Physics - Magnetism Question with Solution | TestHub

When ${\mathrm{e}}^{-}$is accelerated by V volts, then work done by ${\mathrm{e}}^{-}$is equals to kinetic energy of ${\mathrm{e}}^{-}$. or $\mathrm{qV}=\frac{1}{2}{\mathrm{mv}}^2$ Then, the velocity of ${\mathrm{e}}^{-}$is $\mathrm{v}=\sqrt{\frac{2\mathrm{qV}}{\mathrm{m}}}$ or $\mathrm{v}=\mathrm{k}\sqrt{\mathrm{V}}$. When ${\mathrm{e}}^{-}$accelerated by V volts, then $\mathrm{v}=\mathrm{k}\sqrt{\mathrm{V}}$. Force experiences by ${\mathrm{e}}^{-}$in a uniform magnetic field, $\mathrm{F}=\mathrm{qvB}$ or $\mathrm{F}=\mathrm{qk}\sqrt{\mathrm{V}B}$ ...(i) When e accelerated by 2 V volts, then ${\mathrm{v}}_1=\mathrm{k}\sqrt{2\mathrm{V}}$. $\therefore {\mathrm{F}}_1={\mathrm{qv}}_1\mathrm{B}=\mathrm{qk}\sqrt{2\mathrm{VB}}$ Hence, ${\mathrm{F}}_1=\sqrt{2}\mathrm{F}$ [from Eq. (i)]