98 Dr. \Y. F. G. Swann on 



Writing l n for the integral, we find for the potential due to 

 the whole sphere of radius a, 



After some rather cumbersome work involving the usual 

 processes, we obtain I n , and finaliy fi, which, expressed to 

 two terms of a series, is 



a=F(a,)|o-194P 1 (^^+0-123P 3 (^)J+ ...}. 



The horizontal field is 



r^6 • '\ r 6 (16 r° cW J 



At the equator we have 



HccaF(a>). 



(4) The magnetic field due to the rotation of a sphere in which 

 each atom is arranged with its doublet axis j^rpendicular 

 to the axis of rotation. 

 Consider any ring perpendicular to the axis, lying on the 

 sphere of radius y. Let us find the magnetic field due to a 

 set of doublets rotating with this ring, their axes being all 

 perpendicular to the axis of rotation. Let h be the doublet 

 separation, N the number of charge elements per centimetre 

 on the ring containing the inner members of the doublets. 

 Let .?':=// sin a be the radius of the inner ring, and 8H the 

 magnetic potential at the point r . 6 due to the rotation of the 

 inner ring. 

 Then 



8n = 27rNew sin a . w sin 2 aT - KrJ~* . P»'(a) . P»(0). 



Let dashed letters refer to the (niter ring, so that 

 Sn' = 2TrNW sin a', to sin- (a + d*)*?? l r.^i 



where y 4- dy and a-\-dx are written in places tor //and*' 

 respectively. It' AH is the magnetic potential due to the 

 rotation of these two rings, we have, Bince 



N'//' sin «'=Ny sin a, 



An=2*N«*«..rin«|^{8in'«2*- | , / .I" «).P, <>)} 



