Page:Elementary Text-book of Physics (Anthony, 1897).djvu/286

272 due to each face is $$2\pi \sigma$$ (§ 57), and the forces from the two faces act in the same direction. The force upon the pole due to the faces of the cavity is therefore $$4\pi \sigma$$ or $$4\pi I,$$ where $$I$$ is the intensity of magnetization (§ 241). The total force acting on the pole is therefore this quantity is called the magnetic induction within the body. It is manifestly a directed quantity or a vector; in the example considered its direction is the same as that of the force. In bodies, which are not isotropic, the induction and the magnetizing force are not necessarily in the same direction. We will confine our attention to isotropic bodies.

'''247. Tubes of Force of a Magnet. Tubes of Induction.'''—The lines of force in the field outside a bar magnet are curves proceeding from the north to the south pole; these lines of force may be conceived of as existing also within the body of the magnet if the magnetic force within the magnet is determined within the long narrow cylinder already described. On the other hand, a set of tubes may be described, called tubes of indiiction, which are closed tubes, proceeding through the magnet from the south to the north pole and outside the magnet from the north to the south pole. To show this let us suppose the magnet divided into two parts by a section at right angles to its axis, and let us consider a closed surface passing between the two parts and enclosing that part which contains the north pole. The distribution over the end of this part which is exposed by the section is equal to the north pole at the end of the original magnet, and is of opposite sign; so that the flux of force $$\textstyle \sum \displaystyle Fs = 0$$ over the closed surface. Nov/ the force within the cavity formed by the section is directed inward, and is at each point equal to $$4\pi I.$$ If $$a$$ represent the area of the section, the flux of force across that part of the surface contained within the section is $$4\pi Ia,$$ and $$Ia = m,$$ the strength of the pole of the original magnet. The number of tubes of force which pass through the section of the magnet is therefore equal to $$4\pi m,$$ or to the number of tubes of force which proceed from the original pole and pass