Page:A Treatise on Electricity and Magnetism - Volume 2.djvu/338

 

is moderate as compared with the radii of the smaller, the series already given do not converge rapidly. In every case, however, we may find the value of $M$ for two parallel circles by elliptic integrals.

For let $b$ be the length of the line joining the centres of the circles, and let this line be perpendicular to the planes of the two circles, and let $A$ and $a$ be the radii of the circles, then

$M=\iint {\frac {\cos \epsilon }{r}}ds\,ds^{\prime }$,

the integration being extended round both curves. In this case,

$r^{2}=A^{2}+a^{2}+b^{2}-2Aa\cos(\phi -\phi ^{\prime })$,

$\epsilon =\phi -\phi ^{\prime }$, $ds=a\,d\phi$, $ds^{\prime }=A\,d\phi ^{\prime }$,

	$M$	${}=\int _{0}^{2\pi }\int _{0}^{2\pi }{\frac {Aa\cos(\phi -\phi ^{\prime })\,d\phi \,d\phi ^{\prime }}{\sqrt {A^{2}+a^{2}+b^{2}2Aa\cos(\phi -\phi ^{\prime })}}}$
		${}=2\pi {\sqrt {Aa}}\left\{\left(c-{\frac {2}{c}}\right)F+{\frac {2}{c}}E\right\}$,

and $F$ and $E$ are complete elliptic integrals to modulus $c$. From this we get, by differentiating with respect to $b$ and remembering that $c$ is a function of $b$,

If $r_{1}$ and $r_{2}$ denote the greatest and least values of $r$,

$r_{1}^{2}=(A+a)^{2}+b^{2}$, $r_{2}^{2}=(A-a)^{2}+b^{2}$,

and if an angle $\gamma$ be taken such that $\cos \gamma ={\frac {r_{2}}{r_{1}}}$,

where $F_{\gamma }$ and $E_{\gamma }$ denote the complete elliptic integrals of the first and second kind whose modulus is $\sin \gamma$. If $A=a$, $\cot \gamma ={\frac {b}{2a}}$, and

${\frac {dM}{db}}=2\pi \cos \gamma \{2F_{\gamma }-(1+\sec ^{2}\gamma )E_{\gamma }\}$.

The quantity ${\frac {dM}{db}}$ represents the attraction between two parallel circular currents, the current in each being unity.