Electron of mass m and charge q is travelling with a speed v along a circular path of radius r at right angles to a uniform magnetic field of intensity B . If the speed of the electron is doubled and the magnetic field is halved, then the resulting path would have a radius

Step-by-step solution

We know that, radius of charge particle of mass m having charge q moving with speed v in magnetic field perpendicular to it, r=m v / B q...(i) For second case, aligned r^ &= m v^ q B^ = m(2 v) q ( B 2 ) [ Since, v^ =2 v, B^ = B 2 ] &=4 m v q B =4 r [using Eq. (i)] aligned