IN CYLINDRICAL CONDUCTORS, ETC. 
69 
origin and BA as the line of zero 0. Consider the field at the point C external to the 
cylinders situated on AB and due to two current systems flowing in the two cylinders 
parallel to the axes and symmetrical on either side of AB. By symmetry a n of equation 
(17) is zero; and since 6 is zero, we have from (10), using K„ instead of B„, for the first 
system of co-ordinates, 
TV co f t„ \2 n 1 
(34) 
Q = ^+S K„r~> 1 1 - ( 2 .i x . 
for the second system 
Q = U”+ 
/■ 
(35) 
Also, if the currents are equal and similarly directed in the two cylinders, 
K 'n = — K n ; 
Iv 'n — -f-Kw. 
if oppositely directed, 
Further, (34) may be divided into two portions, 
and 
O — y TV w 
bh - — - f ^TTi x» 
Q, = 2 K n r n ~\ 
1 
(36) 
(37) 
the former arising from causes inside the cylinder A, and the latter from causes outside A, 
which in this case are located inside B, and therefore Q 2 has also the value 
K'o “ V, a 
O — y tv' 
'%2 — f ■" n 
2 n 
V 
/n + 
l x»- 
Equating (37) and (38), 
Kb V, TX / a 
2 n 
2 K n r n ~ l = - - K' 
,/n + l 
x»- 
(38) 
(39) 
Putting K 'n — +K n, r' = D—r, expanding the right-hand side of (39) in 
ascending powers of r, and equating coefficients of r n ~ l , a series of equations are 
obtained to' determine K„ in terms of K 0 . Now if I is the total current in either wire, 
K 0 = 21, so that the method yields the values of K„ completely. Thus, in the case where 
the currents are similarly directed, K f n = — Ivn; and on equating coefficients we 
find 
K pi = — ’2\fx + (K]Cs) A (K 2 gt) yu"^ 2 + ... 
K 2 a 2 = — 2l/* 3 +2 (K x a) Abr + 3 (K,« 2 ) / a %+ ... 
K 3 a 3 = — 2l / u :! + 3 (K x a) fdxi + 6 (K/r) nX 2 + ■ 
in which p = a/D and is less than 1/2. 
