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Unformatted text preview: taherisefat (mt23852) – hw 11 – Opyrchal – (121012) 1 This printout should have 15 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A coil is wrapped with 285 turns of wire on the perimeter of a square frame of sides 19 . 3 cm. Each turn has the same area, equal to that of the frame, and the total resistance of the coil is 4 . 59 Ω. A uniform magnetic field is turned on perpendicular to the plane of the coil. If the field changes linearly from 0 to . 391 Wb / m 2 in a time of 0 . 627 s, find the magnitude of the induced emf in the coil while the field is changing. Correct answer: 6 . 62016 V. Explanation: Basic Concept: Faraday’s Law is E = d Φ B dt . Solution: The magnetic flux through the loop at t = 0 is zero since B = 0. At t = . 627 s , the magnetic flux through the loop is Φ B = B A = . 0145644 Wb . Therefore the magnitude of the induced emf is E = N · ΔΦ B Δ t = (285 turns) [( . 0145644 Wb) 0] (0 . 627 s) = 6 . 62016 V E = 6 . 62016 V . 002 10.0 points The plane of a rectangular coil of dimension 5 cm by 8 cm is perpendicular to the direction of a magnetic field B . The coil has 136 turns and a total resistance of 10 . 4 Ω. At what rate must the magnitude of B change in order to induce a current of 0 . 275 A in the windings of the coil? Correct answer: 5 . 25735 T / s. Explanation: Basic Concepts: Faraday’s Law of Induc tion E = d Φ B dt Ohm’s law I = V R Solution: With Ohm’s law, the emf induced in one turn of coil is E 1 = I R n = (0 . 275 A) (10 . 4 Ω) (136 turns) = 0 . 0210294 V , while with Faraday’s law, we get E 1 = d Φ B dt = d ( A · B ) dt = A d B dt . So, the rate at which magnetic field changes will be d B dt = E 1 A = (0 . 0210294 V) (0 . 004 m 2 ) = 5 . 25735 T / s . 003 (part 1 of 2) 10.0 points The counterclockwise circulating current in a solenoid is increasing at a rate of 10 . 4 A / s. The crosssectional area of the solenoid is 3 . 14159 cm 2 , and there are 421 turns on its 17 . 8 cm length. What is the magnitude of the induced E produced by the increasing current? Correct answer: 4 . 08825 mV. Explanation: Faraday’s Law for solenoid: E = N d Φ dt = N A d B dt . taherisefat (mt23852) – hw 11 – Opyrchal – (121012) 2 Magnetic field produced by the changing current is B = μ N I L d B d t = μ N L d I d t . Faraday’s Law for solenoid: E = N d Φ dt = N d ( BA ) dt = N 2 A L μ d I d t . Magnetic field induced by current: B = μ N I L . Thus, the induced E is E = μ N 2 L A d I dt = (1 . 25664 × 10 6 N / A 2 ) (421) 2 17 . 8 cm × (3 . 14159 cm 2 ) (10 . 4 A / s) × parenleftbigg 10 3 mV V parenrightbiggparenleftbigg 1 10 2 m cm parenrightbigg = 4 . 08825 mV ....
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This note was uploaded on 01/31/2011 for the course PHYS 111 taught by Professor Moro during the Spring '08 term at NJIT.
 Spring '08
 moro

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