196 ON MAGNETISM. 



EG x KL - 2 area of triangle EKG ab . sin < : there- 

 fore the momentum of the strain in the direction EG 

 to produce rotation of the magnet is 



W 



, ab . sin <f>. 



A similar momentum is produced by the strain in the 

 direction FH : therefore the whole momentum of rota- 

 tion is 



W.ab . 

 sm (f>. 



Now let the upper suspension-bar be turned round 

 till the magnet is turned to a position at right angles 

 to the magnetic meridian. The momentum of terres- 

 trial horizontal magnetism upon it, by Article 21, 

 supposing it inclined to the magnetic meridian by 

 the angle 0, will be E . B. sin 6: and sin 6 will sensibly 

 = 1, not only when 6 = 90, but also when 6 = 90 4- #, 

 where a; is a small angle (such as we have to consider) 



a: 2 

 which makes sin 6 1 + &c. We shall therefore 



consider the momentum of terrestrial horizontal mag- 

 netism &s=E. J3. And, as this balances the momentum 

 of torsion, we have the equation 



