
Radiation from an Alternating Circular Electric Current. 337 
g’ denoting the longitude, where a is still an integer but « is 
unrestricted, it is readily seen that we obtain 
a=—xaR-! cos 0{C, sin (o¢ + «Vé—KR) —(—)°Cy sin(—o6+«Vi—«KR) } 
+7 
( cos cos (ow + «a sin 6 sin y)dw, 
with corr eae changes in the formule for @ and y¥. 
Now 
cos (w sin Wr) =Jo(wv) + 2Jo(w) cos 2+ 2d 4 (wv) cos drt .. 
i sin (w sin y) =2J, (2) sin + 2J33() sindy... 
hence 
cos (ow + x sin W) =cos o {J (x) +2) o(w) cos 2p + 2S ,(v) cos 4+... } 
— sin ow {2J,(x) sin w+ 2J3(2) sin 3+... }. 
Multiplying by cos and integrating, we obtain 
| cos W cos (oy +a sin p)dp=(—)7 r{J,1(v) +5e41(2)} 
=(— ere 2or ay 
ae 
Also 
+3 ke 
d 
( sin Y sin (oW+a sin y)dw= — == cos (ow+z2 sin W)dy, 
and from the same equations this is readily seen to be 
(=) "20rd (2). 
If now in the expression for the current we write C;=C, 
=(/2, we see that the magnetic force due to the current 
C cos cd! cos xVt is given by “the equations 
==2Crr«aR cos 6 sin ete cos (Vt—R) . oJ. (asin 0)/«a sin 6, 
if o even ; 
or ==—2CK«aR—! cos 6 cos o¢ sin x(Vt—R). od (xa sin 8) /easin 8, 
if o odd; 
B2=2CrKaR-! cos 8 cos of cos «(Vt—R) .J,,(«a sin 8), if o even ; 
or +2Um«aR— cos 6 sin o¢ sin e(Vt=R).Jf(xasin 0), if o odd; 
y being obtainable from 8 by changing the factor cos 6 into 
— sin 0. 
