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A electron $\left(\mathrm{e}^{-}\right)$is accelerated by V volts experiences a force F , when it enters in a uniform magnetic field. What will the force experienced when it is accelerated by 2 V ?
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The correct answer is:
$\sqrt{2} F$
When $\mathrm{e}^{-}$is accelerated by V volts, then work done by $\mathrm{e}^{-}$is equals to kinetic energy of $\mathrm{e}^{-}$.
or $\quad \mathrm{qV}=\frac{1}{2} \mathrm{mv}^2$
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 $\quad \mathrm{F}=\mathrm{qk} \sqrt{\mathrm{V} B}$ ...(i)
When e accelerated by 2 V volts, then $\mathrm{v}_1=\mathrm{k} \sqrt{2 \mathrm{~V}}$.
$\therefore \quad \mathrm{F}_1=\mathrm{qv}_1 \mathrm{~B}=\mathrm{qk} \sqrt{2 \mathrm{VB}}$
Hence, $\quad \mathrm{F}_1=\sqrt{2} \mathrm{~F} \quad$ [from Eq. (i)]
or $\quad \mathrm{qV}=\frac{1}{2} \mathrm{mv}^2$
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 $\quad \mathrm{F}=\mathrm{qk} \sqrt{\mathrm{V} B}$ ...(i)
When e accelerated by 2 V volts, then $\mathrm{v}_1=\mathrm{k} \sqrt{2 \mathrm{~V}}$.
$\therefore \quad \mathrm{F}_1=\mathrm{qv}_1 \mathrm{~B}=\mathrm{qk} \sqrt{2 \mathrm{VB}}$
Hence, $\quad \mathrm{F}_1=\sqrt{2} \mathrm{~F} \quad$ [from Eq. (i)]
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