CHAPTER 30—SOURCES OF THE MAGNETIC FIELD
ActivPhysics
can help with these problems: Activities 13.1–13.3, 13.5
Section 30

1:—The BiotSavart Law
Problem
1.
A wire carries 15 A. You form the wire into a singleturn circular loop with magnetic field
80
T
μ
at the loop center.
What is the loop radius?
Solution
Equation 303, with
x
=
0, gives the magnetic field at the center of a circular loop,
B
I
a
=
μ
0
2
=
(with direction along
the axis of the loop, consistent with the sense of circulation of the current and the righthand rule). Thus, the radius is
a
I
B
=
=
×
=

μ
π
μ
0
7
2
2
2
10
15
80
118
=
=
(
/
)(
) (
)
.
.
N A
A
T
cm
Problem
2.
A singleturn wire loop is 2.0 cm in diameter and carries a 650mA current. Find the magnetic field strength (a) at the
loop center and (b) on the loop axis, 20 cm from the center.
Solution
Equation 303 gives: (a) at the center,
x
B
I
a
=
=
=
×
×
=

0
2
4
10
650
2
1
40 8
0
7
2
,
(
/
)(
) (
)
.
μ
π
μ
=
=
N A
mA
cm
T; (b) on the
axis,
x
B
Ia
x
a
I
=
=
+
=
×
+
=


20
1
20
1
5 09
1
2
0
2
2
2
3 2
1
2
0
2
2
2
3 2
cm,
cm
cm
cm
nT
μ
μ
(
)
(
)
[(
)
(
) ]
.
.
=
=
Problem
3.
A 2.2mlong wire carrying 3.5 A is wound into a tight, loopshaped coil 5.0 cm in diameter. What is the magnetic
field at its center?
Solution
The magnetic field at the center of each of the tightlywound turns is the same (Equation 303 with
x
=
0
), so the net field
is
B
N
I
a
=
(
),
μ
0
2
=
where
N
is the number of turns. If the wire has length
L
and the coil has diameter
D
, then
N
L
D
=
=
π
and
B
L
D
I D
IL
D
=
=
=
×
=

(
)(
)
(
)( .
)( .
) (
)
.
=
=
=
=
π
μ
μ
π
0
0
2
7
2
2
4
10
3 5
2 2
5
123
N/A
A
m
cm
mT.
Problem
4.
What is the current in a long wire if the magnetic field strength 1.2 cm from the wire’s axis is
67
T?
μ
Solution
Equation 305 (or Equation 308) gives the magnetic field strength near a long straight wire,
B
I
r
=
μ
π
0
2
=
, from which we
find
I
rB
=
=
×
=

2
12
67
2
10
4 02
0
7
2
π
μ
μ
=
=
( .
)(
) (
/
)
.
cm
T
N A
A.
Problem
5.
Suppose Earth’s magnetic field arose from a single loop of current at the outer edge of the planet’s liquid core
(core radius 3000 km), concentric with Earth’s center. What current would be necessary to give the observed field
strength of
62
T
μ
at the north pole? (The currents responsible for Earth’s field are more complicated than this
problem suggests.)
Solution
The north pole is on the axis of the hypothetical circular current loop, with radius
a
=
3 0
.
Mm, at a distance
x
R
E
=
=
6 37
.
Mm
from the earth’s center. Equation 303 gives a current of
I
B x
a
a
=
+
=
2
2
2
3 2
0
2
(
)
=
=
μ
2 62
6 37
3 0
4
10
3 0
3 83
2
2
3 2
7
2
(
)[( .
)
( .
)
]
(
/
( .
)
.
T
Mm
Mm
T
m A)
Mm
GA.
μ
π
+
×
⋅
=

=
=
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192
CHAPTER 30
Problem
6.
Earth’s magnetic dipole moment is
8 0
10
22
2
.
.
×
⋅
A
m
What is the magnetic field strength on Earth’s surface at either
pole?
Solution
The magnetic field strength at points on the axis of a magnetic dipole is given by Equation 304. The magnetic poles are
two such points on the Earth’s surface; see Fig. 306. Thus,
B
R
E
=
=
×
F
H
I
K
×
⋅
×
=
×
=


μ
π
μ
0
3
7
2
22
2
6
3
5
2
2
10
8
10
6 37
10
6 20
10
0 62
N
A
A
m
m
T
G
(
)
( .
)
.
.
.
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 Spring '07
 L.ROSE
 Physics, Magnetic Field, Ampere, magnetic field strength

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