Page:A Treatise on Electricity and Magnetism - Volume 2.djvu/68

The second term is zero unless the point (ξ, η, ζ) is included in the magnet, in which case it becomes 4 π (φ) where (φ) is the value of φ at the point ξ, η, ζ. The surface-integral may be expressed in terms of r, the line drawn from (x, y, z) to (ξ, η, ζ), and the angle which this line makes with the normal drawn outwards from dS, so that the potential may be written

$V=\iint {{\frac {1}{r^{2}}}\cos \theta \,dS}+4\pi (\phi ),$

where the second term is of course zero when the point (ξ, η, ζ) is not included in the substance of the magnet.

The potential, V expressed by this equation, is continuous even at the surface of the magnet, where φ becomes suddenly zero, for if we write

$\Omega =\iint {{\frac {1}{r^{2}}}\cos \theta \,dS},$

and if Ω₁ is the value of Ω at a point just within the surface, and Ω₂ that at a point close to the first but outside the surface,

${\begin{aligned}\Omega _{2}&=\Omega 1+4\pi (\phi ),\\{\text{or}}\quad \quad \quad V_{2}&=V_{1}.\end{aligned}}$

The quantity Ω is not continuous at the surface of the magnet.

The components of magnetic induction are related to Ω by the equations

$a=-{\frac {d\Omega }{dx}},\quad b=-{\frac {d\Omega }{dy}},\quad c=-{\frac {d\Omega }{dz}}.$

416.] In the case of a lamellar distribution of magnetism we may also simplify the vector-potential of magnetic induction.

Its x-component may be written

$F=\iiint {\left({\frac {d\phi }{dy}}{\frac {dp}{dz}}-{\frac {d\phi }{dz}}{\frac {dp}{dy}}\right)dxdydz}.$

By integration by parts we may put this in the form of the surface-integral

${\begin{aligned}F=\iint {\phi \left(m{\frac {dp}{dz}}-n{\frac {dp}{dy}}\right)dS},\\{\text{or}}\quad \quad \quad F=\iint {p\left(m{\frac {d\phi }{dz}}-n{\frac {d\phi }{dy}}\right)dS}.\end{aligned}}$

The other components of the vector-potential may be written down from these expressions by making the proper substitutions.

On Solid Angles.

417.] We have already proved that at any point P the potential