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Electromagnetic power radiation

  1. Dec 12, 2003 #1
    Given that the electromagnetic power radiated by a nonrelativistic moving point charge q having acceleration a is:
    [tex]P = \frac {q^2a^2}{6 \pi \epsilon_0 c^3} [/tex]
    (This formula was presented without discussion or explanation in the problems section of my textbook.)

    If a proton is placed in a cyclotron with a radius of 0.500 m and a magnetic field of magnitude 0.350 T, what electromagnetic power is radiated by this proton?

    This looks like one of those dumb problems where you just plug numbers into a formula, but ... what's a in this situation? We are not given the voltage difference between the dees of the cyclotron. So I thought maybe I can use the exit velocity of the proton of the proton given by
    [tex]v = \frac {qBr}{m}[/tex]

    OK. It turns out that was the right approach. Only I went off talking about angular acceleration, when I should be dealing with centripetal acceleration. So, the following lines of tex are garbage. The corrected stuff is in the next post.

    [tex]a = \frac{v^2}{r} = \frac{q^2B^2r^2}{m^2r}[/tex]

    [tex]a^2 = \frac{q^4B^4r^2}{m^4}[/tex]
    Last edited: Dec 12, 2003
  2. jcsd
  3. Dec 12, 2003 #2

    Please just disregard that garbage up above. I've tried editing it several times, but for some reason the board won't let me fix the tex images, so the last 2 lines of tex are nonsense.


    It should say

    [tex]a = \frac{v^2}{r} = \frac{q^2B^2r^2}{m^2r} [/tex]

    and then:

    [tex]a^2 = \frac{q^4B^4r^2}{m^4} [/tex]

    and then putting that into the power equation works.
    Last edited: Dec 12, 2003
  4. Dec 12, 2003 #3

    Doc Al

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    Staff: Mentor

    Well... you saved me the trouble of decoding them!

    That power equation is called Larmor's formula, by the way.
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