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Unformatted text preview: PHYS 4D Solution to HW 2 January 15, 2011 Problem Giancoli 3122 (I) The E field in an EM wave has a peak of 26 . 5 mV/m . What is the average rate at which this wave carries energy across unit area per unit time? Solution: The average rate at which this wave carries energy across unit area per unit time is the average magnitude of the Poynting vector. S = 1 2 cE 2 = 1 2 ( 8 . 85 10 12 C 2 /N m 2 )( 3 10 8 m/s ) (0 . 0265 V/m ) = 9 . 32 10 7 W/m 2 . Problem Giancoli 3124 (II) How much energy is transported across a 1 . 00 cm 2 area per hour by an EM wave whose E field has an rms strength of 32 . 8 mV/m ? Solution: The energy transported across unit area per unit time is the magnitude of the Poynting vector. S = c E 2 rms = U A t U t = c E 2 rms A = ( 3 10 8 m/s )( 8 . 85 10 12 C 2 /N m 2 ) (0 . 0328 V/m ) 2 ( 1 10 4 m 2 ) (3600 s/h ) = 1 . 03 10 6 J/h. Problem Giancoli 3126 (II) If the amplitude of the B field of an EM wave is 2 . 5 10 7 T , (a) what is the amplitude of the E field? Solution: E = cB = ( 3 10 8 m/s ) 2 . 5 10 7 T = 75 V/m (b) What is the average power per unit area of the EM wave? Solution: The average power per unit area is given by the Poynting vector I = E B 2 = 75 V/m 2 . 5 10 7 T 2 (4 10 7 Ns 2 /C 2 ) = 7 . 5 W/m 2 . Problem Giancoli 3128 (II) A 15 . 8 mW laser puts out a narrow beam 2 . 00 mm in diameter. What are the rms values of E and B in the beam? Solution: The power output per unit area is the intensity and also is the magnitude of the Poynting vector. Use Eq. 3119a with rms values S = P A = c E 2 rms E rms = P Ac = . 0158 W (10 3 m ) 2 (3 10 8 m/s ) (8 . 85 10 12...
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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.
 Winter '08
 staff
 Physics, Energy

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