An electron accelerated through potential difference ' \(V\) ' passes through a uniform
transverse magnetic field and experiences a force ' \(\mathrm{F}^{\prime}\). If the accelerating potential
is increased to '2V', the electron in the same magnetic field will experience a force

Kinetic energy \(\quad k_{1}=\mathrm{eV} \quad\) and \(\quad k_{2}=2 \mathrm{eV}\)
\(\therefore \frac{\mathrm{k}_{2}}{\mathrm{k}_{1}}=2 \quad \therefore \mathrm{v}_{1}\) and \(\mathrm{v}_{2}\) are the velocities in the two cases then
\(\begin{aligned} \frac{\mathrm{k}_{2}}{\mathrm{k}_{1}} &=\frac{\mathrm{v}_{2}^{2}}{\mathrm{v}_{1}^{2}} \\ \therefore \frac{\mathrm{v}_{2}}{\mathrm{v}_{1}} &=\sqrt{\frac{\mathrm{k}_{2}}{\mathrm{k}_{1}}}=\sqrt{2} \\ \frac{\mathrm{F}_{2}}{\mathrm{~F}_{1}} &=\frac{\mathrm{e}_{2} \mathrm{~B}}{\mathrm{ev}_{1} \mathrm{~B}}=\frac{\mathrm{v}_{2}}{\mathrm{v}_{1}}=\sqrt{2} \\ \therefore \mathrm{F}_{2} &=\sqrt{2} \mathrm{~F}_{1}=\sqrt{2} \mathrm{~F} \end{aligned}\)