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# 23 - Shie Gary Homework 23 Due 4:00 am Inst Turner This...

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Shie, Gary – Homework 23 – Due: Oct 27 2004, 4:00 am – Inst: Turner 1 This print-out should have 11 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 2) 10 points Given: μ 0 = 1 . 25664 × 10 - 6 N / A 2 . A solenoid has 150 turns of wire uniformly wrapped around an air-filled core, which has a diameter of 11 . 9 mm and a length of 6 . 55 cm. Calculate the self-inductance of the solenoid. Correct answer: 4 . 80104 × 10 - 5 H. Explanation: Given: The number of turns in the solenoid N = 150 turns, the diameter of the solenoid D = 11 . 9 mm, and the length of the solenoid l = 6 . 55 cm. 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. From the text we can use Ampere’s Law to find that 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 ) (150) 2 × π (0 . 0119 m) 2 4 (0 . 0655 m) = 4 . 80104 × 10 - 5 H . 002 (part 2 of 2) 10 points The core is replaced with a soft iron rod that has the same dimensions, but a magnetic per- meability 800 μ 0 . What is the new inductance? Correct answer: 0 . 0384083 H. Explanation: Let L 2 be the inductance of the solenoid with the soft iron core. In the first part of the problem we just need to change μ 0 by μ . Thus L 2 = μ N 2 π D 2 4 = μ μ 0 L 1 = 800 (4 . 80104 × 10 - 5 H) = 0 . 0384083 H . 003 (part 1 of 1) 10 points A 3 . 2 H inductor carries a steady current of 0 . 595 A. When the switch in the circuit is thrown open, the current disappears in 7 . 3 ms.

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