Page:On Faraday's Lines of Force.pdf/62

216 The whole potential is therefore equal to $I\left(r+\frac{k-k\prime}{2k+k'}\frac{a^3}{r^2}\right)\cos\theta=p$, $\frac{dp}{dr}=I\left(1-2\frac{k-k\prime}{2k+k\prime}\frac{a^3}{r^3}\right)\cos\theta$, $\frac{1}{r}\frac{dp}{d\theta}=-I\left(1+\frac{k-k\prime}{2k+k\prime}\frac{a^3}{r^3}\right)\sin\theta,\quad\frac{dp}{d\phi}=0$, $\begin{align} \therefore i^2=\overline{\frac{dp}{dr}}\Bigg\vert^2+\frac{1}{r^2}\overline{\frac{dp}{d\theta}}\Bigg\vert^2+\frac{1}{r^2\sin^2\theta}\overline{\frac{dp}{d\phi}}&\Bigg\vert^2\\ =I^2&\left\{1+\frac{k-k\prime}{2k+k\prime}\frac{a^3}{r^3}(1-3\cos^2\theta)+\overline{\frac{k-k\prime}{2k+k\prime}}\Bigg\vert^2\frac{a^5}{r^5}(1+3\cos^2\theta)\right\}. \end{align}$|undefined

This is the value of the square of the intensity at any point. The moment of the couple tending to turn the combination of balls in the direction of the original force $L=\frac{1}{2}\frac{d}{d\theta}i^2\left(\frac{k-k\prime}{2k+k\prime}a^3\right)$ when $r=b$, $L=\frac{3}{2}I^2\overline{\frac{k-k\prime}{2k+k\prime}}\Bigg\vert^2\frac{a^6}{b^3}\left(1-\frac{k-k\prime}{2k+k\prime}\frac{a^3}{b^3}\right)\sin 2\theta$.|undefined

This expression, which must be positive, since $$b$$ is greater than $$a$$, gives the moment of a force tending to turn the line joining the centres of the spheres towards the original lines of force.

Whether the spheres are magnetic or diamagnetic they tend to set in the axial direction, and that without distinction of north and south. If, however, one sphere be magnetic and the other diamagnetic, the line of centres will set equatoreally [sic]. The magnitude of the force depends on the square of $$(k-k\prime)$$, and is therefore quite insensible except in iron.

V. Two Spheres between the poles of a Magnet.

Let us next take the case of the same balls placed not in a uniform ﬁeld but between a north and a south pole, $$\pm M$$, distant $$2c$$ from each other in the direction of $$x$$.