BIFILAR OR HorIZONTAL ForcE MAGNETOMETER. XXVil 
30. If v be the excess of the angular motion of the arms of the torsion circle, 
or upper extremities of the wire, over wv, that of the lower extremity or magnetic bar 
in moving the latter from the meridian, the equation of equilibrium will be 
2 
+) a . 
m X sin u — W—sin 
m, X, W, a, and J being respectively the magnetic moment of the bar, the hori- 
zontal component of the earth’s magnetic force, the weight suspended, the interval, 
and the length of the wires. The differential of this equation (w = 90°) divided by 
it, gives 
aX 
aie na cotv+¢ (Q+2e—e') 
n being the number of scale divisions from the zero, or scale reading when u=90", 
athe are value in parts of radius of one scale division, ¢ the number of degrees 
Fahrenheit which the temperature of the magnet is above the adopted zero, Q the 
coefficient of the temperature correction for the varying magnetic moment of the 
bar or the value of <" for 1° Fahr., e and e’ the coefficients of expansion for the 
brass of the grooved wheel and silver of the wires. 
31. It is assumed, in the previous investigation, that the suspending wire does 
not act by any inherent elastic force ; that the torsion force depends wholly on the 
length and interval of the two portions of the wire and the angle of twist: it seems 
extremely probable that this condition will not be rigorously sustained, and it is 
very possible that there may be considerable twist in the suspending wire or thread ; 
for this reason, the following methods, which are independent of the angle of torsion, 
were employed to determine the coefficient :— 
32. If the equation of equilibrium for the bifilar magnet at right angles to the 
magnetic meridian be 
Pe TEN EMD Al poeee TERR aI IS CTR 
and if a magnet whose magnetic moment is M be placed with its axis in the mag- 
netic meridian passing through the centre of the bifilar bar, the centres of the two 
bars being at a distance 7, and the resulting angle of deflection be n scale divisions 
=a, the equation of equilibrium will be 
2M pg aes 
m{X+2% (14+ 444) } cos Av=F.. 
For a value of the earth’s horizontal foree X+4X, which would alone have pro- 
duced the deviation av; we have 
m(X+4X) cos av=F"; 
