# Optics problem: electromagnetic theory

1. Nov 21, 2004

### cuti_pie75

If anyone who's good at optics/physics can help me with this electromagnetic part of optics...it'll be great.

so here's the problem:

imagine an electromagnetic wave with it's E-field in the y-direction. Show that (dE/dx)=-(dB/dt) applied to the harmonic wave B

E=Eo cos(kx - wt) B=Bo cos(kx - wt)

yields to the fact that Eo=cBo
in agreement with Ey=cBz

2. Nov 21, 2004

### Galileo

Hi cuti_pie75.

Just work out (dE/dx)=-(dB/dt) with what you're given.
I`ll do dE/dx:

$$\frac{dE}{dx}=\frac{d}{dx} E_0 \cos(kx-\omega t) = -kE_o\sin(kx-\omega t)$$
Differentiate B with respect to t and use the equation (dE/dx)=-(dB/dt).

You also need that $\vec E_0=E_y \hat y$ (this was given).
And assuming the direction of the wave is in the +x-direction. $\vec B_0= B_z \hat z$.

Last edited: Nov 21, 2004
3. Nov 21, 2004

### cuti_pie75

Thank you very much for your fast reply Galileo, everything's much more clearer to me now...but i just have one more question:

i did the derivative for B in respect to t and comes to: kEo = -wBo
so my question is -w/k = c?

anyway, sorry for the bother and thanx again!

4. Nov 21, 2004

### Galileo

Almost, remember the equation is (dE/dx)=-(dB/dt), so you have a minus sign. There is another minus sign coming from the derivative of the cosine and another one from the chain rule (derivative of (kx-wt) with respect to t is -w).
So you get kEo=wBo
w/k indeed equals the speed of the wave (c in this case).
the way I always remember it, is by knowing that a traveling wave always has the form f(x-vt) with v the speed of the wave. kx-wt=k(x-w/kt), so w/k is the speed,