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Unformatted text preview: Version One Homework 4 Juyang Huang 24018 Mar 28, 2008 1 This printout should have 43 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. The due time is Central time. Electromagnetic Wave 01 34:02, trigonometry, numeric, < 1 min, nor mal. 001 (part 1 of 1) 10 points The electric field in an electromagnetic wave (in vacuum) is described by E = E max sin( k x t ) , where E max = 100 N / C and k = 1 10 7 m 1 . The speed of light is 2 . 99792 10 8 m / s. Find the amplitude of the corresponding magnetic wave. Correct answer: 3 . 33564 10 7 T. Explanation: Let : c = 2 . 99792 10 8 m / s and E max = 100 N / C . The relation between the magnitude of the electric field and that of the magnetic field for an electromagnetic wave is given by B = E c . Thus, the magnitude of the magnetic wave is B max = E max c = 100 N / C 2 . 99792 10 8 m / s = 3 . 33564 10 7 T . keywords: Serway CP 21 50 34:03, trigonometry, numeric, > 1 min, nor mal. 002 (part 1 of 1) 10 points The eye is most sensitive to light of wave length 5 . 5 10 7 m, which is in the green yellow region of the visible electromagnetic spectrum. The speed of light is 3 10 8 m / s. What is the frequency of this light? Your answer must be within 0.1%. Correct answer: 5 . 45455 10 14 Hz. Explanation: Let : = 5 . 5 10 7 m and c = 3 10 8 m / s . The frequency is f = c = 3 10 8 m / s 5 . 5 10 7 m = 5 . 45455 10 14 Hz . keywords: Cable Energy Flux short 34:04, trigonometry, numeric, > 1 min, fixed. 003 (part 1 of 1) 10 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 , length and radius a , using the expression 1 S = 1 1 E 1 B . 1. S = a 2 I 2 0 2. S = 0 I 2 a 2 3. S = I 2 a 2 4. S = / I 2 a 2 5. S = cI 2 4 a 2 6. S = I 2 4 c0 7. S = I 2 2 2 a 3 correct Version One Homework 4 Juyang Huang 24018 Mar 28, 2008 2 8. S = / I 2 2 a 2 9. S = I 2 2 a 2 10. None of these. Explanation: The basic expression for the Poynting Vec tor is 1 S = 1 1 E 1 B . Based on Amperes law, B = I 2 a , and Ohms law is E V = I R = I a 2 , since R 0 a 2 . It is easy to see that 1 E and 1 B are perpen dicular to each other. So the energy flux can be written down as S = 1 I a 2 I 2 a = I 2 2 2 a 3 . Note: Also, we have S = E 2 c = 2 I 2 c 2 a 4 S = cB 2 = c I 2 4 2 a 2 , which are not given as choices. keywords: Cosmic Microwave Radiation 34:04, trigonometry, numeric, > 1 min, wordingvariable....
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 Spring '10
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