Cyclotron frequency of a proton

Problem

In a cyclotron a proton ($m = 1{,}67\cdot 10^{-27}$ kg, $q = 1{,}6\cdot 10^{-19}$ C) moves in a field $B = 1$ T perpendicular to the velocity. Derive the cyclotron frequency $f_c = qB/(2\pi m)$ and calculate its value.

Solution

In uniform circular motion the Lorentz force acts as the centripetal force: $|q|vB = \dfrac{mv^2}{r}$, from which the radius $r = \dfrac{mv}{|q|B}$. The period is $T = \dfrac{2\pi r}{v} = \dfrac{2\pi m}{|q|B}$ and the frequency $f_c = 1/T$: $f_c = \frac{qB}{2\pi m} = \frac{1{,}6\cdot 10^{-19}\cdot 1}{2\pi\cdot 1{,}67\cdot 10^{-27}}$ $\ev{f_c \approx 1{,}5\cdot 10^7\;\text{Hz}}$ The cyclotron frequency does not depend on the velocity $v$: it is this property that makes fixed-phase acceleration possible in the cyclotron.

Links

Topics: Magnetism Concepts: Lorentz force · Uniform circular motion