solution12_pdf

solution12_pdf - taherisefat(mt23852 hw 12 Opyrchal(121012...

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taherisefat (mt23852) – hw 12 – Opyrchal – (121012) 1 This print-out should have 11 questions. Multiple-choice questions may continue on the next column or page – Fnd all choices before answering. 001 10.0 points Calculate the magnetic ±ux through a 350 turn, 8 . 01 mH coil when the current in the coil is 11 mA. Correct answer: 2 . 51743 × 10 7 Wb. Explanation: Let : N = 350 , L = 8 . 01 mH = 0 . 00801 H , and I = 11 mA = 0 . 011 A . The inductance of a coil is related to the magnetic ±ux as L = N Φ I Φ = L I N = (0 . 00801 H) (0 . 011 A) 350 = 2 . 51743 × 10 7 Wb . 002 10.0 points An inductor in the form of an air-core solenoid contains 639 turns, is of length 10 . 1 cm, and has a cross-sectional area of 1 . 2 cm 2 . The permeability of free space is 1 . 25664 × 10 6 N / A 2 . What is the magnitude of the uniform rate of change in current through the inductor that induces an emf of 245 μ V? Correct answer: 0 . 401878 A / s. Explanation: Let : N = 639 turns , μ 0 = 1 . 25664 × 10 6 N / A 2 , = 10 . 1 cm , and A = 1 . 2 cm 2 = 0 . 00012 m 2 . The self-inductance of a solenoid is L = μ 0 N 2 A = 1 . 25664 × 10 6 N / A 2 × (639 turns) 2 (0 . 00012 m 2 ) 0 . 101 m = 0 . 000609637 H . The induced emf E is given by |E| = v v v v - L d I dt v v v v d I dt = v v v v -E L v v v v = v v v v - 245 μ V 0 . 000609637 H v v v v = 0 . 401878 A / s . 003 10.0 points A solenoid inductor is 30 cm long and has a cross-sectional area of 5 cm 2 . When the current through the solenoid decreases at a rate of 0 . 625 A / s, the induced emf is 160 μ V. The permeability of free space is 1 . 25664 × 10 6 N / A 2 . ²ind the number of turns per meter of the solenoid. Correct answer: 1165 . 39 m 1 .
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solution12_pdf - taherisefat(mt23852 hw 12 Opyrchal(121012...

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