Conway—— Electro-magnetie Mass. 57 
VI.—Tue Corrricrents oF INDUCTION. 
- As an example of the methods of the preceding Article, let us consider the 
action of a moving electron on a fixed, closed, conducting circuit. 
The activity communicated to the circuit will be the line integral of the 
product of the electric force resolved along the curve and the current at each 
point. The current being of constant magnitude will appear as a factor in 
the result ; the other factor will be the ‘‘ electromotive force ”’ due to the motion. 
It becomes at once clear that we are to reject the parts of the electric force 
which arise from the scalar potential, so that the electromotive force is 
or 0G oT 0 
alt. da + = dy + ie) = 0 ap (tute + iidy + dz) 
x {7 aT [( = 4%) ay + (y — nn + (2 = 21) 21] Ue 
when the speed is slow, this becomes 
@ 
wr | e (ade + ydy + nde)/r. 
If the moving electron forms part of a current, we evidently reproduce the 
ordinary expression for the electromotive force of induction. 
VII.—Newronran Mass. 
Consider first the case of an electric system moving with uniform velocity 
in a straight line. Let P; and P, be any two points of the system at the 
time ¢, then it follows from the results of S that the electric force of P,; and P, 
is equal and opposite to that of P, on P;, so that the activity of the internal 
forces is zero, and the system, if symmetrical about the direction of motion, 
will continue its motion. ‘his represents the Newtonian law of inertia. 
The system will not, however, be in relative equilibrium in general. For 
the system of internal reactions will be in general equivalent to a couple. In 
a similar manner we can treat of the slow motion of any system. We shall 
confine ourselves to the following problem. A system moves with a motion 
of translation: what is the resultant of the internal reactions? For brevity, we 
shall put our results in the quaternion form. Let the vectors to the various 
charges ¢, @,... be denoted by pi, p2,,».... The force exerted by any charge 
e, at the extremity of the vector p is equal to 
eV (p — pr) Vo (p — pr) T(p — pr) IT; 
where o is the acceleration. 
