62 MR. S. BUTTERWORTH ON EDDY-CURRENT LOSSES 
into the cylinder through a small area da about the point a, 6 is eqda/^ir by Poynting’s 
theorem, and the integral rate of flow into unit length of the cylinder is 
~\ .(12) 
47T Jo 
Now, by (9) and (10), e and q have the forms cos [n6 -f- a n ) and cos (nd -f a n ) 
respectively. 
Using these forms in (12), and remembering that 
cos ( nO + a n ) cos (md + a n ) dS = 0, n ^ m, 
Jo 
the integral rate of flow becomes 
j(2 u 0 v 0 + Zu„v„) .(13) 
To determine the products u n v n , the obvious method is to determine the real parts 
of the complex coefficients in (9) and (10). The product u n v n expressed as a function 
of time would then be of the form w n cos wt cos (wt + <p H ), so that the rate of energy 
dissipation in the cylinder would be \w n cos <p n . 
A better method is to make use of the method of conjugates. If U', V' are the 
conjugates* of two complex quantities U, V; u, v their real parts, then 
M = i(U + U'), u = HV+Y'), 
so that 
uv = i{TJ + TJ'){V+Y'). 
Further, if U, V rotate with time in the same sense, U', V' will rotate in the opposite 
sense, so that UV' and VU' will not rotate. The steady portion of uv is thus 
i(UV' + U'V). 
Applying this to the present case, the steady flow into unit length of the cylinder 
is from (9), (10) and (13) 
w = i{B 0 B',( Xl ,- x '„)+i. i « !, *B.B'.(x.-x',)| 1 ' 
in which as before the accents denote conjugates. 
Now is a function of \a and A 2 = -j-4 ttMo, so that, putting 
z 2 = A-akwa 3 ,.( 14 ) 
X n may be written 
Xn = (pn{z)~i^n{z) .. (15) 
from which 
i {Xn — Xn) = 2^ n (2), 
* If U = Ac‘^, then U' = is the conjugate of U. 
