Considering the moments of the friction, we therefore obtain 
- llS) 
^ I ™ = Z„. nrdr \ 2w + ir ^ + r &''. . . (13) 
Qi—Q* L dr dr J 
—HIr d<a d*<o 
— -Tx = 2io + 4r — + r* — 
flip, -P, ) dr dr 
If we put 
2^(p,*—9, a ) 
which is a constant, since H is inversely proportional to r, 
solution of equation (14) is 
Since the angular velocity must be continuous for r = 9, and for 
r=Q s we obtain from the formulae (9), (10) and (16) 
Besides — must be continuous. This can be proved by considering 
dr 
a flat cylindrical space containing a part of the surface bounding 
the space occupied by the discharge. The flat space is so chosen 
that its two parallel and equal surfaces are also parallel to the said 
surface and are situated at an infinitely small distance from that 
surface on either side of it. The force exerted by the magnetic field 
on the ions in the flat space may be neglected, so there must be 
equilibrium between the forces of friction on its two parallel surfaces, 
whence we infer that there must be continuity of — and since to is 
