Chapter 19
Magnetism
Quick Quizzes
1.
(b). The force that a magnetic field exerts on a charged particle moving through it is given
by F = qvB sin θ = qvB⊥ , where B⊥ is the component of the field perpendicular to the
particle’s velocity. Since the particle moves in a straight line, the magnetic force (and
hence B⊥ , since qv ≠ 0 ) must be zero.
2.
(c). The magnetic force exerted by a magnetic field on a charge is proportional to the
charge’s velocity relative to the field. If the charge is stationary, as in this situation, there is
no magnetic force.
3.
(c). The torque that a planar current loop will experience when it is in a magnetic field is
given by τ = BIA sin θ . Note that this torque depends on the strength of the field, the
current in the coil, the area enclosed by the coil, and the orientation of the plane of the coil
relative to the direction of the field. However, it does not depend on the shape of the loop.
4.
(a). The magnetic force acting on the particle is always perpendicular to the velocity of the
particle, and hence to the displacement the particle is undergoing. Under these conditions,
the force does no work on the particle and the particle’s kinetic energy remains constant.
5.
(b). The two forces are an action-reaction pair. They act on different wires and have equal
magnitudes but opposite directions.
141
142
CHAPTER 19
Answers to Even Numbered Conceptual Questions
2. No. The force that a constant magnetic field exerts on a charged particle is dependent on
the velocity of that particle. If the particle has zero velocity, it will experience no magnetic
force and cannot be set in motion by a constant magnetic field.
4. The force exerted on a current-carrying conductor by a magnetic field is F = BI lsin θ ,
where θ is the angle between the direction of the current and the direction of the magnetic
field. Thus, if the current is in a direction parallel (θ = 0 ) or anti-parallel (θ = 180° ) to the
magnetic field, there is no magnetic force exerted on the conductor.
6. Straight down toward the surface of Earth.
8. The magnet causes domain alignment in the iron such that the iron becomes magnetic and
is attracted to the original magnet. Now that the iron is magnetic, it can produce an
identical effect in another piece of iron.
10. The magnet produces domain alignment in the nail such that the nail is attracted to the
magnet. Regardless of which pole is used, the alignment in the nail is such that it is
attracted to the magnet.
12. The magnetic field inside a long solenoid is given by B = µ0 nI = µ0 NI l .
(a) If the length l of the solenoid is doubled, the field is cut in half.
(b) If the number of turns, N, on the solenoid is doubled, the magnetic field is doubled.
14. Near the poles the magnetic field of Earth points almost straight downward (or straight
upward), in the direction (or opposite to the direction) the charges are moving. As a result,
there is little or no magnetic force exerted on the charged particles at the pole to deflect
them away from Earth.
16. The loop can be mounted on an axle that can rotate. The current loop will rotate when
placed in an external magnetic field for some arbitrary orientation of the field relative to
the loop. As the current in the loop is increased, the torque on it will increase.
18. Yes. If the magnetic field is directed perpendicular to the plane of the loop, the forces on
opposite sides of the loop will be equal in magnitude and opposite in direction, but will
produce no net torque on the loop.
20. No. The magnetic field created by a single current loop resembles that of a bar magnet –
strongest inside the loop, and decreasing in strength as you move away from the loop.
Neither is the field uniform in direction – the magnetic field lines loop through the loop.
22. (a) The magnets repel each other with a force equal to the weight of one of them.
(b) The 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 other has its north pole on the bottom side.
(d) Then if either were inverted they would attract each other and stick firmly together.
Magnetism
143
Answers to Even Numbered Problems
(a’) to the left
(b’) into the page
(d’) toward the top
(e’) into the page
(b) All answers are the reverse of those given in (a).
2.
(a)
4.
(a) toward the top of the page
(c) zero force
6.
4.96 × 10 −17 N southward
8.
(a)
7.90 × 10 −12 N
(b)
(b)
(d)
(c’) out of the page
(f’) out of the page
out of the page
into the page
0
10.
806 N
12.
(a) to the left
(d) toward top of page
14.
0.245 T eastward
16.
(a)
(b)
18.
0.109 A toward the right
20.
(a)
22.
4.33 × 10 −3 N ⋅ m
24.
10 N·m, clockwise (as viewed from above the loop)
26.
118 N·m
28.
7.88 × 10 −12 T
30.
0.150 mm
32.
3.11 cm
34.
(a)
36.
675 A, conventional current is downward or negative charges flow upward
38.
(a)
(c)
(b) into the page
(e) into the page
(c) out of the page
(f) out of the page
9.0 × 10 −3 N at 15° above horizontal in northward direction
2.3 × 10 −3 N horizontal and due west
4.73 N
toward the left
40.0 µ T into the page
1.67 µ T out of the page
(b)
(b)
(b)
5.46 N
out of the page
5.00 µ T out of the page
(c)
4.73 N
(c)
lower left to upper right
144
CHAPTER 19
4.00 µ T toward the bottom of the page
6.67 µ T at 77.0° to the left of vertical
40.
(a)
(b)
42.
5.40 cm
44.
(a)
46.
2.70 × 10 −5 N to the left
48.
(a)
919 turns
(b)
11.5 cm
50.
(a)
2.8 µ T
(b)
0.91 mA
52.
(a)
1.79 ns
(b)
3.51 keV
54.
3.92 × 10 −2 T
56.
(a)
30.0 A
(b)
1.60 × 10 − 4 T out of the page
58.
(a)
2.46 N upward
(b)
107 m s 2 upward
60.
4.5 mm
62.
(a)
6.2 m s 2
(b)
0.40 s
64.
(a)
opposite directions
(b)
68 A
66.
0.59 T
68.
(a)
1.79 × 10 −8 s
(b)
35.1 eV
70.
(a)
(c)
1.00 × 10 −5 T
1.60 × 10 −5 T
(b)
(d)
2.00 × 10 − 4 N m , attracted
(b)
2.00 × 10 −4 N m , repelled
8.00 × 10 −5 N toward the first wire
8.00 × 10 −5 N toward the second wire
Magnetism
145
Problem Solutions
19.1
19.2
The direction in parts (a) through (d) is found by use of the right hand rule. You must
remember that the electron is negatively charged and thus experiences a force in the
direction exactly opposite that predicted by the right hand rule for a positively charged
particle.
(a)
horizontal and due east
(b)
horizontal and 30° N of E
(c)
horizontal and due east
(d)
zero force , F = qvB sin θ = qvB sin ( 180° ) = 0
(a) For a positively charged particle, the direction of the force is that predicted by the
right hand rule. These are:
(a’)
in plane of page and to left
(b’)
into the page
(c’)
out of the page
(d’)
in plane of page and toward the top
(e’)
into the page
(f’)
out of the page
(b) For a negatively charged particle, the direction of the force is exactly opposite what
the right hand rule predicts for positive charges. Thus, the answers for part (b) are
reversed from those given in part (a) .
19.3
Since the particle is positively charged, use the right hand rule. In this case, start with
G
the fingers of the right hand in the direction of v and the thumb pointing in the
G
G
direction of F . As you start closing the hand, the fingers point in the direction of B after
they have moved 90°. The results are
(a)
19.4
into the page
(b)
toward the right
(c)
toward bottom of page
G
Hold the right hand with the fingers in the direction of v so that as you close your hand,
G
the fingers move toward the direction of B . The thumb will point in the direction of the
force (and hence the deflection) if the particle has a positive charge. The results are
(a)
toward top of page
(b)
out of the page , since the charge is negative.
(c)
zero force
(d)
into the page
146
CHAPTER 19
19.5
Gravitational force:
Fg = mg = ( 9.11 × 10 −31 kg )( 9.80 m s 2 ) = 8.93 × 10 −30 N downward
Electric force:
Fe = qE = ( −1.60 × 10 −19 C ) ( −100 N C ) = 1.60 × 10 −17 N upward
Magnetic force:
Fm = qvB sin θ = ( −1.60 × 10 −19 C )( 6.00 × 106 m s )( 50.0 × 10 −6 T ) sin ( 90.0° )
= 4.80 × 10 −17 N in direction opposite right hand rule prediction
Fm = 4.80 × 10 −17 N downward
19.6
From F = qvB sin θ , the magnitude of the force is found to be
F = ( 1.60 × 10 −19 C )( 6.2 × 10 6 m s )( 50.0 × 10 −6 T ) sin ( 90.0° ) = 4.96 × 10 −17 N
JG
Using the right-hand-rule (fingers point westward in direction of v , so they move
JG
downward toward the direction of B as you close the hand, the thumb points
southward. Thus, the direction of the force exerted on a proton (a positive charge) is
toward the south .
19.7
The gravitational force is small enough to be ignored, so the magnetic force must supply
the needed centripetal acceleration. Thus,
m
qBr
v2
= qvB sin 90° , or v =
where r = RE + 1000 km=7.38 × 10 6 m
m
r
(1.60 × 10
v=
−19
C )( 4.00 × 10 −8 T )( 7.38 × 106 m )
1.67 × 10 −27 kg
= 2.83 × 107 m s
G
G
G
If v is toward the west and B is northward, F will be directed downward as
required.
Magnetism
19.8
The speed attained by the electron is found from
v=
147
1
mv 2 = q ( ∆V ) , or
2
2 ( 1.60 × 10 −19 C ) ( 2 400 V )
2 e ( ∆V )
=
= 2.90 × 107 m s
m
9.11 × 10 −31 kg
(a) Maximum force occurs when the electron enters the region perpendicular to the
field.
Fmax = q vB sin 90°
= ( 1.60 × 10 −19 C )( 2.90 × 107 m s ) ( 1.70 T ) = 7.90 × 10 −12 N
(b) Minimum force occurs when the electron enters the region parallel to the field.
Fmin = q vB sin 0° = 0
19.9
B=
−27
13
2
F m a ( 1.67 × 10 kg )( 2.0 × 10 m s )
=
=
= 0.021 T
qv qv
(1.60 × 10−19 C )(1.0 × 107 m s )
G
The right hand rule shows that B must be in the − y direction to yield a force in the
G
+x direction when v is in the +z direction.
19.10
The force on a single ion is
F1 = qvB sin θ
= ( 1.60 × 10 −19 C ) ( 0.851 m s ) ( 0.254 T ) sin ( 51.0° ) = 2.69 × 10 −20 N
The total number of ions present is
ions 

N =  3.00 × 10 20
100 cm 3 ) = 3.00 × 10 22
3 (
cm 

Thus, assuming all ions move in the same direction through the field, the total force is
F = N ⋅ F1 = ( 3.00 × 10 22 )( 2.69 × 10 −20 N ) = 806 N
148
CHAPTER 19
19.11
From F = BI L sin θ , the magnetic field is
B=
FL
0.12 N m
=
= 8.0 × 10 −3 T
I sin θ ( 15 A ) sin 90°
G
G
G
The direction of B must be the + z direction to have F in the –y direction when I is
in the +x direction.
19.12
19.13
Hold the right hand with the fingers in the direction of the current so, as you close the
hand, the fingers move toward the direction of the magnetic field. The thumb then
points in the direction of the force. The results are
(a)
to the left
(b)
into the page
(c)
out of the page
(d)
toward top of page
(e)
into the page
(f)
out of the page
Use the right hand rule, holding your right hand with the fingers in the direction of the
current and the thumb pointing in the direction of the force. As you close your hand, the
fingers will move toward the direction of the magnetic field. The results are
(a)
19.14
into the page
(b)
toward the right
(c)
toward the bottom of the page
In order to just lift the wire, the magnetic force exerted on a unit length of the wire must
be directed upward and have a magnitude equal to the weight per unit length. That is,
the magnitude is
F
m
= BI sin θ =   g
l
 l
giving
m g
B= 
 l  I sin θ
To find the minimum possible field , the magnetic field should be perpendicular to the
current (θ = 90.0° ) . Then,

g
g  1 kg   10 2 cm   9.80 m s 2
m
Bmin =  
= 0.500
= 0.245 T



cm  10 3 g   1 m   ( 2.00 A )( 1)
 l  I sin 90.0° 
To find the direction of the field, hold the right hand with the thumb pointing upward
(direction of the force) and the fingers pointing southward (direction of current). Then,
as you close the hand, the fingers point eastward. The magnetic field should be directed
eastward .
Magnetism
19.15
F = BIL sin θ = ( 0.300 T )( 10.0 A )( 5.00 m ) sin ( 30.0° ) = 7.50 N
19.16
(a) The magnitude is
149
F = BIL sin θ = ( 0.60 × 10 -4 T ) ( 15 A )( 10.0 m ) sin ( 90° ) = 9.0 × 10 −3 N
G
G
G
F is perpendicular to B . Using the right hand rule, the orientation of F is found to
be 15° above the horizontal in the northward direction .
(b) F = BIL sin θ = ( 0.60 × 10 -4 T ) ( 15 A )( 10.0 m ) sin ( 165° ) = 2.3 × 10 −3 N
and, from the right hand rule, the direction is horizontal and due west
19.17
G
For minimum field, B should be perpendicular to the wire. If the force is to be
northward, the field must be directed downward .
To keep the wire moving, the magnitude of the magnetic force must equal that of the
kinetic friction force. Thus, BI L sin 90° = µk ( mg ) , or
B=
19.18
µk ( m L ) g
I sin 90°
=
( 0.200 ) (1.00 g cm ) ( 9.80 m s 2 )  1 kg   102 cm 
 3 
 = 0.131 T
( 1.50 A )( 1.00 )
 10 g   1 m 
To have zero tension in the wires, the magnetic force per unit
length must be directed upward and equal to the weight per
unit length of the conductor. Thus,
G
Fm
L
= BI =
mg
, or
L
( m L ) g = ( 0.040
I=
B
ur
Bin
kg m ) ( 9.80 m s 2 )
3.60 T
= 0.109 A
From the right hand rule, the current must be to the right if the force is to be upward
when the magnetic field is into the page.
150
CHAPTER 19
19.19
For the wire to move upward at constant speed, the net force acting on it must be zero.
Thus, BI L sin θ = mg and for minimum field θ = 90° . The minimum field is
2
mg ( 0.015 kg ) ( 9.80 m s )
B=
=
= 0.20 T
IL
( 5.0 A )( 0.15 m )
For the magnetic force to be directed upward when the current is toward the left, B must
be directed out of the page .
19.20
The magnitude of the magnetic force exerted on a current-carrying conductor in a
magnetic field is given by F = BI lsin θ , where B is the magnitude of the field, l is the
length of the conductor, I is the current in the conductor, and θ is the angle the
conductor makes with the direction of the field. In this case,
F = ( 0.390 T )( 5.00 A )( 2.80 m ) sin θ = ( 5.46 N ) sin θ
(a) If θ = 60.0°, then sin θ = 0.866 and F = 4.73 N
(b) If θ = 90.0°, then sin θ = 1.00 and F = 5.46 N
(c) If θ = 120°, then sin θ = 0.866 and F = 4.73 N
19.21
For each segment, the magnitude of the force is given by F = BI L sin θ , and the direction
is given by the right hand rule. The results of applying these to each of the four
segments are summarized below.
Segment
L (m)
θ
F (N)
Direction
ab
0.400
180°
0
_
bc
0.400
90.0°
0.040 0
negative x
cd
0.400 2
45.0°
0.040 0
negative z
0.056 6
parallel to x-z
plane at 45° to
both +x and +z
directions
da
0.400 2
90.0°
y
d
a
z
I
c
ur
B
b
x
Magnetism
19.22
151
The magnitude of the torque is τ = NBIA sin θ , where θ is the angle between the field
and the perpendicular to the plane of the loop. The circumference of the loop is
1.00 m
1
and the area is A = π r 2 = m 2 .
2π r = 2.00 m , so the radius is r =
π
π
1

Thus, τ = ( 1)( 0.800 T ) ( 17.0 × 10 −3 A )  m 2  sin 90.0° = 4.33 × 10 −3 N ⋅ m
π

19.23
The area is A = π ab = π ( 0.200 m )( 0.150 m ) = 0.094 2 m 2 . Since the field is parallel to the
plane of the loop, θ = 90.0° and the magnitude of the torque is
τ = NBIA sin θ
= 8 ( 2.00 × 10 −4 T ) ( 6.00 A ) ( 0.094 2 m 2 ) sin 90.0° = 9.05 × 10 −4 N ⋅ m
The torque is directed to make the left-hand side of the loop move toward you and the
right-hand side move away.
19.24
Note that the angle between the field and the perpendicular to the plane of the loop is
θ = 90.0° − 30.0° = 60.0° . Then, the magnitude of the torque is
τ = NBIA sin θ = 100 ( 0.80 T )( 1.2 A ) ( 0.40 m )( 0.30 m )  sin 60.0° = 10 N ⋅ m
With current in the –y direction, the outside edge of the loop will experience a force
directed out of the page (+z direction) according to the right hand rule. Thus, the loop
will rotate clockwise as viewed from above .
152
CHAPTER 19
19.25
(a) Let θ be the angle the plane of the loop makes
with the horizontal as shown in the sketch at the
right. Then, the angle it makes with the vertical is
φ = 90.0° − θ . The number of turns on the loop is
y
x
L
4.00 m
N=
=
= 10.0
circumference 4 ( 0.100 m )
q
The torque about the z axis due to gravity is
s
τ g = mg  cosθ  , where s = 0.100 m is the length
2

of one side of the loop. This torque tends to rotate
the loop clockwise. The torque due to the magnetic
force tends to rotate the loop counterclockwise about
the z axis and has magnitude τ m = NBIA sin θ . At
s/2
z
normal
line
q
I
f
I
ur
mg
equilibrium, τ m = τ g or NBI ( s2 ) sin θ = mg ( s cosθ ) 2 .
This reduces to
tan θ =
( 0.100 kg ) ( 9.80 m s2 )
mg
=
= 14.4
2NBIs 2 ( 10.0 ) ( 0.010 0 T ) ( 3.40 A )( 0.100 m )
Since tan θ = tan ( 90.0° − φ ) = cot φ , the angle the loop makes with the vertical at
equilibrium is
φ = cot −1 ( 14.4 ) = 3.97° .
(b) At equilibrium,
τ m = NBI ( s2 ) sin θ
= ( 10.0 ) ( 0.010 0 T ) ( 3.40 A )( 0.100 m ) sin ( 90.0° − 3.97° )
2
= 3.39 × 10 −3 N ⋅ m
Magnetism
19.26
153
The resistance of the loop is
R=
ρL
A
(1.70 × 10
=
−8
Ω ⋅ m ) ( 8.00 m )
1.00 × 10
−4
and the current in the loop is I =
m
2
= 1.36 × 10 −3 Ω
∆V
0.100 V
=
= 73.5 A
R 1.36 × 10 −3 Ω
The magnetic field exerts torque τ = NBIA sin θ on the loop, and this is a maximum
when sin θ = 1 . Thus,
τ max = NBIA = ( 1)( 0.400 T )( 73.5 A ) ( 2.00 m ) = 118 N ⋅ m
2
19.27
The magnitude of the force a proton experiences as it moves perpendicularly to a
magnetic field is
F = qvB sin θ = ( + e ) vB sin ( 90.0° ) = evB
This force is always directed perpendicular to the velocity of the proton and will supply
the centripetal acceleration as the proton follows a circular path. Thus,
evB = m
v2
r
or
v=
erB
m
and the time required for the proton to complete one revolution is
T=
2π r
2π r
2π m
=
=
v
erB m
eB
If it is observed that T = 1.00 µ s , the magnitude of the magnetic field is
B=
2π ( 1.67 × 10 −27 kg )
2π m
=
= 6.56 × 10 −2 T
−19
−6
eT
×
×
1.60
10
C
1.00
1
0
s
(
)(
)
154
CHAPTER 19
19.28
Since the path is circular, the particle moves perpendicular to the magnetic field, and the
mv
v2
. But
magnetic force supplies the centripetal acceleration. Hence, m = qvB , or B =
r
qr
the momentum is given by p = mv = 2m ( KE ) , and the kinetic energy of this proton is
 1.60 × 10 −19
KE = ( 10.0 × 106 eV ) 
1 eV

B=
19.29
2m ( KE )
qr
=
J
−12
 = 1.60 × 10 J . We then have

2 ( 1.67 × 10 −27 kg )( 1.60 × 10 −12 J )
(1.60 × 10
−19
C )( 5.80 × 10
10
m)
= 7.88 × 10 −12 T
For the particle to pass through with no deflection, the net force acting on it must be
zero. Thus, the magnetic force and the electric force must be in opposite directions and
have equal magnitudes. This gives
Fm = Fe , or qvB = qE which reduces to v = E B
19.30
The speed of the particles emerging from the velocity selector is v = E B (see Problem
29). In the deflection chamber, the magnetic force supplies the centripetal acceleration,
mv m ( E B ) mE
mv 2
, or r =
so qvB =
=
= 2
r
qB
qB
qB
Using the given data, the radius of the path is found to be
( 2.18 × 10
r=
(1.60 × 10
−26
19.31
-19
kg ) ( 950 V m )
C ) ( 0.930 T )
2
= 1.50 × 10 −4 m = 0.150 mm
From conservation of energy, ( KE + PE ) f = ( KE + PE )i , we find that
1
mv 2 + qVf = 0 + qVi ,
2
or the speed of the particle is
v=
(
2 q Vi − Vf
m
)=
2 ( 1.60 × 10 −19 C ) ( 250 V )
2 q ( ∆V )
=
= 5.66 × 10 4 m s
m
2.50 × 10 -26 kg
The magnetic force supplies the centripetal acceleration giving qvB =
or
r=
mv 2
r
−26
4
mv ( 2.50 × 10 kg )( 5.66 × 10 m s )
=
= 1.77 × 10 −2 m = 1.77 cm
−19
qB
(1.60 × 10 C ) ( 0.500 T )
Magnetism
19.32
Since the centripetal acceleration is furnished by the magnetic force acting on the ions,
mv
mv 2
qvB =
or the radius of the path is r =
. Thus, the distance between the impact
r
qB
points (that is, the difference in the diameters of the paths followed by the U 238 and the
U 235 isotopes) is
∆d = 2 ( r238 − r235 ) =
=
or
19.33
155
2v
( m238 − m235 )
qB
2 ( 3.00 × 10 5 m s )
(1.60 × 10
−19


−27 kg  

( 238 u − 235 u )  1.66 × 10
u  
C ) ( 0.600 T ) 

∆d = 3.11 × 10 −2 m = 3.11 cm
In the perfectly elastic, head-on collision between the α -particle and the initially
stationary proton, conservation of momentum requires that mp vp + mα vα = mα v0 while
(
conservation of kinetic energy also requires that v0 − 0 = − vα − vp
)
or vp = vα + v0 . Using
the fact that mα = 4 mp and combining these equations gives
(
)
(
)
mp ( vα + v0 ) + 4 mp vα = 4 mp v0
and
vα = 3v0 5
or
3
3 5  3
Thus, vα = v0 =  vp  = vp
5
5 8  8
vp = ( 3v0 5 ) + v0 = 8v0 5
After the collision, each particle follows a circular path in the horizontal plane with the
magnetic force supplying the centripetal acceleration. If the radius of the proton’s
trajectory is R, and that of the alpha particle is r, we have
vp2
mp v p mp v p
or
q p v p B = mp
R=
=
R
qp B
eB
and
19.34
qα vα B = mα
vα2
r
r=
or
(
)(
)
4mp 3vp 8
mα vα
3  mp v p  3
=
= 
= R
qα B
4  eB  4
( 2e ) B
Imagine grasping the conductor with the right hand so the fingers curl around the
conductor in the direction of the magnetic field. The thumb then points along the
conductor in the direction of the current. The results are
(a)
toward the left
(b)
out of page
(c)
lower left to upper right
156
CHAPTER 19
19.35
Treat the lightning bolt as a long, straight conductor. Then, the magnetic field is
µ I ( 4π × 10
B= 0 =
2π r
19.36
−7
T ⋅ m A )( 1.00 × 10 4 A )
2π ( 100 m )
= 2.00 × 10 −5 T = 20.0 µ T
Model the tornado as a long, straight, vertical conductor and imagine grasping it with
the right hand so the fingers point northward on the western side of the tornado. (that
is, at the observatory’s location) The thumb is directed downward, meaning that the
conventional current is downward or negative charge flows upward .
The magnitude of the current is found from B = µ0 I 2π r as
I=
19.37
2π rB
µ0
=
2π ( 9.00 × 10 3 m )( 1.50 × 10 −8 T )
4π × 10 -7 T ⋅ m A
= 675 A
From B = µ0 I 2π r , the required distance is
µ0 I ( 4π × 10 T ⋅ m A ) ( 20 A )
=
= 2.4 × 10 −3 m = 2.4 mm
−3
2π B
2π ( 1.7 × 10 T )
-7
r=
19.38
Assume that the wire on the right is wire 1 and that on the left is wire 2. Also, choose the
positive direction for the magnetic field to be out of the page and negative into the page.
(a) At the point half way between the two wires,
µ I
µI 
µ
Bnet = −B1 − B2 = −  0 1 + 0 2  = − 0 ( I1 + I 2 )
2π r
 2π r1 2π r2 
( 4π × 10 T ⋅ m A ) (10.0 A ) = − 4.00 × 10
2π ( 5.00 × 10 m )
−7
=−
-2
−5
T
or Bnet = 40.0 µ T into the page
Bnet = + B1 − B2 =
(b) At point P1 ,
Bnet =
( 4π × 10
µ0
2π
 I1 I 2 
 − 
 r1 r2 
T ⋅ m A )  5.00 A
5.00 A 
 0.100 m − 0.200 m  = 5.00 µ T out of page
2π
−7
157
Magnetism
Bnet = −B1 + B2 =
(c) At point P2 ,
Bnet
( 4π × 10
=
µ0
2π
 I1 I 2 
− + 
 r1 r2 
T ⋅ m A )  5.00 A
5.00 A 
−
+

2π
 0.300 m 0.200 m 
−7
= 1.67 µ T out of page
The distance from each wire to point P is given by
r=
1
2
2
2
( 0.200 m ) + ( 0.200 m ) = 0.141 m
A
ur
BA
ur
BC
P
At point P, the magnitude of the magnetic field
produced by each of the wires is
µ I ( 4π × 10 T ⋅ m A ) ( 5.00 A )
B= 0 =
= 7.07 µ T
2π r
2π ( 0.141 m )
C
ur
BB
−7
B
ur
BD
0.200 mm
19.39
0.200 mm D
Carrying currents into the page, the field A produces
at P is directed to the left and down at –135°, while B
creates a field to the right and down at – 45°. Carrying currents toward you, C produces
a field downward and to the right at – 45°, while D’s contribution is downward and to
the left. The horizontal components of these equal magnitude contributions cancel in
pairs, while the vertical components all add. The total field is then
Bnet = 4 ( 7.07 µ T ) sin 45.0° = 20.0 µ T toward the bottom of the page
19.40
Call the wire carrying a current of 3.00 A wire 1 and the other wire 2. Also, choose the
line running from wire 1 to wire 2 as the positive x direction.
(a) At the point midway between the wires, the field due to
each wire is parallel to the y axis and the net field is
Bnet = + B1 y − B2 y = µ0 ( I1 − I 2 ) 2π r
( 4π × 10
−7
T ⋅ m A)
I1
( 3.00 A − 5.00 A ) = − 4.00 × 10 −6 T
Thus,
Bnet =
or
Bnet = 4.00 µ T toward the bottom of the page
2π ( 0.100 m )
ur
B1
ur
B2
+x
I2
158
CHAPTER 19
(b) At point P, r1 = ( 0.200 m ) 2 and B1 is directed at θ1 = +135° .
ur
B1
The magnitude of B1 is
B1 =
−7
µ0 I1 ( 4π × 10 T ⋅ m A ) ( 3.00 A )
=
= 2.12 µ T
2π r1
2π 0.200 2 m
(
P
ur
B2
r1
)
r2
+x
I1
The contribution from wire 2 is in the –x direction and
has magnitude
I2
µ0 I 2 ( 4π × 10 T ⋅ m A ) ( 5.00 A )
=
= 5.00 µ T
2π r2
2π ( 0.200 m )
−7
B2 =
Therefore, the components of the net field at point P are:
Bx = B1 cos135° + B2 cos180°
= ( 2.12 µ T ) cos 135° + ( 5.00 µ T ) cos180° = −6.50 µ T
and
By = B1 sin135° + B2 sin 180° = ( 2.12 µ T ) sin135° + 0 = +1.50 µ T
Therefore, Bnet = Bx2 + By2 = 6.67 µ T at
B 
 6.50 µ T 
θ = tan −1  x  = tan −1 
 = 77.0°
 By 
 1.50 µ T 


Bx = –6.50 mT
ur
Bnet
q
By = 1.50 mT
JG
or Bnet = 6.67 µ T at 77.0° to the left of vertical
19.41
Call the wire along the x axis wire 1 and the other wire 2. Also, choose the positive
direction for the magnetic fields at point P to be out of the page.
At point P, Bnet = + B1 − B2 =
or
Bnet
( 4π × 10
=
µ0 I1 µ0 I 2 µ0  I 1 I 2 
−
=
 − 
2π r1 2π r2 2π  r1 r2 
T ⋅ m A )  7.00 A 6.00 A 
−7
−

 = +1.67 × 10 T
2π
 3.00 m 4.00 m 
−7
Bnet = 0.167 µ T out of the page
Magnetism
19.42
159
Since the proton moves with constant velocity, the net force acting on it is zero. Thus,
the magnetic force due to the current in the wire must be counterbalancing the weight of
the proton, or qvB = mg where B = µ0 I 2π d . This gives
qvµ0 I
= mg , or the distance the proton is above the wire must be
2π d
qvµ0 I ( 1.60 × 10
d=
=
2π mg
−19
C )( 2.30 × 10 4 m s )( 4π × 10 −7 T ⋅ m A )( 1.20 × 10 −6 A )
2π ( 1.67 × 10 −27 kg )( 9.80 m s 2 )
d = 5.40 × 10 −2 m = 5.40 cm
19.43
(a) From B = µ0 I 2π r , observe that the field is inversely proportional to the distance
from the conductor. Thus, the field will have one-tenth its original value if the
distance is increased by a factor of 10. The required distance is then
r ′ = 10 r = 10 ( 0.400 m ) = 4.00 m
(b) A point in the plane of the conductors and 40.0 cm from the center of the cord is
located 39.85 cm from the nearer wire and 40.15 cm from the far wire. Since the
currents are in opposite directions, so are their contributions to the net field.
Therefore, Bnet = B1 − B2 , or
Bnet =
µ0 I  1 1  ( 4π × 10
 − =
2π  r1 r2 
−7
T ⋅ m A ) ( 2.00 A ) 

1
1
−


2π
 0.398 5 m 0.401 5 m 
= 7.50 × 10 −9 T = 7.50 nT
(c) Call r the distance from cord center to field point P
and 2d = 3.00 mm the distance between centers of
the conductors.
Bnet =
µ0 I  1
1  µ0 I  2d 
−

=


2π  r − d r + d  2π  r 2 − d 2 
7.50 × 10 −10 T =
( 4π × 10
−7
r
2d
P
r–d
r+d
T ⋅ m A ) ( 2.00 A )  3.00 × 10 −3 m 
 2
−6
2 
2π
 r − 2.25 × 10 m 
so r = 1.26 m
The field of the two-conductor cord is weak to start with and falls off rapidly with
distance.
160
CHAPTER 19
(d) The cable creates zero field at exterior points, since a loop in Ampère’s law
encloses zero total current.
19.44
(a)
−7
F µ0 I1I 2 ( 4π × 10 T ⋅ m A ) ( 10.0 A )
=
=
L
2π d
2π ( 0.100 m )
= 2.00 × 10 −4 N m
2
( attraction )
(b) The magnitude remains the same as calculated in (a), but the wires are repelled.
F
Thus, = 2.00 × 10 −4 N m ( repulsion )
L
19.45
In order for the system to be in equilibrium, the repulsive magnetic force per unit length
on the top wire must equal the weight per unit length of this wire.
F µ0 I 1 I 2
=
= 0.080 N m , and the distance between the wires will be
L
2π d
Thus,
d=
µ0 I1 I 2
2π ( 0.080 N m )
( 4π × 10
=
−7
T ⋅ m A ) ( 60.0 A )( 30.0 A )
2π ( 0.080 N m )
= 4.5 × 10 −3 m = 4.5 mm
19.46
The magnetic forces exerted on the top and bottom segments of the rectangular loop are
equal in magnitude and opposite in direction. Thus, these forces cancel, and we only
need consider the sum of the forces exerted on the right and left sides of the loop.
Choosing to the left (toward the long, straight wire) as the positive direction, the sum of
these two forces is
Fnet = +
or Fnet =
µ0 I1 I 2 A
µIIA
µ I I A1
1 
− 0 1 2 = 0 1 2  −

2π c
2π ( c + a )
2π  c c + a 
( 4π × 10
−7
T ⋅ m A ) ( 5.00 A )( 10.0 A )( 0.450 m ) 
1
1

−


2π
 0.100 m 0.250 m 
= + 2.70 × 10 −5 N = 2.70 × 10−5 N to the left
Magnetism
19.47
N
The magnetic field inside a long solenoid is B = µ0 nI = µ0   I . Thus, the required
L
current is
I=
19.48
161
(1.00 × 10−4 T ) ( 0.400 m ) = 3.18 × 10−2 A = 31.8 mA
BL
=
µ0 N ( 4π × 10 -7 T ⋅ m A ) ( 1 000 )
(a) From R = ρ L A , the required length of wire to be used is
( 5.00 Ω ) π ( 0.500 × 10 −3 m ) 4 
2
L=
R⋅A
ρ
=
1.70 × 10 −8 Ω ⋅ m
= 57.7 m
The total number of turns on the solenoid (that is, the number of times this length of
wire will go around a 1.00 cm radius cylinder) is
N=
L
2π r
=
57.7 m
= 919
2π ( 1.00 × 10 −2 m )
(b) From B = µ0 nI , the number of turns per unit length on the solenoid is
n=
B
4.00 × 10 −2 T
=
= 7.96 × 10 3 turns m
µ0 I ( 4π × 10 -7 T ⋅ m A ) ( 4.00 A )
Thus, the required length of the solenoid is
N
919 turns
=
= 0.115 m = 11.5 cm
n 7.96 × 10 3 turns m
19.49
The magnetic field inside the solenoid is
turns  100 cm  

−2
B = µ0 nI1 = ( 4π × 10 −7 T ⋅ m A )  30

  ( 15.0 A ) = 5.65 × 10 T
cm
1
m



Therefore, the magnitude of the magnetic force on any one of the sides of the square
loop is
F = BI 2 L sin 90.0° = ( 5.65 × 10 −2 T ) ( 0.200 A ) ( 2.00 × 10 −2 m ) = 2.26 × 10 −4 N
162
CHAPTER 19
The forces acting on the sides of the loop lie in the plane of the loop, are perpendicular
to the sides, and are directed away from the interior of the loop. Thus, they tend to
stretch the loop but do not tend to rotate it. The torque acting on the loop is τ = 0
19.50
(a) The magnetic force supplies the centripetal acceleration, so qvB = mv 2 r . The
magnetic field inside the solenoid is then found to be
−31
4
mv ( 9.11 × 10 kg )( 1.0 × 10 m s )
B=
=
= 2.847 × 10 −6 T = 2.8 µ T
−19
−2
qr
1.60
×
10
C
2.0
×
10
m
(
)(
)
(b) From B = µ0 nI , the current is the solenoid is found to be
I=
2.847 × 10 −6 T
B
=
µ0 n ( 4π × 10 -7 T ⋅ m A ) ( 25 turns cm )( 100 cm 1 m ) 
= 9.1 × 10 −4 A = 0.91 mA
19.51
When the plane of the coil makes an angle of 35° with the field direction, the
perpendicular to the plane of the coil makes an angle of θ = 90° − 35° = 55° with the
magnetic field.
Thus, the torque exerted on the loop is
2
τ = NBIA sin θ = ( 1)( 0.30 T )( 25 A ) π ( 0.30 m )  sin 55° = 1.7 N ⋅ m
19.52
Since the magnetic force must supply the centripetal acceleration, qvB = mv 2 r or the
radius of the path is r = mv qB .
(a) The time for the electron to travel the semicircular path (of length π r ) is
t=
πr
v
=
π  mv  π m
π ( 9.11 × 10 −31 kg )
=

=
v  qB  qB ( 1.60 × 10 −19 C ) ( 0.010 0 T )
= 1.79 × 10 −9 s = 1.79 ns
163
Magnetism
(b) The radius of the semicircular path is 2.00 cm. From r = mv qB , the momentum of
the electron is p = mv = qBr, and the kinetic energy is
q2 B2 r 2 ( 1.60 × 10
KE = mv =
=
=
2m
2m
1
2
2
( mv )
2
−19
C ) ( 0.010 0 T ) ( 2.00 × 10 −2 m )
2
2
2
2 ( 9.11 × 10 −31 kg )


1 keV
KE = ( 5.62 × 10 −16 J ) 
 = 3.51 keV
−16
 1.60 × 10 J 
Assume wire 1 is along the x axis and wire 2 along the y axis.
(a) Choosing out of the page as the positive field direction, the field at point P is
µ
B = B1 − B2 = 0
2π
−7
 I1 I 2  ( 4π × 10 T ⋅ m A )  5.00 A
3.00 A 
−
 − =


2π
 0.400 m 0.300 m 
 r1 r2 
= 5.00 × 10 −7 T = 0.500 µ T out of the page
T ⋅ m A ) ( 5.00 A )
2π ( 0.300 m )
Bx = B2 =
1
I
= −3.33 × 10 −6 T
µ0 I 2 ( 4π × 10 T ⋅ m A ) ( 3.00 A )
=
= 2.00 × 10 −6 T
2π r
2π ( 0.300 m )
−7
and
0
2
( 4π × 10
=−
+y
0.300 m
µI
By = −B1 = − 0 1
2π r
−7
ur
+z ur
Bx = B2
–B ur
(b) At 30.0 cm above the intersection of the wires, the field
components are as shown at the right, where
u
Br
y =
19.53
The resultant field is
 By 
B = Bx2 + By2 = 3.89 × 10 −6 T at θ =tan −1   = −59.0°
 Bx 
G
or B = 3.89 µ T at 59.0° clockwise from +x direction
=
0
3.
A
+x
I1 = 5.00 A
164
CHAPTER 19
19.54
For the rail to move at constant velocity, the net force acting on it must be zero. Thus, the
magnitude of the magnetic force must equal that of the friction force giving
BIL = µk ( mg ) , or
B=
19.55
µk ( mg )
IL
( 0.100 ) ( 0.200 kg ) ( 9.80 m s2 )
=
= 3.92 × 10 −2 T
( 10.0 A )( 0.500 m )
The magnetic force acting on each type particle supplies the centripetal acceleration for
that particle. Thus, qvB = mv 2 r or r = mv qB .
After completing one half of the circular paths, the two types of particle are separated by
the difference in the diameters of the two paths. Therefore,
∆d = 2 ( r2 − r1 ) =
=
2v
( m2 − m1 )
qB
2 ( 1.00 × 10 5 m s )
(1.60 × 10
−19
( 23.4 − 20.0 ) × 10−27 kg 
C ) ( 0.200 T )
= 2.13 × 10 −2 m = 2.13 cm
19.56
Let the leftmost wire be wire 1 and the rightmost be wire 2.
(a) At point C, B1 is directed out of the page and B2 is into the page. If the net field is
zero, then B1 = B2 , or
r 
µ0 I1 µ0 I 2
 15.0 cm 
=
, giving I1 = I 2  1  = ( 10.0 A ) 
 = 30.0 A
2π r1 2π r2
 5.00 cm 
 r2 
(b) At point A, B1 and B2 are both directed out of the page, so
( 4π × 10 T ⋅ m A ) ( 30.0 A+10.0 A )
µ0
( I1 + I 2 ) =
2π r
2π ( 5.00 × 10 -2 m )
−7
Bnet = B1 +B2 =
= 1.60 × 10 − 4 T out of the page
Magnetism
19.57
165
(a) Since the magnetic field is directed from N to S (that is, from left to right within the
artery), positive ions with velocity in the direction of the blood flow experience a
magnetic deflection toward electrode A. Negative ions will experience a force
deflecting them toward electrode B. This separation of charges creates 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 )
or v =
∆V
160 × 10 −6 V
=
= 1.33 m s
Bd ( 0.040 0 T ) ( 3.00 × 10 −3 m )
(b) The magnetic field is directed from N to S. If the charge carriers are negative
G
moving in the direction of v , the magnetic force is directed toward point B.
Negative charges build up at point B, making the potential at A higher than that at
G
B. If the charge carriers are positive moving in the direction of v , the magnetic force
is directed toward A, so positive charges build up at A. This also makes the
potential at A higher than that at B. Therefore the sign of the potential difference
does not depend on the charge of the ions .
19.58
(a) Since the distance between them is so small in comparison to the radius of
curvature, the hoops may be treated as long, straight, parallel wires. Because the
currents are in opposite directions, the hoops repel each other. The magnetic force
on the top loop is
µ0 I 2 ( 2π r ) µ0 I 2 r
 µ0 I 1 I 2 
=
Fm = 
L =
2π d
d
 2π d 
( 4π × 10
=
−7
T ⋅ m A ) ( 140 A ) ( 0.100 m )
2
1.00 × 10 −3 m
= 2.46 N upward
(b) ΣFy = may = Fm − mg
or ay =
2.46 N
Fm
−g=
− 9.80 m s 2 = 107 m s 2 upward
0.021 kg
m
166
CHAPTER 19
19.59
The magnetic force is very small in comparison to the weight of the ball, so we treat the
motion as that of a freely falling body. Then, as the ball approaches the ground, it has
velocity components with magnitudes of
vx = v0 x = 20.0 m s , and
vy = v02 y + 2ay ( ∆y ) = 0 + 2 ( −9.80 m s 2 ) ( −20.0 m ) = 19.8 m s
The velocity of the ball is perpendicular to the magnetic field and, just before it reaches
the ground, has magnitude v = vx2 + vy2 = 28.1 m s . Thus, the magnitude of the magnetic
force is
Fm = qvB sin θ
= ( 5.00 × 10 −6 C ) ( 28.1 m s )( 0.010 0 T ) sin 90.0° = 1.41 × 10 −6 N
19.60
We are given that the field at points on the axis of the disk varies as B = k h 3 , where k is
a constant and h is the distance from the midplane of the disk. At the surface of the disk,
B = 5.0 × 10 −3 T
and
h=
thickness 1.0 mm
=
= 0.50 mm
2
2
Thus, k = Bh 3 = ( 5.0 × 10 −3 T ) ( 0.50 mm ) . The distance from the center plane of the disk
3
to the height where B = 5.0 × 10 −5 T is then
1
1
 ( 5.0 × 10 −2 T ) ( 0.50 mm )3  3
k
3 3



=
×
h=  =
1.0
10

 ( 0.50 mm ) = 5.0 mm
5.0 × 10 −5 T
B


1
3
The distance of this position above the surface of the disk is
x=h−
thickness
1.0 mm
= 5.0 mm −
= 4.5 mm
2
2
167
Magnetism
First, observe that ( 5.00 cm ) + ( 12.0 cm ) = ( 13.0 cm ) . Thus, the
2
2
2
5.00
I1
triangle shown in dashed lines is a right triangle giving
cm
a
12.0 cm 
 = 67.4° , and β = 90.0° − α = 22.6°
 13.0 cm 
cm
13.0 cm
α = sin −1 
b
At point P, the field due to wire 1 is
( 4π × 10 T ⋅ m A ) ( 3.00 A ) = 12.0 µ T
µI
B1 = 0 1 =
2π r1
2π ( 5.00 × 10 −2 m )
P
12.0
19.61
−7
and it is directed from P toward wire 2, or to the left and at
67.4° below the horizontal. The field due to wire 2 has
magnitude
I2
I1
y
P
a b
a
µ0 I 2 ( 4π × 10 T ⋅ m A ) ( 3.00 A )
=
= 5.00 µ T
2π r2
2π ( 12.00 × 10 −2 m )
−7
B2 =
ur
B1
and at P is directed away from wire 1 or to the right and at
22.6° below the horizontal.
Thus, B1x = −B1 cos 67.4° = −4.62 µ T B1 y = −B1 sin 67.4° = −11.1 µ T
b
I2
B2 x = B2 cos 22.6° = + 4.62 µ T B2 y = − B2 sin 22 .6° = − 1.92 µ T
and
Bx = B1x + B2 x = 0 , while By = B1y + B2 y = −13.0 µ T .
The resultant field at P is
G
B = 13.0 µ T directed toward the bottom of the page
19.62
(a) The magnetic force acting on the wire is directed upward and of magnitude
Fm = BIL sin 90° = BIL
Thus,
ay =
ay =
ΣFy
( 4.0 × 10
m
−3
5.0 × 10
=
Fm − mg
BI
=
− g , or
m
( m L)
T ) ( 2.0 A )
−4
kg m
− 9.80 m s 2 = 6.2 m s 2
x
b
ur
B2
168
CHAPTER 19
(b) Using ∆y = v0 y t +
t=
19.63
2 ( ∆y )
ay
=
1 2
ay t with v0 y = 0 gives
2
2 ( 0.50 m )
= 0.40 s
6.2 m s 2
Label the wires 1, 2, and 3 as shown in Figure 1,
and let B1 , B2 , and B3 respectively represent the
magnitudes of the fields produced by the
currents in those wires. Also, observe that
θ = 45° in Figure 1.
(
B1 = B2 =
( 4π × 10
A
q
)
T ⋅ m A ) ( 2.0 A )
2π ( 0.010 m ) 2
= 28 µ T
q
_
aÖ2
q
B
a
_
aÖ2
At point A, B1 = B2 = µ0 I 2π a 2 or
−7
1
_
aÖ2
C
a
a
_
aÖ2
q
3
2
Figure 1
( 4π × 10 T ⋅ m A ) ( 2.0 A ) = 13 µ T
µ0 I
=
2π ( 3 a )
2π ( 0.030 m )
−7
and B3 =
These field contributions are oriented as shown in Figure 2.
G
G
Observe that the horizontal components of B1 and B2 cancel
G
while their vertical components add to B3 . The resultant
field at point A is then
BA = ( B1 + B2 ) cos 45° + B3 = 53 µ T , or
ur
B2
45°
45°
ur
B3
ur
B1
Figure 2
G
B A = 53 µ T directed toward the bottom of the page
µ0 I ( 4π × 10 T ⋅ m A ) ( 2.0 A )
=
= 40 µ T
2π a
2π ( 0.010 m )
−7
At point B,
B1 = B2 =
µ0 I
and B3 =
= 20 µ T . These contributions are oriented as
2π ( 2a )
shown in Figure 3. Thus, the resultant field at B is
G
G
BB = B3 = 20 µ T directed toward the bottom of the page
ur
B2
ur
ur B1
B3
Figure 3
169
Magnetism
(
)
ur
B2
At point C, B1 = B2 = µ0 I 2π a 2 = 28 µ T while
B3 = µ0 I 2π a = 40 µ T . These contributions are oriented
as shown in Figure 4. Observe that the horizontal
G
G
components of B1 and B2 cancel while their vertical
G
components add to oppose B3 . The magnitude of the
resultant field at C is
ur
B1
45°
45°
ur
B3
Figure 4
BC = ( B1 + B2 ) sin 45° − B3
= ( 56 µ T ) sin 45° − 40 µ T= 0
(a) Since one wire repels the other, the currents must be in opposite directions .
(b) Consider a free body diagram of one of the wires as shown at
the right.
ΣFy = 0 ⇒ T cos 8.0° = mg
or T =
mg
cos 8.0°
ur
Fm
 mg 
ΣFx = 0 ⇒ Fm = T sin 8.0° = 
 sin 8.0°
 cos 8.0° 
or Fm = ( mg ) tan 8.0° . Thus,
I=
d
–
2
µ0 I 2 L
= ( mg ) tan 8.0° which gives
2π d
d ( m L ) g  tan 8.0°
µ0 2π
Observe that the distance between the two wires is
d = 2 ( 6.0 cm ) sin 8.0° = 1.7 cm , so
I=
(1.7 × 10
ur
T
8.0°
6.0 cm
8.0°
8.0°
19.64
−2
m ) ( 0.040 kg m ) ( 9.80 m s 2 ) tan 8.0°
2.0 × 10 −7 T ⋅ m A
= 68 A
d
–
2
ur
mg
170
CHAPTER 19
19.65
Note: We solve part (b) before part (a) for this problem.
(b) Since the magnetic force supplies the centripetal acceleration for this particle,
qvB = mv 2 r or the radius of the path is r = mv qB . The speed of the particle may be
written as v = 2 ( KE ) m , so the radius becomes
r=
2 m ( KE )
qB
=
2 ( 1.67 × 10 −27 kg )( 5.00 × 106 eV )( 1.60 × 10 −19 J eV )
(1.60 × 10
-19
C ) ( 0.050 0 T )
ur
vi
= 6.46 m
1.00 m
Consider the circular path shown at the right and
observe that the desired angle is
a
ur
vf
α = sin −1 
1.00 m 
−1  1.00 m 
 = sin 
 = 8.90°
 r 
 6.46 m 
a
(a) The constant speed of the particle is v = 2 ( KE ) m , so
the vertical component of the momentum as the particle leaves the field is
py = mvy = − mv sin α = − m
(
)
2 ( KE ) m sin α = − sin α
2 m ( KE )
or py = − sin ( 8.90° ) 2 (1.67 × 10 −27 kg )( 5.00 × 10 6 eV )( 1.60 × 10 −19 J eV )
= − 8.00 × 10 −21 kg ⋅ m s
19.66
The force constant of the spring system is found from the elongation produced by the
weight acting alone.
−3
2
F mg ( 10.0 × 10 kg )( 9.80 m s )
k= =
=
= 19.6 N m
x
x
0.50 × 10 −2 m
The total force stretching the springs when the field is turned on is
ΣFy = Fm + mg = kxtotal
Magnetism
171
Thus, the downward magnetic force acting on the wire is
Fm = kxtotal − mg
= ( 19.6 N m ) ( 0.80 × 10 −2 m ) − ( 10.0 × 10 −3 kg )( 9.80 m s 2 )
= 5.9 × 10 −2 N
Since the magnetic force is given by Fm = BIL sin 90° , the magnetic field is
( 12 Ω ) ( 5.9 × 10 −2 N )
Fm
Fm
=
=
= 0.59 T
B=
IL ( ∆V R ) L ( 24 V ) ( 5.0 × 10 −2 m )
19.67
Each turn of wire occupies a length of the solenoid equal to the diameter of the wire.
Thus, the number of turns on the solenoid is
N=
Lsolenoid
75.0 cm
=
= 750
0.100 cm
dwire
The length of wire required to make this number of turns on the solenoid is
l = N ( circumference of solenoid ) = N (π dsolenoid ) = 750π ( 10.0 × 10 −2 m ) = 75.0π m
and this copper wire has a resistance of
R = ρ Cu
4 ( 75.0π m )
l
4l
= ρ Cu
= ( 1.7 × 10 −8 Ω ⋅ m )
= 5.10 Ω
2
2
π dwire
Awire
π ( 0.100 × 10 −2 m )
Near the center of a long solenoid, the magnetic field is given by B = µ0 nI = µ0 NI Lsolenoid .
Thus, if the field at the center of the solenoid is to be B = 20.0 µ T , the current in the
solenoid must be
I=
−3
BLsolenoid ( 20.0 × 10 T ) ( 0.750 m )
=
= 15.9 A
µ0 N
( 4π × 10−7 T ⋅ m A ) ( 750 )
and the power that must be delivered to the solenoid is
P = I 2 R = ( 15.9 A ) ( 5.10 Ω ) = 1.29 × 10 3 W = 1.29 kW
2
172
CHAPTER 19
19.68
(a) A charged particle moving perpendicular to a magnetic field follows a circular arc
with the magnetic force supplying the centripetal acceleration. Thus,
evB = m
v2
r
v=
or
erB
m
The time required for the electron to traverse the semicircular path it follows while
in the field is then
π ( 9.11 × 10 −31 kg )
πr
πm
distance traveled
=
=
=
= 1.79 × 10 −8 s
t=
−19
−3
v
erB m eB ( 1.60 × 10 C )( 1.00 × 10 T )
(b) The maximum penetration of the electron into the field equals the radius of the
semicircular path followed while in the field. Therefore, r = 2.00 cm = 2.00 × 10 −2 m
and
v=
erB ( 1.60 × 10
=
m
−19
C )( 2.00 × 10 −2 m )( 1.00 × 10 −3 T )
9.11 × 10 −31 kg
= 3.51 × 10 6 m s
The kinetic energy is then
KE =
2
1
1
mv 2 = ( 9.11 × 10 −31 kg )( 3.51 × 10 6 m s ) = 5.62 × 10 −18 J
2
2

1 eV
or KE = ( 5.62 × 10 −18 J ) 
−19
 1.60 × 10
19.69

 = 35.1 eV
J
With currents in opposite directions, wires 1 and 2
Wire 3
Wire 1
Wire 2
G
G
repel each another with forces F12 and F 21 as shown
ur
ur
ur
ur
ur
ur
in the sketch at the right. For these wires to be in
F31 F32 F12 F13
F23 F21
equilibrium, wire 3 must exert a force directed to the
1.50 A
right on wire 1 and a force to the left on wire 2. If
wire 3 was between wires 1 and 2, the forces it exerts
I3
4.00 A
on these two wires would be in the same directions,
contrary to what is needed. If wire 3 were to the right
d
20.0 cm
of wire 2, the force exerted on it by wire 2 (having
the larger current and being nearer) would always
exceed that exerted by wire 1. Hence, wire 3 could not be in equilibrium. Thus, we
conclude that wire 3 must be to the left of wire 1 as shown above.
Magnetism
173
(a) For wire 3 to be in equilibrium, we must require that F31 = F32 , or
µ0 I 1 I 3 l
µ0 I 2 I 3 l
=
2π d
2π ( d + 20.0 cm )
Thus,
d=
giving
I 
d + 20.0 cm=  2  d
 I1 
20.0 cm
20.0 cm
=
= 12.0 cm (to the left of wire 1)
( I 2 I1 ) − 1 ( 4.00 A 1.50 A ) − 1
(b) If wires 1 and 2 are to be in equilibrium, wire 3 must repel wire 1 and attract wire 2
as shown above. Hence, the current in wire 3 must be directed downward . The
magnitude of this current can be determined by requiring that wire 1 be in
equilibrium, or that F13 = F12 . This gives
µ0 I1 I 3 l
2π ( 12.0 cm )
=
µ0 I1 I 2 l
2π ( 20.0 cm )
or
 12.0 cm 
I3 = I2 
 = ( 4.00 A )( 0.600 ) = 2.40 A
 20.0 cm 
Note that the same result could have been obtained by requiring that wire 2 be in
equilibrium.
µ0 I1 ( 4π × 10 T ⋅ m A ) ( 5.00 A )
=
= 1.00 × 10 −5 T
2π d
2π ( 0.100 m )
−7
19.70
(a)
B1 =
(b)
F21
= B1 I 2 = ( 1.00 × 10 −5 T ) ( 8.00 A ) = 8.00 × 10 −5 N directed toward wire 1
l
(c)
B2 =
(d)
F12
= B2 I1 = ( 1.60 × 10 −5 T ) ( 5.00 A ) = 8.00 × 10 −5 N directed toward wire 2
l
µ0 I 2 ( 4π × 10 T ⋅ m A ) ( 8.00 A )
=
= 1.60 × 10 −5 T
2π d
2π ( 0.100 m )
−7
174
CHAPTER 19