Self-induction of Wires. 13 



Or, in Cartesian coordinates, let H x and H 2 be the x and y 

 components of the magnetic force H, z being parallel to the 

 current ; then 



Hl =- 2 ^r -g, H 2=2 ^r -^ . (io«) 



express (la), and (8a) is represented by 



T=£2(H 1 2 + H/), 



dSi d€l 



-¥ r w + rt-4 i <•*-**, 



<m\ 



(11a) 



the latter form expressing (9a). 



It will be observed that the mathematical conditions are 

 identical with those existing in St. Tenant's torsion problems. 

 Thus, if a and h are the y and x tangential strain components 

 in the plane x, y in a twisted prism, and y the longitudinal 

 displacement along z, parallel to the length of the prism, we 

 have 



6= - t ^ + £' a =™ + %> • • • ( i2a) 



where t is the twist (Thomson and Tait, Part II. § 706, 

 equation (9)). The corresponding forces are n times as great, 

 if n is the rigidity {loc. cit. equation (10)); so that the energy 

 per unit length is 



■^nS(a 2 + 6 2 ) over section (13a) 



Also, to find 7 we have 



(loc. cit. equations (12) and (18)). Comparing (14 a) with 

 (5a), (12a) with (10a), and (13a) with the first of (11a), we 

 see that there is a perfect correspondence, except, of course, 

 as regards the constants concerned. The lines of tangential 

 stress in the torsion problem and the lines of magnetic force 

 in our problem are identical, and the energy is similarly 

 reckoned. We may therefore make use of all St. Venant's 

 results. 



It will be sufficient here to point out that the ratio of the 

 inductance of wires of different sections is the same as the 

 ratio of their torsional rigidities. Thus, as L = |/x in the case 

 of a round wire, that of a wire of elliptical section, semiaxes 

 a and 6, is ~L = fj,ab/(a 2 + b 2 ) ; when the section is a square, it 

 is '4417/-6; when it is an equilateral triangle, '3627//,, &c. 



