hw14a - aldawsari (ma28683) HW14 Betancourt (16856) 1 This...

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Unformatted text preview: aldawsari (ma28683) HW14 Betancourt (16856) 1 This print-out should have 14 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 29.4-29.7 001 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. 4 . 86cm 6 . 03 cm 34 . 3cm . 0292A Find the total magnetic flux through the loop. Correct answer: 1 . 61613 10 9 Wb. Explanation: Let : c = 4 . 86 cm , a = 6 . 03 cm , b = 34 . 3 cm , and I = 0 . 0292 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 vector B is parallel to d vector A , the magnetic flux through an area element dA is integraldisplay B dA = integraldisplay I 2 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 B = I 2 b integraldisplay a + c c d r r = I b 2 ln r vextendsingle vextendsingle vextendsingle a + c c = I b 2 ln parenleftbigg a + c c parenrightbigg = (0 . 0292 A)(0 . 343 m) 2 ln parenleftbigg a + c c parenrightbigg = (0 . 0292 A)(0 . 343 m) 2 (0 . 806806) = 1 . 61613 10 9 Wb . 002 (part 1 of 2) 10.0 points A 0 . 1481 A current is charging a capacitor that has square plates, 5 . 52 cm on a side. The permitivity of free space is 8 . 85419 10 12 C 2 / N m 2 . If the plate separation is 3 . 6 mm find the time rate of change of electric flux between the plates. Correct answer: 1 . 67265 10 10 V m / s. Explanation: Let : I = 0 . 1481 A and = 8 . 85419 10 12 C 2 / N m 2 . aldawsari (ma28683) HW14 Betancourt (16856) 2 The electric field between the two plates is E = Q A . Then the rate of change of electric flux be- tween the plates is d E dt = d dt ( E A ) = I = . 1481 A 8 . 85419 10 12 C 2 / N m 2 = 1 . 67265 10 10 V m / s . 003 (part 2 of 2) 10.0 points Find the displacement current between the plates. Correct answer: 0 . 1481 A. Explanation: The displacement current is I d = d E dt = I = . 1481 A . 004 (part 1 of 2) 10.0 points A long, straight wire carries a current and lies in the plane of a rectangular loops of wire, as shown in the figure. 2cm 23 cm 8 . 7cm (53A)sin bracketleftBig (207rad / s) t + bracketrightBig 61 loops (turns) Determine the maximum emf , |E| , induced in the loop by the magnetic field created by the current in the straight wire....
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This note was uploaded on 01/27/2011 for the course PHYS 251 taught by Professor Gavrine during the Spring '10 term at Purdue University-West Lafayette.

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hw14a - aldawsari (ma28683) HW14 Betancourt (16856) 1 This...

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