Lemuel physics 2 11 - Wong Lemuel – Homework 11 – Due 11:00 pm – Inst Matt Mackie 1 This print-out should have 16 questions Multiple-choice

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Unformatted text preview: Wong, Lemuel – Homework 11 – Due: Apr 16 2007, 11:00 pm – Inst: Matt Mackie 1 This print-out should have 16 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. NOTE the unusual due TIME: 5pm central, which is 6pm our time. this is meant to allow more time with the solutions before the exam. 001 (part 1 of 1) 10 points A rectangular loop located a distance from a long wire carrying a current is shown in the figure. The wire is parallel to the longest side of the loop. 5 . 5cm 6 . 88 cm 23 . 2cm . 0661A Find the total magnetic flux through the loop. Correct answer: 2 . 48839 × 10- 9 Wb. Explanation: Let : c = 5 . 5 cm , a = 6 . 88 cm , b = 23 . 2 cm , and I = 0 . 0661 A . c a b r dr I From Amp` ere’s law, the strength of the magnetic field created by the current-carrying wire at a distance r from the wire is (see figure.) B = μ I 2 π r , so the field varies over the loop and is directed perpendicular to the page. Since ~ B is parallel to d ~ A , the magnetic flux through an area element dA is Φ ≡ Z B dA = Z μ I 2 π r dA. Note: ~ B is not uniform but rather depends on r , so it cannot be removed from the integral. In order to integrate, we express the area element shaded in the figure as dA = bdr . Since r is the only variable that now appears in the integral, we obtain for the magnetic flux Φ B = μ I 2 π b Z a + c c dr r = μ I b 2 π ln r fl fl fl a + c c = μ I b 2 π ln µ a + c c ¶ = μ (0 . 0661 A)(0 . 232 m) 2 π ln µ a + c c ¶ = μ (0 . 0661 A)(0 . 232 m) 2 π (0 . 811334) = 2 . 48839 × 10- 9 Wb . keywords: 002 (part 1 of 1) 10 points A solenoid 8 . 3 cm in radius and 20 m in length has 14000 uniformly spaced turns and carries a current of 9 . 4 A. Consider a plane circular surface of radius 2 cm located at the center of the solenoid with its axis coincident with the axis of the solenoid. What is the magnetic flux Φ through this circular surface?(1 Wb = 1 T m 2 ) Correct answer: 1 . 03907 × 10- 5 Wb. Wong, Lemuel – Homework 11 – Due: Apr 16 2007, 11:00 pm – Inst: Matt Mackie 2 Explanation: Let : ‘ = 20 m , N = 14000 , I = 9 . 4 A , and r = 2 cm = 0 . 02 m . Basic Concepts: B = μ N I l I = μ nI Φ B = Z ~ B · d ~ A The magnetic field in the solenoid is given by B = μ I N ‘ Call A the area of the plane circular surface. The the magnetic flux through this surface is Φ = B A = μ I N ‘ π r 2 = μ (9 . 4 A)(14000) 20 m π (0 . 02 m) 2 = 1 . 03907 × 10- 5 Wb . The radius of the solenoid is not required since it is smaller that the plane circular surface. keywords: 003 (part 1 of 1) 10 points A coil is wrapped with 119 turns of wire on the perimeter of a circular frame (of radius 32 cm). Each turn has the same area, equal to that of the frame. A uniform magnetic field is directed perpendicular to the plane of the coil. This field changes at a constant rate from 23 mT to 66 mT in 22 ms....
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This note was uploaded on 02/17/2011 for the course PHY 303L taught by Professor Turner during the Spring '08 term at University of Texas at Austin.

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Lemuel physics 2 11 - Wong Lemuel – Homework 11 – Due 11:00 pm – Inst Matt Mackie 1 This print-out should have 16 questions Multiple-choice

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