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Unformatted text preview: Shie, Gary – Oldquiz 3 – Due: Nov 8 2004, 1:00 pm – Inst: Turner 1 This printout should have 32 questions. Multiplechoice 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 π 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 > μ i a 2. B CA < μ i 8 π a 3. B CA = μ i 8 π a 4. B CA = μ i π a 5. B CA = μ i 2 π a 6. B CA = μ i 2 a 7. B CA = μ i 4 a 8. B CA = μ i 6 a 9. B CA = μ i 4 π a correct 10. B CA = μ i a Explanation: Note: 17 48 360 ◦ = 127 . 5 ◦ = 17 24 π . This problem has a direct way for its solu tion (based on the BiotSavart law) and a fast insightful way, based on Ampere’s law and the principle of superposition. Fast Way: Choose the orientation of the xaxis so that CA is along the negative xaxis with origin at A (see figure). θ 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 Ampere’s law, which gives 2 π a B = μ I , or B = μ I 2 π a , with B coming out of the paper. This follows from the standard right hand rule associated with Ampere’s 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 BiotSavart 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 = μ I 2 π a . Therefore B = μ I 4 π a . Direct Way: Set up a coordinate system with the origin at A , such that the segment CA goes along the negative xdirection as in the figure. The distance from the field point O to the wire is a . Consider a certain point P along CA with xcoordinate 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 θ . We find sin θ = a r , cos θ = x r . Dividing gives x = a cot θ , then dx = a sin 2 θ dθ = r 2 a dθ ....
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 Spring '10
 ERSKINE
 Physics, Magnetic Field, shie, Gary – Oldquiz

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