A uniform magnetic field of induction B is confined to a cylindrical region of radius R. The magnetic field is increasing at a constant rate of $\frac{\text{dB}}{\text{dt}}\left(\text{tesla/second}\right)$ . An electron of charge q, placed at the point P on the periphery of the field experiences an acceleration

Consider the cylindrical surface to be a ring of radius R, there will be an induced emf due to changing magnetic field.

${\displaystyle \int \overrightarrow{E}.\overrightarrow{dl}}=-\frac{\text{d}\varphi }{\text{dt}}=-\text{A}\frac{\text{dB}}{\text{dt}}$

$\text{E}\left(2\pi \text{R}\right)=-\text{A}\frac{\text{dB}}{\text{dt}}=-\pi {\text{R}}^{\text{2}}\frac{\text{dB}}{\text{dt}}$

$\text{E}=-\frac{\text{R}}{\text{2}}\frac{\text{dB}}{\text{dt}}$

Force on the electron, $\text{F}=-\text{Ee}$

$\text{F}=-\frac{\text{eR}}{\text{2}}\frac{\text{dB}}{\text{dt}}$

Acceleration $\text{a}=\frac{\text{eR}}{\text{2m}}\frac{\text{dB}}{\text{dt}}$

As field is increasing and is being directed into the paper current is induced in the ring in anticlock wise direction. Hence there will be force towards left on the electron