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(D) The kinetic energy $K$ of an electron accelerated by a potential $V$ is given by $K = eV = \frac{1}{2}mv^2$. Thus,the velocity $v$ is proportional

Vedclass · Accepted Answer

$1: 4$

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(D) The kinetic energy $K$ of an electron accelerated by a potential $V$ is given by $K = eV = \frac{1}{2}mv^2$. Thus,the velocity $v$ is proportional to $\sqrt{V}$.The magnetic force on a charged particle moving in a magnetic field is $F = qvB \sin 	heta$. Assuming the angle $	heta$ remains constant,$F \propto v$.Since $v \propto \sqrt{V}$,it follows that $F \propto \sqrt{V}$.Given $F_{1} \propto \sqrt{V_{1}}$ and $F_{2} = 2F_{1} \propto \sqrt{V_{2}}$.Dividing the two expressions: $\frac{F_{2}}{F_{1}} = \frac{\sqrt{V_{2}}}{\sqrt{V_{1}}} = 2$.Squaring both sides: $\frac{V_{2}}{V_{1}} = 4$,which implies $\frac{V_{1}}{V_{2}} = \frac{1}{4}$.

An electron at rest is accelerated by a potential $V_{1}$ and then enters a uniform magnetic field,experiencing a force $F_{1}$. When the potential is changed to $V_{2}$,the force experienced by the electron becomes $2F_{1}$. The ratio of $V_{1}$ to $V_{2}$ is:

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