Page:Elementary Text-book of Physics (Anthony, 1897).djvu/285

§ 246] of force determined by the contours of these areas will be called unit tubes. Since $$\frac{1}{s} = N$$ is the number of unit tubes which pass through a unit area, we have $$N = F.$$ Hence the magnetic force at a point is equal to the number of unit tubes which pass perpendicularly through unit area at that point.

246. Magnetic Force and Magnetic Induction.—If a body like a piece of iron be brought into a magnetic field it will be magnetized by induction, and will in turn affect the distribution of the tubes of force in the field. To determine the way in which its magnetization is affected, we need some convention upon which we shall measure the force within it. This force is measured within a cavity in the mass of iron, and will have different values for different forms of the cavity. Let us first suppose the cavity cylindrical, with its axis in the direction of the lines of force, and with its bases infinitesimal in comparison with its height. The distribution of magnetism throughout the iron will give rise to free magnetism on the bases of the cylinder, and the force on a unit pole placed at the middle of the cylinder will be due to the original force in the field, to the forces arising from the induced poles, and to the forces exerted by these distributions. These last forces are infinitesimal, in case the bases are infinitesimal in comparison with the length of the cylinder, and may be neglected. The force on the pole is called the magnetic force at the point where the pole is placed.

Let us in the second place suppose the cavity in the shape of a disk, with its faces normal to the lines of force, and infinitely great in comparison with the distance between them. The magnetization of the iron will give rise to a distribution of free magnetism on each of these faces. If we consider the lines of force of the field as passing into the iron from left to right, the distribution on the left-hand face of the cavity is positive and that on the right-hand face negative. The force on a unit pole placed within the cavity is due to the original force $$R$$ in the field, and the forces exerted by the distributions on the faces of the cavity. If we represent by $$\sigma$$ the density of this distribution on either face, the force on the pole