# Determining Electric and Magnetic field given certain conditions

guyvsdcsniper
Homework Statement:
Refer to attached image
Relevant Equations:
comlplex wave equation, I am unsure of my solutions and am looking for some guidance.

a.)The real part of the wave in complex notation can be written as ##\widetilde{A} = A^{i\delta}##. Writing the Complex Wave equation, we have ##\vec E(t) = \widetilde{A}e^{(-kz-\Omega t)} \hat x##. Therefore the real part is ##\vec E(t) =Ae^{(-kz-\Omega t+\delta)} \hat x##. The negative in front of kz indicates it is a left traveling wave.

b.) The unit vector of ##\hat B = \frac{(\hat x - 2\hat z)}{\sqrt{5}}##. I know that ##\hat E## must be perpendicular to ##\hat B##, so simply,
##\hat E = \frac{(\hat x + 2\hat z)}{\sqrt{5}}##

c.) I am not so sure about this problem. I know that ##\vec E = \widetilde{E}_oe^{i(ky-wt)}\hat x##
Griffiths states ##\widetilde{B}_o = \widetilde{E}_o/c## and ##v=c/n##.

So ##\vec B = \frac{c\widetilde{E}_o}{1.7}e^{i(ky-wt)}\hat z##

Staff Emeritus
Homework Helper
a.)The real part of the wave in complex notation can be written as ##\widetilde{A} = A^{i\delta}##. Writing the Complex Wave equation, we have ##\vec E(t) = \widetilde{A}e^{(-kz-\Omega t)} \hat x##. Therefore the real part is ##\vec E(t) =Ae^{(-kz-\Omega t+\delta)} \hat x##. The negative in front of kz indicates it is a left traveling wave.
The problem statement said the wave travels in the ##-x## direction. Your answer isn't sinusoidal. It decays with time. What is ##\delta##?

b.) The unit vector of ##\hat B = \frac{(\hat x - 2\hat z)}{\sqrt{5}}##. I know that ##\hat E## must be perpendicular to ##\hat B##, so simply,
##\hat E = \frac{(\hat x + 2\hat z)}{\sqrt{5}}##
It doesn't look like ##\hat E \cdot \hat B=0##.

• topsquark
guyvsdcsniper
The problem statement said the wave travels in the ##-x## direction. Your answer isn't sinusoidal. It decays with time. What is ##\delta##?
You're right. I should have ##\vec E(t) =Acos{(-kz-\Omega t+\delta)} \hat x##
##\delta## is the phase
So I need to make ##\hat x## be ##\hat - x## as well as account for it is a wave traveling left by the ##-kz## in the ##cos##?

I assumed the ##-kz## in the ##cos## accounted for the negative direction.

Staff Emeritus
Homework Helper
I assumed the ##-kz## in the ##cos## accounted for the negative direction.
In the ##-z## direction, not the ##-x## direction.

$$\vec{B}=A \vec{e}_z \exp(-\mathrm{i} \Omega t-\mathrm{i} k x), \quad \Omega,k>0.$$
• 