Homework12

Homework12 - johnson (rj6247) hw 12 Opyrchal (121014) This...

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johnson (rj6247) – hw 12 – Opyrchal – (121014) 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 210 turn, 11 . 47 mH coil when the current in the coil is 4 mA. Correct answer: 2 . 18476 × 10 7 Wb. Explanation: Let : N = 210 , L = 11 . 47 mH = 0 . 01147 H , and I = 4 mA = 0 . 004 A . The inductance oF a coil is related to the magnetic ±ux as L = N Φ I Φ = L I N = (0 . 01147 H) (0 . 004 A) 210 = 2 . 18476 × 10 7 Wb . 002 10.0 points An inductor in the Form oF an air-core solenoid contains 252 turns, is oF length 20 . 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 180 μ V? Correct answer: 3 . 77813 A / s. Explanation: Let : N = 252 turns , μ 0 = 1 . 25664 × 10 6 N / A 2 , = 20 . 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 × (252 turns) 2 (0 . 00012 m 2 ) 0 . 201 m = 4 . 76427 × 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 - 180 μ V 4 . 76427 × 10 5 H v v v v = 3 . 77813 A / s . 003 10.0 points A solenoid inductor is 15 cm long and has a cross-sectional area oF 4 cm 2 . When the current through the solenoid decreases at a rate oF 0 . 625 A / s, the induced emf is 230 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: 2209 . 24 m 1 .
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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.

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Homework12 - johnson (rj6247) hw 12 Opyrchal (121014) This...

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