Note contribution to b 0 from the current in the

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Note: Contribution to B 0 from the current in the segment CA equals the contribution from the current in the segment AC 0 . Reason: This follows by noticing (from the Biot-Savart law) that the contribution to ~ B 0 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 0 with P and P 0 equally distant from point A . Thus each half of the extended wire gives the desired magnetic field B , and B + B = B 0 = μ 0 I 2 π a . Therefore B = μ 0 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 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
Platt, David – Oldquiz 3 – Due: Nov 13 2005, 4:00 am – Inst: Ken Shih 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 = r 2 a dθ . The contribution to the magnetic field at O from an incremental line segment dx at point P is dB CA = μ 0 4 π i dx r 2 sin θ = μ 0 i 4 π a sin θ. Every segment part of CA gives a contribu- tion to the field which points out of the paper. Thus the total field due to CA is B CA = Z dB CA = μ 0 i 4 π a Z sin θ dθ . The integration starts at infinity, where the angle θ is 0 . The integration ends at A , where the angle is 90 = π/ 2 radians. Thus the integral is Z π/ 2 0 sin θ dθ = - cos θ fl fl fl fl π/ 2 0 = 1 , and so the magnetic field is B CA = μ 0 i 4 π a . 002 (part 2 of 4) 10 points Find the magnitude of the field B AD at O due to the circular arc AD alone. 1. B AD = 13 96 μ 0 i a correct 2. B AD = 5 32 μ 0 i a 3. B AD = 7 32 μ 0 i a 4. B AD = 1 6 μ 0 i a 5. B AD = 3 16 μ 0 i a 6. B AD = 23 96 μ 0 i a 7. B AD = 17 96 μ 0 i a 8. B AD = 5 24 μ 0 i a 9. B AD = 11 48 μ 0 i a 10. B AD = 7 48 μ 0 i a Explanation: We sum up the incremental current contri- butions along the arc: Δ B AD = μ 0 4 π i Δ x r 2 sin θ. In this case, Δ x = a Δ θ which is the formula for the arc length of an arc segment with angle Δ θ and radius a . Note: The r in the previous formula is the distance to the field point, but for the circular arc this is constantly a . Finally, we note that the angle θ is always 90 along the arc. Thus Δ B AD = μ 0 4 π i a Δ θ a 2 sin 90 = μ 0 i 4 π a Δ θ . The arc has an angular span (given in the question) of Δ θ = 13 48 2 π . Thus, the magnitude of the field B AD at O due to the circular arc AD alone is B AD = μ 0 i 4 π a 13 48 2 π = 13 96 μ 0 i a . 003 (part 3 of 4) 10 points
Platt, David – Oldquiz 3 – Due: Nov 13 2005, 4:00 am – Inst: Ken Shih 3 Select the choice which represents the magni- tude and the direction of the resultant mag- netic field O due to the vector sum of all contributions: ~ B = ~ B CA + ~ B AD + ~ B DF .

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