198 THE MAGNETIC CIRCUIT [ART. 61 



of force which divide the total flux between the wires into ten equal 

 parts (not counting the llux within the wires). 



Ans. The circles nearest to 00' cross AB at a distance of 48.4 

 cm. from each other. 



Prob. 14. Referring to the two preceding problems, draw ten 

 equipotential circles which divide the whole m.m.f. of 100 ampere-turns 

 into ten equal parts. 



Ans. The arcs nearest to AB intersect 00' at a distance of 

 32.5 cm. from each other. 



Prob. 16. A telephone line runs parallel to a single-phase power 

 transmission line. The position of one of the telephone wires is fixed; 

 show how to determine the position of the other wire so as to have a 

 minimum of inductive disturbance in the telephone circuit. Hint: 

 The center lines of the two telephone wires must intersect the same line 

 of force due to the power line. 



Prob. 16. A telephone line runs parallel to a 25-cycle, single-phase 

 transmission line. The distances from one of the telephone wires to 

 the power wires are 3.5 m. and 2.7 m.; the distances from the other 

 telephone wire to the power wires are 3.6 and 2.5 m. (in the same order). 

 What voltage is induced in the telephone line per kilometer of its length, 

 when the current in the power line is 100 effective amperes? 



NOTE: In practice, this voltage is neutralized by transposing 

 either line after a certain number of spans. Ans. 0.33 volt. 



61. The Inductance of a Single-phase Line. The inductance of 

 a single-phase line (Fig. 47) can be calculated according to the fun- 

 damental formulae (105) or (106), provided that the permeances 

 of the elementary paths be expressed analytically. However, in 

 this case it is much simpler to use the principle of superposition 

 employed in the preceding article, and to consider the actual flux 

 as the resultant of two fluxes each surrounding concentrically one 

 of the wires and extending to infinity. The fluxes produced by 

 the two component systems are equal and symmetrical with 

 respect to the wires. It is therefore sufficient to calculate the 

 linkages of the loop AB with the flux produced by one of the sys- 

 tems, say that corresponding to A } and to multiply the result by 

 two. 



The flux produced by A and having A as a center link, with 

 the current in the loop AB. These linkages may be divided into 

 the following: 



(a) Linkages within the wire A ; that is, from x =0 to x =a; 



(6) Linkages between the wires A and B, that is, from 



x=a to x=ba', 



