LV. On the Self-Induction and on the Gravity-Potential of a Ring. By W. M. HICKS*. HE self-induction of a finite ring has been considered by Professor Minchin in the March number of the Philosophical Magazine on the assumption that it is the flux through a diaphragm composed of the aperture and half the surface of the ring. In reading this paper it occurred to me that the problem could more naturally be attacked by means of Toroidal Functions. In considering it from this point of view I was led to a conclusion which does not seem to have been heretofore noticed. The expression for the self-induction of a thin circular ring as given by Maxwell is TR(4 log 8R/r-8), and for a ring of finite size as given by Minchin is πR(4 log 8R/r−8) + terms involving the radius (r) of the transverse section. In the finite ring the current-density varies inversely as the distance from the straight† axis. In Maxwell's wire the distribution of current across the transverse section is considered uniform, and it would therefore appear that his formula gives the self-induction so far as it is independent of the thickness, and that this part is independent of the distribution of current-density within the wire itself. As a matter of fact, however, the distribution of current-density has a deciding effect on the value of the second term, and this however small the section of the ring may be. The correct value of the selfinduction, neglecting powers of r/R, is TR(4 log 8R/r-7). The expression TR(4 log 8R/r-8) gives the value of the flux through the aperture only, which to this degree of approximation is independent of the law of current-density. The difference is due to the fact that although the section across the ring may be exceedingly small, yet the forces are correspondingly large, and the interaction of the different parts of the current cannot be neglected even when the ring is extremely thin and of large aperture. E. g. in a ring of 1 metre radius and 1 millim. thick the usually accepted formula gives a value 34 per cent. too small. In the present paper the question is treated by the method usually employed in dealing with problems of electric potential and fluid motion, but which, so far as I know, is new in its application to electromagnetic and gravity potentials. The *Communicated by the Author. † By straight axis is meant the line through the centre of the ring perpendicular to its plane. The circular axis is the locus of the centres of the transverse sections of the ring. functions which naturally come in are Toroidal Functions, the properties of which are developed in two papers in the Transactions of the Royal Society *. These papers are referred to in the present pages as [T. F. i.] and [T. F. ii.]. The special problems considered are: (1) The force-flux function and self-induction for a ring in which the current-density varies inversely as the distance from the straight axis. This is the case of current through a ring consisting of a single turn of round wire. (2) The same quantities when the current-density is constant. This leads to the case of a uniformly wound coil of circular cross section. (3) The method is then exemplified by finding the gravitypotential of a finite ring-a problem which has been recently very fully considered by Mr. Dyson t. 1. The force is symmetrical round the straight axis of the ring. Take this axis for the axis of z and the origin at the centre of the ring, and let p, z denote the cylindrical coordinates of any point. Further, let R denote the radius of the circular axis, that of a cross section, and a that of the critical circle in the curvilinear coordinates used in toroidal functions—i. e. in the system of coaxal circles having Oz for common radical axis and having one circle of the system coinciding with the cross section. The length of a is equal to the length of a tangent line from the centre to the circle. It will be found convenient to express values later in terms of R and the angle subtended at the centre by a cross section: this angle will be denoted by a. The flux function at a point P will be taken as denoting the total flux up through a circle, radius P N (see figure). It is the same as that through a diaphragm stretching from O *Phil. Trans. 1881, Part iii. p. 609, and 1884, Part i. p. 161. Phil. Trans. 1893, A. pp. 43 & 1041. Phil. Mag. S. 5. Vol. 38. No. 234. Nov. 1894. 2 I and bounded by this circle. Let w, w denote the velocities at P parallel to NP and Oz. Then dy dp dy dz dp=flux up through an annulus whose radii are whence 2. Consider a current flowing through an elementary ring whose cross section is dp dz and in which the current-density is σ. Then the circulation taken round this At points where there is no current the right-hand side is of course zero. 3. To find the force-flux we shall require two functions, one () for space outside the ring, and the other (4) for space inside. The conditions which they have to fulfil are as follows: must be finite at all points outside the ring, be zero at the point O, and satisfy the equation must be finite at all points inside the ring, and must satisfy 1 ¥4=ƒ+ √(C—c) ΣTn(B2 cos nv + B, sin nv), where f is any particular solution of the equation (1). Further, at the boundary of the ring and the force are continuous. That is, when u=u, then for all values of v, dy dy y= and = (2) where u is the value of u corresponding to the boundary. These conditions will serve to find completely all cases. and in In order to apply these last conditions it will be necessary to expand ƒ in a series of cosines and sines of multiple angles. In the cases here considered it will be seen later that the sine terms do not enter, and that ƒ is of the form where dashed letters denote differentiation with respect to u, and the functions are now to be regarded as functions of up only. From these, An(R2T„— R„T")=T2F2-T,F2, B1 (RT-R,T) = RF-R2F2. Now R„T„ — R„T2 = (n2 —1) S2 (P„Qn-PnQn) [T. F. ii. 2.] =(n2-4)πS. [T. F. i. 24.] This is the general solution for any case in terms of F. 4. We proceed to apply the foregoing to special cases. Consider, first, the case of a current flowing in a uniform circular ring of any size. The parts of the current all flow in circles, and the equipotential surfaces will be planes through the straight axis. The lengths of conductor which a current uses between two such surfaces will be proportional to the arc between-i. e, to p. Hence, in order that each element of current may produce equal fall of potential between the two planes, the current-density must vary inversely as p, i. e. σ=b/p, where b is some constant. Let I denote the total current; then, integrating over a cross section, it will be found that I=2πb{R-√(R2 —p2)}. This has to be expanded in a series of the form Hence It is clear that there are no terms in sin nv. = = 0 π (C—c)* 472ba2 d cos nv — 1⁄2 (cos n +2v+cos n−2 v) S ✓ (C—c) dv { Qn− & (Qn+2+Qn-2)} [T. F.ii. 1]; |