Derivation of the equation for the cyclotron frequency: Say you have a wire of cross-sectional area A and length L that contains particles, each with charge q and drift velocity v. Let's suppose that the time it takes for one particle to travel the length of the wire is t, so that v = L/t => L = vt. If we say the number of particles in the wire is N and the number of particles per unit volume is n, then n = N/AL, as AL is the volume of the wire. Hence: n = N/Avt => nAvt = N.

Now, the total charge that has travelled a distance L is Q, so Q = Nq => N = Q/q. Therefore: nAvt = Q/q => nAqv = Q/t. Current is defined as being the charge that passes a point in a wire per unit time, so I = Q/t. Finally, then: nAqv = I.

This, ladies and gentlemen, is what we call a breakthrough.

For now (until somebody messages me telling me how to derive this, or does it themselves), I will pull an equation out of thin air! F = BIL, where F is the force on a current-carrying wire, B is the magnetic flux density and L is the length of the wire. But I = nAqv. So: F = BnAqvL. As before, AL is the volume of the wire, and the number of particles per unit volume is n. Hence, the number of particles N = nAL. So, the force on one particle (call it f for the moment) is F/N, where F/N = f = nAqvL/nAL => f = Bqv.

You have all just witnessed a miracle of science. I hope you liked it, because here comes another one.

First things first. That small f looks ridiculous, so let's call it big F. From now on, F will mean the force on one particle. Let's say you have an electron on the floor, in a magnetic field, where the field lines are going straight downwards. For the purposes of illustration, I shall give the electron a kick to the east (not to the right, because then in becomes down and it doesn't work). Using Fleming's Left Hand Rule (Thumb = Force, Index finger = Field direction, Middle finger = Direction of conventional current, i.e. opposite direction to motion of electron), we can figure out that the electron is going to have a force southwards on it, so it will accelerate southwards. In fact, my exquisitely rectangular fingers tell me that the force will always be at right angles to the current, and so also to the motion of the electron (in the opposite direction).

This means that our electron will always have a force perpendicular to its motion; this force will remain constant if B, q and v are constant. If this doesn't make you slap your thigh, read up about circular motion in the weightlessness node. That's right. A constant force perpendicular to its motion means a constant force towards the centre of a circle, so the resultant force on the particle is its centripetal force, which must be m(v^2)/r. Therefore Bqv = m(v^2)/r => Bq = mv/r.

This is where another bit of circular motion comes in: v = rw, so: Bq = mrw/r => Bq = mw. A couple of new equations: w = 2(pi)/T, T = 1/f. Hence: w = 2(pi)/(1/f) => w = 2(pi)f. Substituting: Bq = 2(pi)mf => f = Bq/2(pi)m, where f is the angular frequency. As this is for an electron going round in a full circle, I imagine that the potential difference between the "dees" needs to be reversed twice as often, so that the cyclotron frequency is Bq/(pi)m.