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Unformatted text preview: taherisefat (mt23852) hw 12 Opyrchal (121012) 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 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 flux as L = N I = 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 aircore solenoid contains 639 turns, is of length 10 . 1 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 245 V? Correct answer: 0 . 401878 A / s. Explanation: Let : N = 639 turns , = 1 . 25664 10 6 N / A 2 , = 10 . 1 cm , and A = 1 . 2 cm 2 = 0 . 00012 m 2 . The selfinductance of a solenoid is L = N 2 A = 1 . 25664 10 6 N / A 2 (639 turns) 2 (0 . 00012 m 2 ) . 101 m = 0 . 000609637 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 245 V . 000609637 H vextendsingle vextendsingle vextendsingle vextendsingle = . 401878 A / s . 003 10.0 points A solenoid inductor is 30 cm long and has a crosssectional 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 V....
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 Spring '08
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