10 BELL SYSTEM TECHNICAL JOURNAL 



In the interior of the conductor let us assume a curve 5 defined 

 as parallel, at every point, to the direction of the resultant current. 

 We do not know precisely the path of this curve but we do know that 

 such a curve can be drawn. In the case of wires of uniform cross 

 section it will be approximately parallel to the axis of the wire. Let 

 the cross section of the conductor normal to 5 be denoted by 6". The 

 total current 7s, parallel to s, is then given by 



Is = I = J UsdS. 



Now corresponding to the surface 5 and its element dS, let us define a 

 hypothetical surface X) and Its element da by the equation 



UadS = Ida-, 

 whence 



fusdS = I = if da = J-L, 

 so that XI is always unity. Now multiply the equation 



-Us = Es° -~^ -io:As (9) 



g ds 



by da and integrate over the cross section X 5 we get 



j —da = I Eda — iw I A^da — — I ^da. 

 This can be written as 



or simply 



r{s)I{s) = E(s) - iccAs(s) - ^Hs), 



OS 



rl = E - icoA - ^^^ (10) 



ds 



r is simply the resistance per unit length of the conductor, since 



Cus^ 

 rP = I -^dS = dissipation per unit length due to current Is, 



while E is the mean impressed electric force, parallel to s, averaged 

 over the surface X- 



Now consider the term ioiA ; we have 



A = \ Asda = j da j ^-ydv, u' = uil - r/c) 



