[NOTES/EM-01004]--Cyclotron Motion

Let a charged particle of mass \(m\), charge \(q\) and having velocity \(v\), enter a region of uniform magnetic field \(B\). Assume that the magnetic field is perpendicular to the velocity as shown in Figure

The force due to the magnetic field is $QvB$ and is directed perpendicular to the velocity of the charged particle. Thus the charge particle traces a circle of radius \(R\)and has acceleration \(a=v^2/R\). The second law of motion gives
\begin{equation} \frac{mv^2}{R} = qvb. \end{equation}

{Cyclotron Motion} $$ \therefore~\left(\frac{1}{R}\right) = \left(\frac{qB}{mv}\right) $$ where $R=$ is radius of the circular orbit is given by the cyclotron formula \begin{equation} R = \left(\frac{mv}{qB}\right)\qquad \end{equation} The frequency of the circular motion, \(\omega\), is given by \begin{eqnarray} \omega = \frac{v}{r} &=& \left(\frac{qB}{m}\right)=\\ \omega &=& \left(\frac{qB}{2\pi m}\right) \end{eqnarray} This frequency \(\omega\) is called the cyclotron frequency. For the unversed, a cyclotron is a machine to accelerate a charged particles in which a magnetic field makes them go into a circular orbit.