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

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

$\omega _{0}=-{\frac {I_{2}n}{2a}}{\frac {1}{3k+k'}}$

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

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

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

$EI_{2}=\int _{0}^{2\pi }\omega asin\theta ds$,

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

$ds=asin\theta d\phi$,

$\ =-an'\pi sin^{2}\theta d\theta$,

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

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