Self-induction of Wires. 281 



back upon (57), or (22), and the normal systems (18) of 

 Part I. 



Now, as regards our obtaining the expansions of R' x &c. in 

 powers of p 2 , we have to expand the numerators and the 

 denominators of R/' and R/ in powers of p y perform the 

 divisions, and then separate into odd and even powers. 

 When the wire is solid, the division is merely of J^J (^) by 

 Ji(#), a comparatively easy matter. The solid wire B/ and 

 1/ expansions were given by Lord Rayleigh (Phil. Mag. 

 May 1886). I should mention that my abbreviated notation 

 was suggested by his. But in the tubular case, the work is 

 very heavy, so, on account of possible mistakes, I go only as 

 far as p 2 , or three terms in the quotient. The work does not 

 need to be done separately for the inner and the outer tube, 

 as a simple change converts one B/ or 1/ into the other. 

 Thus, in the case of the inner tube,- we shall have 



R x =Ea[l + n (^Traf) j A -§ ? + /^ - a»(a»-«;) 



a i( a i a l) ' J 

 L/=E 1 (M 1 ^){i-f| + 21og|.^ y } ) . (73) 



where n 2 is written for — p 2 , for the 8.H. application. 



As for L/, it is simply the inductance of the tube per unit 

 length (of the tube only), as may be at once verified by the 

 square of force method. The first correction depends upon 

 p' d . But R/ gives us the first correction to R 1; which is the 

 steady-flow resistance, so it is of some use. To obtain R 2 ' 

 and L 2 ' from these, change R x to R 2 , ^ and h x to /ul 3 and k 3 , 

 a to a 3 , and a x to a 2 . Or, more simply, (72) and (73) being 

 the tube -formulas when the return is outside it, if we simply 

 exchange a and a } we shall get the formulas for the same 

 tube when the return is inside it. 



If the tube is thin, there is little change made by thus 

 shifting the locality of the return. But if aja be large, 

 there is a large change. This will be readily understood by 

 considering the case of a wire whose return is outside it, and 

 of great bulk. Although the steady resistance of the return 

 may be very low, yet the percentage correction will be very 

 large, compared with that for the wire. 



Taking a 1 /a = 2 only, we shall find 



R 1 ' = R 1 [1 + (whafan) 2 x -012] 

 Phil Mag. 8. 5. Vol. 22. No. 136. Sept. 1886. U 



