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Unformatted text preview: dzifanu (sd26397) – hw 10 – opyrchal – (121102) 1 This printout should have 11 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points The segment of wire in the figure carries a current of 2 A, where the radius of the circular arc is 9 cm. The permeability of free space is 1 . 25664 × 10 − 6 T · m / A . 2 A 9 c m O Determine the magnitude of the magnetic field at point O , the origin of the arc. Correct answer: 3 . 49066 μ T. Explanation: Let : I = 2 A and R = 9 cm = 0 . 09 m . For the straight sections d vectors × ˆ r = 0. The quarter circle makes onefourth the field of a full loop B = μ I 8 R into the paper. Or, you can use the equation B = μ o I 4 π θ , where θ = π 2 . Thus the magnetic field is B = μ I 8 R , = (1 . 25664 × 10 − 6 T · m / A) (2 A) 8 (0 . 09 m) = 3 . 49066 μ T , into the paper . 002 10.0 points A conductor consists of a circular loop of radius 0 . 0543 m and straight long sections, as shown. The wire lies in the plane of the paper and carries a current of 8 . 75 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: 0 . 000133477 T. Explanation: Let : μ = 1 . 25664 × 10 − 6 N / A 2 , I = 8 . 75 A , and R = 0 . 0543 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 ) (8 . 75 A) 2 (0 . 0543 m) = . 000133477 T . 003 (part 1 of 2) 10.0 points dzifanu (sd26397) – hw 10 – opyrchal – (121102) 2 A conductor in the shape of a square, whose sides are of length 0 . 439 m, carries a clockwise 20 A current as shown in the figure below. . 439 m 20 A P What is the magnitude of the magnetic field at point P (at the center of the square loop) due to the current in the wire? Correct answer: 51 . 5431 μ T. Explanation: Let : μ = 1 . 25664 × 10 − 6 T m / A , ℓ = 0 . 439 m and I = 20 A . By the BiotSavart law, dB = μ 4 π I dvectors × ˆ r r 2 . Consider a thin, straight wire carring a con stant current I along the xaxis with the y axis pointing towards the center of the square, as in the following figure. y x O r P x I ˆ r ds a θ Let us calculate the total magnetic field at the point P located at a distance a from the wire. An element dvectors is at a distance r from P . The direction of the field at P due to this element is out of the paper, since dvectors × ˆ r is out of the paper. In fact, all elements give a contribution directly out of the paper at...
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 Spring '09
 PRODAN
 Physics, Current, Magnetic Field, Bnet, Bstraight

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