284 Solution of a Geometrical Problem in Magnetism. 



perpendicularly from the axis of the magnetometer-needle); 

 but the pole-distance m— n—2a is only implicitly known. 



Its value may be deduced from the equations (4) and (5); 

 thus 



m — n 2a a , 30 



or — = T = tann -^-, 



m-\-n 21 I 

 where 



a = \/ 



cosh 30 + 1 

 4 cosh 



The latter function is not hard to deduce from the tables. 



Experiments on this plan showed that the virtual distance 

 between the poles soon approaches the full length of the 

 magnet. 



It remains to give the expression for the field under the 

 circumstances. It is 



4M cosh 2 



d 3 (4 cosh 8 0-1)*' 



where M is the moment, or p(n— m). 



It will be noticed that we have ~j at our disposal if we allow 



d to vary, i. e. one degree of freedom. Suppose, therefore, 

 we arrange to simplify the expression for the field by putting 



cosh 2 0=L 

 4 



Then the field = ^J ; 



but at the same time, since 



cosh = —=-) 



cosh 30= 4/5, 



L- 4 / cosh 3(9 + 1 = / 5* + l \* 

 d V 4 C osh V2x5*/ 



V2 + 10/ 



= •850651. 



