Chapter 28
p.1
CHAPTER 28 – Sources of Magnetic Field
1.
The magnetic field of a long wire depends on the distance from the wire:
B
= (
µ
0
/4p)2
I
/
r
= (10
–7
T
·
m/A)2(65 A)/(0.075 m) =
1.7
×
10
–4
T
.
When we compare this to the Earth’s field, we get
B
/
B
Earth
= (1.7
×
10
–4
T)/(5.5
×
10
–5
T) =
3.1
×
.
2.
We find the current from
B
= (
µ
0
/4p)2
I
/
r
;
5.5
×
10
–5
T = (10
–7
T
·
m/A)2
I
/(0.25 m), which gives
I
=
69 A
.
3.
The two currents in the same direction will be attracted with a force of
F
=
I
1
(
µ
0
I
2
/2p
d
)
L
=
µ
0
I
1
I
2
L
/2p
d
= (4p
×
10
–7
T
·
m/A)(35 A)(35 A)(45 m)/2p(0.060 m) =
0.18 N attraction
.
4.
Because the force is attractive, the second current must be in the same direction as the first.
We find the
magnitude from
F
/
L
=
µ
0
I
1
I
2
/2p
d
8.8
×
10
–4
N/m = (4p
×
10
–7
T
·
m/A)(22 A)
I
2
/2p(0.070 m), which gives
I
2
=
14 A upward
.
5.
The magnetic field will be tangent to the circle with the current at the center.
C
E
B
C
D
I
B
E
B
D
6.
The magnetic field produced by the wire must be less than 1.0% of the magnetic field of the Earth.
We
find the current from
B
= (
µ
0
/4p)2
I
/
r
< 0.01
B
Earth
;
(10
–7
T
·
m/A)2
I
/(1.00 m) < 0.010(5.5
×
10
–5
T), which gives
I
<
2.8 A
.

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