Page:Elementary Text-book of Physics (Anthony, 1897).djvu/284

270 in different relations. By the elimination of one of them the other can be obtained in absolute units. In practice the simple conditions assumed in this discussion cannot be obtained, and corrections must be introduced, arising from the departures from these conditions. In the determination of $$MH$$ we must take into account the change of the magnetic moment by the induction of the field, and the facts that the oscillations are not infinitesimal, and that they are affected by the friction of the air and the torsion of the suspension fibre. In the determination of $$\frac{M}{H},$$ we must take into account the induction of one magnet on the other, and the fact that the lengths of the magnets are not negligible in comparison with the distance between them.

245. The Magnetic Field.—Up to this point our discussion has been conducted on the supposition that forces obeying a definite law act directly between magnetic poles. Various phenomena, especially those of magnetic induction and the relation between magnetism and the electrical current, as well as the general tendency of modern speculation in physics, lead us to think that this mode of representing the interactions of magnets is an artificial one, and that the true seat of the magnetic action is in the medium between the magnets. From this point of view it is desirable to have a mode of describing the magnetic field in such a way that the relation between its condition and the magnetic forces exhibited may be expressed in measurable terms. The most useful mode of describing the field is that depending upon the use of lines and tubes of force.

Since the force due to a magnet pole obeys the law of inverse squares, the theorems of §§ 53-57 are immediately applicable.

For the sake of clearness in statement we will define a unit tube of force in the following way: Consider a closed surface which encloses a quantity of free positive magnetism $$m.$$ By § 56, $$\textstyle \sum \displaystyle Fs = 4\pi m,$$ where $$F$$ represents the normal component of the force at each point on the surface and $$s$$ the area of the element of surface over which it acts. Now let us suppose the whole surface divided into $$4\pi m$$ parts, for each of which $$Fs = 1;$$ then the tubes