FORCES ON MAGNETISED BODIES 263 



This system, like the electric system, will maintain each element 

 of a body in which ^ is constant in equilibrium. Tne forces 

 acting on the two sides of an element of surface where /UL changes 

 will not be in equilibrium, and their resultant ought to be equal 

 to the mechanical force acting on the surface, equal, in fact, to the 

 forces deduced by supposing that the imaginary surface layer is 

 acted on by the field existing about it. We shall find that the 

 two systems are not equivalent, element by element, but that they 

 give the same resultant force on the body as a whole. Perhaps 

 this is as much as could be expected. 



The tensions and pressures are in equilibrium on any ele- 



ment of a medium in which JUL is constant, exactly as the tensions and 



KE 2 



pressures -^ are in equilibrium on an element of dielectric. Any 



O7T 



portion of a medium with /* constant is therefore in equilibrium, 

 and if we draw a closed surface S wholly within the medium, the 

 tensions and pressures acting on S from outside form a system in 

 equilibrium. Equally the reactions from inside on the medium 

 outside S are in equilibrium. If we imagine the medium within S 

 to have the same distribution of induction, but some other constant 

 value of ft, we can superpose the stresses due to this on the actual 

 system without destroying equilibrium. Obviously we need not 

 even have the distribution of induction the same in the superposed 

 system so long as it is a possible distribution obeying the tube law, 

 but we shall only require the more limited case. 



The stresses on an element of surface separating 

 media of permeabilities ^ and 

 /z 2 . Let the plane of Fig. 203 be that 

 through the normal to the surface and 

 the axis of a tube in the /^ medium, 

 meeting the surface in an element d$ 

 represented by its trace AB. Let the 

 tube make 1 with the normal. Let 

 the tube be rectangular in section with 

 two sides parallel to the plane of the 

 figure, and consider the equilibrium of 

 the wedge ABC. The forces across AC 

 and BC must give the force across AB. 

 Let N be the normal force outwards across AB and T the tangential 

 force in the direction AB. Since BC = dS cos and AC = dS sin 0, 

 we have, on resolving the assumed tension and pressure, 



^i 2 dS cos 2 0! - 

 OTT 



or 



N = co, 2 - sin" 0) = l cos 



