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Unformatted text preview: hussain (wh4892) – hw 12 – Opyrchal – (121104) 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 Calculate the magnetic flux 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 flux 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 aircore solenoid contains 276 turns, is of length 14 cm, and has a crosssectional 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 = 1 . 25664 × 10 − 6 N / A 2 , ℓ = 14 cm , and A = 1 . 2 cm 2 = 0 . 00012 m 2 . The selfinductance of a solenoid is L = μ N 2 A ℓ = 1 . 25664 × 10 − 6 N / A 2 × (276 turns) 2 (0 . 00012 m 2 ) . 14 m = 8 . 20505 × 10 − 5 H . The induced emf E is given by E = vextendsingle vextendsingle vextendsingle vextendsingle L d I dt vextendsingle vextendsingle vextendsingle vextendsingle d I dt = vextendsingle vextendsingle vextendsingle vextendsingleE L vextendsingle vextendsingle vextendsingle vextendsingle = vextendsingle vextendsingle vextendsingle vextendsingle 130 μ V 8 . 20505 × 10 − 5 H vextendsingle vextendsingle vextendsingle vextendsingle = 1 . 58439 A / s . 003 10.0 points A solenoid inductor is 30 cm long and has a crosssectional 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....
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 Spring '08
 moro
 Inductance, Correct Answer, Inductor, RL circuit

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