W9_parallel_resonance.eps

Week 10: Maxwell’s Equations and Light 347

S

S

C

I

+Q −Q

I

1

Bφ 2

no current through S 2

Figure 138: A simple circuit and pair of surfaces that illustrate how Ampere’s Law is (so far)wrong,
with two completely different currents for the two surfacesS 1 andS 2.

Fortunately, we’ve learned enough at this point to be able to see that Ampere’s Law is

obviously wrong!Consider the following specific example. In figure 138 I’ve drawn a side

view of a humble parallel plate capacitor. At this particular instant, acurrentI(t) is

flowing along the wire on the left, charging up the capacitor so that acharge +Q(t) is

increasing on the left plate.

To this innocuous looking problem we’ll applyAmpere’s Law– specifically to the nice

circular loopCdrawn around the supply wire. This loop is quite far away from the

capacitor, and the electric field the capacitor is making is more or lessconfined to live

between its plates, and the currentIquite obviously goesthroughthe surfaceS 1 stretched

acrossC(and hence goes “throughC”), so we should be quite justified in deducing the

usual: ∮

C

B~·d~ℓ=Bφ 2 πr=μ 0

∫

S 1 /C

J~·ˆndA=μ 0 I (815)

(where recall thatS/Cshould be read as “the open surfaceSbounded by the closed

curveC”) so that

Bφ=

μ 0 I

2 πr (816)

around the circle in the right handed sense. No problem, the field of an infinitely long

straight wire carrying current, the simplest possible situation. Howcould this be wrong?

But wait. When I wrote the right-hand side of Ampere’s Law, Ihappenedto choose the

“easy” surfaceS 1 that stretches straight across the curveC(and an easy curveCthat

lies in a plane). However, there is nothing in themathematicsof Ampere’s Law that

requires me to use that particular surface.

I could choose to use surfaceS 2 /Cinstead.S 2 is just as “bounded by the closed curve

C” asS 1 is. They are topologically equivalent –S 1 is like the film of soap stretched

across a bubble blowing loop, andS 2 is like the bubble as it has been blown out but is

still attached to the loop. The only problem with this is that the current:

I=

∫

S 2 /C

J~·ˆndA= 0(!) (817)

because the surfaceS 2 goes inbetweenthe plates of the capacitor, whereno charge flows!

This is a disaster! Ampere’s Law seems to give us two possible answers. In fact, since

there are an infinite number of surfacesSI could draw bounded byCthat intercept

different parts of the capacitor and wire supplying it, there are an infinite number of

possible answers! But the two answersBφ=μ 0 I/ 2 πr 6 = 0 andBφ= 0 are more than

enough for us to see that we have a serious problem to deal with. Thecurrent on the

right hand side of Ampere’s Law(correctly evaluated as the flux of the current density