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Unformatted text preview: Shie, Gary – Homework 28 – Due: Nov 8 2004, 4:00 am – Inst: Turner 1 This printout should have 13 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 2) 10 points A microwave transmitter emits electromag netic waves of a single wavelength. The max imum electric field 1 . 52 km from the trans mitter is 7 . 71 V / m. Assuming that the trans mitter is a point source and neglecting waves reflected from the Earth, calculate the maxi mum magnetic field at this distance. Correct answer: 2 . 57178 × 10 8 T. Explanation: B = E c = 7 . 71 V / m 2 . 99792 × 10 8 m / s = 2 . 57178 × 10 8 T . 002 (part 2 of 2) 10 points Calculate the total power emitted by the transmitter. Correct answer: 2 . 29058 × 10 6 W. Explanation: I = E max B max 2 μ = (7 . 71 V / m)(2 . 57178 × 10 8 T) 2(1 . 25664 × 10 6 N / A 2 ) = 0 . 0788948 W / m 2 , and P = I A = 4 π r 2 I = 4 π r 2 µ E max B max 2 μ ¶ = 4 π (1520 m) 2 (0 . 0788948 W / m 2 ) = 2 . 29058 × 10 6 W . 003 (part 1 of 2) 10 points The cable is carrying the current I ( t ). Evaluate the electromagnetic energy flux at the surface of a long transmission cable of resistivity ρ , length ‘ and radius a . 1. S = ρ I 2 2 π 2 a 3 correct 2. S = μ ρ I 2 π a 2 3. S = μ I 2 2 π a 2 4. None of these. 5. S = μ ² ρ I 2 π a 2 6. S = ρ ‘ I 2 π a 2 7. S = μ c I 2 4 π a 2 ‘ 8. S = ² μ I 2 2 π a 2 9. S = π a 2 I 2 ρ ‘ 10. S = I 2 4 π μ c ‘ Explanation: We will solve this problem using the basic expression for the Poynting Vector ~ S = 1 μ ~ E × ~ B . Based on Ampere’s law, B = μ I 2 π a . And Ohm’s law tells us E = ρ I π a 2 . It is easy to see that ~ E and ~ B are perpen dicular to each other....
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This note was uploaded on 01/20/2010 for the course PHY 1 taught by Professor Erskine during the Spring '10 term at University of Texas.
 Spring '10
 ERSKINE
 Physics, Work

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