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# enm hw - Assign Ten Homework 9 Due 2:00 pm Inst Richard...

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Assign, Ten – Homework 9 – Due: Dec 11 2003, 2:00 pm – Inst: Richard Saenz 1 This print-out should have 34 questions. Multiple-choice questions may continue on the next column or page – find all choices before making your selection. The due time is Central time. these problem are extra credit homework, but the material will be in the final 001 (part 1 of 2) 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. 6 . 45 cm 4 . 2 cm 18 cm 0 . 0915 A Find the total magnetic flux through the loop. Correct answer: 1 . 65187 × 10 9 Wb. Explanation: Let : c = 6 . 45 cm , a = 4 . 2 cm , b = 18 cm , and I = 0 . 0915 A . c a b r dr I From Amp` ere’s law, we know that the strength of the magnetic field created by the current-carrying wire at a distance r from the wire is (see figure.) B = μ 0 I 2 π r , so the field varies over the loop and is directed perpendicular to the page. Since B is paral- lel to d A , we can express the magnetic flux through an area element dA as Φ B dA = μ 0 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 = b dr . Since r is the only variable that now appears in the integral, we obtain for the magnetic flux Φ B = μ 0 I 2 π b a + c c d r r = μ 0 I b 2 π ln r a + c c = μ 0 I b 2 π ln a + c c = μ 0 (0 . 0915 A)(0 . 18 m) 2 π ln a + c c = μ 0 (0 . 0915 A)(0 . 18 m) 2 π (0 . 50148) = 1 . 65187 × 10 9 Wb . 002 (part 2 of 2) 10 points What is the direction of the magnetic field through the rectangular loop? 1. out of the plane of the paper 2. into the plane of the paper 3. cannot be determined with information given correct

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Assign, Ten – Homework 9 – Due: Dec 11 2003, 2:00 pm – Inst: Richard Saenz 2 Explanation: The direction of the current is not given, hence the absolute direction of the magnetic field cannot be determined, although the mag- netic field is perpendicular to the plane of the paper. For example if the current flows up- ward (downward) the magnetic field would be into (out of) the plane of the paper. 003 (part 1 of 1) 10 points A coil is wrapped with 198 turns of wire on the perimeter of a square frame of sides 34 . 4 cm. Each turn has the same area, equal to that of the frame, and the total resistance of the coil is 1 . 74 . A uniform magnetic field is turned on perpendicular to the plane of the coil. If the field changes linearly from 0 to 0 . 908 Wb / m 2 in a time of 1 . 13 s, find the mag- nitude of the induced emf in the coil while the field is changing. Correct answer: 18 . 8274 V. Explanation: Basic Concept: Faraday’s Law is E = d Φ B dt . Solution: The magnetic flux through the loop at t = 0 is zero since B = 0. At t = 1 . 13 s , the magnetic flux through the loop is Φ B = B A = 0 . 107449 Wb . Therefore the magnitude of the induced emf is E = N · ∆Φ B t = (198 turns) [(0 . 107449 Wb) 0] (1 . 13 s) = 18 . 8274 V |E| = 18 . 8274 V .
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