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(D) The magnetic force $F$ on a charged particle of mass $m$ and charge $q$ moving with velocity $v$ in a uniform magnetic field $B$ is given by $F =

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$\sqrt{2}F$

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(D) The magnetic force $F$ on a charged particle of mass $m$ and charge $q$ moving with velocity $v$ in a uniform magnetic field $B$ is given by $F = Bqv$. When an electron is accelerated through a potential difference $V$,its kinetic energy is given by $\frac{1}{2}mv^2 = eV$. From this,the velocity $v$ is $v = \sqrt{\frac{2eV}{m}}$. Substituting this into the force equation,we get $F = B e \sqrt{\frac{2eV}{m}} = B e \sqrt{\frac{2e}{m}} \sqrt{V}$. This shows that $F \propto \sqrt{V}$. If the potential difference is increased to $2V$,the new force $F'$ will be $F' \propto \sqrt{2V}$. Therefore,$\frac{F'}{F} = \frac{\sqrt{2V}}{\sqrt{V}} = \sqrt{2}$. Thus,$F' = \sqrt{2}F$.

An electron accelerated through a potential difference $V$ passes through 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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