Introductory Physics Volume Two

104 Sources of Magnetic Field 5.2
Example
In this example we will see how to deal with a current density that is
not uniform. Suppose that we have a long wire with a current density
that is greater at the center and drops off to zero at the edge: J(r) =
3I
πR
(1 − r/R). We want to find the magnetic field and from Ampere’s
2
law we find B = µ0I through
2πr
. The difficulty is that since J is not uniform
we cannot use I = JA to find I through . We need to use dI = JdA to
find I through = ∫ JdA. This example will show how to do that.
We need to pick the area elements
so that the current density is uniform
over the entire surface of each element.
Since the current density is a function
of r only, J is constant in regions that
are circles around the center. So we will
use a circular element dA, as pictured in
gray in the figure. The figure shows a
cross section of the wire.
R
dA
r+dr
r
In the limit that dr is very small the area will be simply the length
around the circular area element, 2πr times the width of the element
dr so that dA = 2πr dr. Now we can compute the current inside a loop
of radius r loop by integrating.
∫ rloop
∫ rloop
3I
I through = J dA =
(1 − r/R)2πr dr
πR2 0
= I r2
R 2 (3 − 2 r R
−→ B = µ 0I through
2πr
)
0
= µ 0Ir
2πR 2 (3 − 2 r R
)
⊲ Problem 5.5
Consider a long wire where the current density is not uniform but
instead increases as you approach the center of the wire, so that at a