502-505] Energy of Field 435 



field is produced such that the numbers of tubes of induction which cross 

 circuits 1, 2, 3 ... are 



jffUl 7/ 12 , 7/ 13 , .... 



Similarly, when a unit current flows through 2, let the numbers of tubes 

 of induction be 



The theorem of 446 shews at once that 



L n = L* = jf ^ dads', etc (428). 



If currents i lt i 2 , ... flow through the circuits simultaneously, and if the 

 numbers of tubes of induction which cut the circuits are N 1} N 2 , N 3 , ..., we 

 have 



A-'Iz ;' + Z 1 7 + z'7+' etc} (429) ' 



The energy of the system of currents is 



= i A^'i 2 + Aaiii + |rf + .................. (430). 



504. The energy required to start the single current i in circuit 1 will 

 be \Ln&. We can obtain the value of L n from equation (428) by making 

 the two circuits ds and ds coincide. It is easily found that L n is infinite. 



505. This can be seen in another way. The energy of the current is 



Near to the wire, at a small distance r from it, the force is , so that 



r 



a 2 + /3 2 4- 7 2 = 4i 2 /r\ Thus the energy within a thin ring formed of coaxal 

 cylinders of radii r lt r 2 , bent so as to follow the wire conveying the current 

 will be 



where the integration with respect to r is from ^ to r 2 , that with respect 

 to 6 is from to 2?r, and that with respect to s is along the wire. Integrat- 

 ing we find energy 



per unit length, and on taking r t = 0, we see that this energy is infinite. 



In practice, the circuits which convey currents are not of infinitesimal 

 cross-section, and so may not be treated geometrically as lines in calculating 

 L u . The current will distribute itself throughout the cross-section of the 

 wire, and the energy is readily seen to be finite so long as the cross-section 

 of the wire is finite. 



282 



