# 10 - kavo(ak22862 – HW10 – Janow –(12121 1 This...

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Unformatted text preview: kavo (ak22862) – HW10 – Janow – (12121) 1 This print-out should have 14 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 15.0 points The wire is carrying a current I . x y I I I 180 ◦ O R Find the magnitude of the magnetic field vector B at O due to a current-carrying wire shown in the figure, where the semicircle has radius r , and the straight parts to the left and to the right extend to infinity. 1. B = μ I 2 r 2. B = μ I 4 r correct 3. B = μ I π r 4. B = μ I 8 π r 5. B = μ I r 6. B = μ I 4 π r 7. B = μ I 3 r 8. B = μ I 7 π r 9. B = μ I 2 π r 10. B = μ I 3 π r Explanation: By the Biot-Savart Law, vector B = μ I 4 π integraldisplay dvectors × ˆ r r 2 . Consider the left straight part of the wire. The line element dvectors at this part, if we come in from ∞ , points towards O; i.e. , in the x- direction. We need to find dvectors × ˆ r to use the Biot-Savart Law. However, in this part of the wire, ˆ r is pointing towards O as well, so dvectors and ˆ r are parallel meaning dvectors × ˆ r = 0 for this part of the wire. It is now easy to see that the right part, having a dvectors antiparallel to ˆ r , also gives no contribution to vector B at O . Let us go through the semicircle C. The element dvectors , which is along the wire, will now be perpendicular to ˆ r , which is pointing along the radius towards O . Therefore | dvectors × ˆ r | = ds using the fact that ˆ r is a unit vector. So the Biot-Savart Law gives for the magnitude B of the magnetic field at O B = μ I 4 π integraldisplay C ds r 2 Since the distance r to the element dvectors is con- stant everywhere on the semicircle C, we will be able to pull it out of the integral. The integral is integraldisplay C ds r 2 = 1 r 2 integraldisplay C ds = 1 r 2 L C , where L C = π r is the length of the semicircle. Thus the magnitude of the magnetic field is B = μ I 4 π 1 r 2 π r = μ I 4 r . 002 (part 2 of 2) 15.0 points Note: ˆ i is in x-direction, ˆ j is in y-direction, and ˆ k direction is perpendicular to paper to- wards reader. Determine the direction of the magnetic field vector B at O due to the current-carrying wire. 1. hatwide B = 1 √ 2 parenleftBig ˆ i − ˆ parenrightBig 2. hatwide B = 1 √ 2 parenleftBig ˆ k +ˆ parenrightBig kavo (ak22862) – HW10 – Janow – (12121) 2 3. hatwide B = − ˆ k correct 4. hatwide B = 1 √ 2 parenleftBig ˆ k − ˆ parenrightBig 5. hatwide B = − ˆ 6. hatwide B = 1 √ 2 parenleftBig ˆ i +ˆ parenrightBig 7. hatwide B = − ˆ ı 8. hatwide B = +ˆ 9. hatwide B = +ˆ ı 10. hatwide B = + ˆ k Explanation: We know from Part 1 that the only contri- bution to the magnetic field at O comes from the semicircle. Furthermore, we need only consider the direction of dvectors × ˆ r for one typi- cal segment dvectors . If we go along the semicircle from left to right, and we know that ˆ r is point-...
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10 - kavo(ak22862 – HW10 – Janow –(12121 1 This...

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