TRANSACTIONS OF SECTION A. dao 
stratum. Such a case is represented in Fig. 3, which is drawn to scale for alternate 
currents of period ; of a second in round rods of copper of six centimetres 
diameter. The spaces between the outer circular boundaries and the inner fine 
circles indicate what I have called the ohmic thickness,! being ‘714 of a centimetre 
for copper of resistivity 1611 square centimetres per second. The full solution for 
such a case as that represented in Fig. 3 belongs to the large class of cases inter- 
mediate between I. and II., and could only be arrived at by a kind of transcendent 
mathematics not hitherto worked. But, without working it out, it is easy to see 
how the time-maximum intensity of the current will diminish inwards from the 
surface, and will be, at any point of either of the inner fine circles, about one-half 
or one-third of what it is at the nearest point of the boundary surface ; and that at 
points in the surface, distant from B B’ by one-half or one or two times the ohmic 
thickness, the current intensity will be much smaller than it isat B and B’. 
TII. In Case I. the heat generated per unit of time, per unit of volume, in 
different parts of the conductors, is inversely as the electric resistivity of the sub- 
stance, and directly as the square of the total strength of current at any instant. 
In Case II. the time-average of the heat generated per unit of time, per unit of 
area of the current stratum, is as the time-average of the square of the quantity 
of current per unit breadth, multiplied by the square root of the product of the 
electric resistivity into the magnetic permeability. 
IV. Example of III.: Let the conductor A be a thin flat bar, as shown in the 
diagram (Fig. 4), A’ being a tube surrounding A, or another flat bar like A, or a 
conductor of any form whatever, provided only that 
its shortest distance from A is a considerable Fie, 4. 
multiple of the breadth of A. The thickness of 
A must be sufficiently great to satisfy the condition 
of II., and its breadth must be a large multiple of 
its thickness. (For copper carrying alternating 
currents of frequency 80 periods per second, these 
- conditions will be practically fulfilled by a flat bar 
of 4 ems. thickness and 30 or 40 cms. breadth.) bee 
The current in it is chiefly confined to two strata, 
_ extending to small distances inwards from its two 
sides. (For copper and frequency 80 periods per second, the time-maximum of in- 
tensity of the current at the surface will be about e*, or 7-4 times what it is at a 
distance 1:43 cm. in from the surface.) The quantity of current per unit breadth, 
or, a8 we may call it for brevity, the surface-density of the current in each stratum, 
is determined by the well-known solution of the problem of finding the surface- 
electric-density of an electrified ellipsoid of conductive material undisturbed by any 
other electrified body. The case we have to consider is that of an ellipsoid whose 
longest diameter is infinite, medium diameter the breadth of our flat conductor, 
and least diameter infinitely small. In this case the electric density varies in- 
versely as 4/(OB?—OP*). The graphic construction in the drawing shows 
PQ=./(OB?—OP?), and we conclude that the time-maximum of the surface- 
density of the current varies inversely as PQ. The infinity, which in the electric 
problem we find for electric density of the ideal conductor, is obviated for the 
electric current problem by the proper consideration of the rectangular corners or 
the rounded edge (as the case may be) of our copper bar, which, though exceed- 
ingly interesting, is not included in the present communication. Suffice it to say 
that there will be no infinities, even if the corners be true mathematical angles. 
Y. Example of Cases I. and II.: Let A consist of three circular wires, C, L, 
and I, of copper, lead, and iron respectively. In Case I. the quantities of the 
whole current they will carry, and the quantities of heat generated per unit of 
time in them, will be inversely as their resistivities. In Case II., if the centres of 
the three circular cross-sections form an equilateral triangle, the quantities of heat 
generated in them will be directly as the square roots of the resistivities for 0 and L; 
and for I would be as the square root of the product of the resistivity into the 
magnetic permeability, if the magnetic permeability were constant and the viscous 
1 Collected Papers, vol. iii. Art. cii. section 35, 
