# 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.
$P$
$\omega ={\frac {1}{c}}{\frac {d}{dr}}(rP)$.
We have in the first place, therefore, to find $P$.
$Z$
$Z$
$dS$
$P$
${\frac {dS}{ZP}}$.
| | ${\frac {dS}{c}}\left\{Q_{0}+Q_{1}{\frac {z}{c}}+\mathrm {\&c.} +Q_{i}{\frac {z^{i}}{c^{i}}}+\mathrm {\&c.} \right\}$, |
| or | ${\frac {dS}{z}}\left\{Q_{0}+Q_{1}{\frac {c}{z}}+\mathrm {\&c.} +Q^{i}{\frac {z^{i}}{z^{i}}}+\mathrm {\&c.} \right\}$, |
$z$
$c$
$z$
$c$
$dS=-c^{2}d\mu \,d\phi$
$\phi$
$0$
$2\pi$
$\mu$
$\cos \alpha$
$1$
| | $P=2\pi c\left\{\int _{\mu }^{1}Q_{0}d\mu +\mathrm {\&c.} +{\frac {z^{i}}{c^{i}}}\int _{\mu }^{1}Q_{i}d\mu \right\}$, | (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 $Q_{i}$,
$i(i+1)Q_{i}+{\frac {d}{d\mu }}\left[(1-\mu ^{2}){\frac {dQ_{i}}{d\mu }}\right]=0$.
$\int _{\mu }^{i}Q_{i}d\mu ={\frac {1-\mu ^{2}}{i(i+1)}}{\frac {dQ_{i}}{d\mu }}$
This expression fails when $i=0$, but since $Q_{0}=1$,
$\int _{\mu }^{1}Q_{0}d\mu =1-\mu$
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′) |
$\omega$
$\omega ={\frac {1}{c}}{\frac {d}{dr}}(Pr)$
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′) |
$r$
$c$
$r$
$c$
$r=c$
$\omega$
$\theta$
$\alpha$
$\theta$
$\alpha$
$\omega ^{\prime }=\omega +4\pi$
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.
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 $C$.
$\omega _{1}$
$M=-\iint {\frac {d\omega _{1}}{dr}}dS$
$M=\int _{\mu _{2}}^{1}{\frac {d\omega _{1}}{dr}}2\pi c_{2}^{2}d\mu _{2}$
$4\pi ^{2}\sin ^{2}\alpha _{1}c_{2}^{2}\left\{{\frac {1}{c_{1}}}Q^{\prime }(\alpha _{1})\int _{\mu _{2}}^{1}Q(\alpha _{2})d\mu _{2}+\mathrm {\&c.} +{\frac {c_{2}^{i-1}}{c^{i}}}Q_{i}^{\prime }(\alpha _{1})\int _{\mu _{2}}^{1}Q(\alpha _{2})d\mu _{2}\right\}$
$M=4\pi ^{2}\sin ^{2}\alpha _{1}\sin ^{2}\alpha _{2}c_{2}^{2}\left\{{\frac {1}{2}}{\frac {c_{2}}{c_{1}}}Q_{1}^{\prime }(\alpha _{1})Q_{1}^{\prime }(\alpha _{2})+\mathrm {\&c.} +{\frac {1}{i(i+1)}}{\frac {c_{2}^{i}}{c_{1}^{i}}}Q_{i}^{\prime }(\alpha _{1})Q_{i}^{\prime })a_{2})\right.$
​ 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 $X={\frac {dM}{dx}}$.
$C$
$\theta$
$\theta$
$\Theta$
$\Theta ={\frac {dM}{d\theta }}$.
${\frac {dQ_{i}(\theta )}{d\theta }}=-\sin \theta \,Q_{i}^{\prime }(\theta )$,
$Q_{1}^{\prime }$
| $\Theta =-4\pi ^{2}\sin ^{2}\alpha _{1}\sin ^{2}\alpha _{2}\sin \theta c_{2}\left\{{\frac {1}{2}}\right.$ | ${\frac {c_{2}}{c_{1}}}Q_{1}^{\prime }(\alpha _{1})Q_{1}^{\prime }(\alpha _{2})Q_{1}^{\prime }(\theta )+\mathrm {\&c.}$ |
| | $\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.
$M$
* $1.2.\pi ^{2}{\frac {A^{2}}{C^{3}}}a^{2}\cos \theta$
* $+2.3.\pi ^{2}{\frac {A^{2}B}{C^{5}}}a^{2}b(\cos ^{2}\theta -{\frac {1}{2}}\sin ^{2}\theta )$
* $+3.4.\pi ^{2}{\frac {A^{2}(B^{2}-{\frac {1}{4}}A^{2})}{C^{7}}}a^{2}(b^{2}-{\frac {1}{4}}a^{2})(\cos ^{3}\theta -{\frac {3}{2}}\sin ^{2}\theta \cos \theta )$
* $+\mathrm {\&c.}$
$\theta =0$
$M$
$b$
${\frac {dM}{db}}=\pi ^{2}{\frac {A^{2}a^{2}}{C^{4}}}\left\{2.3{\frac {B}{C}}+2.3.4{\frac {B^{2}-{\frac {1}{4}}A^{2}}{C^{3}}}b+\mathrm {\&c.} \right\}$
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.
$P$
$x$
$y$
${\overline {P}}$
${\overline {P}}xy=\int _{-{\frac {1}{2}}x}^{+{\frac {1}{2}}x}\int _{-{\frac {1}{2}}y}^{+{\frac {1}{2}}y}P\,dxdy$
In this expression ${\overline {P}}$ is the mean value of $P$ within the limits of integration.
$P_{0}$
$P$
$x=0$
$y=0$
$P$
$P=P_{0}+x{\frac {dP_{0}}{dx}}+y{\frac {dP_{0}}{dy}}+{\frac {1}{2}}x^{2}{\frac {d^{2}P}{dx^{2}}}+\mathrm {\&c.}$
$xy$
${\overline {P}}$
| ${\overline {P}}=$ | $P_{0}+{\frac {1}{24}}\left(x^{2}{\frac {d^{2}P_{0}}{dx^{2}}}+y^{2}{\frac {d^{2}P_{0}}{dy^{2}}}\right)$ |
| | $+{\frac {1}{960}}\left(x^{4}{\frac {d^{4}P_{0}}{dx^{4}}}+y^{4}{\frac {d^{4}P_{0}}{dy^{4}}}\right)+{\frac {1}{576}}x^{2}y^{2}{\frac {d^{4}P_{0}}{dx^{2}\,dy^{2}}}+\mathrm {\&c.}$ |
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. |
$G_{0}$
$G_{1}$
$G_{2}$
$\omega$
$r$
$C$
$\omega =-2\pi +2G_{0}-G_{1}rQ_{1}(\theta )-G_{2}r^{2}Q_{2}(\theta )-\mathrm {\&c.}$
$g_{1}$
$g_{2}$
$\omega ^{\prime }$
$r$
$c$
$\omega ^{\prime }=g_{1}{\frac {1}{r^{2}}}Q_{1}(\theta )+g_{2}{\frac {1}{r^{3}}}Q_{2}(\theta )+\mathrm {\&c.}$
$M=G_{1}g_{1}Q_{1}(\theta )+G_{2}g_{2}Q_{2}(\theta )+\mathrm {\&c.}$
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.
$b$
$A$
$a$
$M=\iint {\frac {\cos \epsilon }{r}}ds\,ds^{\prime }$,
$r^{2}=A^{2}+a^{2}+b^{2}-2Aa\cos(\phi -\phi ^{\prime })$,
$\epsilon =\phi -\phi ^{\prime }$, $ds=a\,d\phi$, $ds^{\prime }=A\,d\phi ^{\prime }$,
| $M$ | $=\int _{0}^{2\pi }\int _{0}^{2\pi }{\frac {Aa\cos(\phi -\phi ^{\prime })\,d\phi \,d\phi ^{\prime }}{\sqrt {A^{2}+a^{2}+b^{2}2Aa\cos(\phi -\phi ^{\prime })}}}$ |
| | $=2\pi {\sqrt {Aa}}\left\{\left(c-{\frac {2}{c}}\right)F+{\frac {2}{c}}E\right\}$, |
$c={\frac {\sqrt {Aa|}}{\sqrt {(A+a)^{2}+b^{2}}}}$
$F$
$E$
$c$
$b$
$c$
$b$
${\frac {dM}{db}}={\frac {4\pi bc^{\frac {1}{2}}}{\sqrt {Aa(1-c^{2})^{2}}}}\{E(1+c^{2})-F(1-c^{2})\}$
$r_{1}$
$r_{2}$
$r$
$r_{1}^{2}=(A+a)^{2}+b^{2}$, $r_{2}^{2}=(A-a)^{2}+b^{2}$,
$\gamma$
$\cos \gamma ={\frac {r_{2}}{r_{1}}}$
${\frac {dM}{db}}=\pi {\frac {b\sin \gamma }{\sqrt {Aa}}}\{2F_{\gamma }-(1+\sec ^{2}\gamma )E_{\gamma }\}$
$F_{\gamma }$
$E_{\gamma }$
$\sin \gamma$
$A=a$
$\cot \gamma ={\frac {b}{2a}}$
${\frac {dM}{db}}=2\pi \cos \gamma \{2F_{\gamma }-(1+\sec ^{2}\gamma )E_{\gamma }\}$.
The quantity ${\frac {dM}{db}}$ represents the attraction between two parallel circular currents, the current in each being unity.
Second Expression for $M$.
$M$
$c_{1}={\frac {r_{1}-r_{2}}{r_{1}+r_{2}}}$
$M=4\pi {\sqrt {Aa}}{\frac {1}{\sqrt {c_{1}}}}(F_{c_{2}}-E_{c_{1}})$
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 $\theta$.
$x$
$z$
$x={\frac {1}{2}}a(\sin \theta +\mathrm {cosec} \,\theta )$
${\frac {1}{2}}a(\mathrm {cosec} \,\theta -\sin \theta )$
$c_{1}$
$\sin \theta$
$M=4\pi {\sqrt {Aa}}{\frac {1}{\sqrt {K_{\theta }}}}$
$A={\frac {1}{16\pi ^{2}}}{\frac {M^{2}K_{\theta }}{a}}$
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.
$m=4\pi a$
$M=nm$
$A=x=n^{2}K_{\theta }a$
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'.
$b$
$a$
$M$
${\frac {d^{2}M}{da^{2}}}+{\frac {d^{2}M}{db^{2}}}-{\frac {1}{a}}{\frac {dM}{da}}=0$
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 ${\frac {dV}{dy}}$.
$\int _{0}^{2\pi }a\delta a{\frac {dV}{dy}}\,d\theta$,
$\theta$
$M$
$a$
${\frac {dM}{da}}\delta a$
${\frac {dM}{da}}=\int _{0}^{2\pi }a{\frac {dV}{dy}}\,d\theta$
(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 $\delta b$.
${\frac {dV}{dr}}$
$r$
${\frac {dM}{db}}=-\int _{0}^{2\pi }a{\frac {dV}{dr}}d\theta$
Differentiating equation (2) with respect to $a$, and (3) with respect to $b$, we get
| | ${\frac {d^{2}M}{da^{2}}}$ | $=\int _{0}^{2\pi }{\frac {dV}{dy}}d\theta +\int _{0}^{2\pi }a{\frac {d^{2}V}{dr\,dy}}d\theta$, | $=\int _{0}^{2\pi }{\frac {dV}{dy}}d\theta +\int _{0}^{2\pi }a{\frac {d^{2}V}{dr\,dy}}d\theta$, | (4) |
| | ${\frac {d^{2}M}{db^{2}}}$ | $=-\int _{0}^{2\pi }a{\frac {d^{2}V}{dr\,dy}}d\theta$, | $=-\int _{0}^{2\pi }a{\frac {d^{2}V}{dr\,dy}}d\theta$, | (5) |
| Hence | ${\frac {d^{2}M}{da^{2}}}+{\frac {d^{2}M}{db^{2}}}$ | ${\frac {d^{2}M}{da^{2}}}+{\frac {d^{2}M}{db^{2}}}$ | $=\int _{0}^{2\pi }{\frac {dV}{dy}}d\theta$, | (6) |
| | | | $={\frac {1}{a}}{\frac {dM}{da}}$, by (2). |
Transposing the last term we obtain equation (1).
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.
$A$
$a$
$r$ = ${\sqrt {(A-a)^{2}+b^{2}}}$.
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).
$ds$
$a$
$\rho$
$ds$
$\theta$
$ds$
$ds$
${\frac {1}{\rho ^{2}}}\sin \theta \,ds$.
$a$
$ds$
$c$
$\int _{0}^{\pi }\int _{c}^{2a\sin \theta }{\frac {1}{\rho ^{2}}}\sin \theta \,ds\,d\theta \,d\rho =\{\log 8a-\log c-2\}\,ds$.
If $c$ is small, the surface-integral for the part of the annular space outside the small circle $c$ may be neglected.
$a-c$
$ds$
$M_{ac}=4\pi a\{\log 8a-\log c-2\}$,
$c$
$a$
$a-c$
$A$
$M_{aA}-M_{ac}=4\pi a\{\log c-\log r\}$
$A$
$a$
$M_{Aa}=4\pi a(\log 8a-\log r-2)$
$r$
$a$
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.
$M$
$M=4\pi \left\{A\log {\frac {8a}{r}}+B\right\}$
| 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.
$x$
$y$
| $A_{1}$ | $=1-A_{1}$, | | $B_{1}$ | $=1-2-B_{1}$, | $=1-2-B_{1}$, |
| $A_{3}$ | $=-A_{2}-A_{3}$, | | $B_{3}$ | $={\frac {1}{3}}-{\frac {1}{2}}A_{1}+$ | $A_{2}-B_{2}-B_{3}$, |
| $A_{3}^{\prime }$ | $=-A_{2}^{\prime }-A_{3}^{\prime }$, | | $B_{3}^{\prime }$ | $=$ | $A_{2}^{\prime }-B_{2}^{\prime }-B_{3}^{\prime }$; |
| $(-)^{n}A_{n}=A_{2}+(n-2)A_{3}+{\frac {(n-2)(n-3)}{1.2}}A_{4}+\mathrm {\&c.} +A_{n}$, ||||||
| $(-)^{n}B_{n}=-{\frac {1}{n}}+{\frac {1}{n-1}}A_{1}-{\frac {1}{n-2}}A_{2}+\mathrm {\&c.} +(-1)^{n}A_{n-1}$ ||||||
| | | $+B_{2}+(n-2)B_{3}+{\frac {(n-2)(n-3)}{1.2}}B_{4}+\mathrm {\&c.} +B_{n}$. | $+B_{2}+(n-2)B_{3}+{\frac {(n-2)(n-3)}{1.2}}B_{4}+\mathrm {\&c.} +B_{n}$. | $+B_{2}+(n-2)B_{3}+{\frac {(n-2)(n-3)}{1.2}}B_{4}+\mathrm {\&c.} +B_{n}$. | $+B_{2}+(n-2)B_{3}+{\frac {(n-2)(n-3)}{1.2}}B_{4}+\mathrm {\&c.} +B_{n}$. |
$M$
${\frac {d^{2}M}{dx^{2}}}+{\frac {d^{2}M}{dy^{2}}}-{\frac {1}{a+x}}{\frac {dM}{dx}}=0$,
* $2A_{2}+2A_{2}^{\prime }=A_{1}$,
* $2A_{2}+A_{2}^{\prime }+6A_{3}+2A_{3}^{\prime }=2A_{2}$;
* $n(n-1)A_{n}+(n+1)nA_{n+1}+1.2A_{n}^{\prime }+1.2A_{n+1}^{\prime }=nA_{n}$,
* $(n-1)(n-2)A_{n}^{\prime }+n(n-1)A_{n+1}^{\prime }+2.3A_{n}^{\prime \prime }+2.3A_{n+1}^{\prime \prime }=(n-2)A_{n}^{\prime }$,
&c.;
| $4A_{2}+$ | $A_{1}=$ | $2B_{2}+2B_{2}^{\prime }-B_{1}=4A_{2}^{\prime }$, |
| $6A_{3}+$ | $3A_{2}=$ | $2B_{2}^{\prime }+6B_{3}+2B_{3}^{\prime }=6A_{3}^{\prime }+3A_{2}^{\prime }$, |
| $(2n-1)A_{n}+(2n+2)A_{n+1}$ |||
| | | $=n(n-2)B_{n}+(n+1)nB_{n+1}+1.2B_{n}^{\prime }+1.2B_{n+1}^{\prime }$, |
$M$
| $M=4\pi a\log {\frac {8a}{r}}$ | $\left\{1+{\frac {1}{2}}{\frac {x}{a}}+{\frac {x^{2}+3y^{2}}{16a^{2}}}-{\frac {x^{3}+3xy^{2}}{32a^{3}}}+\mathrm {\&c.} \right\}$ |
| | $+4\pi a\left\{-2-{\frac {1}{2}}{\frac {x}{a}}+{\frac {3x^{2}-y^{2}}{16a^{2}}}-{\frac {x^{3}-6xy^{2}}{48a^{3}}}+\mathrm {\&c.} \right\}$. |
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.
$L=4\pi n^{2}a\left(\log {\frac {8a}{R}}-2\right)$,
$n$
$a$
$R$
$R$
$n$
$R^{2}$
$2\pi an$
$a$
$n$
${\frac {dn}{n}}=2{\frac {dR}{R}}$, and ${\frac {da}{a}}=-2{\frac {dR}{R}}$,
$L$
$\log {\frac {8a}{R}}={\frac {7}{2}}$.
$c$
$\log {\frac {R}{c}}=-{\frac {1}{4}}$,
and $\log {\frac {8a}{c}}={\frac {13}{4}}$,
$a=3.22c$
$3.22$
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.
* ↑ 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.
* ↑ Trans. R. S., Edin., vol. xxv. p. 217 (1869).
* ↑ Werke, Göttingen edition, 1867, vol. v. p. 622.