An electron of mass `'m'` is accelerated through a potential difference of `V` and then it enters a magnetic field of induction`B`. Normal to the lines of force. Then the radius of the circular path is

Options

A. `sqrt(mv)/(cB)`
B. `1/Bsqrt(2mV)/e`
C. `1/Bsqrt(mV)/e`
D. `sqrt(mV)/(2qB^(2))`

Answer

B

Explanation

`1/2mv^(2)=eVtherefore v=sqrt(2eV)/m` and the radius of the circular path `r=(mv)/(eB)=m/(eB)sqrt(2eV)/m=1/Bsqrt(2eVm^(2))/(e^(2)m)=1/Bsqrt(2mV)/e`.