Problem 1
A laser pointer delivers 0.10 mW average power in a beam 0.90 mm in diameter.
Find
(a) the average intensity,
(b) the peak electric field,
(c) the peak magnetic field.
Solution:
SP
1 2
d  157 W/m2.
4
E0  2 0 c S  2S  0 /  0  2S /( 0 c)  344 V/m
B0  E0 c 1.15 T .

Problem 2
At 1.5 km from the transmitter, the peak electric field of a radio wave is 350 mV/m.
(a) What is the transmitter’s power output, assuming it broadcasts uniformly in all
directions?
(b) What is the peak electric field 10 km from the transmitter?

Solution:
2
2
(a) P  4 r (E 0 2 0c)  4.59 kW
2 2
'
(b) Since r E 0  const , E 0  (r r')E 0  52.5 mV/m at a distance of 10 km.

Problem 3
13
of a particular TV station’s signal is 1.0 10
W/m2 when it
The average intensity

arrives at 33-cm –diameter satellite TV antenna.
(a) Calculate a total energy received by the antenna during 6 hours of viewing this
station’s programs.
(b) What are the amplitudes of E and B fields
 of the EM wave?
Solution: This is Problem 53 in Giancoli textbook.
1
(a) E  St d 2  1.8  10 10 J .
4
(b) E0  2 0 c S  2S  0 /  0  2S /( 0 c)  8.7  10 6 V / m.
B0  E0 / c  2.9  10 14 T .