# Electromagnetic wave equation

• Jun 8th 2007, 09:34 AM
totalnewbie
Electromagnetic wave equation
First of all I have to say that translating specific words from native language to english, is not easy. So I hope that you realize what is going on:

What did I do wrong ?

(Traveling waves from: Wave - Wikipedia, the free encyclopedia ).
• Jun 8th 2007, 05:08 PM
Pterid
OK. First, we should just state the form of the 'wave equations'. These have a lot of interesting physics behind them, but for now we'll just state them!

Say the light wave travelling in the $\displaystyle x$-direction, in free space. Then, say the wave is polarised, so the electric field is in the $\displaystyle y$-direction. Then the only non-zero components of the fields are $\displaystyle E_y$ and $\displaystyle B_z$:

$\displaystyle \bold{E} = (0, E_y, 0); \quad \bold{B} = (0, 0, B_z);$
$\displaystyle E_y = E_0 \sin (kx - \omega t);$
$\displaystyle B_z = B_0 \sin (kx - \omega t);$

$\displaystyle B_0 = E_0/c;$
$\displaystyle \omega = c k.$

Now, an explanation of the symbols I've used:
• $\displaystyle k$ is the wavenumber (or wavevector) of the wave. It's defined by $\displaystyle k = \frac{2\pi}{\lambda}$, where $\displaystyle \lambda$ is the wavelength.
• $\displaystyle \omega$ is the angular frequency of the wave. It has units of "radians per second", and relates to the 'true' frequency $\displaystyle \nu$ by $\displaystyle \omega = 2\pi \nu$.
• $\displaystyle c = 3 \times 10^8 \text{m~s}^{-1}$ is the speed of light.
In your question, you have been given the following information:

$\displaystyle E_0 = 1.5\times 10^6 \text{V~m}^{-1}$
$\displaystyle \lambda = 10.6\times 10^{-6} \text{m}$

and so you can calculate

$\displaystyle k = 2\pi / \lambda = 0.59\times 10^6 \text{m}^{-1}$.
$\displaystyle B_0 = E_0 / c = 0.005 T$ ($\displaystyle T$ is a unit, the tesla.)

Put these into the expressions above, and ignoring the units, you get:

$\displaystyle E_y = (1.5 \times 10^6) \sin (kx - c k t)$
$\displaystyle = (1.5 \times 10^6) \sin [k(x - c t)]$
$\displaystyle = (1.5 \times 10^6) \sin [0.59\times 10^6 (x - 3\times 10^8~ t)]$;

$\displaystyle B_z = 0.005 \sin (kx - c k t)$
$\displaystyle = 0.005 \sin [0.59\times 10^6 (x - 3\times 10^8~ t)]$.

- - -

Your answers are written slightly diffierently: they have a minus sign, $\displaystyle (x - ct) \rightarrow (ct - x)$. This doesn't matter. However, I seem to have a different answer for $\displaystyle B_0$ (I don't know why!)

Also, the form used in Wikipedia has an "extra" phase factor $\displaystyle \phi$; I have made $\displaystyle \phi = 0$. You can do this just by choosing the origin of your $\displaystyle xyz$coordinates in the right place!
• Jun 9th 2007, 10:34 AM
totalnewbie
Quote:

Originally Posted by Pterid
This doesn't matter. However, I seem to have a different answer for $\displaystyle B_0$ (I don't know why!)

B_0=E_0/c is in vacuum. Maybe for that reason.
Although you answer for B_0 is close to the original one.
• Jun 9th 2007, 03:49 PM
Pterid
Quote:

Originally Posted by totalnewbie
B_0=E_0/c is in vacuum. Maybe for that reason.
Although you answer for B_0 is close to the original one.

Well, yes, but the $\displaystyle \omega(k)$relation is consistent with a vacuum - and if the problem were "set" in a medium, you would expect the question to tell you something about the medium...

If I were a more arrogant kind of person, I might suggest that the answers had a typo in! :o