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Some electrodynamics questions! (waves)

  • #1

Homework Statement



ok, for one of the prioblems I'm given the equations of an electric field:
Ex(z,t) = E0sin(kz-wt)
Ey(z,t) = E0cos(kz-wt)
for a circularly polarized plane wave of light traveling in the z direction.
I have to show the magnetic field, energy density, and Poynting vector for this wave.

Homework Equations



the energy density is u = .5 [eE2 + B2/u]
the Poynting vector is S = (E x B)/u

The Attempt at a Solution



I believe I can just plug in the values of E and B directly in the two equations above; however, I am having a tough time finding out what B is. In my textbook it says the general formula is b(z, t) = (k x E)/c, but I dont know how to compute this cross product.

any advice?
 

Answers and Replies

  • #2
674
2
k = (0, 0, k)

and...

E = (E_0 sin(kz-wt), E_0 cos(kz-wt), 0)

Then take the cross product of these two vectors. If you don't know how, then you can find examples of cross products all over the web or even the text book.
 
  • #3
k = (0, 0, k)

and...

E = (E_0 sin(kz-wt), E_0 cos(kz-wt), 0)

Then take the cross product of these two vectors. If you don't know how, then you can find examples of cross products all over the web or even the text book.
by k = (0, 0, k), do you mean k = (0, 0, 1) because the direction is in the z direction and the vector itself is k?
 
  • #4
674
2
No, I was writing the vector k in terms of its components. The magnitude of (0,0,k) is k. The magnitude of (0,0,1) is 1. So that wouldn't be a good vector for k.
 
  • #5
No, I was writing the vector k in terms of its components. The magnitude of (0,0,k) is k. The magnitude of (0,0,1) is 1. So that wouldn't be a good vector for k.
so I should do <0, 0, k> x <E_0sin(kz-wt), E_0cos(kz-wt), 0> ?

If so, I can do this...

I got <-kE_0cos(kz-wt), -kE_0sin(kz-wt), 0>
 
Last edited:
  • #6
674
2
The second term should be positive.
 
  • #7
right, I completely forgot that when using cross product you have to take the opposite of the j vector. Thanks for your help :) you were very patient and understanding!
 
  • #8
674
2
No problem. Good luck with the other parts of the problem. Should be easy now that you have B.
 
  • #9
I hate to bother you or anyone else again, but I have another question..

I have two components of both the electric and magnetic field.
to express it as one component, could I just take the magnitude of the components?
i,e. | <a, b, c> | for whatever the values of a b and c are?

The energy density doesnt ask for the vector component of E and B, so would the method I mentioned above be correct?
 
  • #10
674
2
E^2 is just the dot product of the vector E with itself. So E^2 = E . E = E_x^2+E_y^2+E_z^2.
 
  • #11
E^2 is just the dot product of the vector E with itself. So E^2 = E . E = E_x^2+E_y^2+E_z^2.
yeah that's what I did as I tried to explain in the last post

Thanks! :)
 

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