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# solution_pdf26 - tovar(jdt436 homework 26 Turner(59070 This...

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tovar (jdt436) – homework 26 – Turner – (59070) 1 This print-out should have 11 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 10.0 points A solenoid has 133 turns of wire uniformly wrapped around an air-filled core, which has a diameter of 16 mm and a length of 8 . 6 cm. The permeability of free space is 1 . 25664 × 10 6 N / A 2 . Calculate the self-inductance of the solenoid. Correct answer: 5 . 1969 × 10 5 H. Explanation: Let : N = 133 , D = 16 mm = 0 . 016 m , = 8 . 6 cm = 0 . 086 m , and μ 0 = 1 . 25664 × 10 6 N / A 2 . The self-inductance of a solenoid is given by L 1 = N Φ I , where Φ is the total flux inside the solenoid and I is the current in the wire wrapped around the solenoid. By Ampere’s Law, the magnetic field in the solenoid is B = μ 0 N I , where μ = μ 0 is the magnetic permeability of the air core which is the same as free space. The magnetic flux in the solenoid is Φ = B A = B π D 2 4 . Using the above expressions for Φ and B , we obtain for the inductance of the solenoid L 1 = N B A I = μ 0 N 2 π D 2 4 = (1 . 25664 × 10 6 N / A 2 ) (133) 2 × π (0 . 016 m) 2 4 (0 . 086 m) = 5 . 1969 × 10 5 H . 002 (part 2 of 2) 10.0 points The core is replaced with a soft iron rod that has the same dimensions, but a magnetic per- meability of 800 μ 0 . What is the new inductance? Correct answer: 0 . 0415752 H. Explanation: Let : μ = 800 μ 0 . L 2 = μ N 2 π D 2 4 = μ μ 0 L 1 = 800 (5 . 1969 × 10 5 H) = 0 . 0415752 H . 003 (part 1 of 2) 10.0 points The current in a 72 mH inductor changes with time as I = b t 2 - a t . With a = 7 A / s and b = 5 A / s 2 , find the magnitude of the induced emf , |E| , at t = 0 . 4 s. Correct answer: 0 . 216 V. Explanation: Let : L = 72 mH = 0 . 072 H , b = 5 A / s 2 , a = 7 A / s , and t = 0 . 4 s . From Faraday’s Law, the induced emf E is proportional to the rate of change of the magnetic flux, which in turn is proportional to the rate of change of the current. This is expressed as E = L d I dt = L d dt ( b t 2 - a t ) = L (2 b t - a )

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tovar (jdt436) – homework 26 – Turner – (59070) 2 At 0 . 4 s ,the magnitude of the induced emf is |E| = (0 . 072 H) vextendsingle vextendsingle vextendsingle 2(5 A / s 2 )(0 . 4 s) - 7 A / s vextendsingle vextendsingle vextendsingle = 0 . 216 V .
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