Page:Philosophical magazine 21 series 4.djvu/191

coming more numerous towards the right. It may be shown that if the force increases towards the right, the lines of force will be curved towards the right. The effect of the magnetic tensions will then be to draw any body towards the right with a force depending on the excess of its inductive capacity over that of the surrounding medium.

We may suppose that in this figure the lines of force are those surrounding an electric current perpendicular to the plane of the paper and on the right hand of the figure.

These two illustrations will show the mechanical effect on a paramagnetic or diamagnetic body placed in a field of varying magnetic force, whether the increase of force takes place along the lines or transverse to them. The form of the second term of our equation indicates the general law, which is quite independent of the direction of the lines of force, and depends solely on the manner in which the force varies from one part of the field to another.

We come now to the third term of the value of X,

$-\mu \beta {\frac {1}{4\pi }}\left({\frac {d\beta }{dx}}-{\frac {d\alpha }{dy}}\right).$

Here $\mu \beta$ is, as before, the quantity of magnetic induction through unit of area perpendicular to the axis of $y$, and ${\tfrac {d\beta }{dx}}-{\tfrac {d\alpha }{dy}}$ is a quantity which would disappear if $\alpha dx+\beta dy+\gamma dz$ were a complete differential, that is, if the force acting on a unit north pole were subject to the condition that no work can be done upon the pole in passing round any closed curve. The quantity represents the work done on a north pole in travelling round unit of area in the direction from +$x$ to +$y$ parallel to the plane of $xy$. Now if an electric current whose strength is $r$ is traversing the axis of $z$, which, we may suppose, points vertically upwards, then, if the axis of $x$ is east and that of $y$ north, a unit north pole will be urged round the axis of $z$ in the direction from $x$ to $y$, so that in one revolution the work done will be $=4\pi r$. Hence ${\tfrac {1}{4\pi }}\left({\tfrac {d\beta }{dx}}-{\tfrac {d\alpha }{dy}}\right)$ represents the strength of an electric current parallel to z through unit of area; and if we write

${\frac {1}{4\pi }}\left({\frac {d\gamma }{dy}}-{\frac {d\beta }{dz}}\right)=p,\ {\frac {1}{4\pi }}\left({\frac {d\alpha }{dz}}-{\frac {d\gamma }{dx}}\right)=q,\ {\frac {1}{4\pi }}\left({\frac {d\beta }{dx}}-{\frac {d\alpha }{dy}}\right)=r,$

(9)

then p, q, r will be the quantity of electric current per unit of area perpendicular to the axes of x, y, and $z$ respectively.

The physical interpretation of the third term of X, $-\mu \beta r$, is that if $\mu \beta$ is the quantity of magnetic induction parallel to $y$, and $r$ the quantity of electricity flowing in the direction of $z$, the