REACTANCE OF WIRE, SINGLE-PHASE 65 



The total flux in the outer ring of the section is 

 2 iri x dx _ TTJ (p 2 x 2 ) 

 10 10 



This cuts the element dx of the wire and produces a voltage 

 along it equal to 



jwri (p 2 x 2 ) io~ 9 volts per cm. (3) 



This voltage leads the current by 90 in phase at all sec- 

 tions. It is greatest at the center and zero at the surface 

 and so is an unbalanced voltage; it therefore causes a local 

 quadrature current to flow along the center of the wire and 

 return near the surface. 



Let the local current at the element dx be i( X ) per unit 

 area of section. Then the average voltage drop along the 

 wire due to the flux inside it and the resulting local bal- 

 ancing current, is equal to 



juILi = jwri (p 2 - x 2 ) lo" 9 + i (x) r, 



where L\ is the self-inductance of the wire due to the 

 above-mentioned flux inside it, and where r is the specific 

 resistance of the metal in centimeter units, that is, the re- 

 sistance of a centimeter cube of the metal. The current 

 i (x ) adjusts itself so that the drop is the same at all parts 

 of the section. From the last equation, we have 



since / = irlp 2 . 



As i (X ) is a local current in the wire, and does not in- 

 crease or decrease the main current /, its sum when added 

 up all over the section must be zero, and thus 



2 irxi( X ) dx = o, 



