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

by James Clerk Maxwell

CHAPTER XII.

CURRENT-SHEETS.

647.] A current-sheet is an infinitely thin stratum of conducting matter, bounded on both sides by insulating media, so that electric currents may flow in the sheet, but cannot escape from it except at certain points called Electrodes, where currents are made to enter or to leave the sheet.

In order to conduct a finite electric current, a real sheet must have a finite thickness, and ought therefore to be considered a conductor of three dimensions. In many cases, however, it is practically convenient to deduce the electric properties of a real conducting sheet, or of a thin layer of coiled wire, from those of a current-sheet as defined above.

We may therefore regard a surface of any form as a current-sheet. Having selected one side of this surface as the positive side, we shall always suppose any lines drawn on the surface to be looked at from the positive side of the surface. In the case of a closed surface we shall consider the outside as positive. See Art. 294, where, however, the direction of the current is defined as seen from the negative side of the sheet.

The Current-function.

648.] Let a fixed point $A$ on the surface be chosen as origin, and let a line be drawn on the surface from $A$ to another point $P$. Let the quantity of electricity which in unit of time crosses this line from left to right be $\phi$, then $\phi$ is called the Current-function at the point $P$.

The current-function depends only on the position of the point $P$, and is the same for any two forms of the line $AP$, provided this line can be transformed by continuous motion from one form to the other without passing through an electrode. For the two forms of the line will enclose an area within which there is no electrode, and therefore the same quantity of electricity which enters the area across one of the lines must issue across the other.

If $s$ denote the length of the line $AP$, the current across $ds$ from left to right will be ${\frac {d\phi }{ds}}\,ds$.

If $\phi$ is constant for any curve, there is no current across it. Such a curve is called a Current-line or a Stream-line.

649.] Let $\psi$ be the electric potential at any point of the sheet, then the electromotive force along any element $ds$ of a curve will be

$-{\frac {d\psi }{ds}}ds$,

provided no electromotive force exists except that which arises from differences of potential.

If $\psi$ is constant for any curve, the curve is called an Equipotential Line.

650.] We may now suppose that the position of a point on the sheet is defined by the values of $\phi$ and $\psi$ at that point. Let $ds_{1}$ be the length of the element of the equipotential line $\psi$ intercepted between the two current lines $\phi$ and $\phi +d\phi$, and let $ds_{2}$ be the length of the element of the current line $\phi$ intercepted between the two equipotential lines $\psi$ and $\psi +d\psi$. We may consider $ds_{1}$ and $ds_{2}$ as the sides of the element d<p d^\r of the sheet. The electromotive force $-d\psi$ in the direction of $ds_{2}$ produces the current $d\phi$ across $ds_{1}$.

Let the resistance of a portion of the sheet whose length is $ds_{2}$, and whose breadth is $ds_{1}$, be

$\sigma {\frac {ds_{2}}{ds_{1}}}$,

where $\sigma$ is the specific resistance of the sheet referred to unit of area, then

$d\psi =\sigma {\frac {ds_{2}}{ds_{1}}}d\phi$,

whence

${\frac {ds_{1}}{d\phi }}=\sigma {\frac {ds_{2}}{d\psi }}$.

651.] If the sheet is of a substance which conducts equally well in all directions, $ds_{1}$ is perpendicular to $ds_{2}$. In the case of a sheet of uniform resistance $\sigma$ is constant, and if we make $\psi ^{\prime }=\sigma \psi$, we shall have

${\frac {ds_{1}}{ds_{2}}}={\frac {d\phi }{d\psi }}$,

and the stream-lines and equipotential lines will cut the surface into little squares.

It follows from this that if $\phi _{1}$ and $\psi _{1}^{\prime }$ are conjugate functions (Art. 183) of $\phi$ and $\psi ^{\prime }$, the curves $\phi _{1}$ may be stream-lines in the sheet for which the curves $\psi _{1}^{\prime }$ are the corresponding equipotential lines. One case, of course, is that in which $\phi _{1}=\psi ^{\prime }$ and $\psi _{1}^{\prime }=-\phi$. In this case the equipotential lines become current-lines, and the current-lines equipotential lines^[1].

If we have obtained the solution of the distribution of electric currents in a uniform sheet of any form for any particular case, we may deduce the distribution in any other case by a proper transformation of the conjugate functions, according to the method given in Art. 190.

652.] We have next to determine the magnetic action of a current-sheet in which the current is entirely confined to the sheet, there being no electrodes to convey the current to or from the sheet.

In this case the current-function $\phi$ has a determinate value at every point, and the stream-lines are closed curves which do not intersect each other, though any one stream-line may intersect itself.

Consider the annular portion of the sheet between the stream-lines $\phi$ and $\phi +\delta \phi$. This part of the sheet is a conducting circuit in which a current of strength $\delta \phi$ circulates in the positive direction round that part of the sheet for which $\phi$ is greater than the given value. The magnetic effect of this circuit is the same as that of a magnetic shell of strength 5 $ at any point not included in the substance of the shell. Let us suppose that the shell coincides with that part of the current-sheet for which $\phi$ has a greater value than it has at the given stream-line.

By drawing all the successive stream-lines, beginning with that for which $\phi$ has the greatest value, and ending with that for which its value is least, we shall divide the current-sheet into a series of circuits. Substituting for each circuit its corresponding magnetic shell, we find that the magnetic effect of the current-sheet at any point not included in the thickness of the sheet is the same as that of a complex magnetic shell, whose strength at any point is $C+\phi$, where $C$ is a constant.

If the current-sheet is bounded, then we must make $C+\phi =0$ at the bounding curve. If the sheet forms a closed or an infinite surface, there is nothing to determine the value of the constant $C$.

653.] The magnetic potential at any point on either side of the current-sheet is given, as in Art. 415, by the expression

$\Omega =\iint {\frac {1}{r^{2}}}\phi \cos \theta \,dS$,

where $r$ is the distance of the given point from the element of surface $dS$, and $\theta$ is the angle between the direction of $r$, and that of the normal drawn from the positive side of $dS$.

This expression gives the magnetic potential for all points not included in the thickness of the current-sheet, and we know that for points within a conductor carrying a current there is no such thing as a magnetic potential.

The value of $\Omega$ is discontinuous at the current-sheet, for if $\Omega _{1}$ is its value at a point just within the current-sheet, and $\Omega _{2}$ its value at a point close to the first but just outside the current-sheet,

$\Omega _{2}=\Omega -1+4\pi \phi$,

where $\phi$ is the current-function at that point of the sheet.

The value of the component of magnetic force normal to the sheet is continuous, being the same on both sides of the sheet. The component of the magnetic force parallel to the current-lines is also continuous, but the tangential component perpendicular to the current-lines is discontinuous at the sheet. If $s$ is the length of a curve drawn on the sheet, the component of magnetic force in the direction of $ds$ is, for the negative side, ${\frac {d\Omega _{1}}{ds}}$, and for the positive side, ${\frac {d\Omega _{2}}{ds}}={\frac {d\Omega _{1}}{ds}}+4\pi {\frac {d\phi }{ds}}$.

The component of the magnetic force on the positive side therefore exceeds that on the negative side by $4\pi {\frac {d\phi }{ds}}$. At a given point this quantity will be a maximum when $ds$ is perpendicular to the current-lines.

On the Induction of Electric Currents in a Sheet of Infinite Conductivity.

654.] It was shewn in Art. 579 that in any circuit

$E={\frac {dp}{dt}}+Ri$,

where $E$ is the impressed electromotive force, $p$ the electrokinetic momentum of the circuit, $R$ the resistance of the circuit, and $i$ the current round it. If there is no impressed electromotive force and no resistance, then ${\frac {dp}{dt}}=0$, or $p$ is constant.

Now $p$, the electrokinetic momentum of the circuit, was shewn in Art. 588 to be measured by the surface-integral of magnetic induction through the circuit. Hence, in the case of a current-sheet of no resistance, the surface-integral of magnetic induction through any closed curve drawn on the surface must be constant, and this implies that the normal component of magnetic induction remains constant at every point of the current-sheet.

655.] If, therefore, by the motion of magnets or variations of currents in the neighbourhood, the magnetic field is in any way altered, electric currents will be set up in the current-sheet, such that their magnetic effect, combined with that of the magnets or currents in the field, will maintain the normal component of magnetic induction at every point of the sheet unchanged. If at first there is no magnetic action, and no currents in the sheet, then the normal component of magnetic induction will always be zero at every point of the sheet.

The sheet may therefore be regarded as impervious to magnetic induction, and the lines of magnetic induction will be deflected by the sheet exactly in the same way as the lines of flow of an electric current in an infinite and uniform conducting mass would be deflected by the introduction of a sheet of the same form made of a substance of infinite resistance.

If the sheet forms a closed or an infinite surface, no magnetic actions which may take place on one side of the sheet will produce any magnetic effect on the other side.

Theory of a Plane Current-sheet.

656.] We have seen that the external magnetic action of a current-sheet is equivalent to that of a magnetic shell whose strength at any point is numerically equal to $\phi$, the current-function. When the sheet is a plane one, we may express all the quantities required for the determination of electromagnetic effects in terms of a single function, $P$, which is the potential due to a sheet of imaginary matter spread over the plane with a surface-density $\phi$. The value of $P$ is of course where $r$ is the distance from the point ($x,$ $y$, $z$) for which $P$ is calculated, to the point $x^{\prime }$, $y^{\prime }$, $0$ in the plane of the sheet, at which the element $dx^{\prime }\,dy^{\prime }$ is taken.

To find the magnetic potential, we may regard the magnetic shell as consisting of two surfaces parallel to the plane of $xy$ the first, whose equation is $z={\frac {1}{2}}c$, having the surface-density ${\frac {\phi }{c}}$, and the second, whose equation is $z=-{\frac {1}{2}}c$, having the surface-density $-{\frac {\phi }{c}}$.

The potentials due to these surfaces will be

${\frac {1}{c}}P_{\left(z-{\frac {c}{2}}\right)}$ and $-{\frac {1}{c}}P_{\left(z+{\frac {c}{2}}\right)}$.

respectively, where the suffixes indicate that $z-{\frac {c}{2}}$ is put for $z$ in the first expression, and $z+{\frac {c}{2}}$ for $z$ in the second. Expanding

these expressions by Taylor s Theorem, adding them, and then making $c$ infinitely small, we obtain for the magnetic potential due to the sheet at any point external to it,

657.] The quantity $P$ is symmetrical with respect to the plane of the sheet, and is therefore the same when $-z$ is substituted for $z$.

$\Omega$ magnetic potential, changes sign when $-z$ is put for $z$.

At the positive surface of the sheet At the negative surface of the sheet

Within the sheet, if its magnetic effects arise from the magnetization of its substance, the magnetic potential varies continuously from $2\pi \phi$ at the positive surface to $-2\pi \phi$ at the negative surface.

If the sheet contains electric currents, the magnetic force within it does not satisfy the condition of having a potential. The magnetic force within the sheet is, however, perfectly determinate.

The normal component, is the same on both sides of the sheet and throughout its substance. If $\alpha$ and $\beta$ be the components of the magnetic force parallel to $x$ and to $y$ at the positive surface, and $\alpha ^{\prime }$, $\beta ^{\prime }$ those on the negative surface

Within the sheet the components vary continuously from $\alpha$ and $\beta$ to $\alpha ^{\prime }$ and $\beta ^{\prime }$.

The equations

${\frac {dH}{dy}}-{\frac {dG}{dz}}=-{\frac {d\Omega }{dx}}$, ${\frac {dF}{dz}}-{\frac {dH}{dx}}=-{\frac {d\Omega }{dy}}$, ${\frac {dG}{dx}}-{\frac {dF}{dy}}=-{\frac {d\Omega }{dx}}$,

(8)

which connect the components $F$, $G$, $H$ of the vector-potential due to the current-sheet with the scalar potential $\Omega$, are satisfied if we make

We may also obtain these values by direct integration, thus for $F$,

$F$	${}=\iint {\frac {u}{r}}\,dx^{\prime }\,dy^{\prime }=\iint {\frac {1}{r}}{\frac {d\phi }{dy}}\,dx^{\prime }\,dy^{\prime }$,
	${}=\int {\frac {\phi }{r}}\,dx^{\prime }-\iint \phi {\frac {d}{dy^{\prime }}}{\frac {1}{r}}\,dx^{\prime }\,dy^{\prime }$.

Since the integration is to be estimated over the infinite plane sheet, and since the first term vanishes at infinity, the expression is reduced to the second term; and by substituting

${\frac {d}{dy}}{\frac {1}{r}}$ for $-{\frac {d}{dy^{\prime }}}{\frac {1}{r}}$,

and remembering that $\phi$ depends on $x^{\prime }$ and $y^{\prime }$ and not on $x$, $y$, $z$, we obtain

$F$	${}={\frac {d}{dy}}\iint {\frac {\phi }{r}}\,dx^{\prime }\,dy^{\prime }$,
	${}={\frac {dP}{dy}}$, by (1).

If $\Omega ^{\prime }$ is the magnetic potential due to any magnetic or electric system external to the sheet, we may write and we shall then have for the components of the vector-potential due to this system.

658.] Let us now determine the electromotive force at any point of the sheet, supposing the sheet fixed.

Let $X$ and $Y$ be the components of the electromotive force parallel to $x$ and to $y$ respectively, then, by Art. 598, we have

If the electric resistance of the sheet is uniform and equal to $\sigma$,

where $u$ and $v$ are the components of the current, and if $\phi$ is the current-function,

But, by equation (3),

$2\pi \phi =-{\frac {dP}{dz}}$

at the positive surface of the current-sheet. Hence, equations (12) and (13) may be written

$-{}$	${\frac {\sigma }{2\pi }}{\frac {d^{2}P}{dy\,dz}}=-{\frac {d^{2}}{dy\,dt}}(P+P^{\prime })-{\frac {d\psi }{dx}}$,
	${\frac {\sigma }{2\pi }}{\frac {d^{2}P}{dx\,dz}}={\frac {d^{2}}{dx\,dt}}(P+P^{\prime })-{\frac {d\psi }{dy}}$,

(16)

(17)

where the values of the expressions are those corresponding to the positive surface of the sheet.

If we differentiate the first of these equations with respect to $x$, and the second with respect to $y$, and add the results, we obtain

The only value of $\psi$ which satisfies this equation, and is finite and continuous at every point of the plane, and vanishes at an infinite distance, is

Hence the induction of electric currents in an infinite plane sheet of uniform conductivity is not accompanied with differences of electric potential in different parts of the sheet.

Substituting this value of $\psi$, and integrating equations (16), (17), we obtain

Since the values of the currents in the sheet are found by differentiating with respect to $x$ or $y$, the arbitrary function of $z$ and $t$ will disappear. We shall therefore leave it out of account.

If we also write for ${\frac {\sigma }{2\pi }}$, the single symbol $R$, which represents a certain velocity, the equation between $P$ and $P^{\prime }$ becomes 659.] Let us first suppose that there is no external magnetic system acting on the current sheet. We may therefore suppose $P^{\prime }=0$. The case then becomes that of a system of electric currents in the sheet left to themselves, but acting on one another by their mutual induction, and at the same time losing their energy on account of the resistance of the sheet. The result is expressed by the equation the solution of which is

Hence, the value of $P$ on any point on the positive side of the sheet whose coordinates are $x$, $y$, $z$, and at a time $t$, is equal to the value of $P$ at the point $x$, $y$, $(z+Rt)$ at the instant when $t=0$.

If therefore a system of currents is excited in a uniform plane sheet of infinite extent and then left to itself, its magnetic effect at any point on the positive side of the sheet will be the same as if the system of currents had been maintained constant in the sheet, and the sheet moved in the direction of a normal from its negative side with the constant velocity $R$. The diminution of the electromagnetic forces, which arises from a decay of the currents in the real case, is accurately represented by the diminution of the force on account of the increasing distance in the imaginary case.

660.] Integrating equation (21) with respect to $t$, we obtain If we suppose that at first $P$ and $P^{\prime }$ are both zero, and that a magnet or electromagnet is suddenly magnetized or brought from an infinite distance, so as to change the value of $P^{\prime }$ suddenly from zero to $P^{\prime }$, then, since the time-integral in the second member of (24) vanishes with the time, we must have at the first instant

$P=-P^{\prime }$

at the surface of the sheet.

Hence, the system of currents excited in the sheet by the sudden introduction of the system to which $P^{\prime }$ is due is such that at the surface of the sheet it exactly neutralizes the magnetic effect of this system.

At the surface of the sheet, therefore, and consequently at all points on the negative side of it, the initial system of currents produces an effect exactly equal and opposite to that of the magnetic system on the positive side. We may express this by saying that the effect of the currents is equivalent to that of an image of the magnetic system, coinciding in position with that system, but opposite as regards the direction of its magnetization and of its electric currents. Such an image is called a negative image.

The effect of the currents in the sheet on a point on the positive side of it is equivalent to that of a positive image of the magnetic system on the negative side of the sheet, the lines joining corresponding points being bisected at right angles by the sheet.

The action at a point on either side of the sheet, due to the currents in the sheet, may therefore be regarded as due to an image of the magnetic system on the side of the sheet opposite to the point, this image being a positive or a negative image according as the point is on the positive or the negative side of the sheet.

661.] If the sheet is of infinite conductivity, $R=0$, and the second term of (24) is zero, so that the image will represent the effect of the currents in the sheet at any time.

In the case of a real sheet, the resistance $R$ has some finite value. The image just described will therefore represent the effect of the currents only during the first instant after the sudden introduction of the magnetic system. The currents will immediately begin to decay, and the effect of this decay will be accurately represented if we suppose the two images to move from their original positions, in the direction of normals drawn from the sheet, with the constant velocity $R$.

662.] We are now prepared to investigate the system of currents induced in the sheet by any system, $M$, of magnets or electromagnets on the positive side of the sheet, the position and strength of which vary in any manner.

Let $P^{\prime }$, as before, be the function from which the direct action of this system is to be deduced by the equations (3), (9), &c., then ${\frac {dP^{\prime }}{dt}}\delta t$ will be the function corresponding to the system represented by ${\frac {dM}{dt}}\delta t$. This quantity, which is the increment of $M$ in the time $\delta t$, may be regarded as itself representing a magnetic system.

If we suppose that at the time $t$ a positive image of the system ${\frac {dM}{dt}}\delta t$ is formed on the negative side of the sheet, the magnetic action at any point on the positive side of the sheet due to this image will be equivalent to that due to the currents in the sheet excited by the change in $M$ during the first instant after the change, and the image will continue to be equivalent to the currents in the sheet, if, as soon as it is formed, it begins to move in the negative direction of $z$ with the constant velocity $R$.

If we suppose that in every successive element of the time an image of this kind is formed, and that as soon as it is formed it begins to move away from the sheet with velocity $R$, we shall obtain the conception of a trail of images, the last of which is in process of formation, while all the rest are moving like a rigid body away from the sheet with velocity $R$.

663.] If $P^{\prime }$ denotes any function whatever arising from the action of the magnetic system, we may find $P$, the corresponding function arising from the currents in the sheet, by the following process, which is merely the symbolical expression for the theory of the trail of images.

Let $P_{\tau }$ denote the value of $P$ (the function arising from the currents in the sheet) at the point ($x$, $y$, $z+R\tau$), and at the time $t-\tau$, and let $P_{\tau }^{\prime }$ denote the value of $P^{\prime }$ (the function arising from the magnetic system) at the point ($x$, $y$, $-(z+R\tau )$), and at the time $t-\tau$. Thenand equation (21) becomes and we obtain by integrating with respect to $\tau$ from $\tau =0$ to $\tau =\infty$,as the value of the function $P$, whence we obtain all the properties of the current sheet by differentiation, as in equations (3), (9), &c.

664.] As an example of the process here indicated, let us take the case of a single magnetic pole of strength unity, moving with uniform velocity in a straight line.

Let the coordinates of the pole at the time $t$ be

$\xi ={\mathfrak {u}}t$,$\eta =0$, $\zeta =c+{\mathfrak {w}}t$.

The coordinates of the image of the pole formed at the time $t-\tau$ are

$\xi ={\mathfrak {u}}(t-\tau )$, $\eta =0$, $\zeta =-(c+{\mathfrak {w}}(t-\tau )+R\tau )$,

and if $r$ is the distance of this image from the point ($x$, $y$, $z$),

$r^{2}=(x-{\mathfrak {u}}(t-\tau ))^{2}+(z+c+{\mathfrak {w}}(t-\tau )+R\tau )^{2}$.

To obtain the potential due to the trail of images we have to calculate

${\frac {d}{dt}}\int _{0}^{\infty }{\frac {d\tau }{r}}$.

If we write

$Q^{2}={\mathfrak {u}}^{2}+(R-{\mathfrak {w}})^{2}$,

$\int _{0}^{\infty }{\frac {d\tau }{r}}={\frac {1}{Q}}\log\{Qr+{\mathfrak {u}}(x-{\mathfrak {u}}t)+(R-{\mathfrak {w}})(z+c+{\mathfrak {w}}t)\}$,

the value of $r$ in this expression being found by making $\tau =0$.

Differentiating this expression with respect to $t$, and putting $t=0$, we obtain the magnetic potential due to the trail of images,

$\Omega ={\frac {1}{Q}}{\frac {Q{\frac {{\mathfrak {w}}(Z+C)-{\mathfrak {u}}x}{r}}-{\mathfrak {u}}^{2}-{\mathfrak {w}}^{2}+R{\mathfrak {w}}}{Qr+{\mathfrak {u}}x+(R-{\mathfrak {w}})(z+c)}}$.

By differentiating this expression with respect to $x$ or $z$, we obtain the components parallel to $x$ or $z$ respectively of the magnetic force at any point, and by putting $x=0$, $z=c$, and $r=2c$ in these expressions, we obtain the following values of the components of the force acting on the moving pole itself,

$X=-{\frac {1}{4c^{2}}}{\frac {\mathfrak {u}}{Q+R-{\mathfrak {w}}}}\left\{1+{\frac {\mathfrak {w}}{Q}}-{\frac {{\mathfrak {u}}^{2}}{Q(Q+R-{\mathfrak {w}})}}\right\}$,

$Z=-{\frac {1}{4c^{2}}}\left\{{\frac {\mathfrak {w}}{Q}}-{\frac {{\mathfrak {u}}^{2}}{Q(Q+R-{\mathfrak {w}})}}\right\}$.

665.] In these expressions we must remember that the motion is supposed to have been going on for an infinite time before the time considered. Hence we must not take ${\mathfrak {w}}$ a positive quantity, for in that case the pole must have passed through the sheet within a finite time.

If we make ${\mathfrak {u}}=0$, and ${\mathfrak {w}}$ negative, $X=0$, and

$Z={\frac {1}{4c^{2}}}{\frac {\mathfrak {w}}{R+{\mathfrak {w}}}}$,

or the pole as it approaches the sheet is repelled from it. If we make ${\mathfrak {w}}=0$, we find $Q^{2}={\mathfrak {u}}^{2}+R^{2}$,

$X=-{\frac {1}{4c^{2}}}{\frac {{\mathfrak {u}}R}{Q(Q+R)}}$ and $Z={\frac {1}{4c^{2}}}{\frac {{\mathfrak {u}}^{2}}{Q(Q+R)}}$.

The component $X$ represents a retarding force acting on the pole in the direction opposite to that of its own motion. For a given value of $R$, $X$ is a maximum when ${\mathfrak {u}}=1.27\,R$.

When the sheet is a non-conductor, $R=\infty$ and $X=0$.

When the sheet is a perfect conductor, $R=0$ and $X=0$.

The component $Z$ represents a repulsion of the pole from the sheet. It increases as the velocity increases, and ultimately becomes ${\frac {1}{4c^{2}}}$ when the velocity is infinite. It has the same value when $R$ is zero.

666.] When the magnetic pole moves in a curve parallel to the sheet, the calculation becomes more complicated, but it is easy to see that the effect of the nearest portion of the trail of images is to produce a force acting on the pole in the direction opposite to that of its motion. The effect of the portion of the trail immediately behind this is of the same kind as that of a magnet with its axis parallel to the direction of motion of the pole at some time before. Since the nearest pole of this magnet is of the same name with the moving pole, the force will consist partly of a repulsion, and partly of a force parallel to the former direction of motion, but backwards. This may be resolved into a retarding force, and a force towards the concave side of the path of the moving pole.

667.] Our investigation does not enable us to solve the case in which the system of currents cannot be completely formed, on account of a discontinuity or boundary of the conducting sheet.

It is easy to see, however, that if the pole is moving parallel to the edge of the sheet, the currents on the side next the edge will be enfeebled. Hence the forces due to these currents will be less, and there will not only be a smaller retarding force, but, since the repulsive force is least on the side next the edge, the pole will be attracted towards the edge.

Theory of Arago's Rotating Disk.

668.] Arago discovered^[2] that a magnet placed near a rotating metallic disk experiences a force tending to make it follow the motion of the disk, although when the disk is at rest there is no action between it and the magnet.

This action of a rotating disk was attributed to a new kind of induced magnetization, till Faraday^[3] explained it by means of the electric currents induced in the disk on account of its motion through the field of magnetic force.

To determine the distribution of these induced currents, and their effect on the magnet, we might make use of the results already found for a conducting sheet at rest acted on by a moving magnet, availing ourselves of the method given in Art. 600 for treating the electromagnetic equations when referred to moving systems of axes. As this case, however, has a special importance, we shall treat it in a direct manner, beginning by assuming that the poles of the magnet are so far from the edge of the disk that the effect of the limitation of the conducting sheet may be neglected.

Making use of the same notation as in the preceding articles (656–667), we find for the components of the electromotive force parallel to $x$ and $y$ respectively,

$\sigma u={}$	$\gamma {\frac {dy}{dt}}-{\frac {d\psi }{dx}}$,
$\sigma v=-{}$	$\gamma {\frac {dx}{dt}}-{\frac {d\psi }{dy}}$,

(1)

where $\gamma$ is the resolved part of the magnetic force normal to the disk.

If we now express $u$ and $v$ in terms of $\phi$, the current-function, and if the disk is rotating about the axis of $z$ with the angular velocity $\omega$,

Substituting these values in equations (1), we find

	$\sigma {\frac {d\phi }{dy}}=\gamma \omega x-{\frac {d\psi }{dx}}$,
$-{}$	$\sigma {\frac {d\phi }{dx}}=\gamma \omega y-{\frac {d\psi }{dy}}$.

(4)

(5)

Multiplying (4) by $x$ and (5) by $y$, and adding, we obtain Multiplying (4) by $y$ and (5) by $-x$, and adding, we obtain

If we now express these equations in terms of $r$ and $\theta$, where they become

a = y co r 2 - / , (9)

Equation (10) is satisfied if we assume any arbitrary function $\chi$ of $r$ and $\theta$. and make

$\phi ={\frac {d\chi }{d\theta }}$, $\psi =\sigma r{\frac {d\chi }{dr}}$.	(11)
	(12)

Substituting these values in equation (9), it becomes Dividing by $\sigma r^{2}$, and restoring the coordinates $x$ and $y$, this becomes

This is the fundamental equation of the theory, and expresses the relation between the function, $\chi$, and the component, $\gamma$, of the magnetic force resolved normal to the disk.

Let< $Q$ be the potential, at any point on the positive side of the disk, due to imaginary matter distributed over the disk with the surface-density $\chi$.

At the positive surface of the disk Hence the first member of equation (14) becomes But since $Q$ satisfies Laplace s equation at all points external to the disk, and equation (14) becomes Again, since $Q$ is the potential due to the distribution $\chi$, the potential due to the distribution $\phi$, or ${\frac {d\chi }{d\theta }}$, will be ${\frac {dQ}{d\theta }}$. From this we obtain for the magnetic potential due to the currents in the disk, and for the component of the magnetic force normal to the disk due to the currents, If $\Omega _{2}$ is the magnetic potential due to external magnets, and if we write the component of the magnetic force normal to the disk due to the magnets will be We may now write equation (18), remembering that

$\gamma =\gamma _{1}+\gamma _{2}$

Integrating twice with respect to $z$, and writing $R$ for ${\frac {\sigma }{2\pi }}$, If the values of $P$ and $Q$ are expressed in terms of $r$, $\theta$, and $\zeta$, where equation (24) becomes, by integration with respect to $\zeta$,

669.] The form of this expression shews that the magnetic action of the currents in the disk is equivalent to that of a trail of images of the magnetic system in the form of a helix.

If the magnetic system consists of a single magnetic pole of strength unity, the helix will lie on the cylinder whose axis is that of the disk, and which passes through the magnetic pole. The helix will begin at the position of the optical image of the pole in the disk. The distance, parallel to the axis between consecutive coils of the helix, will be $2\pi {\frac {R}{\omega }}$. The magnetic effect of the trail will be the same as if this helix had been magnetized everywhere in the direction of a tangent to the cylinder perpendicular to its axis, with an intensity such that the magnetic moment of any small portion is numerically equal to the length of its projection on the disk.

The calculation of the effect on the magnetic pole would be complicated, but it is easy to see that it will consist of

(1) A dragging force, parallel to the direction of motion of the disk.

(2) A repulsive force acting from the disk.

(3) A force towards the axis of the disk.

When the pole is near the edge of the disk, the third of these forces may be overcome by the force towards the edge of the disk, indicated in Art. 667.

All these forces were observed by Arago, and described by him in the Annales de Chimie for 1826. See also Felici, in Tortolini's Annals, iv, p. 173 (1853), and v. p. 35; and E. Jochmann, in Crelle's Journal, lxiii, pp. 158 and 329; and Pogg. Ann. cxxii, p. 214 (1864). In the latter paper the equations necessary for determining the induction of the currents on themselves are given, but this part of the action is omitted in the subsequent calculation of results. The method of images given here was published in the Proceedings of the Royal Society for Feb. 15, 1872.

Spherical Current-Sheet.

670.] Let $\phi$ be the current-function at any point $Q$ of a spherical current-sheet, and let $P$ be the potential at a given point, due to a sheet of imaginary matter distributed over the sphere with surface-density $\phi$, it is required to find the magnetic potential and the vector-potential of the current-sheet in terms of $P$.
Fig. 39.

Let $a$ denote the radius of the sphere, $r$ the distance of the given point from the centre, and $p$ the reciprocal of the distance of the given point from the point $Q$ on the sphere at which the current-function is $\phi$.

The action of the current-sheet at any point not in its substance is identical with that of a magnetic shell whose strength at any point is numerically equal to the current-function.

The mutual potential of the magnetic shell and a unit pole placed at the point $P$ is, by Art. 410,

$\Omega =\iint \phi {\frac {dp}{da}}\,dS$.

Since $p$ is a homogeneous function of the degree $-1$ in $r$ and $a$,

	$a{\frac {dp}{da}}$	${}+r{\frac {dp}{dr}}=-p$,
or	${\frac {dp}{da}}$	${}=-{\frac {1}{a}}{\frac {d}{dr}}(pr)$,
and	$\Omega$	${}=-\iint {\frac {\phi }{a}}{\frac {d}{dr}}(pr)\,dS$.

Since $r$ and $a$ are constant during the surface-integration,

$\Omega =-{\frac {1}{a}}{\frac {d}{dr}}\left(r\iint \phi p\,ds\right)$.

But if $P$ is the potential due to a sheet of imaginary matter of surface-density $\phi$,

$P=\iint \phi p\,dS$,

and $\Omega$, the magnetic potential of the current-sheet, may he expressed in terms of $P$ in the form

$\Omega =-{\frac {1}{a}}{\frac {d}{dr}}(Pr)$.

671.] We may determine $F$, the $x$-component of the vector-potential, from the expression given in Art. 416,

$F=\iint \phi \left(m{\frac {dp}{d\zeta }}-n{\frac {dp}{d\eta }}\right)\,dS$,

where $\xi$, $\eta$, $\zeta$ are the coordinates of the element $dS$, and $l$, $m$, $n$ are the direction-cosines of the normal. Since the sheet is a sphere, the direction-cosines of the normal are

$l={\frac {\xi }{a}}$,$m={\frac {\eta }{a}}$,$n={\frac {\zeta }{a}}$.

so that

$m{\frac {dp}{d\zeta }}-n{\frac {dp}{d\eta }}$	${}=\left(\eta (z-\zeta )-\zeta (y-\eta )\right){\frac {p^{3}}{a}}$,
	${}=\left(z(\eta -y)-y(\zeta -z)\right){\frac {p^{3}}{a}}$,
	${}={\frac {z}{a}}{\frac {dp}{dy}}-{\frac {y}{a}}{\frac {dp}{dz}}$;

multiplying by $\phi \,dS$, and integrating over the surface of the sphere, we find

$F={\frac {z}{a}}{\frac {dP}{dy}}-{\frac {y}{a}}{\frac {dP}{dz}}$.

Similarly

$G={\frac {x}{a}}{\frac {dP}{dz}}-{\frac {z}{a}}{\frac {dP}{dx}}$. $H={\frac {y}{a}}{\frac {dP}{dx}}-{\frac {x}{a}}{\frac {dP}{dy}}$.

The vector ${\mathfrak {A}}$, whose components are $F$, $G$, $H$, is evidently perpendicular to the radius vector $r$, and to the vector whose components are ${\frac {dP}{dx}}$, ${\frac {dP}{dy}}$, and ${\frac {dP}{dz}}$. If we determine the lines of intersections of the spherical surface whose radius is $r$, with the series of equipotential surfaces corresponding to values of $P$ in arithmetical progression, these lines will indicate by their direction the direction of ${\mathfrak {A}}$, and by their proximity the magnitude of this vector.

In the language of Quaternions,

${\mathfrak {A}}={\frac {1}{a}}V\rho \nabla P$.

672.] If we assume as the value of $P$ within the sphere

$P=a\left({\frac {r}{a}}\right)^{i}Y_{i}$,

where $Y_{i}$ is a spherical harmonic of degree $i$, then outside the sphere

$P^{\prime }=A\left({\frac {a}{r}}\right)^{i+1}Y_{i}$.

The current-function $\phi$ is

$\phi ={\frac {2i+1}{4\pi }}{\frac {1}{a}}AY_{i}$.

The magnetic potential within the sphere is

	$\Omega =-(i+1){\frac {1}{a}}A\left({\frac {a}{r}}\right)^{i}Y_{i}$,
and outside	$\Omega ^{\prime }=i{\frac {1}{a}}A\left({\frac {a}{r}}\right)^{i+1}Y_{i}$.

For example, let it be required to produce, by means of a wire coiled into the form of a spherical shell, a uniform magnetic force $M$ within the shell. The magnetic potential within the shell is, in this case, a solid harmonic of the first degree of the form

$\Omega =Mr\cos \theta$,

where $M$ is the magnetic force. Hence $A=-{\frac {1}{2}}a^{2}M$, and

$\phi ={\frac {3}{8\pi }}Ma\cos \theta$.

The current-function is therefore proportional to the distance from the equatorial plane of the sphere, and therefore the number of windings of the wire between any two small circles must be proportional to the distance between the planes of these circles.

If $N$ is the whole number of windings, and if $\gamma$ is the strength of the current in each winding,

$\phi ={\frac {1}{2}}N\gamma \cos \theta$.

Hence the magnetic force within the coil is

$M={\frac {4\pi }{3}}{\frac {N\gamma }{a}}$.

673.] Let us next find the method of coiling the wire in order to produce within the sphere a magnetic potential of the form of a solid zonal harmonic of the second degree,

12= jl*(f cos 2 0-i).

Here = ~A (f cos 2 0-i).

i .

If the whole number of windings is $N$ the number between the pole and the polar distance $\theta$ is ${\frac {1}{2}}N\sin ^{2}\theta$.

The windings are closest at latitude 45°. At the equator the direction of winding changes, and in the other hemisphere the windings are in the contrary direction.

Let $\gamma$ be the strength of the current in the wire, then within the shell

$\Omega ={\frac {4\pi }{5}}N\gamma {\frac {r^{2}}{a^{2}}}\left({\frac {3}{2}}\cos ^{2}\theta -{\frac {1}{2}}\right)$.

Let us now consider a Conductor in the form of a plane closed curve placed anywhere within the shell with its plane perpendicular to the axis. To determine its coefficient of induction we have to find the surface-integral of ${\frac {d\Omega }{dz}}$ over the plane bounded by the curve, putting $\gamma =1$.

Now 12 = ~ 2 N (z 2 - i (x* 4 /)),

O d

d& STT ,

and - = 2 Nz. dz 5 a 2

Hence, if $S$ is the area of the closed curve, its coefficient of induction is

$M={\frac {8\pi }{5a^{2}}}NSz$.

If the current in this conductor is $\gamma ^{\prime }$, there will be, by Art. 583, a force $Z$, urging it in the direction of $z$, where

$Z=\gamma \gamma ^{\prime }{\frac {dM}{dz}}={\frac {8\pi }{5a^{2}}}NS\gamma \gamma ^{\prime }$,

and, since this is independent of $x$, $y$, $z$, the force is the same in whatever part of the shell the circuit is placed. 674.] The method given by Poisson, and described in Art. 437, may be applied to current-sheets by substituting for the body supposed to be uniformly magnetized in the direction of $z$ with intensity $I$, a current-sheet having the form of its surface, and for which the current-function isThe currents in the sheet will be in planes parallel to that of $xy$, and the strength of the current round a slice of thickness $dz$ will be $I\,dz$. The magnetic potential due to this current-sheet at any point outside it will be At any point inside the sheet it will be The components of the vector-potential are These results can be applied to several cases occurring in practice.

675.] (1) A plane electric circuit of any form.

Let $V$ be the potential due to a plane sheet of any form of which the surface-density is unity, then, if for this sheet we substitute either a magnetic shell of strength $I$ or an electric current of strength $I$ round its boundary, the values of $\Omega$ and of $F$, $G$, $H$ will be those given above.

(2) For a solid sphere of radius $a$,

		$V={\frac {4\pi }{3}}{\frac {a^{3}}{r}}$ when $r$ is greater than $a$,	(5)
	and	$V={\frac {2\pi }{3}}(3a^{2}-r^{2})$ when $r$ is less than $a$.	(6)

Hence, if such a sphere is magnetized parallel to $z$ with intensity $I$, the magnetic potential will be

		$\Omega ={\frac {4\pi }{3}}I{\frac {a^{3}}{r^{3}}}z$ outside the sphere,	(7)
	and	$\Omega ={\frac {4\pi }{3}}Iz$ inside the sphere.	(8)

If, instead of being magnetized, the sphere is coiled with wire in equidistant circles, the total strength of current between two small circles whose planes are at unit distance being $I$, then outside the sphere the value of $\Omega$ is as before, but within the sphere This is the case already discussed in Art. 672.

(3) The case of an ellipsoid uniformly magnetized parallel to a given line has been discussed in Art. 437.

If the ellipsoid is coiled with wire in parallel and equidistant planes, the magnetic force within the ellipsoid will be uniform.

(4) A Cylindric Magnet or Solenoid.

676.] If the body is a cylinder having any form of section and bounded by planes perpendicular to its generating lines, and if $V_{1}$ is the potential at the point ($x$, $y$, $z$) due to a plane area of surface-density unity coinciding with the positive end of the solenoid, and $V_{2}$ the potential at the same point due to a plane area of surface-density unity coinciding with the negative end, then, if the cylinder is uniformly and longitudinally magnetized with intensity unity, the potential at the point ($x$, $y$, $z$) will be If the cylinder, instead of being a magnetized body, is uniformly lapped with wire, so that there are $n$ windings of wire in unit of length, and if a current, $\gamma$, is made to flow through this wire, the magnetic potential outside the solenoid is as before, but within the space bounded by the solenoid and its plane ends

The magnetic potential is discontinuous at the plane ends of the solenoid, but the magnetic force is continuous.

If $r_{1}$, $r_{2}$, the distances of the centres of inertia of the positive and negative plane end respectively from the point ($x$, $y$, $z$), are very great compared with the transverse dimensions of the solenoid, we may write where $A$ is the area of either section.

The magnetic force outside the solenoid is therefore very small, and the force inside the solenoid approximates to a force parallel to the axis in the positive direction and equal to $4\pi n\gamma$.

If the section of the solenoid is a circle of radius $a$, the values of $V_{1}$ and $V_{2}$ may be expressed in the series of spherical harmonics given in Thomson and Tait's Natural Philosophy, Art. 546, Ex. II.,

	$V=2\pi \left\{-rQ_{1}+a{\frac {1}{2}}{\frac {r^{2}}{a}}Q_{2}-{\frac {1\cdot 1}{2\cdot 4}}{\frac {r^{4}}{a^{3}}}Q_{4}+{\frac {1\cdot 1\cdot 3}{2\cdot 4\cdot 6}}{\frac {r^{6}}{a^{5}}}Q_{6}+\mathrm {\&c.} \right\}$ when $r<a$,	(14)
	$V=2\pi \left\{{\frac {1}{2}}{\frac {a^{2}}{r}}-{\frac {1\cdot 1}{2\cdot 4}}{\frac {a^{4}}{r^{3}}}Q_{2}+{\frac {1\cdot 1\cdot 3}{2\cdot 4\cdot 6}}{\frac {a^{6}}{r^{5}}}Q_{4}-\mathrm {\&c.} \right\}$ when $r>a$.	(15)

In these expressions $r$ is the distance of the point ($x$, $y$, $z$) from the centre of one of the circular ends of the solenoid, and the zonal harmonics, $Q_{1}$, $Q_{2}$, &c., are those corresponding to the angle $\theta$ which $r$ makes with the axis of the cylinder.

The first of these expressions is discontinuous when $\theta ={\frac {\pi }{2}}$, but we must remember that within the solenoid we must add to the magnetic force deduced from this expression a longitudinal force $4\pi n\gamma$.

677.] Let us now consider a solenoid so long that in the part of space which we consider, the terms depending on the distance from the ends may be neglected.

The magnetic induction through any closed curve drawn within the solenoid is $4\pi n\gamma A^{\prime }$, where $A^{\prime }$ is the area of the projection of the curve on a plane normal to the axis of the solenoid.

If the closed curve is outside the solenoid, then, if it encloses the solenoid, the magnetic induction through it is $4\pi n\gamma A$, where $A$ is the area of the section of the solenoid. If the closed curve does not surround the solenoid, the magnetic induction through it is zero.

If a wire be wound $n^{\prime }$ times round the solenoid, the coefficient of induction between it and the solenoid is By supposing these windings to coincide with $n$ windings of the solenoid, we find that the coefficient of self-induction of unit of length of the solenoid, taken at a sufficient distance from its extremities, is

Near the ends of a solenoid we must take into account the terms depending on the imaginary distribution of magnetism on the plane ends of the solenoid. The effect of these terms is to make the coefficient of induction between the solenoid and a circuit which surrounds it less than the value $4\pi nA$, which it has when the circuit surrounds a very long solenoid at a great distance from either end.

Let us take the case of two circular and coaxal solenoids of the same length $l$. Let the radius of the outer solenoid be $c_{1}$, and let it be wound with wire so as to have $n_{1}$ windings in unit of length. Let the radius of the inner solenoid be $c_{2}$, and let the number of windings in unit of length be $n_{2}$, then the coefficient of induction between the solenoids, neglecting the effect of the ends, is

		$M$	${}=Gg$,	(18)
where		$G$	${}=4\pi n$,	(19)
and		$g$	${}=\pi c_{2}^{2}ln_{2}$.	(20)

678.] To determine the effect of the positive end of the solenoids we must calculate the coefficient of induction on the outer solenoid due to the circular disk which forms the end of the inner solenoid. For this purpose we take the second expression for $V$, as given n equation (15), and differentiate it with respect to $r$. This gives the magnetic force in the direction of the radius. We then multiply this expression by $2pir^{2}\,d\mu$, and integrate it with respect to $\mu$ from $\mu =0$ to $\mu ={\frac {z}{\sqrt {z_{2}+c_{1}^{2}}}}$. This gives the coefficient of induction with respect to a single winding of the outer solenoid at a distance $z$ from the positive end. We then multiply this by $dz$, and integrate with respect to $z$ from $z=l$ to $z=0$. Finally, we multiply the result by $n_{1}n_{2}$, and so find the effect of one of the ends in diminishing the coefficient of induction.

We thus find for the value of the coefficient of mutual induction between the two cylinders,

where		$a$	${}={\frac {1}{2}}{\frac {c_{2}+l-r}{c_{2}}}+{\frac {1\cdot 3}{2\cdot 4}}\cdot {\frac {1}{2\cdot 3}}{\frac {c_{2}^{2}}{c_{1}^{2}}}\left(1-{\frac {c_{1}^{3}}{r^{3}}}\right)$
			${}+{\frac {1\cdot 3\cdot 5}{2\cdot 4\cdot 6}}\cdot {\frac {1}{4\cdot 5}}{\frac {c_{2}^{4}}{c_{1}^{4}}}\left({\frac {5}{12}}-{\frac {2}{3}}{\frac {c_{1}^{3}}{r^{3}}}+4{\frac {c_{1}^{5}}{r^{5}}}-{\frac {15}{4}}{\frac {c_{1}^{7}}{r^{7}}}\right)+\mathrm {\&c.}$.	(22)

where $r$ is put, for brevity, for ${\sqrt {l^{2}+c_{1}^{2}}}$.

It appears from this, that in calculating the mutual induction of two coaxal solenoids, we must use in the expression (20) instead of the true length $l$ the corrected length $l-2c_{1}a$, in which a portion equal to $ac_{1}$ is supposed to be cut off at each end. When the solenoid is very long compared with its external radius,

679.] When a solenoid consists of a number of layers of wire of such a diameter that there are $n$ layers in unit of length, the number of layers in the thickness $dr$ is $n\,dr$, and we have If the thickness of the wire is constant, and if the induction take place between an external coil whose outer and inner radii are $x$ and $y$ respectively, and an inner coil whose outer and inner radii are $y$ and $z$, then, neglecting the effect of the ends,

That this may be a maximum, $x$ and $z$ being given, and $y$ variable,

This equation gives the best relation between the depths of the primary and secondary coil for an induction-machine without an iron core.

If there is an iron core of radius $z$, then $G$ remains as before, but

	$g$	${}=\pi l\int n^{2}(r^{2}+4\pi \kappa z^{2})\,dr$,	(27)
		${}=\pi ln^{2}\left({\frac {y^{3}-z^{2}}{3}}+4\pi \kappa z^{2}(y-z)\right)$.	(28)

If $y$ is given, the value of $z$ which gives the maximum value of $g$ is When, as in the case of iron, $\kappa$ is a large number, $z={\frac {2}{3}}y,$ nearly. If we now make $x$ constant, and $y$ and $z$ variable, we obtain the maximum value of $Gg$ when The coefficient of self-induction of a long solenoid whose outer and inner radii are $x$ and $y$, and having a long iron core whose radius is $z$, is

680.] We have hitherto supposed the wire to be of uniform thickness. We shall now determine the law according to which the thickness must vary in the different layers in order that, for a given value of the resistance of the primary or the secondary coil, the value of the coefficient of mutual induction may be a maximum.

Let the resistance of unit of length of a wire, such that $n$ windings occupy unit of length of the solenoid, be $\rho n^{2}$.

The resistance of the whole solenoid is

The condition that, with a given value of $R$, $G$ may be a maximum is ${\frac {dG}{dr}}=C{\frac {dR}{dr}}$, where $C$ is some constant.

This gives $n^{2}$ proportional to ${\frac {1}{r}}$, or the diameter of the wire of the exterior coil must be proportional to the square root of the radius.

In order that, for a given value of R, g may be a maximum

Hence, if there is no iron core, the diameter of the wire of the interior coil should be inversely as the square root of the radius, but if there is a core of iron having a high capacity for magnetization, the diameter of the wire should be more nearly directly proportional to the square root of the radius of the layer.

An Endless Solenoid.

681.] If a solid be generated by the revolution of a plane area $A$ about an axis in its own plane, not cutting it, it will have the form of a ring. If this ring be coiled with wire, so that the windings of the coil are in planes passing through the axis of the ring, then, if $n$ is the whole number of windings, the current-function of the layer of wire is $\phi ={\frac {1}{2\pi }}n\gamma \theta$, where $\theta$ is the angle of azimuth about the axis of the ring.

If $\Omega$ is the magnetic potential inside the ring and $\Omega ^{\prime }$ that outside, then

$\Omega -\Omega ^{\prime }=4\pi \phi +C=2n\gamma \theta +C$.

Outside the ring $\Omega ^{\prime }$, must satisfy Laplace's equation, and must vanish at an infinite distance. From the nature of the problem it must be a function of $\theta$ only. The only value of $\Omega ^{\prime }$ which fulfils these conditions is zero. Hence

$\Omega ^{\prime }=0$,$\Omega =2n\gamma \theta +C$.

The magnetic force at any point within the ring is perpendicular to the plane passing through the axis, and is equal to $2n\gamma {\frac {1}{r}}$ where $r$ is the distance from the axis. Outside the ring there is no magnetic force.

If the form of a closed curve be given by the coordinates $z$, $r$, and $\theta$ of its tracing point as functions of $s$, its length from a fixed point, the magnetic induction through the closed curve is

$2n\gamma \int _{0}^{s}{\frac {z}{r}}{\frac {dr}{ds}}ds$

taken round the curve, provided the curve is wholly inside the ring. If the curve lies wholly without the ring, but embraces it, the magnetic induction through it is

$2n\gamma \int _{0}^{s^{\prime }}{\frac {z^{\prime }}{r^{\prime }}}{\frac {dr^{\prime }}{ds^{\prime }}}ds^{\prime }=2n\gamma a$,

where the accented coordinates refer not to the closed curve, but to a single winding of the solenoid.

The magnetic induction through any closed curve embracing the ring is therefore the same, and equal to $2n\gamma a$, where $a$ is the linear quantity $\int _{0}^{s^{\prime }}{\frac {z^{\prime }}{ds^{\prime }}}ds^{\prime }$. If the closed curve does not embrace the ring, the magnetic induction through it is zero.

Let a second wire be coiled in any manner round the ring, not necessarily in contact with it, so as to embrace it $n^{\prime }$ times. The induction through this wire is $2nn^{\prime }\gamma a$, and therefore $M$, the coefficient of induction of the one coil on the other, is $M=2nn^{\prime }a$.

Since this is quite independent of the particular form or position of the second wire, the wires, if traversed by electric currents, will experience no mechanical force acting between them. By making the second wire coincide with the first, we obtain for the coefficient of self-induction of the ring-coil

$L=2n^{2}a$.

1. See Thomson, Camb. and Dub. Math. Journ., vol.iii. p. 286.
2. Annales de Chimie et de Physique, 1826.
3. Exp. Res., 81.