Download Chapter 19 Solutions - Mosinee School District

Chapter 19
Magnetism
Problem Solutions
19.1
Consid er a three-d im ensional coord inate system w ith the xy-plane in the plane of this
page, the +x-d irection tow ard the right ed ge of the page and the +y-d irection tow ard the
top of the page. Then, the z-axis is perpend icular to the page w ith the +z-d irection being
upw ard , out of the page. The m agnetic field is d irected in the + x-d irection, tow ard the
right .
(a) When a proton (positively charged ) m oves in the +y-d irection, the right-hand rule
num ber 1 gives the d irection of the m agnetic for ce as into the page or in the
z-direction .
(b) With v in the
y-direction , the right-hand rule num ber 1 gives the d irection of the
force on the proton as out of the page, in the +z-direction .
(c) When the proton m oves in the +x-d irection, parallel to the m agnetic field , the
m agnitud e of the m agnetic force it experiences is F qvB sin 0
0 , or
there is a zero force in this case.
19.2
(a) For a positively charged particle, the d irection of the force is that pred icted by the
right hand rule num ber one. 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 d irection of the force is exactly opposite w hat
the right hand rule num ber 1 pred icts for positive charges. Thus, the answ ers for
part (b) are reversed from those given in part (a) .
217
218
19.3
CH APTER 19
Since the particle is positively charged , use the right hand rule num ber 1. In this case,
start w ith the fingers of the right hand in the d irection of v and the thum b pointing in
the d irection of F . As you start closing the hand , the fingers point in the d irection of B
after they have m oved 90°. The results are
(a)
19.4
toward the right
(c)
toward bottom of page
H old the right hand w ith the fin gers in the d irection of v so that as you close your hand ,
the fingers m ove tow ard the d irection of B . The thum b w ill point in the d irection of the
force (and hence the d eflection) if the particle has a p ositive charge. The results are
(a)
(c)
19.5
(b)
into the page
toward top of page
180
zero force
(b)
out of the page , since the charge is negative.
(d )
into the page
(a) The proton experiences m axim um force w hen it m oves perpend icular to the
m agnetic field , and the m agnitud e of this m axim um force is
Fmax
(b) amax
qvB sin 90
1.60 10
1.44 10 12 N
1.67 10-27 kg
Fmax
mp
19
C 6.00 106 m s 1.50 T 1
1.44 10
12
N
8.62 1014 m s 2
(c) Since the m agnitud e of the charge of an electron is the sam e as that of a proton, the
force experienced by the electron would have the same magnitude , but w ould be in the
opposite d irection d ue to the negative charge of the electron. The acceleration of the
electron w ould be m uch greater than that of the proton because of the m ass of the
electron is m uch sm aller.
19.6
From F
F
qvB sin , the m agnitud e of the force is foun d to be
1.60 10
19
C 6.2 106 m s 50.0 10
6
T sin 90.0
4.96 10
17
N
Using the right-hand -rule (fingers point w estw ard in d irection of v , so they m ove
d ow nw ard tow ard the d irection of B as you close the hand , the thum b points
southw ard . Thus, the d irection of the force exerted on a proton (a positive charge) is
toward the south .
M agnetism
19.7
219
The gravitational force is sm all enough to be ignored , so the m agnetic force m ust supply
the need ed centripetal acceleration. Thus,
m
v2
r
v
qvB sin 90 , or v
1.60 10
19
qBr
w here r
m
C 4.00 10
1.67 10
27
8
RE 1000 km=7.38 106 m
T 7.38 106 m
2.83 107 m s
kg
If v is toward the west and B is northw ard , F w ill be d irected d ow nw ard as required .
19.8
The speed attained by the electron is found from
v
2e
2 1.60 10
V
m
19
1 2
mv
2
C 2 400 V
9.11 10
31
kg
q
V , or
2.90 107 m s
(a) Maxim um force occurs w hen the electron enters the region perpend icular 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) Minim um force occurs w hen the electron enters the region parallel to the field .
Fmin
q vB sin 0
0
220
19.9
CH APTER 19
The m agnitud e of the m agnetic force acting on the electron is F
the m agnitud e of the m agnetic field is given by
B
me a
ev
9.11 10
31
1.60 10
kg 4.0 1016 m s2
19
1.5 10
7
C 1.5 10 m s
2
q vB sin90
me a , so
T
To d eterm ine the d irection of the field , em ploy a variation of right -hand rule num ber 1.
H old your right hand flat w ith the fingers extend ed in the d irection of the electron’ s
velocity (tow ard the top of the page) and the thum b in the d irection of the m agnetic
force (tow ard the right ed ge of the page). Then, as you close your hand , the fingers w ill
point out of the page after they have m oved 90°. This w ould be the correct d irection for
the m agnetic field if the particle were positively charged. Since the electron is a negative
particle, the actual d irection of the field is opposite that pred icted by the right -hand rule,
or it is d irected into the page (the –z-direction) .
19.10
The force on a single ion is
F1
qvB sin
1.60 10
19
C 0.851 m s 0.254 T sin 51.0
2.69 10
20
N
The total num ber of ions present is
N
3.00 1020
ions
100 cm3
cm3
3.00 1022
Thus, assum ing all ions m ove in the sam e d irection through the field , the total force is
F
N F1
3.00 1022 2.69 10
20
N
806 N
M agnetism
19.11
221
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
4.80 10
Fm
19.12
19.13
1.60 10
17
4.80 10
17
19
C 6.00 106 m s 50.0 10
6
T sin 90.0
N in direction opposite right hand rule prediction
N downward
H old the right hand w ith the fingers in the d irection of the current so, as you close the
hand , the fingers m ove tow ard the d irection of the m ag netic field . The thum b then
points in the d irection 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
From F
B
BI L sin , the m agnetic field is
F L
I sin
0.12 N m
15 A sin 90
The d irection of B m ust be the
the +x-d irection.
8.0 10
3
T
z-direction to have F in the –y-d irection w hen I is in
222
19.14
CH APTER 19
(a)
F
(b)
Neither the direction of the magnetic field nor that of the current is given. Both m ust be
BIL sin
0.28 T 3.0 A 0.14 m sin90
0.12 N
know n before the d irection of the force can be d eterm ined . In this problem , you can
only say that the force is perpen d icular to both the w ire and the field .
19.15
Use the right hand rule num ber 1, hold ing your right hand w ith the fingers in the
d irection of the current and the thum b pointing in the d irection of the force. As you
close your hand , the fingers w ill m ove tow ard the d irection of the m agnetic field . The
results are
(a)
19.16
(b)
into the page
(c)
toward the right
toward the bottom of the page
In ord er to just lift the w ire, the m agnetic force exerted on a unit length of the w ire m ust
be d irected upw ard and have a m agnitud e equal to the w eight per unit length. That is,
the m agnitud e is
F
m
BI sin
g
giving
B
m
g
I sin
To find the m inim um possible field , the m agnetic field should be perpend icular to the
current
90.0 . Then,
Bmin
m
g
I sin90.0
0.500
g 1 kg
cm 103 g
102 cm
1m
9.80 m s2
2.00 A 1
0.245 T
To find the d irection of the field , hold the right hand w ith the thum b pointing upw ard
(d irection of the force) and the fingers pointing southw ard (d irection of current). Then,
as you close the hand , the fingers point eastw ard . The m agnetic field should be d irected
eastward .
19.17
F
BIL sin
19.18
(a) The m agnitud e is
F
0.300 T 10.0 A 5.00 m sin 30.0
BIL sin
7.50 N
0.60 10-4 T 15 A 10.0 m sin 90
9.0 10
3
N
F is perpend icular to B . Using the right hand rule num ber 1, the orientation of F
is found to be 15 above the horizontal in the northward direction .
M agnetism
(b) F
0.60 10-4 T 15 A 10.0 m sin 165
BIL sin
2.3 10
3
223
N
and , from the right hand rule num ber 1, the d irection is horizontal and due west
19.19
For m inim um field , B should be perpend icular to the w ire. If the force is to be
northw ard , the field m ust be d irected downward .
To keep the w ire m oving, the m agnitud e of the m agnetic force m ust equal that of the
kinetic friction force. Thus, BI L sin90
k mg , or
B
19.20
k
mL g
0.200 1.00 g cm 9.80 m s2
I sin 90
1.50 A 1.00
1 kg
103 g
102 cm
1m
0.131 T
To have zero tension in the w ires, the m agnetic force per unit
length m ust be d irected upw ard and equal to the w eight per
unit length of the cond uctor. Thus,
Fm
L
I
BI
mg
, or
L
m L g
0.040 kg m 9.80 m s 2
B
3.60 T
0.109 A
From the right hand rule num ber 1, the current m ust be to the right if the force is to be
upw ard w hen the m agnetic field is into the page.
19.21
(a)
The magnetic force must be directed upward and its magnitude must equal mg ,the w eight
of the w ire. Then, the net force acting on the w ire w ill b e zero and it can m ove
upw ard at constant speed .
(b) The m agnitud e of the m agnetic force m ust be BI L sin
field
90 . Thus,
Bmin
mg
IL
0.015 kg 9.80 m s2
5.0 A 0.15 m
mg , and for m inim um
0.20 T
For the m agnetic force to be d irected upw ard w hen the current is tow ard the left, B
m ust be d irected out of the page .
224
CH APTER 19
(c) If the field exceed s 0.20 T, the upw ard m agnetic force exceed s the d ow nw ard force
of gravity, so the wire accelerates upward.
19.22
The m agnitud e of the m agnetic force exerted on a current-carrying cond uctor in a
m agnetic field is given by F BIL sin , w here B is the m agnitud e of the field , L is the
length of the cond uctor, I is the current in the cond uctor, and
is the angle the
cond uctor m akes w ith the d irection of the field . In this case,
F
19.23
0.390 T 5.00 A 2.80 m sin
(a) If
60.0 , then sin
0.866 and F
(b) If
90.0 , then sin
1.00 and F
5.46 N
(c) If
120 , then sin
0.866 and F
4.73 N
4.73 N
For each segm ent, the m agnitud e of the force is given by F BI L sin , and the d irection
is given by the right hand rule num ber 1. The results of applying these to each of the
four segm ents are sum m arized below .
Segm ent
L (m )
ab
0.400
bc
cd
da
19.24
5.46 N sin
F (N )
Direction
180°
0
_
0.400
90.0°
0.040 0
negative x
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-d irections
0.400 2
90.0°
The m agnitud e of the force is
F
BIL sin
5.0 10
5
N 2.2 103 A 58 m sin 65
5.8 N
and the right-hand rule num ber 1 show s its d irection to be into the page .
19.25
The torque on a cu rrent loop in a m agnetic field is
BIAN sin , and m axim um torque
occurs w hen the field is d irected parallel to the plane of the loop (
90 ). Thus,
max
0.50 T 25 10
3
A
5.0 10
2
m
2
50 sin 90
4.9 10
3
N m
M agnetism
19.26
The m agnitud e of the torque is
NBIAsin , w here is the angle betw een the field
and the perpend icular to the plane of the loop. The circum ference of the loop is
1.00 m
1 2
r2
m .
and the area is A
2 r 2.00 m , so the rad ius is r
1 0.800 T 17.0 10
Thus,
19.27
225
3
1
A
m 2 sin 90.0
4.33 10
3
N m
0.200 m 0.150 m 0.094 2 m2 . Since the field is parallel to the
90.0 and the m agnitud e of the torque is
The area is A
ab
plane of the loop,
NBIA sin
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 d irected to m ake the left-hand sid e of the loop m ove tow ard you and the
right-hand sid e m ove aw ay.
19.28
N ote that the angle betw een the field and the perpend icular to the plane of the loop is
90.0 30.0 60.0 . Then, the m agnitud e of the torque is
NBIAsin
100 0.80 T 1.2 A
0.40 m 0.30 m sin 60.0
10 N m
With current in the –y-d irection, the outsid e ed ge of the loop w ill experience a force
d irected out of the page (+z-d irection) accord ing to the right hand rule num ber 1. Thus,
the loop w ill rotate clockwise as viewed from above .
19.29
(a) The torque exerted on a coil by a uniform m agnetic field is
BIAN sin , w ith
m axim um torque occurring w hen
90 , Thus, the current in the coil m ust be
I
0.15 N m
max
BAN
0.90 T
3.0 10
2
m 5.0 10
(b) If I has the value found above and
BIAN sin
0.90 T 0.56 A
2
0.56 A
m
200
is now 25°, the torque on the coil is
0.030 m 0.050 m
200 sin 25
0.064 N m
226
19.30
CH APTER 19
The resistance of the loop is
R
1.70 10
L
A
8
m 8.00 m
1.00 10
4
m2
V
R
and the current in the loop is I
The m agnetic field exerts torque
sin
1. Thus,
NBIA
max
19.31
1.36 10
0.100 V
1.36 10 3
NBIAsin
3
73.5 A
on the loop, and this is a m axim um w hen
1 0.400 T 73.5 A 2.00 m
2
118 N m
(a) Let be the angle the plane of the loop m akes
w ith the horizontal as show n in the sketch at the
right. Then, the angle it m akes w ith the vertical is
. The num ber of turns on the loop is
90.0
N
L
circumference
4.00 m
4 0.100 m
10.0
The torque about the z-axis d ue to gravity is
s
mg cos , w here s 0.100 m is the length
g
2
of one sid e of the loop. This torque tend s to rotate
the loop clockw ise. The torque d ue to the m agnetic
force tend s to rotate the loop counterclockw ise about
the z-axis and has m agnitud e m NBIAsin . At
equilibrium ,
m
g
or NBI s 2 sin
mg s cos
2.
This red uces to
tan
Since tan
mg
2 NBIs
0.100 kg 9.80 m s2
2 10.0 0.010 0 T 3.40 A 0.100 m
cot , the angle the loop m akes w ith the vertical at
tan 90.0
equilibriu m is
14.4
cot
1
14.4
3.97 .
M agnetism
227
(b) At equilibrium ,
m
NBI s 2 sin
2
10.0 0.010 0 T 3.40 A 0.100 m sin 90.0
3.39 10
19.32
3
3.97
N m
(a) The current in segm ent a-b is in the +y-d irection. Thus, by right-hand rule 1, the
m agnetic force on it is in the + x-direction . Im agine this force being concentrated at
the center of segm ent a-b. Then, w ith a pivot at point a (a point on the x-axis), this
force w ould tend to rotate the cond uctor a-b in a clockw ise d irection about the zaxis, so the d irection of this torque is in the z -direction .
(b) The current in segm ent c-d is in the
y-direction , and the right-hand rule 1 gives the
d irection of the m agnetic force as the x-direction . With a pivot at point d (a point
on the x-axis), this force w ould tend to rotate the cond uctor c-d counter-clockw ise
about the z-axis, and the d irection of this torque is in the z -direction .
(c)
No. The torques d ue to these forces are along the z-axis and cannot cause rotation
about the x-axis. Further, both the forces and the torques are equal in m agnitud e
and opposite in d irection, so they sum to zero and cannot affect the m otion of the
loop.
228
CH APTER 19
(d ) The m agnetic force is perpend icular to both the d irection of the current in b -c (the
+x-d irection) and the m agnetic field . As given by right-hand rule 1, this places it
in the y-z plane at 130° counterclockwise from the +y-axis. The force acting on segm ent
b-c tend s to rotate it counterclockw ise about the x-axis, so the torque is in
the +x-direction.
(e) The loop tend s to rotate counterclockwise about the x-axis.
(f)
IAN
0.900 A
0.500 m 0.300 m
0.135 A m2
1
(g) The m agnetic m om ent vector is perpend icular to the plane of the loop (the x-y
plane), and is therefore parallel to the z-axis. Because the current flow s clockw ise
around the loop, the m agnetic m om ent vector is d irected d ow nw ard , in the
negative z-d irection. This m eans that the angle betw een it and the d irection of the
m agnetic field is
90.0 40.0 130 .
(h)
19.33
B sin
0.135 A m2 1.50 T sin 130
0.155 N m
(a) The m agnetic force acting on the electron provid es the centripetal acceleration,
hold ing the electron in the circular path. Therefore, F q vB sin90 me v2 r , or
me v
eB
r
9.11 10
31
1.60 10
kg 1.5 107 m s
19
C 2.0 10
3
T
0.043 m
4.3 cm
(b) The tim e to com plete one revolution around the orbit (i.e., the period ) is
T
19.34
(a)
F
(b) a
19.35
distance traveled
constant speed
qvB sin
F
m
1.60 10
1.25 10 13 N
1.67 10 27 kg
2 r
v
19
2
0.043 m
1.5 107 m s
1.8 10
8
s
C 5.02 106 m s 0.180 T sin 60.0
1.25 10
13
N
7.50 1013 m s 2
For the particle to pass through w ith no d eflection, the net force acting on it m ust be
zero. Thus, the m agnetic force and the electric force m ust be in opposite d irections and
have equal m agnitud es. This gives
Fm
Fe , or qvB qE w hich red uces to v
E B
M agnetism
19.36
229
The speed of the particles em erging from the velocity selector is v E B (see Problem
35). In the d eflection cham ber, the m agnetic force supplies the centripetal acceleration,
mv m E B
mE
mv 2
so qvB
, or r
qB
qB
qB 2
r
Using the given d ata, the rad ius of the path is found to be
26
kg 950 V m
-19
2
2.18 10
r
1.60 10
19.37
C 0.930 T
1.50 10
From conservation of energy, KE PE
1 2
mv
2
qV f
m
0.150 mm
KE PE i , w e find that
0 qVi , or the speed of the particle is
2 q Vi V f
v
f
4
2q
m
2 1.60 10
V
19
C 250 V
2.50 10-26 kg
m
5.66 104 m s
The m agnetic force supplies the centripetal acceleration giving qvB
or
19.38
mv
qB
r
2.50 10
26
kg 5.66 104 m s
1.60 10
19
C 0.500 T
1.77 10
2
m
mv 2
r
1.77 cm
Since the centripetal acceleration is furnished by the m agnetic force acting on the ions,
mv 2
mv
qvB
or the rad ius of the path is r
. Thus, the d istance betw een the im pact
qB
r
points (that is, the d ifference in the d iam eters of the paths follow ed by the U 238 and the
U 235 isotopes) is
d
2 r238
2v
m238
qB
r235
2 3.00 105 m s
1.60 10
or
d
3.11 10
2
19
C 0.600 T
m
3.11 cm
m235
238 u 235 u 1.66 10
27
kg
u
230
19.39
CH APTER 19
In the perfectly elastic, head -on collision betw een the -particle and the initially
stationary proton, conservation of m om entum requires that mp v p m v m v0 w hile
conservation of kinetic energy also requires that v0
the fact that m
mp v
and
vp
0
v
vp
or v p
v
v0 . Using
4mp and com bining these equations gives
v0
4mp v
3v0 5
v0
4mp v0
or
v
Thus, v
8v0 5
3v0 5
3
v0
5
3 5
vp
5 8
3
vp
8
After the collision, each particle follow s a circular path in the horizontal plane w ith the
m agnetic force supplying the centripetal acceleration. If the rad ius of the proton’ s
trajectory is R, and that of the alpha particle is r, w e have
v2p
mp v p mp v p
q p v p B mp
or
R
R
qp B
eB
and
19.40
q v B m
v2
r
or
4mp 3v p 8
mv
q B
2e B
3 mp v p
4 eB
3
R
4
A charged particle follow s a circular path w hen it m oves perpend icular to the m agnetic
field . The m agnetic force acting on the particle provid es the centripetal acceleration,
hold ing the particle in the circular path. Therefore, F qvB sin90 mv2 r . Since the
kinetic energy is K
mv2 2 , w e rew rite the force as F
for the speed v gives v
19.41
r
2K r , and solving
qvB sin90
2K
.
qBr
(a) Within the velocity selector, the electric and m agnetic field s exert forces in opposite
d irections on charged particles passing through. For particles having a particular
speed , these forces have equal m agnitud es, and the particles pass through w ithout
d eflection. The selected speed is found from Fe qE qvB Fm , giving v E B . In
the d eflection cham ber, the selected particles follow a circular path having a
d iam eter of d 2r 2mv qB . Thus, the m ass to charge ratio for these particles is
m
q
Bd
2v
Bd
2 E B
B2d
2E
0.0931 T
2
0.396 m
2 8 250 V m
2.08 10
7
kg C
M agnetism
231
(b) If the particle is d oubly ionized (i.e., tw o electrons h ave been rem oved from the
neutral atom ), then q 2e and the m ass of the ion is
m
2e
m
q
19
2 1.60 10
C 2.08 10
7
kg C
6.66 10
26
kg
(c) Assum ing this is an elem ent, the m ass of the ion should be roughly equal to the
atom ic w eight m ultiplied by the m ass of a proton (or neutron). This w ould give the
atom ic w eight as
At. wt.
19.42
m
mp
6.66 10
1.67 10
26
27
kg
kg
39.9 suggesting that the elem ent is calcium .
Since the path is circular, the particle m oves perpend icular to the m agnetic field , and the
v2
mv
qvB , or B
m agnetic force supplies the centripetal acceleration. H ence, m
. But
qr
r
the m om entum is given by p mv
KE
19.43
1.60 10
2 1.67 10
2m KE
qr
1.60 10
27
19
12
J . We then have
kg 1.60 10
12
J
10
C 5.80 10 m
7.88 10
12
T
Treat the lightning bolt as a long, straight cond uctor. Then, the m agnetic field is
B
19.44
1.60 10 19 J
1 eV
10.0 106 eV
B
2m KE , and the kinetic energy of this proton is
I
2 r
0
4
10
7
T m A 1.00 104 A
2
100 m
2.00 10
5
T
20.0 T
Im agine grasping the cond uctor w ith the right hand so the fingers curl around the
cond uctor in the d irection of the m agnetic field . The thum b then points along the
cond uctor in the d irection of the current. The results are
(a)
toward the left
(b)
out of page
(c)
lower left to upper right
232
19.45
CH APTER 19
The m agnetic field at d istance r from a long cond ucting w ire is B
B 1.0 10
I
15
2 rB
2
0.040 m 1.0 10
4
10
7
15
T
T mA
A
Mod el the tornad o as a long, straight, vertical cond uctor and im a gine grasping it w ith
the right hand so the fingers point northw ard on the w estern sid e of the tornad o (that is,
at the observatory’ s location.) The thum b is d irected d ow nw ard , m eaning that the
conventional current is downward or negative charge flows upward .
I
2 rB
9.00 103 m 1.50 10
2
4
0
From B
r
19.48
10
2.0 10
The m agnitud e of the current is found from B
19.47
I 2 r . Thus, if
T at r 4.0 cm , the current m ust be
0
19.46
0
0
8
0
T
I 2 r as
675 A
10-7 T m A
I 2 r , the required d istance is
10-7 T m A 20 A
4
I
2 B
0
2
1.7 10
3
T
2.4 10
3
m
2.4 mm
Assum e that the w ire on the right is w ire 1 and that on the left is w ire 2. Also, choose the
positive d irection for the m agnetic field to be out of the page and negative into the page.
(a) At the point half w ay betw een the tw o w ires,
Bnet
or Bnet
B1
7
4
10
2
5.00 10-2 m
T mA
I
2 r2
0 2
0
2 r
10.0 A
I1
I2
4.00 10
5
T
40.0 T into the page
Bnet
(b) At point P1 ,
Bnet
I
2 r1
0 1
B2
4
10
7
B1
B2
0
2
T m A 5.00 A
2
0.100 m
I1
r1
I2
r2
5.00 A
0.200 m
5.00 T out of page
M agnetism
Bnet
(c) At point P2 ,
4
Bnet
10
7
B1
T mA
2
5.00 A
0.300 m
2
I1
r1
0
B2
233
I2
r2
5.00 A
0.200 m
1.67 T out of page
19.49
The d istance from each w ire to point P is given by
r
1
2
0.200 m
2
2
0.200 m
0.141 m
At point P, the m agnitud e of the m agnetic field
prod uced by each of the w ires is
B
4
I
2 r
0
10
7
T m A 5.00 A
2
0.141 m
7.07 T
Carrying currents into the page, the field A prod uces
at P is d irected to the left and d ow n at –135°, w hile B
creates a field to the right and d ow n at – 45°. Carrying currents tow ard you, C
prod uces a field d ow nw ard and to the right at – 45°, w hile D ’ s contribution is
d ow nw ard and to the left. The horizontal com ponents of these equal m agnitud e
contributions cancel in pairs, w hile the vertical com ponen ts all ad d . The total field
is then
Bnet
19.50
4 7.07 T sin 45.0
20.0 T toward the bottom of the page
Call the w ire carrying a current of 3.00 A w ire 1 and the other w ire 2. Also, choose the
line running from w ire 1 to w ire 2 as the positive x-d irection.
(a) At the point m id w ay betw een the w ires, the field d ue to
each w ire is parallel to the y-axis and the net field is
Bnet
B1 y
Thus,
Bnet
or
Bnet
B2 y
I1 I 2 2 r
0
4
10
2
7
T mA
0.100 m
3.00 A 5.00 A
4.00 T toward the bottom of the page
4.00 10
6
T
234
CH APTER 19
(b) At point P, r1
2 and B1 is d irected at
0.200 m
1
135 .
The m agnitud e of B1 is
B1
4
I
2 r1
0 1
7
10
2
T m A 3.00 A
2.12 T
0.200 2 m
The contribution from w ire 2 is in the –x-d irection and
has m agnitud e
B2
4
I
2 r2
0 2
10
7
T m A 5.00 A
2
5.00 T
0.200 m
Therefore, the com ponents of the net field at point P are:
Bx
B1 cos135
B2 cos180
2.12 T cos135
and
By
B1 sin135
Bx2
Therefore, Bnet
1
tan
or Bnet
19.51
5.00 T cos180
Bx
tan
By
B2 sin180
By2
1
6.50 T
2.12 T sin135
0
1.50 T
6.67 T at
6.50 T
1.50 T
77.0
6.67 T at 77.0° to the left of vertical
Call the w ire along the x-axis w ire 1 and the other w ire 2. Also, choose the positive
d irection for the m agnetic field s at point P to be out of the page.
At point P, Bnet
or
Bnet
Bnet
4
B1
10
B2
7
2
I
2 r1
0 1
T mA
I
2 r2
0 2
7.00 A
3.00 m
0.167 T out of the page
0
2
I1
r1
6.00 A
4.00 m
I2
r2
1.67 10
7
T
M agnetism
19.52
235
(a) Im agine the horizontal x-y plane being perpend icular to the page, w ith the positive
x-axis com ing out of the page tow ard you and the positive y-axis tow ard the right
ed ge of the page. Then, the vertically upw ard positive z-axis is d irected tow ard the
top of the page. With the current in the w ire flow ing in the positive x-d irection, the
right-hand rule 2 gives the d irection of the m agnetic field above the wire as being
tow ard the left, or in the y-direction.
(b) With the positively charged proton m oving in the –x-d irection (into the page), righthand rule 1 gives the d irection of the m agnetic force on the proton as being d irected
tow ard the top of the page, or upward, in the positive z-direction.
(c) Since the proton m oves w ith constant velocity, a zero net force acts on it. Thus, the
magnitude of the magnetic force must equal that of the gravitational force .
(d )
Fz
maz
0
qv 0 I
2 d
(e)
d
19.53
Fg or qvB mg w here B
0
I 2 d . This gives
qv 0 I
.
2 mg
mg , or the d istance the proton is above the w ire m ust be d
1.60 10
qv 0 I
2 mg
d
Fm
19
C 2.30 104 m s 4
2
5.40 10
2
m
1.67 10
27
10
7
T m A 1.20 10
kg 9.80 m s
6
A
2
5.40 cm
(a) From B
0 I 2 r , observe that the field is inversely proportional to the d istance
from the cond uctor. Thus, the field w ill have one-tenth its original value if the
d istance is increased by a factor of 10. The required d istance is then
r 10r 10 0.400 m
4.00 m
(b) A point in the plane of the cond uctors and 40.0 cm from the center of the cord is
located 39.85 cm from the nearer w ire and 40.15 cm from the far w ire. Since the
currents are in opposite d irections, so are their contributions to the net field .
Therefore, Bnet B1 B2 , or
Bnet
I 1
2
r1
1
r2
0
7.50 10
9
T
4
10
7
T m A 2.00 A
2
7.50 nT
1
0.398 5 m
1
0.4015 m
236
CH APTER 19
(c) Call r the d istance from cord center to field point P
and 2d 3.00 mm the d istance betw een centers of
the cond uctors.
0
Bnet
2
1
1
r d
7.50 10
so r
I
10
r
4
T
0
d
10
2
7
I
2d
r d2
2
T m A 2.00 A
3.00 10 3 m
r 2 2.25 10 6 m2
2
1.26 m
The field of the tw o-cond uctor cord is w eak to start w ith and falls off rapid ly w ith
d istance.
(d ) The cable creates zero field at exterior points, since a loop in Am père’ s law
encloses zero total current.
19.54
(a) Point P is equid istant from the tw o w ires w hich
carry id entical currents. Thus, the contributions of
the tw o w ires, Bupper and Blower , to the m agnetic
field at P w ill have equal m agnitud es. The
horizontal com ponents of these contributions w ill
cancel, w hile the vertical com ponents ad d . The
resultant field w ill be vertical, in the y-direction.
(b) The d istance of each w ire from point P is
r
x2
d 2 , and the cosine of the angle that
Bupper and Blower m ake w ith the vertical is cos
x r . The m agnitud e of either Bupper
or Blower is Bwire
0 I 2 r and the vertical com ponents of either of these
contributions have values of
Bwire
y
Bwire cos
I x
2 r r
0
Ix
2 r2
0
The m agnitud e of the resultant field at point P is then
BP
2 Bwire
Ix
r2
0
y
x
0
2
Ix
d2
M agnetism
237
(c) The point m id w ay betw een the tw o w ires is the origin (0,0). From the above result
for Part (b), the resultant field at this m id point is BP x 0 0 . This is as expected ,
because right-hand rule 2 show s that at the m id point the field d ue to the upper w ire
is tow ard the right, w hile that d ue to the low er w ire is tow ard the left. Thus, the tw o
field s cancel, yield ing a zero resultant field .
19.55
The force per unit length that one w ire exerts on the other is F
0 I1I 2 2 d , w here d
is the d istance separating the tw o w ires. In this case, the value of this force is
F
4
10
2
7
T m A 3.0 A
6.00 10
2
2
3.0 10
m
5
Nm
Im agine these tw o w ires lying sid e by sid e on a table w ith the tw o currents flow ing
tow ard you, w ire 1 on the left and w ire 2 on the right. Right-hand rule 2 show s the
m agnetic field d ue to w ire 1 at the location of w ire 2 is d irected vertically upw ard . Then,
right-hand rule 1 gives the d irection of the force experienced by w ire 2, w ith its current
flow ing through this field , as being to the left, back tow ard w ire 1. Thus, the force one
w ire exerts on the other is an attractive force.
19.56
(a) The force per unit length that parallel cond uctors exert on each other is
2.0 10 4 N m , I1 5.0 A , and d 4.0 cm , the
F
0 I1I 2 2 d . Thus, if F
current in the second w ire m ust be
I2
2 d F
0 I1
4
2
4.0 10
10
7
2
m
T m A 5.0 A
2.0 10
4
Nm
8.0 A
(b) Since parallel cond uctors carrying currents in the sam e d irection attract e ach other
(see Section 19.8 in the textbook), the currents in these cond uctors w hich repel each
other m ust be in opposite directions.
(c) The result of reversing the d irection of either of the currents w ould be that the
force of interaction would change from a force of repulsion to an attractive force . The
expression for the force per unit length, F
I I 2 d , show s that d oubling
0 1 2
either of the currents w ould double the magnitude of the force of interaction .
238
19.57
CH APTER 19
In ord er for the system to be in equilibrium , the repulsive m agnetic force per unit length
on the top w ire m ust equal the w eight per unit length of this w ire.
Thus,
0 I1 I 2
2 d
F
L
d
0.080 N m , and the d istance betw een the w ires w ill be
2
4.5 10
19.58
3
m
or
T m A 60.0 A 30.0 A
2
0.080 N m
4.5 mm
0 I1 I 2
2 c
Fnet
4
10
II
c a
II
2
0 1 2
2
7
0 1 2
1
c
T m A 5.00 A 10.0 A 0.450 m
5
N
2.70 10
5
BL
0N
2.0 10
4
3
10
T 6.0 10
7
2
m
T m A 30
The m agnetic field insid e of a solenoid is B
turns on this solenoid m ust be
N
BL
0I
9.0 T 0.50 m
4
10
7
1
0.100 m
1
0.250 m
N to the left
The m agnetic field insid e of a solenoid is B
solenoid m ust be
I
1
c a
2
2.70 10
19.60
7
10
The m agnetic forces exerted on the top and bottom segm ents of the rectangular loop are
equal in m agnitud e and opposite in d irection. Thus, these forces cancel, and w e only
need consid er the sum of the forces exerted on the right and left sid es of the loop.
Choosing to the left (tow ard the long, straight w ire) as the positive d irection, the sum of
these tw o forces is
Fnet
19.59
4
II
0.080 N m
0 1 2
T m A 75 A
0
nI
0
N L I . Thus, the current in this
0
N L I . Thus, the num ber of
3.2 A
0
nI
4.8 104 turns
M agnetism
19.61
(a) From R
L
239
L A , the required length of w ire to be used is
5.00
R A
0.500 10
1.7 10
8
3
m
2
4
58 m
m
The total num ber of turns on the solenoid (that is, the num ber of tim es this length of
w ire w ill go around a 1.00 cm rad ius cylind er) is
N
L
2 r
(b) From B
n
58 m
2
0
B
0I
1.00 10
2
9.2 102
m
920
nI , the num ber of turns per unit length on the solenoid is
4.00 10
-7
4
10
2
T
T m A 4.00 A
7.96 103 turns m
Thus, the required length of the solenoid is
L
19.62
N
n
9.2 102 turns
7.96 103 turns m
0.12 m
12 cm
The m agnetic field insid e the solenoid is
B
0
nI1
4
10
7
T m A
30
turns
cm
100 cm
1m
15.0 A
5.65 10
2
T
Therefore, the m agnitud e of the m agnetic force on an y one of the sid es 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
The forces acting on the sid es of the loop lie in the plane of the loop, are perpend icular
to the sid es, and are d irected away from the interior of the loop. Thus, they tend to
stretch the loop but d o not tend to rotate it. The torque acting on the loop is
0
240
19.63
CH APTER 19
(a) The m agnetic force supplies the centripetal acceleration, so qvB mv2 r . The
m agnetic field insid e the solenoid is then fou nd to be
B
(b) From B
I
1.60 10
0
kg 1.0 104 m s
19
C 2.0 10
2
m
2.8 10
6
T
2.8 T
nI , the current is the solenoid is found to be
B
0n
2.8 10
4
8.9 10
19.64
31
9.11 10
mv
qr
4
10
A
-7
T mA
6
T
25 turns cm 100 cm 1 m
0.89 mA
(a) When sw itch S is closed , a total current N I (current I in a
total of N cond uctors) flow s tow ard the right through
the low er sid e of the coil. This results in a d ow nw ard
force of m agnitud e Fm B NI w being exerted on the
coil by the m agnetic field , w ith the requirem ent that the
balance exert a upw ard force F mg on the coil to
bring the system back into balance.
In ord er for the m agnetic force to be d ow nw ard , the right -hand rule num ber 1
show s that the m agnetic field m ust be d irected out of the page tow ard the read er.
For the system to be restored to balance, it is necessary that
Fm
F
or
B NI w mg , giving
B
mg
NIw
(b) The m agnetic field exerts forces of equal m agnitud es and opposite d irections on the
tw o sid es of the coil. These forces cancel each other and do not affect the balance of the
coil. H ence the d im ension of the sizes is not need ed .
(c)
B
mg
NIw
20.0 10
3
kg 9.80 m s 2
50 0.30 A 5.0 10
2
m
0.26 T
M agnetism
19.65
(a) The m agnetic field at the center of a circular current loop of rad ius R and carrying
current I is B
0 I 2R . The d irection of the field at this center is given by righthand rule num ber 2. Taking out of the page (tow ard the read er) as positive the net
m agnetic field at the com m on center of these coplanar loops is
Bnet
B2
I
2r2
B1
(b) To have Bnet
I
2r1
0 2
6
5.2 10
T
0 1
4
10
7
T mA
2
3.0 A
5.0 A
2
9.0 10 m 12 10 2 m
5.2 T into the page
0 , it is necessary that I2 r2
I2
r1
I1
r2
19.66
241
3.0 A
12 cm
5.0 A
I1 r1 , or
7.2 cm
Since the m agnetic force m ust supply the centripetal acceleration, qvB mv2 r or the
rad ius of the path is r mv qB .
(a) The tim e for the electron to travel the sem icircular path (of length
r
v
t
mv
v qB
1.79 10
9
9.11 10
m
qB
s
1.60 10
19
31
kg
C 0.010 0 T
1.79 ns
(b) The rad ius of the sem icircular path is 2.00 cm . From r
the electron is p = mv = qBr, and the kinetic energy is
KE
KE
1
2
mv
2
mv
5.62 10
r ) is
2
2m
16
J
q2 B2 r 2
2m
1.60 10
1 keV
1.60 10 16 J
19
C
2
0.010 0 T
2 9.11 10
3.51 keV
mv qB , the m om entum of
31
2
2.00 10
kg
2
m
2
242
19.67
CH APTER 19
Assum e w ire 1 is along the x-axis and w ire 2 along the y-axis.
(a) Choosing out of the page as the positive field d irection, the field at point P is
B
B1
0
B2
2
7
5.00 10
I1
r1
T
4
I2
r2
10
7
T mA
5.00 A
0.400 m
2
3.00 A
0.300 m
0.500 T out of the page
(b) At 30.0 cm above the intersection of the w ires, the field
com ponents are as show n at the right, w here
By
I
2 r
0 1
B1
4
10
2
and
Bx
B2
7
T m A 5.00 A
3.33 10
0.300 m
4
I
2 r
0 2
10
7
2
6
T
T m A 3.00 A
0.300 m
2.00 10
6
T
The resultant field is
B
or B
19.68
Bx2
By2
3.89 10
6
T at
=tan
1
By
Bx
59.0
3.89 T at 59.0 clockwise from +x direction
For the rail to m ove at constant velocity, the net force acting on it m ust be zero. Thus, the
m agnitud e of the m agnetic force m ust equal that of the friction force giving
BIL
k mg , or
B
k
mg
IL
0.100 0.200 kg 9.80 m s 2
10.0 A 0.500 m
3.92 10
2
T
M agnetism
19.69
243
(a) Since the m agnetic field is d irected from N to S (that is, from left to right w ithin the
artery), positive ions w ith velocity in the d irection of the blood flow experience a
m agnetic d eflection tow ard electrod e A . N egative ions w ill experience a force
d eflecting them tow ard electrod e B. This separation of charges creates an electric
field d irected from A tow ard B. At equilibrium , the electric force caused by this
field m ust balance the m agnetic force, so
qvB qE q
or v
V
Bd
V d
160 10
6
V
0.040 0 T 3.00 10
3
m
1.33 m s
(b) The m agnetic field is d irected from N to S. If the charge carriers are negative
m oving in the d irection of v , the m agnetic force is d irected tow ard point B.
N egative charges build up at point B, m aking the potential at A higher than that at
B. If the charge carriers are positive m oving in the d irection of v , the m agnetic force
is d irected tow ard A , so positive charges build up at A . This also m akes the
potential at A higher than that at B. Therefore the sign of the potential d ifference
does not depend on the charge of the ions .
19.70
(a) The m agnetic force acting on the w ire is d irected upw ard and of m agnitud e
Fm
ay
Thus,
ay
(b) Using
t
4.0 10
BIL sin90
Fy
m
3
5.0 10
Fm
BIL
mg
BI
m L
m
T 2.0 A
4
kg m
9.80 m s 2
y
v0 y t
1 2
a y t w ith v0 y
2
2
y
2 0.50 m
ay
6.2 m s2
g , or
0 gives
0.40 s
6.2 m s 2
244
19.71
CH APTER 19
Label the w ires 1, 2, and 3 as show n in Figure 1,
and let B1 , B2 , and B3 respectively rep resent the
m agnitud es of the field s prod uced by the
currents in those w ires. Also, observe that
45 in Figure 1.
At point A , B1
B1
and B3
B2
4
B2
I
3a
4
a 2 or
I 2
10
2
0
2
0
7
T m A 2.0 A
0.010 m
10
7
2
2
T m A 2.0 A
0.030 m
28 T
13 T
These field contributions are oriented as show n in Figure 2.
Observe that the horizontal com ponents of B1 and B2 cancel
w hile their vertical com ponents ad d to B 3 . The resultant
field at point A is then
B1 B2 cos45
BA
53 T directed toward the bottom of the page
At point B,
and B3
B1
I
2a
0
2
B2
I
2 a
0
B3
53 T , or
BA
4
10
2
7
T m A 2.0 A
0.010 m
40 T
20 T . These contributions are oriented as
show n in Figure 3. Thus, the resultant field at B is
BB
B3
20 T directed toward the bottom of the page
M agnetism
At point C, B1
B2
0
I 2
28 T w hile
a 2
B3
40 T . These contributions are oriented
0I 2 a
as show n in Figure 4. Observe that the horizontal
com ponents of B1 and B2 cancel w hile their vertical
com ponents ad d to oppose B 3 . The m agnitud e of the
resultant field at C is
BC
B1
B2 sin 45
B3
56 T sin 45
19.72
40 T= 0
(a) Since one w ire repels the other, the currents m ust be in opposite directions .
(b) Consid er a free bod y d iagram of one of the w ires as show n at
the right.
Fy
T cos8.0
mg
mg
cos8.0
or T
Fx
or Fm
I
0
0
Fm
mg tan8.0 . Thus,
d
mg
cos8.0
T sin 8.0
I 2L
2 d
0
sin 8.0
mg tan 8.0 w hich gives
m L g tan8.0
2
0
Observe that the d istance betw een the tw o w ires is
d
I
2 6.0 cm sin8.0
1.7 10
2
1.7 cm , so
m 0.040 kg m 9.80 m s 2 tan8.0
2.0 10
7
T mA
68 A
245
246
19.73
CH APTER 19
N ote: We solve part (b) before part (a) for this problem .
(b) Since the m agnetic force supplies the centripetal acceleration for this particle,
qvB mv2 r or the rad ius of the path is r mv qB . The speed of the particle m ay be
w ritten as v
2 KE m , so the rad ius becom es
2 1.67 10
2 m KE
r
27
kg 5.00 106 eV 1.60 10
19
J eV
1.60 10-19 C 0.050 0 T
qB
6.46 m
Consid er the circular path show n at the right and
observe that the d esired angle is
sin
1
1.00 m
r
sin
1.00 m
6.46 m
1
8.90
(a) The constant speed of the particle is v
2 KE m , so
the vertical com ponent of the m om entum as the particle leaves the field is
py
mv y
or p y
mv sin
sin 8.90
2 KE m sin
2 1.67 10
8.00 10
19.74
m
21
27
sin
2 m KE
kg 5.00 106 eV 1.60 10
19
J eV
kg m s
The force constant of the spring system is found from the elongation prod uced by the
w eight acting alone.
F
x
k
mg
x
10.0 10
3
kg 9.80 m s 2
0.50 10
2
m
19.6 N m
The total force stretching the springs w hen the field is turned on is
Fy
Fm
mg
kxtotal
M agnetism
247
Thus, the d ow nw ard m agnetic force acting on the w ire is
Fm
kxtotal
mg
19.6 N m 0.80 10
5.9 10
2
2
m
10.0 10
B
19.75
kg 9.80 m s 2
N
Since the m agnetic force is given by Fm
Fm
IL
3
5.9 10
2
N
24 V 5.0 10
2
m
12
Fm
V R L
BIL sin90 , the m agnetic field is
0.59 T
The m agnetic force is very sm all in com parison to the w eight of the ball, so w e treat the
m otion as that of a freely falling bod y. Then, as the ball approaches the ground , it has
velocity com ponents w ith m agnitud es of
vx
v0 x
20.0 m s , and
v02y
vy
2a y
y
0 2
9.80 m s 2
20.0 m
19.8 m s
The velocity of the ball is perpend icular to the m agnetic field and , just before it reaches
the ground , has m agnitud e v
vx2
v y2
28.1 m s . Thus, the m agnitud e of the m agnetic
force is
Fm
qvB sin
5.00 10
19.76
(a)
B1
(b)
F21
(c)
B2
(d )
F12
I
2 d
0 1
4
6
C 28.1 m s 0.010 0 T sin 90.0
10
7
2
0.100 m
B1 I 2
1.00 10
I
2 d
4
B2 I1
1.60 10
0 2
T m A 5.00 A
10
5
7
T 8.00 A
8.00 10
T m A 8.00 A
2
0.100 m
5
T 5.00 A
1.00 10
5
5
5
6
N
T
N directed toward wire 1
1.60 10
8.00 10
1.41 10
5
T
N directed toward wire 2