Page:On Faraday's Lines of Force.pdf/64

218 and a the radius of the sphere. Let $$I$$ be the undisturbed magnetic intensity of the ﬁeld into which the sphere is introduced, and let its direction-cosines be $$l, m, n$$.

Let us now take the case of a homogeneous sphere whose coefﬁcient is $$k$$, placed in a uniform magnetic ﬁeld Whose intensity is $$lI$$ in the direction of $$x$$. The resultant potential outside the sphere would be

$p'=lI\left(1+\frac{k_1-k'}{2k_1+k'}\frac{a^3}{r^3}\right) \ x$,

and for internal points

$p_1=lI\frac{k_1-k'}{2k_1+k'}\ x.$

So that in the interior of the sphere the magnetization is entirely in the direction of $$x$$. It is therefore quite independent of the coefﬁcients of resistance in the directions of $$x$$ and $$y$$, which may be changed from $$k_1$$ into $$k_2$$ and $$k_3$$ without disturbing this distribution of magnetism. We may therefore treat the sphere as homogeneous for each of the three components of $$I$$, but we must use a different. coefﬁcient for each. We ﬁnd for external points

$p'=I\left\{lx+my+nz+\left(\frac{k_1-k'}{2k_1+k'}lx+\frac{k_2-k'}{2k_2+k'}my+\frac{k_3-k'}{2k_3+k'}nz+\right)\frac{a^3}{r^3}\right\}$,

and for internal points

$p_1=I\left(\frac{3k_1}{2k_1+k'}lx+\frac{3k_2}{2k_2+k'}my\frac{3k_3}{2k_3+k'}nz\right)$.

The external effect is the same as that which would have been produced if the small magnet whose moments are

$\frac{k_1-k'}{2k_1+k'}lIa^3, \ \frac{k_2-k'}{2k_2+k'}mIa^3, \ \frac{k_3-k'}{2k_3+k'}nIa^3, \ $

had been placed at the origin with their directions coinciding with the axes of $$x, y, z$$. The effect of the original force $$I$$ in turning the sphere about the axis of $$x$$ may be found by taking the moments of the components of that force on these equivalent magnets. The moment of the force in the direction of $$y$$ acting on the third magnet is

$\frac{k_3-k'}{2k_3+k'}mnI^2a^3,$

and that of the force in $$z$$ on the second magnet is

$-\frac{k_2-k'}{2k_2+k'}mnI^2a^3.$