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
=
0,
B
=
μ
0
I
=
2
a
=
(4
!
"
10
#
7
N/A
2
)(650 mA)
=
(2
"
1 cm)
=
40.8
T;
(b) on the axis,
x
=
20 cm,
B
=
1
2
0
Ia
2
(
x
2
+
a
2
)
!
3
=
2
=
1
2
0
I
(1 cm)
2
"
[(20 cm)
2
+
(1 cm)
2
]
!
3
=
2
=
5.09 nT.
Problem
7. A singleturn current loop carrying 25 A produces a magnetic field of 3.5 nT at a point on its axis 50
cm from the loop center. What is the loop area, assuming the loop diameter is much less than 50 cm?
Solution
If the radius of the loop is assumed to be much smaller than the distance to the field point
(
a
ø
x
=
50 cm),
then
Equation 304 for the field on the axis of a magnetic dipole can be used to find
=
2
x
3
B
=
0
.
The
magnetic moment of a singleturn loop is
=
IA
,
therefore
A
=
=
I
=
2
x
3
B
=
0
I
=
(3.5 nT)(50 cm)
3
=
(2
"
10
#
7
N/A
2
)(25 A)
=
0.875 cm
2
.
Problem
8. Two identical current loops are 10 cm in diameter and carry 20A currents. They are placed 1.0 cm
apart, as shown in Fig. 3047. Find the magnetic field strength at the center of either loop when their
currents are in (a) the same and
(b) opposite directions.
FIGURE 3047 Problem 8.
Solution
The magnetic field strength at the center of either loop is the magnitude of the vector sum of the fields due
to its own current and the current in the other loop. Equation 303 gives
B
self
=
0
I
=
2
a
and
B
other
=
0
2
=
2(
a
2
+
x
2
)
3
=
2
.
When the currents are in the same directions, the fields are parallel and
B
=
B
self
+
B
other
while if the currents are in opposite directions,
B
=
B
self
!
B
other
.
Numerically,
B
self
=
"
10
#
7
)(20)T
=
(0.1)
=
251
T,
and
B
other
=
B
self
(1
+
x
2
=
a
2
)
!
3
=
2
=
(251
T)[1
+
(1 cm
=
5 cm)
2
]
!
3
=
2
=
237
so (a)
B
self
+
B
other
=
488
T,
and (b)
B
self
!
B
other
=
14.4
T.
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View Full DocumentProblem
10. A single piece of wire is bent so that it includes a circular loop of radius
a
, as shown in Fig. 3048. A
current
I
flows in the direction shown. Find an expression for the magnetic field at the center of the
loop.
FIGURE 3048 Problem 10.
Solution
The field at the center is the superposition of fields due to current in the circular loop and straight sections
of wire. The former is
μ
0
I
=
2
a
out of the page (Equation 303 at
x
=
0
for CCW circulation), and the latter
is
0
I
=
2
!
a
out of the page (Equation 305 at
y
=
a
for the very long, straight sections). Their sum is
B
=
(1
+
)
0
I
=
2
a
out of the page.
Problem
12. Four long, parallel wires are located at the corners of a square 15 cm on a side. Each carries a current
of 2.5 A, with the top two currents into the page in Fig. 3049 and the bottom two out of the page. Find
the magnetic field at the center of the square.
FIGURE 3049 Problem 12 Solution.
Solution
The magnetic field from each wire has magnitude
0
I
=
2
(
a
=
2
)
(from Equation 305, with
y
=
a
=
2,
the
distance
from a corner of a square of side
a
to the center). The righthand rule gives the field direction along one of
the diagonals, as shown superposed on Fig. 3049, such that the fields from currents at opposite corners are
parallel. The net field is
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 Winter '10
 Hirsch
 Magnetic Field, th e cen, magnetic f ield, th e loop

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