• # question_answer 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. Then, the radius of the circular path is: A) $\sqrt{(2\,eV/m)}$                 B) $\sqrt{(2Vm/e{{B}^{2}})}$C) $\sqrt{(2Vm/eB)}$                 D) $\sqrt{(2Vm/{{e}^{2}}B)}$

$Bev=\frac{m{{v}^{2}}}{r}$or $r=\frac{mv}{Be}$ As,$mv=\sqrt{2mT},$where $T=K.E.$ So, $r=\frac{\sqrt{2mT}}{Be}$ As the electron has been accelerated from rest through a potential difference of V volt, then$T=eV$ $r=\frac{\sqrt{2mVe}}{Be}=\sqrt{\frac{2mV}{{{B}^{2}}e}}$ Hence, the correction option is [b].