An electron accelerated through potential difference \(V\) passes through a uniform transverse magnetic field and experiences a force \(F\). If the accelerating potential is increased to \(2 \mathrm{~V}\), the electron in the same magnetic field will experience a force

The magnetic force on charged particle \((m, q)\) passing through a uniform transverse magnetic field. \(B\) is given by,
\(F=B q v\)
The kinetic energy gained during acceleration through a potential difference \(V\) is,
\(\begin{aligned} & \frac{1}{2} m v^2=e V \\ & \Rightarrow v=\sqrt{\frac{2 e V}{m}}\end{aligned}\)
Therefore, the magnetic force on the charged particle is directly proportional to the square-root of the potential difference applied during acceleration phase:
\(F \propto \sqrt{V}\)
Now, if the potential difference used to accelerate the electron is doubled:
\(F^{\prime} \propto \sqrt{2 V}\)
\(\therefore \frac{F^{\prime}}{F}=\sqrt{2}\)