258 Dr. J. W. Nicholson on the 



The Single Wire. 



We consider first the distribution of vectors in and about 

 a single wire conveying an alternating current, whose dis- 

 tribution is undisturbed by the presence of a second wire. 

 All vectors are independent of orientation, and the impressed 

 electric force may be supposed equal at all points of the 

 wire, the resistivity (<r) of the latter being constant. Let 

 ~Ee tpi be the line integral of this force per unit length. Then 

 the total force, inductive and non-inductive, in the wires is 



^t 

 = E-cp AJ {kr), 



since H does not depend on 6. The time factor is ignored 

 throughout. 



The corresponding force outside is 



B=-^=i-ipBKo(i*r), 



and by continuity, if a be the radius of the wire, 



E — ip AJ (ka) = — ip BK {ilia). 



The tangential magnetic force (/3) at the surface is also 

 continuous, and therefore, if p. be the permeability of the 

 wire, 



-AkJ '(ka) = ihBk '(iha), 

 A 6 



we deduce that 



A= — - K '(iha)/(K '(iha)Jo(ka)--Jj- J '{ka) K {tha)). (15) 



up /X/it 



B= J ^-J 'M/(K '(Lha)J Q (ka) -~J '(ka)-K (i7ia)), (16) 



and the vector potential in the wire is 



H = AJ (*r), (17) 



and outside is 



H = BK o (J»0 (18) 



The current crossing any section is the line integral of 

 magnetic force round the section, divided by 47r, and becomes 



OT=*-j-27ra(/3) r=a 



= AkaJ / (ka)/2p, .... (19) 



