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 $\frac{dB}{dt}$(tesla/second). An electron of charge ‘q’. Placed at the point P on the periphery of the field experiences an acceleration $\frac{eR}{nm}\frac{dB}{dt}.$ Then ‘$n$’ is equal to ……….

 

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

$\int \overline{E}.\overline{dl}=\frac{d\phi }{dt}=-A\frac{dB}{dt}\Rightarrow E\left(2\pi R\right)=A\frac{dB}{dt}=-\pi R^2\frac{dB}{dt}\Rightarrow E=\frac{E}{2}\frac{dB}{dt}$

$\therefore$force on the electron. $F=Ee=-\frac{eR}{2}\frac{dB}{dt}\Rightarrow acceleration=\frac{1}{2}\frac{eR}{m}\frac{dB}{dt}$

As the field is increasing being directed inside the paper, hence there will be anticlockwise induced current (in order to oppose the cause) in the ring (assumed). Hence there will be a force towards left on the electron