W9_parallel_resonance.eps

Week 7: Sources of the Magnetic Field 243

On your homework, I’ll ask you to do them again (without looking, before you are done) and will
throw you a few simple enough variants of the problems.

7.6: Applications of Ampere’s Law

There are basically four problem geometries where Ampere’s Law canbe used to find the field. As
was the case with Gauss’s Law for Electricity, each of them has an associatedsymmetrythat permits
the path integral on the left to be evaluated “once and for all”, so that solving the problem amounts
ot finding the total current through the Amperian Loop (topical equivalent of the Gaussian Surface)
in question.

Those categories are:

a) Infinitely long straight wire, or cylinder, or cylindrical shell, or anything else where the current

has cylindrical symmetry. These examples will be like the argument weused to justify Ampere’s

Law above, only backwards.

b) Infinitely long solenoid.

c) Toroidal solenoid (which also has cylindrical symmetry, but in a different way).

d) Infinite plane sheet of current (which may or may not be a “thick”sheet).

Example 7.6.1: Cylindrical Current Density – Infinitely Long Thin Wire

The simplest example of this is the infinitely long straightthinwire, so we’ll do that just as a
warm-up.

Take an infinitely long, straight wire carrying a currentI. We knowfrom symmetry(not just
because we used the fact to sort-of-derive Ampere’s Law in the first place) that the magnetic field
is constant in magnitude on and tangent to a circle of radiusrbecause the problem doesn’t change
as we walk around the wire. We therefore choose the Amperian PathCto be acircle of radiusr
and do so once and for all for this kind (symmetry) of problem. Then:
∮

C

B~·d~ℓ = μ 0 Ithru C

Bt

∮

C

dℓ = μ 0 I

Bt 2 πr = μ 0 I

Bt =

μ 0 I

2 πr

(537)

Big surprise. We find that the field tangent to the circle at all pointsBtis exactly what we know
it to be as we more or less invert our “derivation” of Ampere’s Law. The oneimportantlesson to
take from this is that the left hand side and concluding algebra for this little mini-derivation will
never change! Foreverycylindrical problem we willalwaysuse a circular Amperian Path and the
left hand integral will (becauseBis constant on and tangent to the circle)alwaysevaluate toBt 2 πr.

The right hand side, on the other hand, we may have to work for. Specifically, we will often have
to work to find the actual current through theparticularCwe have drawn.

Still, this seems a hell of a lot easier than setting up and evaluating theBiot-Savart integral we
did earlier this week. Maybe thereissomething useful in here, after all!

This is more apparent in the next example.