Since we just argued that
B
is a constant at constant radius, we can pull it out of the integral all
together.

B

I
ds
=
μ
0
i
⇒

B

=
μ
0
i
2
πr
which is exactly the equation we derived from the BiotSavart law.
10.4
Problems
Problem 29.24
A current is set up in a wire loop consisting of a semicircle of radius 4
cm
, a smaller concentric
semicircle, and two radial straight lengths, all in the same plane. Figure 30 shows the arrange
ment but is not drawn to scale. The magnitude of the magnetic field produced at the center
of curvature is 47
.
25
μT
. The smaller semicircle is then flipped over (rotated) until the loop is
again entirely in the same plane. The magnetic field produced at the (same) center of curvature
how has magnitude 15
.
75
μT
, and its direction is reversed. What is the radius of the smaller
semicircle?
Figure 30: Figure for problem 29.24
We first need to recognize that the straight portions of wire do not contribute to the magnetic
field. We can see this from the BiotSavart law because
d~s
and
~
r
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 Spring '08
 Any
 Physics, Addition, Current, Magnetic Field, Elementary algebra, Radius of curvature

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