12 Mr. 0. Heaviside on the 



that the magnetic force must be tangential to the boundary, 

 and therefore have no normal component ; or, if N be the unit 

 vector normal drawn outward, 



-FN = hN (2a) 



is the boundary condition. This gives F, when it is remem- 

 bered that F must have no convergence within the wire. 



In another form, since we have h circular about the axis, 

 and of intensity 2iTrT , at distance r from it, the current- 

 density being T ; or 



h = 27rr Vkr, (3a) 



if r is the vector distance from the axis in a plane perpendi- 

 cular to it, and k a unit vector parallel to the current ; we 

 have 



hN = (2<7rr )(NVkr) = (27rF )(rVNk) 



if s be length measured along the bounding curve, in the 

 direction of the magnetic force. The boundary condition (2 a) 

 therefore becomes, in terms of the magnetic potential, 



__dn_ p d(r 2 ) 

 dp Y ° ds 



which, with V 2 ^ = 0, finds the magnetic potential. Here 

 p 1 is length measured along the normal to the boundary 

 outward. 



Or we may use the vector-potential A. It is parallel to 

 the current, and consists of two parts ; thus, 



A=A'-(/x7rI> 2 )k, (6a) 



where the second part on the right side is, except as regards 

 a constant, what it would be if the boundary were circular, its 

 curl being ph.. To find A', let its tensor be A'; then 



V 2 A' = 0, and A^/wiy, .... (7a) 



the latter being the boundary condition, expressing that A is 

 zero at the boundary. Comparing with (5a), we see that 

 (7a) is the simpler. 



The magnetic energy per unit length of rod, say T, is 



T=2^H 2 /87r = 2^(h + F) 2 /87r, . . . (8a) 



the summation extending over the section. But 2FH = 0, 

 because F is polar and H is closed ; so that 



T = 2/ih 2 /87r-S / u<F 2 /87r 

 -^h787r + 2/*hF/8/x (9a) 



= *r ^, (5a) 



