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Unformatted text preview: lim (kl9356) – Homework 15 – Weathers – (22202) 1 This print-out should have 10 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 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. 3 . 61cm 2 . 34 cm 32 . 9cm . 0653A Find the total magnetic flux through the loop. Correct answer: 2 . 14701 × 10 − 9 Wb. Explanation: Let : c = 3 . 61 cm , a = 2 . 34 cm , b = 32 . 9 cm , and I = 0 . 0653 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 vector B is parallel to d vector A , the magnetic flux through an area element dA is Φ ≡ integraldisplay B dA = integraldisplay μ I 2 π r dA . Note: vector 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 I 2 π b integraldisplay a + c c d r r = μ I b 2 π ln r vextendsingle vextendsingle vextendsingle a + c c = μ I b 2 π ln parenleftbigg a + c c parenrightbigg = μ (0 . 0653 A)(0 . 329 m) 2 π ln parenleftbigg a + c c parenrightbigg = μ (0 . 0653 A)(0 . 329 m) 2 π (0 . 499683) = 2 . 14701 × 10 − 9 Wb . 002 10.0 points A solenoid 6 . 6 cm in radius and 69 m in length has 11100 uniformly spaced turns and carries a current of 9 . 7 A. Consider a plane circular surface of radius 1 . 8 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 . 99595 × 10 − 6 Wb....
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