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Unformatted text preview: Patel, Kinal – Homework 14 – Due: Mar 14 2008, 7:00 pm – Inst: Weathers 1 This printout should have 20 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points A solenoid 7 . 6 cm in radius and 69 m in length has 6100 uniformly spaced turns and carries a current of 1 . 1 A. Consider a plane circular surface of radius 2 . 4 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: 2 . 21134 × 10 7 Wb. Explanation: Let : ‘ = 69 m , N = 6100 , I = 1 . 1 A , and r = 2 . 4 cm = 0 . 024 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 (1 . 1 A)(6100) 69 m π (0 . 024 m) 2 = 2 . 21134 × 10 7 Wb . The radius of the solenoid is not required since it is smaller that the plane circular surface. keywords: 002 (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. 7 . 41cm 5 . 12 cm 30 . 3cm . 0383A Find the total magnetic flux through the loop. Correct answer: 1 . 2192 × 10 9 Wb. Explanation: Let : c = 7 . 41 cm , a = 5 . 12 cm , b = 30 . 3 cm , and I = 0 . 0383 A . c a b r dr I From Amp` ere’s law, the strength of the magnetic field created by the currentcarrying 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 Patel, Kinal – Homework 14 – Due: Mar 14 2008, 7:00 pm – Inst: Weathers 2 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 . 0383 A)(0 . 303 m) 2 π ln µ a + c c ¶ = μ (0 . 0383 A)(0 . 303 m) 2 π (0 . 525295) = 1 . 2192 × 10 9 Wb . keywords: 003 (part 1 of 2) 10 points A 0 . 1553 A current is charging a capacitor that has circular plates, 11 . 19 cm in radius. The plate separation is 4 . 53 mm. The permitivity or free space is 8 . 85 × 10 12 and the permeability of free space is 1 . 25664 × 10 6 T · m / A....
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 Spring '08
 Weathers
 Work, Magnetic Field, patel

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