A Treatise on Electricity and Magnetism/Part IV/Chapter XIV

by James Clerk Maxwell

CHAPTER XIV.

CIRCULAR CURRENTS.

Magnetic Potential due to a Circular Current.

694.] The magnetic potential at a given point, due to a circuit carrying a unit current, is numerically equal to the solid angle subtended by the circuit at that point; see Arts. 409, 485.

When the circuit is circular, the solid angle is that of a cone of the second degree, which, when the given point is on the axis of the circle, becomes a right cone. When the point is not on the axis, the cone is an elliptic cone, and its solid angle is numerically equal to the area of the spherical ellipse which it traces on a sphere whose radius is unity.

This area can be expressed in finite terms by means of elliptic integrals of the third kind. We shall find it more convenient to expand it in the form of an infinite series of spherical harmonics, for the facility with which mathematical operations may be performed on the general term of such a series more than counterbalances the trouble of calculating a number of terms sufficient to ensure practical accuracy.

Fig. 46.

For the sake of generality we shall assume the origin at any point on the axis of the circle, that is to say, on the line through the centre perpendicular to the plane of the circle.

Let $O$ (Fig. 46) be the centre of the circle, $C$ the point on the axis which we assume as origin, $H$ a point on the circle.

Describe a sphere with $C$ as centre, and $CH$ as radius. The circle will lie on this sphere, and will form a small circle of the sphere of angular radius $a$.

Let

$CH=c$,

$OC=b=c\cos a$,

$OH=a=c\sin \alpha$.

Let $A$ be the pole of the sphere, and $Z$ any point on the axis, and let $CZ=z$.

Let $R$ be any point in space, and let $CR=r$, and $ACR=\theta$.

Let $P$ be the point when $CR$ cuts the sphere.

The magnetic potential due to the circular current is equal to that due to a magnetic shell of strength unity bounded by the current. As the form of the surface of the shell is indifferent, provided it is bounded by the circle, we may suppose it to coincide with the surface of the sphere.

We have shewn in Art. 670 that if $P$ is the potential due to a stratum of matter of surface-density unity, spread over the surface of the sphere within the small circle, the potential due to a magnetic shell of strength unity and bounded by the same circle is

We have in the first place, therefore, to find $P$.

Let the given point be on the axis of the circle at $Z$, then the part of the potential at $Z$ due to an element $dS$ of the spherical surface at $P$ is This may be expanded in one of the two series of spherical harmonics,

		⁠	${\frac {dS}{c}}\left\{Q_{0}+Q_{1}{\frac {z}{c}}+\mathrm {\&c.} +Q_{i}{\frac {z^{i}}{c^{i}}}+\mathrm {\&c.} \right\}$,
	or		${\frac {dS}{z}}\left\{Q_{0}+Q_{1}{\frac {c}{z}}+\mathrm {\&c.} +Q^{i}{\frac {z^{i}}{z^{i}}}+\mathrm {\&c.} \right\}$,

the first series being convergent when $z$ is less than $c$, and the second when $z$ is greater than $c$. and integrating with respect to $\phi$ between the limits $0$ and $2\pi$, and with respect to $\mu$ between the limits $\cos \alpha$ and $1$, we find

			$P=2\pi c\left\{\int _{\mu }^{1}Q_{0}d\mu +\mathrm {\&c.} +{\frac {z^{i}}{c^{i}}}\int _{\mu }^{1}Q_{i}d\mu \right\}$,		(1)
	or	⁠	$P^{\prime }=2\pi {\frac {c^{2}}{z}}\left\{\int _{\mu }^{1}Q_{0}d\mu +\mathrm {\&c.} +{\frac {c^{i}}{z^{i}}}\int _{\mu }^{1}Q_{i}d\mu \right\}$.		(1′)

By the characteristic equation of $Q_{i}$,

$i(i+1)Q_{i}+{\frac {d}{d\mu }}\left[(1-\mu ^{2}){\frac {dQ_{i}}{d\mu }}\right]=0$.

This expression fails when $i=0$, but since $Q_{0}=1$,

As the function ${\frac {dQ_{i}}{d\mu }}$ occurs in every part of this investigation we shall denote it by the abbreviated symbol $Q_{i}^{\prime }$. The values of $Q_{i}^{\prime }$ corresponding to several values of $i$ are given in Art. 698.

We are now able to write down the value of $P$ for any point $R$, whether on the axis or not, by substituting $r$ for $z$, and multiplying each term by the zonal harmonic of $\theta$ of the same order. For $P$ must be capable of expansion in a series of zonal harmonics of $\theta$ with proper coefficients. When $\theta =0$ each of the zonal harmonics becomes equal to unity, and the point $R$ lies on the axis. Hence the coefficients are the terms of the expansion of $P$ for a point on the axis. We thus obtain the two series

			$P$	${}=2\pi c\left\{1-\mu +\mathrm {\&c.} +{\frac {1-\mu ^{2}}{i(i+1)}}{\frac {r^{i}}{c^{i}}}Q_{i}^{\prime }(\alpha )Q_{i}(\theta )\right\}$,		(4)
			or⁠ $P^{\prime }$	${}=2\pi {\frac {c^{2}}{r}}\left\{1-\mu +\mathrm {\&c.} +{\frac {1-\mu ^{2}}{i(i+1)}}{\frac {c^{i}}{r^{\prime }}}Q_{i}^{\prime }(\alpha )Q_{i}(\theta )\right\}$.		(4′)

695.] We may now find $\omega$, the magnetic potential of the circuit, by the method of Art. 670, from the equation

We thus obtain the two series

		$\omega$	${}=-2\pi \left\{1-\cos \alpha +\mathrm {\&c.} +{\frac {\sin ^{2}\alpha }{i}}{\frac {r^{i}}{c^{i}}}Q_{i}^{\prime }(\alpha )Q_{i}(\theta )+\mathrm {\&c.} \right\}$,		(6)
		or⁠$\omega ^{\prime }$	${}=2\pi \sin ^{2}\alpha \left\{{\frac {1}{2}}{\frac {c^{2}}{r^{2}}}Q_{1}^{\prime }(\alpha )Q_{1}(\theta )+\mathrm {\&c.} +{\frac {1}{i+1}}{\frac {c^{i+1}}{r^{i+1}}}Q_{i}^{\prime }(\alpha )Q_{i}(\theta )\right\}$.		(6′)

The series (6) is convergent for all values of $r$ less than $c$, and the series (6′) is convergent for all values of $r$ greater than $c$. At the surface of the sphere, where $r=c$, the two series give the same value for $\omega$ when $\theta$ is greater than $\alpha$, that is, for points not occupied by the magnetic shell, but when $\theta$ is less than $\alpha$, that is, at points on the magnetic shell,

If we assume $O$, the centre of the circle, as the origin of coordinates, we must put $\alpha ={\frac {\pi }{2}}$, and the series become

		$\omega ={}$	$-2\pi \left\{1+{\frac {r}{c}}Q_{1}(\theta )+\mathrm {\&c.} +(-)^{8}{\frac {1.3.(2s-1)}{2.4.2s}}{\frac {r^{2s+1}}{c^{2s+1}}}Q_{2s+1}(\theta )\right\}$,		(8)
⁠
		$\omega ^{\prime }={}$	$\;\;2\pi \left\{{\frac {1}{2}}{\frac {c^{2}}{r^{2}}}Q_{1}(\theta )+\mathrm {\&c.} +(-)^{8}{\frac {1.3\ldots (2s+1)}{2.4\ldots (2s+2)}}{\frac {c^{2s+2}}{r^{2s+2}}}Q_{2s+1}(\theta )\right\}$,		(8′)

where the orders of all the harmonics are odd^[1].

On the Potential Energy of two Circular Currents.

Fig. 47. 696.] Let us begin by supposing the two magnetic shells which are equivalent to the currents to be portions of two concentric spheres, their radii being $c_{1}$ and $c_{2}$, of which $c_{1}$ is the greater (Fig. 47). Let us also suppose that the axes of the two shells coincide, and that $a_{1}$ the angle subtended by the radius of the first shell, and $a_{2}$ the angle subtended by the radius of the second shell at the centre $C$.

Let $\omega _{1}$ be the potential due to the first shell at any point within it, then the work required to carry the second shell to an infinite distance is the value of the surface-integral extended over the second shell. Hence or, substituting the value of the integrals from equation (2), Art. 694,

Fig. 48. 697.] Let us next suppose that the axis of one of the shells is turned about $C$ as a centre, so that it now makes an angle $\theta$ with the axis of the other shell (Fig. 48). We have only to introduce the zonal harmonics of $\theta$ into this expression for $M$, and we find for the more general value of $M$,

	$M=4\pi ^{2}\sin 2\alpha _{1}\sin ^{2}\alpha _{2}c_{2}\left\{{\frac {1}{2}}{\frac {c_{2}}{c_{1}}}Q_{1}^{\prime }\right.$	$(\alpha _{1})Q_{1}^{\prime }(\alpha _{2})Q_{1}(\theta )+\mathrm {\&c.}$
		$\left.+{\frac {1}{i(i+1)}}{\frac {c_{2}^{i}}{c_{1}^{i}}}Q_{i}^{\prime }(\alpha _{1})Q_{i}^{\prime }(\alpha _{2})Q_{i}(\theta )\right\}$.

This is the value of the potential energy due to the mutual action of two circular currents of unit strength, placed so that the normals through the centres of the circles meet in a point $C$ in an angle $\theta$, the distances of the circumferences of the circles from the point $C$ being $c_{1}$ and $c_{2}$, of which $c_{1}$ is the greater.

If any displacement $dx$ alters the value of $M$, then the force acting in the direction of the displacement is $X={\frac {dM}{dx}}$.

For instance, if the axis of one of the shells is free to turn about the point $C$, so as to cause $\theta$ to vary, then the moment of the force tending to increase $\theta$ is $\Theta$, where Performing the differentiation, and remembering that where $Q_{1}^{\prime }$ has the same signification as in the former equations,

	$\Theta =-4\pi ^{2}\sin ^{2}\alpha _{1}\sin ^{2}\alpha _{2}\sin \theta c_{2}\left\{{\frac {1}{2}}\right.$	${\frac {c_{2}}{c_{1}}}Q_{1}^{\prime }(\alpha _{1})Q_{1}^{\prime }(\alpha _{2})Q_{1}^{\prime }(\theta )+\mathrm {\&c.}$
		$\left.{}+{\frac {1}{i(i+1)}}{\frac {c_{2}^{i}}{c_{1}^{i}}}Q_{i}^{\prime }(\alpha _{1})Q_{i}^{\prime }(\alpha _{2})Q_{i}^{\prime }(\theta )\right\}$.

698.] As the values of $Q_{i}^{\prime }$ occur frequently in these calculations the following table of values of the first six degrees may be useful. In this table $\mu$ stands for $\cos \theta$, and $\nu$ for $\sin \theta$.

$Q_{1}^{\prime }=1$,

$Q_{2}^{\prime }=2\mu$,

$Q_{3}^{\prime }={\frac {3}{2}}(5\mu ^{2}-1)=6(\mu ^{2}-{\frac {1}{4}}\nu ^{2})$,

$Q_{4}^{\prime }={\frac {5}{2}}\mu (7\mu ^{2}-3)=10\mu (\mu ^{2}-{\frac {3}{4}}\nu ^{2})$,

$Q_{5}^{\prime }={\frac {15}{8}}(21\mu ^{4}-14\mu ^{2}+1)=15(\mu ^{4}-{\frac {3}{2}}\mu ^{2}\nu ^{2}+{\frac {1}{8}}\nu ^{4})$,

$Q_{6}^{\prime }={\frac {21}{8}}\mu (33\mu ^{4}-30\mu ^{2}+5)=21\mu (\mu ^{4}-{\frac {5}{2}}\mu ^{2}\nu ^{2}+{\frac {5}{8}}\nu ^{4})$.

699.] It is sometimes convenient to express the series for $M$ in terms of linear quantities as follows:—

Let $a$ be the radius of the smaller circuit, $b$ the distance of its plane from the origin, and $c={\sqrt {a^{2}+b^{2}}}$.

Let $A$, $B$, and $C$ be the corresponding quantities for the larger circuit.

The series for $M$ may then be written,

	$M={}$	$1.2.\pi ^{2}{\frac {A^{2}}{C^{3}}}a^{2}\cos \theta$
		${}+2.3.\pi ^{2}{\frac {A^{2}B}{C^{5}}}a^{2}b(\cos ^{2}\theta -{\frac {1}{2}}\sin ^{2}\theta )$
		${}+3.4.\pi ^{2}{\frac {A^{2}(B^{2}-{\frac {1}{4}}A^{2})}{C^{7}}}a^{2}(b^{2}-{\frac {1}{4}}a^{2})(\cos ^{3}\theta -{\frac {3}{2}}\sin ^{2}\theta \cos \theta )$
		${}+\mathrm {\&c.}$

If we make $\theta =0$, the two circles become parallel and on the same axis. To determine the attraction between them we may differentiate $M$ with respect to $b$. We thus find

700.] In calculating the effect of a coil of rectangular section we have to integrate the expressions already found with respect to $A$, the radius of the coil, and $B$, the distance of its plane from the origin, and to extend the integration over the breadth and depth of the coil.

In some cases direct integration is the most convenient, but there are others in which the following method of approximation leads to more useful results.

Let $P$ be any function of $x$ and $y$, and let it be required to find the value of ${\overline {P}}$ where

In this expression ${\overline {P}}$ is the mean value of $P$ within the limits of integration.

Let $P_{0}$ be the value of $P$ when $x=0$ and $y=0$, then, expanding $P$ by Taylor's Theorem, Integrating this expression between the limits, and dividing the result by $xy$, we obtain as the value of ${\overline {P}}$,

${\overline {P}}={}$	$P_{0}+{\frac {1}{24}}\left(x^{2}{\frac {d^{2}P_{0}}{dx^{2}}}+y^{2}{\frac {d^{2}P_{0}}{dy^{2}}}\right)$
	${}+{\frac {1}{960}}\left(x^{4}{\frac {d^{4}P_{0}}{dx^{4}}}+y^{4}{\frac {d^{4}P_{0}}{dy^{4}}}\right)+{\frac {1}{576}}x^{2}y^{2}{\frac {d^{4}P_{0}}{dx^{2}\,dy^{2}}}+\mathrm {\&c.}$

In the case of the coil, let the outer and inner radii be $A+{\frac {1}{2}}\xi$, and $A-{\frac {1}{2}}\xi$ respectively, and let the distance of the planes of the windings from the origin lie between $B+{\frac {1}{2}}\eta$ and $B-{\frac {1}{2}}\eta$, then the breadth of the coil is $\eta$, and its depth $\xi$, these quantities being small compared with $A$ or $C$.

In order to calculate the magnetic effect of such a coil we may write the successive terms of the series as follows:—

$G_{0}=\pi {\frac {B}{C}}\left(1+{\frac {1}{24}}{\frac {2A^{2}-B^{2}}{C^{4}}}\xi ^{2}-{\frac {1}{8}}{\frac {A^{2}}{C^{4}}}\eta ^{2}\right)$,

$G_{1}=2\pi {\frac {A^{2}}{C^{3}}}\left(1+{\frac {1}{24}}\left({\frac {2}{A^{2}}}-15{\frac {B^{2}}{C^{4}}}\right)\xi ^{2}+{\frac {1}{8}}{\frac {4B^{2}-A^{2}}{C^{4}}}\eta ^{2}\right)$,

$G_{2}=3\pi {\frac {A^{2}B}{C^{5}}}\left(1+{\frac {1}{21}}\left({\frac {2}{A^{2}}}-{\frac {25}{C^{2}}}+{\frac {35A^{2}}{C^{4}}}\right)\xi ^{2}+{\frac {5}{24}}{\frac {4B^{2}-3A^{2}}{C^{4}}}\eta ^{2}\right)$,

$G_{3}=4\pi {\frac {A^{2}(B^{2}-{\frac {1}{4}}A^{2})}{C^{7}}}+{\frac {\pi }{24}}{\frac {\xi ^{2}}{C^{11}}}\left\{C^{4}(8B^{2}-12A^{2})+35A^{2}B^{2}(5A^{2}-4B^{2})\right\}$

${}+{\frac {\pi }{24}}{\frac {\eta ^{2}}{C^{11}}}\left\{3A^{2}C^{2}(5A^{2}-44B^{2})+63A^{2}B^{2}(4B^{2}-A^{2})\right\}$,

&c., &c.;

$g_{1}=\pi a^{2}$	${}+{\frac {1}{12}}\pi \xi ^{2}$,
$g_{2}=2\pi a^{2}b$	${}+{\frac {1}{6}}\pi b\xi ^{2}$,
$g_{2}=3\pi a^{2}(b^{2}-{\frac {1}{4}}a^{2})$	${}+{\frac {\pi }{8}}\xi ^{2}(2b^{2}-3a^{2})+{\frac {\pi }{4}}\eta ^{2}a^{2}$,
&c., &c.

The quantities $G_{0}$, $G_{1}$, $G_{2}$, &c. belong to the large coil. The value of $\omega$ at points for which $r$ is less than $C$ is The quantities $g_{1}$, $g_{2}$, &c. belong to the small coil. The value of $\omega ^{\prime }$ at points for which $r$ is greater than $c$ is The potential of the one coil with respect to the other when the total current through the section of each coil is unity is

To find $M$ by Elliptic Integrals.

701.] When the distance of the circumferences of the two circles 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, thenthe integration being extended round both curves. In this case,

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

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 $r$, 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

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

Second Expression for $M$.

An expression for $M$, which is sometimes more convenient, is got by making $c_{1}={\frac {r_{1}-r_{2}}{r_{1}+r_{2}}}$ in which case

To draw the Lines of Magnetic Force for a Circular Current.

702.] The lines of magnetic force are evidently in planes passing through the axis of the circle, and in each of these lines the value of $M$ is constant.

Calculate the value of $K_{\theta }={\frac {\sin \theta }{(F_{\sin \theta }-E_{\sin \theta })^{2}}}$ from Legendre's tables for a sufficient number of values of $\theta$.

Draw rectangular axes of $x$ and $z$ on the paper, and, with centre at the point $x={\frac {1}{2}}a(\sin \theta +\mathrm {cosec} \,\theta )$, draw a circle with radius ${\frac {1}{2}}a(\mathrm {cosec} \,\theta -\sin \theta )$. For all points of this circle the value of $c_{1}$ will be $\sin \theta$. Hence, for all points of this circle,

Now $A$ is the value of $x$ for which the value of $M$ was found. Hence, if we draw a line for which $x=A$, it will cut the circle in two points having the given value of $M$.

Giving $M$ a series of values in arithmetical progression, the values of $A$ will be as a series of squares. Drawing therefore a series of lines parallel to $z$, for which $x$ has the values found for $A$, the points where these lines cut the circle will be the points where the corresponding lines of force cut the circle.

If we put $m=4\pi a$, and $M=nm$, then

We may call $n$ the index of the line of force.

The forms of these lines are given in Fig. XVIII at the end of this volume. They are copied from a drawing given by Sir W. Thomson in his paper on 'Vortex Motion^[2]'.

703.] If the position of a circle having a given axis is regarded as defined by $b$, the distance of its centre from a fixed point on the axis, and $a$, the radius of the circle, then $M$, the coefficient of induction of the circle with respect to any system whatever 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).

Coefficient of Induction of Two Parallel Circles when the Distance between the Arcs is Small compared with the Radius of either Circle.

704.] We might deduce the value of $M$ in this case from the expansion of the elliptic integral already given when its modulus is nearly unity. The following method, however, is a more direct application of electrical principles.

First Approximation.

Let $A$ and $a$ be the radii of the circles, and b the distance between their planes, then the shortest distance between the arcs is

$r$ = ${\sqrt {(A-a)^{2}+b^{2}}}$.

Fig. 49. We have to find $M_{1}$, the magnetic induction through the circle $A$, due to a unit current in $a$ on the supposition that $r$ is small compared with $A$ or $a$.

We shall begin by calculating the magnetic induction through a circle in the plane of $a$ whose radius is $a-c$, $c$ being a quantity small compared with $a$ (Fig. 49).

Consider a small element $ds$ of the circle $a$. At a point in the plane of the circle, distant $\rho$ from the middle of $ds$, measured in a direction making an angle $\theta$ with the direction of <$ds$, the magnetic force due to $ds$ is perpendicular to the plane, and equal to

${\frac {1}{\rho ^{2}}}\sin \theta \,ds$.

If we now calculate the surface-integral of this force over the space which lies within the circle $a$, but outside of a circle whose centre is $ds$ and whose radius is $c$, we find it

$\int _{0}^{\pi }\int _{c}^{2a\sin \theta }{\frac {1}{\rho ^{2}}}\sin \theta \,ds\,d\theta \,d\rho =\{\log 8a-\log c-2\}\,ds$.

If $c$ is small, the surface-integral for the part of the annular space outside the small circle $c$ may be neglected.

We then find for the induction through the circle whose radius is $a-c$, by integrating with respect to $ds$,

$M_{ac}=4\pi a\{\log 8a-\log c-2\}$,

provided $c$ is very small compared with $a$. Since the magnetic force at any point, the distance of which from a curved wire is small compared with the radius of curvature, is nearly the same as if the wire had been straight, we can calculate the difference between the induction through the circle whose radius is $a-c$, and the circle $A$ by the formula Hence we find the value of the induction between $A$ and $a$ to be approximately, provided $r$ is small compared with $a$.

705.] Since the mutual induction between two windings of the same coil is a very important quantity in the calculation of experimental results, I shall now describe a method by which the approximation to the value of $M$ for this case can be carried to any required degree of accuracy.

We shall assume that the value of $M$ is of the form

where	⁠	$A={}$		$a+A_{1}x+A_{2}{\frac {x^{2}}{a}}+A_{2}^{\prime }{\frac {y^{2}}{a}}+A_{3}+A_{3}^{\prime }{\frac {xy^{2}}{a^{2}}}+\mathrm {\&c.}$,	⁠
and		$B={}$	$-2$	$a+B_{1}x+B_{2}{\frac {x^{2}}{a}}+B_{2}^{\prime }+{\frac {y^{2}}{a}}+B_{3}{\frac {x^{3}}{a^{2}}}+B_{3}^{\prime }{\frac {xy^{2}}{a^{2}}}+\mathrm {\&c.}$,

where $a$ and $a+x$ are the radii of the circles, and $y$ the distance between their planes.

We have to determine the values of the coefficients $A$ and $B$. It is manifest that only even powers of y can occur in these quantities, because, if the sign of $y$ is reversed, the value of $M$ must remain the same.

We get another set of conditions from the reciprocal property of the coefficient of induction, which remains the same whichever circle we take as the primary circuit. The value of $M$ must therefore remain the same when we substitute $a+x$ for $a$, and $-x$ for $x$ in the above expression.

We thus find the following conditions of reciprocity by equating the coefficients of similar combinations of $x$ and $y$,

$A_{1}$	${}=1-A_{1}$,	⁠	$B_{1}$	${}=1-2-B_{1}$,
$A_{3}$	${}=-A_{2}-A_{3}$,	⁠	$B_{3}$	${}={\frac {1}{3}}-{\frac {1}{2}}A_{1}+{}$	$A_{2}-B_{2}-B_{3}$,
$A_{3}^{\prime }$	${}=-A_{2}^{\prime }-A_{3}^{\prime }$,		$B_{3}^{\prime }$	${}={}$	$A_{2}^{\prime }-B_{2}^{\prime }-B_{3}^{\prime }$;
$(-)^{n}A_{n}=A_{2}+(n-2)A_{3}+{\frac {(n-2)(n-3)}{1.2}}A_{4}+\mathrm {\&c.} +A_{n}$,
$(-)^{n}B_{n}=-{\frac {1}{n}}+{\frac {1}{n-1}}A_{1}-{\frac {1}{n-2}}A_{2}+\mathrm {\&c.} +(-1)^{n}A_{n-1}$
		${}+B_{2}+(n-2)B_{3}+{\frac {(n-2)(n-3)}{1.2}}B_{4}+\mathrm {\&c.} +B_{n}$.

From the general equation of $M$, Art. 703, we obtain another set of conditions,

$2A_{2}+2A_{2}^{\prime }=A_{1}$,

$2A_{2}+A_{2}^{\prime }+6A_{3}+2A_{3}^{\prime }=2A_{2}$;

$n(n-1)A_{n}+(n+1)nA_{n+1}+1.2A_{n}^{\prime }+1.2A_{n+1}^{\prime }=nA_{n}$,

$(n-1)(n-2)A_{n}^{\prime }+n(n-1)A_{n+1}^{\prime }+2.3A_{n}^{\prime \prime }+2.3A_{n+1}^{\prime \prime }=(n-2)A_{n}^{\prime }$,

&c.;

$4A_{2}+{}$	$A_{1}={}$	$2B_{2}+2B_{2}^{\prime }-B_{1}=4A_{2}^{\prime }$,
$6A_{3}+{}$	$3A_{2}={}$	$2B_{2}^{\prime }+6B_{3}+2B_{3}^{\prime }=6A_{3}^{\prime }+3A_{2}^{\prime }$,
$(2n-1)A_{n}+(2n+2)A_{n+1}$
		${}=n(n-2)B_{n}+(n+1)nB_{n+1}+1.2B_{n}^{\prime }+1.2B_{n+1}^{\prime }$,

Solving these equations and substituting the values of the coefficients, the series for $M$ becomes

$M=4\pi a\log {\frac {8a}{r}}$	$\left\{1+{\frac {1}{2}}{\frac {x}{a}}+{\frac {x^{2}+3y^{2}}{16a^{2}}}-{\frac {x^{3}+3xy^{2}}{32a^{3}}}+\mathrm {\&c.} \right\}$
	$+4\pi a\left\{-2-{\frac {1}{2}}{\frac {x}{a}}+{\frac {3x^{2}-y^{2}}{16a^{2}}}-{\frac {x^{3}-6xy^{2}}{48a^{3}}}+\mathrm {\&c.} \right\}$.

To find the form of a coil for which the coefficient of self-induction is a maximum, the total length and thickness of the wire being given.

706.] Omitting the corrections of Art. 705, we find by Art. 673 where $n$ is the number of windings of the wire, $a$ is the mean radius of the coil, and $R$ is the geometrical mean distance of the transverse section of the coil from itself. See Art. 690. If this section is always similar to itself, $R$ is proportional to its linear dimensions, and $n$ varies as $R^{2}$. Since the total length of the wire is $2\pi an$, $a$ varies inversely as $n$. Hence and we find the condition that $L$ may be a maximum

If the transverse section of the coil is circular, of radius $c$, then, by Art. 692, or, the mean radius of the coil should be $3.22$ times the radius of the transverse section of the coil in order that such a coil may have the greatest coefficient of self-induction. This result was found by Gauss^[3].

If the channel in which the coil is wound has a square transverse section, the mean diameter of the coil should be $3.7$ times the side of the square section.

1. The value of the solid angle subtended by a circle may be obtained in a more direct way as follows.—

   The solid angle subtended by the circle at the point $Z$ in the axis is easily shewn to be

   $\omega =2\pi \left(1-{\frac {z-c\cos \alpha }{HZ}}\right)$.

   Expanding this expression in spherical harmonics, we find

   	$\omega$	${}=2\pi \left\{(\cos \alpha -1)+(Q_{1}(\alpha )\cos \alpha -Q_{0}(\alpha )){\frac {z}{c}}+\mathrm {\&c.} +(Q_{1}(\alpha )\cos \alpha -Q_{i-1}(\alpha )){\frac {z^{i}}{c^{i}}}+\mathrm {\&c.} \right\}$,
   	$\omega ^{\prime }$	${}=2\pi \left\{(Q_{0}(\alpha )\cos \alpha -Q_{a}(\alpha )){\frac {c}{z}}+\mathrm {\&c.} +(Q_{i}(\alpha )\cos \alpha -Q_{i+1}(\alpha )){\frac {c^{i+1}}{z^{i+1}}}+\mathrm {\&c.} \right\}$,

   for the expansions of $\omega$ for points on the axis for which $z$ is less than $c$ or greater than $c$ respectively. Remembering the equations (42) and (43) of Art. 132 (vol. i. p. 165), the coefficients in these equations are evidently the same as those we have now obtained in a more convenient form for computation.

2. Trans. R. S., Edin., vol. xxv. p. 217 (1869).
3. Werke, Göttingen edition, 1867, vol. v. p. 622.