This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: terry (ect328) – homework 26 – Turner – (59130) 1 This printout should have 11 questions. Multiplechoice 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 airfilled 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 selfinductance 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 selfinductance 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...
View
Full
Document
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.
 Fall '09
 Turner

Click to edit the document details