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An electron accelerated through a potential difference ‘V’ passes through a uniform transverse magnetic field and experience a force F. If the accelerating potential is increased to 2V, the electron in the same magnetic field will experience a force.

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$\sqrt{2} F$

$\frac{F}{2}$

answer is C.

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Detailed Solution

$F = Bqv; F = Bq \sqrt{\frac{2 qV}{m}} (∵ qV = \frac{1}{2} {mv}^{2})$
$\Rightarrow Fα \sqrt{v} \frac{F_{1}}{F_{2}} = \sqrt{\frac{V_{1}}{V_{2}}} = \sqrt{\frac{V}{2 V}} \Rightarrow F_{2} = \sqrt{2} F$

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