An electron accelerated through a potential difference $V$ enters a uniform transverse magnetic field and experiences 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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$F=B q v$
But $\frac{1}{2} m v^{2}=e V$ or $v=\sqrt{\frac{2 e V}{m}}$
$\therefore F=B q \sqrt{\frac{2e v}{m}}$
$\Rightarrow F \propto \sqrt{V}$ and
$F \propto \sqrt{2 V}$
$\frac{F^{\prime}}{F}=\sqrt{2}$ or $F^{\prime}=\sqrt{2} F$
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