1204 THE BELL SYSTEM TECHNICAL JOURNAL, NOVEMBER 1952 



which yield, on rearrangement, 



(p2/pi) Lpi ^^ "" PU Pi ' 



1,11 2pl 



log (p2 



1 



log (p2/pl) 



2 "T 1,2 

 Pi 



— P2j 



(A14) 



— P2 J (0 — P2) 



Subtracting the second of equations (A14) from the first and solving the 

 resulting equation for pi , we get 



.2 1,2 



(A15) 



(A16) 



(A17) 



and on making use of (A 15) we obtain finally 



. PI = 0.392966, (A18) 



P2 = 0.822646. 



Substituting these values, with a = 0, into (A12), we get for the minimum 

 value of xi , when juq/m ^ 1, 



2 25.905iu ..^_. 



Xi = rr- • (A19) 



Appendix III 



POWER DISSIPATION IN A HOLLOW CONDUCTING CYLINDER 



Consider a hollow cylinder of inner radius pi , outer radius p2 , and high 

 conductivity gi . Denote the total current flowing in the positive ^-direction 

 inside the radius pi by /i and the total current inside the radius p2 by h ; 

 then the current carried by the conducting cylinder is just I2 — h , and the 

 net return current outside the cylinder is —I2. We assume the current 

 distribution to be independent of the coordinate angle 0, but the radial dis- 

 tribution of the currents inside and outside the given cylinder is of no 

 importance. 



General expressions for the field components in the conducting cylinder 



