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Unformatted text preview: gonzales (pag757) – Homework 12 – Hoffmann – (56745) 1 This printout should have 15 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Consider the electromagnetic wave pattern as shown in the figure below. B E What is the direction in which the wave is traveling? 1. fromlefttoright. 2. undetermined, since the figure shows a standing wave. 3. fromrighttoleft. correct Explanation: The direction of traveling is the direction of vector E × vector B ; i.e. , fromrighttoleft. 002 (part 1 of 3) 10.0 points A plane electromagnetic sinusoidal wave of frequency 18 . 7 MHz travels in free space. The speed of light is 2 . 99792 × 10 8 m / s. Determine the wavelength of the wave. Correct answer: 16 . 0317 m. Explanation: Let : c = 2 . 99792 × 10 8 m / s , and f = 18 . 7 MHz = 1 . 87 × 10 7 Hz . The speed of light is c = λ f λ = c f = 2 . 99792 × 10 8 m / s 1 . 87 × 10 7 Hz = 16 . 0317 m . 003 (part 2 of 3) 10.0 points Find the period of the wave. Correct answer: 5 . 34759 × 10 − 8 s. Explanation: The period T of the wave is the inverse of the frequency: T = 1 f = 1 1 . 87 × 10 7 Hz = 5 . 34759 × 10 − 8 s . 004 (part 3 of 3) 10.0 points At some point and some instant, the electric field has has a value of 205 N / C. Calculate the magnitude of the magnetic field at this point and this instant. Correct answer: 6 . 83806 × 10 − 7 T. Explanation: Let : E = 205 N / C . The magnitudes of the electric and the mag netic fields are related by E B = c , B = E c = 205 N / C 2 . 99792 × 10 8 m / s = 6 . 83806 × 10 − 7 T . 005 10.0 points The cable is carrying the current I ( t ). Evaluate the electromagnetic energy flux S at the surface of a long transmission cable of resistivity ρ , length ℓ and radius a , using the expression vector S = 1 μ vector E × vector B . gonzales (pag757) – Homework 12 – Hoffmann – (56745) 2 1. S = μ ǫ ρ I 2 π a 2 2. S = ǫ μ I 2 2 π a 2 3. S = ρ I 2 2 π 2 a 3 correct 4. S = μ c I 2 4 π a 2 ℓ 5. None of these. 6. S = μ ρ I 2 π a 2 7. S = ρ ℓ I 2 π a 2 8. S = I 2 4 π μ c ℓ 9. S = μ I 2 2 π a 2 10. S = π a 2 I 2 ρ ℓ Explanation: The basic expression for the Poynting Vec tor is vector S = 1 μ vector E × vector B ....
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This note was uploaded on 01/27/2012 for the course PHY 303L taught by Professor Turner during the Spring '08 term at University of Texas.
 Spring '08
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