Page:Elementary Text-book of Physics (Anthony, 1897).djvu/281

area. Let the origin (Fig. 77) be taken half-way between the two faces of the shell, and let the shell stand perpendicular to the x axis. Let $a$ represent the area of the shell, supposed infinitesimal, $2l$ the thickness of the shell, and $d$ the intensity of magnetization. The volume of this infinitesimal magnet is $2al,$ and, from the definition of intensity of magnetization, $2ald$ is its magnetic moment. The potential at the point $P$ is then given by equation (81), since $l$ is very small. We have $V={\frac {M}{r^{2}}}\cos \theta ={\frac {2ald}{r^{2}}}\cos \theta .$ Now $a\,\cos \theta$ is the projection of the area of the shell upon a plane through the origin normal to the radius vector $r,$ and, since $a$ is infinitesimal, ${\frac {a\,\cos \theta }{r^{2}}}$ is the solid angle $\omega$ bounded by the lines drawn from $P$ to the boundary of the area $a.$ The potential then becomes $V=2ld\omega =j\omega ,$ since $2ld$ is what has been called the strength of the shell.

The same proof may be extended to any number of contiguous areas making up a finite magnetic shell. The potential due to such, a shell is then $\textstyle \sum \displaystyle j\omega .$ If the shell be of uniform strength, the potential due to it becomes $\textstyle \sum \displaystyle j\omega ,$ and is got by summing the elementary solid angles. This sum is the solid angle $\Omega ,$ bounded by the lines drawn from the point of which the potential is required to the boundary of the shell. The potential due to a magnetic shell of uniform strength is therefore

$j\Omega .$

(84)

It does not depend on the form of the shell, but only on the angle subtended by its contour. At a point very near the positive face of a flat shell, so near that the solid angle subtended by the shell equals $2\pi ,$ the potential is $2\pi j;$ at a point in the plane of the shell outside its boundary, where the angle subtended is zero, the potential is zero; and near the other or negative face of the shell it is $-2\pi j$ The whole work done, then, in moving a unit magnet pole from a point very near one face to a point very near the other