Introductory Physics Volume Two

100 Sources of Magnetic Field 5.1
Now we can write out the Biot-Savart integral in terms of the
parameterization.
⃗B(r f ) = µ ∫
0I ⃗dl × ˆr
4π r 2
= µ ∫
0I ⃗dl × ⃗r
4π r 3
= µ ∫ drs ⃗
0I
dt
× [⃗r f − ⃗r s (t)]
4π |⃗r f − ⃗r s (t)| 3 dt
This gives us a system by which we can find the field at any point due
to any current that we can parameterize. Of course it is very rare that
the integral has an analytic solution, but one can use a computer to
evaluate the integral numerically once you have used this system to
write out the integral.
In order to see exactly what all this means let us do an example.
Example
Suppose that we have a current of 15 amps going through a section of
wire that follows the curve below, which has the following parameterization.
⃗r s (t) = (−t + t 3 )î + t 2 ĵ
Where the distance is in meters, and t goes from -0.9 to 0.9.
We want to find the field at the point ⃗r f = 0î + 1ĵ. In preparation for
plugging into the Biot-Savart law we compute the following quantities.
⃗r f − ⃗r s (t) = (t − t 3 )î + (1 − t 2 )ĵ = (1 − t 2 )(tî + 1ĵ)
|⃗r f − ⃗r s (t)| 2 = (1 − t 2 ) 2 (t 2 + 1)
|⃗r f − ⃗r s (t)| 3 = (1 − t 2 ) 3 (t 2 + 1) 3/2
and
⃗ dr s
dt = (−1 + 3t2 )î + 2tĵ