Ch23_h1-solutions

# Ch23_h1-solutions - pan(zp695 Ch23 h1 chiu(56565 This...

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pan (zp695) – Ch23 h1 – chiu – (56565) 1 This print-out should have 28 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 10.0 points Amagnet icfe ldnearthe±oorpo intsupand is increasing. Looking down at the ±oor, does the non-Coulomb electric feld curl clockwise or counterclockwise? 1. Clockwise correct 2. Not enough inFormation 3. Counterclockwise Explanation: ²rom the Maxwell’s equation, we see that change in the magnetic feld d ± B dt in this case is down so the electric feld ± E NC curls clockwise. 002 10.0 points ldnearthece i l ingpo intsdown and is decreasing. Looking up at the ceil- ing, does the non-Coulomb electric feld curl clockwise or counterclockwise? 1. Counterclockwise correct 2. Clockwise 3. Not enough inFormation Explanation: ²rom the Maxwell’s equation, we see that change in the magnetic feld d ± B dt in this case is toward the ±oor so the electric feld ± E NC curls counterclockwise. 003 (part 1 of 2) 10.0 points On a circular path oF radius 11 cm in air around a solenoid with increasing magnetic feld, the emF is 35 V. What is the magnitude oF the non-Coulomb electric feld on this path? Correct answer: 50 . 6402 V / m. Explanation: Here we will use the defnition oF emF writ- ten in terms oF the electric feld. Since the value oF potential is given, we can fnd out the magnitude oF the electric feld. We begin by writing down the equation For the emF: emF = ± C ± E NC · d ± ² = ² ² ² ± E NC ² ² ² ± C d ± ² =2 π r ² ² ² ± E NC ² ² ² =35V . So, rearranging this expression, we obtain ² ² ² ± E NC ² ² ² = 35 V 2 π r = 35 V 2 π (11 cm) = 50 . 6402 V / m . 004 (part 2 of 2) 10.0 points Awirewithresistance5 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 fgure out the current ±ow: I = Δ V R = emF R = 35 V 5 Ω = 7A . 005 10.0 points Aun i Fo rmmagn e t i cf e ldo F3Tpo in t s29 degrees away From the perpendicular to the plane oF a rectangular loop oF wire 0 . 2mby 0 . 4masinthefgurebelow. ˆ n 29

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pan (zp695) – Ch23 h1 – chiu – (56565) 2 What is the magnetic fux on this loop? Correct answer: 0 . 209909 T · m 2 . Explanation: We can write Φ mag = ± S ± B · ˆ ndA = ± S ² ² ² ± B ² ² ² cos 29 dA = ² ² ² ± B ² ² ² cos 29 ± S dA = ² ² ² ± B ² ² ² (cos 29 ) A =(3T)(cos29 )(0 . 2m)(0 . 4m) = 0 . 209909 T · m 2 . 006 (part 1 of 6) 10.0 points Aconvent iona lcurrent I runs through a coil in the direction shown in the Fgure below. A single loop o± copper wire is near the coil. The loop and the coil are stationary. The point P is located on the - x axis inside the copper loop. Power Supply + - y x z Loop P Coil I In this initial state (constant current in coil), what is the direction o± the magnetic Feld at the center o± the copper loop, due to the current in the coil?
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Ch23_h1-solutions - pan(zp695 Ch23 h1 chiu(56565 This...

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