440 THE ELECTROMAGNETIC FIELD. [PT. III. CH. XI. 



F ffl\ 



dy 



over any surface bounded by the current. 

 But by the equations 225 (14), 



(14) T = 1 [{{L cos (nx) + M cos (ny) + N cos (nz)} dS, 



or the electrokinetic energy of a system of currents is equal to one- 

 half the sum of the strengths of each current multiplied by the total 

 flux of magnetic force through its own circuit in the positive direc- 

 tion. The part of the flux due to the current itself constitutes the 

 term ^LI 2 , while for any two currents 1 and 2, the portions con- 

 sisting of one -half the strength of either times the flux through 

 its circuit due to the other current, being eo.ua! to the two 

 middle terms of (8), are equal. We may consequently express the 

 mutual kinetic energy of two currents as the strength of either 

 multiplied by the flux through its circuit of the magnetic force due 

 to the other. 



227. Mechanical Forces. We may deduce the mechanical 

 forces acting on conductors carrying currents from the expressions 

 found in 219 (7). Calling the forces per unit of volume H, H, Z, 

 and writing for Ida; the value in terms of the current density udr, 

 we have 



(I) jjJB*r = 



The first terms in the first and second integrals destroy each 

 other. The second terms may be written respectively, since the 

 accented quantities are independent of the unaccented, 



and 



