Some electrodynamics questions (waves)

In summary, the conversation was about finding the magnetic field, energy density, and Poynting vector for a circularly polarized plane wave of light traveling in the z direction, given the equations for the electric field. The solution involved finding the cross product of the vector k and the vector E, and taking the magnitude of the components to express them as one component for the energy density calculation. The dot product of the vector E with itself was also used to find the magnitude.
  • #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 don't know how to compute this cross product.

any advice?
 
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  • #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 textbook.
 
  • #3
nickjer said:
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 textbook.

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
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
nickjer said:
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
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
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 doesn't ask for the vector component of E and B, so would the method I mentioned above be correct?
 
  • #10
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
nickjer said:
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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