Page:A Treatise on Electricity and Magnetism - Volume 2.djvu/127

$\zeta =\alpha +{\frac {\pi }{2}}-\lambda ,$

(1)

$\zeta '=\alpha -{\frac {\pi }{2}}+\lambda ,$

(2)

Hence

$\lambda ={\frac {\pi }{2}}+{\frac {1}{2}}(\zeta '-\zeta ).$

(3)

Next, hang the stirrup to the suspension fibre, and place the magnet in it, adjusting it carefully so that y may be vertical and upwards, then the moment of the force tending to increase α is

$MH\sin m\sin \left(\delta -\alpha -{\frac {\pi }{2}}+l\right)-\tau (\alpha -\beta -\gamma ).$

(4)

But if ζ is the observed azimuth of the line of collimation

$\zeta =\alpha +{\frac {\pi }{2}}-\lambda ,$

(5)

so that the force may be written

$MH\sin m\sin(\delta -\zeta +l-\lambda )-\tau \left(\zeta +\lambda -{\frac {\pi }{2}}-\beta -\gamma \right).$

(6)

When the apparatus is in equilibrium this quantity is zero for a particular value of ζ.

When the apparatus never comes to rest, but must be observed in a state of vibration, the value of ζ corresponding to the position of equilibrium may be calculated by a method which will be described in Art. 735.

When the force of torsion is small compared with the moment of the magnetic force, we may put δ - ζ + l - λ for the sine of that angle.

If we give to β, the reading of the torsion circle, two different values, β₁ and β₂, and if ζ₁ and ζ₂ are the corresponding values of ζ,

$MH\sin m(\zeta _{1}-\zeta _{2})\tau (\zeta _{1}-\zeta _{2}-\beta _{1}+\beta )2),\,$

(7)

or, if we put

${\frac {\zeta _{1}-\zeta _{2}}{\zeta _{1}-\zeta _{2}-\beta _{1}+\beta _{2}}}=\tau '{\text{ then }}\tau =MH\sin m\tau ',$

(8)

and equation (7) becomes, dividing by M H sin m,

$\delta -\zeta +l-\lambda -\tau '\left(\zeta +\lambda -{\frac {\pi }{2}}-\beta -\gamma \right)=0.$

(9)

If we now reverse the magnet so that y is downwards, and adjust the apparatus till y is exactly vertical, and if ζ' is the new value of the azimuth, and δ' the corresponding declination,

$\delta '-\zeta '-l+\lambda -\tau '\left(\zeta '+\lambda +{\frac {\pi }{2}}-\beta -\gamma \right)=0.$

(10)

whence

${\frac {\delta +\delta '}{2}}={\frac {1}{2}}(\zeta +\zeta ')+{\frac {1}{2}}\tau '(\zeta +\zeta '-2(\beta +\gamma )).$

(11)