A proton and an alpha particle, initially at rest, are accelerated through the same potential difference. They then enter a region of uniform magnetic field perpendicular to their velocities, where they move in circular orbits. What is the ratio of the magnetic moment of the alpha particle's orbit to that of the proton's orbit?

Step-by-step solution

The kinetic energy of a particle accelerated through a potential difference V is K = qV . The magnetic moment of a charged particle moving in a circular orbit is given by M = IA . The time period of revolution is T = 2 m qB , so the equivalent current is I = q T = q^2 B 2 m . The radius of the circular path is r = 2mK qB , so the area of the orbit is A = r^2 = ( 2mK qB )^2 = 2 m K q^2 B^2 . Therefore, the magnetic moment is M = I A = ( q^2 B 2 m ) ( 2 m K q^2 B^2 ) = K B = qV B . For a proton, the charge is q , so