Page:A Treatise on Electricity and Magnetism - Volume 2.djvu/340

 

of magnets or currents, is subject to the following equation

To prove this, let us consider the number of lines of magnetic force cut by the circle when $a$ or $b$ is made to vary.

(1) Let $a$ become $a+\delta a$, $b$ remaining constant. During this variation the circle, in expanding, sweeps over an annular surface in its own plane whose breadth is $\delta a$.

If $V$ is the magnetic potential at any point, and if the axis of $y$ be parallel to that of the circle, then the magnetic force perpendicular to the plane of the ring is ${\frac {dV}{dy}}$.

To find the magnetic induction through the annular surface we have to integrate

$\int _{0}^{2\pi }a\delta a{\frac {dV}{dy}}\,d\theta$,

where $\theta$ is the angular position of a point on the ring. But this quantity represents the variation of $M$ due to the variation of $a$, or ${\frac {dM}{da}}\delta a$. Hence

(2) Let $b$ become $b+\delta b$, $a$ remaining constant. During this variation the circle sweeps over a cylindric surface of radius $a$ and length $\delta b$.

The magnetic force perpendicular to this surface at any point is ${\frac {dV}{dr}}$ where $r$ is the distance from the axis. Hence

Differentiating equation (2) with respect to $a$, and (3) with respect to $b$, we get

	⁠	${\frac {d^{2}M}{da^{2}}}$	${}=\int _{0}^{2\pi }{\frac {dV}{dy}}d\theta +\int _{0}^{2\pi }a{\frac {d^{2}V}{dr\,dy}}d\theta$,		⁠	(4)
		${\frac {d^{2}M}{db^{2}}}$	${}=-\int _{0}^{2\pi }a{\frac {d^{2}V}{dr\,dy}}d\theta$,			(5)
Hence		${\frac {d^{2}M}{da^{2}}}+{\frac {d^{2}M}{db^{2}}}$		${}=\int _{0}^{2\pi }{\frac {dV}{dy}}d\theta$,		(6)
				${}={\frac {1}{a}}{\frac {dM}{da}}$, by (2).

Transposing the last term we obtain equation (1).