An electron of 40 eV energy is revolving in a circular path in a magnetic field of 910^-5T.Determine (i)speed of the electron (ii)radius of the circular path .
ans is (i03.7510^6m s^-1,(ii)23.7cm

Question

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

Vara · Accepted Answer

(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

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

An electron of 40 eV energy is revolving in a circular path in a magnetic field of 910^-5T.Determine (i)speed of the electron (ii)radius of the circular path .
ans is (i03.7510^6m s^-1,(ii)23.7cm