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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 ε 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 = cε E 2 rms = ∆ U A ∆ t ∆ U ∆ t = cε 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 μ = 75 V/m × 2 . 5 × 10 7 T 2 × (4 π × 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 = cε E 2 rms ⇒ E rms = √ P Acε = √ . 0158 W π (10 3 m ) 2 (3 × 10 8 m/s ) (8 . 85 × 10 12...
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 Winter '08
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 Physics, Energy, Poynting vector, Erms, Problem Giancoli

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