Field of Circular Currents. 431 



capacity. Formula? for the capacity K have been found by 

 W. M. Hicks * and F. W. Dyson f. 



When the radius of the ring is large compared with the 

 radius of the circular section, they both agree in giving the 

 formula as 



K=7ra/{log(8a/r)}. 



Hence, if the resistivity of the ring be zero, we have 



KL = (27ra) 2 {l-2/log(8«/r)}. . . . (25) 



When 100r = «, this equation becomes 



KL = 0'7(27ra) 2 , approximately. 



It will be seen, therefore, that we can only use the equation 

 KL = (2-7ra) 2 when r/a is an exceedingly small fraction. 

 The corresponding equation KL = Z 2 , for two long parallel 

 cylinders of infinite conductivity is strictly true (see A. Russell, 

 •^Alternating Currents/ vol. i. p. 141). 



5. Applications to Hydrodynamics. 



Lord Kelvin has shown (' Papers on Electrostatics and 

 Magnetism,' p. 444) that the problem of findiug the velocity 

 .at any point of an incompressible fluid near a vortex filament 

 is, from the point of view of mathematical analysis, the same 

 as the problem of finding the magnetic force near a current 

 filament. The strength m of the vortex corresponds to the 

 strength i of the current and the fluid velocity to the magnetic 

 force. If r be the radius of the cross section (supposed 

 circular) of a vortex ring and co the angular velocity of 

 points on the ring round the circular axis, the strength J m 

 of the vortex is generally defined to be irr 2 co. In the 

 electrical problem the magnetic force tangential to the ring 

 is 2ijr, and in the hydrodynamical problem the corresponding 

 velocity is cor. Hence to convert the formula? for the 

 magnetic force at a point into the corresponding formula? for 

 the fluid velocity at that point we must multiply by cor 2 /2i, 

 that is, by ml'liri. For instance, we see from the construction 

 given in fig. 2 that the velocity at a point P in the fluid near 

 a vortex ring can be found as follows : — The component 

 velocity at right angles to AP (fig. 2) equals (2??ia/7rr 1 r 2 )E, 

 and the component at right angles to OP equals 



m(F-E)/(7rr 1 cos6>), 



the modulus of E and F being (1 — r 2 2 /r ] 2 ) 1/2 . At points on 



* Phil. Trans, vol. clxxii. p. 643 (1881). 

 f F. W. Dyson, p. 68, I. c. ante. 

 X It is sometimes defined as 2v.r' 2 <o. 



