solution12_pdf

solution12_pdf - hussain(wh4892 hw 12 Opyrchal(121104 This...

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hussain (wh4892) – hw 12 – Opyrchal – (121104) 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 400 turn, 8 . 06 mH coil when the current in the coil is 11 mA. Correct answer: 2 . 2165 × 10 7 Wb. Explanation: Let : N = 400 , L = 8 . 06 mH = 0 . 00806 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 . 00806 H) (0 . 011 A) 400 = 2 . 2165 × 10 7 Wb . 002 10.0 points An inductor in the Form oF an air-core solenoid contains 276 turns, is oF length 14 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 130 μ V? Correct answer: 1 . 58439 A / s. Explanation: Let : N = 276 turns , μ 0 = 1 . 25664 × 10 6 N / A 2 , = 14 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 × (276 turns) 2 (0 . 00012 m 2 ) 0 . 14 m = 8 . 20505 × 10 5 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 - 130 μ V 8 . 20505 × 10 5 H v v v v = 1 . 58439 A / s . 003 10.0 points A solenoid inductor is 30 cm long and has a cross-sectional area oF 6 cm 2 . When the current through the solenoid decreases at a rate oF 0 . 625 A / s, the induced emf is 150 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.
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solution12_pdf - hussain(wh4892 hw 12 Opyrchal(121104 This...

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