Page:Elementary Text-book of Physics (Anthony, 1897).djvu/283

§ 244] giyen by $$MH \, \sin \phi = MH \, \sin \phi,$$ if the oscillations are always very small. If $$I$$ represent the moment of inertia of the magnet, we have (§ 39) $$MH \, \phi = I \alpha,$$ where $$\alpha$$ is the angular acceleration. Now since the motion of each particle of the magnet is simple harmonic, and since the linear motions of the particles are proportional to their angular motions, we have $$\alpha = \frac{4\pi^2}{T^2}\phi$$ and by substituting this value of $$\alpha,$$ we obtain The moment of inertia $$I$$ may be either computed directly from the magnet itself, if it be of symmetrical form, or it may be determined experimentally by the method given in § 37, which applies in this case. The horizontal intensity is then determined relative to the magnetic moment of the assumed standard magnet.

This result may be used to give an absolute measure of $$H$$ by combining with it the result of another observation which gives an independent relation between $$M$$ and $$H.$$ In the arrangement of the apparatus two magnets are used : one, the deflected magnet, is so suspended as to turn freely in the horizontal plane; and the other, the deflecting magnet, the one of moment $$M$$ used in the last operation, is carried upon a bar which can be set at right angles to the magnetic meridian. The centre of the deflected magnet is in the prolongation of the axis of the deflecting magnet. The deflected magnet makes an angle with the magnetic meridian determined by the equality between the two couples acting on the deflected magnet, one arising from the action of the earth's magnetism, and the other from that of the deflecting magnet. This latter has been already discussed in § 243.

The couple due to the deflecting magnet is given by $$\frac{2MM'}{r^3} \cos \theta,$$ and that due to the earth's magnetism by $$M' H \, \sin \theta.$$ We have then  Equations (85) and (86) contain the unknown quantities $$M$$ and $$H$$