23 - Shie, Gary Homework 23 Due: Oct 27 2004, 4:00 am Inst:...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 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: = 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 I , where 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 Amperes Law to find that the magnetic field in the solenoid is B = N I , where = is the magnetic permeability of the air core which is the same as free space. The magnetic flux in the solenoid is = B A = B D 2 4 . Using the above expressions for and B , we obtain for the inductance of the solenoid L 1 = N B A I = N 2 D 2 4 = (1 . 25664 10- 6 N / A 2 )(150) 2 (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 . 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 by . Thus L 2 = N 2 D 2 4 = 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....
View Full Document

This note was uploaded on 01/20/2010 for the course PHY 1 taught by Professor Erskine during the Spring '10 term at University of Texas at Austin.

Page1 / 5

23 - Shie, Gary Homework 23 Due: Oct 27 2004, 4:00 am Inst:...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online