Page:On Faraday's Lines of Force.pdf/69

Rh Within the shell $$\omega$$ cannot become inﬁnite; therefore $$\omega=C_1$$ is the solution, and outside $$a$$ must vanish at an inﬁnite distance, so that

$\omega=\frac{C_2}{r^2}$

is the solution outside. The magnetic quantity within the shell is found by last article to be

$-2I_2\frac{n}{6a}\frac{3}{2k+k'}=a_1=\frac{d\beta_0}{dz}-\frac{d\gamma_0}{dy}=2C_1$

therefore within the sphere

$\omega_0=-\frac{I_2n}{2a}\frac{1}{3k+k'}$

Outside the sphere we must determine $$\omega$$ so as to coincide at the surface with the internal value. The external value is therefore

$\omega=-\frac{I_2n}{2a}\frac{1}{3k+k'}\frac{a^3}{r^3}$

where the shell containing the currents is made up of $$n$$ coils of wire, conducting a current of total quantity $$I_2$$.

Let another wire be coiled round the shell according to the same law, and let the total number of coils be $$n'$$; then the total electro-tonic intensity $$EI_2$$round the second coil is found by integrating

$EI_2=\int_{0}^{2\pi} \omega a sin \theta ds $,

along the whole length of the wire. The equation of the wire is

$cos \theta=\frac{\phi}{n'\pi}$,

where $$n'$$ is a large number; and therefore

$ds=a sin \theta d\phi$, $\ =-an'\pi sin^2 \theta d\theta$,

$\therefore EI_2=\frac{4\pi}{3}\omega a^2n'=-\frac{2\pi}{3}ann'I\frac{1}{3k+k'}$

$$E$$ may be called the electro-tonic coefﬁcient for the particular wire.