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Unformatted text preview: tovar (jdt436) homework 29 Turner (59070) 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 What current is required in the windings of a long solenoid that has 630 turns uni- formly distributed over a length of 0 . 548 m in order to produce a magnetic field of magnitude 0 . 000285 T at the center of the solenoid? The permeability of free space is 4 10 7 T m / A . Correct answer: 197 . 276 mA. Explanation: Let : N = 630 , L = 0 . 548 m , B = 0 . 000285 T , and = 4 10 7 T m / A . The magnetic field inside a long solenoid is B = n I = parenleftbigg N L parenrightbigg I . Thus the required current is I = B L N = (0 . 000285 T) (0 . 548 m) (4 10 7 T m / A) (630) 1000 mA 1 A = 197 . 276 mA . 002 10.0 points A single piece of wire is bent into the shape of Texas, with a total area of 3 . 41 cm 2 . This Texas shaped loop is perpendicular to a mag- netic field which increases uniformly in mag- nitude from 0 . 415 T to 2 . 19 T in a time of 2 s. The wire has a total resistance of 4 . What is the current? Correct answer: 0 . 0756594 mA. Explanation: Basic Concepts: Faradays Law of Induc- tion: E =- d B dt The particular shape of the wire is unimpor- tant, only the area enclosed by the wire mat- ters. The magnetic flux through the loop is given by = B A . d dt = d dt ( B A ) = A d B dt = A B t = A B 2- B 1 t = (0 . 000341 m 2 )(0 . 8875 T / s) = 0 . 000302637 V . From Faradays Law: E =- d dt =- . 000302637 V From Ohms Law: I = E R = . 000302637 V 4 = 0 . 0756594 mA 003 10.0 points A toroid having a rectangular cross section ( a = 1 . 61 cm by b = 1 . 94 cm) and inner radius 6 . 28 cm consists of N = 420 turns of wire that carries a current I = I sin t , with I = 64 . 8 A and a frequency f = 41 . 5 Hz. A loop that consists of N = 21 turns of wire links the toroid, as in the figure....
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