Page:A Treatise on Electricity and Magnetism - Volume 2.djvu/131

Taking the arithmetical mean of (6) and (7),

${\frac {1}{4}}{\frac {H}{M}}{r_{1}}^{3}(\tan \theta _{1}-\tan \theta _{2})=1+A_{2}{\frac {1}{{r_{1}}^{2}}}+A_{4}{\frac {1}{{r_{1}}^{4}}}+\&{\text{c.}}$

(8)

Now remove $M$ to the west side of the suspended magnet, and place it with its centre at the point marked $2s_{0}-s$ on the scale. Let the deflexion when the axis is in the first position be $\theta _{3}$, and when it is in the second $\theta _{4}$, then, as before,

${\frac {1}{4}}{\frac {H}{M}}{r_{2}}^{3}(\tan \theta _{3}-\tan \theta _{4})=1+A_{2}{\frac {1}{{r_{2}}^{2}}}+A_{4}{\frac {1}{{r_{2}}^{4}}}+\&{\text{c.}}$

(9)

Let us suppose that the true position of the centre of the suspended magnet is not $s$ but $s_{0}+\sigma$, then

$r_{1}=r-\sigma$, $r_{2}=r+\sigma ,$

(10)

and

${\frac {1}{2}}({{r_{1}}^{n}}+{{r_{2}}^{n}})=r^{n}(1+{\frac {n(n-1)}{2}}{\frac {\sigma ^{2}}{r^{2}}}+\&{\text{c.}});$

(11)

and since ${\theta _{2}}{r_{2}}$ may be neglected if the measurements are carefully made, we are sure that we may take the arithmetical mean of ${r_{1}}^{n}$ and ${r_{2}}^{n}$ for $r^{n}$.

Hence, taking the arithmetical mean of (8) and (9),

${\frac {1}{8}}{\frac {H}{M}}r^{3}(\tan \theta _{1}-\tan \theta _{2}+\tan \theta _{3}-\tan \theta _{4})=1+{A_{2}}{\frac {1}{r^{2}}}+\&{\text{c.,}}$

(12)

or, making

${\frac {1}{4}}(\tan \theta _{1}-\tan \theta _{2}+\tan \theta _{3}-\tan \theta _{4})=D,$ ${\frac {1}{2}}{\frac {H}{M}}Dr^{3}=1+{A_{2}}{\frac {1}{r^{2}}}+\&{\text{c.}}$

(13)

454.] We may now regard $D$ and $r$ as capable of exact determination.

The quantity $A_{2}$ can in no case exceed $2L^{2}$, where $L$ is half the length of the magnet, so that when $r$ is considerable compared with $L$ we may neglect the term in $A_{2}$ and determine the ratio of $H$ to $M$ at once. We cannot, however, assume that $A_{2}$ is equal to $2L^{2}$, for it may be less, and may even be negative for a magnet whose largest dimensions are transverse to the axis. The term in $A_{4}$, and all higher terms, may safely be neglected.

To eliminate $A_{2}$, repeat the experiment, using distances $r_{1},r_{2},r_{3},$ &c., and let the values of $D$ be $D_{1},D_{2},D_{3}$, &c., then

$D_{1}={\frac {2M}{H}}({\frac {1}{{r_{1}}^{3}}}+{\frac {A_{2}}{{r_{1}}^{5}}}),$ $D_{2}={\frac {2M}{H}}({\frac {1}{{r_{2}}^{3}}}+{\frac {A_{2}}{{r_{2}}^{5}}}),$
&c.	&c.