Impulsive Motion of Electrified Systems. 143 



By symmetry 2E]E 2 = 0, and 2E 2 E 3 =0, and hence, by (43). 



M/= -^XE/Eo^ -^(E, cos + E 3 sin0)E«^O. 



The momentum at right angles to the direction of motion is 

 thus given by 



M 3 '^= -^^Ei'Es^^SCEr-Es 2 ) sin 6 cos 6 



= (N -MO sin cos (45) 



When #=-^7r, the component o£ the momentum at right 

 angles to the direction o£ motion vanishes, and hence N is 

 the resultant momentum in that case. 



From (45) we see that the resultant momentum is inclined 

 to the direction of motion at an angle i/r, where 



tan yfr = M 3 7M,'= (BT— MJ sin 6 cos 0/(M 1 cos 2 6 + N sin 2 6). 



I£ T' be the magnetic energy for anv value of 6. and T and 

 S be the magnetic energies when the system moves along 

 and at right angles to its axis, we have, by (25), 



T'=|M/z/, T^^Mrt^ S=±Nti. 

 Tims 



T' = Tcos 2 + Ssin 2 (46) 



and 



S=« 2 U> 8 -iT (47) 



§ 30. We may illustrate the formulae of § 29 by the results 

 for a charged disk of radius b. Here we have, by § 26, for 

 very small speeds 



U = ttQ 2 4K^ T = iw 2 UoAr. 

 Hence 



S — u 2 JJ jv 2 — JT = |/rU v 2 , 



and thus the magnetic energy of the disk moving slowly in 

 its own plane is half as great again as when it moves along 

 its axis at the same speed, as we found in § 27. 



Now N-M 1 = 2(S-T)/m=1wU /« 2 and hence, when the 

 axis of the disk makes an angle 6 with the direction of 

 motion, 



M^Mi-f- (N-MO sin 2 0=i(^ 2 )II o (2 + sin 2 0) 3 



M 3 '=i(w/i> 2 )tj o sin0cos'0. 



Thus the resultant momentum is inclined to the direction of 

 motion at an angle i/r, where 



. sin cos 

 tanifr= - 



2 + sur 6 



