Inductance of Two Parallel Wires. 269 



A potential 



H = BK (tAr), (74) 



leavino- the first wire, reaches the second as 



H = BW'P) + B ^ OsJsQip) cos s4> 

 where 



6 S = 2i*K s {ihc). 



If the reflected and internal potentials be respectively 



H = c ~K Q (dip) + 2 X f«K s (tfy>) cos 5c/> ") 



r> • ( 75 ) 

 H = dj b (k'p) + 2, cZ,J (k'p) cos *<£ ) 



the surface conditions readily give 



c. = *B0Aj/&, d* = i,B0 t /hbp„ . . (76) 



where 



8 1 i = J,(Aa)j:(fci)-^J.(&)J/(ifco), . . (77) 



a s = K/(iAa)J s (£a) - -~ J,'(*a)K,(*/m), . (78) 



and (S 2 S , ft) are corresponding functions for the second wire. 

 The potential 



00 



H = c K (ihp)-\-X l CsK s (ihp) cos s<f> 



leaves the second wire. But, if the argument of the functions 

 K be the in all cases, 



K 1 {ihp)cos<j> = KA^O + ^^-OlK.+i + K.-ijJA/^cos^, 

 K 2 (J <i o)cos2(^=K 2 J ar) + 2l 1 (-0 s {K s+2 + K s _ 2 }J 6 .(^)cos^, 



and so on, 



Thus the potential becomes 



H = A J (/<r) + 2* A s J s (hr) cos &0, . . (79) 

 where 



A -C K + C 1 K 1 + C 2 K 2 + . .. 



A^-tCo^Ki-tdCKa+KoJ-ACKa+Ks)... 



A 2 = -C .2K 2 -C 1 (K,-i-K 1 )-C 2 (K 4 + K )... . (80) 



