Field of a Circular Current. 355 



each other, since PQ = PQ'. If yjr is the angle QOA, 

 a = radius of circle, i = current strength, the components 

 of ids along Q'Q being equal and in the same sense, the two 

 elements of current at Q' and Q conspire in giving a vector 



potential — pn . <fo perpendicular to the plane PON. 



Hence the total vector potential at P is perpendicular to 

 the plane PON. If, therefore, OA is the axis of a, the 

 perpendicular at to the plane of the circle the axis of z 9 

 and the diameter at perpendicular to AB the axis of y, 

 the components of the vector potential being, as usual, denoted 

 by F, G, H, the only component existing is G ; but, by 

 taking the components of the vector potential at a point 

 indefinitely close to P in the direction of the axis of y, we 

 easily find that 



d¥ G 



Hence if X, Y, Z are the components of the force of the 

 current per unit magnetic pole at P, since this force is the 

 curl of the vector potential, we have 



ay 7 da ct 



where « ( = ON) and 7 ( = NP) are the coordinates of P. 

 If along the line of force at P the increments of the co- 

 ordinates are Aa, Ay, we have 



A«__X 

 Ay"" Z* 



Hence along this line we have 



-7-Aa + - 1 - Ay + — Aa= 0, 

 dot dy ' a 



i.e., G . a — constant along the line of force. 



We shall therefore calculate the vector potential, G, at P. 

 Evidently 



»-«■•* 



COS yfr . dfty 



(" cos yjr 



Jo vWa 2 + y 2 



+ y — 2aot cos ty 



I —5 D) d^ 



