# Page:A Treatise on Electricity and Magnetism - Volume 2.djvu/55

399.]

MAGNETIC FORCE IN A CAVITY.

23

that the magnetization of the part removed may be regarded as everywhere parallel to the axis of the cylinder, and of constant magnitude I, the force on a magnetic pole placed at the middle point of the axis of the cylindrical hollow will be compounded of two forces. The first of these is that due to the distribution of magnetic matter on the outer surface of the magnet, and throughout its interior, exclusive of the portion hollowed out. The components of this force are α, β, γ, derived from the potential by equations (1). The second is the force R, acting along the axis of the cylinder in the direction of magnetization. The value of this force depends on the ratio of the length to the diameter of the cylindric cavity.

398.] Case I. Let this ratio be very great, or let the diameter of the cylinder be small compared with its length. Expanding the expression for R in terms of ${\displaystyle |frac{a}{b}}$, it becomes

 ${\displaystyle R=4\pi I\left\{{\frac {1}{2}}{\frac {a^{2}}{b^{2}}}-{\frac {3}{8}}{\frac {a^{4}}{b^{4}}}+{\And }c.\right\},}$ (3)

a quantity which vanishes when the ratio of b to a is made infinite. Hence, when the cavity is a very narrow cylinder with its axis parallel to the direction of magnetization, the magnetic force within the cavity is not affected by the surface distribution on the ends of the cylinder, and the components of this force are simply α, β, γ, where

 ${\displaystyle \alpha =-{\frac {dV}{dx}},\quad \beta =-{\frac {dV}{dy}},\quad \gamma =-{\frac {dV}{dz}}.}$ (4)

We shall define the force within a cavity of this form as the magnetic force within the magnet. Sir William Thomson has called this the Polar definition of magnetic force. When we have occasion to consider this force as a vector we shall denote it by ${\displaystyle {\mathfrak {A}}}$.

399.] Case II. Let the length of the cylinder be very small compared with its diameter, so that the cylinder becomes a thin disk. Expanding the expression for R in terms of ${\displaystyle {\frac {b}{a}}}$, it becomes

 ${\displaystyle R=4\pi I\left\{1-{\frac {b}{a}}+{\frac {1}{2}}{\frac {b^{3}}{a^{3}}}-\And \!{c}.\right\},}$ (5)

the ultimate value of which, when the ratio of a to b is made ${\displaystyle 4\pi I}$.

Hence, when the cavity is in the form of a thin disk, whose plane is normal to the direction of magnetization, a unit magnetic pole