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Practice Homework 16 2010

Practice Homework 16 2010 - choe(dec865 Practice HW 16...

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choe (dec865) – Practice HW 16 Solutions – Weathers – (22202) 1 This print-out should have 10 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A coil is wrapped with 207 turns of wire on the perimeter of a square frame of sides 20 cm. Each turn has the same area, equal to that of the frame, and the total resistance of the coil is 4 . 66 Ω. A uniform magnetic field is turned on perpendicular to the plane of the coil. If the field changes linearly from 0 to 0 . 737 Wb / m 2 in a time of 0 . 798 s, find the magnitude of the induced emf in the coil while the field is changing. Correct answer: 7 . 64707 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 = 0 . 798 s , the magnetic flux through the loop is Φ B = B A = 0 . 02948 Wb . Therefore the magnitude of the induced emf is E = N · ΔΦ B Δ t = (207 turns) [(0 . 02948 Wb) 0] (0 . 798 s) = 7 . 64707 V |E| = 7 . 64707 V . 002 10.0 points The two-loop wire circuit is 117 . 543 cm wide and 78 . 362 cm high. The wire circuit in the figure is located in a magnetic field whose magnitude varies with time according to the expression B = (0 . 001 T / s) t and its direction is into the page. Assume The resistance per length of the wire is 0 . 0519 Ω / m. B B P Q 39 . 181 cm 78 . 362 cm 78 . 362 cm When the magnetic field is 0 . 5 T, find the magnitude of the current through middle leg PQ of the circuit. Correct answer: 1258 . 22 μ A. Explanation: Let : = 0 . 78362 m , A l = 1 2 2 = 0 . 61406 m 2 / s , A r = 2 = 1 . 22812 m 2 / s , δ = 0 . 0519 Ω / m , and d B dt = d dt α t = α = 0 . 001 T / s . Basic Concept: Faraday’s Law is E = d Φ B dt , where Φ B = A B . Ohm’s Law is V = I R . Solution: The instantaneous value of the magnetic field ( B = 0 . 5 T) is not germane to this problem.
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choe (dec865) – Practice HW 16 Solutions – Weathers – (22202) 2 B B P Q 2 I r I l I PQ The resistance for the wire is proportional to the length of the wire. For a length of 78 . 362 cm, the resistance is R = δ ℓ = (0 . 0519 Ω / m) (0 . 78362 m) = 0 . 0406699 Ω . Using Ohm’s law for the right perimeter and left perimeter of the left loops in the circuit, we have E r = 4 ℓ δ I r = 4 R I r , and (1) E l = 3 ℓ δ I l = 3 R I l . (2) Note: When the magnetic field changes with time, there is an induced emf in both the right-hand side and the left-hand side. From Faraday’s law, the magnitude of the induced emf in the loops equals E r = A r d B dt = 2 d B dt ,
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