Drill 8.5: A small-loop antenna of radius /100 is excited by a current i = 1 cos(2 ×
109t)A. (a) Determine the maximum power density radiated by this antenna at a distance
of 100 m. (b) What is the time-averaged power density at the point P(100,/4,/2)? (c)
Find the radiation resistance.
Solution: We have (8.63): Pmax 
 2 o2 I o2 S 2  2
32o 2 r 2
c 3 108 m / s
Here,   
 0.3m
f
109 / s
Pmax
Pmax
 2 109 4 10 7 H 1A   0.3m  2 2

1





32 120 
s
m
 100  0.3m  100m 
nW
 18.4 2
m
nW

P  r ,  ,    Pmax sin 2   18.4 2
m

2
nW
 2  
 sin    9.2 2
m

4
2
(8.65) Rrad
2
2


 S 
4    100 
 320  2   320 
  3.1m
2



 


4
Written by Stuart M. Wentworth. Copyright John Wiley and Sons 2007