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 due to unit current is 

 2 . AK B. Denoting the angles OAK and OBK by a. and /3 we 

 have for the magnetic potential 



U = 2.AK B = 2(2y + cL + {3). 



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 parallel to the infinite sides. 

 Now from the triangles OAK , 0BK , 



sin a = - sin (a + 7 — 0), sin ft = (jj sin (/3 + 7 4- 0). 

 Hence, expanding (Todhunter, Plane Trigonometry, Chapter 



XXI.), 



a = ^sin(7-</>) + -^sin2(7-(^)+-^sin3(7-^) + ..., 



/3 = ^sin( 7 + (/)) + -gsin2(7 + </)) + g^sin3(7 + 0)+.... 



Thus, for unit current in S, 



U = 2(2y+a + /3) 



= 2 [27 + p (Ui cos — v t sin 0) + J p 2 (u 2 cos 20 — v 2 sin 20) + . . .] 



-(8), 



where u n = sin n 7 (— + ^J , < = cosw 7 (- -^J . 



But the potential at if is equal to the potential at ^T > the pro- 

 jection of K. It then only remains to express (8) in terms of 

 r, 6, 0, the coordinates of K, by substituting for p its value 

 r sin 6. We thus obtain for the potential at (r, 0, 0) due to unit 

 current in S, 



U = 2 [27 + r sin 6 (u^ cos — v 1 sin 0) 



+ \r* sin 2 # (w 2 cos 20 — v 2 sin 20) + . . .] (9). 



We have thus obtained the expansion contemplated in (5) in 

 terms of the Sectorial Harmonics 



r n sin w 6 cos n0, r n sin w 6 sin ??0. 



Comparing the coefficients of r n in (5) and (9) we have 



2 



Y n = - sin' 1 (w M cos ?20 — v n sin n<f>) (10), 



except when n = 0. In this case, F = 47. 



