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Some electrodynamics questions! (waves)

  1. May 12, 2009 #1
    1. The problem statement, all variables and given/known data

    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.

    2. Relevant equations

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

    3. 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?
     
  2. jcsd
  3. May 12, 2009 #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.
     
  4. May 12, 2009 #3
    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?
     
  5. May 12, 2009 #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.
     
  6. May 12, 2009 #5
    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: May 12, 2009
  7. May 12, 2009 #6
    The second term should be positive.
     
  8. May 12, 2009 #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!
     
  9. May 12, 2009 #8
    No problem. Good luck with the other parts of the problem. Should be easy now that you have B.
     
  10. May 12, 2009 #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?
     
  11. May 12, 2009 #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.
     
  12. May 12, 2009 #11
    yeah that's what I did as I tried to explain in the last post

    Thanks! :)
     
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