418 



a> = 0. In the foriner case Hiis will not be Irne. Let us suppose 

 that the electron goes through the ring once, in the positive direction, 

 and let A and B be two positions, before and after the enconntei-, 

 both far away from the ring. Then, whatever be these positions, 

 provided only that they do not coincide, we can choose the surface 

 a in such a way that it is not intersected by the path of the 

 particle from A to B, and that to = at the point A. It is easily 

 seen that then the final value will be tu = 4.t. 

 Bij integration of (8) one finds 



a k e 



» = », + -CU, ........ (9) 



4:7tcQ 



if «3', is the angular velocity which the ring may have had befoie 

 the encounter. 



^ 6. We have next to consider the motion of the electron. The 

 rotation of the ring constitutes a magnetic current 



i — ak&. . . . . . . . . (10) 



giving rise to an electric field that is easily determined if we sup- 

 pose it not to differ appreciably from the field that would exist if 

 i were constant. The calculation, exactly similar to that of the 

 magnetic field due to an electric cui-rent (the vector potential b is 

 first determined and then d =: — rot b) leads to the result 



i do) i da> i öa> 



AjtcOa; 4 ;7r c oy 4 rr c Oz 



from which, combined with (10) and (9), we can deduce that the 

 force <?d acting on the electron depends on a potential 



a Ice a^k^ e* 



If we wanted exactly to determine the motion we should also 

 have to take into account the force with which, owing to its velo- 

 city, the electron is acted on by the magnetic field that is due to 

 the ring and to stationary magnetic charges eventually existing in 

 the atom, and so the problem would become very difficult. Since, 

 however, the latter force does no work, we can write down the 

 equation of energy 



i m v' = ^ m v/ — t|?, (13) 



(Vj the initial velocity at a point where to = 0) and this is sufficient 

 for some interesting conclusions. 



