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An electron is accelerated from rest through a potential difference $V$ . This electron experiences a force $F$ is a uniform magnetic field. On increasing the potential difference to $V^{1}$ , the force experienced by the electron in the same magnetic field becomes $6 F$ . Then the ratio $(V^{1} / V)$ is equal to

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Detailed Solution

$K = \frac{1}{2} m v^{2} = e V ∴ v = \sqrt{\frac{2 e V}{m}}$

Also, $F = q v B$
$\Rightarrow F = e (\sqrt{\frac{2 e V}{m}}) B \Rightarrow F \propto \sqrt{V}$
$∴ \frac{F}{6 F} = \sqrt{\frac{V_{1}}{V_{2}}}$ $= \sqrt{\frac{V}{V^{1}}}$
$∴ \frac{V^{1}}{V} = 36$ .

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