80 



EXPLORATION GEOPHYSICS 

 H -^^* 



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Case II. Magnet broadside: Position II of Gauss. (See Figure 15.) 



Fig. 15. — Magnetic force at a point due to a bar magnet: 

 Position II of Gauss. 



By the figure, from similar triangles, 



Hn r ^ r H 



Hence : Hp = — g- 



but r = W2 + F 



so that Hp = 



2m I 



((/2 + /2)3/2 



As before, M = 2ml and / may be neglected, hence : 



M 



Hn — 



d^ 



(22) 



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Case III. Magnet in any position : Position III of Gauss. (See Figure 16.) 



We can replace magnet NS by two magnets, one end on to P and the 



other perpendicular to the first. Then if the original magnet has a moment 



* If small in comparison to d, I may be neglected. 



