W9_parallel_resonance.eps

Week 10: Maxwell’s Equations and Light 355

Let us graph the fields on an arbitrary coordinate system and applyAmpere’s Law and

Faraday’s Law (only) to our graph. E~andB~have many components each, of course,

and can be varying with respect to both position and time, so we needto simplify a bit

to make sense of things. We will then imagine that either our distant source created only

x-directed electric fields andy-directed magnetic fields or that, equivalently, we are only

consideringExandBycomponents in particular of a more complicated field. Since the

fields satisfy the superposition principle, any results we get for thispair of components

can be generalized to any actual directions we like.

The graph is shown in figure 141, along with two dashed curves (bounding the shaded

surfaces) to which we will apply Ampere’s and Faraday’s Laws. We will assume that

Ex(z, t) is a function ofzandtonly – it may vary with respect toxoryas well, but

for the moment we’ll ignore any such variation^98. Similarly we will assumeBy(z, t) only.

Our graph is a snapshot at some particular timet, so we don’t bother writingtin on

the figure (but it is really there). I’m sorry if it is a bit confusing to constantly ignore

variation with respect to this or that variable – if/when you take multivariate calculus

you’ll learn once and for all how to deal with this sort of thing and encode it into the

notion of thepartial derivativebut for the moment we’re working our way towards a

resultthat should be expressed in partial derivatives without actually using them or

their (honestly, much simpler) notation.

Now let us apply Faraday’s Law to the small differential loop in thex-zplane. This

loop has an area ∆A= ∆x∆z, and we need to define aright handed normalto the

loop in they-direction (parallel toB~). That means that we need to go around the loop

counterclockwiseas drawn in the page. Then:

∮

E~·dℓ = −d

dt

∫

∆A

B~·nˆdA

0 ·∆z+Ex(z+ ∆z)∆x− 0 ·∆z−Ex(z)∆x = −

d

dt

(By∆A)

(Ex(z+ ∆z)−Ex(z)) ∆x = −dBy

dt

∆x∆z

(Ex(z+ ∆z)−Ex(z))

∆z

= −dBy

dt

(847)

where we do the loop piecewise and get no contribution when we go in thezdirection

(becauseE~is in thex-direction perpendicular toz). If we take the limit ∆z→0 of the

left hand side this is just thedefinitionof the derivative and we get^99 :

dEx

dz

=−dBy

dt

Let’s do exactly the same thing for Ampere’s Law, this time using the more lightly

shaded surface and curve in they-zplane with area ∆A= ∆y∆z. Again we must go

around it so that the right handed normal is parallel toE~in thexdirection, or again

counterclockwise as seen on the page from above. Theonlyterm on the right is the

(^98) It isn’t too difficult to imagine how such a field could be produced by (say) a distant oscillating electric dipole in
the−zdirection, actually.
(^99) Technically, this should be expressed aspartialderivatives:∂E∂zx=−∂B∂ty, but since we cleverly arranged it so
thatExis a function of only one spatial coordinate andxandtare independent, it doesn’t matter in this case.