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Unformatted text preview: brown (twb493) Ch23 h1 chiu (56565) 1 This printout should have 28 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 10.0 points A magnetic field near the floor points up and is increasing. Looking down at the floor, does the nonCoulomb electric field curl clockwise or counterclockwise? 1. Not enough information 2. Counterclockwise 3. Clockwise correct Explanation: From the Maxwells equation, we see that change in the magnetic field d vector B dt in this case is down so the electric field vector E NC curls clockwise. 002 10.0 points A magnetic field near the ceiling points down and is decreasing. Looking up at the ceil ing, does the nonCoulomb electric field curl clockwise or counterclockwise? 1. Counterclockwise correct 2. Not enough information 3. Clockwise Explanation: From the Maxwells equation, we see that change in the magnetic field d vector B dt in this case is toward the floor so the electric field vector E NC curls counterclockwise. 003 (part 1 of 2) 10.0 points On a circular path of radius 8 cm in air around a solenoid with increasing magnetic field, the emf is 35 V. What is the magnitude of the nonCoulomb electric field on this path? Correct answer: 69 . 6303 V / m. Explanation: Here we will use the definition of emf writ ten in terms of the electric field. Since the value of potential is given, we can find out the magnitude of the electric field. We begin by writing down the equation for the emf: emf = contintegraldisplay C vector E NC d vector = vextendsingle vextendsingle vextendsingle vector E NC vextendsingle vextendsingle vextendsingle contintegraldisplay C d vector = 2 r vextendsingle vextendsingle vextendsingle vector E NC vextendsingle vextendsingle vextendsingle = 35 V . So, rearranging this expression, we obtain vextendsingle vextendsingle vextendsingle vector E NC vextendsingle vextendsingle vextendsingle = 35 V 2 r = 35 V 2 (8 cm) = 69 . 6303 V / m . 004 (part 2 of 2) 10.0 points A wire with resistance 5 is placed along the path. What is the current in the wire? Correct answer: 7 A. Explanation: Now we use Ohms Law to figure out the current flow: I = V R = emf R = 35 V 5 = 7 A . 005 10.0 points A uniform magnetic field of 3 T points 28 degrees away from the perpendicular to the plane of a rectangular loop of wire 0 . 2 m by . 4 m as in the figure below. n 28 brown (twb493) Ch23 h1 chiu (56565) 2 What is the magnetic flux on this loop? Correct answer: 0 . 211907 T m 2 . Explanation: We can write mag = integraldisplay S vector B ndA = integraldisplay S vextendsingle vextendsingle vextendsingle vector B vextendsingle vextendsingle vextendsingle cos 28 dA = vextendsingle vextendsingle vextendsingle vector B vextendsingle vextendsingle vextendsingle cos 28 integraldisplay S dA = vextendsingle vextendsingle vextendsingle vector B vextendsingle vextendsingle vextendsingle...
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This note was uploaded on 12/01/2011 for the course PHY 303L taught by Professor Turner during the Fall '08 term at University of Texas at Austin.
 Fall '08
 Turner
 Physics

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