(38) 



262 Dr. J. W. Nicholson on the 



Since the electric force R is continuous at the surface, H 

 is also continuous (for this disturbance due to induction), 

 and therefore 



D J (k'b) = OK (iJib) + B<9 J (hb\ 

 and because the magnetic force is also continuous, 



-h'DJ ' (k'b) = iJiCK ' (ihb)+Bh0 J ' {hb). 



Adopting the notation 



k 

 S x = J (Jed) J J (ha) j J (ha) J Q ' (ka) J 



S 2 = J (k'b) J ' (hb) - h - J (Kb) J ' (hb) j ' 



it appears that 



C=,B0 o S 2 //3 -) 



D = t B0 o /M/3J' { } 



where the numerator of D has been simplified by the 

 reduction 



*K ' (M) J (Jib) -K (lAft) J ' (Jib) 

 = i (Jo' (Aft) Y (Aft)- J (Aft) Y ' (Aft)) 

 1_ 

 ~ Aft' 



by the usual property of the Hankel solution Y . 



The second wire, therefore, throws out a disturbance of 

 vector potential 



H=^M 2 K„0fy) (40) 



Potential independent of Orientation. 



In the calculation just made, the terms of a transformed 

 series of! Bessel functions involving orientation were neglected. 

 We now trace the series of compensating disturbances which 

 may be regarded as thrown off by the wires, and form their 

 sum under the same condition. This leads to a very accurate 

 result if the wires are not very close together. 



A vector potential H = BK (ihr) leaving the first wire 

 has, for its main term near the second, a magnitude 

 ~B6 J (7ip), and causes (1) an inside potential 



