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MIDTERM 3

# MIDTERM 3 - Version 017/AABAB midterm 03 Turner(59130 This...

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Version 017/AABAB – midterm 03 – Turner – (59130) 1 This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Use Lenz’s law to answer the following ques- tion concerning the direction of induced cur- rents. R v S N a b We introduce the following notations: A) The direction of the induced magnetic field within the solenoid is A1 : to the left A2 : to the right A3 : upward A4 : downward A5 : into the paper B) The direction of the induced current through the resistor R is B1 : from a through R to b B2 : from b through R to a What is correct when the bar magnet is moved to the left? 1. A1, B1 2. A3, B2 3. A4, B2 4. A2, B1 correct 5. A5, B2 6. A2, B2 7. A4, B1 8. A3, B1 9. A5, B1 10. A1, B2 Explanation: The magnetic field inside the coil points to the right. When the magnet moves to the left, the magnetic flux through the coils decreases, so the induced current must produce a magnetic field pointing to the right. 002 (part 1 of 2) 10.0 points Consider two radial legs (extending to in- finity) and a connecting 12 23 π circular arc car- rying a current I as shown below. x y I I 12 23 π I I O r What is the magnitude of the magnetic field B 0 (at the origin O ) due to the current through this path? 1. B 0 = 4 23 μ 0 I π r + μ 0 I 2 π r 2. B 0 = 3 23 μ 0 I r + μ 0 I 2 π r correct 3. B 0 = 3 23 μ 0 I π r + μ 0 I 2 π r 4. B 0 = 4 23 μ 0 I π r + μ 0 I 2 r 5. B 0 = 4 23 μ 0 I π r + μ 0 I 4 π r 6. B 0 = 4 23 μ 0 I r + μ 0 I 2 π r

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Version 017/AABAB – midterm 03 – Turner – (59130) 2 7. B 0 = 3 23 μ 0 I r + μ 0 I 4 π r 8. B 0 = 3 23 μ 0 I π r + μ 0 I 2 r 9. B 0 = 3 23 μ 0 I π r + μ 0 I 4 π r 10. B 0 = 4 23 μ 0 I r + μ 0 I 4 π r Explanation: Note: The magnetic field at B 0 for the entire path points in the same direction. The two straight wire segments produce the same magnetic field at B 0 as a single long straight wire. Using Amp´ ere’s law, for the magnetic field a distance r from a straight wire, we have contintegraldisplay vector B · dvectors = μ 0 I contintegraldisplay B ds = μ 0 I B contintegraldisplay ds = μ 0 I B 2 π r = μ 0 I , so B 0 = μ 0 I 2 π r . (1) However, around the arc we will use the Biot-Savart law, where | dvectors × ˆ r | = ds = r dθ . The magnetic field at at the center of an arc with a current I is B 0 = μ 0 I 4 π integraldisplay dvectors × ˆ r r 2 = μ 0 I 4 π r 2 integraldisplay ds = μ 0 I 4 π r 2 integraldisplay r dθ = μ 0 I 4 π r integraldisplay 12 23 π 0 = μ 0 I 4 π r θ vextendsingle vextendsingle vextendsingle vextendsingle 12 23 π 0 = μ 0 I 4 π r parenleftbigg 12 23 π 0 parenrightbigg = 3 23 μ 0 I r . (2) The magnet field at B O for the entire path is the sum of Eqs. 2 and 1. B 0 = 3 23 μ 0 I r + μ 0 I 2 π r . 003 (part 2 of 2) 10.0 points What is the direction of the magnetic field vector B 0 at point O due to the current through the path?
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MIDTERM 3 - Version 017/AABAB midterm 03 Turner(59130 This...

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