Forced Vibrations of Electromagnetic Systems. 449 



excess of the electric over the magnetic energy at any point 

 is independent of the time. Both decrease at an equal rate; 

 the magnetic energy to zero, the electric energy to that of 

 the final steady displacement ce/47r. 



The above E and H solutions are fundamental, because 

 all electromagnetic disturbances due to impressed force depend 

 solely upon, and come from, the lines of curl of the impressed 

 force. From them, by integration, we can find the disturb- 

 ances due to any collection of rectilinear filaments of f. Thus, 

 to find the H due to a plane sheet of parallel uniformly dis- 

 tributed filaments, of surface-density/, we have, by (349), at 

 distance a from the plane, on either side, 



H - f ft = J- r S in- - y 1 



where the limits are + (v 2 t 2 — a 2 ) . Therefore 



H=//2/*» 



after the time t = a/v ; before then, H is zero. 



Similarly, a cylindrical sheet of longitudinal f produces 



fa C dO 



~ 27T/A17J (vH 2 -b* 



,11 



where b is the distance of the point where H is reckoned from 

 the element a dd of the circular section of the sheet, a being 

 its radius. The limits have to be so chosen as to include all 

 elements of/ which have had time to produce any effect at 

 the point in question. When the point is external and vt 

 exceeds a+r the limits are complete, viz. to include the whole 

 circle. The result is then, at distance r from the axis of the 

 cylinder, 



H== f a l/* v f 1.3.v 1.3.5.7 ^4.3 

 (i*f -^-fjy L + 2*|2 2 + 2 4 |4 2 3 1 . 2 



. 1.3.5.7.9.11.* 3 6.5.4 



where 



2 6 |6 2 5 1 . 2 . 3 



+ ...], (355) 



x=(2ar)*(v 2 t 2 -a 2 -r>)-K 



This formula begins to operate when a?=l, or vt = a + r. As 

 time goes on, x falls to zero, leaving only the first term. 



[To be continued.] 



