Chapter 29: Magnetic Fields Due to Currents
Problem 3
A surveyor is using a magnetic compass 6.1 m below a power line in which there is a steady
current of 100 A. (a) What is the magnetic field at the site due to the power line? (b) Will
this interfere seriously with the compass reading? The horizontal component of Earth's
magnetic field at the site is 20
μ
T.
Answer
(a) 3.28
μ
T. (b) Yes.
Solution
(a)
The magnetic field at a distance r from the center of a long straight wire carrying
current i is
B
=
μ
o
i
2
π
r
Therefore, 20 ft from the power line the magnetic field due to the power line will be
B
=
4
π
×
10
−
7
Tm/A *100 A
2
π
*6.1
m
=
3.28
μ
T
.
(b)
This is a significant field compared to the field of Earth. If the two fields were
perpendicular to each other then the resultant field would make an angle of arctan(3.28/20)
= 9.3
°
with the earth's field  meaning that the compass would be off by 9.3
°
.
Problem 7
Two long straight wires are parallel and 8.0 cm apart. They are to carry equal currents such
that the magnetic field halfway between them has magnitude 300
μ
T. (a) Should the
currents be in the same or opposite directions? (b) How much current is needed?
Answer
(a) opposite; (b) 30 A.
Solution
(a)
Referring to Diagram 297, if the currents are in the same direction then the
magnetic field contributions at a point halfway between the two wires will be opposite
directions and add up to zero since the two contributions will have the same size.
(b)
If the two currents are in opposite directions then the total magnetic field at the
midpoint will be twice the field due to one wire.
B
=
2
μ
o
i
2
π
r
⎛
⎝
⎜
⎞
⎠
⎟
=
μ
o
i
π
r
Therefore the current in each wire must be
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292
Chapter 29: Magnetic Fields Due to Currents
i
=
π
rB
μ
o
=
π
4.00 cm
(
)
∗
300
μ
T
(
)
4
π
×
10
7
T
−
m
A
=
30 A
Problem 11
In Fig. 2942, a current I = 10 A is set up in a long
hairpin conductor formed by bending a piece of wire
into a semicircle of radius R = 5.0 mm. Point b is
midway between the straight sections and so distant
from the semicircle that each straight section cam be
approximated as being an infinite wire. What are the
(a) magnitude and (b) direction (into or out of the
page) of
B
at a and the (c) magnitude and (d)
direction of
B
at b?
i
Figure 2942
Answer
(a) 1.03 mT, (b) out of the page. (c) 0.80 mT, (d) out of the page.
Solution
(a)
The
B
field at a is due to two "half infinite" long wires and the semicircle. The field
due to each of these pieces is directed out of the figure so the magnitude of the field is just
the sum of the magnitudes of the three contributions
B = B
∞
/2
+ B
∞
/2
+ B
semicircle
B
=
μ
o
i
4
π
R
+
μ
o
i
4
π
R
+
μ
o
i
4R
=
μ
o
i
4R
2
π
+
1
⎛
⎝
⎞
⎠
B
=
4
π ×
10
−
7
T  m/A
(
)
*
10 A
(
)
4 *
5.0 mm
(
)
2
π
+
1
⎛
⎝
⎞
⎠
B = 1.03 mT.
(b)
The right hand rule reveals that the magnetic field is directed out of the page.
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 Spring '09
 C
 Physics, Current, Power, Magnetic Field, Ampere

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