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

§ 243] 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 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