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

by James Clerk Maxwell

CHAPTER VIII.

EXPLORATION OF THE FIELD BY MEANS OF THE SECONDARY CIRCUIT.

585.] We have proved in Arts. 582, 583, 584 that the electromagnetic action between the primary and the secondary circuit depends on the quantity denoted by M, which is a function of the form and relative position of the two circuits.

Although this quantity M is in fact the same as the potential of the two circuits, the mathematical form and properties of which we deduced in Arts. 423, 492, 521, 539 from magnetic and electromagnetic phenomena, we shall here make no reference to these results, but begin again from a new foundation, without any assumptions except those of the dynamical theory as stated in Chapter VII.

The electrokinetic momentum of the secondary circuit consists of two parts (Art. 578), one, Mi₁, depending on the primary current i₁, while the other, Ni₂, depends on the secondary current i₂. We are now to investigate the first of these parts, which we shall denote by p, where

$p=Mi_{1}.\,$

(1)

We shall also suppose the primary circuit fixed, and the primary current constant. The quantity p, the electrokinetic momentum of the secondary circuit, will in this case depend only on the form and position of the secondary circuit, so that if any closed curve be taken for the secondary circuit, and if the direction along this curve, which is to be reckoned positive, be chosen, the value of p for this closed curve is determinate. If the opposite direction along the curve had been chosen as the positive direction, the sign of the quantity p would have been reversed.

586.] Since the quantity p depends on the form and position of the circuit, we may suppose that each portion of the circuit contributes something to the value of p, and that the part contributed by each portion of the circuit depends on the form and position of that portion only, and not on the position of other parts of the circuit.

This assumption is legitimate, because we are not now considering a current the parts of which may, and indeed do, act on one an other, but a mere circuit, that is, a closed curve along which a current may flow, and this is a purely geometrical figure, the parts of which cannot be conceived to have any physical action on each other.

We may therefore assume that the part contributed by the element ds of the circuit is Jds, where J is a quantity depending on the position and direction of the element ds. Hence, the value of p may be expressed as a line-integral

$p=\int {J\,ds},$

(2)

where the integration is to be extended once round the circuit.

587.] We have next to determine the form of the quantity J. In the first place, if ds is reversed in direction, J is reversed in sign. Hence, if two circuits ABCE and AECD have the arc AEG common, but reckoned in opposite directions in the two circuits, the sum of the values of p for the two circuits ABCE and AECD will be equal to the value of p for the circuit ABCD, which is made up of the two circuits.

For the parts of the line-integral depending on the arc AEC are equal but of opposite sign in the two partial circuits, so that they destroy each other when the sum is taken, leaving only those parts of the line-integral which depend on the external boundary of ABCD.

In the same way we may shew that if a surface bounded by a closed curve be divided into any number of parts, and if the boundary of each of these parts be considered as a circuit, the positive direction round every circuit being the same as that round the external closed curve, then the value of p for the closed curve is equal to the sum of the values of p for all the circuits. See Art. 483.

588.] Let us now consider a portion of a surface, the dimensions of which are so small with respect to the principal radii of curvature of the surface that the variation of the direction of the normal within this portion may be neglected. We shall also suppose that if any very small circuit be carried parallel to itself from one part of this surface to another, the value of p for the small circuit is not sensibly altered. This will evidently be the case if the dimensions of the portion of surface are small enough compared with its distance from the primary circuit.

If any closed curve be drawn on this portion of the surface, the value of p will be proportional to its area.

For the areas of any two circuits may be divided into small elements all of the same dimensions, and having the same value of p. The areas of the two circuits are as the numbers of these elements which they contain, and the values of p for the two circuits are also in the same proportion.

Hence, the value of p for the circuit which bounds any element dS of a surface is of the form IdS, where I is a quantity depending on the position of dS and on the direction of its normal. We have therefore a new expression for p,

$p=\iint {I\,dS},$

(3)

where the double integral is extended over any surface bounded by the circuit.

589.] Let ABCD be a circuit, of which AC is an elementary portion, so small that it may be considered straight. Let APB and CQB be small equal areas in the same plane, then the value of p will be the same for the small circuits APB and CQB, or

$p(APB)=p(CQB).\,$

Hence

${\begin{aligned}p(APBQCD)&=p(ABQCD)+p(APB),\\&=p(ABQCD)+p(CQB),\\&=p(ABCD),\end{aligned}}$

or the value of p is not altered by the substitution of the crooked line APQC for the straight line AC, provided the area of the circuit is not sensibly altered. This, in fact, is the principle established by Ampère's second experiment (Art. 506), in which a crooked portion of a circuit is shewn to be equivalent to a straight portion provided no part of the crooked portion is at a sensible distance from the straight portion.

If therefore we substitute for the element ds three small elements, dx, dy, and dz, drawn in succession, so as to form a continuous path from the beginning to the end of the element ds, and if Fdx, Gdy, and Hdz denote the elements of the line-integral corresponding to dx, dy, and dz respectively, then

$Jds=Fdx+Gdy+Hdz.\,$

(4)

590.] We are now able to determine the mode in which the quantity J depends on the direction of the element ds. For, by (4),

$J=F{\frac {dx}{ds}}+G{\frac {dy}{ds}}+H{\frac {dz}{ds}}.$

(5)

This is the expression for the resolved part, in the direction of ds, of a vector, the components of which, resolved in the directions of the axes of x, y, and z are F, G and H respectively.

If this vector be denoted by ${\mathfrak {A}}$, and the vector from the origin to a point of the circuit by ρ, the element of the circuit will be dρ, and the quaternion expression for J will be

$-S{\mathfrak {A}}\,d\rho .$

We may now write equation (2) in the form

$p=\int \left(F{\frac {dx}{ds}}+G{\frac {dy}{ds}}+H{\frac {dz}{ds}}\right)ds,$

(6)

or

$p=-\int S{\mathfrak {A}}\,d\rho .$

(7)

The vector ${\mathfrak {A}}$ and its constituents F, G, H depend on the position of ds in the field, and not on the direction in which it is drawn. They are therefore functions of x, y, z, the coordinates of ds, and not of l, m, n, its direction-cosines.

The vector ${\mathfrak {A}}$ represents in direction and magnitude the time-integral of the electromotive force which a particle placed at the point (x, y, z) would experience if the primary current were suddenly stopped. We shall therefore call it the Electrokinetic Momentum at the point (x, y, z). It is identical with the quantity which we investigated in Art. 405 under the name of the vector-potential of magnetic induction.

The electrokinetic momentum of any finite line or circuit is the line-integral, extended along the line or circuit, of the resolved part of the electrokinetic momentum at each point of the same.

591.] Let us next determine the value of p for the elementary rectangle ABCD, of which the sides are dy and dz, the positive direction being from the direction of the axis of y to that of z.

Let the coordinates of 0, the centre of gravity of the element, be x₀, y₀, z₀, and let G₀, H₀, be the values of G and of H at this point.

The coordinates of A, the middle point of the first side of the rectangle, are y₀ and z − ½dz. The corresponding value of G is

$G=G_{0}-{\frac {1}{2}}{\frac {dG}{dz}}dz+\And c.,$

(8)

and the part of the value of p which arises from the side A is approximately

$G_{0}dy-{\frac {1}{2}}{\frac {dG}{dz}}dydz.$

(9)

Similarly, for B,

$H_{0}dz+{\frac {1}{2}}{\frac {dH}{dy}}dydz.$

For C,

$-G_{0}dy-{\frac {1}{2}}{\frac {dG}{dz}}dydz.$

For D,

$-H_{0}dz+{\frac {1}{2}}{\frac {dH}{dy}}dydz.$

Adding these four quantities, we find the value of p for the rectangle

$p=\left({\frac {dH}{dy}}-{\frac {dG}{dz}}\right)dydz.$

(10)

If we now assume three new quantities, a, b, c, such that

${\begin{aligned}a&={\frac {dH}{dy}}-{\frac {dG}{dz}},\\b&={\frac {dF}{dz}}-{\frac {dH}{dx}},\\c&={\frac {dG}{dx}}-{\frac {dF}{dy}},\end{aligned}}$

(A)

and consider these as the constituents of a new vector ${\mathfrak {B}}$, then, by Theorem IV, Art. 24, we may express the line-integral of ${\mathfrak {A}}$, round any circuit in the form of the surface-integral of ${\mathfrak {B}}$, over a surface bounded by the circuit, thus

$p=\int {\left(F{\frac {dx}{ds}}+G{\frac {dy}{ds}}+H{\frac {dz}{ds}}\right)ds}=\iint {(la+mb+nc)dS},$

(11)

or

$p=\int {T{\mathfrak {A}}\cos \epsilon \,ds}=\iint {T{\mathfrak {B}}\cos \eta \,dS},$

(12)

where ε is the angle between ${\mathfrak {A}}$ and ds, and η that between ${\mathfrak {B}}$ and the normal to dS, whose direction-cosines are l, m, n, and $T{\mathfrak {A}}$, $T{\mathfrak {B}}$ denote the numerical values of ${\mathfrak {A}}$ and ${\mathfrak {B}}$

Comparing this result with equation (3), it is evident that the quantity I in that equation is equal to ${\mathfrak {B}}\cos \eta$, or the resolved part of ${\mathfrak {B}}$ normal to dS.

592.] We have already seen (Arts. 490, 541) that, according to Faraday s theory, the phenomena of electromagnetic force and induction in a circuit depend on the variation of the number of lines of magnetic induction which pass through the circuit. Now the number of these lines is expressed mathematically by the surface-integral of the magnetic induction through any surface bounded by the circuit. Hence, we must regard the vector ${\mathfrak {B}}$ and its components a, b, c as representing what we are already acquainted with as the magnetic induction and its components.

In the present investigation we propose to deduce the properties of this vector from the dynamical principles stated in the last chapter, with as few appeals to experiment as possible.

In identifying this vector, which has appeared as the result of a mathematical investigation, with the magnetic induction, the properties of which we learned from experiments on magnets, we do not depart from this method, for we introduce no new fact into the theory, we only give a name to a mathematical quantity, and the propriety of so doing is to be judged by the agreement of the relations of the mathematical quantity with those of the physical quantity indicated by the name.

The vector ${\mathfrak {B}}$, since it occurs in a surface-integral, belongs evidently to the category of fluxes described in Art. 13. The vector ${\mathfrak {A}}$, on the other hand, belongs to the category of forces, since it appears in a line-integral.

593.] We must here recall to mind the conventions about positive and negative quantities and directions, some of which were stated in Art. 23. We adopt the right-handed system of axes, so that if a right-handed screw is placed in the direction of the axis of x, and a nut on this screw is turned in the positive direction of rotation, that is, from the direction of y to that of z, it will move along the screw in the positive direction of x.

We also consider vitreous electricity and austral magnetism as positive. The positive direction of an electric current, or of a line of electric induction, is the direction in which positive electricity moves or tends to move, and the positive direction of a line of magnetic induction is the direction in which a compass needle points with the end which turns to the north. See Fig. 24, Art. 498, and Fig. 25, Art. 501.

The student is recommended to select whatever method appears to him most effectual in order to fix these conventions securely in his memory, for it is far more difficult to remember a rule which determines in which of two previously indifferent ways a statement is to be made, than a rule which selects one way out of many.

594.] We have next to deduce from dynamical principles the expressions for the electromagnetic force acting on a conductor carrying an electric current through the magnetic field, and for the electromotive force acting on the electricity within a body moving in the magnetic field. The mathematical method which we shall adopt may be compared with the experimental method used by Faraday^[1] in exploring the field by means of a wire, and with what we have already done at Art. 490, by a method founded on experiments. What we have now to do is to determine the effect on the value of p, the electrokinetic momentum of the secondary circuit, due to given alterations of the form of that circuit.

Let AA', BB' be two parallel straight conductors connected by the conducting arc C, which may be of any form, and by a straight conductor AB, which is capable of sliding parallel to itself along the conducting rails AA' and BB'.

Let the circuit thus formed be considered as the secondary circuit, and let the direction ABC be assumed as the positive direction round it.

Let the sliding piece move parallel to itself from the position AB to the position A'B'. We have to determine the variation of p, the electrokinetic momentum of the circuit, due to this displacement of the sliding piece.

The secondary circuit is changed from ABC to A'B'C, hence, by Art, 587,

$p(A'B'C')-p(ABC)=p(A'AB'B).\,$

(13)

We have therefore to determine the value of p for the parallelogram AA'B'B. If this parallelogram is so small that we may neglect the variations of the direction and magnitude of the magnetic induction at different points of its plane, the value of p is, by Art. 591, ${\mathfrak {B}}\cos \eta \cdot AA'B'B$, where ${\mathfrak {B}}$ is the magnetic induction, and η the angle which it makes with the positive direction of the normal to the parallelogram AA'B'B.

We may represent the result geometrically by the volume of the parallelepiped, whose base is the parallelogram AA'B'B, and one of whose edges is the line AM, which represents in direction and magnitude the magnetic induction ${\mathfrak {B}}$. If the parallelogram is in the plane of the paper, and if AM is drawn upwards from the paper, the volume of the parallelepiped is to be taken positively, or more generally, if the directions of the circuit AB, of the magnetic induction AM, and of the displacement AA', form a right-handed system when taken in this cyclical order.

The volume of this parallelepiped represents the increment of the value of p for the secondary circuit due to the displacement of the sliding piece from AB to A'B'.

Electromotive Force acting on the Sliding Piece.

595.] The electromotive force produced in the secondary circuit by the motion of the sliding piece is, by Art. 579,

$E=-{\frac {dp}{dt}}.$

(14)

If we suppose AA' to be the displacement in unit of time, then AA' will represent the velocity, and the parallelepiped will represent ${\frac {dp}{dt}}$, and therefore, by equation (14), the electromotive force in the negative direction BA.

Hence, the electromotive force acting on the sliding piece AB, in consequence of its motion through the magnetic field, is represented by the volume of the parallelepiped, whose edges represent in direction and magnitude the velocity, the magnetic induction, and the sliding piece itself, and is positive when these three directions are in right-handed cyclical order.

Electromagnetic Force acting on the Sliding Piece.

596.] Let i₂ denote the current in the secondary circuit in the positive direction ABC, then the work done by the electromagnetic force on AB while it slides from the position AB to the position A'B' is (M' – M)i_li₂, where M and M' are the values of M₁₂ in the initial and final positions of AB. But (M' – M)i₁ is equal to p' – p, and this is represented by the volume of the parallelepiped on AB, AM, and AA'. Hence, if we draw a line parallel to AB to represent the quantity AB . i₂, the parallelepiped contained by this line, by AM, the magnetic induction, and by AA', the displace ment, will represent the work done during- this displacement.

For a given distance of displacement this will be greatest when the displacement is perpendicular to the parallelogram whose sides are AB and AM. The electromagnetic force is therefore represented by the area of the parallelogram on AB and AM multiplied by i₂, and is in the direction of the normal to this parallelogram, drawn so that AB, AM, and the normal are in right-handed cyclical order.

On Four Definitions of a Line of Magnetic Induction.

597.] If the direction AA', in which the motion of the sliding piece takes place, coincides with AM, the direction of the magnetic induction, the motion of the sliding piece will not call electromotive force into action, whatever be the direction of AB, and if AB carries an electric current there will be no tendency to slide along AA'.

Again, if AB, the sliding piece, coincides in direction with AM, the direction of magnetic induction, there will be no electromotive force called into action by any motion of AB, and a current through AB will not cause AB to be acted on by mechanical force.

We may therefore define a line of magnetic induction in four different ways. It is a line such that

(1) If a conductor be moved along it parallel to itself it will experience no electromotive force.

(2) If a conductor carrying a current be free to move along a line of magnetic induction it will experience no tendency to do so.

(3) If a linear conductor coincide in direction with a line of magnetic induction, and be moved parallel to itself in any direction, it will experience no electromotive force in the direction of its length.

(4) If a linear conductor carrying an electric current coincide in direction with a line of magnetic induction it will not experience any mechanical force.

On General Equations of Electromotive Force.

598.] We have seen that E, the electromotive force due to induction acting on the secondary circuit, is equal to $-{\frac {dp}{dt}}$, where

$p=\int {\left(F{\frac {dx}{ds}}+G{\frac {dy}{ds}}+H{\frac {dz}{ds}}+\right)ds}.$

(1)

To determine the value of E, let us differentiate the quantity under the integral sign with respect to t, remembering that if the secondary circuit is in motion, x, y and z are functions of the time. We obtain

${\begin{aligned}E=&-\int {\left({\frac {dF}{dt}}{\frac {dx}{ds}}+{\frac {dG}{dt}}{\frac {dy}{ds}}+{\frac {dH}{dt}}{\frac {dz}{ds}}\right)ds}\\&-\int {\left({\frac {dF}{dx}}{\frac {dx}{ds}}+{\frac {dG}{dx}}{\frac {dy}{ds}}+{\frac {dH}{dx}}{\frac {dz}{ds}}\right){\frac {dx}{dt}}ds}\\&-\int {\left({\frac {dF}{dy}}{\frac {dx}{ds}}+{\frac {dG}{dy}}{\frac {dy}{ds}}+{\frac {dH}{dy}}{\frac {dz}{ds}}\right){\frac {dy}{dt}}ds}\\&-\int {\left({\frac {dF}{dx}}{\frac {dx}{ds}}+{\frac {dG}{dz}}{\frac {dy}{ds}}+{\frac {dH}{dz}}{\frac {dz}{ds}}\right){\frac {dz}{dt}}ds}\\&-\int {\left(F{\frac {d^{2}x}{dsdt}}+G{\frac {d^{2}y}{dsdt}}+H{\frac {d^{2}z}{dsdt}}\right)ds}.\end{aligned}}$

(2)

Now consider the second term of the integral, and substitute from equations (A), Art. 591, the values of ${\frac {dG}{dx}}$ and ${\frac {dH}{dx}}$. This term then then becomes,

$-\int {\left(c{\frac {dy}{ds}}-b{\frac {dz}{ds}}+{\frac {dF}{dx}}{\frac {dx}{ds}}+{\frac {dF}{dy}}{\frac {dy}{ds}}+{\frac {dF}{dz}}{\frac {dz}{ds}}\right){\frac {dx}{dt}}ds},$

which we may write

$-\int {\left(c{\frac {dy}{ds}}-b{\frac {dz}{ds}}+{\frac {dF}{ds}}\right){\frac {dx}{dt}}ds},$

Treating the third and fourth terms in the same way, and collecting the terms in ${\frac {dx}{ds}}$, ${\frac {dy}{ds}}$, and ${\frac {dz}{ds}}$, remembering that

$\int {\left({\frac {dF}{ds}}{\frac {dx}{dt}}+F{\frac {d^{2}x}{dsdt}}\right)ds}=F{\frac {dx}{dt}},$

(3)

and therefore that the integral, when taken round the closed curve, vanishes,

${\begin{aligned}E&=\int {\left(c{\frac {dy}{dt}}-b{\frac {dz}{dt}}+{\frac {dF}{dt}}\right){\frac {dx}{ds}}ds}\\&+\int {\left(a{\frac {dz}{dt}}-c{\frac {dx}{dt}}+{\frac {dG}{dt}}\right){\frac {dy}{ds}}ds}\\&+\int {\left(b{\frac {dx}{dt}}-a{\frac {dy}{dt}}+{\frac {dH}{dt}}\right){\frac {dz}{ds}}ds}.\end{aligned}}$

(4)

We may write this expression in the form

$E=\int {\left(P{\frac {dx}{ds}}+Q{\frac {dy}{ds}}+R{\frac {dz}{ds}}\right)ds},$

(5)

where

$\left.{\begin{aligned}P&=c{\frac {dy}{dt}}-b{\frac {dz}{dt}}-{\frac {dF}{dt}}-{\frac {d\Psi }{dt}},\\Q&=a{\frac {dz}{dt}}-c{\frac {dx}{dt}}-{\frac {dG}{dt}}-{\frac {d\Psi }{dt}},\\R&=b{\frac {dx}{dt}}-a{\frac {dy}{dt}}-{\frac {dH}{dt}}-{\frac {d\Psi }{dt}}.\end{aligned}}\right\}{\begin{matrix}{\text{Equations of }}\\{\text{Electromotive}}\\{\text{Force.}}\end{matrix}}$

(B)

The terms involving the new quantity $\Psi$ are introduced for the sake of giving generality to the expressions for $P$, $Q$, $R$. They disappear from the integral when extended round the closed circuit. The quantity $\Psi$ is therefore indeterminate as far as regards the problem now before us, in which the total electromotive force round the circuit is to be determined. We shall find, however, that when we know all the circumstances of the problem, we can assign a definite value to $\Psi$, and that it represents, according to a certain definition, the electric potential at the point $x,y,z$.

The quantity under the integral sign in equation (5) represents the electromotive force acting on the element $ds$ of the circuit.

If we denote by $T{\mathfrak {F}}$, the numerical value of the resultant of P, Q, and R, and by ε, the angle between the direction of this resultant and that of the element ds, we may write equation (5),

$E=\int {T{\mathfrak {F}}\cos \epsilon \,ds}.$

(6)

The vector $T{\mathfrak {F}}$ is the electromotive force at the moving element ds. Its direction and magnitude depend on the position and motion of ds, and on the variation of the magnetic field, but not on the direction of ds. Hence we may now disregard the circumstance that ds forms part of a circuit, and consider it simply as a portion of a moving body, acted on by the electromotive force $T{\mathfrak {F}}$. The electromotive force at a point has already been defined in Art. 68. It is also called the resultant electrical force, being the force which would be experienced by a unit of positive electricity placed at that point. We have now obtained the most general value of this quantity in the case of a body moving in a magnetic field due to a variable electric system.

If the body is a conductor, the electromotive force will produce a current; if it is a dielectric, the electromotive force will produce only electric displacement.

The electromotive force at a point, or on a particle, must be carefully distinguished from the electromotive force along- an arc of a curve, the latter quantity being- the line-integral of the former. See Art. 69.

599.] The electromotive force, the components of which are defined by equations (B), depends on three circumstances. The first of these is the motion of the particle through the magnetic field. The part of the force depending on this motion is expressed by the first two terms on the right of each equation. It depends on the velocity of the particle transverse to the lines of magnetic induction. If ${\mathfrak {G}}$ is a vector representing the velocity, and ${\mathfrak {B}}$ another representing the magnetic induction, then if ${\mathfrak {F}}_{1}$ is the part of the electromotive force depending on the motion,

${\mathfrak {F}}_{1}=V\cdot {\mathfrak {G\,B}},$

(7)

or, the electromotive force is the vector part of the product of the magnetic induction, multiplied by the velocity, that is to say, the magnitude of the electromotive force is represented by the area of the parallelogram, whose sides represent the velocity and the magnetic induction, and its direction is the normal to this parallelogram, drawn so that the velocity, the magnetic induction, and the electromotive force are in right-handed cyclical order.

The third term in each of the equations (B) depends on the time-variation of the magnetic field. This may be due either to the time-variation of the electric current in the primary circuit, or to motion of the primary circuit. Let ${\mathfrak {E}}_{2}$ be the part of the electro motive force which depends on these terms. Its components are

$-{\frac {dF}{dt}},\quad -{\frac {dG}{dt}},{\text{ and }}-{\frac {dH}{dt}},$

and these are the components of the vector, $-{\frac {d{\mathfrak {A}}}{dt}}$ or $-{\dot {\mathfrak {A}}}$.^[2] Hence,

${\mathfrak {E}}_{2}=-{\dot {\mathfrak {A}}},$

(8)

The last term of each equation (B) is due to the variation of the function $\Psi$ in different parts of the field. We may write the third part of the electromotive force, which is due to this cause,

${\mathfrak {E}}_{3}=-\Delta \Psi ,\,$

(9)

The electromotive force, as defined by equations (B), may therefore be written in the quaternion form,

${\mathfrak {E}}=V\cdot {\mathfrak {GB}}-{\dot {\mathfrak {A}}}-\Delta \Psi .$

(10)

On the Modification of the Equations of Electromotive Force when the Axes to which they are referred are moving in Space.

600.] Let x', y', z' be the coordinates of a point referred to a system of rectangular axes moving in space, and let x, y, z be the coordinates of the same point referred to fixed axes.

Let the components of the velocity of the origin of the moving system be u, v, w, and those of its angular velocity ω₁, ω₂, ω₃, referred to the fixed system of axes, and let us choose the fixed axes so as to coincide at the given instant with the moving ones, then the only quantities which will be different for the two systems of axes will be those differentiated with respect to the time. If ${\frac {\delta x}{\delta t}}$ denotes a component velocity of a point moving in rigid connexion with the moving axes, and ${\frac {dx}{dt}}$ and ${\frac {dx'}{dt'}}$ that of any moving point, having the same instantaneous position, referred to the fixed and the moving axes respectively, then

${\frac {dx}{dt}}={\frac {\delta x}{\delta t}}+{\frac {dx'}{dt}},$

(l)

with similar equations for the other components.

By the theory of the motion of a body of invariable form,

$\left.{\begin{aligned}{\frac {\delta x}{\delta t}}=u+\omega _{2}z-\omega _{3}y,\\{\frac {\delta y}{\delta t}}=v+\omega _{3}z-\omega _{1}z,\\{\frac {\delta z}{\delta t}}=w+\omega _{1}z-\omega _{2}x.\end{aligned}}\right\}$

(2)

Since F is a component of a directed quantity parallel to x, if ${\frac {dF'}{dt}}$ be be the value of ${\frac {dF}{dt}}$ referred to the moving axes,

${\frac {dF'}{dt}}={\frac {dF}{dx}}{\frac {\delta x}{\delta t}}+{\frac {dF}{dy}}{\frac {\delta y}{\delta t}}+{\frac {dF}{dz}}{\frac {\delta z}{\delta t}}+G\omega _{3}-H\omega _{2}+{\frac {dF}{dt}}.$

(3)

Substituting for ${\frac {dF}{dy}}$ and ${\frac {dF}{dz}}$ their values as deduced from the equations (A) of magnetic induction, and remembering that, by (2),

${\frac {d}{dx}}{\frac {\delta x}{\delta t}}=0,\quad {\frac {d}{dx}}{\frac {\delta y}{\delta t}}=\omega _{3},\quad {\frac {d}{dx}}{\frac {\delta z}{\delta t}}=-\omega _{2},$

(4)

${\begin{aligned}{\frac {dF'}{dt}}={\frac {dF}{dx}}{\frac {\delta x}{\delta t}}+F{\frac {d}{dx}}{\frac {\delta x}{\delta t}}+{\frac {dG}{dx}}{\frac {\delta y}{\delta t}}+G{\frac {d}{dy}}{\frac {\delta y}{\delta t}}+{\frac {dH}{dx}}{\frac {\delta z}{\delta t}}+H{\frac {d}{dx}}{\frac {\delta z}{\delta t}}\\-c{\frac {\delta y}{\delta t}}+b{\frac {\delta z}{\delta t}}+{\frac {dF}{dt}}.\end{aligned}}$

(5)

If we now put

$-\Psi '=F{\frac {\delta x}{\delta t}}+G{\frac {\delta y}{\delta t}}+H{\frac {\delta z}{\delta t}},$

(6)

${\frac {dF'}{dt}}=-{\frac {d\Psi '}{dx}}-c{\frac {\delta y}{\delta t}}+b{\frac {\delta z}{\delta t}}+{\frac {dF}{dt}}.$

(7)

The equation for P, the component of the electromotive force parallel to x, is, by (B),

$P=c{\frac {\delta y}{\delta t}}-b{\frac {\delta z}{\delta t}}-{\frac {dF}{dt}}-{\frac {d\Psi }{dx}},$

(8)

referred to the fixed axes. Substituting the values of the quantities as referred to the moving axes, we have

$P'=c{\frac {\delta y'}{\delta t}}-b{\frac {\delta z'}{\delta t}}-{\frac {dF'}{dt}}-{\frac {d(\Psi +\Psi ')}{dx}},$

(8)

for the value of P referred to the moving* axes.

601.] It appears from this that the electromotive force is expressed by a formula of the same type, whether the motions of the conductors be referred to fixed axes or to axes moving in space, the only difference between the formulae being that in the case of moving axes the electric potential Ψ must be changed into Ψ + Ψ'.

In all cases in which a current is produced in a conducting circuit, the electromotive force is the line-integral

$E=\int {\left(P{\frac {dx}{ds}}+Q{\frac {dy}{ds}}+R{\frac {dz}{ds}}\right)ds}.$

(10)

taken round the curve. The value of Ψ disappears from this integral, so that the introduction of Ψ' has no influence on its value. In all phenomena, therefore, relating to closed circuits and the currents in them, it is indifferent whether the axes to which we refer the system be at rest or in motion. See Art. 668.

On the Electromagnetic Force acting on a Conductor which carries an Electric Current through a Magnetic Field.

602.] We have seen in the general investigation, Art. 583, that if x₁ is one of the variables which determine the position and form of the secondary circuit, and if X₁ is the force acting on the secondary circuit tending to increase this variable, then

$X_{1}={\frac {dM}{dx_{1}}}i_{1}i_{2}.$

(1)

Since $i_{1}$ is independent of $x_{1}$, we may write

$Mi_{1}=p=\int {\left(F{\frac {dx}{ds}}+G{\frac {dy}{ds}}+H{\frac {dz}{ds}}\right)ds},$

(2)

and we have for the value of $X_{1}$,

$X_{1}=i_{2}{\frac {d}{dx_{1}}}\int {\left(F{\frac {dx}{ds}}+G{\frac {dy}{ds}}+H{\frac {dz}{ds}}\right)ds}.$

(3)

Now let us suppose that the displacement consists in moving every point of the circuit through a distance δx in the direction of x, δx being any continuous function of s, so that the different parts of the circuit move independently of each other, while the circuit remains continuous and closed.

Also let X be the total force in the direction of x acting on the part of the circuit from s = 0 to s = s, then the part corresponding to the element ds will be ${\frac {dX}{ds}}ds$. We shall then have the following expression for the work done by the force during the displacement,

$\int {{\frac {dX}{ds}}\delta x\,ds}=i_{2}\int {{\frac {d}{d\delta x}}\left(F{\frac {dx}{ds}}+G{\frac {dy}{ds}}+H{\frac {dz}{ds}}\right)\delta x\,ds}$

(4)

where the integration is to be extended round the closed curve, remembering that δx is an arbitrary function of s. We may there fore perform the differentiation with respect to δx in the same way that we differentiated with respect to t in Art. 598, remembering that

${\frac {dx}{d\delta x}}=1,\quad {\frac {dy}{d\delta x}}=0\quad {\text{and }}{\frac {dz}{d\delta x}}=0.$

(5)

We thus find

$\int {{\frac {dX}{ds}}\delta x\,ds}=i_{2}\int {\left(c{\frac {dy}{ds}}-b{\frac {dz}{ds}}\right)\delta x\,ds}+\int {{\frac {d}{ds}}(F\delta x)ds}.$

(6)

The last term vanishes when the integration is extended round the closed curve, and since the equation must hold for all forms of the function δx, we must have

${\frac {dX}{ds}}=i_{2}\left(c{\frac {dy}{ds}}-b{\frac {dz}{ds}}\right)$

(7)

an equation which gives the force parallel to x on any element of the circuit. The forces parallel to y and z are

${\frac {dY}{ds}}=i_{2}\left(a{\frac {dz}{ds}}-c{\frac {dx}{ds}}\right)$

(8)

${\frac {dZ}{ds}}=i_{2}\left(b{\frac {dx}{ds}}-a{\frac {dy}{ds}}\right)$

(9)

The resultant force on the element is given in direction and magnitude by the quaternion expression $i_{2}Vd\rho {\mathfrak {B}}$, where i₂ is the numerical measure of the current, and dρ and ${\mathfrak {B}}$ are vectors representing the element of the circuit and the magnetic induction, and the multiplication is to be understood in the Hamiltonian sense.

603.] If the conductor is to be treated not as a line but as a body, we must express the force on the element of length, and the current through the complete section, in terms of symbols denoting the force per unit of volume, and the current per unit of area.

Let X, Y, Z now represent the components of the force referred to unit of volume, and u, v, w those of the current referred to unit of area. Then, if S represents the section of the conductor, which we shall suppose small, the volume of the element ds will be S ds, and $u={\frac {i_{2}}{S}}{\frac {dx}{ds}}$. Hence, equation (7) will become

${\frac {XSds}{ds}}=S(vc-wb),$

(10)

or

$\left.{\begin{aligned}X&=vc-wb,\\Y&=wa-uc,\\Z&=ub-va.\end{aligned}}\right\}{\begin{matrix}{\text{(Equations of}}\\{\text{Electromagnetic}}\\{\text{Force.)}}\end{matrix}}$

(C)

Here X, Y, Z are the components of the electromagnetic force on an element of a conductor divided by the volume of that element; u, v, w are the components of the electric current through the element referred to unit of area, and a, b, c are the components of the magnetic induction at the element, which are also referred to unit of area.

If the vector ${\mathfrak {F}}$ represents in magnitude and direction the force acting on unit of volume of the conductor, and if ${\mathfrak {C}}$ represents the electric current flowing through it,

${\mathfrak {F}}=V.{\mathfrak {C\,B}}.$

(11)

1. Exp. Res., 3082, 3087, 3113.
2. Implementing Errata from Page 1 of this volume.