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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
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
[...]... (16)(4.00 A) ˆ j 2π ˆ 3.72 10 6 Ti 1 .28 105 Tˆ j ˆ j c) F I l B I lB ˆ I lB i 1 1 x 1 y ˆ 1.00 A 0.010 m [(3.72 106 T) ˆ (1 .28 10 5 T) i ] j ˆ 3.72 108 Tˆ 1 .28 10 7 Ti ; F 1.33 10 7 N, 16.2 counterclockwise j from +x-axis 28. 55: a) If the magnetic field at point P is zero, then from Figure (28. 46) the current I 2 must be out of the page,... currents is B 80 B 80 (0.0267 T) 0.0263 T 28. 40: a) B 28. 41: a) If K m 1400 B 2πr B 2π (0.0290 m) (0.350 T) K m μ0 NI I 2πr K m μ0 N μ0 (1400)(500) 0.0725 A b) If K m 5200 I 1400 I part(a) 0.0195 A 5200 2πrB 2π (0.2500 m) (1.940 T) K m μ0 NI Km 2021 2πr μ0 NI μ0 (500) ( 2.400 A) b) X m K m 1 2020 28. 42: a) B 28. 43: a) The magnetic field from the solenoid... center, Bc 0 NI 2a At a distance x from the center, 28. 27: a) Bx μ0 NI 2a , so I Bx a3 0 NIa 2 NI 0 2 2 2 32 ( x a 2 )3 2 Bc 2( x a ) 2a Bx 1 Bc means 2 a3 2 ( x a 2 )3 2 a3 1 2 2 32 (x a ) 2 ( x 2 a 2 )3 4a 6 , with a 0.024 m, so x 0.0184 m NI 0 (600) (0.500 A) 28. 28: a) From Eq (29-17), Bcenter 0 9.42 10 3 T... 4 T 2 2 32 2 2 32 2( x a ) 2((0.080 m) (0.020 m) ) 28. 29: B( x) μ0 NIa 2 2 B( x) ( x 2 a 2 )3 2 N 2( x 2 a 2 )3 2 μ0 Ia 2 2(6.39 104 T) (0.06 m) 2 (0.06 m) 2 μ0 (2.50 A) (0.06 m) 2 28. 30: B dl μ I 0 encl 32 69 3.83 10 4 T m I encl 305 A b) 3.83 10 4 since dl points opposite to B everywhere 28. 31: We will travel around the loops in the counterclockwise... ( I1 I 2 ) B μ0 ( I 1 I 2 ) 2πr 28. 34: Using the formula for the magnetic field of a solenoid: μ NI μ0 (600) (8.00 A) B μ0 nI 0 0.0402 T (0.150 m) L μ0 NI BL (0.0270 T) (0.400 m) N 716 turns L μ0 I 0 (12.0 A) N 716 turns n 1790 turns m 0.400 m L b) The length of wire required is 2 rN 2 ( 0 0140 m ) (116 ) 63 m 28. 35: a) B 28. 36: B 0 I N L BL 0 N (0.150 T) (1.40... ( N L) I , so I BL μ0 N 237 A a) B 28. 37: 28. 38: Outside a toroidal solenoid there is no magnetic field and inside it the magnetic μ NI field is given by B 0 2πr a) r = 0.12 m, which is outside the toroid, so B = 0 NI 0 (250) (8.50 A) b) r = 0.16 m B 0 2.66 10 3 T 2r 2 (0.160 m) c) r = 0.20 m, which is outside the toroid, so B = 0 28. 39: B 0 NI 0 (600) (0.650 A) ... opposite directions b) Doubling the currents makes the force increase by a factor of four to F 2.40 10 5 N 28. 22: a) F 2 (0.0250 m) F 0 I1 I 2 F 2r I2 (4.0 10 5 N m) 8.33 A L 2r L 0 I1 0 (0.60 A) b) The two wires repel so the currents are in opposite directions 28. 23: 28. 24: There is no magnetic field at the center of the loop from the straight sections The magnetic field from... (5200)(1.13 10 3 T) 5.88 T b) 2 J N m C m 2 28. 44: Am T N s C m s 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 1 1 C 1.55 10 5 K A T m 0 (slope) 0 (5.13) 28. 46: The magnetic field of charge q at the location of... 0 0 , 2 2 2R 4R 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: 1 B ( B1 B2 ) 0 I 1 I 2 , 2 2R 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... I encl I B dl B 2r 0 I encl 0 I B 0 2r 28. 69: a) I JdA αrrdrdθ α 2π R r 2 dr A r2 r2 28. 70: a )r a I encl I r I 2 B dl B 2πr 0 I encl μ0 I 2 a a A a μ Ir B 0 2 2πa μI When r a, B 0 which is just what was found from Exercise 28. 32, part (a) 2πa A r 2 b2 b) b r c I encl I .
ckwisecounterclo2.16,N1033.1;
ˆ
T1 028. 1
ˆ
T1072.3
]
ˆ
T)1 028. 1(
ˆ
T)1072.3[(m010.0A00.1
778
56
Fij
ij
from +x-axis.
28. 55: a) If the magnetic.
.
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