Determination of the Earth's Horizontal Magnetic Foice. 409 



This would hardly be sensible supposing X measured as usual to five 

 significant figures. It might happen however that owing to as}- mmetry 

 in the stirrup the C.G. of the magnet— with scale and lens — had to lie, 

 in the absence of vertical force, at a distance d, comparable with 1 mm. 

 on one side of the suspension fibre. Under such circumstances the 

 change in the moment of inertia when the vertical force altered from 

 0*44 to would exceed that calculated above in the ratio 



As perfect mechanical symmetry must be the exception, we see that 

 this source of uncertainty is likely to be considerably more serious 

 than might be supposed from our first example. 



The above considerations may suggest that shifting the collimator 

 magnet in its stirrup is a mistake. If, however, this is not done when 

 required to keep the magnet horizontal, the magnet and appendages 

 must tilt over slightly. This not only alters the moment of inertia, 

 but likewise prevents the magnetic couple in the vibration experiment 

 from having its proper value. 



The source of trouble we have just been considering is distinctly 

 exceptional in this respect, that it is most serious in magnets of large 

 magnetic moment. 



Law of Action between Magnets. 



§ 45. The formula 



couple = 2mm'r z (\ +P?- 2 + Q?- 4 + ) 



assumes that the centre of the mirror magnet lies on the prolongation 

 of the axis of the collimator magnet, and that the axes of the two 

 magnets are perpendicular to one another. With the magnets in the 

 position supposed, it is not difficult to calculate the values of P and Q, 

 if we can regard a magnet as consisting of two equal and opposite 

 " poles," the product of whose distance into the strength of either con- 

 stitutes the magnetic moment. Thus if 2 A. and 2 A/ be the distances 

 between the poles of the collimator and mirror magnets respectively, 

 I find 



couple = 



2mm V- 3 { 1 + (2X2 - 3 x'2) r -2 + s ( 8 >4 _ 40 A 2 A'2 + 15 A'*) r~ 4 + } 



2d : 0-013. 



so that 



P = 2X2- 



(15) , 



(16) . 



"When r = 30 cm. 



Pr-2 = (2X2-3X'2)/900 



Qr-4 = (8A 4 -40A2A'2+15A' 4 )10- 4 /216 



(17). 



