196 THE MAGNETIC CIRCUIT [ART. GO 



as the resultant of the fluxes due to the systems A and B. Accord- 

 ing to eq. (115), the flux density due to A, at a distance x from A, 

 is .#1 =/n/27r:r, so that the flux due to A which crosses CP is 



' = f l Bidx 



J AC 



(117) 



This flux is directed to the left, looking from C to P. By analogy, 

 the flux due to the system B is 



^^(^TOI^fo/^Qmaxwells/cm., . . . (118) 



and is directed to the right, looking from C to P. The condition 

 that no flux crosses CP is, that C^' is equal to $2, or 



or 



ri/r 2 =^C/5C=Const ..... (119) 



This is the equation of a line of force in " bipolar " co-ordinates; 

 the curve is such that the ratio of r\ to r 2 remains constant, How- 

 ever, this constant is different for each line of force, because each 

 line has its own point C. 



Eq. (119) may be proved to represent a circle, by selecting an 

 origin, say at A, and substituting for 7*1 and r 2 their values in terms 

 of the rectangular co-ordinates x and y. The following proof by 

 elementary geometry leads to the same result. Produce AP and 

 lay off PD=PB=r 2 . According to eq. (119), BD is parallel to 

 CP, and consequently PC bisects the angle APE =a>. Let the 

 point C' lie on the same line of force with C; then no flux passes 

 through PC', and by analogy with eq. (119) we have 



r l /r 2 =AC'/BC'= Const ..... (120) 



By plotting PD' =r 2 (not shown in figure) along PA, in the oppo- 

 site direction from PD, and connecting D' to B, one can show as 

 before that PC' bisects the angle BPD = 180-w. But the 

 bisectors of two supplementary angles are perpendicular to each 

 other; consequently, CPC' is a right angle, and the point P lies 

 on a semicircle erected on the diameter CC'. This semicircle is the 

 line of force itself, because all the points, such as P, which are deter- 

 mined by C and C' must lie on it. Another semicircle below the 

 line AB closes the line of force. 



