the Structure of the Electric Field. 307 



If S is the area of the cross-section of the tube at AB the 

 area of the constricted portion is S sin #, where is the 

 angle between AB and AC, 



. AA' 



sin 6 





AC " y^ + cV 



_CT 



ut 



approximately when t is la roe. 



Thus the area of the constricted portion of the tube is 



S — , so that if F is the value of the electric force at AB its 



value in the portion AC will be F — . If co is the solid an^le 



1 ct ° 



of the tube and r = OA, Fr 2 a>=2ire } hence the electric force 



'"Ji'jre iit 

 in AC equals ;2 — , or since ct = r approximately, the force 



in AC is 



2ire u 

 red c 2 t' 



The lines of electric force in this portion are moving 

 approximately at right angles to themselves with velocity c, 

 hence they give rise to a magnetic force at right angles to 

 the plane of the paper, and equal to 



cor ct 



Let us now calculate the energy in the portion A( !. The 

 •energy per unit volume is equal to 



or/* c*t" 

 The length of AC is approximately equal to ut and the 



CT 



cross-section is r 2 a>x— , hence the volume of the portion 

 AC is r' z oo x ct, and the energy is equal to 



n o 



TTt~ " 



co t V 



and there is an equal amount transmitted along the other 

 half of the tube, hence the amount of energy travelling out 

 as radiation is equal to 



ire- ur 



(O c 6 t' 



X 2 



