# Uniform circular motion for proton and 2 particles

1. Nov 28, 2014

### cseil

[Note by mentor: This thread does not use the homework template because it was moved here from a non-homework forum.]

Hello,
I have a problem with this exercise. I don't know if my solution is right! Could you check it, please?

A proton with kinetic energy Kp is moving in a uniform magnetic field on a circular orbit.
I have to calculate the energy of two particles moving on the same orbit knowing that:
particle 1 : q=+e, m=2u
particle 2: q=+2e, m=4u

$Kp = 1/2 mv^2$
$v = qBr/m$ (uniform circular motion).

So

$Kp = 1/2 q^2B^2r^2/m$

proton: $Kp = \frac{1}{2} \frac{e^2B^2r^2}{u}$

particle 1: $K = \frac{1}{2} \frac{e^2B^2r^2}{2u} = Kp/2$

particle 2: $K = \frac {1}{2} \frac{4e^22B^2r^2}{4u} = Kp$

Is that right?
Thank you

Last edited by a moderator: Nov 28, 2014
2. Nov 28, 2014

### ehild

Yes, I think it is right.

3. Nov 28, 2014

### cseil

Thank you.
I'm sorry but I didn't see the Homework section, won't happen again!