﻿of finding the Forces acting in Magnetic Circuits. 97 



force per unit area at a point about 0. Draw OQ perpen- 

 dicular to OP and in the plane of the paper. Then the 

 pressural forces lie in a semicircle of which OQ is a radius, 

 and whose plane contains OQ. Since the pressures are 

 symmetrical with respect to OQ, OQ is their resultant, and 



by the theory this is equal to OP — so that OQ is the vector 

 representing the pressures. A force represented by OM equal 

 and parallel to PQ is therefore the resultant force, and clearly 

 in this case is a repulsion whose magnitude along the normal 



produced is 



OP.v/2 



cos 



(?-«> 



an expression which gives no normal component at all when 

 # = 7r/4. The force is therefore an attraction or repulsion 

 according as 6 is less or greater than 7r/4 — and a shear at this 

 point. I tried to observe this, but could not get the lines to 

 leave the surface at the exact angle. However, the above 

 way of looking at the matter is convenient when filings are 

 used to trace the direction of the induction. This expression 

 has been pointed out to me by Mr. Pollock as identical with 

 that given by Maxwell's general formula in § 643 for the 

 special case here considered. 



It is now evident why it was that Bosanquet got results 

 differing from those calculated from the formula for normal 

 inductions, because, as filings show, a very small gap is 

 sufficient to produce a marked spreading of the field. 



§ 14. By observing the distribution of filings about different 

 air-gaps, it appeared probable to me that the following pro- 

 position might be true — as referring to bars of different 



Phil Mag. S. 5. Vol. 38. No. 230. July 1894. H 



