An electron accelerated through potential difference $V$ passes through a uniform transverse magnetic field…
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
$\frac{F}{2}$
$3 F$
F
$\sqrt{2} F$
Solution
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:
$F^{\prime} \propto \sqrt{2 V}$
$\therefore \frac{F^{\prime}}{F}=\sqrt{2}$