Page:Scientific Memoirs, Vol. 2 (1841).djvu/88

If we divide the equation (II.) by (III.) we obtain

${\frac {mM}{mT}}={\frac {FR^{3}\cdot \mathrm {tang} \,v}{2F}},$

whence the independence of the deflection $v$, both of the magnetism of the needle $m$, and of the moment of rotation $F$, is evident of itself, and we have the following simple result:

${\frac {M}{T}}={\frac {R^{3}\mathrm {tang} \,v}{2}}$

(IV.)

"The determination of the intensity of the magnetism of the globe is therefore reduced to two principal operations.

"I. To observe the time of vibration of a needle $N\;S$, and to deduce from thence the moment of rotation which the terrestrial magnetism exerts on it."

 

This moment of rotation will be expressed by the product $M\;T$, and calculated by the equation (I.)

$T={\frac {C}{Mt^{2},}}\;\;{\mbox{or }}MT={\frac {C}{t^{2}}}$

in which $C$ represents the moment of inertia of the bar, multiplied by the number, $\pi ^{2}$, or 9·8696 … and divided by the double of the space of the fall of a heavy body in the unit of time.

 

"II. A second needle, $n\;s$, being suspended: its position is observed; first, when subject to the influence of the earth's magnetism alone; and secondly, after $N\;S$ has been placed at a considerable distance, as represented in the figure. Then calculate from the difference between the two positions, or from the deflection, what fraction of the force of the earth's magnetism, the magnetic force of the needle $N\;S$, corresponds to at the selected distance. An equal fraction of the moment of rotation, found in (I.,) gives the moment of rotation which the needle $N\;S$ at that distance would impart to a similar one; this result, multiplied by the cube of the distance, gives the reduced moment of rotation; the square root of this gives the force of the needle $N\;S$ in absolute measure: and finally, the number found in (I.) divided by this square root, gives the expression for the absolute measure of the earth's magnetism."