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23-1 - brown(twb493 Ch23 h1 chiu(56565 This print-out...

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brown (twb493) – Ch23 h1 – chiu – (56565) 1 This print-out should have 28 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0points A magnetic field near the floor points up and is increasing. Looking down at the floor, does the non-Coulomb electric field curl clockwise or counterclockwise? 1. Not enough information 2. Counterclockwise 3. Clockwise correct Explanation: From the Maxwell’s 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 10.0points A magnetic field near the ceiling points down and is decreasing. Looking up at the ceil- ing, does the non-Coulomb electric field curl clockwise or counterclockwise? 1. Counterclockwise correct 2. Not enough information 3. Clockwise Explanation: From the Maxwell’s 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(part1of2)10.0points 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 non-Coulomb 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 π 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 π r = 35 V 2 π (8 cm) = 69 . 6303 V / m . 004(part2of2)10.0points A wire with resistance 5 Ω is placed along the path. What is the current in the wire? Correct answer: 7 A. Explanation: Now we use Ohm’s Law to figure out the current flow: I = Δ V R = emf R = 35 V 5 Ω = 7 A . 005 10.0points 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 0 . 4 m as in the figure below. ˆ n 28

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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 Φ mag = integraldisplay S vector B · ˆ n dA = 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 (cos 28 ) A = (3 T)(cos 28 )(0 . 2 m)(0 . 4 m) = 0 . 211907 T · m 2 .
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