Tài liệu Physics exercises_solution: Chapter 28 doc

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28.1: For a charge with velocity
,
ˆ
)sm108.00(
6
jv 

the magnetic field produced at
a position
r
away from the particle is
.
ˆ
4
2
0
r
qμ rv
B




So for the cases below:
a)
4
1
2
0
,
ˆ

ˆˆ
ˆ
)m0.500(  rkrvir

.
ˆˆ
)T101.92(
ˆ
)m0.50(
)sm108.0)(C10(6.0
4
ˆ
4
0
5
2
66
0
2
0
0
kkkkB B
π
μ
r
qv
π
μ






b)
.00
ˆˆ
ˆ
)m0.500(
 Brvjr

c)
.
4
1
,
ˆ
ˆˆ
ˆ
)m0.500(
2
0
 rirvkr

.
ˆˆ
4
0
0
0
iiB

2
B
r
qv



d)
0
2
2
2
1
,
ˆ
ˆˆ
ˆ
)m0.500(
ˆ
)m0.500(
rr  irvkjr

22
ˆ
2
ˆ
2
2
ˆ
4

000
iBB
r
qv

ii
B
2


28.2:











22
0
total
4 d
vq
d
qv
BBB



page.theintoT,1038.4
)m120.0(
)sm100.9)(C100.3(
)m120.0(
)sm105.4)(C100.8(
4
4
2
66
2
66
0















B

μ
B

28.3:
3
0
4
r
q
π
μ rv
B





a)
;
ˆ
, iriv rv 



0,



B0rv



b)
;
ˆ
,
ˆ
jriv rv 


m0.500,
ˆ
 rvrkrv


kB
ˆ
)T01.31(sonegative,is
T101.31
)m0.500(
)sm106.80)(C104.80)(CsN101
4
6
6
2
56227
2
0
















q
r
vq
π
μ
B
c)
);
ˆ
ˆ
)(m0.500(,
ˆ
jiriv 


v m0.7071,
ˆ
)m0.500(  rvkrv



 
kB
ˆ
)T104.62(;T104.62
)m0.7071(
)sm106.80)(m0.500(C)104.80)(CsN101
4
77
3
56227
3
0














B
rrvq

π
μ
B
d)
;
ˆ
,
ˆ
kriv rv 


m0.500,
ˆ
 rvrjrv


jB
ˆ
T)101.31(;T101.31
m)0.500(
)sm106.80)(C104.80)(CsN101
4
66
2
56227
2
0














B
r
vq
π
μ
B
28.4:
a) Following Example 28.1 we can find the magnetic force between the charges:
down).points
charge
lowertheonforcetheanduppointschargeuppertheonforce(theN101.69
)m0.240(
m
109.00)(sm104.50)(C103.00)(C108.00(
)AmT10(
4
3
2
6666
7

2
0








r
vvqq
π
μ
F
B
The Coulomb force between the charges is
N3.75)CmN108.99(
2
212
m)(0.240
C100)(8.00)(3.0
229



2
21
r
qq

kF
(the force on the upper
charge points up and the force on the lower charge points down).
The ratio of the Coulomb force to the magnetic force is
21
2
vv
c



3
N101.69
N3.75
102.22
3
.
b) The magnetic forces are reversed when the direction of only one velocity is
reversed but the magnitude of the force is unchanged.
28.5:
The magnetic field is into the page at the origin, and the magnitude is
page.theintoT,101.64
m)0.400(
)sm108.0)(C101.5(
)m0.300(
)sm102.0)(C104.0(
4
4
6
2

56
2
56
0
22
0



























B
π
B
r
vq
r
qv
π
μ
BBB

28.6: a)
2
0
4
;
πd
qvμ
Bqq
q


into the page;
2
0
4
πd
vqμ

B
q



out of the page.
(i)
)2(42
2
0
dπ
qvμ
B
v
v


into the page.
(ii)
.0




Bvv
(iii)
2
0
4
2

d
qv
Bvv




out of the page.
b)
2
2
0
)2(4 dπ
vvqμ
q
q




BvF

and is attractive.
c)
25
0000
2
0
2
2

2
0
)sm103.00(
)2(4
,
)2(4





εμvvεμ
F
F
d
πε
q
F
d
π
vvqμ
F
C
B
CB
.101.00
6

28.7: a)
.

ˆ
)0.500(
ˆ
)0.866(
ˆ
)150(sin
ˆ
)150(cos
ˆ
sin
ˆ
cos
ˆ
jijijir  θθ
b)
kkjiirl
ˆ
)m105.00(
ˆ
)0.500()
ˆ
)0.500(
ˆ
)0.866(()
ˆ
(
ˆ
3
 dldld


c)
kk
rl
B
ˆ
)m1.20(
)m0.500)(m0.010)(A125(
4
ˆ
m)0.500(
4
ˆ
4
2
0
2
0
2
0
π
μ
r
dlI
π
μ
r
dI
π
μ
d 




.
ˆ
)T104.3(
8
kB

 d
28.8:
The magnetic field at the given points is:
2
0
2
0
2
0
6
2
0
2
0
6
2
0
2
0
6
2

0
2
0
sin
4
.0
)0(sin
4
sin
4
T.102.00
)m0.100(
)m0.000100(A)200(
4
sin
4
T.100.705
)m0.100(2
45sin)m0.000100(A)(200
4
sin
4
T.102.00
)m0.100(
)m0.000100(A)200(
4
sin
4
r
θdlI

π
μ
dB
r
dlI
π
μ
r
θdlI
π
μ
dB
π
μ
r
θdlI
π
μ
dB
π
μ
r
θdlI
π
μ
dB
π
μ
r
θdlI

π
μ
dB
e
d
c
b
a












T.10545.0
3
2
)m100.0(3
)m00100.0(A)200(
4
6
2
0




e
e
dB
dB


28.9:
The wire carries current in the z-direction. The magnetic field of a small piece of
wire
2
0
ˆ
4
r
dI
π
μ
d
rl
B



at different locations is therefore:
a)
jrlir
ˆ
ˆ

ˆ
ˆ
)m00.2( 


.
ˆ
1000.5
m)00.2(
90sinm)105()A00.4(
4
ˆ
sin
4
11
2
4
0
2
0
jjB T
π
μ
r
θdlI
π
μ
d






b)
.
ˆ
ˆ
ˆ
ˆ
)m00.2( irljr 


.
ˆ
T1000.5
ˆ
)m00.2(
)90(sinm)105()A00.4(
4
ˆ
sin
4
11
2
4
0
2
0
i
iiB








π
μ
r
θdlI
π
μ
d

c)
)
ˆˆ
(
2
1
ˆ
ˆˆ
)m00.2(
ˆ
)m00.2(
ijrljir 


)

ˆˆ
T(1077.1
)
ˆˆ
(
2
1
m)2.00(m)00.2(
m)10(5.0A)00.4(
4
)
ˆˆ
(
2
1sin
4
11
22
4
0
2
0
ij
ijijB








π
μ
r
θdlI
π
μ
d
d)
0
ˆ
ˆ
ˆ
)m00.2(  rlkr

28.10: a) At
,
3
4
3
8
223
1
2
1
2
:
2
000
πd

Iμ
dπ
Iμ
ddπ
Iμ
B
d
x

















in the
j
ˆ
direction.
b) The position

2
d
x

is symmetrical with that of part (a), so the magnetic
field there is
d
I
B


3
4
0

, in the
j
ˆ
direction.
28.11:
a) At the point exactly midway between the wires, the two magnetic fields are in
opposite directions and cancel.
b) At a distance a above the top wire, the magnetic fields are in the same
direction and add up:
kkkkkB
ˆ
3
2
ˆ
)3(2

ˆ
2
ˆ
2
ˆ
2
000
2
0
1
0
a
I
a
I
a
I
r
I
r
I












.
c) At the same distance as part (b), but below the lower wire, yields the same
magnitude magnetic field but in the opposite direction:
kB
ˆ
3
2
0
πa
Iμ

.
28.12:
The total magnetic field is the vector sum of the constant magnetic field and the
wire’s magnetic field. So:
a) At (0, 0, 1 m):
.
ˆ
)T100.1(
ˆ
)m00.1(2
)A00.8(
ˆ
)T1050.1(
ˆ
2
7
0

6
0
0
iiiiBB






r
I
b) At (1 m, 0, 0):




46.8atT,102.19
ˆ
T)10(1.6
ˆ
T)1050.1(
ˆ
)m00.1(2
)A00.8(
ˆ
)T1050.1(
ˆ
2
666

0
6
0
0
θ
π
μ
πr
Iμ
kiB
kikBB
from x to z.
c) At (0, 0, – 0.25 m):
iiiBB
ˆ
)m25.0(2
)A00.8(
ˆ
)T1050.1(
ˆ
2
0
6
0
0
π
μ
πr
I





.
ˆ
T)109.7(
6
i


28.13:
.
)(
2
4)(4)(4
2122
0
21222
0
2322
0
axx
a
π
Iμ
yxx
y
π
Ixμ
yx

xdy
π
Iμ
B
a
a
a
a









28.14: a)
A.110
T)10(5.50m)040.0(22
2
0
4
0
00
0





μ
π
μ
πrB
I
πr
Iμ
B
b)
T,1075.2
2
m)0.080(so,
2
4
00


B
rB
πr
Iμ
B

T.10375.1
4
m)160.0(
4
0



B
rB
28.15: a)
,T1090.2
m)(5.502
A)800(
2
5
00


π
μ
πr
Iμ
B
to the east.
b) Since the magnitude of the earth’s magnetic filed is
5
1000.5


T, to the north,
the total magnetic field is now
o
30
east of north with a magnitude of
5
1078.5



T. This
could be a problem!
28.16:
a) B = 0 since the fields are in opposite directions.
b)









baba
ba
rrπ
I
πr
Iμ
πr
Iμ
BBB
11
222
000


T6.67T1067.6

m0.2
1
m0.3
1
2
)A(4.0)ATm014(
6
7
















π
c)
Note that
aa
rB
and

bb
rB
θB
θBθBB
a
ba
cos2
coscos



tan
22
m)(0.05m)20.0(:04.14
20
5

a
rθθ

04.14cos
)m(0.05m)02.0(2
)A(4.0)ATm104(
2
cos
2
2
22
7
0










θ
r
I
B
a
,T53.7T1053.7
6



to the left.
28.17:
The only place where the magnetic fields of the two wires are in opposite
directions is between the wires, in the plane of the wires.
Consider a point a distance x from the wire carrying
2
I
= 75.0 A.
tot
B
will be

zero where
21
BB 
.
A0.75A,0.25;)m400.0(
2)m400.0(2
2112
2010



IIxIxI
πx
Iμ
xπ
Iμ
x
= 0.300 m;
0
tot

B
along a line 0.300 m from the wire carrying 75.0 A amd 0.100 m
from the wire carrying current 25.0 A.
b) Let the wire with
0.25
1
I
A be 0.400 m above the wire with
2

I
= 75.0 A.
The magnetic fields of the two wires are in opposite directions in the plane of the wires
and at points above both wires or below both wires. But to have
21
BB 
must be closer
to wire #1 since
1
I
<
2
I
, so can have
0
tot
B
only at points above both wires.
Consider a point a distance x from the wire carrying
0.25
1
I
A.
tot
B
will be
zero where
.
21
BB


m200.0);m400.0(
)m400.0(22
12
2010



xxIxI
x
π
Iμ
πx
Iμ
0
tot
B
along a line 0.200 m from the wire carrying 25.0 A and 0.600 m from the wire
carrying current
0.75
2
I
A.
28.18:
(a) and (b) B = 0 since the magnetic fields due to currents at opposite corners of
the square cancel.
(c)
left. the toT,100.4
45cos
m)210.0(2

A)(100)ATm104(
4
m20.10cm210cm)(10cm)(10
45cos
2
445cos4
45cos45cos45cos45cos
4
7
22
0

























π
B
r
πr
Iμ
B
BBBBB
a
dcba
28.19:

321
,, BBB




⊙
m200.0;
2
0
 r
r
I
B



for each wire
T1000.2,T1080.0T,1000.1
5
3
5
2
5
1

 BBB
Let ⊙ be the positive z-direction.
A0.20A,0.8A,0.10
321
 III
T100.2)(
0
T1000.2,T1080.0T,1000.1
6
3214
432z1
5
z3
5
z2
5
z1






zzzz
zzz
BBBB
BBBB
BBB
To give
4
B
in the

direction the current in wire 4 must be toward the bottom of the
page.
A0.2
)AmT10(2
T)100.2(m)200.0(
)2(
so
2
7
6
0
4
4
0
4







πμ
rB
I
πr
Iμ
B
28.20: On the top wire:
,
42
11
2
2
0
2
0
d
I
dd
I
L
F













upward.
On the middle wire, the magnetic fields cancel so the force is zero.
On the bottom wire:
,
42
11
2
2
0
2
0
d
I
dd
I
L
F













downward.
28.21:
We need the magnetic and gravitational forces to cancel:
g
I
h
h
LI
Lg
λ22
λ
2
0
2
0





28.22: a)
,N1000.6
)m400.0(2
)m20.1()A00.2()A00.5(

2
6
0210






r
LII
F
and the force is
repulsive since the currents are in opposite directions.
b) Doubling the currents makes the force increase by a factor of four to
.N1040.2
5
F
28.23:
.A33.8
)A60.0(
)m0250.0(2
)mN100.4(
2
2
0
5
10
2
210









I
r
L
F
I
r
II
L
F
b) The two wires repel so the currents are in opposite directions.
28.24:
There is no magnetic field at the center of the loop from the straight sections.
The magnetic field from the semicircle is just half that of a complete loop:
,
422
1
2
1
00
loop
R
I

R
I
BB









into the page.
28.25:
As in Exercise 28.24, there is no contribution from the straight wires, and now we
have two oppositely oriented contributions from the two semicircles:
,
22
1
)(
21
0
21
II
R
BBB










into the page. Note that if the two currents are equal, the magnetic field goes to zero at
the center of the loop.
28.26:
a) The field still points along the positive x-axis, and thus points into the loop
from this location.
b)
If the current is reversed, the magnetic field is reversed. At point P the field would
then point into the loop.
c) Point the thumb of your right hand in the direction of the magnetic moment, under
the given circumstances, the current would appear to flow in the direction that your
fingers curl (i.e., clockwise).
28.27: a)
A77.2
)800()AmT104(
)T0580.0()m024.0(22
so,2
7
0
0




πNμ
aB

IaNI
μB
x
x
b) At the center,
.2
0
aNIB
c


At a distance x from the center,
m0184.0so,m024.0with ,4)(
2
1
)(
means
)()(2)(2
6322
2322
3
2
1
2322
3
2322
3
0
2322
2

0

































xaaax
ax
a
BB
ax
a
B
ax
a
a
NI
ax
NIa
B
cx
cx

28.28: a) From Eq. (29-17),
.T1042.9
)m020.0(2
)A500.0()600(
2
3
00
center





a
NI
B
b) From Eq. (29-16),
.T1034.1
))m020.0()m080.0((2
)m020.0()A500.0()600(
)m08.0(
)(2
)(
4
2322
2
0
2322
2
0







B
ax
NIa
xB

28.29:
2
0
2322
2322
2
0
)()(2
)(2
)(
Iaμ
axxB
N
ax
NIa
μ
xB







69
)m06.0()A50.2(
)m06.0()m06.0()T1039.6(2
2
0
23224





μ
28.30:



.A305mT1083.3
encl
4
encl0
IIμdlB

b)
4
1083.3


since
l

d
points opposite to
B

everywhere.
28.31:
We will travel around the loops in the counterclockwise direction.

a)
.00
encl


lB

dI
b)



.mT1003.5)A0.4(A0.4
6
01encl
μdII lB

c)
)A0.2(A2.0A0.6A0.4
021encl
μdIII 

lB


.mT1051.2
6


d)

.mT1003.5)A0.4(A0.4
6
0321encl



μdIIII lB

Using Ampere’s Law in each case, the sign of the line integral was determined by
using the right-hand rule. This determines the sign of the integral for a counterclockwise
path.
28.32:
Consider a coaxial cable where the currents run in OPPOSITE directions.
a) For
.
2
2,
0
00
πr
Iμ
BIμπrBIμdIIbra
encl


lB

b) For
,
c

r

the enclosed current is zero, so the magnetic field is also
zero.
28.33:

Consider a coaxial cable where the currents run in the SAME direction.
a) For
.
2
2,
10
10101encl
πr
Iμ
BIμrBIμdIIbra 


lB

b) For
)(2)(,
21021021encl
IIμrBIIμdIIIcr 


lB


.

2
)(
210
πr
IIμ
B


28.34:
Using the formula for the magnetic field of a solenoid:
.T0402.0
)m150.0(
)A00.8()600(
00
0

μ
L
NI
μ
nIμB
28.35: a)
turns716
)A0.12(
)m400.0()T0270.0(
00
0


Iμ

BL
N
L
NI
μ
B

.mturns1790
m400.0
turns716

L
N
n
b) The length of wire required is
.m63)116()m0140.0(22




rN
28.36:
L
N
IB
0


N
BL

I
0



A8.41
)4000)(ATm104(
)m40.1()T150.0(
7





28.37: a)
A1072.3
)2(
so,
2
6
0
0

π
Br
I
πr
Iμ
B


b)
A10492
2
so
2
5
0
0
 .
N
μ
aB
I,
a
NI
μ
B
x
x
c)
A237so,)(
000
 NμBLIILNnIμB

28.38:
Outside a toroidal solenoid there is no magnetic field and inside it the magnetic
field is given by
.
2
0

πr
NIμ
B 
a) r = 0.12 m, which is outside the toroid, so B = 0.
b) r = 0.16 m
.T1066.2
)m160.0(2
)A50.8()250(
2
3
00






r
NI
B
c) r = 0.20 m, which is outside the toroid, so B = 0
28.39:
.T1011.1
)m070.0(2
)A650.0()600(
2
3
00







r
NI
B
28.40: a)
.T0267.0
)m060.0(2
)A25.0()400()80(
22
00







r
NIK
r
NI
B
m
b) The fraction due to atomic currents is
.T0263.0T)0267.0(
80
79

80
79


BB
28.41: a) If
 BK
m
1400

)500)(1400(
)T350.0()m0290.0(22
2
00
0
μ
π
NμK
B
πr
I
πr
NIμK
m
m
.A0725.0
b)
.A0195.0
5200
1400

5200If
part(a)m
 IIK
28.42: a)
.2021
)A400.2()500(
)T940.1()m2500.0(22
2
00
0

μ
π
NIμ
πrB
K
πr
NIμK
B
m
m
b)
.20201X 
mm
K
28.43:
a) The magnetic field from the solenoid alone is:
(i)
.T1013.1A)15.0()m6000(
3

0
1
000

 BμnIμB
(ii) But
m.A104.68T)1013.1(
5199
1
63
0
0
0




MB
K
M
m

(iii)
T.5.88T)1013.1)(5200(
3
0


BKB
m

b)
28.44:
 
.mA
s
mC
mCsN
mN
T
J
2
2

























28.45:
The material does obey Curie’s Law because we have a straight line for temperature
against one over the magnetic susceptibility. The Curie constant from the graph is
m.TAK1055.1
)13.5(
1
)slope(
1
5
00


C
28.46: The magnetic field of charge
q

at the location of charge q is into the page.
jki
rv
iBvF
ˆ
sin
4
)

ˆ
(
sin
4
ˆ
)(
ˆ
4
ˆ
)(
2
0
2
0
2
0






















r
θqq
π
μ
r
θvq
π
qv
r
q
π
μ
qvq




where

is the angle between
.
ˆ
and

r


v
jF
ˆ
5.0
4.0
)m500.0(
s)m1050.6)(sm1000.9)(C1000.5)(C1000.8(
4
2
4466
0



















μ
.
ˆ
)N1049.7(
8
jF


28.47:
m)045.0(
A)s)(2.50m1000.6)(C1060.1(
22
419
00










π
μ
πr
Iμ

qvqvBF

.N1007.1
19

Let the current run left to right, the electron moves in the opposite direction,
below the wire, then the magnetic field at the electron is into the page, and the electron
feels a force upward, toward the wire, by the right-hand rule (remember the electron is
negative).
28.48: (a)







πr
I
m
ev
m
qvB
m
F
a
2
sin
0


)m020.0)(2)(kg1011.9(
)A25)(ATm104)(sm000,250)(C106.1(
31
717
π
π
a






,sm101.1
213

away from the wire.
b) The electric force must balance the magnetic force.
eE = eVB
r
i
vvBE


2
0


m)(0.0202
A)Tm/A)(2510m/s)(4000,250(

7
π
π




,N/C5.62

away from the wire.
(c)
N10)m/skg)(9.81011.9(
29231 
mg

gravity.neglectcanweso,10
N10N/C)C)(62.5106.1(
grav
12
el
1719
el
FF
eEF



28.49:
Let the wire connected to the 25.0


resistor be #2 and the wire connected to
the 10.0

resistor be #1. Both
21
and II
are directed toward the right in the figure, so at
the location of the proton


12
and IisI
⊙
T1080.4andT1020.3T,1000.8
m.0.0250rwith,
2
and
2
5
21
5
2
5
1
20
2
10
1




BBBBB
πr
I
B
πr
Iμ
B

and in the direction ⊙.
Force is to the right.
N105.00T)10m/s)(4.8010C)(65010602.1(
185319 
 qvBF
28.50:
The fields add








2/322
2
0
121
)(2
22

xR
IR
BBBB

T1075.5
]m)125.0(m)20.0[(
m)A)(0.2050.1(Tm/A)104(
6
2/322
27






π


sinBqvF




90sinT)10m/s)(5.75C)(2400106.1(
619
N,102.21
21

perpendicular to the line

ab
and to the velocity.
28.51:
a)
Along the dashed line,
2
1
and BB
are in opposite directions.
If the line has slope
21
then00.1 rr 
and
.0so,
tot21
 BBB
28.52: a)
001
ˆˆˆ
4
ˆ
4
000
2
0
2
00
zyx
vvv
r

q
π
μ
r
q
π
μ
kji
rv
B





m/s.521
)C1020.7(
m)T)(0.251000.6(4
T1000.6
||
4
and00
4
ˆ
T)1000.6(
ˆˆ
4
3
0
26

0
6
0
2
0
00
2
0
6
00
2
0












π
v
v
r
q
π

vv
r
q
π
μ
vv
r
q
π
μ
z
zyy
yz
jkj
m/s.607m/s)521()m/s800(And
22
2
0
2
0
2
00

zyx
vvvv
b)
 
ik
kji
rv

B
ˆˆ
4
010
ˆˆˆ
4
ˆ
4
)0,m250.0,0(
00
2
0
000
2
0
2
0
0
zxzyx
vv
r
q
π
μ
vvv
r
q
πr
q
π

μ




.T109.2m/s800
m)250.0(
)C1020.7(
4
||
4
0)m,250.0,0(
6
2
3
0
0
2
0





π
μ
v
r
q
π

μ
B
28.53:
Choose a cube of edge length
L
, with one face on the y-z plane. Then:
,0
ˆ
0
0
3
000

  

B
a
LB
dA
a
LB
d
a
xB
dd
LxLxLx
AiABAB
so the only possible field is a zero field.
28.54:
a)

b)
i
r
I
ˆ
2
2
20
2









B
)
ˆ
cos
ˆ
(sin
2
3
30
3
jiB












πr
Iμ
And so


























jiB
ˆ
cos
ˆ
sin
2
3
3
3
3
2
20
θ
r
I
θ
r
I
r
I
π
μ

 
jiB
jiB
ˆ
)16(
ˆ
)3.3312(
2
ˆ
)8.0(
m)050.0(
ˆ
)6.0(
m)050.0(m)030.0(2
323
0
3320
III
π
μ
III
π
μ




































.ji
ji

ˆ
T1028.1
ˆ
T1072.3
ˆ
A)(16)(4.00
ˆ
A))0(33.3)(2.0A)00.4)(12(
2
56
0



π
μ
c)
ijBlF
ˆ
ˆ
111 yx
lBIlBII 









ckwisecounterclo2.16,N1033.1;
ˆ
T1028.1
ˆ
T1072.3
]
ˆ
T)1028.1(
ˆ
T)1072.3[(m010.0A00.1
778
56




Fij
ij
from +x-axis.
28.55: a) If the magnetic field at point P is zero, then from Figure (28.46) the current
2
I
must be out of the page, in order to cancel the field from
1
I
. Also:

 
.A00.2
m)50.1(

m)500.0(
A00.6
22
1
2
12
2
20
1
10
21

r
r
II
πr
Iμ
πr
Iμ
BB
b) Given the currents, the field at Q points to the right and has magnitude

.T1013.2
m50.1
A00.2
m500.0
A00.6
22
6
0

2
2
1
10




















π
μ
r
I
r
I

π
μ
B
Q
c) The magnitude of the field at S is given by the sum of the squares of the two
fields because they are at right angles. So:
.T101.2
m80.0
A00.2
m60.0
A00.6
22
6
2
0
2
2
2
2
1
10
2
2
2
1








































π
μ
r
I
r
I
π
μ
BBB
S
28.56:
a)
b) At a position on the x-axis:

 
,
ax
π
Iaμ
B
ax
a
ax
π
Iμ
πr
Iμ

B
net
22
0
2222
00
net
sin
2
2





in the positive x-direction , as shown at left.
c)
d) The magnetic field is a maximum at the origin, x = 0.
e) When
.,
2
0
πx
Iaμ
Bax 
28.57:
a)
b) At a position on the x-axis:

 

,
cos
2
2
22
0
net
2222
00
net
axπ
Ixμ
B
ax
x
ax
π
Iμ
θ
πr
Iμ
B




in the negative y-direction, as shown at left.
c)
d) The magnetic field is a maximum when:

 
 
axxax
ax
Cx
ax
C
dx
dB





222
2
22
2
22
2
2
0
e) When
,,
π
0
x
I
Bax



which is just like a wire carrying current 2I.
28.58:
a) Wire carrying current into the page, so it feels a force downward from the
other wires, as shown at right.

 
   
   
 
m.N1011.1
40006000
4000006
5
22
2
0
22
0















m.m.π
m.A.μ
L
F
axπ
Iaμ
IIB
L
F

b) If the wire carries current out of the page then the forces felt will be the
opposite of part (a) . Thus the force will be
,mN1011.1
5

upward.
28.59: The current in the wires is




.A0.90500.0V0.45R






I
The currents
in the wires are in opposite directions, so the wires repel. The force each wire exerts on
the other is







 
N378.0
m0150.0
m50.3A0.90AN102
2
2
27
0






πr
LIIμ
F

To hold the wires at rest, each spring exerts a force of 0.189 N on each wire.





mN8.37m0050.0N189.0so  xFkkxF
28.60:
a) Note that the Earth’s magnetic field exerts no force on wire B, since the
current in wire B is parallel to the Earth’s magnetic field. Thus, for equilibrium, the
remaining two forces that act on wire B must cancel. Assuming that the length of wire B
is
L
and that wire A carries a current
I
we obtain
0
m)(0.1002
)AA)(3.0(1.0
05002
01
00

π
Lμ
m).π(
A)L.I(μ
So
A1.5
m0.100
m0.050

A)0.3(
I
b)
Note that the force on wire B that is generated by wire C is to the right. Thus, if the
current in wire C is increased, wire B will slide to the right.
28.61:
The wires are in equilibrium, so:
mgθTyθTFx


cos:andsin:

lB
mg
mgTIlBF


tan
Itan sin
.m108.36)]sin(6.00m)0400.0(2[And
tan2tan2
2
But
3
00
0



r

l
μ
θ
mgπr
I
Il
μ
θ
mgπr
I
r
I
B


A.2.23
)tan(6.00)sm(9.80m)kg(0.0125m)1036.8(2
0
23




μ
π
I
28.62:
The forces on the top and bottom segments cancel, leaving the left and right sides:
.
ˆ

)N10(7.97
ˆ
m0.026
1
m0.100
1
2
A)m)(14.0A)(0.200(5.00
ˆ
11
2
ˆ
22
ˆ
)(
ˆ
)(
5
0
wire0wire0wire0
iiF
iiiiFFF
































π
μ
rrπ
IlIμ
πr
Iμ
πr

Iμ
IlIlBIlB
lrrl
rl
rl
28.63: a)
θμB
x
IaN
μ
ax
IaN
μ
Bax sin||||and
2)(2
3
2
0
2/322
2
0



Bμτ
3
22
0
3
2

0
2
sin
sin
2
)(
x
aaI
πINN
θ
x
IaN
μ
AINτ














.
2

cos
cos
2
)(cos)b
3
22
0
3
2
0
2
x
aaII
μNN
x
IaN
μ
aINθμBU















Bμ
c) Having
a
x

allows us to simplify the form of the magnetic field, whereas
assuming
ax


means we can assume that the magnetic field from the first loop is
constant over the second loop.
28.64:
page.theofout,1
4
11
22
1
00






















b
a
a
I
ba
I
μ
BBB
ba

28.65: a) Recall for a single loop:
.
2/322
2
0
)(2 ax
aI
B




Here we have two loops, each of
N
turns, and measuring the field along the x-axis from between them means that the
""x
in the formula is different for each case:
Left coil:
.
))2((22
2322
2
0
aax
NIaμ
B
a
xx
l


Right coil:
.
))2((22
2322
2
0
aax
NIa

μ
B
a
xx
r


So the total field at a point x from the point between them is:
.
))2((
1
))2((
1
2
23222322
2
0












aaxaax

NIa
μ
B
b) Below left: Total magnetic field. Below right: Magnetic field from right coil.
c) At point












23222322
2
0
))2((
1
))2((
1
2
0,
aaaa
NIa
μ
BxP