22
자기력과 자기장
CHAPTER OUTLINE
22.1
Historical Overview
22.2
The Magnetic Field
22.3
Motion of a Charged Particle in a Uniform Magnetic Field
22.4
Applications Involving Charged Particles Moving in a Magnetic Field
22.5
Magnetic Force on a Current-Carrying Conductor
22.6
Torque on a Current Loop in a Uniform Magnetic Field
22.7
The Biot–Savart Law
22.8
The Magnetic Force Between Two Parallel Conductors
22.9
Ampère’s Law
22.10
The Magnetic Field of a Solenoid
22.11
Context Connection: Remote Magnetic Navigation for Cardiac
Catheter Ablation Procedures
* An asterisk indicates an item new to this edition.
ANSWERS TO OBJECTIVE QUESTIONS
*OQ22.1
Answers (d) and (e). The force that a magnetic field exerts on a
moving charge is always perpendicular to both the direction of the
field and the direction of the particle’s motion. Since the force is
perpendicular to the direction of motion, it does no work on the
particle and hence does not alter its speed. Because the speed is
unchanged, both the kinetic energy and the magnitude of the linear
momentum will be constant.
*OQ22.2
Answers (b) and (c). In each case, electric charge is moving.
*OQ22.3
Answers (c) and (e). The magnitude of the magnetic force
357
358
Chapter 22
experienced by a charged particle in a magnetic field is given by
FB = q vBsin θ , where v is the speed of the particle and θ is the angle
between the direction of the particle’s velocity and the direction of
the magnetic field. If either v = 0 [choice (e)] or sin θ = 0 [choice (c)],
this force has zero magnitude.
*OQ22.4
Answer (a). The contribution made to the magnetic field at point P
by the lower wire is directed out of the page, while the contribution
due to the upper wire is directed into the page. Since point P is
equidistant from the two wires, and the wires carry the same
magnitude currents, these two oppositely directed contributions to
the magnetic field have equal magnitudes and cancel each other.
*OQ22.5
Answer (a) and (c). The magnetic field due to the current in the
vertical wire is directed into the page on the right side of the wire
and out of the page on the left side. The field due to the current in the
horizontal wire is out of the page above this wire and into the page
below the wire. Thus, the two contributions to the total magnetic
field have the same directions at points B (both out of the page) and
D (both contributions into the page), while the two contributions
have opposite directions at points A and C. The magnitude of the
total magnetic field will be greatest at points B and D where the two
contributions are in the same direction, and smallest at points A and
C where the two contributions are in opposite directions and tend to
cancel.
(a) Yes, as described by F = qE. (b) No, because, as described by
FB = qv × B , when v = 0, FB = 0.
(c) Yes. F = qE does not depend upon velocity. (d) Yes, because the
velocity and magnetic field are perpendicular. (e) No, because the
wire is uncharged. (f) Yes, because the current and magnetic field are
perpendicular. (g) Yes. (h) Yes.
*OQ22.7
*OQ22.6
*OQ22.7
Answer (c). FB = qv × B and î × (− k̂) = ĵ.
*OQ22.8
Answer (a). According to the right-hand rule, the magnetic field at
point P due to the current in the wire is directed out of the page, and
the magnitude of this field is given by Equation 22.21: B = μ0 I 2π r .
*OQ22.9
Answer (c). The magnetic force must balance the weight of the rod.
From equation 22.10,
FB = IL × B → FB = ILBsin θ
For maximum current, θ = 90°, and we have
ILBsin 90° = mg
→
2
mg ( 0.050 0 kg ) ( 9.80 m/s )
I=
=
= 4.90 A
LB
(1.00 m ) ( 0.100 T )
Magnetic Forces and Magnetic Fields
359
*OQ22.12 Ranking AA > AC > AB. The torque exerted on a single turn coil
*OQ21.10
carrying current I by a magnetic field B is τ = BIA sin θ . The normal
perpendicular to the plane of each coil is also perpendicular to the
direction of the magnetic field (i.e., θ = 90°). Since B and I are the
same for all three coils, the torques exerted on them are proportional
to the area A enclosed by each of the coils. Coil A is rectangular with
2
the largest area AA = (1 m)(2 m) = 2 m . Coil C is triangular with area
1
AC = ( 1 m ) ( 3 m ) = 1.5 m 2 . By inspection of the figure, coil B
2
encloses the smallest area.
SOLUTIONS TO END-OF-CHAPTER PROBLEMS
Section 22.2
P22.1
The Magnetic Field
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.
ANS. FIG. P22.1
(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 .
P22.2
See ANS. FIG. P22.2 for right-hand rule diagrams for each of the
situations.
(a)
(b)
(c)
(d)
up
out of the page, since the charge is negative.
no deflection
into the page
ANS. FIG. P22.2
360
P22.3
Chapter 22
First find the speed of the electron.
( ΔK + ΔU )i = ( ΔK + ΔU ) f
(a)
1
mv 2 = eΔV :
2
2 ( 1.60 × 10−19 C ) ( 2 400 J C )
2eΔV
=
m
v=
→
9.11 × 10
−31
kg
= 2.90 × 107 m s
FB, max = qvB = ( 1.60 × 10−19 C ) ( 2.90 × 107 m s ) ( 1.70 T )
= 7.90 × 10−12 N
(b)
P22.4
FB, min = 0 occurs when v is either parallel to or anti-parallel to
B.
FB = qv × B
î
ĵ k̂
v × B = +2 −4 +1 = ( 12 − 2 ) î + ( 1 + 6 ) ĵ + ( 4 + 4 ) k̂ = 10î + 7 ĵ + 8k̂
+1 +2 −3
v × B = 102 + 7 2 + 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
Section 22.3
P22.5
(a)
Motion of a Charged Particle
in a Uniform Magnetic Field
We begin with
or
But
qvB =
mv 2
R
qRB = mv.
L = mvR = qR B.
2
Therefore, R =
L
=
qB
4.00 × 10−25 J ⋅ s
= 0.050 0 m
(1.60 × 10−19 C)(1.00 × 10−3 T )
= 5.00 cm
(b)
P22.6
L
4.00 × 10−25 J ⋅ s
=
= 8.78 × 106 m s .
Thus, v =
−31
mR ( 9.11 × 10 kg ) ( 0.050 0 m )
By conservation of energy for the proton-electric field system in the
process that set the proton moving, its kinetic energy is
E = 21 mv 2 = eΔV
so its speed is v =
2eΔV
.
m
Magnetic Forces and Magnetic Fields
361
Now Newton’s second law for its circular motion in the magnetic field
is
mv 2
∑ F = ma which becomes R = evB sin 90.
so
B =
mv
m 2eΔV
2mΔV
= 1
=
.
R
eR
eR
m
e
and
−27
6
1
⎛
⎞ 2 ( 1.67 × 10 kg ) ( 10.0 × 10 V )
B=⎜
⎝ 5.80 × 1010 m ⎟⎠
1.60 × 10−19 C
= 7.88 × 10−12 T
Section 22.4
P22.7
Applications Involving Charged Particles
Moving in a Magnetic Field
In the velocity selector: v =
E 2 500 V m
=
= 7.14 × 10 4 m/s
0.035 0 T
B
In the deflection chamber:
−26
4
mv ( 2.18 × 10 kg ) ( 7.14 × 10 m/s )
=
r=
=
qB
(1.60 × 10−19 C)( 0.035 0 T )
P22.8
0.278 m
FB = Fe
so
qvB = qE,
where v =
2K
and K is kinetic energy of the electron.
m
E = vB =
2K
B=
m
2 ( 750 eV ) ( 1.60 × 10−19 J/eV )
9.11 × 10−31 kg
( 0.015 0 T )
= 244 kV/m
*P22.9
(a)
FB = qvB =
mv 2
R
−19
v qBR qB ( 1.60 × 10 C ) ( 0.450 T )
=
=
ω= =
= 4.31 × 107 rad/s
1.67 × 10−27 kg
R mR
m
(b)
−19
qBR ( 1.60 × 10 C ) ( 0.450 T ) ( 1.20 m )
=
v=
= 5.17 × 107 m/s
m
1.67 × 10−27 kg
362
Chapter 22
Section 22.5
P22.10
P22.18
(a)
Magnetic Force on a Current-Carrying Conductor
The magnetic field is perpendicular to all line elements d s on the
ring, so the magnetic force dF = Ids × B on each element has
magnitude I d s × B = IdsB and is radially inward and upward, at
angle θ above the radial line. The radially inward components
IdsB cos θ tend to squeeze the ring but all cancel out because
forces on opposite sides of the ring cancel in pairs. The upward
components IdsB sin θ all add to I ( 2π r ) Bsin θ .
(a)
magnitude: 2π rIB sin θ
(b)
direction: up, away from magnet
ANS.
ANS. FIG.
FIG. P22.10
P22.18
P22.11
P22.19
*P22.12
(a)
FB = ILBsin 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)
The magnetic force must be upward to lift
the wire. For current in the south direction,
the magnetic field must be east to
produce an upward force, as shown by the
right-hand rule in the figure.
(b)
FB = ILBsin θ
with
mg = ILBsin θ
so
ANS. FIG. P22.12
FB = Fg = mg
m
g = IBsin θ
L
→
B=
m g
L I sin θ
Magnetic Forces and Magnetic Fields
B=
Section 22.6
P22.25
P22.13
(a)
363
⎞
⎛ 0.500 × 10−3 kg ⎞ ⎛
9.80 m/s 2
m g
=⎜
= 0.245 T
−2
⎟
⎜
L I sin θ ⎝ 1.00 × 10 m ⎠ ⎝ ( 2.00 A ) sin 90.0° ⎟⎠
Torque on a Current Loop
in a Uniform Magnetic Field
τ = NBAI sin φ
τ = 100 ( 0.800 T ) ( 0.400 × 0.300 m 2 ) × ( 1.20 A ) sin 60°
τ = 9.98 N ⋅ m
(b)
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, clockwise as seen looking down from a position
on the positive y axis.
ANS. FIG. P22.13
P22.27
P22.14
(a)
2 π r = 2.00 m
→
(b)
r = 0.318 m
2
μ = IA = ( 17.0 × 10−3 A ) ⎡⎣π ( 0.318 ) m 2 ⎤⎦ = 5.41 mA ⋅ m 2
τ =μ×B
τ = ( 5.41 × 10−3 A ⋅ m 2 ) ( 0.800 T ) = 4.33 mN ⋅ m
Section 22.7
P22.15
The Biot–Savart Law
For leg
, ds × r̂ = 0, so there is no contribution to the field from this
contribution to the field from this
segment. For leg
the
wire is only semi-infinite;
thus,
1⎛ μ I ⎞ μ I
B = ⎜ 0 ⎟ = 0 into the paper
2 ⎝ 2 π x ⎠ 4π x
ANS. FIG. P22.15
364
Chapter 22
*P22.16
We can think of the total magnetic field as the superposition of the
μI
field due to the long straight wire, having magnitude 0 and
2π R
directed into the page, and the field due to the circular loop, having
μI
magnitude 0 and directed into the page. The resultant magnetic
2R
field is:
−7
⎛
1 ⎞ μ0 I ⎛
1 ⎞ ( 4π × 10 T ⋅ m/A ) ( 1.00 A )
B = ⎜1+ ⎟
= ⎜1+ ⎟
⎝
π ⎠ 2R ⎝
π⎠
2 ( 0.150 m )
= 5.52 × 10−6 = 5.52 μT into the page
P22.17
Treat the magnetic field as that produced in the center of a ring of
μI
radius R carrying current I: from Equation 22.24, the field is B = 0 .
2R
Δq
e
ev
=
=
, so the
The current due to the electron is I =
Δt 2π R v 2π R
magnetic field is
B=
μ0 I μ0 ⎛ ev ⎞ μ0 ev
=
⎜
⎟=
2R 2R ⎝ 2π R ⎠ 4π R 2
−19
6
⎛ 4π × 10−7 T ⋅ m/A ⎞ ( 1.60 × 10 C ) ( 2.19 × 10 m/s )
=⎜
= 12.5 T
2
⎟⎠
−11
4π
⎝
5.29
×
10
m
(
)
P22.18
(a)
First, we need to determine the
magnetic field due to a finitelength wire. Let’s start by
considering a length element d s
located a distance r from P, as
shown in ANS. FIG. P22.18(a). The
direction of the magnetic field at
point P due to the current in this
element is out of the page because
ANS. FIG. P22.18(a)
ds × r is out of the page. In fact,
because all the current elements
I ds lie in the plane of the page,
they all produce a magnetic field
directed out of the page at point P.
Therefore, the direction of the
magnetic field at point P is out of
the page and we need only find the
ANS. FIG. P22.18(b)
magnitude of the field. We place the
origin at O and let point P be along the positive y axis, with k̂
being a unit vector pointing out of the page. Evaluating the cross
Magnetic Forces and Magnetic Fields
365
product in the Biot-Savart law gives
⎡
⎛π
⎞⎤
ds × r = ds × r k̂ = ⎢ dx sin ⎜ − θ ⎟ ⎥ k̂ = ( dx cos θ ) k̂
⎝
⎠⎦
2
⎣
Substituting into Equation 22.20 gives
μ I dx cos θ
k̂
dB = dBk̂ = 0
2
4π
r
[1]
From the geometry of ANS. FIG. P22.18(a), we express r in terms
of θ:
r=
a
cos θ
[2]
Notice that tan θ = –x/a from the right triangle in ANS. FIG.
P22.41(a)
necessary because
because dds is located at aa
P22.18(a) (the negative sign is necessary
negative value of x) and solve for x:
x:
x = −a tan θ
Then
2
dx = −a sec 2 θ dθ = −
adθ
2
cos θ
[3]
Substituting [2] and [3] into the magnitude of the field from
equation [1]:
2
μ0 I ⎛ adθ ⎞ ⎛ cos θ ⎞
μI
dB = −
cos θ = − 0 cos θ dθ
⎜⎝
2 ⎟⎜
2
⎟
4π cos θ ⎠ ⎝ a ⎠
4π a
[4]
We then integrate equation [4] over all length elements on the
wire, where the subtending angles range from θ1 to θ2 as defined
in ANS. FIG. P22.18(b):
B=−
μ0 I
4π a
∫
θ2
θ1
cos θ dθ =
μ0 I
( sin θ1 − sin θ 2 )
4π a
From Figure
Figure P22.18,
P22.41, we note that θ1 = 45.0°, θ2 = –45.0°, and a =
From
[5]
.
2
negative value of x) and solve for x:
ds is located at a
ANS. FIG. P22.18(c)
ANS. FIG.
P22.41(c)
x) and solve for x:
Also,
from
ANS.
FIG.
P22.18(c),
each
side
produces
a field
into the
Notice that tan θ = –x/a from the
right
triangle
in ANS.
FIG.
page.
The(the
fournegative
sides altogether
produce because ds is located at a
P22.41(a)
sign is necessary
366
Chapter 22
Bcenter = 4B = 4
B=
μ0 I
( sin θ1 − sin θ 2 )
4π a
=
μ0 I
[ sin 45.0° − sin ( −45.0°)]
π 2
=
2 μ0 I ⎡ 2 ⎤ 2 2 μ0 I
=
π ⎢⎣ 2 ⎥⎦
π
2 2 ( 4π × 10−7 T ⋅ m/A ) ( 10.0 A )
π ( 0.400 m )
= 2 2 × 10−5 T = 28.3 μT into the page
(b)
For a single circular turn with 4 = 2 π R ,
μ I μ π I ( 4π × 10 T ⋅ m/A ) π ( 10.0 A )
B= 0 = 0 =
4
2R
4 ( 0.400 m )
−7
= 24.7 μT into the page
Section 22.8
P22.19
The Magnetic Force Between
Two Parallel Conductors
To the right of the long, straight wire, current I1 creates a magnetic
field into the page. 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 (from Equation 22.27):
μ I I ⎛ 1
μ I I ⎡ −a ⎤
1⎞
F = F1 + F2 = 0 1 2 ⎜
− ⎟ î = 0 1 2 ⎢
⎥ î
2π ⎝ c + a c ⎠
2π ⎣ c ( c + a ) ⎦
( 4π × 10−7 N A 2 ) ( 5.00 A ) ( 10.0 A ) ( 0.450 m )
F=
2π
⎛
⎞
−0.150 m
×⎜
î
⎝ ( 0.100 m ) ( 0.250 m ) ⎟⎠
F = −2.70 × 10−5 î N = −27.0 × 10−6 î N = −27.0î μN
(
)
(
ANS. FIG. P22.19
)
Magnetic Forces and Magnetic Fields
P22.20
367
Let both wires carry current in the x direction, the first at y = 0 and the
second at y = 10.0 cm.
(a)
μ0 I
4π × 10−7 T ⋅ m A ) ( 5.00 A )
(
B=
k̂ =
k̂
2π r
2 π ( 0.100 m)
B = 1.00 × 10−5 T out of the page
ANS. FIG. P22.20(a)
(b)
FB = I 2 × B = ( 8.00 A ) ⎡⎣( 1.00 m ) î × ( 1.00 × 10−5 T ) k̂ ⎤⎦
( )
= ( 8.00 × 10−5 N ) − ĵ
ANS. FIG. P22.20(b)
FB = 8.00 × 10−5 N toward the first wire
(c)
μ0 I
4π × 10−7 T ⋅ m A ) ( 8.00 A )
(
B=
− k̂ =
− k̂
2π r
2π ( 0.100 m )
( )
( )
( )
= ( 1.60 × 10−5 T ) − k̂
ANS. FIG. P22.20(c)
B = 1.60 × 10−5 T into the page
(d)
FB = I1 × B = ( 5.00 A ) ⎡( 1.00 m ) î × ( 1.60 × 10−5 T ) − k̂ ⎤
⎣
⎦
( )
= ( 8.00 × 10−5 N ) + ĵ
( )
ANS. FIG. P22.20(d)
FB = 8.00 × 10−5 N towards the second wire
P22.21
Carrying oppositely directed currents, wires 1 and 2 repel each other.
368
Chapter 22
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 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 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.
ANS. FIG. P22.21
(b)
For the equilibrium of wire 3 we have
μ0 ( 1.50 A ) I 3
μ0 ( 4.00 A ) I 3
=
2π d
2π ( 20.0 cm + d )
F1 on 3 = F2 on 3 :
1.50(20.0 cm + d) = 4.00d
d=
30.0 cm
= 12.0 cm to the left of wire 1
2.50
(a)
Thus the situation is possible in just one way.
(c)
For the equilibrium of wire 1,
μ0 I 3 ( 1.50 A ) μ0 ( 4.00 A ) ( 1.50 A )
=
2π ( 12.0 cm )
2π ( 20.0 cm )
I3 =
12
( 4.00 A ) = 2.40 A down
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 equalmagnitude forces that 1 exerts on 3 and that 3 exerts on 1.
Section 22.9
*P22.22
(a)
Ampère’s Law
From Ampère’s law, the magnetic field at point a is given by
μI
Ba = 0 a , where Ia is the net current through the area of the
2π ra
circle of radius ra. In this case, Ia = 1.00 A out of the page (the
current in the inner conductor), so
( 4π × 10 T ⋅ m A )(1.00 A )
2π ( 1.00 × 10 m )
−7
Ba =
−3
= 200 μT toward top of page
Magnetic Forces and Magnetic Fields
(b)
369
μ0 Ib
, where Ib is the net current through
2π rb
the area of the circle having 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,
Similarly at point b: Bb =
( 4π × 10 T ⋅ m A )( 2.00 A )
2π ( 3.00 × 10 m )
−7
Bb =
−3
= 133 μT toward bottom of page
P22.48
P22.23
P22.24
−3
2π rB 2π ( 1.00 × 10 m ) ( 0.100 T )
From ∫ B ⋅ d = μ0 I, I =
=
= 500 A .
μ0
4π × 10−7 T ⋅ m A
By Ampère’s law, the field at the position of the wire at distance r from
the center is due to the fraction of the other 99 wires that lie within the
radius r.
∫ B ⋅ ds = μ0 I:
⎡ ⎛ π r 2 ⎞⎤
B ⋅ 2 π r = μ 0 ⎢99I ⎜ 2 ⎟⎥
⎣ ⎝ π R ⎠⎦
→
B=
μ 0 ( 99I ) ⎛ r 2 ⎞ μ 0 ( 99I ) ⎛ r ⎞
⎜ ⎟
⎜ ⎟=
2π r ⎝ R2 ⎠
2π R ⎝ R ⎠
ANS. FIG. P22.24
The field is proportional to r, as shown in the graph. This field points
tangent to a circle of radius r and exerts a force F = I × B on the wire
toward the center of the bundle. The magnitude of the force per unit
length is
⎡ μ 0 ( 99) I ⎛ r ⎞⎤
μ 0 ( 99) I 2 ⎛ r ⎞
F
= IBsin θ = I ⎢
⎜ ⎟⎥ sin 90° =
⎜ ⎟
2π R ⎝ R ⎠
⎣ 2 π R ⎝ R ⎠⎦
( 4π × 10
=
−7
T ⋅ m/A ) ( 99) ( 2.00 A )
2 π ( 0.500 × 10
(a)
6.34 × 10−3 N m
−2
m)
2
(0.400) = 6.34 × 10−3
N m
370
Chapter 22
(b)
Referring to ANS. FIG. P22.24, the field is clockwise, so at the
position of the wire, the field is downward, and the force is
inward toward the center of the bundle .
(c)
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 .
P22.25
P22.53
(a)
(b)
Binner =
Bouter
Section 22.10
P22.26
(a)
μ 0 NI ( 4π × 10
=
2π r
−7
μ NI ( 4π × 10
=
= 0
2π r
T ⋅ m A ) ( 900) (14.0 × 103 A )
2 π ( 0.700 m)
−7
= 3.60 T
T ⋅ m A ) ( 900 ) ( 14.0 × 103 A )
2π ( 1.30 m )
= 1.94 T
The Magnetic Field of a Solenoid
The field produced by the solenoid in its interior is given by
⎛ 30.0 ⎞
B = μ 0 nI −î = ( 4π × 10−7 T ⋅ m A ) ⎜ −2 ⎟ (15.0 A ) −î
⎝ 10 m ⎠
B = − ( 5.65 × 10−2 T ) î
( )
( )
The force exerted on side AB of the square current loop is
FB
= IL × B = ( 0.200 A )
( )
AB
( )
× ⎡( 2.00 × 10−2 m ) ĵ × ( 5.65 × 10−2 T ) − î ⎤
⎣
⎦
(F )
B AB
= ( 2.26 × 10−4 N ) k̂
ANS. FIG. P22.26
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 of the loop .
371
Magnetic Forces and Magnetic Fields
(b)
From the above result, it is seen that the net torque exerted on the
square loop by the field of the solenoid is zero . More formally,
the magnetic dipole moment of the square loop is given by
2
μ = IA = ( 0.200 A ) ( 2.00 × 10−2 m ) − î = −80.0 μA ⋅ m 2 î
( )
The torque exerted on the loop is then
τ = μ × B = −80.0 μA ⋅ m 2 î × −5.65 × 10−2 T î = 0
(
P22.27
P22.56
(a)
) (
)
Let the axis of the solenoid lie along the y axis from y = − to y =
0. We will determine the field at position y = x: this point will be
inside the solenoid if − < x < 0 and outside if x < − or x > 0. 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
⎛N ⎞
of turns per length is ⎜ ⎟ . So the number of turns in the ring is
⎝⎠
⎛N ⎞
⎛ N⎞
⎜ ⎟ dy and the current in the ring is I ring = I ⎜ ⎟ dy. Now, we use
⎝ ⎠
⎝⎠
Equation 22.23 for the field created by one ring:
Bring =
μ 0 I ring a 2
2 ⎡⎣( x − y ) + a 2 ⎤⎦
32
2
where x – y is the distance from the center of the ring, at location
y, to the field point (note that y is negative, so x – y = x + |y|).
Each ring creates a field in the same direction, along the y axis, so
the whole field of the solenoid is
B = ∑ Bring = ∑
all rings
=
μ0 I ring a 2
0
2 ⎡( x − y ) + a 2 ⎤
⎣
⎦
dy
2
32
→∫
−
μ0 I ( N ) a 2 dy
2 ⎡( x − y ) + a 2 ⎤
⎣
⎦
2
32
μ0 INa 2 0
∫
2 − ⎡( x − y )2 + a 2 ⎤ 3 2
⎣
⎦
To perform the integral we change variables to
u=x–y
→
dy = –du
then
B=−
μ0 INa 2 x
du
∫
2 x + ( u2 + a 2 )3 2
and then use the table of integrals in the appendix:
μ INa 2
u
B=− 0
2
2 a u2 + a 2
x
=−
x+
μ0 IN ⎡
x
⎢
−
2
2 ⎢ x + a 2
⎣
⎤
⎥
( x + )2 + a 2 ⎥⎦
x+
372
Chapter 22
=
(b)
μ0 IN ⎡
⎢
2 ⎢
⎣
x+
( x + )2 + a 2
−
⎤
⎥
x 2 + a 2 ⎥⎦
x
If is much larger than a and x = 0,
we have B ≅
⎤ μ IN
μ 0 IN ⎡ ⎢ 2 + 0⎥ = 0 .
2 ⎣ 2
⎦
This is just half the magnitude of the field deep within the
solenoid. We would get the same result by substituting x = − to
describe the other end.
P22.57
P22.28
The magnetic field at the center of a solenoid is B = μ0
N
I, so
1.00 × 10−4 T ) ( 0.400 m )
(
B
I=
=
= 31.8 mA
μ0 n ( 4π × 10−7 T ⋅ m A ) ( 1 000 )
Section 22.11
P22.29
(a)
From Equation 22.16, we have
τ =μ×B
so
(b)
Context Connection: Remote Magnetic Navigation
for Cardiac Catheter Ablation Procedures
τ = ( 0.10 A ⋅ m 2 ) ( 0.080 T ) sin 30° = 4.0 mN ⋅ m .
The potential energy of a system of a magnetic moment in a
magnetic field is given by Equation 22.17:
U = − μ ⋅ B = μBcos φ = ( 0.10 A ⋅ m 2 ) ( 0.080 T ) cos 30°
= −6.9 mJ
Additional Problems
*P22.30
We may regard the sheet as being
composed of filaments of current J s ds
directed out of the page. According to the
Biot-Savart law, the field contribution at a
point has the direction ds × r̂, where r̂
points
from
the current
to theThe resulting field from both
filament
+z and
–x field filament
components.
point.
Consider
thethe
field
contributions
at
an reasoning, the magnetic
filaments
points
+z-direction.
similar
equidistant
fromin
the normal
(and P).By
The
arbitrary
point
P to
of the
sheet.
field
any
point
onthe
theright
left+z
side
of+x
thefield
sheet
points in the
–z-direction.
upperatfilament
contributes
and
components,
but
the lower
Draw
a
line
normal
to
the
sheet
that
passeswithin the sheet. Also, the
These
same
arguments
hold
for
any
point
filament +z and –x field components. The resulting field from both
through
P. Consider
thethat
contributions
toof filaments that lie on the
same
reasoning
for any pair
filaments
pointsshows
in the +z-direction.
By similar
reasoning, the magnetic
the
field
at
P
from
two
filaments
that
lie
sameat
vertical
line,on
thethe
magnetic
at sheet
a point
midway
between
them
field
any point
left sidefield
of the
points
in the
–z-direction.
ANS.
FIG.
P22.30
A
NS.
FIG.
P22.64
along
the
samethe
vertical
lineno
and
are
is
zero.
Thus,
field
has
horizontal
component
within
the
sheet.
These same arguments hold for any point within the sheet. Also, the
equidistant from the normal (and P). The
same reasoning shows that for any pair of filaments that lie on the
upper filament contributes +z and +x field components, but the lower
same vertical line, the magnetic field at a point midway between them
is zero. Thus, the field has no horizontal component within the sheet.
Magnetic Forces and Magnetic Fields
*P22.64
373
We may +z
regard
thefield
sheetcomponents.
as being
filament
and –x
The resulting field from both
s similar reasoning, the magnetic
composedpoints
of filaments
current J s dBy
filaments
in the +ofz-direction.
field
at any
the left
side of the
sheet points in the –z-direction.
directed
outpoint
of theonpage.
According
to the
These
same
arguments
hold
for
any
point
Biot-Savart law, the field contribution at a within the sheet. Also, the
same
shows that
any pair
point reasoning
has the direction
ds ×for
r̂, where
r̂ of filaments that lie on the
same
vertical
line,
the
magnetic
field
at
points from the current filament to the a point midway between them
is
zero.Consider
Thus, thethe
field
hascontributions
no horizontalatcomponent
within the sheet.
point.
field
an
Therefore,
each Pfilament
of current
arbitrary point
to the right
of the creates
sheet. a contribution to the total
field
is parallel
to the current
Drawthat
a line
normal to
to the
the sheet
sheet and
that perpendicular
passes
direction.
createthe
field
lines straight
through P.They
Consider
contributions
toup to the right of the sheet
and
straight
to the
left of the
sheet.
the field
at Pdown
from two
filaments
that
lie
along Ampère’s
the same vertical
line and
aresuggested rectangle,
From
law applied
to the
equidistant
from
the
normal
(and
P). The
= μand
B ⋅ ds = μ 0contributes
I : B ⋅ 2 + 0+z
upper∫ filament
0 J s +x field components, but the lower
Therefore the field is uniform in space, with the magnitude
B=
P22.31
(a)
μ0 J s
2
The net force is the Lorentz force given by
F = qE + qv × B = q E + v × B
F = ( 3.20 × 10−19 )
(
(
)
) (
) (
)
⎡ 4î − 1ĵ − 2k̂ + 2 î + 3 ĵ − 1k̂ × 2 î + 4 ĵ + 1k̂ ⎤ N
⎣
⎦
Carrying out the indicated operations, we find:
F = 3.52 î − 1.60 ĵ × 10−18 N
(
(b)
P22.66
P22.32
)
⎛
⎛F ⎞
θ = cos −1 ⎜ x ⎟ = cos −1 ⎜
⎝ F⎠
⎜⎝
axis.
⎞
⎟ = 24.4° below the +x
( 3.52 )2 + (1.60)2 ⎟⎠
3.52
(a)
If the
carriers
are negative,
carry(!).
current
x
Model
the charge
two wires
as straight
paralleltowires
From in
thethe
treatment
of this situation in the chapter text (refer to Equation 22.27), we have
(a)
FB =
FB
μ0 I 2 2π a
( 4π × 10
=
−7
2
T ⋅ m/A ) (140 A ) ⎡⎣( 2 π ) ( 0.100 m)⎤⎦
2 π (1.00 × 10−3 m)
= 2.46 N upward
374
Chapter 22
ANS. FIG. P22.32
(b)
Equation 22.23, Bx =
μ0 Ia 2
2 ( a2 + x2 )
3/2
is the expression for
the magnetic field produced a distance x above the
center of a loop. 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
mloop
= 107 m s 2 upward