oldq3 - Shie, Gary Oldquiz 3 Due: Nov 8 2004, 1:00 pm Inst:...

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Unformatted text preview: Shie, Gary Oldquiz 3 Due: Nov 8 2004, 1:00 pm Inst: Turner 1 This print-out should have 32 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. The due time is Central time. 001 (part 1 of 4) 10 points Consider the current segment shown where a current i is flowing through a straight segment from C to A , then an arc from A to D , and a straight segment from D to F . 17 24 D F i A C i r O Both segments AC and DF extend to in- finity. The circular arc AD is 17 48 of a circle with a radius a . The arc is centered at O . The magnitude ( k ~ B CA k B CA ) of the magnetic field at O due to the CA segment alone is given by 1. B CA > i a 2. B CA < i 8 a 3. B CA = i 8 a 4. B CA = i a 5. B CA = i 2 a 6. B CA = i 2 a 7. B CA = i 4 a 8. B CA = i 6 a 9. B CA = i 4 a correct 10. B CA = i a Explanation: Note: 17 48 360 = 127 . 5 = 17 24 . This problem has a direct way for its solu- tion (based on the Biot-Savart law) and a fast insightful way, based on Amperes law and the principle of superposition. Fast Way: Choose the orientation of the x-axis so that CA is along the negative x-axis with origin at A (see figure). P P A a O C C From C to A a current I flows, and imag- ine this current continuing from A to C . At O , the magnetic field B from the current in the extended wire ( CAC ) can be found from Amperes law, which gives 2 a B = I , or B = I 2 a , with B coming out of the paper. This follows from the standard right hand rule associated with Amperes law. Note: Contribution to B from the current in the segment CA equals the contribution from the current in the segment AC . Reason: This follows by noticing (from the Biot-Savart law) that the contribution to ~ B from a current through the line element d centered about P is the same as the contribu- tion from an equal incremental line element d centered about P with P and P equally distant from point A . Thus each half of the extended wire gives the desired magnetic field B , and B + B = B = I 2 a . Therefore B = I 4 a . Direct Way: Set up a coordinate system with the origin at A , such that the segment CA goes along the negative x-direction as in the figure. The distance from the field point O to the wire is a . Consider a certain point P along CA with x-coordinate x . The distance from O to this particular point is called r . Denote the (counterclockwise) angle between Shie, Gary Oldquiz 3 Due: Nov 8 2004, 1:00 pm Inst: Turner 2 the line OP and the horizontal by . We find sin = a r , cos =- x r . Dividing gives x =- a cot , then dx = a sin 2 d = r 2 a d ....
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This note was uploaded on 01/20/2010 for the course PHY 1 taught by Professor Erskine during the Spring '10 term at University of Texas at Austin.

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oldq3 - Shie, Gary Oldquiz 3 Due: Nov 8 2004, 1:00 pm Inst:...

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