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
rB
and
bb
rB
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
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