PROFESSOR CLERK MAXWELL OX THE ELECTROMAGNETIC FIELD. 
47 3 
If the electromotive force is of the form E sin^£, as in the case of a coil revolving in 
a magnetic field, then 
X—~ sin (pt — a), 
where g 2 =R 2 -|-L 2 _p 2 , and tan «=-,(-■ 
Case of two Circuits. 
(37) Let K be the primary circuit and S the secondary circuit, then we have a case 
similar to that of the induction coil. 
The equations of currents are those marked A and B, and we may here assume 
L, M, N as constant because there is no motion of the conductors. The equations 
then become 
B *'+ L 5+ M ;M| 
'Sy+M§+N|=0.{ 
To find the total quantity of electricity which passes, we have only to integrate these 
equations with respect' to t\ then if^ 0 ,y 0 be the strengths of the currents at time 0, 
and Wi , y x at time t, and if X, Y be the quantities of electricity passed through each 
circuit during time t, 
X= jr {^+L(ff 0 — ^) 7%-?/,)}, 
(14*) 
Y =4{ M(^-^)+N(y 0 -^)}. 
When the circuit R is completed, then the total currents up to time t , when t is 
great, are found by making 
then 
tf 0 =0, y o 0 , y, = 0; 
X=.,(i-|), Y=-~s,. (15*) 
The value of the total counter-current in It is therefore independent of the secondary 
circuit, and the induction current in the secondary circuit depends only on M, the 
coefficient of induction between the coils, S the resistance of the secondary coil, and 
the final strength of the current in It. 
When the electromotive force f ceases to act, there is an extra current in the pri- 
mary circuit, and a positive induced current in the secondary circuit, whose values are 
equal and opposite to those produced on making contact. 
(38) All questions relating to the total quantity of transient currents, as measured 
by the impulse given to the magnet of the galvanometer, may be solved in this way 
without the necessity of a complete solution of the equations. The heating effect of 
