16 Mr. 0. Heaviside on the 



T= JLC 2 , which (25a) becomes when t=0 ; this gives 



L=4tt/*22 —*- - -. (26a) 



(ma) 2 (n&) 2 j ^) 2 +^W 2 } 



The lines of magnetic current are also the lines of equal 

 electric-current density. That is, a line drawn in the plane x, y 

 through the points where T has the same value is a line of 

 magnetic current. For, if s he any line in the plane x, y, 



-j-= component of /^H perpendicular to s, 



so that H is parallel to s, when dE/ds=0. The transfer of 

 energy is, as usual, perpendicular to the lines of magnetic 

 force and electric force. 



The above expression (26a) for L may be summed up 

 either with respect to ma or to nb, but not to both, by any 

 way I know. Thus, writing it 



(ma) 2 {nbfa. , N2 b . ,, ' 



v 7 v y T (nb) 2 + -(maY 

 o\ / a v ' 



we may effect the second summation, with respect to nb, re- 

 garding ma as constant in every term. Use the identity 



l_ x e Ki-x)- e -Ki-x -) _ 2 cos(wive/2Z) 



"7? A»(e« + 6-W) "~2 ^(i7r/20 2 {(^7r/20 2 + A 2 [' 



where i has the values 1, 3, 5, &c. Take x=0, iir/2l=nb, 

 h = (b/a)(ma), 1=1, and apply to (27a), giving 



L= 4 ^ 2 ^ i ^ o^n 1 -f • ss • >) +> -t ( J / (28a) 



where the quantity in the \ } is the value of the second 2 in 

 (27a). The first part of (28a) is again easily summed up. 

 and the result is 



_ ail lc^ 1 e 2 « (wa) — 1 ) t Q . 



in which summation, we may repeat, ma has the values 



i,^ JL } -rjr^ & c . The quantities a and & may be exchanged ; 

 2 2 2 



that is, ajb changed to b/a, without altering the value of L. 

 This follows by effecting the ma summation in (26a) instead 

 of the nb, as was done. 



