This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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 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 currentcarrying 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....
View
Full
Document
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.
 Spring '08
 Weathers
 Work

Click to edit the document details