# Page:A Dynamical Theory of the Electromagnetic Field.pdf/25

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PROFESSOR CLERK MAXWELL ON THE ELECTROMAGNETIC FIELD.

Electromotive Force in a Circuit.

(63) Let $\xi$ be the electromotive force acting round the circuit A, then

 $\xi =\int \left(P{\frac {dx}{ds}}+Q{\frac {dy}{ds}}+R{\frac {dz}{ds}}\right)ds$ (32)

where $ds$ is the element of length, and the integration is performed round the circuit.

Let the forces in the field be those due to the circuits A and B, then the electromagnetic momentum of A is

 $\int \left(F{\frac {dx}{ds}}+G{\frac {dy}{ds}}+H{\frac {dz}{ds}}\right)ds=Lu+Mv$ (33)

where $u$ and $v$ are the currents in A and B, and

 $\xi =-{\frac {d}{dt}}(Lu+Mv)$ (34)

Hence, if there is no motion of the circuit A,

 $\left.{\begin{array}{l}P=-{\frac {dF}{dt}}-{\frac {d\Psi }{dx}},\\\\Q=-{\frac {dG}{dt}}-{\frac {d\Psi }{dy}},\\\\R=-{\frac {dH}{dt}}-{\frac {d\Psi }{dz}}.\end{array}}\right\}$ (35)

where $\Psi$ is a function of $x,y,z$ , and $t$ , which is indeterminate as far as regards the solution of the above equations, because the terms depending on it will disappear on integrating round the circuit. The quantity $\Psi$ can always, however, be determined in any particular case when we know the actual conditions of the question. The physical interpretation of $\Psi$ is, that it represents the electric potential at each point of space.

Electromotive Force on a Moving Conductor.

(64) Let a short straight conductor of length a, parallel to the axis of $x$ , move with a velocity whose components are ${\tfrac {dx}{dt}},{\tfrac {dy}{dt}},{\tfrac {dz}{dt}}$ , and let its extremities slide along two parallel conductors with a velocity ${\tfrac {ds}{dt}}$ . Let us find the alteration of the electromagnetic momentum of the circuit of which this arrangement forms a part.

In unit of time the moving conductor has travelled distances ${\tfrac {dx}{dt}},{\tfrac {dy}{dt}},{\tfrac {dz}{dt}}$ along the directions of the three axes, and at the same time the lengths of the parallel conductors included in the circuit have each been increased by ${\tfrac {ds}{dt}}$ .

Hence the quantity

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