158 Prof. J. J. Thomson on the 



angles both to itself and the direction in which it is moving, 

 and whose magnitude is Air times the strength of the tube 

 multiplied by its velocity at right angles to its direction. 

 The direction of the force is such that the magnetic force 

 and rotation from the direction of motion to that of the tube 

 are related like translation and rotation in a right-angled 

 screw. 



Let us first consider the case where all the tubes are moving 

 with the same velocity in a field whose magnetic permeability 

 is unity. 



The energy in the magnetic field per unit volume is 



i(« 2 + /3 2 + 7 2 ), 

 or substituting for a, ft, y their values from (1), 



2*{(hv -gwf+(fic-huy+ (gu-fvy\ . 



The momentum U per unit volume parallel to x is the 

 differential coefficient of this expression with respect to u, i. e. 



4ir\g(gu-fo)-h(fw-hu)\, 



TJ^gy-h/3. 



Similarly, if V, W are the components of the momentum 

 parallel to y and z, we have 



or 



W=f/3-g«. J ■••■(*) 



Thus the momentum per unit volume possessed by the 

 moving tube is at right angles to the tube and to the magnetic 

 force produced by it, and equals the product of the strength 

 of the tube and the magnetic force. 



The electromotive intensity parallel to x produced by the 

 moving tube can be got by differentiating the expression for 

 the Kinetic Energy with respect to /; in this way we obtain 

 the following expressions for X, Y, Z ; the components of the 

 electromotive intensity : — 



X = w/3—vy } ^ 



Y = wy—wa, y (3) 



Z = vu-u{3. j 



Thus the electromotive intensity produced by the motion of 

 the tube is equal to the product of the velocity of the tube and 

 the magnetic force produced by it ; and is at right angles to 



