HW 26 - terry (ect328) – homework 26 – Turner –...

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Unformatted text preview: 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 – find all choices before answering. 001 (part 1 of 2) 10.0 points A solenoid has 111 turns of wire uniformly wrapped around an air-filled 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 , ℓ = 7 cm = 0 . 07 m , and μ = 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 N I ℓ , where μ = μ 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 N 2 π D 2 4 ℓ = (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 . What is the new inductance? Correct answer: 0 . 0312694 H. Explanation: Let : μ = 800 μ . L 2 = μ N 2 π D 2 4 ℓ = μ μ L 1 = 800(3 . 90868 × 10 − 5 H) = . 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 , find 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 . 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 ) terry (ect328) – homework 26 – Turner – (59130) 2 At 1 . 2 s ,the magnitude of the induced...
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This note was uploaded on 02/17/2010 for the course PHY 59130 taught by Professor Turner during the Fall '09 term at University of Texas at Austin.

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HW 26 - terry (ect328) – homework 26 – Turner –...

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