Electron of mass $m$ and charge $e$ is traveLling with a speed $v$ along circular path of radius $r$

Question

Electron of mass $m$ and charge $e$ is traveLling with a speed $v$ along circular path of radius $r$ at right angles to a uniform magnetic field $B$. If the speed of the electron is doubled and the magnetic field is halved the resulting path would have a radius\

Professor PC Thomas Classes · Answer

A): Here centripetal force is provided by magnetic force\ $	herefore \frac{\mathrm{mv}^{2}}{\mathrm{r}}=\mathrm{qvB}$\ $r=\frac{m v}{q B}=\frac{m v}{e B}($ Here $q=e)$\ Now when speed of electron is doubled $v^{\prime}=2 v$.\ Magnetic field is halved $B^{\prime}=\frac{B}{2}$\ $r^{\prime}=\frac{m v^{\prime}}{e B^{\prime}}=\frac{m}{e} \frac{2 v}{B / 2}=\frac{4 m v}{e B}=4 r$\