1884.] Theory of the Magnetic Balance of Hughes. 321 



compensator NS, which has been turned through an angle OMS = 

 in order to . balance the magnetic force exerted on the suspended 

 needle at B by the specimen of iron or steel placed at A. It is 

 required to determine the relation between the angle of turning 0, 

 and the magnetic force thereby brought to bear along the axis AX at 

 B tending to thrust back the suspended needle to its zero point. Let 

 the length of the compensator be called 21, and the distance of B from 

 M=r. It is evident that, in general, the resultant magnetic force at 

 B due to the compensator will not be along the axis AX, but may be 

 resolved into a part acting at right angles to that axis, and a part 

 parallel to it. The former part will be parallel to the magnetic 

 meridian, and, therefore, when the needle is at zero, this part will not 

 tend to move the needle to either side. Its only action is to increase ■ 

 or diminish (according to circumstances) the directive force of the 

 earth's magnetism upon the needle, and render it either more or less 

 sensitive. The other part is that which acts along the axis and 

 balances the magnetic force of the iron or steel bar at A. When the 

 compensator lies at right angles to AX this component of the force 

 vanishes by symmetry : consequently the zero of the scale lies in this 

 line. If the compensator be turned through an angle 0, the com- 

 ponent of its magnetic force in the line AX will increase to a 

 maximum when 0=90°. The values of the force for different angles 

 may be calculated as follows : — 



Call BS = r', 



BN=r", 



angle SBM=«, 



angle NBM=/3. 



Then the forces along BS and BN resolved along AB will give the 

 following resultant F : — 



Jcos*_cos/3\ . 

 I r' 2 r" 2 j ' 



m being the strength of the pole of the compensator. 

 But cos a= 



COS /3: 



0-2 + Z 2 -2rZ s in0)*' 

 r + 1 sin 



(r 2 + Z 2 + 2rZsin0)* 7 

 and r ' = (r 3 + 1 2 — 2r I sin 0) * ; 



r"=O 2 + Z 2 + 2rZsin0)*. 



Whence F-=m( r-Zsinft, r + l sin? j 



l( r 2 + ^_2 r Zsin0)t (r 2 + V + 2rl sin 0)1 J V 75 



z2 



