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Unformatted text preview: johnson (rj6247) – hw 10 – Opyrchal – (121014) 1 This printout should have 12 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A conductor consists of a circular loop of radius 0 . 136 m and straight long sections, as shown. The wire lies in the plane of the paper and carries a current of 4 . 94 A. I R Find the magnitude of the magnetic field at the center of the loop. The permeability of free space is 1 . 25664 × 10 − 6 N / A 2 . Correct answer: 3 . 00875 × 10 − 5 T. Explanation: Let : μ = 1 . 25664 × 10 − 6 N / A 2 , I = 4 . 94 A , and R = 0 . 136 m . We can think of the total magnetic field as the superposition of the field due to the long straight wire having magnitude B straight = μ I 2 π R and directed into the page and the field due to the circular loop having magnitude B loop = μ I 2 R , also directed into the page. Using appropriate right hand rules, both B fields are in the same direction, so the resultant magnetic field is B = B straight + B loop = parenleftbigg 1 + 1 π parenrightbigg μ I 2 R = parenleftbigg 1 + 1 π parenrightbigg (1 . 25664 × 10 − 6 N / A 2 ) (4 . 94 A) 2 (0 . 136 m) = 3 . 00875 × 10 − 5 T . 002 (part 1 of 2) 10.0 points The set up is shown in the figure, where 19 A is flowing in the wire segments, AB = CD = 41 m, and the wire segment arc has a radius 35 m subtending an angle of 90 ◦ . The permeability of free space is 1 . 25664 × 10 − 6 T · m / A . 41m 19A 1 9 A 41 m 19 A 3 5 m A B C D O Find the magnitude of the magnetic field at O due to the current segment ABCD . Correct answer: 8 . 52718 × 10 − 8 T. Explanation: Let : I = 19 A , μ = 1 . 25664 × 10 − 6 T · m / A , and a = 35 m . Along CD dvectors and ˆ r are antiparallel, so again dvectors × ˆ r = 0 . Therefore that segment of the current also creates no magnetic field at O . Along BC dvectors is perpendicular to ˆ r so  dvectors × ˆ r  = ds = a dθ . Also dvectors × ˆ r is in the same direction for all dvectors along BC , while r = a , so the magnitude of the magnetic field at O due to ABCD is B = μ 4 π I integraldisplay π/ 2 a dθ a 2 = μ 4 π a I θ vextendsingle vextendsingle vextendsingle π/ 2 = μ I 8 a = (1 . 25664 × 10 − 6 T · m / A) (19 A) 8 (35 m) = 8 . 52718 × 10 − 8 T . johnson (rj6247) – hw 10 – Opyrchal – (121014) 2 003 (part 2 of 2) 10.0 points Find the magnitude of the magnetic field at O due to the current segment AB alone. Correct answer: 0 T. Explanation: The BiotSavart law is d vector B = μ 4 π I dvectors × ˆ r r 2 for the magnetic field due to a small current segment, I dvectors, at a point a distance r from the segment. ˆ r is the unit vector pointing from the segment to the point O. Along AB dvectors is parallel to ˆ r , so dvectors × ˆ r = 0 . Therefore d vector B =0 and vector B = integraldisplay d vector B = 0 T ....
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This note was uploaded on 03/11/2010 for the course PHYS 121 taught by Professor Opyrchal during the Spring '08 term at NJIT.
 Spring '08
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