An electron of 40 eV energy is revolving in a circular path in a magnetic field of 9*10^-5T.Determine (i)speed of the electron (ii)radius of the circular path . ans is (i03.75*10^6m s^-1,(ii)23.7cm Share with your friends Share 14 Vara answered this (i)Kinetic energy of electron, E=12mv2Herem=mass of the electronv=Speed of the electronv=2Em =240×1.6×10-199.1×10-31 =3.75×106 m/s(ii)Centriepetal force =Force due to magnetic fieldmv2r=BqvHere,B=Strength of the magnetic fieldr=Radius of the circular pathHence,mv=Bqrr=mvBq =9.1×10-313.75×1069×10-51.6×10-19 =0.237 m =23.7 cm 15 View Full Answer