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Unformatted text preview: Reilly (mar3978) – hw19 – turner – (56725) 1 This print-out should have 17 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Consider the set up shown in the figure where a solenoid has a steadily increasing magnetic flux which generates identical in- duced emf s for the two cases illustrated. B B B B A #1 B #2 i i Case 1: Case 1: Two identical light bulbs are in series. The corresponding electrical power consumed by bulb 1 and bulb 2 are P 1 and P 2 , respectively. B B B B A #1 B #2 D C O i 2 i 1 Case 2: Case 2: 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. The corresponding electrical power consumed by bulb 1 and bulb 2 are P ′ 1 and P ′ 2 , respectively. What is the ratio of P ′ 1 P 1 ? It may be helpful to first find an expression for P 1 and then write down the loop equation for ACODA in case 2. 1. P ′ 1 P 1 = 2 2. P ′ 1 P 1 = 0 3. P ′ 1 P 1 = 1 8 4. P ′ 1 P 1 = 8 5. P ′ 1 P 1 = 1 3 6. P ′ 1 P 1 = 1 2 7. P ′ 1 P 1 = 4 8. P ′ 1 P 1 = 3 9. P ′ 1 P 1 = 1 4 10. P ′ 1 P 1 = 1 correct Explanation: Let E and R be the induced emf and resis- tance of the light bulbs, respectively. For case 1, 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 . Hence, for case 1, the power consumed by bulb 1 is P 1 = parenleftbigg E 2 R parenrightbigg 2 R = E 2 4 R . For case 2, the loop equation for ACODA is E 2 − I 1 R = 0 . Solving for I 1 yields I 1 = E 2 R . Hence the power dissipated by bulb 2 for case 2 is P ′ 1 = parenleftbigg E 2 R parenrightbigg 2 R = E 2 4 R . Reilly (mar3978) – hw19 – turner – (56725) 2 This is identical to the expression for P 1 . Therefore, P ′ 1 P 1 = 1 . keywords: 002 (part 1 of 3) 10.0 points Four long, parallel conductors carry equal cur- rents of I . An end view of the conductors is shown in the figure. Each side of the square has length of ℓ . A B D C × P x y ℓ Which diagram correctly denotes the direc- tions of the magnetic fields from each conduc- tor at the point P ? The current direction is out of the page at points A , B , and D indi- cated by the dots and into the page at point C indicated by the cross. 1. P B A B B B C B D 2. P B A B B B C B D 3. P B A B B B C B D 4. P B A B B B C B D 5. P B A B B B C B D correct Explanation: The direction of the magnetic field due to each wire is given by the right hand rule. Place the thumb of the right hand along the direction of the current; your fingers now curl in the direction of the magnetic field’s circular path: A B D C × P B A B B B D B C 003 (part 2 of 3) 10.0 points At point P , as far as the magnitudes are concerned, B A = B B = B C = B D ≡ B i , where B i is introduced to represent any of the four magnetic fields....
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This note was uploaded on 11/30/2010 for the course PHY 60230 taught by Professor Turner,j during the Spring '10 term at University of Texas.
- Spring '10