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Unformatted text preview: Rodriguez, Brandon – Homework 16 – Due: Oct 17 2006, 11:00 am – Inst: Andrew J Rader 1 This print-out should have 14 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. Please notice that for your homework to be considered towards your grade, it needs to be submitted one hour before the corresponding recitation starts. Work submitted after this time, but before the DUE DATE on top of this page, will be accepted but not graded. PLEASE REMEMBER THAT YOU MUST CARRY OUT YOUR CALCULA- TIONS TO AT LEAST THREE SIGNIFI- CANT FIGURES. YOUR ANSWER MUST BE WITHIN ONE PERCENT OF THE CORRECT RESULT TO BE MARKED AS CORRECT BY THE SERVER. 001 (part 1 of 1) 6 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 . 06cm 7 . 19 cm 25 . 3cm . 0887A Find the total magnetic flux through the loop. Correct answer: 4 . 57433 × 10- 9 Wb. Explanation: Let : c = 4 . 06 cm , a = 7 . 19 cm , b = 25 . 3 cm , and I = 0 . 0887 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 ~ B is parallel to d ~ A , the magnetic flux through an area element dA is Φ ≡ Z B dA = Z μ 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 = bdr . 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 Z a + c c dr r = μ I b 2 π ln r fl fl fl a + c c = μ I b 2 π ln µ a + c c ¶ = μ (0 . 0887 A)(0 . 253 m) 2 π ln µ a + c c ¶ = μ (0 . 0887 A)(0 . 253 m) 2 π (1 . 01919) = 4 . 57433 × 10- 9 Wb . Rodriguez, Brandon – Homework 16 – Due: Oct 17 2006, 11:00 am – Inst: Andrew J Rader 2 keywords: 002 (part 1 of 2) 4 points A 0 . 0568 A current is charging a capacitor that has square plates, 5 . 59 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 . 33 mm find the time rate of change of electric flux between the plates. Correct answer: 6 . 41504 × 10 9 V m / s. Explanation: Let : I = 0 . 0568 A and ² = 8 . 85419 × 10- 12 C 2 / N m 2 . The electric field between the two plates is E = Q ² A . Then the rate of change of electric flux be- tween the plates is d Φ E dt = d dt ( E A ) = I ² = . 0568 A 8 . 85419 × 10- 12 C 2 / N m 2 = 6 . 41504 × 10 9 V m / s ....
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This note was uploaded on 10/18/2010 for the course PHYSICS 251 taught by Professor Pedon during the Spring '08 term at IUPUI.
- Spring '08