PROFESSOR HELE-SHAW AND MR. ALFRED HAY 
322 
p ^ is the intensity-component in the direction of S.s due to one of the cylinders. 
Thus the problem of finding the magnetic potential function is reduced to that of 
determining the intensity due to a solid cylinder of attracting or repelling matter. The 
components of intensity due to this latter cylinder may be shown to be 
X = 
Y = 
47 rcibp 
/=s=— 
\f ar — O'¬ 
COS 9, 
sin 9. 
We are thus led to adopt tentatively the somewhat more general forms 
(A cosh u — B sinh u) cos 9 
and (C cosh -u — L) sinh u) sin 6 , 
where A, B, C, and 1 ) are constants, as the typical magnetic potential functions for 
magnetised elliptic cylinders and confocal elliptic cylindric shells. 
In the above expressions, A, B, C and D will all be different for the space included 
between two confocal bounding surfaces; A = B and C = I) for the external space 
extending to infinity. And B = 0, 0 = 0 for the space inside the innermost bounding 
surface, which includes the foci. 
If we assume the permeability to be constant, then we need only consider the two 
standard cases of magnetisation along the two axes of the ellipse (1), as any inter¬ 
mediate direction may he obtained by a proper superposition of the two standard 
cases. 
We shall in the first place consider the case of a solid cylinder whose bounding 
surface is the ellipse of reference (1), the impressed field H being along the major axis 
of the ellipse. 
If V„, V; and V„ stand for the impressed potential, the induced potential inside, 
and that outside the cylinder respectively, then 
Y 0 = — Hcc = — H v/o. 3 — lr cos 6 cosh u, 
and we assume 
V; = Ax = A v/— lr . cos 6 cosh V, 
V^Bv 7 ( i 2 — /A . < J ~" cos 9. 
In order that these functions—which satisfy Laplace's equation—may afford a 
solution of the problem, they must satisfy the conditions of (f) continuity of potential 
and (2) continuity of normal induction- -two conditions which enable us to determine 
the constants A and B. 
The first condition (when u = tanh 1 -) gives 
aA — (a — b)B = 0 
