Page:A Treatise on Electricity and Magnetism - Volume 2.djvu/92

Case of a Sphere in which the Coefficients of Magnetization are Different in Different Directions.

435.] Let α, β, γ be the components of magnetic force, andA, B, C those of the magnetization at any point, then the most general linear relation between these quantities is given by the equations

${\begin{aligned}A&=r_{1}\alpha +p_{3}\beta +q_{2}\gamma ,\\B&=q_{3}\alpha +r_{2}\beta +p_{1}\gamma ,\\C&=p_{2}\alpha +q_{1}\beta +r_{3}\gamma ,\\\end{aligned}}$

(1)

where the coefficients r, p, q are the nine coefficients of magnetization.

Let us now suppose that these are the conditions of magnetization within a sphere of radius a, and that the magnetization at every point of the substance is uniform and in the same direction, having the components A, B, C.

Let us also suppose that the external magnetizing force is also uniform and parallel to one direction, and has for its components X, Y, Z.

The value of V is therefore

$V=-(Xx+Yy+Zz),\,$

(2)

and that of Ω' the potential of the magnetization outside the sphere is

$\Omega '=(Ax+By+Cz){\frac {4\pi a^{3}}{3r^{3}}}.$

(3)

The value of Ω, the potential of the magnetization within the sphere, is

$\Omega ={\frac {4\pi }{3}}(Ax+By+Cz).$

(4)

The actual potential within the sphere is V + Ω, so that we shall have for the components of the magnetic force within the sphere

${\begin{matrix}\alpha =X-{\frac {4}{3}}\pi A,\\\beta =Y-{\frac {4}{3}}\pi B,\\\gamma =Z-{\frac {4}{3}}\pi C.\end{matrix}}$

(5)

Hence

${\begin{aligned}(1+{\frac {4}{3}}\pi r_{1})A+{\frac {4}{3}}\pi p_{3}B+{\frac {4}{3}}\pi q_{2}C&=r_{1}X+p_{3}Y+q_{2}Z,\\{\frac {4}{3}}\pi q_{3}A+(1+{\frac {4}{3}}\pi r_{2})B+{\frac {4}{3}}\pi p_{1}C&=q_{3}X+r_{2}Y+p_{1}Z,\\{\frac {4}{3}}\pi p_{2})A+{\frac {4}{3}}\pi q_{1}B+(1+{\frac {4}{3}}\pi r_{3}C&=p_{2}X+q_{1}Y+r_{3}Z.\end{aligned}}$

(6)

Solving these equations, we find

${\begin{aligned}A=r_{1}'X+p_{3}'Y+q_{2}'Z,\,\\B=q_{3}'X+r_{2}'Y+p_{1}'Z,\,\\C=p_{2}'X+q_{1}'Y+r_{3}'Z,\,\end{aligned}}$

(7)