246 



Mr. S. H. Burbury on the 



hasfthe same value at all points on that circle. Let (j> be the 

 angle between the plane SPS X and a fixed plane through SS'. 

 Then we may express an element of volume of the ellipsoidal 

 shell at P as follows : — 



Fig. 1. 



In the figure let SP, SP' be tw T o neighbouring values of 

 r, making the angle PSP' = d0 with each other, and let PP' 

 be the corresponding elementary arc of the inner ellipse. 

 Let SQ, SQ' be the corresponding values of r, and QQ / the 

 corresponding arc for the outer ellipse. If P'R be drawn at 

 right angles to ISP, P'R = rd0. An element of volume 

 of the ellipsoidal shell is the area PQQ'P' multiplied by 

 rsiuOdtj). But the area PQQ'P' is equal to rdQ multiplied 

 by PQ, that is by vdt + ti cos 0dt. We may therefore take 

 for an element of volume of the ellipsoidal shell 



r* sin dd d<j> vdtfl + "cos 0\=dco. 



8. I shall now make the hypothesis that -, the eccentricity 

 of the ellipse, is very small, so that its square and higher 

 powers maybe neglected. Let us write for brevity - =k. 



9. We have now to find the electric force at b 7 due to the 

 change of magnetic force H in the element of volume at P. 

 To this end let us replace H at P by a small circular electric 

 current. Suppose about P in the plane SPS' a circle de- 

 scribed of very small radius a, and in that circle an electric 

 current i, such that the magnetic force due to it at P is H. 



Then - has the dimensions of H, and dri has the dimensions 

 a 



of Hdco. We may therefore write, since a is an arbitrary 



radius subject to a numerical factor, 



aH = B.r 2 sin dd d<f> vdt(l + k cos 6) , 

 and ^ ^jj 



a 2 -r = — r i 3, sin 6 dd d<p vdt (1 + k cos 6) . 



