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

by James Clerk Maxwell

CHAPTER XIII.

PARALLEL CURRENTS.

Cylindrical Conductors.

682.] In a very important class of electrical arrangements the current is conducted through round wires of nearly uniform section, and either straight, or such that the radius of curvature of the axis of the wire is very great compared with the radius of the transverse section of the wire. In order to be prepared to deal mathematically with such arrangements, we shall begin with the case in which the circuit consists of two very long parallel conductors, with two pieces joining their ends, and we shall confine our attention to a part of the circuit which is so far from the ends of the conductors that the fact of their not being infinitely long does not introduce any sensible change in the distribution of force.

We shall take the axis of $z$ parallel to the direction of the conductors, then, from the symmetry of the arrangements in the part of the field considered, everything will depend on $H$, the component of the vector-potential parallel to $z$.

The components of magnetic induction become, by equations (A),

⁠	$a={\frac {dH}{dy}}$,	(1)
	$b=-{\frac {dH}{dx}}$,	(2)
	$c=0$.

For the sake of generality we shall suppose the coefficient of magnetic induction to be $\mu$, so that $a=\mu \alpha$, $b=\mu \beta$, where $\alpha$ and $\beta$ are the components of the magnetic force.

The equations (E) of electric currents, Art, 607, give

$u=0$,⁠$v=0$,⁠$4\pi w={\frac {d\beta }{dx}}-{\frac {d\alpha }{dy}}$.

(3)

683.] If the current is a function of $r$, the distance from the axis of $z$, and if we write

$x=r\cos \theta$,⁠ and ⁠$y=r\sin \theta$,

(4)

and $\beta$ for the magnetic force, in the direction in which $\theta$ is measured perpendicular to the plane through the axis of $z$, we have

$4\pi w={\frac {d\beta }{dr}}+{\frac {1}{r}}\beta ={\frac {1}{r}}{\frac {d}{dr}}(\beta r)$.

(5)

If $C$ is the whole current flowing through a section bounded by a circle in the plane $xy$, whose centre is the origin and whose radius is $r$,

$C=\int _{0}^{r}2\pi rw\,dr={\frac {1}{2}}\beta r$.

(6)

It appears, therefore, that the magnetic force at a given point due to a current arranged in cylindrical strata, whose common axis is the axis of $z$, depends only on the total strength of the current flowing through the strata which lie between the given point and the axis, and not on the distribution of the current among the different cylindrical strata.

For instance, let the conductor be a uniform wire of radius $a$, and let the total current through it be $C$, then, if the current is uniformly distributed through all parts of the section, $w$ will be

constant, and

$C=\pi wa^{2}$.

(7)

The current flowing through a circular section of radius $r$, $r$ being less than $a$, is $C^{\prime }=\pi wr^{2}$. Hence at any point within the wire,

	⁠	$\beta ={\frac {2C^{\prime }}{r}}=2C{\frac {r}{a^{2}}}$.	(8)
Outside the wire		$\beta =2{\frac {C}{r}}$.	(9)

In the substance of the wire there is no magnetic potential, for within a conductor carrying an electric current the magnetic force does not fulfil the condition of having a potential.

Outside the wire the magnetic potential is

$\Omega =2C\theta$.

(10)

Let us suppose that instead of a wire the conductor is a metal tube whose external and internal radii are $a_{1}$ and $a_{2}$, then, if $C$ is the current through the tubular conductor,

$C=\pi w(a_{1}^{2}-a_{2}^{2})$.

(11)

The magnetic force within the tube is zero. In the metal of the tube, where $r$ is between $a_{1}$ and $a_{2}$,

$\beta =2C{\frac {1}{a_{1}^{2}-a_{2}^{2}}}\left(r-a{\frac {a_{2}^{2}}{r}}\right)$,

(12)

and outside the tube,

$\beta =2{\frac {C}{r}}$,

(13)

the same as when the current flows through a solid wire.

684.] The magnetic induction at any point is $b=\mu \beta$, and since, by equation (2),

⁠	$b$	${}=-{\frac {dH}{dr}}$,	(14)
	$H$	${}=-\int \mu \beta \,dr$.	(15)

The value of $H$ outside the tube is

$A-2\mu _{0}C\log r$,

(16)

where $\mu _{0}$ is the value of $\mu$, in the space outside the tube, and $A$ is a constant, the value of which depends on the position of the return current. In the substance of the tube,

$H=A-2\mu _{0}C\log a_{1}+{\frac {\mu C}{a_{1}^{2}-a_{2}^{2}}}\left(a_{1}^{2}-r^{2}+2a_{2}^{2}\log {\frac {r}{a_{1}}}\right)$.

(17)

In the space within the tube $H$ is constant, and

$H=A-2\mu _{0}C\log a_{1}+\mu C\left(1+{\frac {2a_{2}^{2}}{a_{1}^{2}-a_{2}^{2}}}\log {\frac {a_{2}}{a_{1}}}\right)$.

(18)

685.] Let the circuit be completed by a return current, flowing in a tube or wire parallel to the first, the axes of the two currents being at a distance $b$. To determine the kinetic energy of the system we have to calculate the integral

$T={\frac {1}{2}}\iiint Hw\,dx\,dy\,dz$.

(19)

If we confine our attention to that part of the system which lies between two planes perpendicular to the axes of the conductors, and distant $l$ from each other, the expression becomes

$T={\frac {1}{2}}l\iint Hw\,dx\,dy$.

(20)

If we distinguish by an accent the quantities belonging to the return current, we may write this

${\frac {2T}{l}}=\iint Hw^{\prime }\,dx^{\prime }\,dy^{\prime }+\iint H^{\prime }w\,dx\,dy+\iint Hw\,dx\,dy+\iint H^{\prime }w^{\prime }\,dx^{\prime }\,dy^{\prime }$.

(21)

Since the action of the current on any point outside the tube is the same as if the same current had been concentrated at the axis of the tube, the mean value of $H$ for the section of the return current is $A-2\mu _{0}C\log b$, and the mean value of $H^{\prime }$ for the section of the positive current is $A^{\prime }-2\mu _{0}C^{\prime }\log b$.

Hence, in the expression for $T$, the first two terms may be written

$AC^{\prime }-2\mu _{0}CC^{\prime }\log b$⁠ and ⁠$A^{\prime }C-2\mu _{0}CC^{\prime }\log b$.

Integrating the two latter terms in the ordinary way, and adding the results, remembering that $C+C^{\prime }=0$, we obtain the value of the kinetic energy $T$. Writing this ${\frac {1}{2}}LC^{2}$, where $L$ is the coefficient of self-induction of the system of two conductors, we find as the value of $L$ for unit of length of the system

⁠	${\frac {L}{l}}=2\mu _{0}\log {\frac {b^{2}}{a_{1}a_{1}^{\prime }}}$	${}+{\frac {1}{2}}{\frac {a_{1}^{2}-3a_{2}^{2}}{a_{1}^{2}-a_{2}^{2}}}+{\frac {4a_{2}^{4}}{(a_{1}^{2}-a_{2}^{2})^{2}}}\log {\frac {a_{1}}{a_{2}}}$
		${}+{\frac {1}{2}}\mu ^{\prime }{\frac {a_{1}^{\prime 2}-3a_{2}^{\prime 2}}{a_{1}^{\prime 2}-a_{2}^{\prime ^{2}}}}+{\frac {4a_{2}^{\prime 4}}{(a_{1}^{\prime 2}-a_{2}^{\prime 2})^{2}}}\log {\frac {a_{1}^{\prime }}{a_{2}^{\prime }}}$.	(22)

If the conductors are solid wires, $a_{2}$ and $a_{2}^{\prime }$ are zero, and

${\frac {L}{l}}=2\mu _{o}\log {\frac {b^{2}}{a_{1}a_{1}^{\prime }}}+{\frac {1}{2}}(\mu +\mu ^{\prime })$.

(23)

It is only in the case of iron wires that we need take account of the magnetic induction in calculating their self-induction. In other cases we may make $\mu _{0}$, $\mu$, and $\mu ^{\prime }$ all equal to unity. The smaller the radii of the wires, and the greater the distance between them, the greater is the self-induction.

To find the Repulsion, $X$, between the Two Portions of Wire.

686.] By Art. 580 we obtain for the force tending to increase $b$,

$X$	${}={\frac {1}{2}}{\frac {dL}{db}}C^{2}$,
	${}=2\mu _{0}{\frac {l}{b}}C^{2}$,

(24)

which agrees with Ampère's formula, when $\mu _{0}=1$, as in air.

687.] If the length of the wires is great compared with the distance between them, we may use the coefficient of self-induction to determine the tension of the wires arising from the action of the current.

If $Z$ is this tension,

⁠	⁠	$Z$	${}={\frac {1}{2}}{\frac {dL}{dl}}C^{2}$,
			${}=C^{2}\left\{\mu _{0}\log {\frac {c^{2}}{a_{1}a_{1}^{\prime }}}+{\frac {\mu }{2}}\right\}$.		(25)

In one of Ampère's experiments the parallel conductors consist of two troughs of mercury connected with each other by a floating bridge of wire. When a current is made to enter at the extremity of one of the troughs, to flow along it till it reaches one extremity of the floating wire, to pass into the other trough through the floating bridge, and so to return along the second trough, the floating bridge moves along the troughs so as to lengthen the part of the mercury traversed by the current.

Professor Tait has simplified the electrical conditions of this experiment by substituting for the wire a floating siphon of glass filled with mercury, so that the current flows in mercury throughout its course.

Fig. 40.

This experiment is sometimes adduced to prove that two elements of a current in the same straight line repel one another, and thus to shew that Ampère's formula, which indicates such a repulsion of collinear elements, is more correct than that of Grassmann, which gives no action between two elements in the same straight line; Art. 526.

But it is manifest that since the formulae both of Ampère and of Grassmann give the same results for closed circuits, and since we have in the experiment only a closed circuit, no result of the experiment can favour one more than the other of these theories.

In fact, both formulae lead to the very same value of the repulsion as that already given, in which it appears that $b$, the distance between the parallel conductors is an important element.

When the length of the conductors is not very great compared with their distance apart, the form of the value of $L$ becomes somewhat more complicated.

688.] As the distance between the conductors is diminished, the value of $L$ diminishes. The limit to this diminution is when the wires are in contact, or when $b=a_{1}+a_{2}$. In this case

$L=2l\log \left({\frac {(a_{1}+a_{2})^{2}}{a_{1}a_{2}}}+{\frac {1}{2}}\right)$.

(26)

This is a minimum when $a_{1}=a_{2}$, and then

	$L$	${}=2l(\log 4+{\frac {1}{2}})$,	⁠
		${}=2l(1.8863)$,
		${}=3.7726l$.		(27)

This is the smallest value of the self-induction of a round wire doubled on itself, the whole length of the wire being $2l$.

Since the two parts of the wire must be insulated from each other, the self-induction can never actually reach this limiting value. By using broad flat strips of metal instead of round wires the self-induction may be diminished indefinitely.

On the Electromotive Force required to produce a Current of Varying Intensity along a Cylindrical Conductor.

689.] When the current in a wire is of varying intensity, the electromotive force arising from the induction of the current on itself is different in different parts of the section of the wire, being in general a function of the distance from the axis of the wire as well as of the time. If we suppose the cylindrical conductor to consist of a bundle of wires all forming part of the same circuit, so that the current is compelled to be of uniform strength in every part of the section of the bundle, the method of calculation which we have hitherto used would be strictly applicable. If, however, we consider the cylindrical conductor as a solid mass in which electric currents are free to flow in obedience to electromotive force, the intensity of the current will not be the same at different distances from the axis of the cylinder, and the electromotive forces themselves will depend on the distribution of the current in the different cylindric strata of the wire.

The vector-potential $H$, the density of the current $w$, and the electromotive force at any point, must be considered as functions of the time and of the distance from the axis of the wire.

The total current, $C$, through the section of the wire, and the total electromotive force, $E$, acting round the circuit, are to be regarded as the variables, the relation between which we have to find.

Let us assume as the value of $H$,

$H=S+T_{0}+T_{1}r^{2}+\mathrm {\&c.} +T_{n}r^{2n}$,

(1)

where $S$, $T_{0}$, $T_{1}$, &c. are functions of the time. Then, from the equation

${\frac {d^{2}H}{dr^{2}}}+{\frac {1}{r}}{\frac {dH}{dr}}=-4\pi w$,

(2)

we find

$-\pi w=T_{1}+\mathrm {\&c} +n^{2}T_{n}r^{2n-2}$.

(3)

If $\rho$ denotes the specific resistance of the substance per unit of volume, the electromotive force at any point is $\rho w$, and this may be expressed in terms of the electric potential and the vector potential $H$ by equations (B), Art. 598,

$\rho w=-{\frac {d\Psi }{dz}}-{\frac {dH}{dt}}$,

(4)

or

$-\rho w={\frac {d\Psi }{dz}}+{\frac {dS}{dt}}+{\frac {dT_{0}}{dt}}+{\frac {dT_{1}}{dt}}r^{2}+\mathrm {\&c.} +{\frac {dT_{n}}{dt}}r^{2n}$.

(5)

Comparing the coefficients of like powers of $r$ in equations (3) and (5),

⁠		$T_{1}$	${}={\frac {\pi }{\rho }}\left({\frac {d\Psi }{dz}}+{\frac {dS}{dt}}+{\frac {dT_{0}}{dt}}\right)$,		(6)
		$T_{2}$	${}={\frac {\pi }{\rho }}{\frac {dT_{1}}{dt}}$,		(7)
		$T_{n}$	${}={\frac {\pi }{\rho }}{\frac {1}{n^{2}}}{\frac {dT_{n-1}}{dt}}$.		(8)

Hence we may write

${\frac {dS}{dt}}=-{\frac {d\Psi }{dz}}$,

(9)

$T_{0}=T$, ⁠ $T_{1}={\frac {\pi }{\rho }}{\frac {dT}{dt}}$, … ⁠ $T_{n}={\frac {\pi ^{n}}{\rho ^{n}}}{\frac {1}{(|{\underline {n}})^{2}}}{\frac {d^{n}T}{dt^{n}}}$.

(10)

690.] To find the total current $C$, we must integrate $w$ over the section of the wire whose radius is $a$,

$C=2\pi \int _{0}^{a}wr\,dr$

(11)

Substituting the value of $\pi w$ from equation (3), we obtain

$C=-(T_{1}a^{2}+\mathrm {\&c.} +nT_{n}a^{2n})$.

(12)

The value of $H$ at any point outside the wire depends only on the total current $C$, and not on the mode in which it is distributed within the wire. Hence we may assume that the value of $H$ at the surface of the wire is $AC$, where $A$ is a constant to be determined by calculation from the general form of the circuit. Putting $H=AC$ when $r=a$, we obtain

$AC=S+T_{0}+T_{1}a^{2}+\mathrm {\&c.} +T_{n}a^{2n}$.

(13)

If we now write ${\frac {\pi a^{2}}{\rho }}=\alpha$, $\alpha$ is the value of the conductivity of unit of length of the wire, and we have

		$C$	${}=-\left(\alpha {\frac {dT}{dt}}+{\frac {2\alpha ^{2}}{1^{2}2^{2}}}{\frac {d^{2}T}{dt^{2}}}+\mathrm {\&c.} +{\frac {n\alpha ^{n}}{(|{\underline {n}})^{2}}}{\frac {d^{n}T}{dt^{n}}}+\mathrm {\&c.} \right)$,		(14)
		$AC-S$	${}=T+\alpha {\frac {dT}{dt}}+{\frac {\alpha ^{2}}{1^{2}2^{2}}}{\frac {d^{2}T}{dt^{2}}}+\mathrm {\&c.} +{\frac {\alpha ^{n}}{(|{\underline {n}})^{2}}}{\frac {d^{n}T}{dt^{n}}}+\mathrm {\&c.}$		(15)

Eliminating $T$ from these two equations, we find

	$\alpha \left(A{\frac {dC}{dt}}-{\frac {dS}{dt}}\right)+C+{\frac {1}{2}}\alpha {\frac {dC}{dt}}-{\frac {1}{12}}\alpha ^{2}$	${\frac {d^{2}C}{dt^{2}}}+{\frac {1}{48}}\alpha ^{3}{\frac {d^{3}C}{dt^{3}}}$
		${}-{\frac {1}{160}}\alpha ^{4}{\frac {d^{4}C}{dt^{4}}}+\mathrm {\&c.} =0$.	⁠	(16)

If $l$ is the whole length of the circuit, $R$ its resistance, and $E$ the electromotive force due to other causes than the induction of the current on itself,

${\frac {dS}{dt}}=-{\frac {E}{l}}$, ⁠ $\alpha ={\frac {l}{R}}$,

(17)

$E=RC+l(A+{\frac {1}{2}}){\frac {dC}{dt}}-{\frac {1}{12}}{\frac {l^{2}}{R}}{\frac {d^{2}C}{dt^{2}}}+{\frac {1}{48}}{\frac {l^{3}}{R^{2}}}{\frac {d^{3}C}{dt^{3}}}-{\frac {1}{180}}{\frac {l^{4}}{R^{3}}}{\frac {d^{4}C}{dt^{4}}}+\mathrm {\&c}$.

(18)

The first term, $RC$, of the right-hand member of this equation expresses the electromotive force required to overcome the resistance according to Ohm's law.

The second term, $l(A+{\frac {1}{2}}){\frac {dC}{dt}}$, expresses the electromotive force which would be employed in increasing the electrokinetic momentum of the circuit, on the hypothesis that the current is of uniform strength at every point of the section of the wire.

The remaining terms express the correction of this value, arising from the fact that the current is not of uniform strength at different distances from the axis of the wire. The actual system of currents has a greater degree of freedom than the hypothetical system, in which the current is constrained to be of uniform strength throughout the section. Hence the electromotive force required to produce a rapid change in the strength of the current is somewhat less than it would be on this hypothesis.

The relation between the time-integral of the electromotive force and the time-integral of the current is

$\int E\,dt=R\int C\,dt+l(A+{\frac {1}{2}})C-{\frac {1}{12}}{\frac {l^{2}}{R}}{\frac {dC}{dt}}+\mathrm {\&c}$.

(19)

If the current before the beginning of the time has a constant value $C_{0}$, and if during the time it rises to the value $C_{1}$, and remains constant at that value, then the terms involving the differential coefficients of $C$ vanish at both limits, and

$\int E\,dt=R\int C\,dt+l(A+{\frac {1}{2}})(C_{1}-C_{0})$,

(20)

the same value of the electromotive impulse as if the current had been uniform throughout the wire.

On the Geometrical Mean Distance of Two Figures in a Plane.^[1]

691.] In calculating the electromagnetic action of a current flowing in a straight conductor of any given section on the current in a parallel conductor whose section is also given, we have to find the integral

$\iiiint \log r\;dx\,dy\,dx^{\prime }\,dy^{\prime }$,

where $dx\,dy$ is an element of the area of the first section, $dx^{\prime }\,dy^{\prime }$ an element of the second section, and $r$ the distance between these elements, the integration being extended first over every element of the first section, and then over every element of the second. If we now determine a line $R$, such that this integral is equal to

$A_{1}A_{2}\log R$,

where $A_{1}$ and $A_{2}$ are the areas of the two sections, the length of $R$ will be the same whatever unit of length we adopt, and whatever system of logarithms we use. If we suppose the sections divided into elements of equal size, then the logarithm of $R$, multiplied by the number of pairs of elements, will be equal to the sum of the logarithms of the distances of all the pairs of elements. Here $R$ may be considered as the geometrical mean of all the distances between pairs of elements. It is evident that the value of $R$ must be intermediate between the greatest and the least values of $r$. If $R_{A}$ and $R_{B}$ are the geometric mean distances of two figures, $A$ and $B$, from a third, $C$, and if $R_{A+B}$ is that of the sum of the two figures from $C$, then

$(A+B)\log R_{A+B}=A\log R_{A}+B\log R_{B}$.

By means of this relation we can determine $R$ for a compound figure when we know $R$ for the parts of the figure.

692.]

Examples.

(1) Let $R$ be the mean distance from the point $O$ to the line $AB$. Let $OP$ be perpendicular to $AB$, then

$AB(\log R+1)=AP\log OA+PB\log OB+OP\;{\widehat {AOB}}$.

Fig. 41.

(2) For two lines (Fig. 42) of lengths $a$ and $b$ drawn perpendicular to the extremities of a line of length $c$ and on the same side of it.

⁠	$ab(2\log R+3)={}$	$(c^{2}-(a-b)^{2})\log {\sqrt {c^{2}+(a-b)^{2}}}+c^{2}\log c$	⁠
		${}+(a^{2}-c^{2})\log {\sqrt {a^{2}+c^{2}}}+(b^{2}-c^{2})\log {\sqrt {b^{2}+c^{2}}}$
		${}-c(a-b)\tan ^{-1}{\frac {a-b}{c}}+ac\tan ^{-1}{\frac {a}{c}}+bc\tan ^{-1}{\frac {b}{c}}$.

Fig. 42.

(3) For two lines, $PQ$ and $RS$ (Fig. 43), whose directions intersect at $O$.

$PQ.RS(2\log R+3)$	${}=\log PR(2OP.OR\sin ^{2}O-PR^{2}\cos O)$
	${}+\log QS(2OQ.OS\sin ^{2}O-QS^{2}\cos O)$
	${}-\log PS(2OP.OS\sin ^{2}O-PS^{2}\cos O)$
	${}-\log QR(2OQ.OR\sin ^{2}O-QR^{2}\cos O)$

Fig. 43.

(4) For a point $O$ and a rectangle $ABCD$ (Fig. 44). Let $OP$, $OQ$, $OR$, $OS$, be perpendiculars on the sides, then

$AB.AD(2\log R+3)$	${}=2.OP.OQ\log OA+2.OQ.OR\log OB$
	${}+2.OR.OS\log OC+2.OS.OP\log OD$
	${}+OP^{2}.{\widehat {DOA}}+OQ^{2}.{\widehat {AOB}}$
	${}+OR^{2}.{\widehat {BOC}}+OS^{2}.{\widehat {COD}}$.

Fig. 44.

(5) It is not necessary that the two figures should be different, for we may find the geometric mean of the distances between every pair of points in the same figure. Thus, for a straight line of length $a$,

		$\log R$	${}=\log a-{\frac {3}{2}}$,
or		$R$	${}=ae^{-{\frac {3}{2}}}$,
		$R$	${}=0.22313a$.

(6) For a rectangle whose sides are $a$ and $b$,

	$\log R=\log {\sqrt {a^{2}+b^{2}}}-{\frac {1}{6}}{\frac {a^{2}}{b^{2}}}\log$	${\sqrt {1+{\frac {b^{2}}{a^{2}}}}}-{\frac {1}{6}}{\frac {b^{2}}{a^{2}}}\log {\sqrt {1+{\frac {a^{2}}{b^{2}}}}}$
		${}+{\frac {2}{3}}{\frac {a}{b}}\tan ^{-1}{\frac {b}{a}}+{\frac {2}{3}}{\frac {b}{a}}\tan ^{-1}{\frac {a}{b}}-{\frac {25}{12}}$.

When the rectangle is a square, whose side is $a$,

$\log R$	${}=\log a+{\frac {1}{3}}\log 2+{\frac {\pi }{3}}-{\frac {25}{12}}$,
$R$	${}=0.44705a$.

(7) The geometric mean distance of a point from a circular line is equal to the greater of the two quantities, its distance from the centre of the circle, and the radius of the circle.

(8) Hence the geometric mean distance of any figure from a ring bounded by two concentric circles is equal to its geometric mean distance from the centre if it is entirely outside the ring, but if it is entirely within the ring

$\log R={\frac {a_{1}^{2}\log a_{1}-a_{2}^{2}\log a_{2}}{a_{1}^{2}-a_{2}^{2}}}-{\frac {1}{2}}$,

where $a_{1}$ and $a_{2}$ are the outer and inner radii of the ring. $R$ is in this case independent of the form of the figure within the ring.

(9) The geometric mean distance of all pairs of points in the ring is found from the equation

$\log R=\log a_{1}-{\frac {a_{2}^{4}}{(a_{1}^{2}-a_{2}^{2})^{2}}}\log {\frac {a_{1}}{a_{2}}}+{\frac {1}{4}}{\frac {3a_{2}^{2}-a_{1}^{2}}{a_{1}^{2}-a_{2}^{2}}}$.

For a circular area of radius $a$, this becomes

		$\log R$	${}=\log a-{\frac {1}{4}}$,
or		$R$	${}=ae^{-{\frac {1}{4}}}$,
		$R$	${}=0.7788a$.

For a circular line it becomes

$R=a$.

693.] In calculating the coefficient of self-induction of a coil of uniform section, the radius of curvature being great compared with the dimensions of the transverse section, we first determine the geometric mean of the distances of every pair of points of the section by the method already described, and then we calculate the coefficient of mutual induction between two linear conductors of the given form, placed at this distance apart.

This will be the coefficient of self-induction when the total current in the coil is unity, and the current is uniform at all points of the section.

But if there are $n$ windings in the coil we must multiply the coefficient already obtained by $n^{2}$, and thus we shall obtain the coefficient of self-induction on the supposition that the windings of the conducting wire fill the whole section of the coil.

But the wire is cylindric, and is covered with insulating material, so that the current, instead of being uniformly distributed over the section, is concentrated in certain parts of it, and this increases the coefficient of self-induction. Besides this, the currents in the neighbouring wires have not the same action on the current in a given wire as a uniformly distributed current.

The corrections arising from these considerations may be determined by the method of the geometric mean distance. They are proportional to the length of the whole wire of the coil, and may be expressed as numerical quantities, by which we must multiply the length of the wire in order to obtain the correction of the coefficient of self-induction.

Fig. 45. Let the diameter of the wire be $d$. It is covered with insulating material, and wound into a coil. We shall suppose that the sections of the wires are in square order, as in Fig. 45, and that the distance between the axis of each wire and that of the next is $D$, whether in the direction of the breadth or the depth of the coil. $D$ is evidently greater than $d$.

We have first to determine the excess of self-induction of unit of length of a cylindric wire of diameter $d$ over that of unit of length of a square wire of side $D$, or

$2\log {\frac {R\;\;\mathrm {for\;the\;square} }{R\;\;\mathrm {for\;the\;circle} }}$

${}=2\left(\log {\frac {D}{d}}+{\frac {4}{3}}\log 2+{\frac {\pi }{3}}-{\frac {11}{6}}\right)$

${}=2\left(\log {\frac {D}{d}}+0.1380606\right)$.

The inductive action of the eight nearest round wires on the wire under consideration is less than that of the corresponding eight square wires on the square wire in the middle by 2 × (.01971).

The corrections for the wires at a greater distance may be neglected, and the total correction may be written

$2(\log _{\epsilon }{\frac {D}{d}}+0.11835)$.

The final value of the self-induction is therefore

$L=n^{2}M+2l(\log _{\epsilon }{\frac {D}{d}}+0.11835)$,

where $n$ is the number of windings, and $l$ the length of the wire, $M$ the mutual induction of two circuits of the form of the mean wire of the coil placed at a distance $R$ from each other, where $R$ is the mean geometric distance between pairs of points of the section. $D$ is the distance between consecutive wires, and $d$ the diameter of the wire.

1. Trans. R. S. Edin., 1871–2.