Page:A Treatise on Electricity and Magnetism - Volume 2.djvu/379

 

Let $M$ be the magnetic moment, and $A$ the moment of inertia of the magnet and suspended apparatus,

If the time of the passage of the current is very small, we may integrate with respect to $t$ during this short time without regarding the change of $\theta$, and we find

This shews that the passage of the quantity $Q$ produces an angular momentum $MGQ\cos \theta _{0}$ in the magnet, where $\theta _{0}$ is the value of $\theta$ at the instant of passage of the current. If the magnet is initially in equilibrium, we may make $\theta _{0}=0$.

The magnet then swings freely and reaches an elongation $\theta _{1}$. If there is no resistance, the work done against the magnetic force during this swing is $MH(1-\cos \theta _{1})$.

The energy communicated to the magnet by the current is

Equating these quantities, we find

	⁠	⁠	$\left.{\overline {\frac {d\theta }{dt}}}\right|^{2}$	${}=2{\frac {MH}{A}}(1-\cos \theta _{1})$,	⁠	⁠	(4)
whence			${\frac {d\theta }{dt}}$	${}=2{\sqrt {\frac {MH}{A}}}\sin {\frac {1}{2}}\theta _{1}$
				${}={\frac {MG}{A}}Q$ by (3).			(5)

But if $T$ be the time of a single vibration of the magnet,

	⁠	⁠	$T=\pi {\sqrt {\frac {A}{MH}}}$,	⁠	⁠	(6)
and we find			$Q={\frac {H}{G}}{\frac {T}{\pi }}2\sin {\frac {1}{2}}\theta _{1}$,			(7)

where $H$ is the horizontal magnetic force, $G$ the coefficient of the galvanometer, $T$ the time of a single vibration, and $\theta _{1}$ the first elongation of the magnet.

749.] In many actual experiments the elongation is a small angle, and it is then easy to take into account the effect of resistance, for we may treat the equation of motion as a linear equation.

Let the magnet be at rest at its position of equilibrium, let an angular velocity $\nu$ be communicated to it instantaneously, and let its first elongation be $\theta _{1}$.