16
KINETIC ENERGY OF ELECTRICITY IN MOTION [1]
plane and consider it as part of an infinitely long system of
straight wires, whose thickness is the same as that of the
layer. For the position of any element is unchanged rela-
tively to its neighbours, and the action of distant portions on
each other is zero.
We shall first determine the self-inductance II of a single
layer. Let the length of the wires be S, their radius R, the
distance of two neighbouring ones q, their number n. Further,
let a be the self-inductance of a single wire, am the mutual in-
ductance of two wires distant mq; then we have
αo=
= 2S
2S
R
= 2S
2S
25 (log 3-4). 4, 28(log 25-1)
By counting we find
II = 2)a, —
2na-2(2n-1)a, + 2(2n- 2)α-... 2α2n-17
and introducing the values of the a's
II = 4Sn+ log
q 1
R η
-
-
12n-1.32n-3...(2n − 3)³. (2n − 1)
22n−2.42n−4…..(2n − 2)²
4Sn{} +log+log-
-
Hence the quotient II/S here has a perfectly determinate
value, which may be called the self-inductance per unit length.
The logarithm involved in the above expression cannot well
be directly calculated for large values of n; for such values
we must use an approximation. For this purpose we split up
the expression into
n log
123252...(2n − 3)²(2n − 1)
--
2242...(2n-2)²
22 44
66
+ log
·
(2n − 2)2n-2
-
1.3 3252 5373' ' ' (2n — 3)”¯¹(2n − 1)n−¹
-
the two parts of which will be evaluated separately. The first
may be written
1
n-1
=
nΣmlog
1
(2m − 1)(2m+1)
(2m)2
n-1
=
nmlog 1
-
1
+m²
1 [n is really the number of double wires, positive and negative counting as
one.-TR.]