Page:A Dynamical Theory of the Electromagnetic Field.pdf/15

 Rh If the electromotive force is of the form $$E\sin{pt}$$, as in the case of a coil revolving in a magnetic field, then

$x=\frac{E}{\rho}\sin{(pt-\alpha)},$

where $$\rho^2=R^2+L^2p^2$$, and $$\tan{\alpha}=\tfrac{Lp}{R}$$.

Case of two Circuits.

(37) Let $$R$$ 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

{{MathForm2|(13*)|$$\left.\begin{matrix}Rx+L\frac{dx}{dt}+M\frac{dy}{dt}=\xi\\Sy+M\frac{dx}{dt}+N\frac{dy}{dt}=0\end{matrix}\right\}.$$}}

To find the total quantity of electricity which passes, we have only to integrate these equations with respect to $$t$$; then if $$x_{0}$$, $$y_{0}$$ be the strengths of the currents at time $$0$$, and $$x_{1}$$, $$y_{1}$$ at time $$t$$, and if $$X$$, $$Y$$ be the quantities of electricity passed through each circuit during time $$t$$,

{{MathForm2|(14*)|$$\left.\begin{matrix}X=\frac{1}{R}\left\{\xi t+L\left(x_{0}-x_1\right)+M(y_{0}-y_{1}) \right\}\\ Y=\frac{1}{S}\left\{M\left(x_{0}-x_1\right)+N(y_{0}-y_{1}) \right\}\end{matrix}\right\}$$}}

When the circuit $$R$$ is completed, then the total currents up to time $$t$$, when $$t$$ is great, are found by making

$x_{0}=0,\ x_{1}=\frac{\xi}{R},\ y_{0}=0,\ y_{1}=0;$

then

The value of the total counter-current in $$R$$ 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 $$x_{1}$$ the final strength of the current in $$R$$. When the electromotive force $$\xi$$ ceases to act, there is an extra current in the primary 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 galvonometer, may be solved in this way without the necessity of a complete solution of the equations. The heating effect of