An electron of mass m and charge e is accelerated by a potential difference V. It then enters a uniform magnetic field B applied perpendicular to its path. The radius of the circular path of the electron is

$\text{Where}\text{v}\text{is}\text{the}\text{Velocity}\text{of}\text{the}\text{electron}\text{.}\text{if}\text{r}\text{is}\text{the}\text{radius}\text{of}\text{the}\text{path}$

$\text{We}\text{}\text{have}$

$\frac{{\text{mv}}^{\text{2}}}{\text{r}}\text{=}\text{Bev}\text{or}\text{r=}\frac{\text{mv}}{\text{Be}}$

$\text{The}\text{velocity}\text{v}\text{is}\text{given}\text{by}$

$\frac{\text{1}}{\text{2}}{\text{mv}}^{\text{2}}\text{=}\text{ev}$

${\text{v}}^{\text{2}}=\frac{\text{2eV}}{\text{m}}\Rightarrow \text{v}=\sqrt{\frac{\text{2eV}}{\text{m}}}$

$\text{r}=\frac{\text{m}}{\text{Be}}\sqrt{\frac{\text{2eV}}{\text{m}}}\text{=}\sqrt{\frac{{\text{m}}^2\text{2eV}}{{\text{B}}^{\text{2}}{\text{e}}^2\text{m}}}$

$\text{r}=\sqrt{\frac{\text{2Vm}}{{\text{B}}^{\text{2}}\text{e}}}\Rightarrow \text{r}={\left(\frac{\text{2Vm}}{{\text{B}}^{\text{2}}\text{e}}\right)}^{\text{1/2}}$