1. Dec 12, 2003

### gnome

Given that the electromagnetic power radiated by a nonrelativistic moving point charge q having acceleration a is:
$$P = \frac {q^2a^2}{6 \pi \epsilon_0 c^3}$$
(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
$$v = \frac {qBr}{m}$$

Edited:
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.

$$a = \frac{v^2}{r} = \frac{q^2B^2r^2}{m^2r}$$

$$a^2 = \frac{q^4B^4r^2}{m^4}$$

Last edited: Dec 12, 2003
2. Dec 12, 2003

### gnome

OK.

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.

Sorry...

It should say

$$a = \frac{v^2}{r} = \frac{q^2B^2r^2}{m^2r}$$

and then:

$$a^2 = \frac{q^4B^4r^2}{m^4}$$

and then putting that into the power equation works.

Last edited: Dec 12, 2003
3. Dec 12, 2003

### Staff: Mentor

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

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