Self-induction of Wires, 283 



may be treated as linear conductors, subject to the equations 

 (71), with no corrections, except under extreme circumstances. 

 The L may be taken as L , except in the case of iron. 



I will now give the S.H. solution in the general case, 

 subject to (58). Let there be any distribution of e (longitu- 

 dinal, and of uniform intensity over cross sections) . Expand 

 it in the Fourier series appropriate to the terminal conditions 

 at z—Q and I. For definiteness, let wire and return be joined 

 direct, without any terminal resistances. Then, e sin nt being 

 e at distance z 3 the proper expansion is 



#o = <?oo + 0oi cos m \ z + ^02 cos ?n 2 z + . . . , 



where m x = 7r/Z, m 2 = 2tt[1 3 &c. [It should be remembered that 

 e is the e 1 —e 2 of (54) and (53). Shifting impressed force 

 from the wire to the return, with a simultaneous reversal of 

 its direction, makes no difference in e. Thus two ^s directed 

 the same way in space, of equal amounts, and in the same 

 plane z = constant, one in the inner, the other in the outer 

 conductor, cancel. This will clearly become departed from 

 as the distance of the return from the wire is increased.] Then, 

 in the equation 



e m = (R'm+ L»O w + (m 2 /Sp)C TO 



= E' TO (J m + (JJ m -m*/8n*)pC m , 



we know e m ; whilst B/ m and I/ m are constants. The com- 

 plete solution is obtained by adding together the separate 

 solutions for e 00 , <? 01 , &c, and is 



fl = l/ goosin(tt£ — fl ) 2 y e 0m sm(nt — m ) cos mz \ ,_„. 



I \ (W + L' 2 n 2 )* [B& + (L' w -m 2 /Sn 2 ) 2 n 2 ]* J A } 



where the summation includes all the m's, and 



tan m = (I/ m — m 2 /Sw 2 )w-H R' m . 



A practical case is, no impressed force anywhere except at 

 z=0, one end of the line, where it is V sinrc£. Then, 

 imagining it to be YJzi from 2 = to z = z^ and zero else- 

 where, and diminishing z x indefinitely, the expansion required 



Y 0i z 1 =(Y ll)(l + 2%co&iwz/l), 



j going from 1, 2, . . . to co . This makes the current solu- 

 tion become 



r- y ojT sin(nf— fl ) g y sin (nt-d m ) cos mz \ ( 

 I l(R' 2 + L'Vp {B' 2 m + (L' m -m 2 /Sn 2 )V}U ,V - } 



If the line is short, neglect the summation altogether, unless 



U2 



