Electromagnetic power of a proton leaving a cyclotron

[JHans]

The electromagnetic power radiated by a nonrelativistic particle with charge q moving with acceleration a is:

$$<br /> P = \frac{q^2 a^2}{6 \pi \epsilon_0 c^3}<br />$$

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 does this proton radiate just before leaving the cyclotron?


I've identified that the unknown variable for the electromagnetic power equation is the acceleration of the particle, a. To find this, I've identified that the velocity of a particle in a cyclotron is:

$$<br /> v = \frac{q B r}{m}<br />$$

It makes sense to me that by differentiating,

$$<br /> a = \frac{dv}{dt}<br />$$

$$<br /> \frac{dv}{dt} = \frac{qB}{m} \frac{dr}{dt}<br />$$

The problem is that I don't know what the rate of change of the cyclotron's radius (with respect to time) is. Am I going in the right direction, or am I making things more difficult than they need to be?

 

[JHans]

Okay... I may have come to an equation for the acceleration of a particle in a cyclotron... I'm a little dubious of its validity, though.

A charged particle in a cyclotron experiences a force (specifically, a centripetal one that does not alter speed, but does alter velocity) of magnitude:

$$<br /> F = m a<br />$$

This force is caused by the motion of a charged particle in a uniform magnetic field, so equivalently:

$$<br /> F = q v B<br />$$

And thus,

$$<br /> a = \frac{q v B}{m}<br />$$

Now, because this is a centripetal force, this is a particle in uniform circular motion. So:

$$<br /> F = m a = m \frac{v^2}{r} = q v B<br />$$

$$<br /> v = \frac{q B r}{m}<br />$$

This can be substituted into the earlier equation for acceleration:

$$<br /> a = \frac{q v B}{m} = \frac{q B}{m} \frac{q B r}{m} = \frac{q^2 B^2 r}{m^2}<br />$$

This would seemingly reduce the acceleration a particle in a cyclotron to a function of its charge, mass, distance from the center of the cyclotron, and the strength of the magnetic field. Could I then substitute this into the equation for the electromagnetic power and enter in my known values?