CHAP. XI] INDUCTANCE OF TRANSMISSION LINES 197 



From eqs. (119) and (120) the following expressions are obtained 

 for the radius R of the line of force : 



. or 



...... (122) 



so that the line of force can be easily drawn for a given C or C". 



To prove that the equipotential lines are also circles we proceed 

 as follows. The line AB is evidently an equipotential line, because 

 it is perpendicular to all the lines of force. The difference of mag- 

 netic potent al or the m.m.f. between AB and P, contributed by 

 the system A, is equal to i(d\/2n) ampere-turns, where the angle 

 BAP is denoted by 0\. This is because the m.m.f. due to the sys- 

 tem A, taken around a complete circle concentric with A, is equal 

 to i, and is distributed uniformly along the circle, for reasons of 

 symmetry. Or else it follows directly from eq. (115). By 

 analogy, the difference of magnetic potential between AB and P, 

 due to the system B, is i(0 2 /2;r). Thus the total difference of 

 potential between A B and P is 



(i/*KKx-*>)- . . (123) 



This shows that the m.m.f. between any two points in the field is 

 proportional to the difference in the angles cu at which the line AB 

 is visible from these points. For any two points on the same 

 equipotential line M^ is the same, so that the equation of such a 

 line is 



a) const ........ (124) 



This represents the arc of a circle passing through A and B, and 

 corresponding to the inscribed angle w. 



Prob. 12. A single-phase transmission line consists of two conductors 

 1 cm. in diameter, and spaced 100 cm. between the centers. Draw 

 the curve of flux density distribution (pqst in Fig. 47) for an instantaneous 

 value of the current equal to 100 amp. 



Ana. z-50.0 25.0 0.5 -0.5 -50.0 -worn. 



B- 1.60 2.1380.40 -79.60 -0.53 maxw./sq. cm. 

 Prob. 13. For the transmission line in problem 12 draw the lines 



