hw19-solutions - howard(cah3459 hw19 turner(56705 This...

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howard (cah3459) – hw19 – turner – (56705) 1 This print-out should have 17 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 10.0 points Consider the set up shown in the fgure where a solenoid has a steadily increasing magnetic ±ux 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 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 ±ux. 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 frst fnd an expression For P 1 and then write down the loop equation For ACODA in case 2. 1. P ± 1 P 1 =4 2. P ± 1 P 1 =8 3. P ± 1 P 1 =2 4. P ± 1 P 1 = 1 2 5. P ± 1 P 1 =0 6. P ± 1 P 1 =3 7. P ± 1 P 1 =1 correct 8. P ± 1 P 1 = 1 3 9. P ± 1 P 1 = 1 8 10. P ± 1 P 1 = 1 4 Explanation: Let E and R be the induced emf and resis- tance oF the light bulbs, respectively. ²or case 1, since the two bulbs are in series, the equivalent resistance is simply R eq = R + R 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 = ± E 2 R ² 2 R = E 2 4 R . ²or case 2, the loop equation For is E 2 - I 1 R . Solving For I 1 yields I 1 = E 2 R . Hence the power dissipated by bulb 2 For case 2 is P ± 1 = ± E 2 R ² 2 R = E 2 4 R .
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howard (cah3459) – hw19 – turner – (56705) 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 ±gure. Each side of the square has length of + . A BD C × P x y + Which diagram correctly denotes the direc- tions of the magnetic ±elds 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 correct 5. P B A B B B C B D Explanation: The direction of the magnetic ±eld 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 ±ngers now curl in the direction of the magnetic ±eld’s circular path: A 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 ±elds. Find B i .
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howard (cah3459) – hw19 – turner – (56705) 3 1. B i = μ 0 I 2 2 π+ 2. B i = μ 0 I 2 3. B i = μ 0 I 2 correct 4. B i = μ 0 I + 2 π 5. B i = μ 0 I 2 2 Explanation: By Ampere’s law, the line integral around any closed path is μ 0 I , with each path defned by a circle with radius r = + cos 45 = + 2 2 .
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This note was uploaded on 11/28/2010 for the course PHY 56705 taught by Professor Turner during the Spring '10 term at University of Texas.

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hw19-solutions - howard(cah3459 hw19 turner(56705 This...

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