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hw16b - aljabr(faa335 Hw16 Ross(89251 This print-out should...

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aljabr (faa335) – Hw16 – Ross – (89251) 1 This print-out should have 14 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. 7 . 89 cm 4 . 89 cm 39 . 2 cm 0 . 0179 A Find the total magnetic flux through the loop. Correct answer: 6 . 7682 × 10 10 Wb. Explanation: Let : c = 7 . 89 cm , a = 4 . 89 cm , b = 39 . 2 cm , and I = 0 . 0179 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 0 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 0 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 0 I 2 π b integraldisplay a + c c d r r = μ 0 I b 2 π ln r vextendsingle vextendsingle vextendsingle a + c c = μ 0 I b 2 π ln parenleftbigg a + c c parenrightbigg = μ 0 (0 . 0179 A)(0 . 392 m) 2 π ln parenleftbigg a + c c parenrightbigg = μ 0 (0 . 0179 A)(0 . 392 m) 2 π (0 . 482285) = 6 . 7682 × 10 10 Wb . 002 (part 1 of 2) 10.0 points A 0 . 0643 A current is charging a capacitor that has square plates, 3 . 23 cm on a side. The permitivity of free space is 8 . 85419 × 10 12 C 2 / N m 2 . If the plate separation is 2 . 35 mm find the time rate of change of electric flux between the plates. Correct answer: 7 . 2621 × 10 9 V m / s. Explanation: Let : I = 0 . 0643 A and ǫ 0 = 8 . 85419 × 10 12 C 2 / N m 2 . The electric field between the two plates is E = Q ǫ 0 A .
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aljabr (faa335) – Hw16 – Ross – (89251) 2 Then the rate of change of electric flux be- tween the plates is d Φ E dt = d dt ( E A ) = I ǫ 0 = 0 . 0643 A 8 . 85419 × 10 12 C 2 / N m 2 = 7 . 2621 × 10 9 V m / s . 003 (part 2 of 2) 10.0 points Find the displacement current between the plates. Correct answer: 0 . 0643 A. Explanation: The displacement current is I d = ǫ 0 d Φ E dt = I = 0 . 0643 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. 7 . 3 cm 12 cm 26 cm (21 A) sin bracketleftBig (219 rad / s) t + δ bracketrightBig 140 loops (turns) Determine the maximum emf , |E| , induced in the loop by the magnetic field created by the current in the straight wire. Correct answer: 32 . 551 mV. Explanation: Let : N = 140 , ω = 219 rad / s , = 26 cm , a = 7 . 3 cm , b = 12 cm , and I 0 = 21 A . a b r dr I = I 0 sin( ω t + δ ) Magnetic field near a long wire is B = μ 0 I 2 π r . Faraday’s Law is E = - d Φ B dt . The magnitude of the magnetic field is B = μ 0 I 2 π r .
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