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Unformatted text preview: aljabr (faa335) – Hw13 – Ross – (89251) 1 This print-out should have 9 questions. Multiple-choice 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 6 A, where the radius of the circular arc is 5 cm. The permeability of free space is 1 . 25664 × 10 − 6 T · m / A . 6 A 5 c m O Determine the magnitude of the magnetic field at point O , the origin of the arc. Correct answer: 18 . 8496 μ T. Explanation: Let : I = 6 A and R = 5 cm = 0 . 05 m . For the straight sections d vectors × ˆ r = 0. The quarter circle makes one-fourth 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) (6 A) 8 (0 . 05 m) = 18 . 8496 μ T , into the paper . 002 (part 1 of 2) 10.0 points A conductor in the shape of a square, whose sides are of length 0 . 825 m, carries a counter- clockwise 21 . 5 A current as shown in the figure below. . 825 m 21 . 5 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: 29 . 4842 μ T. Explanation: Let : μ = 1 . 25664 × 10 − 6 T m / A , ℓ = 0 . 825 m and I = 21 . 5 A . By the Biot-Savart law, dB = μ 4 π I dvectors × ˆ r r 2 . Consider a thin, straight wire carring a con- stant current I along the x-axis 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 P . Therefore, we have only to determine the aljabr (faa335) – Hw13 – Ross – (89251) 2 magnitude of the field at P . In fact, taking the origin at O and letting P be along the positive y axis, with ˆ k being a unit vector pointing out of the paper, we see that dvectors × ˆ r = ˆ k | dvectors × ˆ r | = ˆ k ( dx sin θ ) ....
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This note was uploaded on 01/27/2011 for the course PHYS 251 taught by Professor Gavrine during the Spring '10 term at Purdue.
- Spring '10