Page:A Treatise on Electricity and Magnetism - Volume 2.djvu/309

Similarly

$G={\frac {x}{a}}{\frac {dP}{dz}}-{\frac {z}{a}}{\frac {dP}{dx}}$. $H={\frac {y}{a}}{\frac {dP}{dx}}-{\frac {x}{a}}{\frac {dP}{dy}}$.

The vector ${\mathfrak {A}}$, whose components are $F$, $G$, $H$, is evidently perpendicular to the radius vector $r$, and to the vector whose components are ${\frac {dP}{dx}}$, ${\frac {dP}{dy}}$, and ${\frac {dP}{dz}}$. If we determine the lines of intersections of the spherical surface whose radius is $r$, with the series of equipotential surfaces corresponding to values of $P$ in arithmetical progression, these lines will indicate by their direction the direction of ${\mathfrak {A}}$, and by their proximity the magnitude of this vector.

In the language of Quaternions,

${\mathfrak {A}}={\frac {1}{a}}V\rho \nabla P$.

672.] If we assume as the value of $P$ within the sphere

$P=a\left({\frac {r}{a}}\right)^{i}Y_{i}$,

where $Y_{i}$ is a spherical harmonic of degree $i$, then outside the sphere

$P^{\prime }=A\left({\frac {a}{r}}\right)^{i+1}Y_{i}$.

The current-function $\phi$ is

$\phi ={\frac {2i+1}{4\pi }}{\frac {1}{a}}AY_{i}$.

The magnetic potential within the sphere is

	$\Omega =-(i+1){\frac {1}{a}}A\left({\frac {a}{r}}\right)^{i}Y_{i}$,
and outside	$\Omega ^{\prime }=i{\frac {1}{a}}A\left({\frac {a}{r}}\right)^{i+1}Y_{i}$.

For example, let it be required to produce, by means of a wire coiled into the form of a spherical shell, a uniform magnetic force $M$ within the shell. The magnetic potential within the shell is, in this case, a solid harmonic of the first degree of the form

$\Omega =Mr\cos \theta$,

where $M$ is the magnetic force. Hence $A=-{\frac {1}{2}}a^{2}M$, and

$\phi ={\frac {3}{8\pi }}Ma\cos \theta$.

The current-function is therefore proportional to the distance from the equatorial plane of the sphere, and therefore the number of windings of the wire between any two small circles must be proportional to the distance between the planes of these circles.