# Angular frequency of orbiting charge in electric and magnetic fields

1. Mar 18, 2013

### JamesOza

1. The problem statement, all variables and given/known data

A particle of mass $m$ and charge $-q$ moves in a circular orbit of radius $R$ about a fixed charge $Q$. The angular frequency for the orbit is given by $$\omega_0^2 = \frac{qQ}{4 \pi \epsilon_0 m R^3}$$ A uniform magnetic field of magnitude $B$ in a direction perpendicular to the plane of the orbit is turned on. As a result, the angular frequency is changed to $\omega_0 + d\omega$. Assuming that $B$ is sufficiently small so that products of $B$ and $d\omega$ can be neglected, calculate $d\omega$.

3. The attempt at a solution

This problem has frustrated me for days, particularly as I know the answer to be $$d\omega = \frac{qB}{2m}$$ I have tried using the Lorentz force with motion in a circle to try and obtain the answer, but end up nowhere. There must be some mathematical trickery, possibly with approximations, that I’m missing. Any help and hints with how to start and proceed with this problem will be greatly appreciated.

Last edited: Mar 18, 2013
2. Mar 18, 2013

### voko

This seems to be a rather straight-forward application of the Lorentz force. Show what you have done.

3. Mar 18, 2013

### JamesOza

I agree it should be a straight forward application of the Lorentz force. I can see that the given answer for $d\omega$ is half the angular frequency if there was no electric field. Here’s how I started, $$m \omega^2 R = q(E + \omega RB)$$ $$\omega^2 = \frac{qQ}{4 \pi \epsilon_0 m R^3} + \frac{q \omega B}{m}$$ However substituting $\omega = \omega_0 + d\omega$ and noticing that the 1st term on the RHS is $\omega_0^2$ doesn’t lead me to the solution. What am I missing? Thanks.

4. Mar 18, 2013

### voko

Expand the LHS. What's left of the equation when you apply your observation about the RHS?

5. Mar 18, 2013

### JamesOza

Solved it! I’d expanded the LHS several times, and every time I did I neglected to ignore $dw^2$. Thanks for your help voko, it was reassuring to know I was on the right track.