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Unformatted text preview: Anderson, Matthew – Homework 10 – Due: Apr 8 2008, noon – Inst: Vitaly 1 This print-out should have 14 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 2) 15 points The wire is carrying a current I . x y I I I 180 ◦ O R Find the magnitude of the magnetic field ~ 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 π r 2. B = μ I π r 3. B = μ I 4 π r 4. B = μ I 2 r 5. B = μ I 7 π r 6. B = μ I 8 π r 7. B = μ I 4 r correct 8. B = μ I 3 π r 9. B = μ I 3 r 10. B = μ I r Explanation: By the Biot-Savart Law, ~ B = μ I 4 π Z d~s × ˆ r r 2 . Consider the left straight part of the wire. The line element d~s at this part, if we come in from ∞ , points towards O; i.e. , in the x- direction. We need to find d~s × ˆ r to use the Biot-Savart Law. However, in this part of the wire, ˆ r is pointing towards O as well, so d~s and ˆ r are parallel meaning d~s × ˆ r = 0 for this part of the wire. It is now easy to see that the right part, having a d~s antiparallel to ˆ r , also gives no contribution to ~ B at O . Let us go through the semicircle C. The element d~s , which is along the wire, will now be perpendicular to ˆ r , which is pointing along the radius towards O . Therefore | d~s × 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 Z C ds r 2 Since the distance r to the element d~s is con- stant everywhere on the semicircle C, we will be able to pull it out of the integral. The integral is Z C ds r 2 = 1 r 2 Z 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 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 ~ B at O due to the current-carrying wire. 1. b B = +ˆ ı 2. b B =- ˆ Anderson, Matthew – Homework 10 – Due: Apr 8 2008, noon – Inst: Vitaly 2 3. b B = 1 √ 2 ‡ ˆ k- ˆ · 4. b B =- ˆ ı 5. b B = +ˆ 6. b B = 1 √ 2 ‡ ˆ k +ˆ · 7. b B =- ˆ k correct 8. b B = 1 √ 2 ‡ ˆ i- ˆ · 9. b B = + ˆ k 10. b B = 1 √ 2 ‡ ˆ i +ˆ · 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 d~s × ˆ r for one typi- cal segment d~s . If we go along the semicircle from left to right, and we know that ˆ r is point- ing in towards O , the right hand rule tells us that the field resulting from this element is into the paper. Since this holds for every el- ement on the semicircle, the total field is also pointing into the paper.pointing into the paper....
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- Spring '08
- Work, Magnetic Field