

Impulsive Motion of Electrified Systems. 129 



(VuE) 8 =h 2 (E 2 2 -4-E 3 2 ), and hence 



H 2 = KV(E 2 2 + E 3 2 ) {2'd) 



But, by (22)] the component of m parallel to 11 is 



m 1 -=(^K 2 /4.7r)(E 2 -E 1 2 ).u= At B 2 /477 ? ,. J . . (24) 



and hence M,=2T/« (25) 



The other components of m have the values 



m 2 = -(fAV/iir) E 2 E x u, m 3 = -(/LtK 2 /47r) E 3 E L m. (26) 



In special cases, as when the system has an axis of sym- 

 metry parallel to u, the components M 2 and M 3 will vanish, 

 and then we have simply 



M=2T/u (27) 



§ 15. When M and I are both parallel to u, we can at once 

 write down P, the momentum in the pulse. For, by (20), 

 P = I — M, and hence, in this case, by (19) and (27), 



V = 2W /u-2T/u (28) 



Thus, by (13), P = 2(U-U )/h (29) 



If we can calculate P and W for the pulse, we can deduce 

 from them not only the electric and the magnetic energies 

 of the system in steady motion but its momentum as well. 

 For, by (29), (28), and (27), 



U=j^P + Uo (30) 



T = W~UI\ . . . , . (31) 



M.= 2W/h-P (32) 



§ 16. The impulse of the force required to suddenly stop 

 a system, which is moving along an axis of symmetry, is 

 easily found. For since both E and H in the pulse change 

 signs, when the system is stopped instead of being started, 

 we see that the momentum in the pulse is the same in both 

 cases. Hence, before the stopping, the momentum is M, 

 while at a great time after the stopping the momentum is P, 

 since the Maxwell stress over an infinite surface enclosing 

 the pulse vanishes, because E and H at points outside the 

 pulse ultimately vary as l/? i2 . Hence, if I' be the impulse, 

 in the opposite direction to u, which is required to reduce the 

 system to rest, 



I' = H-P = 2(T-U + U )/m. . . . (33) 

 Phil Mag. S. 6. Vol. 13. No. 73. Jan. 1007. K 



