of Electric Oscillations along Parallel Wires. 61 L 



§ 7. Case of Two nearly equal Wires, 



Let a x = a, a 2 = a{l + e), where e is a small fraction whose 

 square may be neglected; then 



log a 2 = log a+e, 

 A 2 = A 1 — €. 



Taking the case where k 2 a is large, / 2 =/ 1 (l — e). 

 The general equation becomes for this case 



(A-£)[A-e-£(l-e) 



=b 2 . . . . (17; 



Regarding this as a quadratic in 4 — A and solving, we get 



i 2 =A + ie(A-l)±B(l + ie). . . . (18) 



(a) The lower sign gives the result for two opposite wires 



l=(A-B)(l + ie)-le 

 c' 



-Iog|+*«(lo|l|-l> • • • (19) 



It is of interest to compare the magnitudes of the currents 

 in the two wires. The magnetic force just outside the wire 



i k 2 

 is D — h-. the total current across the section is therefore 



P ca ik, 2 

 equal to D- — 2 , and the currents at corresponding points of 



the two wires are in the ratio D : : D 2 . It is easy to prove that 

 the ratio of the surface-charges has the same value. Referring 

 to equation (14) we find 



A-l 

 D 2 _ A c 1 B ~i e (A-B-l) 



l>i ~ B B 



_, ,K-) 



= '-*« \ a ■ ■ ■ (20) 



Examination shows that the small quantity has a positive 

 real part, therefore the modulus of D 2 is smaller than that of 

 I),, or the larger wire carries less current and surface-charge 

 than the other. 



Looking at the matter from a more physical standpoint, we 



