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Homework 13 Solutions

Homework 13 Solutions - lim(kl9356 Homework 13...

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lim (kl9356) – Homework 13 – Weathers – (22202) 1 This print-out should have 11 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 3) 10.0 points An infinite wire is bent as shown in the figure. The current is I . It consists of 3 segments AB , BDE , and EF . AB is vertical with A extending to infinity. EF is horizontal with F extending to infinity. BDE is one-quarter of a circular arc. We use the Roman numerals I, II, III and IV to denote the direction of a vector in the xy plane, which begins at the origin and is pointing into the quadrants I, II, III and IV respectively. B O D C E F A O y x I II III IV The direction of the magnetic vector at O due to the current segment BD , that due to DE and that due to EF are given by respectively (“in, out” mean “into, or out of”the xy plane) 1. B BD = in, B DE = out, B ED = out 2. B BD = in, B DE = out, B ED = in 3. B BD = out, B DE = out, B ED = out correct 4. I, II, B ED = in 5. B BD = in, B DE = in, B ED = out 6. B BD = out, B DE = in, B ED = out 7. B BD = in, B DE = in, B ED = in 8. IV, IV, B ED = out 9. B BD = out, B DE = in, B ED = in 10. B BD = out, B DE = out, B ED = in Explanation: The Biot-Savart Law is given by d vector B = μ 0 4 π I d vector l × ˆ r r 2 . Since both d vector l and ˆ r lie in the xy plane, their cross product must be perpendicular to this plane. Apply the Biot-Savart Law to each line element of the wire, d vector B is found always to direct out the xy plane at O . 002 (part 2 of 3) 10.0 points What is the magnitude of vector B at the center of the arc due to the arc BDE alone? 1. B = μ 0 I 2 r 2. B = μ 0 I 2 π r 3. B = μ 0 I 4 r 4. B = μ 0 I 8 r correct 5. B = μ 0 π I r 6. B = μ 0 2 π I r 7. B = μ 0 I 8 π r 8. B = μ 0 I π r 9. B = μ 0 I r 10. B = μ 0 I 4 π r Explanation: Note: The distance r from a current el- ement of BDE to O is a constant and the current element I d vector l is always perpendicular to ˆ r . Hence, the magnitude of the vector B at the center of the arc due to the arc BDC alone is

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lim (kl9356) – Homework 13 – Weathers – (22202) 2 B BDE = μ 0 4 π I r 2 integraldisplay dl = μ 0 4 π I r 2 π 2 r = μ 0 I 8 r . 003 (part 3 of 3) 10.0 points Let r = 0 . 069 m and I = 0 . 4 A. The permeability of free space is 1 . 25664 × 10 6 N / A 2 . Find the magnitude of vector B at O due to the entire wire ABDEF ? Correct answer: 2 . 07003 × 10 6 T. Explanation: Let : μ 0 = 1 . 25664 × 10 6 N / A 2 , I = 0 . 4 A , and r = 0 . 069 m . First calculate the magnetic field of a straight wire carrying a current(See figure be- low). I r s y y O θ B From the Biot-Savart Law, the contribution to the magnetic field at O due to a current element I y at y is given by B = μ 0 4 π I y s 2 sin θ = μ 0 I 4 π θ r sin θ . Upon integration, the magnetic field con- tributed by a current segment from θ 1 to θ 2 is given by B = μ 0 I 4 π r (cos θ 1 cos θ 2 ) For the wire segment AB , we set θ 1 = π 2 and θ 2 = π . Hence B AB = μ 0 I 4 π r parenleftBig cos π 2 cos π parenrightBig = μ 0 I 4 π r .
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