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velocity of the motor becomes very great the current becomes very small. In the limit no level surface cuts the circuit, all converging on the motor. That is, all the energy passes into the motor when it is transformed into work, and the efficiency of the arrangement is perfect, though the rate of doing work is infinitely slow.

(6.) Induced currents.

It is not so easy to form a mental picture of the movement of energy which takes place when the field is changing and induced currents are created. But we can see in a general way how these currents are accounted for. When there is a steady current in a field there is corresponding to it a definite distribution of energy. If there is a secondary circuit present, so long as the primary current is constant, there is no E.M.I. in the secondary circuit for it is all at the same potential. The energy neither moves into nor out of it, but streams round it somewhat as a current of liquid would stream round a solid obstacle. But if the primary current changes there is a redistribution of the energy in the field. While this takes place there will be a temporary E.M.I. set up in the conducting matter of the secondary circuit, energy will move through it, and some of the energy will there be transformed into heat or work, that is, a current will be induced in the secondary circuit.

(7.) The electromagnetic theory of light.

The velocity of plane waves of polarised light on the electromagnetic theory may be deduced from the consideration of the flow of energy. If the waves pass on unchanged in form with uniform velocity the energy in any part of the system due to the disturbance also passes on unchanged in amount with the same velocity. If this velocity be v, then the energy contained in unit volume of cubical form with one face in a wave front will all pass out through that face in $1/v$th of a second. Let us suppose that the direction of propagation is straightforward, while the displacements are up and down; then the magnetic intensity will be right and left. If ${\mathfrak {E}}$ be the E.M.I. and ${\mathfrak {H}}$ the M.I. within the volume, supposed so small that the intensities may be taken as uniform through the cube, then the energy within it is $K{\mathfrak {E}}^{2}/8\pi +\mu {\mathfrak {H}}^{2}/8\pi$. The rate at which energy crosses the face in the wave front is ${\mathfrak {EH}}/4\pi$ per second, while it takes $1/v$th of a second for the energy in the cube to pass out.

Then

${\frac {\mathfrak {EH}}{4\pi v}}={\frac {K{\mathfrak {E}}^{2}}{8\pi }}+{\frac {\mu {\mathfrak {H}}^{2}}{8\pi }}$

(1)

Now, if we take a face of the cube perpendicular to the direction of displacement, and therefore containing the M.I., the fine-integral of the M.I. round this face is equal to $4\pi \times$ current through the face. If we denote distance in the direction of propagation