A DYNAMICAL THEORY OF THE ELECTROMAGNETIC FIELD. 565 



PART IV. 



MECHANICAL ACTIONS IN THE FIELD. 



Mechanical Force on a Moveable Conductor. 



(76) We have shewn ( 34 & 35) that the work done by the electro- 



magnetic forces in aiding the motion of a conductor is equal to the product 



of the current in the conductor multiplied by the increment of the electro- 

 magnetic momentum due to the motion. 



Let a short straight conductor of length a move parallel to itself in the 

 direction of x, with its extremities on two parallel conductors. Then the incre- 

 ment of the electromagnetic momentum due to the motion of a will be 



fdFdx dGdy dHdz\ g 

 \dx ds dx ds dx ds) 



That due to the lengthening of the circuit by increasing the length of the 

 parallel conductors will be 



fdFdx dFdy dF_dz\ 8 

 \dx ds dy ds dz ds) 



The total increment is 



8 Wf/ 1 ^ M% */d^ 'dH\\ 



\ds\dx dy) ds\dz dx ) J ' 



which is by the equations of Magnetic Force (B), p. 556, 



dz 



Let X be the force acting along the direction of x per unit of length of 

 the conductor, then the work done is XaSx. 



Let C be the current in the conductor, and let p', q, r' be its com- 

 ponents, then 



XaSx = CaSx (-^ py - ~ i 



