Answer. Cos 0, + k cos 02 = constant, where k is the ratio of the poles' strengths; and 01, 0, have the same meanings as in Question 2.

4. What current must pass through a solenoid having ten turns per centimetre of length in order that the field strength in its interior may be 5 units?

Answer. 1.25 amperes.

5. Show that if two coils are so situated that all the lines of force due to a current passing through one coil interlinks the other, then the coefficient mutual induction of the two coils is the square root of the product of the coefficient of self-induction of the two coils.

6. What is the energy of charge of a condenser whose capacity is 5 microfarads when its potential is 1 volt? What is the charge on the condenser? Answers. 25 ergs; 5 x 10-7 coulombs.

7. What is the specific inductive capacity of the dielectric when the mutual force exerted by two quantities of electricity, each equal to 100 coulombs, situated 10 centimetres apart, is 0.5 dyne?

Answer. 2.

8. Show that lines of force always cut the equipotential surfaces at right angles.

9. Show that the self-induction of a coil is proportional to the square of its number of turns.

CHAPTER II.

Induced Electromotive Forces-Faraday's Law-Induced Currents-Lenz's Law -Self and Mutual Induction-Energy of a Magnetic Field due to Electric Currents-Currents in Inductive Circuits.

INDUCED ELECTROMOTIVE FORCES.

12. Induced E.M.F.S.-Faraday showed experimentally that if a conductor is moved in a magnetic field, or if there is by any means produced a relative motion between the conductor and the lines of magnetic force, so that the conductor cuts the lines of force, an electromotive force is induced in it, and also that the rate of cutting the lines of force is a measure of the E.M.F. induced. Thus, if dN lines of force are cut in an infinitely small time dt by a moving conducting wire, the E.M.F. e generated is given by

where k is some constant determined by the units employed. If, further, we take as the unit of E.M.F. that which is induced when one centimetre length of wire cuts one line of force per second, k = 1, and the above equation becomes

The negative sign is prefixed because the E.M.F. is always induced in such a direction as to oppose the cause which produces it.

Equation (2) simply means that the E.M.F. induced in the conductor is numerically equal, at each instant, to the rate of which it is cutting lines of magnetic force.

The C.G.S. unit of electromotive force upon which equation (2) is based is so small that for practical purposes it is convenient

to take as the unit of E.M.F. that which is induced in a conductor which cuts 108 C.G.S. lines of force per second. This practical unit is called a Volt.

The direction of the induced E.M.F. depends not only on the direction of motion of the conductor, but also on the direction of the magnetic field. The following is an easily remembered rule for finding the direction of the E.M.F. when the directions of motion and of the field are given.

Suppose that the directions of motion. and of the field are given by the lines OM, OF respectively (Fig. 1), then the induced E.M.F. is given in direction by the line OE drawn at right angles to both OM and OF in such a way that beginning at E and going round in a counter-clockwise direction, the order is EMF

M

FIG. 1.

INDUCED CURRENTS.

13. If the conductor which is moving in a magnetic field forms a closed circuit, the induced E.M.F. will produce an electric current, and, neglecting self-induction, the value of the current at any instant is given by

dN
dt

where r is the total resistance of the circuit, and N is the number of magnetic lines threading through the circuit at that instant. Here we suppose that the current i is due solely to the induced E.M.F.

E.M.F. of Self-induction.-If a closed circuit carries a current the number of lines of force of self-induction is Li d(Li)

(§ 6). If, by any means, i is varied, an E.M.F. equal to dt

is induced, or, if L is considered constant.

This is called the E.M.F. of self-induction, and is that E.M.F. which is induced in a conductor when the magnetic field due to the current flowing in it varies.

E.M.F. of Mutual Induction.-If two closed circuits A and B, carry respectively currents i and is, the number of lines of force linking A due to mutual induction is Mi2 (see § 7, Chap. I.), and the number linking B due to mutual induction is Mi. If the two currents vary there will be E.M.F.s due to mutual induction in A and B respectively equal to

or if the coefficient of mutual induction, M, is taken to be constant, these become respectively—

which mean that the induced E.M.F. of mutual induction in either coil is numerically equal to the rate at which that coil cuts the lines of magnetic force due to the current flowing in the other.

ENERGY OF A MAGNETIC FIELD DUE TO ELECTRIC CURRENTS.

14. CASE I. Field due to a single electric circuit, If a conducting wire carries an electric current, a magnetic field is produced. Suppose that at that time t after the circuit is made the value of the electric current is i and that the coefficient of self-induction of the circuit is L.

The magnitude of the E.M.F. which opposes the growth of the current is

Lai

dt'

by § 13, equation (3)

and the rate at which work is being done which is the product of the corresponding instantaneous values of the current and E.M.F. is

If I is the maximum value of the current (when the steady state is attained) the total work done against the counter E.M.F. is the sum of the products

where St is a small element of time corresponding to the value i

of the current. Thus the work, W, expended in creating the magnetic field is given by

Energy of the Magnetic Field.--This is the energy expended in driving the current against the counter E.M.F. of self-induction from the instant at which the circuit is made to the time when the current attains its maximum value, and it has its equivalent in the potential energy stored up in the magnetic field.

CASE II. Field due to currents in two mutually inductive circuits.

If I and I are the maxima values of the two currents in the two circuits respectively, the energy expended in driving the currents against their respective E.M.F.s of self-induction will (by Case I.) be

where L1 and L2 are their respective coefficients of self-induction. There will, however, be opposing E.M.F.s in each circuit due to mutual induction. If M be the coefficient of mutual induction of the two circuits, and i and is be the instantaneous values of their respective currents, the opposing E.M.F. in circuit 1 due to mutual induction is (§ 13)

The rate at which work is being done against mutual induction in the two circuits taken together is therefore—

and the whole energy expended in driving the currents against