Page:A Treatise on Electricity and Magnetism - Volume 2.djvu/67

magnetization must be such that a system of surfaces can be drawn cutting them at right angles. This condition is expressed by the well-known equation

$A\left({\frac {dC}{dy}}-{\frac {dB}{dz}}\right)+B\left({\frac {dA}{dz}}-{\frac {dC}{dx}}\right)+C\left({\frac {dB}{dx}}-{\frac {dA}{dy}}\right)=0.$

Forms of the Potentials of Solenoidal and Lamellar Magnets.

414.] The general expression for the scalar potential of a magnet is

$V=\iiint {\left(A{\frac {dp}{dx}}+B{\frac {dp}{dy}}+C{\frac {dp}{dz}}\right)dxdydz},$

where p denotes the potential at (x, y, z) due to a unit magnetic pole placed at ξ, η, ζ, or in other words, the reciprocal of the distance between (ξ, η, ζ), the point at which the potential is measured, and (x, y, z), the position of the element of the magnet to which it is due.

This quantity may be integrated by parts, as in Arts. 96, 386.

$V=\iint {p(Al+Bm+Cn)dS}-\iiint {p\left({\frac {dA}{dx}}+{\frac {dB}{dy}}+{\frac {dC}{dz}}\right)dxdydz},$

where l, m, n are the direction-cosines of the normal drawn out wards from dS, an element of the surface of the magnet.

When the magnet is solenoidal the expression under the integral sign in the second term is zero for every point within the magnet, so that the triple integral is zero, and the scalar potential at any point, whether outside or inside the magnet, is given by the surface-integral in the first term.

The scalar potential of a solenoidal magnet is therefore completely determined when the normal component of the magnetization at every point of the surface is known, and it is independent of the form of the solenoids within the magnet.

415.] In the case of a lamellar magnet the magnetization is determined by φ, the potential of magnetization, so that

$A={\frac {d\phi }{dx}},\quad B={\frac {d\phi }{dy}},\quad C={\frac {d\phi }{dz}}.$

The expression for V may therefore be written

$V=\iiint {\left({\frac {d\phi }{dx}}{\frac {dp}{dx}}+{\frac {d\phi }{dy}}{\frac {dp}{dy}}+{\frac {d\phi }{dz}}{\frac {dp}{dz}}\right)dxdydz}.$

Integrating this expression by parts, we find

$V=\iint {\phi \left(l{\frac {dp}{dx}}+m{\frac {dp}{dy}}+n{\frac {dp}{dz}}\right)dS}-\iiint {\phi \left({\frac {d^{2}p}{dx^{2}}}+{\frac {d^{2}p}{dy^{2}}}+{\frac {d^{2}p}{dz^{2}}}\right)dxdydz}.$