solutions 14 - Patel, Kinal Homework 14 Due: Mar 14 2008,...

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Unformatted text preview: Patel, Kinal Homework 14 Due: Mar 14 2008, 7:00 pm Inst: Weathers 1 This print-out should have 20 questions. Multiple-choice 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 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 = N I l I = nI B = Z ~ B d ~ A The magnetic field in the solenoid is given by B = I N Call A the area of the plane circular surface. The the magnetic flux through this surface is = B A = I N r 2 = (1 . 1 A)(6100) 69 m (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` eres law, the strength of the magnetic field created by the current-carrying wire at a distance r from the wire is (see figure.) B = I 2 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 Z B dA = Z I 2 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 B = I 2 b Z a + c c dr r = I b 2 ln r fl fl fl a + c c = I b 2 ln a + c c = (0 . 0383 A)(0 . 303 m) 2 ln a + c c = (0 . 0383 A)(0 . 303 m) 2 (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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This note was uploaded on 10/24/2010 for the course PHYS PHYS 2220 taught by Professor Weathers during the Spring '08 term at North Texas.

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solutions 14 - Patel, Kinal Homework 14 Due: Mar 14 2008,...

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