574 MAGNETIC FIELD. 



1046. If the direction of the field is unknown, the measure- 

 ment of the deflections and that of the oscillations, in the two 

 cases, give the direction of the field, and its value as a function 

 of the terrestrial field. For let a be the angle of the field F with 

 the meridian, 8 the deflection, N and N the numbers of oscillations 

 which correspond to the original direction and to the deflected 

 needle, R the resultant of the forces F and H, we have, neg- 

 lecting the induced magnetisation, 



F R H 



sin 8 sin a sin (a - 8) ' 

 N 2 R 



from which is deduced 



COt a = COt 8 - 



1047. The method of deflections may also be utilised even 

 when the fields to be compared are parallel to the terrestrial field. 



Let us assume that the needle of a declinometer of moment M is 

 supported by a bifilar suspension, and deflected through an angle 0. 

 The equation of equilibrium is 



(7) HMsin0 = Csin(o>-0). 



to being the torsion of the system from the meridian. 



In these conditions, if the needle is displaced through an angle 8, 

 the couple which tends to restore it to its primitive position is 



P = HM sin (0 + 8) -C sin (a- 0-8) 

 -T-g sm(0 + 8)sin(tt-0)-sm0sin(tt-0-8) . 



Replacing the products of the sines by sums of cosines, we find 

 finally, 



sm sin (w - 0) 



