Quiz_2_sol

# Quiz_2_sol - 1.0 × 10-13 W/m 2 when it arrives at 33-cm...

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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 4 π d 2 = 157 W/m 2 . 344 ) /( 2 / 2 2 0 0 0 0 0 = = = = c S S S c E ε μ V/m B 0 = E 0 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: (a) P = 4 r 2 ( E 0 2 2 0 c ) = 4 .59 kW (b) Since r 2 E 0 2 = const , E 0 ' = ( rr ' ) E 0 = 52 .5 mV/m at a distance of 10 km. Problem 3 The average intensity of a particular TV station’s signal is
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Unformatted text preview: 1 .0 × 10-13 W/m 2 when it 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. (a) . 10 8 . 1 4 1 10 2 J d t S E-× = = (b) . / 10 7 . 8 ) /( 2 / 2 2 6 m V c S S S c E-× = = = = . 10 9 . 2 / 14 T c E B-× = =...
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## This note was uploaded on 09/19/2011 for the course PHYS 4d taught by Professor Staff during the Winter '08 term at UCSD.

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