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Unformatted text preview: johnson (rj6247) – hw 11 – Opyrchal – (121014) 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 224 turns of wire on the perimeter of a square frame of sides 21 . 6 cm. Each turn has the same area, equal to that of the frame, and the total resistance of the coil is 2 . 42 Ω. A uniform magnetic field is turned on perpendicular to the plane of the coil. If the field changes linearly from 0 to . 237 Wb / m 2 in a time of 1 . 16 s, find the magnitude of the induced emf in the coil while the field is changing. Correct answer: 2 . 13524 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 = 1 . 16 s , the magnetic flux through the loop is Φ B = B A = . 0110575 Wb . Therefore the magnitude of the induced emf is E = N · ΔΦ B Δ t = (224 turns) [( . 0110575 Wb) 0] (1 . 16 s) = 2 . 13524 V E = 2 . 13524 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 49 . 6 turns and a total resistance of 7 . 21 Ω. At what rate must the magnitude of B change in order to induce a current of 0 . 104 A in the windings of the coil? Correct answer: 3 . 77944 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 . 104 A) (7 . 21 Ω) (49 . 6 turns) = 0 . 0151177 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 . 0151177 V) (0 . 004 m 2 ) = 3 . 77944 T / s . 003 (part 1 of 2) 10.0 points The clockwise circulating current in a solenoid is increasing at a rate of 4 . 88 A / s. The cross sectional area of the solenoid is 3 . 14159 cm 2 , and there are 464 turns on its 12 . 8 cm length. What is the magnitude of the induced E produced by the increasing current? Correct answer: 3 . 24045 mV. Explanation: Faraday’s Law for solenoid: E = N d Φ dt = N A d B dt . Magnetic field produced by the changing current is B = μ N I L johnson (rj6247) – hw 11 – Opyrchal – (121014) 2 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 ) (464) 2 12 . 8 cm × (3 . 14159 cm 2 ) (4 . 88 A / s) × parenleftbigg 10 3 mV V parenrightbiggparenleftbigg 1 10 2 m cm parenrightbigg = 3 . 24045 mV ....
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This note was uploaded on 03/11/2010 for the course PHYS 121 taught by Professor Opyrchal during the Spring '08 term at NJIT.
 Spring '08
 Opyrchal
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