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

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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 – fnd 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 SN ab We introduce the Following notations: A) The direction oF the induced magnetic feld 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 feld inside the coil points to the right. When the magnet moves to the leFt, the magnetic ±ux through the coils decreases, so the induced current must produce a magnetic feld pointing to the right. 002 (part 1 oF 2) 10.0 points Consider two radial legs (extending to in- fnity) 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 feld 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 2. B 0 = 3 23 μ 0 I r + μ 0 I 2 correct 3. B 0 = 3 23 μ 0 I + μ 0 I 2 4. B 0 = 4 23 μ 0 I + μ 0 I 2 r 5. B 0 = 4 23 μ 0 I + μ 0 I 4 6. B 0 = 4 23 μ 0 I r + μ 0 I 2
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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 + μ 0 I 2 r 9. B 0 = 3 23 μ 0 I + μ 0 I 4 10. B 0 = 4 23 μ 0 I r + μ 0 I 4 Explanation: Note: The magnetic feld at B 0 For the entire path points in the same direction. The two straight wire segments produce the same magnetic feld at B 0 as a single long straight wire. Using Amp´ ere’s law, For the magnetic feld a distance r From a straight wire, we have ± , B · d,s = μ 0 I ± B ds = μ 0 I B ± ds = μ 0 I B 2 = μ 0 I, so B 0 = μ 0 I 2 . (1) However, around the arc we will use the Biot-Savart law, where | d,s × ˆ r | = ds = r dθ . The magnetic feld at at the center oF an arc with a current I is B 0 = μ 0 I 4 π ² d,s × ˆ r r 2 = μ 0 I 4 2 ² ds = μ 0 I 4 2 ² r dθ = μ 0 I 4 ² 12 23 π 0 = μ 0 I 4 θ ³ ³ ³ ³ 12 23 π 0 = μ 0 I 4 ´ 12 23 π - 0 µ = 3 23 μ 0 I r . (2) The magnet feld 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 . 003 (part 2 oF 2) 10.0 points What is the direction oF the magnetic feld , B 0 at point O due to the current through the path? 1. B is down the page. 2. B is zero. 3. B is up the page. 4. B is into the page. correct 5. B is out oF the page. 6. B is to the leFt. 7. B is to the right. Explanation: The right-hand rule dictates that magnetic feld is into the page. 004 10.0 points A conducting bar moves as shown near a long wire carrying a constant I = 39 A current.
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This note was uploaded on 11/28/2010 for the course PHY 56705 taught by Professor Turner during the Spring '10 term at University of Texas at Austin.

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

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