68 
MR. S. BUTTERWORTH ON EDDY-CURRENT LOSSES 
The energy dissipation due to a uniform field H is got by putting K : = H, 
K> = K 3 = ... = 0, giving 
W 2 = iH 2 aVlW2..(31) 
If the field is non-uniform, a comparison of (29) with (17) shows that the various 
harmonic terms in the field produce terms of equal importance in the expression for 
the eddy-current losses. 
The examples of the last section give, for the single thin wire, 
and for the pair of wires, 
a-D 2 
W = 
iy-a 4 
I 2 \/R 0 w/2 or 
lVR„»/2. .... (33) 
according as the currents flow in opposite or the same direction in the two wires. 
As regards (32), it is seen that the uniform field theory may be applied if we take as 
the uniform field the field at the point w r kere the tangent plane through the wire 
touches the surface of the cylinder. 
(B). Eddy-Current Losses in Two Parallel Cylinders Carrying 
Equal Currents. 
(7) If the field acting on the cylinder is due to currents in neighbouring cylinders, 
then, because of the distortion of the current distribution in these cylinders, the external 
field acting on the cylinder under consideration is itself variable with frequency, and 
the assumption that this field is that which would occur if all eddy-currents are absent 
will lead to wrong results. The case of two similar parallel cylinders carrying equal 
currents may be solved by considerations of symmetry. 
(8) Let the cylinders each have radius a, and let the distance of their centres be D. 
Take two systems of cylindrical co-ordinates (fig. 2), the first system (r, 0 A ) having 
A as origin and AB as the line of zero 0, and the second system (/, d B ) having B as 
