274 Prof. R. Clausius on the Determination of the 



§3. 



Now, in order to deduce from the preceding formulae the 

 ponderomotive force exerted upon a current-element by a closed 

 current, we first form from the potential function the poten- 

 tials of the closed current upon the two electricities flowing in 

 the current-element. From these we get, by the above-stated 

 operation, the components, in any one direction (e. g. the x 

 direction), of the forces which the two electricities undergo ; 

 and the sum of these two components is the respective force- 

 component which refers to the entire current-element. 



Suppose, then, given, in the point x, y, z, an element of 

 current ds, in which the electricities lids and —lids flow in 

 opposite directions with the velocities c and c : . Now, by my 

 fundamental law we first employ for the potential function the 

 value 



dx 



n=m cF 



given in equation (17 a), and obtain for the quantity lids of 



positive electricity: — 



dx 

 Potential = lids SH 2 itj 



Force-component = lids (^- % H, — -*■ —J ■ 



1 \bx* x dt dt / 



In these we must put 



dx __ d# ^x 



dt ~ dt ~fts 



dt " ~dt ±c ^s > 



by which the expressions are changed into: — 



Potential =MsTR x {^ + c gf); 



t? 7 7 f^ vu fi x a. ~& x \ d H * BH^-i 



Force-comp. =^[^2^^ +^j- ^- -^J- 



In precisely the same way we get for the quantity — lids of 

 negative electricity (for which we must bring into use the 

 velocity of flow — c x ): — 



Potential -hdsXB. x (^ -<?i|f), 



