Page:A Treatise on Electricity and Magnetism - Volume 2.djvu/93

where

${\begin{aligned}&D'r_{1}'=r_{1}+{\frac {4}{3}}\pi (r_{3}r_{1}-p_{2}q_{2}+r_{1}r_{2}-p_{3}q_{3})+\left({\frac {4}{3}}\pi \right)^{2}D,\\&D'p_{1}'=p_{1}-{\frac {4}{3}}\pi (q_{2}q_{3}-p_{1}r_{1}),\\&D'q_{1}'=q_{1}-{\frac {4}{3}}\pi (p_{2}p_{3}-q_{1}r_{1}),\\&{\And }c.,\end{aligned}}$

(8)

where D is the determinant of the coefficients on the right side of equations (6), and D' that of the coefficients on the left.

The new system of coefficients p', q', r' will be symmetrical only when the system p, q, r is symmetrical, that is, when the coefficients of the form p are equal to the corresponding ones of the form q.

436.] The moment of the couple tending to turn the sphere about the axis of x from y towards z is

${\begin{aligned}L&={\frac {4}{3}}\pi a^{3}(ZB-YC)\\&={\frac {4}{3}}\pi a^{3}\left[p_{1}'Z^{2}-q_{1}'Y^{2}+(r_{2}'-r_{3}')YZ+X(q_{3}'Z-p_{2}'Y)\right].\end{aligned}}$

(9)

If we make

$X=0,\quad Y=F\cos \theta ,\quad Z=F\sin \theta ,$

this corresponds to a magnetic force F in the plane of yz, and inclined to y at an angle θ. If we now turn the sphere while this force remains constant the work done in turning the sphere will be $\int _{0}^{2\pi }{L\,d\theta }$ in each complete revolution. But this is equal to

${\frac {2}{3}}\pi a^{3}F^{2}(p_{1}'-q_{1}').$

)10

Hence, in order that the revolving sphere may not become an inexhaustible source of energy, p₁' = q₁', and similarly q₁' = q₁'.

These conditions shew that in the original equations the coefficient of B in the third equation is equal to that of C in the second, and so on. Hence, the system of equations is symmetrical, and the equations become when referred to the principal axes of magnetization,

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

(11)

The moment of the couple tending to turn the sphere round the axis of x is

$L={\frac {4}{3}}\pi a^{3}{\frac {r_{2}-r_{3}}{(1+{\frac {4}{3}}\pi r_{2})(1+{\frac {4}{3}}\pi r_{3})}}YZ.$

(12)