Page:Electromagnetic effects of a moving charge.djvu/6

But I was not then aware that the case u<v admitted of being presented in a nearly equally-simple form. That such is the fact is rather surprising, for it is very exceptional to arrive at simple results, and these now in question are sufficiently free from complexity to take a place in text-books of electricity.

Let the axis of z be the line of motion of the charge q at speed u. Everything is symmetrical with respect to this axis. The lines of electric force are radial out from the charge. Those of magnetic force are circles about the axis. The two forces are perpendicular. Having thus settled the directions, it only remains to specify their intensities at any point P distant r from the charge, the line r making an angle θ with the axis. Let E be the intensity of the electric, and H of the magnetic force. Then, if c is the permittivity and μ the inductivity, such that μcv²=1, we have

$\scriptstyle {(u<v)}{\begin{cases}\scriptstyle {cE={\frac {{\frac {q}{r^{2}}}\left(1-{\frac {u^{2}}{v^{2}}}\right)}{\left(1-{\frac {u^{2}}{v^{2}}}\sin ^{2}\theta \right)^{\frac {3}{2}}}},\quad \cdots \cdots \cdots \cdots \cdots \cdot \cdot }&\scriptstyle {({\text{A}})}\\\scriptstyle {H=cEu\ \sin \theta .\quad \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots }&\scriptstyle {({\text{B}})}\end{cases}}$

That (A), (B) represent the complete solution may be proved by subjecting them to the proper tests. Premising that the whole system is in steady motion at speed u, we have to satisfy the two fundamental laws of electromagnetism:—

(1). (Faraday's law). The electromotive force of the field [or voltage] in any circuit equals the rate of decrease of the induction through the circuit (or the magnetic current × -4π).

(2). (Maxwell's law). The magnetomotive force of the field [or gaussage] in any circuit equals the electric current × 4π through the circuit.

Besides these, there is continuity of the displacement to be attended to. Thus:—

(3). (Maxwell). The displacement outward through any surface equals the enclosed charge.

Since (A) and (B) satisfy these tests, they are correct. And since no unrealities are involved, there is no room for misinterpretation.

When u/v is very small, we have, approximately,

$\scriptstyle {cE={\frac {q}{r^{2}}},\quad \quad H={\frac {qu}{r^{2}}}\sin \theta }$,

representing Prof. J. J. Thomson's solution—that is, the lines of displacement radiate uniformly from the charge, and the magnetic force is that of the corresponding displacement-currents together with the moving charge regarded as a current-element of moment qu. Instantaneous action through the medium is involved—that is, to make the solution quite correct.

That the lines of electric force should remain straight as the speed of the charge is increased is itself a rather remarkable result. Examining