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Unformatted text preview: marout (ssm979) – ohw15 – turner – (56725) 1 This printout should have 15 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A charged particle is projected with its initial velocity parallel to a uniform magnetic field. What is the resulting path? 1. parabolic arc. 2. straight line parallel to the field. correct 3. straight line perpendicular to the field. 4. circular arc. 5. spiral. Explanation: The force on a moving charge due to a magnetic field is given by vector F = qvectorv × vector B . If vectorv and vector B are parallel, then vectorv × vector B = 0 . Hence the force on the particle is zero, and the particle continues to move in a straight line parallel to the field. 002 10.0 points In the diagram below, the resistances are 2 . 7 Ω and 2 . 35 Ω. The current through R 1 is 3 . 97 A. A B R 1 2 R Find the potential difference between points A and B . Correct answer: 20 . 0485 V. Explanation: Let : R 1 = 2 . 7 Ω , R 2 = 2 . 35 Ω , and I = 3 . 97 A . Currents through elements in a series combi nation are the same. The equivalent resis tance is R = R 1 + R 2 , so V = I R = I ( R 1 + R 2 ) = 3 . 97 A (2 . 7 Ω + 2 . 35 Ω) = 20 . 0485 V . 003 10.0 points The figure represents two possible ways to connect two lighbulbs X and Y to a battery. Bulb X has less resistance than bulb Y . Y X A X Y B Which bulb has the most current running through it? 1. Bulb X in B 2. Bulb X in A correct 3. Bulb Y in B 4. Bulb Y in A Explanation: In A , a parallel circuit, the voltage across X is the same as Y , but X has less resistance, so it has more current running through it. marout (ssm979) – ohw15 – turner – (56725) 2 In B , a series circuit, the current is the same through both bulbs. 004 (part 1 of 3) 10.0 points 9 . 7 V 1 . 7 V 3 . 8 V I 1 1 . 4 Ω 2 . 9 Ω I 2 5 . 9 Ω I 3 9 . 7 Ω Find the current I 1 in the 1 . 4 Ω resistor at the bottom of the circuit between the two power supplies. Correct answer: 1 . 1341 A. Explanation: E 1 E 2 E 3 I 1 R A R B I 2 R C I 3 R D At a junction (Conservation of Charge) I 1 + I 2 I 3 = 0 . (1) Kirchhoff’s law on the large outside loop gives ( R A + R B ) I 1 + R D I 3 = E 1 + E 2 . (2) Kirchhoff’s law on the righthand small loop gives R C I 2 + R D I 3 = E 3 . (3) Let : R A = 1 . 4 Ω , R B = 2 . 9 Ω , R C = 5 . 9 Ω , R D = 9 . 7 Ω , E 1 = 9 . 7 V , E 2 = 1 . 7 V , and E 3 = 3 . 8 V . Using determinants, I 1 = vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle 1 1 E 1 + E 2 R D E 3 R C R D vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle 1 1 1 R A + R B R D R C R D vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle Expanding along the first row, the numera tor is D 1 = vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle...
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This note was uploaded on 10/24/2010 for the course PHY 56735 taught by Professor Turner,j during the Fall '10 term at University of Texas at Austin.
 Fall '10
 TURNER,J

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