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.
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...
Homework Statement
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...