Question

Can anyone help me translate the equation for an electromagnetic wave into its parts?

Answer 1

I'll be back in a bit but let me know if you're adept at reading these kinds of equations

Answer 2

Excuse the horrific file name ^^'

Answer 3

lol

Answer 4

damn ok ill try my best

Answer 5

i dont think i will be of any help.... damn... i... dont even know were to start, my math skills are useless

Answer 6

is there anything else maybe?

Answer 7

That's what I'm stuck on atm, so no just this

Answer 8

man if only i had your knowledge i could try and help but i take basic math
.

Answer 9

If I was good at it I'd have done it by now =/

Answer 10

can you determine the angular frequency \(\omega\)?

Answer 11

for a wave oscillation with the angular frequency \(\omega\),
one of the components in the complex exponential (or trig. function), must be \(\omega t\)

Answer 12

Is it 9.42 X 10^15?

Answer 13

and with SI units?

Answer 14

Radians per second?

Answer 15

yep

Answer 16

the other term in the complex exponential (or trig. function), is k•r,

where k is the wave vector, and r is position vector

Answer 17

Hey I do still need help with this but I'm extremely tired. Will you be around tomorrow?

Answer 18

👀

Answer 19

Oh fun haven't worked with these in a bit! :P

The direction of oscillation will act in the direction of the amplitude (normalized)

\[\large \frac{-6 \hat{i} + 3\sqrt{5}\hat{j}}{|(-6)^2 + (3\sqrt{5})^2|}\]

Scalar Amplitude would just be \[\large |A| = (10^4 \frac{V}{m})\sqrt{(-6)^2 + (3\sqrt{5})^2}\]

The direction of propagation works with the complex exponent, we know it is in the form
\(\large e^{i[\vec{k} \cdot \vec{r}-\omega t]}\) and we know the position vector \(\large \vec{r} = x\hat{x} + y\hat{\hat{y}}\) we can see that \(\large \vec{k} = \frac{1}{3}\pi \times 10^7(\sqrt{5}\hat{i} + 2\hat{j})\)

The propagation number 'k' is \(\large k = \sqrt{\vec{k} \cdot \vec{k}}\) which here means \(\large \sqrt{(\pi \times 10^7) \times (\pi \times 10^7)}\) or just \(\large \pi \times 10^7\)

We know the wavelength can be found by \(\large \lambda = \frac{2\pi}{k}\) so I'm not gonna help you with that :P

You already found the angular frequency to be 9.42 x 10^15

So we can use that to solve for the frequency since we know \(\large f = \frac{\omega}{2\pi}\)

And finally the speed would be the product of the wavelength and the frequency \(\large v = \lambda f\)