70 
MR. S. BUTTER WORTH ON EDDY-CURRENT LOSSES 
Solving by successive approximations to the order (W 4 , 
Kl« — — 2I/X {l+^Xi + yW 4 (xY + Xs) + •••} 
K 2 a 2 = -2 I m -( 1+2 m 2 Xi +...) 
Kyr = — 2l / u s (1 + ...) 
( 40 ) 
In the expression for the eddy-current losses (18) the moduli of the complex quantities 
K 15 K 2j K 3 are required, which, from (40) with x?l = <p n — i\ts n , are 
(Kj)' — — {1 + 2/U0, + m‘ (20 2 + 30," — \fs{) + 
(K 2 )- = ^-(1+4^'^!+...) 
( K 3 F = j |(1 + ...) 
(41) 
Substituting in (18), the energy dissipation per unit length in either cylinder is 
given by 
W = 0)1" ~ 11 + ( 20, + ^ y^) + /W 4 (202 + 3 /y + 30," — \\r{ + ) j 
When the currents are in opposite directions a similar treatment gives 
W = ml 2 [| +i^. + W, I 1 +/(3^-* a -2^-^*+i|5 /J J 
In (42) and (43) the term 
0 ) 1 " (— + 1^2 
• (42) 
• (43) 
is due to the ordinary skin effect. Since 2o ojz 2 = iR 0 , this term may be written 
|Tt 0 I 2 {1+F (z)} in which F (z) {= p0- 2 } is plotted in fig. 3 up to 2 — 5. When z is 
greater than 5, then, by (21c), 
F(s) = (v/22-3)/4.(44) 
This is shown in fig. 3 by the broken line A. 
The next term, /PA,, is due to the proximity of the two cylinders when they are so 
far apart that the quantity in { } may be regarded as practically unity. It would 
have been the term obtained by assuming the current in the second cylinder as 
concentrated on the axis and producing a uniform field of strength 2 1/D on the first 
cylinder. Including the proximity effect to this order we may write 
