Forced Vibrations of Electromagnetic System} 



,-s V _j (Jl«-.yftl«) - (Joa-yGoa)47rKs/^ 



[ } ^~ dor ^ / 7r S a-J 0a (Joa-y^oa)47rK S /cp /l 



(Out) E = JutJor-yaOr) y 



same denominator' 

 from which H may be got by the E/H operator. 



445 



(333) 



The external sheet, say/ 2 , produces 

 (in) ill - — — - — - — - — -/ 2 , 



(out)E = (J 0r -j/G 0r ) 



^/J 



(334) 



where the unwritten denominators are as in the first of (333). 

 Observe that when J la = 0, f\ produces no external field (in 

 tube or beyond it). It is then only/ 2 that operates in the 

 tube and beyond. 



Now take/ 2 =« and/ 1= -e in (333) and (334) and add the 

 results. We then obtain (317), (318) ; and it is now J 0o = 

 that makes the external field vanish, instead of J la = when 

 fi alone is operative. 



Haying treated this problem of a tube in some detail, the 

 other examples may be very briefly considered, although they 

 too admit of numerous singularities. 



60. Circular Impressed Force in Conducting-tube. — The tube 

 being as before, let the impressed force e (per unit length) 

 act circularly in it instead of longitudinally, and let e be a 

 function of t only, so that we have an inner and an outer 

 cylindrical surface of longitudinally directed curl of e. H is 

 evidently longitudinal and E circular, so that we now require 

 to use the (314) operator. 



At the tube E a is continuous, this being the tensor of the 

 force of the flux on either side, and H is discontinuous thus, 



H(b) — H (out) = 47rK (e + E a ) , 

 or 



1 C H 



-{ 



1 + 



4ttK\E 



(out)-?(in))}E.. . (335) 



Substituting the (314) operator, with y = inside, and y 

 undetermined outside, and using the conjugate property (307), 

 we obtain 



. (Jla— ^/Gla)Jor Or J la (J 0r ~ */Gor) 



H 



On; 



or 



(out) 



/^Jla(Jla-*/Gla)+ -~^ 



e, (336) 



Trap 



E, 



(J„ 



■yQ,.)J„.orJ,.(J„.-yQ, r ) «. (337) 



same denominator 

 Phil. Mag. S. 5. Vol. 26. No. 162. Nov. 1888. 2 H 



