A uniform magnetic field of induction `B` is confined to a cyclindrical region of radius `R`. The magnetic field is increasing at a constant rate of `dB//dt` (tesla`//` second). A charge `e` of mass `m`, placed at the point `P` on the periphery of the fixed experiences an acceleration :

Options

A. `1/2(eR)/m(dB)/(dt)` towards left
B. `1/2(eR)/m(eB)/(dt)` towards right
C. `(eR)/m(dB)/(dt)` towards left
D. zero

Answer

A

Explanation

If we consider the cylindrical surface to be a ring of radius R, there will be an induced electric field due to time varying magnetic field.
`intvecE.vec(dl)=-(d phi)/(dt)=(AdB)/(dt)`
`impliesE(2piR)=A(dB)/(dt)=piR^(2)(dB)/(dt)`
`E=R/2(dB)/(dt)`
So force on the electron
`F=-Ee=-(eR)/2(dB)/(dt)`
Acceleration `=1/2(eR)/m(dB)/(dt)`
Force will be towards left on electron.