In cylindrical conductors, etc. 
81 
By (49) the eddy-current loss in the cylinder is 
W = y (H 1 2 + H 2 2 +2H 1 H 2 cos 0 ) 
= y (a 2 I 1 2 + /3 2 I 2 2 +2a/3l 1 I 2 COS 0) .(55) 
in which 
y = iR 0 d 2 G (z). 
Now, instead of the cylinder being present, suppose the two circuits carrying the 
s 
currents I 15 I 2 to be linked by the resistance system shown in fig. 8. The rate of 
dissipation of energy by this system is 
W' = i {R 1 I 1 2 +RJ 2 2 +R 3 (I 1 -I 2 ) 2 } 
= i(R 1 +R 3 )I 1 2 +i(R 2 +R3)I 2 2 -R 3 I 1 I 2 ,.(56) 
W' is identical with W if the resistances have the values 
B] = 2a (a + /3 COS 0) y, R 2 = 2/3 (/3 -f a COS 0) y 
R 3 = — 2 (a/3 cos 0 ) y 
In the case Ijl = I 2 , no current flows through R 3 , so that the potential differences 
produced by the eddy losses in the cylinder are then such that they may be represented 
by the resistances R 15 R 2 in series with the respective coils. 
Applying to a single layer system, suppose we require the resistance to be added to 
a wire s which would represent the contribution of s to the eddy losses in the whole 
system. Let the wire s be the coil (I), another wire r be the cylinder, and the 
remaining wires be the coil (II). al is the field acting on r due to the current in 
s, /3l is the field acting on r due to the currents in the remaining wires, so that 
(a-f-/3 cos 0 ) I is the nett field acting on r resolved in the direction of al. Since in 
. the single layer system the fields due to individual wires are collinear, (a-j -/3 cos 0 ) I 
is the total field in which the wire r is situated. It is to be regarded as positive 
when the field due to s is the same sense as the total field. 
VOL. ccxxii.—A. 
N 
