HW 26 - terry(ect328 homework 26 Turner(59130 This...

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terry (ect328) – homework 26 – Turner – (59130) 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 (part 1 oF 2) 10.0 points A solenoid has 111 turns oF wire uniFormly wrapped around an air-flled core, which has a diameter oF 15 mm and a length oF 7 cm. The permeability oF Free space is 1 . 25664 × 10 6 N / A 2 . Calculate the selF-inductance oF the solenoid. Correct answer: 3 . 90868 × 10 5 H. Explanation: Let : N = 111 , D = 15 mm = 0 . 015 m , = 7 cm = 0 . 07 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 ±ux inside the solenoid and I is the current in the wire wrapped around the solenoid. By Ampere’s Law, the magnetic feld 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 ±ux 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 ) (111) 2 × π (0 . 015 m) 2 4 (0 . 07 m) = 3 . 90868 × 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 . 0312694 H. Explanation: Let : μ = 800 μ 0 . L 2 = μ N 2 π D 2 4 = μ μ 0 L 1 = 800(3 . 90868 × 10 5 H) = 0 . 0312694 H . 003 (part 1 oF 2) 10.0 points The current in a 102 mH inductor changes with time as I = b t 2 - a t . With a = 16 A / s and b = 2 A / s 2 , fnd the magnitude oF the induced emf , |E| , at t = 1 . 2 s. Correct answer: 1 . 1424 V. Explanation: Let : L = 102 mH = 0 . 102 H , b = 2 A / s 2 , a = 16 A / s , and t = 1 . 2 s . ²rom ²araday’s Law, the induced emf E is proportional to the rate oF change oF the magnetic ±ux, 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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terry (ect328) – homework 26 – Turner – (59130) 2 At 1 . 2 s ,the magnitude of the induced emf is |E| = (0 . 102 H) v v v 2(2 A / s 2 )(1 . 2 s) - 16 A / s v v v = 1 . 1424 V .
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HW 26 - terry(ect328 homework 26 Turner(59130 This...

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