Page:A Treatise on Electricity and Magnetism - Volume 2.djvu/145

Let $p$ be the perpendicular from the centre of inertia on the plane on which the axis rolls, then $p$ will be a function of $\theta$, whatever be the shape of the rolling surfaces. If both the rolling sections of the ends of the axis are circular,

where $a$ is the distance of the centre of inertia from the line joining the centres of the rolling sections, and $\alpha$ is the angle which this line makes with the line of collimation.

If $M$ is the magnetic moment, $m$ the mass of the magnet, and $g$ the force of gravity, $I$ the total magnetic force, and $i$ the dip, then, by the conservation of energy, when there is stable equilibrium,

must be a maximum with respect to $\theta$, or

$MI\sin(\theta +\lambda -i)$	${}=-mg{\frac {dp}{d\theta }}$,
	${}=-mga\cos(\theta +\alpha )$,

(3)

if the ends of the axis are cylindrical.

Also, if $T$ be the time of vibration about the position of equilibrium,

where $A$ is the moment of inertia of the needle about its axis of rotation.

In determining the dip a reading is taken with the dip circle in the magnetic meridian and with the graduation towards the west.

Let $\theta _{1}$ be this reading, then we have

The instrument is now turned about a vertical axis through 180°, so that the graduation is to the east, and if $\theta _{2}$ is the new reading,

Taking (6) from (5), and remembering that $\theta _{1}$ is nearly equal to $i$ and $\theta _{2}$ nearly equal to $\pi -i$, and that $\lambda$ is a small angle, such that $mga\lambda$ may be neglected in comparison with $MI$,

Now take the magnet from its bearings and place it in the deflexion apparatus, Art. 453, so as to indicate its own magnetic moment by the deflexion of a suspended magnet, then

where $D$ is the tangent of the deflexion.