## Electric Charge in a Magnetic Field

1) For a particle of mass m and a charge of q moving in a circular path in a magnetic field B, show that its kinetic energy is proportional to r^2, the square of the radius of the curvature of the path.

I started off with the definition of KE: (mv)^2/2 and the fact that a = v^2/r, where F = m(v^2/r).

Then, I thought;

mv^2 = 2*KE and mv^2 = F*r so r = (2KE)/F so r^2 = (4KE^2)/ F^2, but this is probably wrong. What should I do if so?

Thanks.

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 What do you know about the strength of magnetic fields in relation to distance R?
 If B is constant, then r is constant??

## Electric Charge in a Magnetic Field

Well typically one would say that backwards, since B is a function of r. What is B as a function of R?

What does B have to be equal to to put the particle in uniform circular motion? Think about the force involved in circular motion, and the force of a magnetic field.

 Quote by Soaring Crane 1) For a particle of mass m and a charge of q moving in a circular path in a magnetic field B, show that its kinetic energy is proportional to r^2, the square of the radius of the curvature of the path. I started off with the definition of KE: (mv)^2/2 and the fact that a = v^2/r, where F = m(v^2/r). Then, I thought; mv^2 = 2*KE and mv^2 = F*r so r = (2KE)/F so r^2 = (4KE^2)/ F^2, but this is probably wrong. What should I do if so? Thanks.
Remember:
F=mv²/r
and F=Bqv

also v=(2pi.r)/t