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# Solution_pdf 20 - malsam(wgm329 – hw20 – turner –(56705 1 This print-out should have 15 questions Multiple-choice questions may continue on

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Unformatted text preview: malsam (wgm329) – hw20 – turner – (56705) 1 This print-out should have 15 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A solenoid with circular cross section pro- duces a steadily increasing magnetic flux through its cross section. There is an octago- nally shaped circuit surrounding the solenoid as shown. The increasing magnetic flux gives rise to a counterclockwise induced emf E . Initial Case: The circuit consists of two identical light bulbs of equal resistance R con- nected in series, leading to a loop equation E - 2 iR = 0 . B B B B X Y i i Figure 1: The corresponding electrical power con- sumed by bulb X and bulb Y are P X and P Y , respectively. Primed ′ Case: Let the points C and D be on the symmetry line of the diagram. Connect points C and D by a wire, which equally divides the magnetic flux as in figure 2. B B B B X Y D C A i 2 i 1 Figure 2: The corresponding electrical power con- sumed by bulb X and bulb Y are P ′ X and P ′ Y , respectively. What are the ratios P ′ X P X and P ′ Y P Y , respec- tively? 1. P ′ X P X = 0 and P ′ Y P Y = 0 2. P ′ X P X = 4 and P ′ Y P Y = 1 2 3. P ′ X P X = 2 and P ′ Y P Y = 1 4 4. P ′ X P X = 1 and P ′ Y P Y = 1 correct 5. P ′ X P X = 2 and P ′ Y P Y = 0 6. P ′ X P X = 4 and P ′ Y P Y = 1 4 7. P ′ X P X = 0 and P ′ Y P Y = 1 2 8. P ′ X P X = 4 and P ′ Y P Y = 0 9. P ′ X P X = 0 and P ′ Y P Y = 1 4 10. P ′ X P X = 2 and P ′ Y P Y = 1 2 Explanation: Let E and R be the induced emf and resis- tance of the light bulbs, respectively. For the first case, since the two bulbs are in series, the equivalent resistance is simply R eq = R + R = 2 R and the current through the bulbs is i = E 2 R , so, the power consumed by bulb X is P X = parenleftbigg E 2 R parenrightbigg 2 R = E 2 4 R , For the second (primed) case, the loop equa- tion for XCADX is E 2- iR = 0 . Solving for i yields i = E 2 R , malsam (wgm329) – hw20 – turner – (56705) 2 so the power dissipated by bulb X is P ′ X = parenleftbigg E 2 R parenrightbigg 2 R = E 2 4 R . Thus P ′ X P X = E 2 4 R E 2 4 R = 1 Using a similar argument, we have P ′ Y P Y = 1 002 (part 1 of 2) 10.0 points A long, straight wire carries a current and lies in the plane of a rectangular loops of wire, as shown in the figure. 2 . 9cm 11 cm 9 . 8cm (56A)sin bracketleftBig (190rad / s) t + δ bracketrightBig → 130 loops (turns) Determine the maximum emf , |E| , induced in the loop by the magnetic field created by the current in the straight wire. Correct answer: 42 . 4873 mV. Explanation: Let : N = 130 , ω = 190 rad / s , ℓ = 9 . 8 cm , a = 2 . 9 cm , b = 11 cm , and I = 56 A ....
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## This note was uploaded on 11/22/2010 for the course PHY 303L taught by Professor Turner during the Spring '08 term at University of Texas at Austin.

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