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Math Help - Electromagnetic wave equation

  1. #1
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    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 ).
    Attached Thumbnails Attached Thumbnails Electromagnetic wave equation-412.jpg  
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  2. #2
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    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 x-direction, in free space. Then, say the wave is polarised, so the electric field is in the y-direction. Then the only non-zero components of the fields are E_y and B_z:

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

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

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

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

    and so you can calculate

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

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

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

    B_z = 0.005 \sin (kx - c k t)
    = 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, (x - ct) \rightarrow (ct - x). This doesn't matter. However, I seem to have a different answer for B_0 (I don't know why!)

    Also, the form used in Wikipedia has an "extra" phase factor \phi; I have made \phi = 0. You can do this just by choosing the origin of your xyzcoordinates in the right place!
    Last edited by Pterid; June 8th 2007 at 05:10 PM. Reason: cleanup
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  3. #3
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    Quote Originally Posted by Pterid View Post
    This doesn't matter. However, I seem to have a different answer for 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.
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  4. #4
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    Quote Originally Posted by totalnewbie View Post
    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 \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!
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