14 Mr. A. Schuster's Electrical Notes. 



current at the origin, which current is necessary to complete 

 the circuit as shown. 



The values of F, G, H in (10) satisfy the condition 



dx dy ' dz ' 



and are therefore solutions of the problem. A comparison of 

 (10) and (9) also shows that V 2 F' = V 2 F, agreeing with the 

 conclusion previously arrived at (7). 



In the expressions (9) the flow at the origin is taken to be 

 parallel to the axis of X. The equations show that the Vector 



Potential may be expressed as the resultant of a vector — ^ 



directed towards the origin, and another along the axis 



of X. If, in the general case, X ; fju, v are the direction- 

 cosines of the axis of flow, the components of the Vector 

 Potential are 



F=— 2wf"- +^(A^ + ^ + v^)l> 

 G = -27r^ + ^(\x + w + vz)\, 

 H=-27r^~ + ^(\^ + /xy + v^J- 



(11) 



We are now prepared to deal with the problem indicated at 

 the beginning of this paper. Imagine a magnet having an 

 intensity of magnetization at a point f , 77, f, measured by its 

 components A, B, C. We wish to find the vector potential 

 for a system of conductors flowing along the lines of induction 

 of such a magnet ; and having therefore a current potential 

 identical with the magnetic potential due to A, B, C. The 

 result is obtained directly from (11), by multiplying the right- 

 hand side with Idf; drj d£, taking account of the change of 

 origin and integrating throughout the space of the magnet 

 We find in this way 



+ CC*-f)]}df«Mfc • • (12) 



with corresponding expressions for G, and £L 



For some purposes it may be more useful to expand the 

 current potential in terms of solid harmonics, and we are 



