Page:A Treatise on Electricity and Magnetism - Volume 2.djvu/80

But since α, β, γ, are derived from the potential U, we may write the second member –κ dU.

Hence, if κ is constant throughout the substance, the first member must also be a complete differential of a function of x, y and z, which we shall call φ, and the equation becomes

$d\phi =-\kappa dU,\,$

(4)

where

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

(5)

The magnetization is therefore lamellar, as defined in Art. 412.

It was shewn in Art. 386 that if ρ is the volume-density of free magnetism,

$\rho =-\left({\frac {dA}{dx}}+{\frac {dB}{dy}}+{\frac {dC}{dz}}\right),$

which becomes in virtue of equations (3),

$\rho =-\kappa \left({\frac {d\alpha }{dx}}+{\frac {d\beta }{dy}}+{\frac {d\gamma }{dz}}\right).$

But, by Art. 77,

${\frac {d\alpha }{dx}}+{\frac {d\beta }{dy}}+{\frac {d\gamma }{dz}}=4\pi \rho .\,$

Hence

$(1+4\pi \kappa )\rho =0,\,$

whence

$\rho =0\,$

(6)

throughout the substance, and the magnetization is therefore solenoidal as well as lamellar. See Art. 407.

There is therefore no free magnetism except on the bounding surface of the body. If ν be the normal drawn inwards from the surface, the magnetic surface-density is

$\sigma =-{\frac {d\phi }{d\nu }}.$

(7)

The potential Ω due to this magnetization at any point may therefore be found from the surface-integral

$\Omega =\iint {{\frac {d\sigma }{r}}dS}.$

(8)

The value of Ω will be finite and continuous everywhere, and will satisfy Laplace s equation at every point both within and without the surface. If we distinguish by an accent the value of Ω outside the surface, and if ν' be the normal drawn outwards, we have at the surface

$\Omega '=\Omega ;\,$

(9)