ELECTROKINETICS Electrokinetics is that part of electrical science which deals with the properties of electric currents. The chief effects pro duced by electric currents are magnetic effects, heating effects and chemical effects. The chemical effects are discussed in the article ELECTROLYSIS.
Electric currents in conductors may be classified as uni directional, or direct, i.e., flowing always in the same direction; or as alternating, i.e., reversing their direction at regular inter vals. We shall deal first with the former type.
It is found that the magnetic moment of a coil is proportional to the cross section of the coil, to the number of turns of wire in it and to the strength of the current. If, then, M denotes the mag netic moment of the coil, N the number of turns of wire in it, A the cross section of the coil in square centimetres and C the current, we have M = kNA C, where k is a constant. The electro magnetic unit of current is defined so as to make the constant k equal to unity when the coil is in a vacuum, so that M = NA C when C is expressed in electromagnetic units of current. Thus the electromagnetic unit of current is defined to be a current such that a small plane circuit, in a vacuum, round which it is flowing has unit magnetic moment per square centimetre of the area of the circuit. The magnetic field due to the circuit in air is practically the same as in a vacuum. Thus a coil of N turns, wound uniformly on a straight cylinder, will have a magnetic moment NAC, if A is the mean area of the turns of wire.
If the length of the coil is 1 cm. then its pole strength is NAC/l. The magnetic potential (see MAGNETISM) at any point is defined to be the work in ergs required to bring a unit north pole to the point from a great distance. The potential due to a pole m at a distance r cm. from it in vacuo is equal to m/r, just as the electro static potential due to a charge E is equal to E/r. The potential due to a small magnet of moment M at any point may be calculated easily. In fig.
let AOB be the small magnet, and let its north pole be at A and its south pole at B. Let 0A = OB =1. Consider the mag netic potential at P, and let OP = r, and let the angle POA between OP and the axis of the magnet be denoted by 0. If the pole at A is of strength m, and that at B of strength — m, then the potential at P is equal to m/PA — ni/PB. If 1 is small compared with r, this is m/(r—lcos6) —m/(r+lcos9), or But 21n is the moment of the magnet, so that, denoting this by M, we have for the magnetic potential at P. The magnetic potential due to a small plane coil of wire having area A square centimetres and one turn is therefore at a point at a distance r from the centre of the coil in a direction making an angle 0 with the axis of the coil. The axis of the coil is a line drawn through its centre perpendicular to its plane. It is drawn in the direction in which a cork screw, turning with the current, would advance, or in the direction from the south to the north pole of the equivalent magnet. The solid angle of a cone may be defined to be numerically equal to the area which the cone cuts off on the surface of a sphere of unit radius with its centre at the vertex of the cone. The total solid angle round a point is thus equal to 47r. The area which a cone of solid angle a cuts off on a sphere of radius r is therefore equal to are, for the area cut off varies as the square of the distance from the vertex.
The solid angle which a small plane circuit, of area A, tends at a point P at a distance r from it, in a direction making an angle 0 with the axis of the circuit, is therefore equal to A For A cos+ is the area of the projection of A on to the surface of a sphere with its centre at P and radius r. Denoting this solid angle by a, we see that the netic potential at P due to the circuit is equal to Ca. We can now show that the magnetic potential, at a point, due to a circuit of any size and shape is equal to the current flowing round the circuit plied by the solid angle which the circuit subtends at the point. In fig. i S, let ABC be a circuit of any size and shape, with a r current C flowing round it, as indicated by , the arrowheads. Divide the area of the cuit up into a large number of small areas, as shown, and suppose that the current C flows round all these small areas in the same direction. This quires equal and opposite currents C along all the dividing lines as indicated by the arrowheads. But two equal and opposite currents along a line are equivalent to zero current, and so produce no netic field. The field due to the circuit ABC is therefore the same as that due to all the small circuits into which it may be divided. The magnetic potential at a point due to each of the small cuits is equal to Ca, where a is the solid angle which the small circuit subtends at the point, so that the potential due to the whole circuit is equal to the current multiplied by the sum of all the small solid angles, which sum is equal to the solid angle sub tended by the whole circuit. The magnetic potential at any point, due to any circuit carrying a current C in a vacuum, is therefore equal to Ca, where a denotes the solid angle subtended by the circuit at the point. The potential in air is practically the same as in a vacuum. If the potential at a point A is PA, and PB at another point B near to A, then, if the component of the mag netic field along AB is denoted by F, we have F• A B = — PA, or if PA - PB = 6P and AB= 8x, then F=-- -- 6P/ 6x. Thus the field strength in any direction is equal to the rate of decrease of the potential in that direction.
Now consider the magnetic field along the axis of a circular circuit, of radius r, carrying a current C. At a point on the axis at a distance x from the centre of the circuit, the solid angle sub tended by the circuit is equal to that of a circular cone of semi angle 0 such that cos9 = x/ . This solid angle is equal to 27r(r — cos8), so that the magnetic potential at the point is given by P= 27rC[I The field F along the axis 2 is therefore given by F= — P = 7rC2 • At the centre of ax the circuit x = o, so that F= 27rC/r.
Now let us investigate the work in ergs required to take a unit pole round a current C. In fig. i6, let A and B be the points at which the circuit carrying the current C cuts the plane of the paper. Let the closed curve be the path along which the unit pole is moved round the current and let its successive positions be the points numbered i, 2, 3, 4,. • • • At each point the solid angle subtended by the circuit is indicated so as to show how it varies as the pole goes round the curve. At i and 2 the solid angle is less than 2 7r. At 3 and 4 it is between 2 7r and 47r. At 5 and 6 it is between 47r and 67r. At 7 and 8 it is between 67r and 87r. At 9 and io it is between 87r and I 7r. Thus we see that, as the pole goes round and round the closed curve threading through the circuit, the solid angle in creases by 47r during each com plete circuit of the curve. The work in ergs required to take the unit pole once round the closed curve is therefore equal to 47rC.
This result enables the magnetic field to be calculated easily in several important cases.
In the case of the field due to a long straight wire, we see by symmetry that the lines of force are circles in planes ular to the wire. If F denotes the strength of the field at a distance r from the wire, the work required to take a unit pole round the circle of radius r is 27rrF, so that we have 27rrF = 47rC, or F= 2 C,/r, where C is the current. The field inside a long straight wire with circular section of radius a, at a distance r from the axis is given by so that F= Now consider the field inside a uniformly wound, ring-shaped coil, having N turns of wire. In fig. 17, let the axis of the coil be at 0, and describe a circle of radius r with its centre on the axis and in a plane perpendicular to the axis, so that the circle is inside the coil as shown. By symmetry the lines of force are circles round the axis. Let F be the field strength on the circle of radius r, inside the coil, so that we have 27rrF = 47rNC, where C is the current, for the total current through the plane of the circle is NC. Hence F= 2NC/r. This result is true for a shaped coil of any cross section, provided it is uniformly wound.
The number of turns of wire per centimetre on the coil is N/2irr. Denoting this by n we have F = 2C2lrrn/r = 47rCn. If the length of the ring-shaped coil is very large, any short portion of it will be approximately straight, so that we see that the field inside a uniformly wound straight coil or solenoid, at any point not near the ends of the coil, is equal to 47rCn, where C is the current and n the number of turns per centimetre. The lines of force due to a straight uniform coil are shown in fig. 18.
They pass through the coil, come out at or near one end, and go in at the other, f orm ing closed curves. The field inside the coil is nearly uniform except near the ends.
Outside the coil the field is like that due to a bar magnet with poles of strength nCA, where A is the area of the cross section of the coil in square centimetres.
An indefinitely long coil gives a field in side equal to 47rCn. If we consider a length I of such a coil, we may regard the field in it as made up of the field due to the length t and that due to the rest of the coil.
The field due to the rest will be approxi mately that due to a pole of strength nCA at one end of the length 1, and one of strength —nCA at the other end. Thus, if F denotes the field inside a coil of length 1, at a distance x from one end, then we have This expression is approximately correct, provided and are small fractions. The foregoing equations are all for the magnetic field in vacuo due to circuits carrying a current C expressed in electromagnetic units of current. The fields in air, or in any nearly non-magnetic medium, are nearly the same as in vacuo. For magnetic fields in magnetic media see MAG