Page:On Faraday's Lines of Force.pdf/71

Rh Putting $$\frac{1}{2}\tau$$ for $$N\left(\frac{n^2}{R}+\frac{n'^2}{R'}\right)$$ we ﬁnd

$I=\frac{F}{R}(1-\epsilon^{-\frac{2t}{\tau}})$,|undefined

$I'=-F\frac{n'}{R'n}\epsilon^{-\frac{2t}{\tau}}$|undefined

The primary current increases very rapidly from $$0$$ to $$\frac{F}{R}$$, and the secondary commences at $$-\frac{F}{R}\frac{n'}{n}$$ and speedily vanishes, owing to the value of $$\tau$$ being generally very small.

The whole work done by either current in heating the wire or in any other kind of action is found from the expression

$\int_{0}^{\infty}I^2R dt.$

The total quantity of current is

$\int_{0}^{\infty}I dt.$

For the secondary current we ﬁnd

$\int_{0}^{\infty}I'^2R' dt=\frac{F^2n'^2}{R'n^2}\frac{\tau}{4},\ \ \int_{0}^{\infty}I' dt=\frac{Fn'}{R'n}\frac{\tau}{2}$.

The work done and the quantity of' the current are therefore the same as if a current of quantity $$I'=\frac{Fn'}{2R'n}$$ had passed through the wire for a time $$\tau$$, where

$\tau =2N\left(\frac{n^2}{R}+\frac{n'^2}{R'}\right)$

This method of considering a variable current of short duration is due to Weber, whose experimental methods render the determination of the equivalent current a matter of great precision.

Now let the electro-motive force F suddenly cease while the current in the primary wire is $$I_0$$ and in the secondary =0 Then we shall have for the subsequent time

$I=I_0\epsilon^{-\frac{2t}{\tau}},\ \ I'=\frac{I_0}{R'}\frac{Rn'}{n}\epsilon^{-\frac{2t}{\tau}}$|undefined