10 The Rev. H. Luoyp on the Determination of the Intensity 
XM’ sin &. 
Hence the equation of equilibrium is 
2m . M3 : M;\ 1 
in w% = — 2—+3(5cosy— 1 =I. 
xsin& = — sin y 1+ ( a + 3( Va ) =a 
There are two cases of this solution which demand our consideration. 
In the method of Gauss, the deflecting magnet is perpendicular to the mag- 
netic meridian, and therefore y= 90°—w. In this case, then, the preceding 
equation becomes 
2M M; M3 Ae ines SI 
< = —i1 (2—— = 15 sin’ w -) —. 
aioe D* | +. M M 1 M 
Accordingly, the term containing the fifth power of the distance is composed of 
two parts, one of which is constant, while the other varies with the angle of de- 
flection ; so that, if there were no means of determining @ priori the values of 
/ 
. M3 Mz é Bie : s : 
the ratios, —, —, three equations of condition would be, in strictness, required for 
M’ M 
the determination of the three unknown quantities ;—namely, the coefficient of 
the inverse cube of the distance, and the two parts of the coefficient of the 
inverse fifth power. However, the distance beg greater than four times the 
length of the magnet, the angle of deflection, w, is always small, and the term 
involving the square of its sine may be neglected in comparison with the others. 
Accordingly, if we make, for abridgment, 
2M M, 
— — Q, 2 = —_ = h, 
Xx 
the expression for the tangent of the angle of deflection is reduced to the form 
Q h 
tan uw = — (1 =) 5 
ace Cat 
In the method of deflection employed by Professor Lamont, the deflecting bar 
is always perpendicular to the suspended bar. In this case, therefore, y= 90’, 
and the equation of equilibrium is reduced to 
l M M 
: 2M f M M3\ 1 
; = -— Siewetvtes t psc Se 
X sin &@ 53 1+ ( 3 ) i 
