Pqr

PQR represent a unit tube of force of the magnetic field of circuit. Along this tube µFa = i, where F is the field strength the tube, a the cross section of the tube and the permeability of the medium. If s is the length of a short ment of the tube, then the work required to take a unit once round the tube is equal to the sum of the products one for each element of the tube. This work is equal 47rI, so that, denoting the sum of all products Fs by 1Fs, we have EFs = The number N of unit tubes pass through the circuit is equal to SI, that, if we take the sum of the Fs for all the unit tubes, we get /Fs where /Fs now denotes the of the products Fs for all the of all the unit tubes in the field. energy W = is therefore given by W Since µFa= i at any section of any unit tube, may multiply each of the products Fs by µFa, F and a the values at the element in question and so obtain W = But sa is the volume of the element of the unit tube, so that it appears that the energy W can be regarded as distributed through out the magnetic field with energy density This result may be compared with the analogous result in electrostatics, that the electrostatic energy may be regarded as distributed throughout the electric field with energy density where K is the specific inductive capacity and E the electric field strength.

Induced Electromotive Force in Conductors Moving in

Magnetic Field.—If the number N of unit tubes of magnetic force passing through a circuit is altered by moving the circuit, or part of it, an induced electromotive force equal to — dN/dt is produced, just as when the circuit is kept fixed and the mag netic field altered. Suppose part of a circuit consists of a straight wire, and that it is in a magnetic field of strength F perpendicular to the wire. If the wire is moved in a direction perpendicular to itself and to the field with a velocity v, the number of unit tubes it crosses per second is v,uF per centimetre length of the wire. There will therefore be an electromotive force per centimetre in the wire equal to vµF. The existence of this electromotive force in the moving wire may also be deduced from the force on a cur rent in a magnetic field. We may suppose that the current con sists of electricity moving along the wire. If E is the amount of electricity per centimetre which moves in the wire, and v its velocity, the current I is given by I = Ev. The force on the wire per centimetre due to a magnetic field F at right angles to the wire is ,uFI =µFEv. The force on unit charge in the wire is therefore ,uFv. If then the wire is made to move across the magnetic field with velocity v, there will be a force on unit charge in it equal to µFv along the length of the wire. The induced electromotive force per centimetre in the wire is therefore equal to µFv.

Absolute Determination of Resistance and Potential Difference.—The resistance of a conductor in electromagnetic units may be found by means of a method due to L. Lorenz (1873). A circular metal disc is mounted on a shaft, so that it can be made to rotate about an axis through its centre and perpendicular to its plane. A magnetic field, perpendicular to the plane of the disc and symmetrical about the axis of rotation, is produced by means of a circular coil of wire, concentric with the disc, through which a current is passed. The current is also passed through the resistance to be determined, and the ends of the resistance are connected to two brushes, one making contact with the circumference of the disc and the other with the shaft close to the disc. If R is the resistance and I the current, both in electromagnetic units, then the potential difference between the ends of the resistance is equal to IR. When the disc is rotating in the magnetic field there is an induced electromotive force in it which tends to produce a potential difference between the two brushes. If the induced electromotive force is equal to the po tential difference IR between the ends of the resistance, there will be no current in the wires connecting the resistance to the brushes. The speed of rotation of the disc is adjusted until the deflection of a galvanometer, in series with the resistance and one of the brushes, is zero.

Let

denote the strength of the magnetic field at a distance r from the centre of the disc, a the radius of the shaft and b that of the disc. The induced electromotive force between distances r and r+dr from the centre of the disc is equal to where w is the angular velocity of the disc, for cor is the velocity of the disc at the radial distance r. The potential difference between the brushes, when no current is flowing through them is therefore b • w a f rdr, and this is equal to IR. Now is proportional to I, so that if is the value of when I = r, then Hr =III',.. Also, if n is the number of revolutions per second made by the The number N' of unit tubes of magnetic force passing through the disc due to unit current in the coil is equal to J H'r2irrdr, a so that we have R = nN'. N' can be calculated from the di mensions of the coil and disc, so that R can be found in this way. This method is only suitable for the determination of very small resistances. For example, suppose b= 20 cm., a= i cm., and that the average value of is 200. Then, if n is ioo revolutions per second, we have R= z oo X Zoo X r(400-1), or R= 2.51 X i o' elec tromagnetic units of resistance, or o-0251 ohm, since one ohm is equal to electromagnetic units of resistance. When a resistance has been found accurately in this way, other resistances can be found by comparing them with it, and so standards of resistance can be made.

The value of a potential difference, in electromagnetic units of potential difference, can be found by comparing it with the poten tial difference between the ends of a wire of known resistance, through which a known current is passing. The current can be measured in electromagnetic units by means of an electro dynamometer. In this way it has been found that the potential difference due to one normal Weston cell is equal to 1•0183 volts at For practical purposes, the ampere is often defined as a current which deposits I • i i 8o milligrams of silver in one second, This definition is based on experimental measurements of the amount of silver deposited by currents measured with an electro dynamometer.

Mutual Induction.

Consider the case of two circuits A and B near together. If some of the unit tubes of magnetic force due to the circuit A pass through the circuit B, when the current in A is changing there will be an induced electromotive force in B. For example, suppose a battery and key included in the circuit A; then, on closing the key, the current in A rises rapidly from zero to a constant value. While the current in A is rising, there is an induced current in B, in the opposite direction to that in A. This current in B stops when the current in A becomes constant. If the key in A is now opened, the current in A very suddenly stops, and there is a momentary current in B in the same di rection as that in A.

The mutual induction between two circuits A and B is defined to be the number of unit tubes of magnetic force which pass through A due to a unit current in B. It can be shown that the number of unit tubes which pass through A, due to unit current in B, is equal to the number which pass through B due to unit current in A. To show this, consider the work required to start the currents, and let the current in A be and that in B be 12. The work done against the electromotive force of self-induction, when a current is started in A with the circuit B open, so that o, is equal to where is the self-induction of A. If we suppose that a current 12 is started in B, while the current in A is kept constant, the work required will be together with the work required to keep the current in A constant, If M is the number of unit tubes through A due to unit current in B, this work since is supposed kept diconstant. Thus the work required to start in A, and then 12 in B is If now we suppose the current 12 started with = o, and then started with 12 constant, we get, for the work to start the currents, where M' is the number of unit tubes through B due to unit current in A. The two results for the work must be equal so that M = W. The work required to start the currents is supposed to be the energy of the magnetic field as in the case of a single circuit. If the electromotive force applied to the circuit A is denoted by E, we have and /2R2= — d12 - M d12, dt dt where and R2 are the resistances of the two circuits.

Suppose that, at t=o, the currents and 12 are both zero, and that a constant electromotive force E is applied to the circuit A. At the start and 12 are zero, so that, when t= o, di dl dt di di and therefore d11 — when t=o. Thus, the pres dt ence of the second circuit causes the current in A to increase more rapidly at the start, as though the self-induction of A were reduced from to If S1S2, the initial value of is infinite, as if S were zero. The initial value of is — — AP). The current in the primary circuit A will become equal to after a time, and then the current in the other or secondary circuit will be zero. The quantity of electricity which flows past any point in the primary circuit in a long or = Et and in the same way R2Q2= — where Q2 is the quantity of electricity, flowing past any point in the secondary circuit, and I'1= is the final value of so that — If, when the primary current has attained the constant value it is then rapidly reduced to zero, by breaking the circuit or otherwise, we have 0 f 0 or Q2 = Thus, when the primary current is stopped, the flow in the secondary is equal but opposite to that when the primary current is started.

If the number of unit tubes of magnetic force

N passing through a circuit, of resistance R and self-induction S due to magnets or currents in other circuits, is changed from N, to N = and that when N = N2. If N is constant and equal to N, for some time after and to N2 for some time before then I will be zero at both and 12, so that RQ = N2, where Q is the amount of electricity which flows past any point in the circuit when N is changed from to N2. The previous result Q2 = — agrees with the equation Q = for, when the primary current changes from o to the number of unit tubes passing through the secondary changes from o to If alternating potential differences and are applied to the primary secondary circuits respectively, we have — Si —dt — M dt I2R2=E2cospt—S2 - M dt Let us suppose that the resistances and R2 are small, and the frequency p/27r large, so that and can be neglected. In this case we get If = the secondary current is zero, so that an electro motive force E1 Sl cospt in the secondary circuit is just suffi cient to stop the induced current in the secondary due to the primary circuit. In the same way if E23/1 / S2, the primary current is zero. For both primary and secondary currents to be zero, we must have E2 = E1 M/S1 and E2 M/S2, which is only sinpt possible when = If E2 = o we have Sl — M2/S2 p and /2= sinpt • Thus the primary current is the same S2 S1 p as that due an electromotive force acting on a single circuit of small resistance and self-induction — The secondary circuit therefore diminishes the apparent inductance of the primary circuit.

Suppose that the primary circuit is a coil of turns of wire, and the secondary circuit a coil, of turns, which practically coincides with the primary coil, so that all the unit tubes of force go through both coils. Let be the number of unit tubes going through the two coils due to unit current flowing round them. Then, if a current is passed through the primary coil, we have and in the same way 12 12 S2, and also = or 12 = so that Si = S2 = N1 and M = Hence = S1S2. Also = Thus the electromotive force in the secondary coil, necessary to stop the secondary current, is equal to and the electro motive force in the primary coil, necessary to stop the primary current, is so that, if = both currents will be zero, approximately, when all the unit tubes pass through both coils, so that = and the resistances and R2 are small.