HW4Solutions

# HW4Solutions - Version One – Homework 4 – Juyang Huang...

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Unformatted text preview: Version One – Homework 4 – Juyang Huang – 24018 – Mar 28, 2008 1 This print-out should have 43 questions. Multiple-choice 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- ω 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 : λ = 5 . 5 × 10- 7 m and c = 3 × 10 8 m / s . The frequency is f = c λ = 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 ρ , length and radius a , using the expression 1 S = 1 μ 1 E × 1 B . 1. S = π a 2 I 2 ρ 0 2. S = ρ 0 I 2 π a 2 3. S = μ ρ I 2 π a 2 4. S = μ / ρ I 2 π a 2 5. S = μ cI 2 4 π a 2 6. S = I 2 4 π μ c0 7. S = ρ I 2 2 π 2 a 3 correct Version One – Homework 4 – Juyang Huang – 24018 – Mar 28, 2008 2 8. S = / μ I 2 2 π a 2 9. S = μ I 2 2 π a 2 10. None of these. Explanation: The basic expression for the Poynting Vec- tor is 1 S = 1 μ 1 E × 1 B . Based on Ampere’s law, B = μ I 2 π a , and Ohm’s law is E ≡ V = I R = ρ I π a 2 , since R ≡ ρ 0 π 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 μ ρ I π a 2 μ I 2 π a = ρ I 2 2 π 2 a 3 . Note: Also, we have S = E 2 μ c = ρ 2 I 2 μ cπ 2 a 4 S = cB 2 μ = cμ I 2 4 π 2 a 2 , which are not given as choices. keywords: Cosmic Microwave Radiation 34:04, trigonometry, numeric, > 1 min, wording-variable....
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HW4Solutions - Version One – Homework 4 – Juyang Huang...

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