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Unformatted text preview: Rodriguez, Brandon – Homework 15 – Due: Oct 12 2006, 11:00 am – Inst: Andrew J Rader 1 This printout should have 11 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. The due time is Central time. Please notice that for your homework to be considered towards your grade, it needs to be submitted one hour before the corresponding recitation starts. Work submitted after this time, but before the DUE DATE on top of this page, will be accepted but not graded. PLEASE REMEMBER THAT YOU MUST CARRY OUT YOUR CALCULA TIONS TO AT LEAST THREE SIGNIFI CANT FIGURES. YOUR ANSWER MUST BE WITHIN ONE PERCENT OF THE CORRECT RESULT TO BE MARKED AS CORRECT BY THE SERVER. 001 (part 1 of 1) 6 points A 65 turns circular coil with a radius 5 . 85 cm and a resistance 6 . 07 Ω is placed in a magnetic field directed perpendicular to the plane of the coil. The magnitude of the magnetic field varies in time according to the expression B = a 1 t + a 2 t 2 , where a 1 = 0 . 0898 T / s, a 2 = 0 . 096 T / s 2 are constants, time t is in seconds and field B is in Tesla. Calculate the magnitude of the induced E in the coil at t = 6 . 27 s. Correct answer: 0 . 904042 V. Explanation: Basic Concepts: Faraday’s Law of Induc tion E = d Φ B dt Note: The resistance of the coil is unnecessary when calculating the induced E . The area of the circular coil is: A = π r 2 = π (0 . 0585 m) 2 = 0 . 0107513 m 2 , from Faraday’s law, we get E = n d Φ B dt = n dAB dt = nA dB dt = (65 turns)(0 . 0107513 m 2 ) × (1 . 29364 T / s) = . 904042 V E = 0 . 904042 V . keywords: 002 (part 1 of 1) 6 points A circular wire loop of radius 9 . 2 m lies in a plane perpendicular to a uniform magnetic field of magnitude 0 . 4 T. In 0 . 1 s the wire is reshaped into a square but remains in the same plane. What is the magnitude of the average in duced emf in the wire during this time? Correct answer: 228 . 254 V. Explanation: Basic Concepts: Faraday’s Law of Induc tion: E = d Φ B dt When the loop is reshaped, the length of the wire remain unchanged. So we can get the side length of the square a = 2 π r 4 = 14 . 4513 m and the area difference is Δ A = Δ A r Δ A a = π r 2 a 2 = 57 . 0636 m 2 . Then using Faraday’s Law, we can get the average emf. E = d Φ B dt = ΔΦ B t = B Δ A t = 228 . 254 V . Rodriguez, Brandon – Homework 15 – Due: Oct 12 2006, 11:00 am – Inst: Andrew J Rader 2 keywords: 003 (part 1 of 2) 5 points In the arrangement shown in the figure, the resistor is 3 Ω and a 5 T magnetic field...
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This note was uploaded on 10/18/2010 for the course PHYSICS 251 taught by Professor Pedon during the Spring '08 term at IUPUI.
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