II
99
INDUCTION IN ROTATING SPHERES
From these equations and from the equations determining
Xo (equations (6) and (7), p. 94) we get x. in the general
case, and therefore in our particular case we have
Xe=
4πθη(n+1)
2η + 1 + 4πθη
i Ρ
iw
-A (R)". (F,(R) sin io + F(R) cos is} P₂-
k n+1R
and
If now by 'we mean that current-function which
Xe together would produce in the unmagnetised sphere, then
in the magnetic mass they will produce the current-function
y'. = (1+4π0)¥\",
and the condition for the stationary state becomes
4 = 40+4'..
is to be formed in exactly the same way as before, so that
also is known. If we substitute the values of Y, Yo, yo
in the last equation and equate coefficients of cos iw and
sin iw, we get for fi, f2, F1, and F, these equations
f₁(p)=
(2n+1)(1+4π0) _ wi 4πơn(1 +4π0)Ƒ₂(R)
2n+1+ 4πθη k 2n+1+4π0n
wi
+
(1+4π0)F₂(p),
ως 4πθη(1 + 4πθ)F, (Ε)
wi
-
[(1+4π0)F,(p).
k
k
If we put
wi
ƒ₂(P) = T 2n+1+4π0n
F(R)
Απίω(1 + 4πθ) = με,
ƒ₁(p) = $₁(µp) = $1(0),
ƒ₂(p) = $₂(µp) = $₂(0),
then 1,, are given by precisely the same differential equa-
tions as before (p. 71). Since we are dealing with a solid