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: aljabr (faa335) Hw16 Ross (89251) 1 This printout should have 14 questions. Multiplechoice 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 . 89cm 4 . 89 cm 39 . 2cm . 0179A 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` 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 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 . 0179 A)(0 . 392 m) 2 ln parenleftbigg a + c c parenrightbigg = (0 . 0179 A)(0 . 392 m) 2 (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 = 8 . 85419 10 12 C 2 / N m 2 . The electric field between the two plates is E = Q A . aljabr (faa335) Hw16 Ross (89251) 2 Then the rate of change of electric flux be tween the plates is d E dt = d dt ( E A ) = I = . 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 = d E dt = I = . 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 . 3cm 12 cm 26cm (21A)sin bracketleftBig (219rad / s) t + 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....
View
Full
Document
This note was uploaded on 01/27/2011 for the course PHYS 251 taught by Professor Gavrine during the Spring '10 term at Purdue UniversityWest Lafayette.
 Spring '10
 Gavrine

Click to edit the document details