Question

A particle of mass 'm' with charge 'q' moving with a uniform speed 'v' normal to a uniform magnetic field 'B', describe a circular path of radius 'r'. Derive expression for
(i) time period of revolution
(ii) kinetic energy of the particle

Answer 1

(i) Motion of charged particle in perpendicular magnetic field: The magnetic
force on charged particle $qvB\sin {90}^o$ provides the necessary centripetal force for a
circular path, so
$qvB=\frac{m{v}^2}{r}\Rightarrow v=\frac{qBr}{m}$
But $v=\frac{2\pi r}{T}$ where $T$ is time period
$\frac{2\pi r}{T}=\frac{qBr}{m}\Rightarrow T=\frac{2\pi m}{qB}$
(ii) Kinetic energy of charged particle:
$KE=\frac{1}{2}m{v}^2\Rightarrow \frac{1}{2}m\frac{q^2{B}^2{r}^2}{m^2}=\frac{q^2{B}^2{r}^2}{2m}$