Practice HW 16 solution - song(shs546 – Practice HW 16...

Unformatted text preview: song (shs546) – 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 0.0 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 0.0 points The two-loop wire circuit is 43 . 7559 cm wide and 29 . 1706 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 . 113 Ω / m. B B P Q 14 . 5853 cm 29 . 1706 cm 29 . 1706 cm When the magnetic field is 0 . 5 T, find the magnitude of the current through middle leg PQ of the circuit. Correct answer: 215 . 122 μ A. Explanation: Let : ℓ = 0 . 291706 m , A l = 1 2 ℓ 2 = 0 . 0850924 m 2 / s , A r = ℓ 2 = 0 . 170185 m 2 / s , δ = 0 . 113 Ω / m , and dB dt = d dt αt = 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. song (shs546) – Practice HW 16 Solutions – weathers – (22202) 2 B B P Q ℓ 2 ℓ ℓ I r I l I PQ The resistance for the wire is proportional to the length of the wire. For a length of 29 . 1706 cm, the resistance is R = δ ℓ = (0 . 113 Ω / m) (0 . 291706m) = 0 . 0329628 Ω ....
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