THE INVERSE SQUARE LAW 217 



Tube of force. If a small closed curve is drawn at any place 

 in a field, and if a line offeree is drawn through every point of it, the 

 lines so drawn will enclose a tubular space termed a tube of force. 



Let us apply Gauss's theorem to the closed 

 surface in air or other non-magnetised medium, 

 formed by a portion of a tube of force, and any 

 two cross-sections of it at right angles to the lines 

 of force. Thus, let S x , S 2 , Fig. 164, be the two 

 cross-sections, and let H x , H 2 be the intensities 

 at S 15 S 2 . The normal intensity over the curved 

 sides of the tube is everywhere zero. At Sj it 

 is H x and at S 2 it is H 2 , negative because H 2 

 is inwards. We have, therefore, 



H! x S x HL X S 2 = 4-7T X magnetisation 

 within the tube = O FIG. 164. 



That is, the product H x S, or the flux of force, 



is constant throughout the tube so long as it does not contain 



magnetisation. 



Unit tubes. If we draw tubes so that for each, intensity X 

 cross-section = 1, that is, so that there is unit flux of force along 

 it, each is termed a unit tul>e. 



A pole m sends out l^m unit tubes, for if we draw a 

 sphere of radius r round m as centre the intensity on the surface of 

 the sphere is m/r* 9 and the total flux of force through the sphere 

 is 4xr 2 w/r 2 = 4-7TW. There are, therefore, 4>7rm tubes passing 

 through the surface. 



If we imagine one line of force along each unit tube the 

 number of lines of force passing through any area will be equal 

 to the total flux of force through the area. 



The number of unit tubes or of lines of force passing through 

 unit area perpendicular to them is equal to 1 /cross-section of 

 unit tube or is equal to the intensity of the field. 



Potential. If we define the potential at any point in the 

 field as the work done in bringing a unit positive pole to the point 

 from a distance so great that the field is negligible, we can show 

 (sec Electricity, p. 42) that 



where every element of magnetisation dm is to be divided by its 

 distance r from the point. 



If the unit pole is moved a distance dx in any direction x to 



a point where the potential is V + T- dx, the work done is -^ dr. 

 So that the force on the unit pole opposing the motion is 



