228 MAGNETISM 



2C 1 

 the intensity of which is , where C is the current and r the 



distance from the axis of the wire, the forces on the pole, though 



equal, will not be opposite. Let 

 O, Fig. 176, be the centre of the 

 wire, NS the small magnet. The 



. /-, 2C//t 



forces on N and S are , 



tangential to the circular line of 

 force through NS. We may take 

 the sides of the triangle ONS to 

 represent the two forces and their 

 FlQ 176 resultant, each side being per- 



pendicular to the force which it 

 represents. Hence the resultant F is given by 



= NS : ON = / : r. 



^, 



Then J< = 



r* 



and it is directed inwards towards O and again towards tin- 

 stronger part of the field. 



We can obtain an expression for the force acting in any field 

 by finding the potential energy of the magnet with respect to the 

 field. Suppose that the magnet is brought from a great distance, 

 where the field is zero, to its actual position, and let it move always 

 so that the South pole follows in the track of the North pole. 

 The total work done in moving the South pole is equal and opposite 

 to the work done in moving the North pole up to the point finally 

 occupied by the South pole, and the potential energy will 

 therefore be the work none in moving the North pole the 

 extra distance equal to the length of the magnet. Let H 

 be the intensity of the field, then the potential energy will be 

 V= -mlH= -MH. 



The force in any direction x will be 



d\ _ dH 



~~" ~~5 WL 



dr ac 



It is therefore the greatest in the direction in which H increases 

 most rapidly, or the magnet tends to move to the strongest part 

 of the field. 



If the magnet be turned end for end, and be constrained to 

 keep in that direction, the forces on the poles are just reversed, and 

 the magnet tends to move in the direction of most rapid decrease 

 of H or into the weakest part of the field. 



