E=cB and electromagnetic energy density

[davidbenari]

I have to questions that I can't seem to understand.

I've seen in my textbook the relation E=cB being deduced from Faradays law applied to some type of rectangular loop. The problem is this is done with regards to a planar wave. So is this relationship general for all EM waves? If so, how? And could someone show me a simple way to prove this to myself? If the velocity of a wave is not c but some other (because of refraction or whatever) then is the relation E=vB?

Another question I have is about electromagnetic energy density. Is the formula for EM energy density $U=\frac{\epsilon E^2}{2} + \frac{B^2}{2\mu}$ applied using the values for the magnetic and electric fields at a specific point? That is: similar to a problem with mass density, we assume small infinitesimal volumes in the neighborhood of a plane having the same energy density? (since the E and B values all have the same value at those points)

 

[mfb]

Planar waves satisfy the maxwell equations only if that relation is true (with magnitudes, not with vectors).

Energy density in "physical" settings is continuous, so it does not vary wildly between points nearby, yes.

 

[davidbenari]

I know how to derive E=cB for planar waves, but I was wondering why this relationship holds true for sinusoidal waves too?

Also, if the energy density is $u=\frac{\epsilon E^2}{2} + \frac{B^2}{2\mu}$ could I say that the average energy density is $u_{average}=\frac{\epsilon E^2}{4} + \frac{B^2}{4\mu}$ taking averages of the cosine^2 functions?

Thanks.

 

[mfb]

I would be surprised if it is true for every point in space for every vacuum solution.

Also, if the energy density is $u=\frac{\epsilon E^2}{2} + \frac{B^2}{2\mu}$ could I say that the average energy density is $u_{average}=\frac{\epsilon E^2}{4} + \frac{B^2}{4\mu}$ taking averages of the cosine^2 functions?

At least for planar waves this is possible if you replace E and B by their peak values.

 

[davidbenari]

Can I apply the derivation I've seen of E=cB for planar waves to sinusoidal waves too?

Namely: Imagine a rectangular loop with its vertical side $a$ parallel to the polarisation of the electric field (the B field is therefore creating flux across this loop). I place this rectangular loop in such a way that half of it is immersed in the field, and the other half in pure vacuum since the wave hasn't reach there yet. Both halves have length $cdt$

Applying faradays law we say the line integral yields $-Ea$ and $\frac{d\Phi}{dt}$ is $frac{(B+dB)acdt}{dt}$ which is just $Bacdt$ ignoring the infinitesimal element. Therefore we obtain again E=cB.

Can I do this? Ignore the infinitesimal element?

 

[davidbenari]

I have the feeling I don't really understand what a plane wave is. Wikipedia includes sinusoidal waves too. I thought plane waves were this gigantic constant phase planes that travel in some direction…

 

[tech99]

I know how to derive E=cB for planar waves, but I was wondering why this relationship holds true for sinusoidal waves too?

Also, if the energy density is $u=\frac{\epsilon E^2}{2} + \frac{B^2}{2\mu}$ could I say that the average energy density is $u_{average}=\frac{\epsilon E^2}{4} + \frac{B^2}{4\mu}$ taking averages of the cosine^2 functions?

Thanks.

 

[tech99]

If you want to work out energy density over the duration of a sine wave you should use the first formula using the RMS values of the fields, as in normal engineering practice.

 

[Khashishi]

E = cB is only true for some special cases. It certainly isn't true in general. (Consider two vertically polarized beams crossing each other at right angles.)
$U=\frac{\epsilon E^2}{2} + \frac{B^2}{2\mu}$ is true much more generally, for arbitrary fields which contain both propagating waves and static fields which surround sources.