Page:On Faraday's Lines of Force.pdf/67

Rh are electric currents, the magnetic effects cannot be represented by a potential. Let $$p', p$$, be the external and internal potentials,

$p'=I\frac{a^3}{r^2} cos \theta ,\ \ p_1=Ar cos \theta$,

and since there is no permanent magnetism, $$\frac{dp'}{dr}=\frac{dp_1}{dr}$$ when $$r=a$$,

$A=-2I$

If we draw any closed curve cutting the shell at the equator, and at some other point for which $$\theta$$ is known, then the total magnetic intensity round this curve will be $$3Ia cos \theta$$, and as this is a measure of the total electric current which ﬂows through it, the quantity of the current at any point may be found by differentiation. The quantity which ﬂows through the element $$td\theta$$ is $$-3Ia\ sin \theta d\theta$$, so that the quantity of the current referred to unit of area of section is

$-3I\frac{a}{t} sin \theta.$

If the shell be composed of a wire coiled round the sphere so that the number of coils to the inch varies as the sine of $$\theta$$, then the external effect will be nearly the same as if the shell had been made of a uniform conducting substance, and the currents had been distributed according to the law we have just given.

If a wire conducting a current of strength $$I_2$$ be wound round a sphere of radius a so that the distance between successive coils measured along the axis of $$x$$ is $$\frac{2a}{n}$$, then there will be $$n$$ coils altogether, and the value of $$I_1$$ for the resulting electro-magnet will be

$I_1=\frac{n}{6a}I_2$

The potentials, external and internal, will be

$p'=I_2\frac{n}{6}\frac{a^2}{r^2} cos \theta, \ \ p_1=-2I_2\frac{n}{6}\frac{r}{a} cos \theta$.

The interior of the shell is therefore a uniform magnetic ﬁeld.