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HW12-Solutions - Version PREVIEW HW12 feng(58230 This...

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Version PREVIEW – HW12 – feng – (58230) 1 This print-out should have 18 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. Poynting Vector 001 10.0 points A radio transmitter in space transmits isotropically with an average power of 630 kW. What is the average magnitude of the Poynting vector 45 miles from the radio trans- mitter? Correct answer: 9 . 56299 × 10 6 W / m 2 . Explanation: Let : P = 630 kW = 6 . 3 × 10 5 W , and d = 45 miles = 72405 m . The average value of the Poynting vector is the intensity of the wave; intensity is power per unit area. Since the wave is isotropical, the intensity is the same in all directions, and therefore the power distributes uniformly over a surface 4 π d 2 at a distance d around the transmitter. This means that ( S ) = I = P 4 π d 2 = 6 . 3 × 10 5 W 4 π (72405 m) 2 = 9 . 56299 × 10 6 W / m 2 . The conversion from miles to meters is given by d m = 1609 m / mi × d mi . Poynting Vector for a Wire 002 10.0 points A long, straight wire of resistance 2 Ω, radius 0 . 5 mm, and length 2 m carries a constant current 2 . 6 A. Calculate the Poynting vector at the surface of the wire. Correct answer: 2151 . 77 W / m 2 . Explanation: Let : R = 2 Ω , a = 0 . 5 mm = 0 . 0005 m , = 2 m , and I = 2 . 6 A . The surface area of the wire is A = 2 π (0 . 0005 m) (2 m) = 0 . 00628319 m 2 . Let us find the magnitude of the electric field vector E along the wire. If V is the potential difference across the ends of the wire, then V = IR and E = V = I R . Applying Amp` ere’s law, we find that that the magnetic field at the surface of the wire is B = μ 0 I 2 π a . The vectors vector E and vector B are mutually perpendic- ular, so | vector E × vector B | = E B . Hence, the Poynting vector vector S has a magnitude S = E B μ 0 = 1 μ 0 I R μ 0 I 2 π a = I 2 R 2 π a ℓ = I 2 R A = (2 . 6 A ) 2 (2 Ω) 0 . 00628319 m 2 = 2151 . 77 W / m 2 . Form this result, we see that S A = I 2 R , that is, the rate SA at which electromagnetic energy flows into the wire equals the power I 2 R ( i.e., rate of energy) dissipated as Joule heat.
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Version PREVIEW – HW12 – feng – (58230) 2 Cable Energy Flux 003 (part 1 of 2) 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 and radius a , using the expression vector S = 1 μ 0 vector E × vector B . 1. S = μ 0 ǫ 0 ρ I 2 π a 2 2. S = μ 0 ρ I 2 π a 2 3. S = ρ ℓ I 2 π a 2 4. S = μ 0 c I 2 4 π a 2 5. S = μ 0 I 2 2 π a 2 6. S = ρ I 2 2 π 2 a 3 correct 7. None of these. 8. S = π a 2 I 2 ρ ℓ 9. S = I 2 4 π μ 0 c ℓ 10. S = ǫ 0 μ 0 I 2 2 π a 2 Explanation: The basic expression for the Poynting Vec- tor is vector S = 1 μ 0 vector E × vector B . Based on Ampere’s law, B = μ 0 I 2 π a , and Ohm’s law is E V = I R = ρ I π a 2 , since R ρ ℓ π a 2 . It is easy to see that vector E and vector B are perpen- dicular to each other.
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