An electron accelerated through potential difference ' $V$ ' passes through a uniform transverse magnetic…
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
$3 \mathrm{~F}$
$\mathrm{F}$
$\sqrt{2} \mathrm{~F}$
$\frac{\mathrm{F}}{2}$
Solution
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
$\begin{aligned} \frac{\mathrm{k}_{2}}{\mathrm{k}_{1}} &=\frac{\mathrm{v}_{2}^{2}}{\mathrm{v}_{1}^{2}} \\ \therefore \frac{\mathrm{v}_{2}}{\mathrm{v}_{1}} &=\sqrt{\frac{\mathrm{k}_{2}}{\mathrm{k}_{1}}}=\sqrt{2} \\ \frac{\mathrm{F}_{2}}{\mathrm{~F}_{1}} &=\frac{\mathrm{e}_{2} \mathrm{~B}}{\mathrm{ev}_{1} \mathrm{~B}}=\frac{\mathrm{v}_{2}}{\mathrm{v}_{1}}=\sqrt{2} \\ \therefore \mathrm{F}_{2} &=\sqrt{2} \mathrm{~F}_{1}=\sqrt{2} \mathrm{~F} \end{aligned}$
.