB = 1.60 ×10−19 C 6.20 ×106 m s 50.0 ×10−6 T sin 90.0°

= 4.96 × 10−17 N

(b) F = mv2 so FIG. P29.9
*P29.10 (a) r B
1.67 ×10−27 kg 6.20 ×106 m s 2
( )( )r = mv2 = = 1.29 km
F
4.96 ×10−17 N

The horizontal velocity component of the electrons is given

by (1/2)mv 2 = | q |V.
x

vx = 2qV = 2(1.6 × 10−19 C) 2 500 J/C = 2.96 × 107 m/s
m 9.11× 10−31 kg

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

Its vertical deflection is y = (1/2) gt 2 = (1/2) 9.8 m/s2 (1.18 × r
10−8 s)2 = 6.84 × 10−16 m down , which is unobservably
small. W
q SN
(b) The magnetic force is in the direction –north × down =
–west = east. The beam is deflected into a circular path with E
FIG. P29.10
radius

r= mv = 9.11 × 10−31 kg 2.96 × 107m/s = 8.44 m.
qB 1.6 × 10−19 C 20 × 10–6N ⋅ s/C ⋅ m

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

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

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

Magnetic Fields 155

P29.11 q (∆V ) = 1 mv2 or v = 2q(∆V )
so
2 m 2m(∆V )
Also, qvB = mv2
r = mv = m 2q(∆V ) = qB2
r
qB qB m

Therefore, rp2 = 2mp (∆V )

eB2

( )rd2 )
= 2md (∆V ) = 2 2mp ( ∆V = 2 ⎛ 2mp (∆V )⎞ = 2rp2
⎝⎜
qd B2 eB2 eB2 ⎠⎟

( )rα2
and = 2mα (∆V ) = 2 4mp (∆V ) = 2 ⎛ 2mp (∆V ) ⎞ = 2rp2
The conclusion is: (2e) B2 ⎜⎝ ⎠⎟
qα B2 eB2

rα = rd = 2rp

P29.12 For each electron, q vB sin 90.0° = mv2 and v = eBr .
rm

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

K = 1 mv12i +0 = 1 mv12f + 1 mv22 f
2 2 2

1 ⎛ e2 B2 R12 ⎞ 1 ⎛ e2 B2 R22 ⎞ e2 B2
( )K 2 ⎜ m2 ⎟ 2 ⎜ m2 ⎟ 2m
= m ⎝ ⎠ + m ⎝ ⎠ = R12 + R22

( ( ) )K C ⋅ m)2
= e 1.60 × 10−19 C (0.044 0 N ⋅ s ⎡⎣(0.010 0 m)2 + (0.024 0 m )2 ⎤ = 115 keV
⎦
2 9.11× 10−31 kg

P29.13 (a) We begin with qvB = mv2
or R
But
qRB = mv
Therefore,
L = mvR = qR2B

4.00 × 10−25 J ⋅ s
1.60 × 10−19C 1.00 × 10−3 T
( )( )R = L = = 0.050 0 m = 5.00 cm
qB

4.00 × 10−25 J ⋅ s

9.11 × 10−31 kg (0.050
(b) Thus, ( )v = L = 0 m) = 8.78 × 106 m s
mR

P29.14 1 mv2 = q(∆V ) so v = 2q(∆V )

2 m

r = mv so r = m 2q(∆V ) m
qB
qB

r2 = m ⋅ 2 ( ∆V ) and (r′)2 = m′ ⋅ 2(∆V )
q q′
B2 B2

m = qB 2 r 2 and ( m′) = ( q′) B2 (r′)2 so m′ = q′ ⋅ (r′)2 = ⎜⎛⎝ 2e ⎞⎠⎟⎜⎝⎛ 2R ⎟⎞⎠2 = 8
(∆V ) m q e R
2(∆V ) 2 r2

156 Chapter 29

P29.15 E = 1 mv2 = e∆V
2

and evB sin 90° = mv2
R

B = mv = m 2e∆V = 1 2m∆V
eR eR m R e

( )( )B = 1
5.80 ×1010 m
2 1.67 ×10−27 kg 10.0 ×106 V = 7.88 ×10−12 T
1.60 ×10−19 C

P29.16 (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° = mv2 FIG. P29.16(a)
r

( ) ( )v = ω = q B = 1.60 ×10−19 C 10−3 N ⋅ s C ⋅ m = 1.76 ×108 rad s
( )r m
9.11× 10−31 kg

The time for one half revolution is,

from ∆θ = ω ∆ t

∆t = ∆θ = π rad s = 1.79 × 10−8 s
ω 1.76 × 108 rad

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

( )Then v = ω r = 1.76 ×108 s−1 (0.02 m) = 3.51×106 m s

and

( )( )K = 1 mv2 = 1 9.11 × 10−31 kg 2 5.62 × 10−18 J 5.62 × 10−18 J ⋅ e
22 1.60 × 10−19 C
3.51 × 106 m s = =

= 35.1 eV

Section 29.3 Applications Involving Charged Particles Moving in a Magnetic Field

P29.17 FB = Fe qvB = qE
so
where v = 2K and K is kinetic energy of the electron.
m

( )E = vB =
2K B = 2(750) 1.60 ×10−19 (0.015 0) = 244 kV m
m
9.11 × 10−31

Magnetic Fields 157

P29.18 K = 1 mv2 = q(∆V ) so v = 2q(∆V )

2 m

FB = qv × B = mv2 r = mv = m 2q(∆V ) m = 1 2m(∆V )
r
qB q B B q

( )(a)
r238 = 2 238 ×1.66 ×10−27 2 000 ⎛⎝⎜ 1 ⎠⎞⎟ = 8.28 × 10−2 m = 8.28 cm
1.60 ×10−19 1.20

(b) r235 = 8.23 cm

r238 = m238 = 238.05 = 1.006 4
r235 m235 235.04
The ratios of the orbit radius for different ions are independent of ∆V and B.

P29.19 In the velocity selector: v = E = 2 500 V m = 7.14 × 104 m s
B 0.035 0 T

) )( (r = mv =
)(qB
In the deflection chamber: 2.18 × 10−26 kg 7.14 × 104 m s = 0.278 m

1.60 × 10−19 C (0.035 0 T)

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

∑F = ma:

q vB sin 90° = mv2
r

q B = m v = mω
r

( ( ) )(a)
ω= q B 1.60 × 10−19 C (0.8 N ⋅ s C⋅ m) ⎛ kg ⋅ m ⎞ = 7.66 × 107 rad s
= ⎜⎝ N ⋅ s2 ⎟⎠
m 1.67 × 10−27 kg

( )(b) ⎝⎜⎛1 1 ⎟⎞⎠
v =ωr = 7.66 ×107 rad s (0.350 m) rad = 2.68 ×107 m s

K = 1 mv2 = 1 1.67 ×10−27 kg 2.68 ×107 m s 2 ⎝⎛⎜1.61×e1V0−19J ⎞⎠⎟ =
22
( )( )(c) 3.76 ×106 eV

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

3.76 × 106 eV = 3.13 × 103 revolutions

2(600 eV)

(e) θ = ω t t = θ = 3.13 ×103 rev ⎜⎝⎛ 2π rad ⎠⎟⎞ = 2.57 ×10−4 s
ω 7.66 ×107 rad s 1 rev

P29.21 (a) FB = qvB = mv2
(b) R

( )ω = v = qBR = qB =
R mR m
1.60 ×10−19 C (0.450 T) = 4.31×107 rad s

1.67 ×10−27 kg

( )v = qBR =
m
1.60 ×10−19 C (0.450 T)(1.20 m) = 5.17 ×107 m s

1.67 ×10−27 kg

158 Chapter 29

*P29.22 (a) The path radius is r = mv/qB, which we can put in terms of energy E by (1/2)mv2 = E.
v = (2E/m)1/2 so r = (m/qB) (2E/m)1/2 = (2m)1/2(qB)−1E1/2

Then dr/dt = = (2m)1/2(qB)−1(1/2) E −1/2 dE/dt = 2m 2m q2B∆V = 1 ∆V
qB2 qBr π m r π B

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

(c) dr = 1 ∆V = 600 V C m N m = 682 m/s
dt r π B 0.35 m π 0.8 N s V C

(d) ∆r = dr T = 1 ∆V 2π m = 2∆Vm = 2 × 600 V 1.67 ×10−27kg C2 m2 Nm = 55.9 µm
dt r πB qB rqB2 0.35 m 1.6 ×10−19C 0.82 N2 s2 VC

P29.23 θ = tan−1 ⎛ 25.0 ⎞ = 68.2° and R = 1.00 cm = 1.08 cm
⎝⎜ 10.0 ⎟⎠ sin 68.2°

Ignoring relativistic correction, the kinetic energy of the electrons is

1 mv2 = q∆V so v = 2q∆V = 1.33 × 108 m s
2m

From Newton’s second law, mv2 = qvB, we find the magnetic field
R

)) ))(( ((B =
mv = 9.11 × 10−31 kg 1.33 × 108 m s = 70.1 mT
qR 1.60 × 10−19 C 1.08 × 10−2 m
FIG. P29.23

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

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

(c) In a charged macroscopic object most of the atoms are uncharged. A molecule never

has all of its atoms ionized. Any atom other than hydrogen contains neutrons and so has

more mass per charge if it is ionized than hydrogen does. The greatest charge-to-mass
ratio Thomson could expect was then for ionized hydrogen, 1.6 × 10−19 C/1.67 × 10−27 kg,
smaller than the value e/m he measured, 1.6 × 10−19 C/9.11 × 10−31 kg, by 1 836 times. The

particles in his beam could not be whole atoms, but rather must be much smaller in mass.

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

v= 200 ⋅1.6 ×10−19 C 1 J/C = 5.93 ×106 m/s. The time to travel 40 cm is
9.11×10−31 kg

0.4 m/(5.93 × 106 m/s) = 6.75 × 10−8 s. If it is fired horizontally it will fall vertically by
(1/2)gt 2 = (1/2)(9.8 m/s2)( 6.75 × 10−8 s)2 = 2.23 × 10−14 m, an immeasurably small

amount. An electron with higher energy falls by a smaller amount.

Magnetic Fields 159

Section 29.4 Magnetic Force Acting on a Current-Carrying Conductor
P29.25 FB = ILB sinθ with FB = Fg = mg

mg = ILB sinθ so m g = IB sinθ
L

FIG. P29.25

I = 2.00 A and m = (0.500 g cm) ⎛ 100 cm m⎞ = 5.00 × 10−2 kg m
L ⎝⎜ 1 000 g kg ⎠⎟

( )Thus
5.00 ×10−2 (9.80) = (2.00) B sin 90.0°

B = 0.245 Tesla with the direction given by right-hand rule: eastward .

( )P29.26 FB = I × B = (2.40 A)(0.750 m) ˆi × (1.60 T)kˆ = −2.88ˆj N

P29.27 (a) FB = ILB sinθ = (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)sin120° = 4.73 N

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

I B sin 90° = mg I = m g = 2.4 ×10−3 kg 9.8 m/s2C m = 840 A
B m 28 ×10−6N s

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

) ( ) ( ) ( )(P29.29 The rod feels force FB = I d × B = Id kˆ × B −ˆj = IdB ˆi . B

( ) ( )The work-energy theorem is d
Ktrasn + Krot + ∆E = Ktrans + Krot I

i f

0 + 0 + Fs cosθ = 1 mv2 + 1 Iω2 L
2 2 y

1 1 ⎛ 1 ⎞ ⎛ v⎞ 2 3 mv2 x
2 2 ⎝ 2 ⎠ ⎝ R⎠ 4 z
IdBL cos 0º = mv2 + mR2 and IdBL =

v= 4IdBL = 4 ( 48.0 A)(0.120 m)(0.240 T)(0.450 m) = 1.07 ms FIG. P29.29
3m
3(0.720 kg)

160 Chapter 29

P29.30 ) ( ) ( ) ( )(The rod feels force FB = I d × B = Id kˆ × B −ˆj = IdB ˆi .

( ) ( )The work-energy theorem is
Ktrans + Krot + ∆E = Ktrans + Krot f

i

0 + 0 + Fs cosθ = 1 mv2 + 1 Iω 2
22

1 1 ⎛ 1 ⎞ ⎛ v⎞ 2 4 IdBL
2 2 ⎝ 2 ⎠ ⎝ R⎠ 3m
IdBL cos 0º = mv2 + mR 2 and v=

P29.31 The magnetic force on each bit of ring is
Id s × B = IdsB radially inward and upward,
at angle q above the radial line. The radi-
ally inward components tend to squeeze
the ring but all cancel out as forces. The
upward components IdsB sinθ all add to

I 2πrB sinθ up .

FIG. P29.31

*P29.32 (a) For each segment, I = 5.00 A and B = 0.020 0 N A ⋅ m ˆj.

Segment −0.400 m ˆj (FB = I × B)
ab 0.400 m kˆ
bc 0

( )(40.0 mN) −ˆi FIG. P29.32

( )cd
−0.400 m ˆi + 0.400 m ˆj (40.0 mN) −kˆ

( )da 0.400 m ˆi − 0.400 m kˆ (40.0 mN) kˆ + ˆi

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

Magnetic Fields 161

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

( ) ( )B = (52.0 µT) cos 60.0° −kˆ + (52.0 µT)sin 60.0° −ˆj
( ) ( )The current then has equivalent length: L′ = 1.40 m −kˆ + 0.850 m ˆj

( ) ( )FB = I L′ × B = (0.035 0 A) 0.850ˆj − 1.40kˆ m × −45.0ˆj − 26.0kˆ 10−6 T
( ) ( )FB = 3.50 × 10−8 N −22.1ˆi − 63.0ˆi = 2.98 × 10−6 N −ˆi = 2.98 µN west

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

P29.34 (a) 2π r = 2.00 m so r = 0.318 m

( )µ = IA = 17.0 ×10−3 A ⎡⎣π (0.318)2 m2 ⎤⎦ = 5.41 mA ⋅ m2
( )(b) τ = µ × B so τ = 5.41×10−3 A ⋅ m2 (0.800 T) = 4.33 mN ⋅ m

P29.35 τ = NBAI sinφ

τ = 100(0.800 T)(0.400 × 0.300 m2 )(1.20 A)sin 60°

τ = 9.98 N ⋅ m

Note that f 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.

FIG. P29.35

P29.36 Choose U = 0 when the dipole moment is at θ = 90.0° to the field. The field exerts torque of mag-
nitude m Bsinq on the dipole, tending to turn the dipole moment in the direction of decreasing q.
According to the mechanical equations ∆U = – ∫dW and dW = t dq , the potential energy of the

dipole-field system is given by

θ or U = −µ ⋅B .

∫U − 0 = µB sinθ dθ = µB(− cosθ ) θ = −µB cosθ + 0
90.0 º
90.0 º

P29.37 (a) 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

( )( )where its energy is Umin = −µB cos 0° = − 9.70 ×10−3 A ⋅ m2 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

( )( )where its energy is Umax = −µB cos180° = + 9.70 ×10−3 A ⋅ m2 55.0 ×10−6 T = +5.34 ×10−7 J.

(b) Umin + W = Umax: ( )W = Umax − Umin = + 5.34 × 10−7 J − −5.34 × 10−7 J = 1.07 µJ

162 Chapter 29

P29.38 (a) τ = µ × B , so τ = µ × B = µB sinθ = NIAB sinθ

A)⎡⎣π (0.050 0 ⎤
( )τ max )2 ⎦
= NIAB sin 90.0° = 1(5.00 m 3.00 ×10−3 T = 118 µN ⋅ m

(b) U = −µ ⋅ B, so −µB ≤ U ≤ +µB

)Since µB = (NIA) B = 1(5.00 A) ⎡⎣π (0.050 0 m)2 ⎤⎦(3.00 × 10−3 T = 118 µJ,

the range of the potential energy is: −118 µJ ≤ U ≤ +118 µJ .

*P29.39 For a single-turn circle, r = 1.5 m/ 2p. The magnetic moment is
m = N I A = 1 (30 × 10–3 A) p (1.5 m / 2p )2 = 5.37 × 10–3 A ⋅ m2
For a single-turn square, = 1.5 m/4. The magnetic moment is
m = N I A = 1 (30 × 10–3 A) (1.5 m /4)2 = 4.22 × 10–3 A ⋅ m2
For a single-turn triangle, = 1.5 m/3 = 0.5 m. The magnetic moment is
m = N I A = 1 (30 × 10–3 A) (1/2) 0.5 m (0.52 – 0.252)1/2 m = 3.25 × 10–3 A ⋅ m2
For a flat compact circular coil with N turns, r = 1.5 m/ 2pN. The magnetic moment is

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

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

P29.40 (a) τ = µ × B = NIAB sinθ

( )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 N ⋅ m (3 600 rev min)⎝⎛ 2π rad⎞ ⎛ 1 min⎞ = 0.241 W
1 rev ⎠ ⎝ 60 s ⎠

(c) In one half revolution the work is

W = Umax −Umin = −µB cos180° − (−µB cos 0°) = 2µB

( )= 2NIAB = 2 6.40 ×10−4 N ⋅ m = 1.28 ×10−3 J

( )In one full revolution, W = 2 1.28 ×10−3 J = 2.56 ×10−3 J .

(d) Pavg = W = 2.56 × 10−3 J = 0.154 W
∆t
(1/60) s
The peak power in (b) is greater by the factor π .
2

Magnetic Fields 163

Section 29.6 The Hall Effect

( )( )( )( )P29.41 B = nqt (∆VH ) = 8.46 ×1028 m−3 1.60 ×10−19 C 5.00 ×10−3 m 5.10 ×10−12 V
I 8.00 A
B = 4.31×10−5 T = 43.1µT

P29.42 (a) ∆VH = IB so nqt = B = 0.080 0 T = 1.14 × 105 T V
nqt I ∆VH 0.700 × 10−6 V

Then, the unknown field is B = ⎛ nqt ⎞ ( ∆VH )
⎝ I ⎠

) )( (B = 1.14 × 105 T V 0.330 × 10−6 V = 0.037 7 T = 37.7 mT

(b) nqt = 1.14 ×105 T V so n = (1.14 ×105 T V) I
I qt

0.120 A
1.60 × 10−19 C 2.00 × 10−3 m
( ) ( )( )n = 1.14 ×105 T V = 4.29 × 1025 m−3

Additional Problems

*P29.43 (a) Define vector h to have the downward direction of the J
current, and vector L to be along the pipe into the page as

shown. The electric current experiences a magnetic force .

( )I h × B in the direction of L

(b) The sodium, consisting of ions and electrons, flows along the FIG. P29.43

pipe transporting no net charge. But inside the section of

length L, electrons drift upward to constitute downward

electric current J × (area) = J Lw.

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

164 Chapter 29

P29.44 (a) At the moment shown in Figure 29.10, the particle must be v
moving upward in order for the magnetic force on it to be
+
into the page, toward the center of this turn of its B

spiral path. Throughout its motion it circulates clockwise. FIG. P29.44(a)

(b) After the particle has passed the middle of the bottle and moves v
into the region of increasing magnetic field, the magnetic force B
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 F
reverses the particle’s motion along the axis.

FIG. P29.44(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 + NS
plane and therefore a magnetic dipole moment IA = q A
T FIG. P29.44(d)
in the –x direction. It is like a little bar magnet with its

N pole on the left.

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

FIG. P29.44(e)

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

r= mv = 1.67 ×10−27 kg 3×107 m/s C m = 12.5 km . It will not hit the Earth, but perform
qB 1.6 ×10−19 C 25 ×10–6 N s

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

P29.46 ∑ Fy = 0: +n − mg = 0
∑ Fx = 0:
−µkn + IBd sin 90.0° = 0

( )B =
µk mg = 0.100 (0.200 kg) 9.80 ms2 = 39.2 mT
Id (10.0 A)(0.500 m)

Magnetic Fields 165

P29.47 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

qvB = mv2
r

and r = mv
qB

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

K = 1 mv2 FIG. P29.47
2
( )( )2 5.00 × 106 eV 1.60 × 10−19 J eV = 3.10 × 107 m s
v = 2K =
m 1.67 × 10−27 kg

) )( (r = mv =
)(qB
Now, 1.67 × 10−27 kg 3.10 × 107 m s

1.60 × 10−19 C (0.050 0 N ⋅ s C ⋅ m) = 6.46 m

(b) We can most conveniently do part (b) first. From the figure observe that

sinα = 1.00 m = 1 m
r 6.46 m

α = 8.90°

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

( )( )− 1.67 ×10−27 kg 3.10 ×107 m s sin 8.90° = −8.00 ×10−21 kg ⋅ m s

*P29.48 (a) If B = Bxˆi + Byˆj + Bzkˆ , ( ) ( )FB = qv × B = e viˆi × Bxˆi + Byˆj + Bzkˆ = 0 + evi Bykˆ − evi Bz ˆj.

Since the force actually experienced is FB = Fi ˆj, observe that

Bx could have any value , By = 0 , and Bz = − Fi .
evi

( )FB ⎛ Fi kˆ ⎠⎟⎞
(b) If v = −viˆi, then = qv × B = e −vi ˆi × ⎜⎝ Bx ˆi + 0ˆj − evi = −Fi ˆj

( )(c) ⎛ Bx ˆi 0ˆj Fi kˆ ⎠⎟⎞
If q = −e and v = −viˆi, then FB = qv × B = −e −vi ˆi × ⎜⎝ + − evi = +Fi ˆj

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

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

( )F = qE + qv × B = q E + v × B

( ) ( ) ( )( )F = 3.20 × 10−19 ⎣⎡ 4ˆi − 1ˆj − 2kˆ + 2ˆi + 3ˆj − 1kˆ × 2ˆi + 4ˆj + 1kˆ ⎤⎦ N

Carrying out the indicated operations, we find:

( )F = 3.52ˆi − 1.60ˆj × 10−18 N .

⎛ Fx ⎞ ⎛ 3.52 ⎞
⎝⎜ F ⎟⎠ ⎜ ⎟
(b) θ = cos−1 = cos−1 ⎜ (3.52)2 + (1.60)2 ⎟ = 24.4°
⎝ ⎠

166 Chapter 29

*P29.50 (a) The field should be in the +z direction, perpendicular to the final as well as to the initial
velocity, and with ˆi × kˆ = −ˆj as the direction of the initial force.

)) )(( ((b)
r = mv = 1.67 × 10−27 kg 20 × 106 m s 0.696 m
qB
1.60 × 10−19 C (0.3 N ⋅ s C ⋅ m) =

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

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

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

the friction force: FB = BIL or B = FB .
IL
I
∑When the wire is just able to move, Fy = n + FB cosθ − mg = 0

so n = mg − FB cosθ

and f = µ (mg − FB cosθ ) FIG. P29.51

Also, ∑ Fx = FB sinθ − f = 0
so FB sinθ = f :
FB sinθ = µ (mg − FB cosθ ) and FB = µmg
sinθ + µ cosθ

We minimize B by minimizing FB: dFB = ( µ mg ) ( cosθ − µ sinθ = 0 ⇒ µ sinθ = cosθ
dθ
sinθ + µ cosθ )2

Thus, θ = tan−1 ⎛ 1 ⎞ = tan−1 (5.00) = 78.7° for the smallest field, and
⎜ µ ⎟
⎝ ⎠

B = FB = ⎛ µg ⎞ (m L)
IL ⎜⎝ I ⎠⎟
sinθ + µ cosθ

( )Bmin ⎤
= ⎢⎡(0.200) 9.80 m s2 ⎥ 0.100 kg m = 0.128 T
⎦⎥
⎣⎢ 1.50 A sin 78.7° + (0.200) cos 78.7°

Bmin = 0.128 T pointing north at an angle of 78.7° below the horizontal

P29.52 Let vi represent the original speed of the alpha particle. Let vα and vp represent the particles’
speeds after the collision. We have conservation of momentum 4mpvi = 4mpvα + mpvp and the
relative velocity equation vi − 0 = vp − vα. Eliminating vi ,

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

For the proton’s motion in the magnetic field,

∑ F = ma ev p B sin 90° = mp v2p eBR = vp
R mp
For the alpha particle,

2evα B sin 90° = 4mp vα2 rα = 2mp vα rα = 2mp 3 v p = 2mp 3 eBR = 3R
rα eB eB 8 eB 8 mp 4

Magnetic Fields 167

P29.53 Let ∆x1 be the elongation due to the weight of the wire and
let ∆x2 be the additional elongation of the springs when the
magnetic field is turned on. Then Fmagnetic = 2k∆x2 where k is
the force constant of the spring and can be determined from
k = mg . (The factor 2 is included in the two previous
equa2ti∆oxn1s since there are 2 springs in parallel.) Combining

these two equations, we find

Fmagnetic = ⎛ mg ⎞ ∆x2 = mg∆x2 ; but FB = I L × B = ILB FIG. P29.53
2 ⎝⎜ 2∆x1 ⎟⎠ ∆x1

(( ))Therefore, where
I = 24.0 V = 2.00 A, B= mg∆x2 (0.100)(9.80) 3.00 × 10−3 = 0.588 T .
12.0 Ω IL∆x1 = (2.00)(0.050 0) 5.00 × 10−3

P29.54 Suppose the input power is I ~ 1 A = 100 A

120 W = (120 V) I :

Suppose ω = 2 000 rev min ⎜⎛1 min ⎟⎞ ⎜⎛ 2π rad ⎟⎞ ~ 200 rad s
and the output power is ⎝ 60 s ⎠ ⎝ 1 rev ⎠

20 W = τω = τ (200 rad s) τ ~ 10−1 N ⋅ m

Suppose the area is about (3 cm) × (4 cm), 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 N ⋅ m ~ N (1 C s) 10−3 m2 10−1 N ⋅s C ⋅ m

giving N ~ 103

P29.55 The sphere is in translational equilibrium; thus µB
θ
fs − Mg sinθ = 0 (1) I

fs

The sphere is in rotational equilibrium. If torques are taken θ Mg

about the center of the sphere, the magnetic field produces
a clockwise torque of magnitude µB sinθ, and the frictional
force a counterclockwise torque of magnitude fs R, where R
is the radius of the sphere. Thus:

fs R − µB sinθ = 0 (2) FIG. P29.55

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

µB = MgR. (3)

)Now
µ = NIπ R2. Thus (3) gives I = Mg = (0.08 kg)(9.80 m s2 = 0.713 A .
π NBR
π (5)(0.350 T)(0.2 m)

The current must be counterclockwise as seen from above.

168 Chapter 29

P29.56 Call the length of the rod L and the tension in each wire alone T Then, at equilibrium:
.
2

∑ Fx = T sinθ − ILB sin 90.0° = 0 or T sinθ = ILB

∑ Fy = T cosθ − mg = 0, or T cosθ = mg

tanθ = ILB = IB or B = (m L) g tanθ = λ g tanθ
mg
(m L)g II

P29.57 ∑ F = ma or qvB sin 90.0° = mv2

r
∴ the angular frequency for each ion is v = ω = qB = 2π f and

rm

( ( ) )∆ω
= ω12 − ω14 = ⎛ 1 − 1 ⎞ = 1.60 ×10−19 C (2.40 T) ⎝⎛⎜12.10 − 1 ⎞
qB ⎜ m14 ⎟ 1.66 ×10−27 14.0 ⎠⎟
⎝ m12 ⎠ kg u u u

∆ω = 2.75 ×106 s−1 = 2.75 Mrad/s

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

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

p = vxT = (v cos 85.0°)⎛⎜ 2π m ⎞ FIG. P29.58
Bq ⎟
⎝ ⎠

( ) ( () )p =
5.00 ×106 (cos 85.0°)(2π ) 9.11×10−31 = 1.04 ×10−4 m

0.150 1.60 ×10−19

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

spiral: r = mv⊥ = mv sin 85.0°
Bq Bq

( )( ( ) )r =
9.11×10−31 5.00 ×106 (sin 85.0°) = 1.89 ×10−4 m

(0.150) 1.60 ×10−19

P29.59 τ = IAB where the effective current due to the orbiting electrons is I = ∆q = q
∆t T
and the period of the motion is
The electron’s speed in its orbit is found by requiring keq2 = mv2 or T = 2π R
v
R2 R
v = q ke
mR

Substituting this expression for v into the equation for T, we find T = 2π mR3
q 2 ke

(( ))( ( ))T = 2π 3
9.11 × 10−31 5.29 × 10−11
1.60 × 10−19 2 8.99 × 109 = 1.52 × 10−16 s

⎛⎝⎜ q ⎞⎟⎠ 1.60 × 10−19 5.29 ×10−11 2 (0.400) =
( )Therefore, τ = T AB = 1.52 × 10−16 π 3.70 ×10−24 N ⋅ m .

Magnetic Fields 169

P29.60 (a) ( )( )K = 1 mv2 = 6.00 MeV = 6.00 ×106 eV 1.60 ×10−19 J eV 'x x x x x
2 Bin = 1.00 T
K = 9.60 ×10−13 J
xxxxx

x 45°x x x x x
45°x x x x x
( )2 9.60 ×10−13 J R xxxxx

v = 1.67 ×10−27 kg = 3.39 ×107 m s xxxxx
v 45.0°x x x x x
mv2
FB = qvB = R so

FIG. P29.60

( )( )R = mv =
( )qB
1.67 ×10−27 kg 3.39 ×107 m s = 0.354 m

1.60 ×10−19 C (1.00 T)

Then, from the diagram, x = 2R sin 45.0° = 2 (0.354 m)sin 45.0° = 0.501 m .

(b) From the diagram, observe that θ′ = 45.0° .

P29.61 (a) 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

qvB = qE = q ⎛⎝⎜ ∆V ⎞⎟⎠
d

∆V (160 ×10−6 V) FIG. P29.61
Bd (0.040 0 T)(3.00 ×10−3 m)
or v = = = 1.33 ms

(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
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.

P29.62 (a) See graph to the right. The Hall 120
voltage is directly proportional to
the magnetic field. A least-square 100
fit to the data gives the equation
of the best fitting line as: 80

( )∆VH = 1.00 ×10−4 V T B ∆VH ( m V) 60

(b) Comparing the equation of the 40 0.2 0.4 0.6 0.8 1.0 1.2
line which fits the data best to B (T)
20

0
0

∆VH = ⎛ 1 ⎞ B FIG. P29.62
⎝⎜ nqt ⎟⎠

continued on next page

170 Chapter 29

I = 1.00 ×10Ϫ4 V T , or t= I .
nqt nq 1.00 ×10Ϫ4 V
( )observe that: T

Then, if I = 0.200 A, q = 1.60 ×10Ϫ19 C, and n = 1.00 ×1026 mϪ3 , the thickness of the
sample is

0.200 A = 1.25 × 10−4 m =
1.60 × 10Ϫ19 C
( )( )( )t = 1.00 × 10 m26 Ϫ3 0.125 mm
1.00 × 10Ϫ4 V T

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

according to qvB = mv2 , so the radius of the path is
r
r = mv .
qB

(a) When r = h = mv , that is, when v = qBh , the
qB m

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

(2h, 0, 0) with velocity v f = −vˆj . FIG. P29.63

(b) When v < qBh the particle will move in a smaller semicircle of radius r = mv < h.
, qB
m

It will leave the field at (2r, 0, 0) with velocity v f = −vˆj .

(c) When v > qBh , the particle moves in a circular arc of radius r = mv > h, centered at
m qB
(r, 0, 0). The arc subtends an angle given by θ = sin−1 ⎛ h⎞ . It will leave the field at the
⎝ r⎠

point with coordinates [r (1− cosθ ), h, 0] with velocity v f = v sinθ ˆi + v cosθ ˆj .

P29.64 (a) I = ev µ = IA = ⎛ ev ⎞ π r 2 = 9.27 ×10−24 A ⋅ m2
2πr ⎜⎝ 2πr ⎟⎠

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

(b) Because the electron is (–), its [conventional] current is clockwise, as seen FIG. P29.64
from above, and µ points downward .

*P29.65 (a) The torque on the dipole τ = µ × B has magnitude mB sin q ≈ mBq, proportional to the
angular displacement if the angle is small. It is a restoring torque, tending to turn the dipole
toward its equilibrium orientation. Then the statement that its motion is simple harmonic is
true for small angular displacements.

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

where w = ( m B/I )1/2 is the angular frequency and f = w /2p = 1 µB is the frequency
2π I
in hertz.

(c) The equilibrium orientation of the needle shows the direction of the field. In a stronger

field, the frequency is higher. The frequency is easy to measure precisely over a wide range

of values. The equation in part (c) gives
0.680 Hz = (1/2p) (m/I)1/2 (39.2 mT)1/2 and 4.90 Hz = (1/2p) (m/I)1/2 (B2)1/2. We square
and divide to find B2 /39.2 mT = (4.90 Hz/0.680 Hz)2 so B2 = 51.9(39.2 mT) = 2.04 mT .

Magnetic Fields 171

ANSWERS TO EVEN PROBLEMS

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

P29.4 (a) 86.7 f N (b) 51.9 Tm s2

P29.6 (a) 7.90 pN (b) 0

P29.8 By = −2.62 mT. Bz = 0. Bx may have any value.

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

P29.12 115 keV (b) 35.1 eV
P29.14 m′ = 8
P29.16 m
(a) 17.9 ns

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

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

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

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

( )P29.26 −2.88ˆj N

P29.28 840 A east
P29.30
P29.32 ⎛ 4IdBL ⎞ 1 2
⎝ 3m ⎠

( ) ( ) ( )(a) Fab = 0; Fbc = 40.0 mN −ˆi ; Fcd = 40.0 mN −kˆ ; Fda = (40.0 mN) ˆi + kˆ (b) The forces

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

P29.34 (a) 5.41 mA ⋅ m2 (b) 4.33 mN ⋅ m

P29.36 See the solution.

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

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

172 Chapter 29

P29.42 (a) 37.7 mT (b) 4.29 × 1025 m3

P29.44 See the solution.

P29.46 39.2 mT

P29.48 (a) Bx is indeterminate. By = 0. Bz = −Fi (b) −Fi ˆj (c) +Fi ˆj
P29.50 evi

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

P29.52 3R
P29.54 4
B ~ 10−1 T; τ ~ 10−1 N ⋅ m; I ~ 1 A; A ~ 10−3 m2; N ~ 103

P29.56 λ g tanθ
I

P29.58 (a) 0.104 mm; (b) 0.189 mm

P29.60 (a) 0.501 m (b) 45.0°

P29.62 (a) See the solution. Empirically, ∆VH = (100 µ V T) B (b) 0.125 mm

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

30

Sources of the Magnetic Field

CHAPTER OUTLINE ANSWERS TO QUESTIONS

30.1 The Biot-Savart Law *Q30.1 Answers (b) and (c).
30.2 The Magnetic Force Between
*Q30.2 (i) Magnetic field lines line in horizontal planes and
Two Parallel Conductors go around the wire clockwise as seen from above. East
30.3 Ampère’s Law of the wire the field points horizontally south. Answer (b).
30.4 The Magnetic Field of a Solenoid (ii) The same. Answer (b).
30.5 Gauss’s Law in Magnetism
30.6 Magnetism in Matter
30.7 The Magnetic Field of the Earth

*Q30.3 (i) Answer (f). (ii) Answer (e).

Q30.4 The magnetic field created by
wire 1 at the position of wire 2 is
into the paper. Hence, the magnetic FIG. Q30.4
force on wire 2 is in direction
down × into the paper = to the right,
away from wire 1. Now wire 2
creates a magnetic field into the
page at the location of wire 1, so
wire 1 feels force up × into the
paper = left, away from wire 2.

*Q30.5 Newton’s third law describes the relationship. Answer (c).

*Q30.6 (a) No (b) Yes, if all are alike in sign. (c) Yes, if all carry current in the same direction. (d) no

Q30.7 Ampère’s law is valid for all closed paths surrounding a conductor, but not always convenient.
There are many paths along which the integral is cumbersome to calculate, although not
impossible. Consider a circular path around but not coaxial with a long, straight current-
carrying wire.

Q30.8 The Biot-Savart law considers the contribution of each element of current in a conductor to
determine the magnetic field, while for Ampère’s law, one need only know the current passing
through a given surface. Given situations of high degrees of symmetry, Ampère’s law is more
convenient to use, even though both laws are equally valid in all situations.

Q30.9 Apply Ampère’s law to the circular path labeled 1 in the picture. Since there is
no current inside this path, the magnetic field inside the tube must be zero. On
the other hand, the current through path 2 is the current carried by the conductor.
Therefore the magnetic field outside the tube is nonzero.

FIG. Q30.9
173

174 Chapter 30

*Q30.10 (i) Answer (b). (ii) Answer (d), according to B = µ0 NI . (iii) Answer (b). (iv) Answer (c).

*Q30.11 Answer (a). The adjacent wires carry currents in the same direction.

*Q30.12 Answer (c). The magnetic flux is ΦB = BA cosθ. Therefore the flux is maximum when the field
is perpendicular to the area of the loop of wire. The flux is zero when there is no component of

magnetic field perpendicular to the loop—that is, when the plane of the loop contains the x axis.

*Q30.13 Zero in each case. The fields have no component perpendicular to the area.

*Q30.14 (a) Positive charge for attraction. (b) Larger. The contributions away from + and toward – are in
the same direction at the midpoint. (c) Downward (d) Smaller. Clockwise around the left-hand
wire and clockwise around the right-hand wire are in opposite directions at the midpoint.

*Q30.15 (a) In units of m0(ampere/cm), the field of the straight wire, from m0I/2p r, is 3/(2p 2) = 0.75/p.
As a multiple of the same quantity, Nm0I/2r gives for (b) 10 × 0.3/2 × 2 = 0.75.
(c)
Nm0I/ gives 1000 × 0.3/200 = 1.5 times m0(ampere/cm), which is also
(d) 1.5 × 4p × 10–7/0.01 T = 0.19 mT.
(e)
The field is zero at the center.

1 mT is larger than 0.19 mT, so it is largest of all.
The ranking is then e > c > b > a > d.

*Q30.16 Yes. Either pole of the magnet creates field that turns atoms inside the iron to align their
magnetic moments with the external field. Then the nonuniform field exerts a net force on each
atom toward the direction in which the field is getting stronger.

A magnet on a refrigerator door goes through the same steps to exert a strong normal force on the
door. Then the magnet is supported by a frictional force.

Q30.17 Magnetic domain alignment within the magnet and then within the first piece of iron creates an
external magnetic field. The field of the first piece of iron in turn can align domains in another
iron sample. A nonuniform magnetic field exerts a net force of attraction on magnetic dipoles
aligned with the field.

Q30.18 The shock misaligns the domains. Heating will also decrease magnetism.

Q30.19 The north magnetic pole is off the coast of Antarctica, near the south geographic pole. Straight up.

*Q30.20 (a) The third magnet from the top repels the second one with a force equal to the weight of the top
two. The yellow magnet repels the blue one with a force equal to the weight of the blue one.

(b) The rods (or a pencil) prevents motion to the side and prevents the magnets from rotating
under their mutual torques. Its constraint changes unstable equilibrium into stable.

(c) Most likely, the disks are magnetized perpendicular to their flat faces, making one face a
north pole and the other a south pole. One disk has its north pole on the top side and the
adjacent magnets have their north poles on their bottom sides.

(d) If the blue magnet were inverted, it and the yellow one would stick firmly together. The
pair would still produce an external field and would float together above the red magnets.

Sources of the Magnetic Field 175

SOLUTIONS TO PROBLEMS

Section 30.1 The Biot-Savart Law

P30.1 B = µ0 I = µ0q (v 2π R) = 12.5 T

2R 2R

( )B = µ0I =
2πr
P30.2 4π ×10−7 (1.00 A) 2.00 ×10−7 T
2π (1.00 m) =

P30.3 (a) B= 4µ0 I ⎛⎝⎜cos π − cos 3π ⎟⎞⎠ where a= 2
4π a 4 4

is the distance from any side to the center.

B = 4.00 × 10−6 ⎛ 2+ 2 ⎞ = 2 2 × 10−5 T = 28.3 µT into the paper
⎜ ⎟
0.200 ⎝ 2 2 ⎠
FIG. P30.3

(b) For a single circular turn with 4 = 2π R,

( )B = µ0 I = µ0π I = 4π 2 ×10−7 (10.0) = 24.7 µT into the paper
2R 4 4 (0.400)

P30.4 We can think of the total magnetic field as the superposition of the field due to the long straight

wire (having magnitude µ0 I and directed into the page) and the field due to the circular loop
2π R

(having magnitude µ0 I and directed into the page). The resultant magnetic field is:
2R

B = ⎛⎜⎝1 + 1 ⎟⎞⎠ µ0 I (directed into the page) .
π 2R

P30.5 For leg 1, d s × rˆ = 0, so there is no contribution to the field from
this segment. For leg 2, the wire is only semi-infinite; thus,

B = 1 ⎛ µ0I ⎞ = µ0 I into the paper
2 ⎜⎝ 2π x ⎠⎟ 4π x

FIG. P30.5

176 Chapter 30

P30.6 Along the axis of a circular loop of radius R,

µ0 IR2
x2 + R2
( )B = 32
2

or B ⎡ 1 ⎤3 2
B0 =⎢ ⎥
R )2 +1⎦⎥
⎣⎢ ( x

where B0 ≡ µ0 I
2R
FIG. P30.6

x R B B0
0.00 1.00
1.00 0.354
2.00 0.0894
3.00 0.0316
4.00 0.0143
5.00 0.00754

P30.7 Wire 1 creates at the origin magnetic field

B1 = µ0I right hand rule = µ0 I1 = µ0 I1 ˆj
2π r 2π a 2π a

(a) If the total field at the origin is 2µ0 I1 ˆj = µ0 I1 ˆj + B2 then the second wire must create field
2π a 2π a

according to B2 = µ0 I1 ˆj = µ0 I2 .
2π a
2π (2a)

Then I2 = 2I1 out of the paper .

( )(b) B1 + B2 = 2µ0 I1 −ˆj = µ0 I1 ˆj + B2 .
The other possibility is 2π a 2π a

( )Then B2 = 3µ0 I1 −ˆj = µ0I2 I2 = 6I1 into the paper
2π a
2π (2a)

P30.8 Every element of current creates magnetic field in the same direction, into the page, at the center of the
arc. The upper straight portion creates one-half of the field that an infinitely long straight wire would
create. The curved portion creates one quarter of the field that a circular loop produces at its center.
The lower straight segment also creates field 1 µ0 I .

2 2πr

The total field is

B = ⎛ 1 µ0 I + 1 µ0 I + 1 µ0 I ⎞ into the page = µ0 I ⎛ 1 + 1 ⎞ into the plane of the paper
⎜⎝ 2 2πr 4 2r 2 2πr ⎠⎟ 2r ⎜⎝ π 4 ⎟⎠

= ⎛ 0.284 15µ0 I ⎞ into the page
⎝⎜ r ⎠⎟

Sources of the Magnetic Field 177

P30.9 (a) Above the pair of wires, the field out of the page of the 50 A

( )current will be stronger than the −kˆ field of the 30 A

current, so they cannot add to zero. Between the wires, both

produce fields into the page. They can only add to zero below
the wires, at coordinate y = − y . Here the total field is

B = µ0I + µ0I :
2πr 2πr

µ0 ⎡ 50 A + 30 A ⎤ FIG. P30.9
2π + 0.28 y ⎥
=( ) ( )0⎢⎢⎣( −kˆ kˆ ⎥⎦
y m)

50 y = 30 ( y + 0.28 m)

50 (−y) = 30 (0.28 m − y)

−20y = 30 (0.28 m) at y = −0.420 m

(b) At y = 0.1 m the total field is B = µ0 I + µ0I :
2π r 2π r

4π × 10−7 T ⋅ m A ⎛ 50 A + 30 A ⎞
2π 0.10 m ⎠⎟
⎜⎝ (0.28 − 0.10)
( ) ( ) ( )B −kˆ −kˆ −kˆ
= m = 1.16 × 10−4 T

The force on the particle is

( ) ( )( )( ) ( )F = qv × B = −2 ×10−6 C 150 ×106 m s ˆi × 1.16 ×10−4 N ⋅ s C ⋅ m −kˆ
( )= 3.47 ×10−2 N −ˆj

(c) We require ( ) )(Fe = 3.47 × 10−2 N +ˆj = qE = −2 × 10−6 C E
So
E = −1.73 × 104 ˆj N C

*P30.10 We use the Biot-Savart law. For bits of wire along the straight-line
sections, d s is at 0° or 180° to rˆ, so d s × rˆ = 0. Thus, only the

curved section of wire contributes to B at P. Hence, d s is tangent to
the arc and rˆ is radially inward; so d s × rˆ = ds 1sin 90° ⊗ = ds ⊗ .
All points along the curve are the same distance r = 0.600 m from the
field point, so

µ0 I d s × rˆ µ0 I µ0 I
4π r2 4π r2 4π r2
∫ ∫ ∫B= dB = = ds = s FIG. P30.10

all current

where s is the arc length of the curved wire,

s = rθ = (0.600 m ) ( 30.0°) ⎜⎝⎛ 2π ⎞ = 0.314 m
360° ⎠⎟

Then, B (= 10−7 T⋅m A) (3.00 A) (0.314 m) B = 262 nT into the page
(0.600 m)2

178 Chapter 30

*P30.11 Label the wires 1, 2, and 3 as shown in Figure (a) and let the

magnetic field created by the currents in these wires be B1,

B2 , and B3 , respectively.

At point A: B1 = B2 = µ0 I and B3 = µ0I .
2π a
( )(a) 2 2π (3a)

The directions of these fields are shown in Figure (b). Figure (a)
Figure (b)
Observe that the horizontal components of B1 and B2
cancel while their vertical components both add onto B3.

Therefore, the net field at point A is:

BA = B1 cos 45.0° + B2 cos 45.0° + B3

= µ0 I ⎡ 2 cos 45.0° + 1⎤
2π a ⎢⎣ 2 3⎦⎥

( ( ) )BA =
4π ×10−7 T ⋅ m A (2.00 A) ⎡ 2 cos 45° + 1⎤
⎢⎣ 2 3⎦⎥
2π 1.00 ×10−2 m

BA = 53.3 µT toward the bottom of the page

(b) At point B : B1 and B2 cancel, leaving

BB = B3 = µ0 I

2π (2a)

BB = (4π ×10−7 T ⋅ m A)(2.00 A)
2π ((2) 1.00 ×10−2 m)
Figure (c)

= 20.0 µT toward the bottom of the page FIG. P30.11

B1 = B2 = µ0 I = µ0 I
2π a 2π a
(c) ( )At point C: 2 and B3 with the directions shown in
*P30.12 (a)
Figure (c). Again, the horizontal components of B1 and B2 cancel. The vertical
components both oppose B3 giving

( )BC= 2 ⎡ 2 π µ0 I 2 cos ⎤ − µ0 I = µ0 I ⎡ 2 cos 45.0° − ⎤ = 0
⎢ a 45.0°⎥⎦⎥ 2π a 2π a ⎢⎣ 2 1⎥⎦
⎢⎣

The upward lightning current creates field in counterclockwise horizontal circles.
µ0 I righthand rule = 4π ×10−7 T m 20 ×103 A north = 8.00 ×10–5 T north
2πr 2π A 50 m

20 kA

50 m F
21.4 mm

vB

N

WE

S

FIG. P30.12

continued on next page

Sources of the Magnetic Field 179

The force on the electron is
F = qv × B = −1.6 × 10−19 C (300 m/s west) × 8 × 10−5 (N s/C m) north

= −3.84 × 10−21 N down sin 90° = 3.84 × 10−21 N up

(b) r= mv = 1.6 9.11 × 10−31 kg 300 m/s m = 2.14 × 10−5 m. This distance is negligible
qB × 10–19 C 8 × 10–5 N s/C

compared to 50 m, so the electron does move in a uniform field.

(c) w = qB/m = 2pN/t

( )N =
qBt 1.6 ×10–19 C 8 ×10–5 N s/C m 60 ×10–6 s 134 rev
= 2π 9.11×10−31 kg =
2πm

*P30.13 (a) We use Equation 30.4 in the chapter text for the field created by

a straight wire of limited length. The sines of the angles appear- q2
a
ing in that equation are equal to the cosines of the complementary LP
2 q1
angles shown in our diagram. For the distance a from the wire to
the field point we have tan 30° = a , a = 0.288 7L. One wire FIG. P30.13(a)

L2
contributes to the field at P

B = µ0 I (cosθ1 − cosθ2 ) = 4π µ0 I 7L) (cos 30° − cos150°)
4π a
(0.288 I

= µ0 I (1.732) = 1.50µ0 I
4π (0.288 7L) πL

Each side contributes the same amount of field in the same direction,
which is perpendicularly into the paper in the picture. So the total field is

3⎝⎜⎛1.5π0Lµ0 I ⎞ = 4.50µ0 I .
⎟ πL
⎠

(b) As we showed in part (a), one whole side of the triangle creates

field at the center µ0 I (1.732) . Now one-half of one Pb
a
4π a
nearby side of the triangle will be half as far away from FIG. P30.13(b)

point Pb and have a geometrically similar situation. Then it

creates at Pb field µ0 I (1.732) = 2µ0 I (1.732) . The two
4π (a 2)
4π a

half-sides shown crosshatched in the picture create at Pb field

2 ⎛ 2µ 0 I (1.732 ) ⎞ = 4µ0 I (1.732) = 6µ0 I . The rest of the triangle
⎜ ⎟ 4π (0.288 7L) πL
⎝ 4π a ⎠

will contribute somewhat more field in the same direction,

so we already have a proof that the field at Pb is stronger .

180 Chapter 30

P30.14 Apply the equation from the chapter text for the field created by a straight wire of
limited length, three times:

B= µ0 I ⎛ d ⎞ toward you + µ0 I ⎛ a+ a ⎞
⎜cos 0 − ⎟ ⎜ d2 + a2 ⎟ away from you
4π a ⎝ d2 + a2 ⎠ 4π d ⎝ d2 + a2 ⎠

+ µ0 I ⎛ −d ⎞
⎜ − cos180°⎟ toward you
4π a ⎝ d2 + a2 ⎠

( )µ0 I a2 + d2 − d a2 + d2

B = away from you
2π ad a2 + d2

P30.15 Take the x-direction to the right and the y-direction up in the plane of the I1 5.00 cm
paper. Current 1 creates at P a field 13.0 cm P
B2
( )B1 B1
= µ0 I = 2.00 × 10−7 T ⋅ m (3.00 A)
2π a A (0.050 0 m) 12.0 cm

B1 = 12.0 µT downward and leftward, at angle 67.4° below the –x axis.
Current 2 contributes

( )2.00 ×10−7 T ⋅ m (3.00 A) I2
B2 = A (0.120 m) clockwise perpendicular to 12.0 cm FIG. P30.15

B2 = 5.00 µT to the right and down, at angle –22.6°

( ) ( )Then, B = B1 + B2 = (12.0 µT) −ˆi cos 67.4° − ˆjsin 67.4° + (5.00 µT) ˆi cos 22.6° − ˆjsin 22.6°

B = (−11.1 µT) ˆj − (1.92 µT) ˆj = (−13.0 µT) ˆj

*P30.16 We apply B = (m 0/2p ) m /x3 to the center of a face, where B = 40 000 m T and x = 0.6 mm, and also
to the exterior weak-field point, where B = 50 m T and x is the unknown.
Then Bx3 = Bx3 40 000 m T (0.6 mm)3 = 50 m T (d + 0.6 mm)3

d = (40 000/50)1/3 0.6 mm – 0.6 mm = 4.97 mm

The strong field does not penetrate your painful joint.

Section 30.2 The Magnetic Force Between Two Parallel Conductors

P30.17 By symmetry, we note that the magnetic forces on the top and bottom
segments of the rectangle cancel. The net force on the vertical segments of
the rectangle is (using an equation from the chapter text)

F = F1 + F2 = µ0 I1I2 ⎛ c 1 a − 1⎞ ˆi = µ0 I1I2 ⎛ −a ⎞ ˆi
2π ⎝ + c⎠ 2π
⎝⎜ c(c + a)⎠⎟

)( 4π × 10−7 N A2 (5.00 A)(10.0 A)(0.450 m) ⎛ −0.150 m ⎞ ˆi
2π ⎝⎜ (0.100 m)(0.250 m)⎟⎠
F=

( )F = −2.70 × 10−5 ˆi N

or F = 2.70 × 10−5 N toward the left

FIG. P30.17

Sources of the Magnetic Field 181

P30.18 Let both wires carry current in the x direction, the first at y = 0 y
and the second at y = 10.0 cm. I2 = 8.00 A
y = 10.0 cm
( )(a)
B = µ0 I kˆ = 4π × 10−7 T⋅ m A (5.00 A) kˆ I1 = 5.00 A
2π r 2π (0.100 m) x

B = 1.00 × 10−5 T out of the page z

FIG. P30.18(a)

( )( ) ( )(b) FB = I2 × B = (8.00 A)⎡⎣(1.00 m) ˆi × 1.00 ×10−5 T kˆ ⎤⎦ = 8.00 ×10−5 N −ˆj

FB = 8.00 × 10−5 N toward the first wire

( ) ( ) ( )( ) ( )(c)
B = µ0I −kˆ = 4π ×10−7 T⋅ m A (8.00 A) −kˆ = 1.60 ×10−5 T −kˆ
2π r 2π (0.100 m)

B = 1.60 × 10−5 T into the page

( ) ( )) )( ((d) FB = I1 × B = (5.00 A) ⎣⎡(1.00 m) ˆi × 1.60 × 10−5 T −kˆ ⎤⎦ = 8.00 × 10−5 N +ˆj

FB = 8.00 × 10−5 N towards the second wire

P30.19 To attract, both currents must be to the right. The attraction is described by

F = I2 B sin 90° = I2 µ0 I
2πr

F 2π r ⎛ 2π (0.5 m) ⎞
µ0 I1 ⎝⎜⎜
4π ×10−7 N ⋅ s C ⋅ m (20 A) ⎟⎟⎠ = 40.0 A
( ) ( )So
I2 = = 320 ×10−6 N m FIG. P30.19

Let y represent the distance of the zero-field point below the upper wire.

Then B = µ0 I + µ0I 0 = µ0 ⎛ 20 A (away) + 40 A ⎞
2π r 2π r ⎜
2π ⎝ y (0.5 m − y) (toward)⎟
⎠

20 (0.5 m − y) = 40y 20 (0.5 m) = 60y

y = 0.167 m below the upper wire

*P30.20 Carrying oppositely directed currents, wires 1 and 2 repel Wire 3 Wire 1 Wire 2
each other. If wire 3 were between them, it would have

to repel either 1 or 2, so the force on that wire could not

be zero. If wire 3 were to the right of wire 2, it would feel I3 1.50 A 4.00 A
a larger force exerted by 2 than that exerted by 1, so the

total force on 3 could not be zero. Therefore wire 3 must d 20 cm
be to the left of both other wires as shown. It must carry

downward current so that it can attract wire 2. We answer

part (b) first. FIG. P30.20

(b) For the equilibrium of wire 3 we have

F1 on 3 = F2 on 3 µ0 (1.50 A) I3 = µ0 (4 A) I3
2π d 2π (20 cm + d)

1.5(20 cm + d) = 4d d = 30 cm = 12.0 cm to the left of wire 1
2.5

continued on next page

182 Chapter 30

(a) Thus the situation is possible in just one way.

(c) For the equilibrium of wire 1,

µ0 I3 (1.5 A) = µ0 (4 A) (1.5 A) I3 = 12 4 A = 2.40 A down
2π (12 cm) 2π (20 cm) 20

We know that wire 2 must be in equilibrium because the forces on it are equal in magnitude
to the forces that it exerts on wires 1 and 3, which are equal because they both balance the
equal-magnitude forces that 1 exerts on 3 and that 3 exerts on 1.

*P30.21 The separation between the wires is

a = 2 (6.00 cm)sin 8.00° = 1.67 cm

(a) Because the wires repel, the currents are in

opposite directions .

(b) Because the magnetic force acts horizontally,

FB = µ0 I 2 = tan 8.00° FIG. P30.21
Fg 2π amg
I 2 = mg2π a tan 8.00° so I = 67.8 A

µ0

(c) Smaller. A smaller gravitational force would be pulling down on the wires and so tend-

ing to pull the wires together. Then a smaller magnetic force is required to keep the wires

apart.

*P30.22 (a) If the conductors are wires, F = I B sin 90° and B = m0I1/2p r give F = I I m /2p r
2 1 2 0

r = I1I2µ0 = (10 A)2 2 ×10−7 T m 0.5 m = 10.0 µm
2πF 1 N A

(b) For a force of ordinary size in a tabletop mechanics experiments to act on ordinary-to-large
size currents, the distance between them must be quite small, but the situation is physically
possible. If we tried to use wires with diameter 10 m m on a tabletop, they would feel the
force only momentarily after we turn on the current, until they melt. We can use wide, thin
sheets of copper, perhaps plated onto glass, and perhaps with water cooling, to have the
forces act continuously. A practical electric motor must use coils of wire with many turns.

Section 30.3 Ampère’s Law

P30.23 Each wire is distant from P by

(0.200 m)cos 45.0° = 0.141 m

Each wire produces a field at P of equal magnitude:

( )BA
= µ0 I = 2.00 × 10−7 T ⋅ m A (5.00 A) = 7.07 µT
2π a
(0.141 m)

Carrying currents into the page, A produces at P a field FIG. P30.23
of 7.07 µT to the left and down at –135º, while B
creates a field to the right and down at –45º. Carrying
currents toward you, C produces a field downward and
to the right at –45º, while D’s contribution is down-
ward and to the left. The total field is then

4 (7.07 µT)sin 45.0° = 20.0 µT toward the bottom of the page.

Sources of the Magnetic Field 183

P30.24 Let the current I be to the right. It creates a field B = µ0 I at the proton’s location. And we have
2π d

( ) ( ) ( )a balance between the weight of the proton and the magnetic force mg −ˆj + qv −ˆi × µ0 I kˆ = 0
2π d
at a distance d from the wire

) ) ) )( ( ( (d = qvµ0I = 1.60 × 10−19 C
) )( (2π mg
2.30 × 104 m s 4π × 10−7 T ⋅ m A 1.20 × 10−6 A = 5.40 cm
2π 1.67 × 10−27 kg 9.80 m s2

P30.25 From Ampère’s law, the magnetic field at point a is given by Ba = µ0 Ia , where Ia is the net
2π ra

current through the area of the circle of radius ra. In this case, Ia = 1.00 A out of the page
(the current in the inner conductor), so

(4π × 10−7 T ⋅ m A)(1.00 A) = 200 µT toward top of page
( )Ba = 2π 1.00 × 10−3 m

Similarly at point b: Bb = µ0 Ib , where Ib is the net current through the area of the circle having
2π rb
radius rb .

Taking out of the page as positive, Ib = 1.00 A − 3.00 A = −2.00 A, or Ib = 2.00 A into the page.
Therefore,

(4π × 10−7 T ⋅ m A)(2.00 A) = 133 µT toward bottom of page
( )Bb = 2π 3.00 × 10−3 m

P30.26 (a) In B = µ0 I , the field will be one-tenth as large at a ten-times larger distance: 400 cm .
2πr

( )(b) B = µ0 I kˆ + µ0 I −kˆ so
2πr1 2πr2

B = 4π × 10−7 T ⋅ m (2.00 A) ⎛ 1 − 1 ⎞ = 7.50 nT
0.398 0.401 5 m ⎟⎠
2π A ⎜⎝ 5 m

(c) Call r the distance from cord center to field point and 2d = 3.00 mm the distance between
conductors.

B = µ0 I ⎛ r 1 d − r 1 ⎞ = µ0 I 2d
2π ⎜⎝ − + d ⎠⎟ 2π r2 − d2

( )3.00 ×10−3 m
( )7.50 ×10−10 T = 2.00 ×10−7 T ⋅ m A (2.00 A) r2 − 2.25 ×10−6 m2 so r = 1.26 m

The field of the two-conductor cord is weak to start with and falls off rapidly with distance.

(d) The cable creates zero field at exterior points, since a loop in Ampère’s law encloses

zero total current. Shall we sell coaxial-cable power cords to people who worry about
biological damage from weak magnetic fields?

184 Chapter 30

P30.27 (a) One wire feels force due to the field of the other ninety-nine.

( ) ( )B = µ0I0r =
( )2π R2
4π ×10−7 T⋅ m A (99)(2.00 A) 0.200 ×10−2 m

2π 0.500 ×10−2 m 2

= 3.17 ×10−3 T

This field points tangent to a circle of radius 0.200 cm and exerts FIG. P30.27
force F = I × B toward the center of the bundle, on the single
hundredth wire:

( )F = IB sinθ = (2.00 A) 3.17 ×10−3 T sin 90° = 6.34 mN m

FB = 6.34 ×10−3 N m inward

(b) B ∝ r, so B is greatest at the outside of the bundle. Since each wire carries the same
current, F is greatest at the outer surface .

( ) ( )(a)
P30.28 Binner = µ0 NI = 4π ×10−7 T ⋅ m A (900) 14.0 ×103 A = 3.60 T
P30.29 2π r 2π (0.700 m)

( ) ( )(b)
Bouter = µ0 NI = 2 ×10−7 T ⋅ m A (900) 14.0 ×103 A = 1.94 T
2π r
1.30 m

We assume the current is vertically upward.

(a) Consider a circle of radius r slightly less than R. It encloses no current so from

͛ B ⋅ d s = µ0 Iinside B(2πr) = 0

we conclude that the magnetic field is zero .

(b) Now let the r be barely larger than R. Ampère’s law FIG. P30.29(a)

becomes B (2π R) = µ0 I,

so B = µ0 I
2π R

The field’s direction is tangent to the wall of the cylinder in a counterclockwise sense .

(c) Consider a strip of the wall of width dx and length . Its width is so
small compared to 2π R that the field at its location would be essentially

unchanged if the current in the strip were turned off.

The current it carries is Is = Idx up.
2π R
The force on it is

F = Is × B = Idx ⎛ µ0I ⎞ up × into page = µ0I 2 dx
2π R ⎜⎝ 2π R⎠⎟ radially inward.
4π 2 R2
FIG. P30.29(c)

The pressure on the strip and everywhere on the cylinder is

P = F = µ0 I 2 dx = µ0 I 2 inward
A 4π 2 R2 dx
(2π R)2

The pinch effect makes an effective demonstration when an aluminum can crushes itself as
it carries a large current along its length.

Sources of the Magnetic Field 185

( )I = 2πrB = 2π
µ0
P30.30 From ͛ B ⋅ d = µ0 I, 1.00 ×10−3 (0.100) = 500 A

4π ×10−7

P30.31 Use Ampère’s law, ͛ B ⋅ d s = µ0 I. For current density J, this becomes

͛ B ⋅ d s = µ0 ∫ J ⋅ dA.

(a) For r1 < R, this gives r1 FIG. P30.31

B2πr1 = µ0 ∫ (br)(2πrdr) and

0

B= µ0br12 (for r1 < R or inside the cylinder)
3

(b) When r2 > R, Ampère’s law yields ∫(2πr2 ) B= µ0 R (br ) (2π rdr ) = 2πµ0bR3 ,
0 3

or B= µ0bR3 (for r2 > R or outside the cylinder)
3r2

*P30.32 (a) See Figure (a) to the right.
(c)
We choose to do part (c) before part (b). At a point on the

z axis, the contribution from each wire has magnitude

B = µ0I and is perpendicular to the line
2π a2 + z2

from this point to the wire as shown in Figure (b). Combining (Currents are into the
paper) Figure (a)
fields, the vertical components cancel while the horizontal

components add, yielding

( )⎛ µ0 I z2 ⎞ = π µ0 I z2 ⎛ z z2 ⎞ = π µ0 Iz
a2 + sin θ ⎟⎠ a2 + ⎜⎝ a2 + ⎟⎠ a2 + z2
By = 2 ⎝⎜ 2π

By = 4π × 10−7 Tm8Az so B = 32 ×10−7 z T ⋅ m ˆj At a distance z
π ⎡⎣(0.03 m)2 + z2 ⎦⎤ 9 ×10−4 m2 + z2 above the plane
of the conductors
(b) Substituting z = 0 gives zero for the field. We can see this from
cancellation of the separate fields in either diagram. Figure (b)
Taking the limit z → ∞ gives 1/z → 0, as we should expect.
FIG. P30.32

(d) The condition for a maximum is:

( )dBy
( ) ( ) ( )dz
= −µ0 Iz (2z) + µ0 I =0 or µ0 I a2 − z2
π a2 + z2 π a2 + z2 2 = 0
π a2 + z2 2

Thus, along the z axis, the field is a maximum at d = a = 3.00 cm .

(e) The value of the maximum field is B = 32 ×10−70.03 m T ⋅ m ˆj = 53.3 ˆj mT .
9 ×10−4 m2 + 9 ×10−4 m2

186 Chapter 30

*P30.33 Js = I . Each filament of current creates a contribution to the
total field that goes counterclockwise around that filament’s
location. Together, they create field straight up to the right of
the sheet and straight down to the left of the sheet.
From Ampère’s law applied to the suggested rectangle,

͛B⋅ds = µ0I

B ⋅ 2 + 0 = µ0 Js Therefore the field is uniform in
space, with the magnitude B = µ0 Js .

2

FIG. P30.33

Section 30.4 The Magnetic Field of a Solenoid

*P30.34 In the expression B = Nm0I/ for the field within a solenoid with radius much less than 20 cm, all
we want to do is increase N.
(a) Make the wire as long and thin as possible without melting when it carries the 5-A current.
Then the solenoid can have many turns.
(b) As small in radius as possible with your experiment fitting inside. Then with a smaller
circumference the wire can form a solenoid with more turns.

(( ) )P30.35
B = µ0 N I I= B = 1.00 × 10−4 T 0.400 m
so µ0n = 31.8 mA
4π × 10−7 T⋅m A 1 000

P30.36 Let the axis of the solenoid lie along the y–axis from y = 0 to y = . We will determine the field
at y = a. This point will be inside the solenoid if 0 < a < and outside if a < 0 or a > . We

think of solenoid as formed of rings, each of thickness dy. Now I is the symbol for the current in

each turn of wire and the number of turns per length is ⎛ N ⎞ . So the number of turns in the ring
⎝⎜ ⎠⎟

is ⎛ N ⎞ dy and the current in the ring is I ring = I ⎛ N ⎞ dy. Now we use the result derived in the
⎜⎝ ⎠⎟ ⎜⎝ ⎟⎠

chapter text for the field created by one ring:

= µ0 Iring R2
x2 + R2 3 2
( )Bring 2

where x is the name of the distance from the center of the ring, at location y, to the field point
x = a − y. Each ring creates field in the same direction, along our y–axis, so the whole field of the
solenoid is

µ0 Iring R2 µ0I (N ) dyR2 µ0 INR2 dy
x2 + R2 3 2 2 (a − y)2 2
+ R2 3 2 (a − y)2 + R2
∑ ∑ ∫ ∫B = Bring = 2 = = 32

( ) ( ) ( )all rings 0 0 2

To perform the integral we change variables to u = a − y.

∫ ( )B = µ0 INR2 a−
2a
−du
u2 + R2 3 2

continued on next page

Sources of the Magnetic Field 187

and then use the table of integrals in the appendix:

(a) B = µ0 INR2 −u a− µ0 IN ⎡ a − a− ⎤
⎢ ⎥
= 2 ⎣⎢ a2 + R2 (a − )2 + R2 ⎦⎥
2 R2 u2 + R2 a

(b) If is much larger than R and a = 0,

we have B ≅ µ0 IN ⎡ − − ⎤ = µ0 IN .
2 ⎢0 ⎥ 2
⎣ 2 ⎦

This is just half the magnitude of the field deep within the solenoid. We would get the same
result by substituting a = to describe the other end.

P30.37 The field produced by the solenoid in its interior is
given by

( ) ( )( )B = µ0nI −ˆi = 4π ×10−7 T ⋅ m A ⎛⎝⎜1300−2.0m ⎠⎟⎞(15.0 A) −ˆi

( )B = − 5.65×10−2 T ˆi

The force exerted on side AB of the square current
loop is

( ) ( )( ) ( )FB AB = I L × B = (0.200 A) ⎣⎡ 2.00 × 10−2 m ˆj × 5.65 × 10−2 T −ˆi ⎦⎤

( ) ( )FB kˆ
= 2.26 ×10−4 N

AB

Similarly, each side of the square loop experiences a

force, lying in the plane of the loop, of

226 µN directed away from the center .

FIG. P30.37

From the above result, it is seen that the net torque exerted on the square
loop by the field of the solenoid should be zero. More formally, the magnetic dipole moment of
the square loop is given by

( )( )µ = IA = (0.200 A) 2.00 ×10−2 m 2 −ˆi = −80.0 µA ⋅ m2ˆi
( ) ( )The torque exerted on the loop is then τ = µ × B = −80.0 µA ⋅ m2ˆi × −5.65 × 10−2 Tˆi = 0

*P30.38 The number of turns is N = 75 cm = 750. We assume that the solenoid is long enough to qualify
0.1 cm

as a long solenoid. Then the field within it (not close to the ends) is B = N µ0 I so

)(I = B =
)(N µ0
8 × 10−3 T (0.75 m ⋅ A)

750 4π × 10−7 T ⋅ m = 6.37 A

The resistance of the wire is

( )R = ρ wire
( )A
= 1.7 × 10−8 Ω ⋅ m 2π (0.05 m)750 = 5.10 Ω

π 0.05 × 10−2 m 2

The power delivered is

P = I ∆V = I 2R = (6.37 A)2 (5.10 Ω) = 207 W

The power required would be smaller if wire were wrapped in several layers.

188 Chapter 30

Section 30.5 Gauss’s Law in Magnetism

( )∫ ( )P30.39 (a) ΦB = B ⋅ dA = B ⋅ A = 5ˆi + 4ˆj+ 3kˆ T ⋅ 2.50 ×10−2 m 2 ˆi

ΦB = 3.12 × 10−3 T ⋅ m2 = 3.12 × 10−3 Wb = 3.12 mWb

͛(b)
( )ΦB = B ⋅ dA = 0 for any closed surface (Gauss’s law for magnetism)

total

P30.40 (a) ( )ΦB flat = B ⋅ A = Bπ R2 cos (180 − θ ) = −Bπ R2 cosθ

(b) The net flux out of the closed surface is zero: ( )ΦB flat ( )+ ΦB curved = 0.

( )ΦB = Bπ R2 cosθ

curved

P30.41 (a) ΦB = B ⋅ A = BA where A is the cross-sectional area of the solenoid.
(b)
⎛ µ0 NI ⎞
( )ΦB= ⎝⎜ ⎟⎠ πr2 = 7.40 µWb

⎛ µ0 NI ⎞
( )ΦB ⎜⎝ ⎟⎠ ⎣⎡π ⎦⎤
= B⋅ A = BA = r22 − r12

( ) ( )ΦB⎡ ⎤
=⎢ 4π ×10−7 T⋅ m A ( 300) (12.0 A) ⎥ π ⎡⎣(8.00)2 − ( 4.00)2 ⎤ 10−3 m 2 = 2.27 µWb
m) ⎦⎥ ⎦
⎣⎢ (0.300

*P30.42 The field can be uniform in magnitude. Gauss’s law for magnetism implies that magnetic field
lines never start or stop. If the field is uniform in direction, the lines are parallel and their density
stays constant along any one bundle of lines. Thus the magnitude of the field has the same value
at all points along a line in the direction of the field. The magnitude of the field could vary over a
plane perpendicular to the lines, or it could be constant throughout the volume.

Section 30.6 Magnetism in Matter

P30.43 The magnetic moment of one electron is taken as one Bohr magneton mB. Let x represent the
number of electrons per atom contributing and n the number of atoms per unit volume. Then

nxm is the magnetic moment per volume and the magnetic field (in the absence of any currents
B
in wires) is B = µ0nxµB = 2.00 T.

x= B = 2.00 T 4π ×10−7 T ⋅ m A = 2.02
µ0µBn 9.27 ×10−24 N ⋅ m T
( )( )( )Then
8.50 ×1028 m−3

Sources of the Magnetic Field 189

Section 30.7 The Magnetic Field of the Earth

( )Bh
P30.44 (a) = Bcoil = µ0 NI = 4π × 10−7 (5.00)(0.600) 12.6 µT
(b) 2R
=
0.300

Bh = B sin φ → B = Bh = 12.6 µT = 56.0 µT
sin φ sin 13.0°

= 8.00 ×1022 A ⋅ m2 = 8.63 × 10 45 FIG. P30.44
9.27 ×10−24 A ⋅ m2
P30.45 (a) Number of unpaired electrons .

Each iron atom has two unpaired electrons, so the number of iron atoms required is

( )1

2
8.63 × 1045 .

(b) Mass = ( )( )4.31×1045 atoms 7 900 kg m3 = 4.01×1020 kg

( )8.50 ×1028 atoms m3

*P30.46 (a) Gravitational field and magnetic field are vectors; atmospheric pressure is a scalar.
(b)
At 42° north latitude, 76° west longitude, 260 m above sea level: Gravitational field is
(c) 9.803 m/s2 down. From the Coast and Geodetic Survey, the magnetic field is 54 m T at 12°
west of geographic north and 69° below the horizontal. Atmospheric pressure is 98 kPa.

The atmosphere is held on by gravitation, but otherwise the effects are all separate. The
magnetic field could be produced by permanent magnetization of a cold iron-nickel deposit
within the Earth, so it need not be associated with present-day action of gravitation.

Additional Problems

P30.47 Consider a longitudinal filament of the strip of width dr as
P30.48 shown in the sketch. The contribution to the field at point P
due to the current dI in the element dr is

dB = µ0dI
2πr

where dI = I ⎛ dr ⎞
⎜⎝ w ⎠⎟

dB = b+w µ0 Idr kˆ = µ0 I ⎛⎝1 + w⎞ FIG. P30.47
b 2π wr 2π w b⎠
∫ ∫B = ln kˆ

µ0 IR2 = µ0 I ( )( )I = 25 2 BR = 25 2 7.00 ×10−5 T 6.37 ×106 m
R2 + R2 25 2 R ( )µ0 4π ×10−7 T⋅ m A
( )B = 32
2

so I = 2.01 × 109 A toward the west

190 Chapter 30

P30.49 At a point at distance x from the left end of the bar, B
xθ
current I2 creates magnetic field B= µ0 I2 to I1
h2 + x2 FIG. P30.49
2π

the left and above the horizontal at angle q where h
θ
tanθ = x . This field exerts force on an element of the I2
h

rod of length dx

dF = I1 ×B= I1 µ0 I2dx sin θ
2π h2 + x2
right hand rule

= µ0 I1I2dx x into the page
2π h2 + x2 h2 + x2

( ) ( )dF = µ0I1I2xdx −kˆ
2π h2 + x2

The whole force is the sum of the forces on all of the elements of the bar:

∫ ( ) ( ) ( ) ∫ ( ) ( )F −kˆ
= µ0 I1I2 xdx −kˆ = µ0 I1I2 −kˆ 2xdx = µ0 I1I2 ln h2 + x2
2π h2 + x2 4π h2 + x2 4π
x=0 0

0

( ) ( ) ( )= µ0I1I2 −kˆ −kˆ
4π
⎡⎣ln h2 + 2 10−7 N (100 A) (200 A) ln ⎢⎡ ( 0.5 cm)2 + (10 cm)2 ⎤
− ln h2 ⎦⎤ = (0.5 cm)2 ⎥
A2 ⎣ ⎦

( ) ( )= 2 ×10−3 N −kˆ ln 401 = 1.20 ×10−2 N −kˆ

P30.50 Suppose you have two 100-W headlights running from a 12-V battery, with the whole

200 W = 17 A current going through the switch 60 cm from the compass. Suppose
12 V

the dashboard contains little iron, so µ ≈ µ0 . Model the current as straight. Then,

( )B = µ0I =
2π r
4π × 10−7 17 ~10−5 T

2π (0.6)

If the local geomagnetic field is 5 ×10−5 T , this is ~10−1 times as large, enough to affect the
compass noticeably.

P30.51 B= µ0 IR2
P30.52 x2 + R2
( )On the axis of a current loop, the magnetic field is given by 2 32

where in this case I = q ω ) . The magnetic field is directed away from the center,

with a magnitude of (2π

( ) ( )B =
4π µ0ω R2q 32 = µ0 (20.0)(0.100)2 10.0 × 10−6 = 1.43 × 10−10 T
x2 + R2 4π ⎡⎣(0.050 0)2 + (0.100)2 ⎦⎤3 2

µ0 IR2
x2 + R2
On the axis of a current loop, the magnetic field is given by ( )B = 32
2

= q µ0ω R2q
4π x2 + R2
where in this case I (2π ω ) . Therefore, ( )B = 32

when x = R B= µ0ω R2q = µ0 qω
2 ( )then 2.5 5π R
4π R5 2 3 2

4

Sources of the Magnetic Field 191

P30.53 (a) Use twice the equation for the field created by a current loop

µ0 IR2
x2 + R2
( )Bx = 2 32

If each coil has N turns, the field is just N times larger.

N µ0 IR2 ⎡ 1 1 ⎤
2 ⎢ x2 + R2 ⎥
( )B = Bx1 + Bx2 = ⎢ 32 + ⎡⎣( R x)2 + ⎦⎤3 2 ⎥ FIG. P30.53
⎣⎢ ⎥⎦
− R2

= N µ0 IR2 ⎡ 1 32 + 1 ⎤
⎢ x2 + R2 ⎥
( ) ( )B
2 ⎢⎣ 2R2 + x2 − 2xR 3 2 ⎦⎥

dB N µ0 IR2 ⎣⎢⎡− 3 −5 2 − 3 −5 2 (2x − 2R)⎦⎤⎥
dx 2 2 2
( ) ( )(b) (2 x)
= x2 + R2 2R2 + x2 − 2xR

Substituting x = R and canceling terms, dB = 0 .
2 dx

( ) ( ) ( )d2B

dx 2
= −3Nµ0 IR2 ⎡ x2 + R2 −5 2 − 5x2 x2 + R2 −7 2 + 2R2 + x2 − 2xR −5 2
2 ⎣⎢

( )−5(x − R)2 −7 2⎤
2R2 + x2 − 2xR ⎦⎥

Again substituting x = R and canceling terms, d2B = 0 .
2 dx 2

P30.54 “Helmholtz pair” → separation distance = radius

= 2µ0 IR2 = µ0 IR2 = µ0 I 1 turn
R 2 2 + R2 ⎦⎤3 2 + 1 3 2 R3 1.40 R
( ) [ ]B2 ⎡⎣ 1 for
4

( )For N turns in each coil, B = µ0 NI =
1.40 R
4π × 10−7 100 (10.0) 1.80 × 10−3 T
1.40 (0.500) =

P30.55 Consider first a solid cylindrical rod of radius R carrying cur-
rent toward you, uniformly distributed over its cross-sectional
area. To find the field at distance r from its center we consider kˆ
a circular loop of radius r:

͛ B ⋅ d s = µ I0 inside B = µ0 Jr B = µ0J kˆ × r P B2
2 2 B1 r1
B2πr = µ0πr2J
r2 a

Now the total field at P inside the saddle coils is the field due kˆ
to a solid rod carrying current toward you, centered at the head
of vector a, plus the field of a solid rod centered at the tail of
vector a carrying current away from you.

( )B1+ B2 = µ0 J kˆ × r1 + µ0 J −kˆ × r2 FIG. P30.55
2 2

Now note a + r1 = r2

B1 + B2 = µ0 J kˆ × r1 − µ0 J kˆ × (a + r1 ) = µ0 J a × kˆ = µ0 Ja down in the diagram
2 2 2 2

192 Chapter 30

P30.56 From Problem 33, the upper sheet creates field y ++
z ++
( )B = µ0 Js kˆ above it and µ0 Js −kˆ below it. w
22 x ––
Consider a patch of the sheet of width w parallel to ––
the z axis and length d parallel to the x axis. The
charge on it σ wd passes a point in time d , so d

v FIG. P30.56

the current it constitutes is q = σ wdv and the linear current density is Js = σ wv = σ v. Then
td w

the magnitude of the magnetic field created by the upper sheet is 1 µ0σ v. Similarly, the lower
2
sheet in its motion toward the right constitutes current toward the left. It creates magnetic field

( )1µ0σv−kˆ above it and 1 µ0σ vkˆ below it. We choose to write down answer (b) first.
2
2

(b) Above both sheets and below both, their equal-magnitude fields add to zero .

(a) Between the plates, their fields add to

( )µ0σ v −kˆ = µ0σ v away from you horizontally .

( )(c) 1 −kˆ
The upper plate exerts no force on itself. The field of the lower plate, 2 µ0σ v

will exert a force on the current in the w- by d-section, given by

( )I 1 1 2 v 2 wdˆj
× B = σ wvdˆi × 2 µ0σ v −kˆ = 2 µ0σ

The force per area is 1 µ0σ 2v2wd ˆj = 1 µ0σ 2 v2 up .
2 wd 2

σ −ˆj = wσ 2
w 2∈0
( ) ( )(d) −ˆj .
The electrical force on our section of the upper plate is qElower = σ 2∈0

The electrical force per area is wσ 2 down = σ 2 down. To have 1 µ 0σ 2 v 2 = σ2
2∈0 2 2∈0
we require 2∈0 w

v= 1 = 1

µ0 ∈0 ( )4π ×10−7 (Tm A)(N TAm) 8.85 ×10−12 C2 Nm2 (As C)2

= 3.00 ×108 m s

This is the speed of light, not a possible speed for a metal plate.

Sources of the Magnetic Field 193

*P30.57 Model the two wires as straight parallel wires (!). From the treatment
of this situation in the chapter text we have

(a) FB = µ0 I 2
2π a

( )4π × 10−7 (140)2 (2π )(0.100)
( )FB = 2π 1.00 × 10−3

= 2.46 N upward

(b) The magnetic field at the center of the loop or on
its axis is much weaker than the magnetic field
just outside the wire. The wire has negligible
curvature on the scale of 1 mm, so we model the
lower loop as a long straight wire to find the field
it creates at the location of the upper wire.

(c) aloop = 2.46 N − mloop g = 107 m s2 upward FIG. P30.57
mloop

P30.58 (a) In dB = µ0 Id s × rˆ, the moving charge constitutes a bit of current as in I = nqvA. For a
4πr2

positive charge the direction of ds is the direction of v, so dB = µ0 nqA(ds) v × rˆ. Next,
4πr2

A (ds) is the volume occupied by the moving charge, and nA (ds) = 1 for just one charge. Then,

B = µ0 qv × rˆ
4πr2

( )(( ) )( )(b)
B= 4π ×10−7 T ⋅ m A 1.60 ×10−19 C 2.00 ×107 m s 3.20 ×10−13 T
4π 1.00 ×10−3 2 sin 90.0° =

( )( )( )(c) FB = q v × B = 1.60 ×10−19 C 2.00 ×107 m s 3.20 ×10−13 T sin 90.0°

FB = 1.02 × 10−24 N directed away from the first proton

8.99 ×109 N ⋅ m2 C2 1.60 ×10−19 C 2
1.00 ×10−3 2
( ( )( ) )(d)Fe= qE= keq1q2 =
r2

Fe = 2.30 × 10−22 N directed away from the first proton

Both forces act together. The electrical force is stronger by two orders of magnitude. It is
productive to think about how it would look to an observer in a reference frame moving
along with one proton or the other.

194 Chapter 30

( )P30.59 (a)
B = µ0I = 4π ×10−7 T⋅ m A (24.0 A) = 2.74 ×10−4 T
2πr
2π (0.017 5 m)

( )( )(b) At point C, conductor AB produces a field 1 2.74 ×10−4 T −ˆj , conductor DE
2

( )( )produces a field of 1 2.74 ×10−4 T −ˆj , BD produces no field, and AE produces
2

( )negligible field. The total field at C is 2.74 ×10−4 T −ˆj .

FB = I × B = (24.0 A) 0.035 0 mkˆ ×⎡⎣5 2.74 ×10−4 T ⎤
( ) ( )( ) ( )(c) ⎦
−ˆj = 1.15 ×10−3 N ˆi

( ) ( )∑(d)
F 1.15 × 10−3 N ˆi 0.384 m s2 ˆi
a = m = 3.0 × 10−3 kg =

(e) The bar is already so far from AE that it moves through nearly constant magnetic field. The
force acting on the bar is constant, and therefore the bar’s acceleration is constant .

( )(f) 2 = vi2 + s2 (1.30 m), v f = (0.999 m s) ˆi
v f 2ax = 0+2 0.384 m so

P30.60 Each turn creates field at the center µ0 I . Together they create field B
2R I

µ0 I ⎛ 1 + 1 +…+ 1 ⎞ 4π × 10−7 TmI ⎛ 1 + 1 +…+ 1⎞ 1 FIG. P30.60
2 ⎜ R2 R50 ⎟= 2A ⎝⎜ 5.05 5.15 9.95 ⎟⎠10−2
⎝ R1 ⎠ m

= µ0 I (50 m) 6.93 = 347µ0 I m

P30.61 The central wire creates field B = µ0 I1 counterclockwise. The curved portions of the loop feels no
2π R

force since × B = 0 there. The straight portions both feel I × B forces to the right, amounting

to FB = I2 2L µ0 I1 = µ0 I1I2 L to the right .
2π R πR

P30.62 (a) From an equation in the chapter text, the magnetic field produced by one loop at the center

( )of
the second loop is given by B = µ0 IR2 = µ0I π R2 = µ0 µ where the magnetic
2x3 2π x3 2π x3

( )moment of either loop is µ = I π R2 . Therefore,

( )Fx 2
µ0 µ 3µ0 π R2 I 3π µ0 I 2 R4
= µ dB = µ ⎛ 2π ⎞ ⎛ 3 ⎞ = 2π x4 = 2 x4 .
dx ⎝⎜ ⎠⎟ ⎜⎝ x4 ⎟⎠

4π ×10−7 T⋅ m A (10.0 A)2 5.00 ×10−3 m 4
( ( ) )( )(b)Fx= 3π µ0I 2R4 = 3π = 5.92 ×10−8 N
2 x4 2 5.00 ×10−2 m 4

Sources of the Magnetic Field 195

P30.63 By symmetry of the arrangement, the magnitude of the net magnetic

field at point P is B = 8B0x where B0 is the contribution to the field

L
due to current in an edge length equal to . In order to calculate B0 ,
2
we use the Biot-Savart law and consider the plane of the square to be

the yz-plane with point P on the x-axis. The contribution to the magnetic

field at point P due to a current element of length dz and located a

distance z along the axis is given by the integral form of the Biot-Savart

law as FIG. P30.63

=∫B0 µ0 I d × rˆ
4π r2

From the figure we see that

( ) ( ( ) )r = x2 + L2 4 + z2 and d × rˆ = dz sinθ = dz
L2 4 + x2
L2 4 + x2 + z2

By symmetry all components of the field B at P cancel except the components along x
(perpendicular to the plane of the square); and

L2
L2 4 + x2
( )B0x = B0 cos φ
where cosφ =

∫Therefore, B0x = µ0 I L2 sin θ cos φdz and B = 8B0x
4π 0 r2

Using the expressions given above for sinθ cosφ, and r, we find

( )( ) ( )B = µ0IL2
2π x2 + L2 4 x2 + L2 2

P30.64 There is no contribution from the straight portion of the
wire since d s × rˆ = 0. For the field of the spiral,

dB = µ0 I (d s × rˆ )
r
(4π) 2

∫ ∫ ( )B = µ0 I 2π d s sinθ rˆ = µ0 I 2π ⎡ ⎛ 3π ⎞⎤ 1
4π θ=0 r2 4π θ=0 2dr ⎢⎣sin ⎜⎝ 4 ⎠⎟⎦⎥ r2

∫ ( )B = µ0 I 2π r−2dr = − µ0 I r−1 2θ
4π θ=0 4π θ=0

Substitute r = eθ : B = − µ0 I ⎣⎡e−θ ⎤⎦20 π = − µ0 I ⎡⎣e−2 π − e0 ⎦⎤ = FIG. P30.64
4π 4π
( )µ0 I 1− e−2π out of the page.

4π

196 Chapter 30

P30.65 Consider the sphere as being built up of little rings of radius r, dr v
centered on the rotation axis. The contribution to the field from
each ring is r
x
µ0 r 2 dI where dI = dQ = ωdQ R dx
x2 + r2 t 2π dx
( )dB = 32
2

dQ = ρdV = ρ (2πrdr)(dx)

= µ0ρω r3drdx where Q
2 x2 + r2 3 2
( )dB ρ = (4 3)π R3

FIG. P30.65

R2 −x2 µ0 ρω
∫ ∫ ( )+R

B=
x=−R r=0 2 r 3drdx
x2 + r2 3 2

Let v = r2 + x2 , dv = 2rdr, and r2 = v − x2.

+R

∫ ∫ ( ) ∫ ∫ ∫B =
R2 µ0ρω v − x2 dv dx = µ 0 ρω ⎡R R2 v−1 2dv − x2 R2 v−3 ⎤
2v3 4 ⎢ v=x2 2 d v⎥ dx
2x=−R v=x2 2
⎣⎢x=−R v=x2 ⎥⎦

µ0 ρω R ⎡⎣⎢2 R2 R2 ⎤ µ0 ρω R⎡ ⎛ 1 1 ⎞⎤
4 x=− x2 x2 ⎦⎥ 4 ⎢2 ⎜⎝⎜ R x ⎟⎟⎠⎥⎦⎥dx
∫ ( ) ∫ ( )B v1 2 2x2 v−1 2 + 2x2 −
= + dx = x=−R ⎣⎢ R− x

R

R ⎡ x2 ⎤ 2µ 0 ρω R ⎡ x2 ⎤
⎢2 + 2R⎥dx = ⎢2 − 4x + 2R⎥dx
∫ ∫B = µ0ρω −4 x
4 −R⎣ R ⎦ 4 0⎣ R ⎦

B = 2µ0 ρω ⎛ 2R3 − 4R2 ⎞ µ0 ρω R 2
⎜ + 2R2 ⎟ =
4 ⎝ 3R 2 ⎠3

P30.66 Consider the sphere as being built up of little rings of radius r, v
centered on the rotation axis. The current associated with each dr
rotating ring of charge is
r
dI = dQ = ω ⎡⎣ρ (2π rdr ) (dx )⎦⎤ x
t 2π R

The magnetic moment contributed by this ring is

dµ = A (dI ) = πr2 ω ⎣⎡ρ (2π rdr ) (dx)⎦⎤ = πωρ r 3drdx
2π

FIG. P30.66

∫ ∫ ∫ ( ) ∫ ( )µ = πωρ ⎢ r3dr⎥dx = πωρ 4
⎡+R R2 −x2 ⎤ R2 − x2 +R R2 − x2 2
+R

x=−R ⎢⎣ r=0 ⎥⎦ 4 dx = πωρ 4 dx

x=− R x=− R

∫ ( )µ = πωρ +R πωρ ⎡ (2R) ⎛ 2R3 ⎞ 2R5 ⎤
R4 − 2R2x2 + x4 dx = ⎢R4 − 2R2 ⎜ ⎟+ ⎥
4 x=−R 4 ⎣ ⎝3⎠ 5⎦

µ = πωρ R5 ⎛ − 4 + 2 ⎞ = πωρ R5 ⎛16 ⎞ = 4 πωρ R5 up
4 ⎝⎜2 3 5 ⎠⎟ 4 ⎝⎜15 ⎠⎟ 15

Sources of the Magnetic Field 197

P30. 67 Note that the current I exists in the conductor with a Bs P1 –B1
current density J = I , where –B2
A

A = π ⎡ a2 − a2 − a2 ⎤ = π a2 r
⎣⎢ 4 4 ⎥⎦ 2 Bs

Therefore J = 2I . r 2 + (a 2)2
π a2
a/2 θ
r
To find the field at either point P1 or P2 , find Bs P2
which would exist if the conductor were solid, using
a/2
Ampère’s law. Next, find B1 and B2 that would be
a −B1' θ θ −B2'

due to the conductors of radius that could occupy FIG. P30.67
2

the void where the holes exist. Then use the

superposition principle and subtract the field that

would be due to the part of the conductor where the

holes exist from the field of the solid conductor.

( )(a) Jπ (a 2)2 Jπ (a 2)2
At point P1 , Bs = µ0J π a2 , B1 = µ0 , and B2 = µ0 .
2πr 2π (r − (a 2)) 2π (r + (a 2))

B = Bs − B1 − B2 = µ Jπ a2 ⎢⎡1 − 1 − 1 ⎤
2π ⎣⎢ r
4 (r − (a 2)) 4(r +(a 2))⎥⎦⎥

= µ0 ( 2 I ) ⎡ 4r2 − a2 − 2r2 ⎤ µ0 I ⎡ 2r 2 − a2 ⎤
( ( ))B ⎢ ⎥= πr ⎢⎣ 4r2 − a2 ⎥⎦ directed to the left
2π ⎢⎣ 4r r2 − a2 4 ⎥⎦

( )(b) µ0J π (a 2)2
At point P2 , Bs = µ0J π a2 and B1′ = B2′ = 2π r2 + (a 2)2 .
2πr

The horizontal components of B1′ and B2′ cancel while their vertical components add.

( ) ( ) ( )B ⎛ ⎞
= Bs − B1′ cosθ − B2′ cosθ = µ0J π a2 − 2 ⎜ µ0 Jπa 24 ⎟ r
2π r ⎜ π r2 + a2 4 ⎟ r2 + a2 4
⎝ 2 ⎠

( )B = µ0J π a2 ⎡ r2 ⎤ = µ0 (2I ) ⎡ − 2r2 ⎤
⎢1 − ⎥ ⎢1 4r2 + ⎥
( ( ))2π r ⎣⎢ 2 r2 + a2 4 ⎥⎦ 2πr ⎣ a2 ⎦

= µ0 I ⎡ 2 r 2 + a2 ⎤ directed toward the top of the page
πr ⎢ 4 r 2 + a2 ⎥
⎣ ⎦

198 Chapter 30

ANSWERS TO EVEN PROBLEMS

P30.2 200 nT
P30.4
P30.6 ⎝⎜⎛1 + 1⎞ µ0 I into the page
P30.8 π ⎠⎟ 2R
P30.10
See the solution.

⎛1 + 1⎞ µ0 I into the page
⎜⎝ π 4 ⎠⎟ 2r

262 nT into the page

P30.12 (a) 3.84 × 10−21 N up. See the solution. (b) 21.4 mm. This distance is negligible compared to 50
m, so the electron does move in a uniform field. (c) 134 revolutions.

P30.14 ( )µ0 I a2 + d2 − d a2 + d2
into the page
2π ad a2 + d2

P30.16 4.97 mm

P30.18 (a) 10.0 µT out of the page (b) 80.0 µN toward wire 1 (c) 16.0 mT into the page
(d) 80.0 µN toward wire 2

P30.20 (a) It is possible in just one way. (b) wire 3 must be 12.0 cm to the left of wire 1, carrying
(c) current 2.40 A down

P30.22 (a) 10.0 mm. (b) Yes. If we tried to use wires with diameter 10 mm, they would feel the force
only momentarily after we turn on the current, until they melt. We can use wide, thin sheets of
copper, perhaps plated onto glass, and perhaps with water cooling, to have the forces act
continuously.

P30.24 5.40 cm

P30.26 (a) 400 cm (b) 7.50 nT (c) 1.26 m (d) zero

P30.28 (a) 3.60 T (b) 1.94 T

P30.30 500 A
P30.32
P30.34 (a) See the solution. (b) zero; zero (c) B = 32 ×10−7 z T⋅ m ˆj (d) at d = 3.00 cm (e) 53.3 ˆj mT
9 ×10−4 m2 + z2

(a) Make the wire as long and thin as possible without melting when it carries the 5-A current.
Then the solenoid can have many turns. (b) As small in radius as possible with your
experiment fitting inside. Then with a smaller circumference the wire can form a solenoid with
more turns.

P30.36 (a) µ0 IN ⎡ a − a− ⎤ (b) See the solution.
P30.38 ⎢ ⎥
2 ⎢⎣ a2 + R2 (a − )2 + R2 ⎦⎥

207 W

P30.40 (a) −Bπ R2 cosθ (b) Bπ R2 cosθ

Sources of the Magnetic Field 199

P30.42 The field can be uniform in magnitude. Gauss’s law for magnetism implies that magnetic field
lines never start or stop. If the field is uniform in direction, the lines are parallel and their density
stays constant along any one bundle of lines. Thus the magnitude of the field has the same value
at all points along a line in the direction of the field. The magnitude of the field could vary over a
plane perpendicular to the lines, or it could be constant throughout the volume.

P30.44 (a) 12.6 µT (b) 56.0 µT

P30.46 (a) Gravitational field and magnetic field are vectors; atmospheric pressure is a scalar. (b) At
42° north latitude, 76° west longitude, 260 m above sea level: Gravitational field is 9.803 m/s2
down. From the Coast and Geodetic Survey, the magnetic field is 54 mT at 12° west of geographic
north and 69° below the horizontal. Atmospheric pressure is 98 kPa. The atmosphere is held on
by gravitation, but otherwise the effects are all separate. The magnetic field could be produced by
permanent magnetization of a cold iron-nickel deposit within the Earth, so it need not be associ-
ated with present-day action of gravitation.

P30.48 2.01 GA west
P30.50 ~10−5 T , enough to affect the compass noticeably

P30.52 µ0 qω

2.5 5π R

P30.54 1.80 mT
P30.56
P30.58 (a) µ0σ v horizontally away from you (b) 0 (c) 1 µ0σ 2 v2 up (d) 3.00 ×108 m s
2

(a) See the solution. (b) 3.20 × 10−13 T (c) 1.02 × 10−24 N away from the first proton

(d) 2.30 × 10−22 N away from the first proton

P30.60 347µ0 I m perpendicular to the coil

P30.62 (a) See the solution. (b) 59.2 nN

P30.64 (m0 I/4p)(1 – e−2p) out of the plane of the paper
P30.66 4 πωρR5 upward
15