402 Mr Searle, On the Coefficient of Mutual Induction, etc. 
appear to circulate in a direction contrary to that of the hands of 
a watch, then the magnetic potential at K 0 due to unit current is 
2 . AK 0 B. Denoting the angles 0AK Q and OBK 0 by a and ft we 
have for the magnetic potential 
U 0 =2.AK 0 B = 2(2y + a + ft). 
On account of the infinite length of the two parallel sides of S, 
the potential has the same value at all points on a straight line 
through K 0 parallel to the infinite sides. 
Now from the triangles OAK 0 , OBK 0 , 
sin a = - sin (a + 7 — <£), sin /5 = ? sin (ft + 7 4- </>). 
a 0 
Hence, expanding (Todhunter, Plane Trigonometry, Chapter 
XXI.), 
a = ^ sin (?-<*>) + “ sin 2 ( 7 - <£) +| £- s sin 3 (y- tf>) + ... , 
j 2 1 ^ 
0 = 1 sin (7 + 4>) + 2 p sin 2 (7 + <£) + g p sin 3 (7 + <f>) + . . . . 
Thus, for unit current in S, 
U 0 = 2(2y+ a + ft) 
= 2 [2<y + p (u-l cos cf) — Vi sin </>) + J P 2 (^2 cos 2</> — v 2 sin 2 <£) + . . .] 
( 8 ), 
where u n = sin ray f + -t) , v n = cosny(~ n -~f 
But the potential at K is equal to the potential at K 0 , the pro- 
jection of K. It then only remains to express (8) in terms of 
r, 6, (f>, the coordinates of K, by substituting for p its value 
r sin 6. We thus obtain for the potential at (r, 6, </>) due to unit 
current in S, 
U = 2 \2<y + r sin 6 (u x cos <f> — v x sin <f>) 
+ sin- 0 ( u 2 cos 2(f) — v 2 sin 2 <£) + . . .] (9). 
We have thus obtained the expansion contemplated in (5) in 
terms of the Sectorial Harmonics 
r n sin n 0 cos ncf), r n sin w 0 sin ncf>. 
Comparing the coefficients of r n in (5) and (9) we have 
2 
F n = - sin n 0 (u n cos n(j> — v n sin n(f>) 
n 
except when n = 0. In this case, F 0 = 4y. 
( 10 ), 
