Page:Electromagnetic effects of a moving charge.djvu/4

 There are several ways of arriving at the conclusion that a moving charge must be regarded as an electric current; but, when that is admitted, we are very far from knowing what its magnetic effect is. No cut-and-dried statement of it can be made, because it varies according to circumstances. The magnetic field, whatever it be in a given case, is not that of a current-element (supposing the charge to be at a point), for that is anti-Maxwellian, but is that of the actual system of electric current, which is variable.

Thus, in the case of motion at a speed which is a small fraction of that of light, the magnetic field (as found by Prof. J. J. Thomson) is the same as that of Ampère's current-element represented by ρu; that is, a current-element whose direction is that of u and whose moment is ρu, if u is the tensor of u (understanding by "moment," current-density &times; volume); but the true current to correspond bears the same relation to the current-element as the induction of an elementary magnet bears to its magnetic moment. The magnetic energy due to the motion of a charge q upon a sphere of radius a in a medium of inductivity μ, at a speed u which is only a very small fraction of that of light, is expressed by $$\scriptstyle{\frac{1}{3}\mu q^{2}u^{2}/a}$$. But if the speed be not a small fraction of that of light, the result is very different. Increasing the speed of the charge causes not merely greater magnetic force but changes its distribution altogether, and with it that of the electric field. It is no use discussing the potential. There is not one. The magnetic field tends to concentrate itself towards the equatorial plane, or plane through the charge perpendicular to the line of motion. When the speed equals that of light itself this process is complete, and the is simply a plane wave (electromagnetic).

Since a charge at a point gives infinite values, it is more convenient to distribute it. Let it be, first, of linear density q along a straight line AB, moving in its own line at the speed of light. Then the field is contained between the parallel planes through A and B perpendicular to AB, and is completely given by

$\scriptstyle{E/\mu v=H=2qu/r}$,

where E and H are the intensities of the electric and magnetic forces at distance r from AB. The lines of E radiate uniformly from AB in all directions parallel to the planes; those of H are everywhere perpendicular to those of E, or are circles centred upon AB. Outside this electromagnetic wave there is no disturbance. I should remark that the above is a description of the exact solution. It is, of course, nothing like the supposed field of a current-element AB.

To still further realize, we may substitute a cylindrical distribution for the linear, and then, again, terminate the lines of E on another cylindrical surface between the bounding planes. To find the resulting distributions of E and H (always perpendicular) may be done by