Page:A Treatise on Electricity and Magnetism - Volume 2.djvu/61

Remembering that ${\frac {dp}{dx}}=-{\frac {dp}{d\xi }}$, and that the integral

$\iiint {A\left({\frac {d^{2}p}{dx^{2}}}+{\frac {d^{2}p}{dy^{2}}}+{\frac {d^{2}p}{dz^{2}}}\right)dxdydz}$

has the value –4π(A) when the point (ξ, η, ζ) is included within the limits of integration, and is zero when it is not so included, (A) being the value of (A) at the point (ξ, η, ζ), we find for the value of the x-component of the magnetic induction,

${\begin{aligned}a&={\frac {dH}{d\eta }}-{\frac {dG}{d\zeta }}\\&=\iiint {\left\{A\left({\frac {d^{2}p}{dyd\eta }}+{\frac {d^{2}p}{dzd\zeta }}\right)-B{\frac {d^{2}p}{dxd\eta }}-C{\frac {d^{2}p}{dxd\zeta }}\right\}dxdydz}\\&=-{\frac {d}{d\xi }}\iiint {\left\{A{\frac {dp}{dx}}+B{\frac {dp}{dy}}+C{\frac {dp}{dz}}\right\}dxdydz}\\&\quad \quad -\iiint {A\left({\frac {d^{2}p}{dx^{2}}}+{\frac {d^{2}p}{dy^{2}}}+{\frac {d^{2}p}{dz^{2}}}\right)dxdydz}.\end{aligned}}$

(24)

The first term of this expression is evidently $-{\frac {dV}{d\xi }}$, or α, the component of the magnetic force.

The quantity under the integral sign in the second term is zero for every element of volume except that in which the point (ξ, η, ζ) is included. If the value of A at the point (ξ, η, ζ) is (A), the value of the second term is 4π(A), where (A) is evidently zero at all points outside the magnet.

We may now write the value of the x-component of the magnetic induction

$a=\alpha +4\pi (A)\,$

(25)

an equation which is identical with the first of those given in Art. 400. The equations for b and c will also agree with those of Art. 400.

We have already seen that the magnetic force ${\mathfrak {H}}$ is derived from the scalar magnetic potential V by the application of Hamilton's operator $\nabla$, so that we may write, as in Art. 17,

${\mathfrak {H}}=-\nabla V,$

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and that this equation is true both without and within the magnet.

It appears from the present investigation that the magnetic induction ${\mathfrak {B}}$ is derived from the vector-potential ${\mathfrak {A}}$ by the application of the same operator, and that the result is true within the magnet as well as without it.

The application of this operator to a vector-function produces,