Electric Waves through Tubes. 131 



like manner 



v- = -7- sin0+ ~^r A cos0= — i&sin(?z — 1)6 J W _,(W 

 ay ar r ad v y v ' 



-~±ksm(n+l)0 J n +i(kr). . (34) 



These forms show directly that dRjdx, dll/dy satisfy the 

 fundamental equation (2). They apply when n is equal to 

 unity or any greater integer. When n = 0, we have 



R=J (*r), (35) 



g = -W 1 (^)cos^ d ^ = -kJ x (kr)smO. . (36) 



The expressions for the electromotive intensity are some- 

 what simpler when the resolution is circumferential and radial: 



circumf. component =Q cos 6 — P sin 6= -jj ~r Q 



= -^J„(fo-) S in«0, . . (37) 



... A ~ . Q im dR 



radial component = r cos t/ + (Jsinp=p — 



= l -^J n '(kr)cosnd. . . . (38) 



If n = 0, the circumferential component vanishes. 

 Also for the magnetization 

 ore. comp. ot mag. —b cos u — a sin u= . , 2 — -j— 



= ^~^ikr)^n6, . . (39) 



rad. comp. ot mag. = a cos V + 6 sin t/= . , 2 — ^ 



= ^+^>J.(fe.) B mntf. . . (40) 



The smallest value of k for vibrations of this class belongs 

 to the series n=0, and is such that kr= 2*404, r being the 

 radius of the cylinder. 



For the vibrations of the second class E = 0, and by (21), 



c= J n (kr) cos n0, (41) 



k being subject to the boundary condition 



J„W)=0 (42) 



