PHY303LHW10S - rosen(arr956 – Homework 10 – Chiu –(58295 1 This print-out should have 27 questions Multiple-choice questions may continue on

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Unformatted text preview: rosen (arr956) – Homework 10 – Chiu – (58295) 1 This print-out should have 27 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A powerful electromagnet has a field of 1 . 56 T and a cross sectional area of 0 . 724 m 2 . Now we place a coil of 176 turns with a total resistance of 2 . 45 Ω around the electromagnet and turn off the power to the electromagnet in 0 . 0154 s. What will be the induced current in the coil? Correct answer: 5268 . 52 a. Explanation: Basic Concepts: Faraday’s Law of Induc- tion: E = − d Φ B dt Ohm’s law: I = V R For coil of many turns, the emf is the sum of induced emf in each turn. Besides, when calculating the induced emf in the coil, we actually are try to get a average emf. For one turn of coil, the induced emf is: E 1 = − d Φ B dt = ΔΦ B Δ t = Δ B · A Δ t = 73 . 3403 V So, the emf in the whole coil is: E = n ·E 1 = 176 turns × 73 . 3403 V = 12907 . 9 V Apply Ohm’s law, we get the current as fol- lowing: I = V R = E R = 12907 . 9 V 2 . 45 Ω = 5268 . 52 a 002 10.0 points A coil with N 1 = 14 turns and radius r 1 = 5 . 18 cm surrounds a long solenoid of radius r 2 = 1 . 15 cm and N 2 ℓ 2 = n 2 = 1300 m − 1 (see the figure below). The current in the solenoid changes as I = I sin( ω t ), where I = 5 A and ω = 120 rad / s. R a b E ℓ 2 ℓ 1 Outside solenoid has N 1 turns Inside solenoid has N 2 turns A 1 A 2 What is the magnitude of the induced emf , E AB , across the 14 turn coil at t = 1000 s? Correct answer: 0 . 00475199 V. Explanation: Faraday’s Law for solenoid E = − N d Φ dt = − N d ( BA ) dt . Magnetic field induced by solenoid B = μ nI . Faraday’s Law for solenoid E = − N d Φ dt = − N d ( BA ) dt . Magnetic field induced by inner solenoid B = μ nI . rosen (arr956) – Homework 10 – Chiu – (58295) 2 So the induced emf is E = − N 1 d Φ dt = − N 1 d ( B A 2 ) dt = − N 1 A 2 dB dt = − N 1 A 2 d ( μ n 2 I ) dt = − μ N 1 n 2 A 2 dI dt = . 00475199 V . 003 10.0 points A long solenoid has a coil made of fine wire inside it and coaxial with it. I Outside solenoid has n turns per meter Inside coil has N turns r R Given a varying current I in the outer solenoid, what is the emf induced in the inner loop? 1. E = − π R 2 μ nN dI dt correct 2. E = − π Rμ N dI dt 3. E = − π r 2 μ n dI dt 4. E = − π r 2 μ nN dI dt 5. E = − π Rμ n dI dt 6. E = − π r μ N dI dt 7. E = − π R 2 μ n dI dt 8. E = − π r μ n dI dt 9. E = − π Rμ nN dI dt 10. E = − π r μ nN dI dt Explanation: The magnetic field of a solenoid is B = μ nI . The magnetic flux is Φ B = B · A = ( μ nI ) ( π R 2 ) , so the emf is E = − d Φ B dt = − π R 2 μ nN dI dt ....
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This note was uploaded on 04/13/2009 for the course PHY 303L taught by Professor Turner during the Spring '08 term at University of Texas at Austin.

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PHY303LHW10S - rosen(arr956 – Homework 10 – Chiu –(58295 1 This print-out should have 27 questions Multiple-choice questions may continue on

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