An α−particle and a proton are moving in circular paths in the same uniform magnetic field with same kinetic energy. Then the ratio of magnetic moment due to the rotation of α−particle to that of proton is

(4)12 mv2=K.E.=E   ⇒ v=2EmAngular  velocity ω=QBm∴   r=Radius  of  circular  path=vω=mQB.  2Em = 2mE1QBi = Equivalent current due to the moving charge.=Qω2π=Q2B2πm∴  M=Magnetic  dipole  moment  =πr2 ×  i=π×2EmQ2B2×  Q2B2πm⇒   M=EB∴      MαMρ=1