68
IJ
INDUCTION IN ROTATING SPHERES
V₁
2π wax / R²
p²
=
n+ 1x Ow₁ 2n+1
y
2n+3
27.2n+3
(2n+1)(2n+3)p²n+1
2π wax R2
ρ
=
n+1 xow, 2n+1
2n+3
2n+1
W₁
2p2n+3
(2n+1)(2n+3)p²n.
These values are got by a simple integration; for u, v, w are
products of p" by spherical surface harmonics. The potential
of each infinitely thin layer is known inside and outside it,
and an integration with respect to p leads to the given values.
Hence follow the currents of the second induction
2π
Ղաջ
(@)² dx" /_ R²
=
2r2n+3
n+1 K wx 2n+1 2n+3 (2n+1)(2n+3)p2n+1
2π (w)²αx"
R2
V2
n+1 x dw 2n+1
მა
y
p²
22n+8
2n+3 (2n+1)(2n+3)p²
2n+1
2π W
R2
22n+3
W2
-
=
General
solution.
n+1 K Ow, 2n+1 2n+3 (2n+1)(2n+3)p²n+¹
In this way the calculation may be continued as far as
may be desired, but the results continually increase in compli-
cation; hence we now proceed to the exact solution of the
problem.
We have seen that the currents are always perpendicular
to the radius, and may therefore again make use of the
current-function.
Let f (p)ƒ be any function of p whatever, and let
=
y = p.f.Xn
be the current-function of a system of currents flowing in the
sphere.
The current-densities are
=
όχι
f
Jwz
v =
= f
w =
>
Jwy
-ཚ".
Let F (p) = F be a second function of p, which is given in
terms of ƒ by the equation
P
4π 1
F(p)=
2n+2
2n+1 p2n+1
2ƒ (a)da+
P
R
[p²=+³af(a)da}.