homework16 - Rodriguez Brandon Homework 16 Due 11:00 am...

Info icon This preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
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 . 06 cm 7 . 19 cm 25 . 3 cm 0 . 0887 A 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 0 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 0 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 = 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 Z a + c c d r r = μ 0 I b 2 π ln r fl fl fl a + c c = μ 0 I b 2 π ln a + c c = μ 0 (0 . 0887 A)(0 . 253 m) 2 π ln a + c c = μ 0 (0 . 0887 A)(0 . 253 m) 2 π (1 . 01919) = 4 . 57433 × 10 - 9 Wb .
Image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
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 0 = 8 . 85419 × 10 - 12 C 2 / N m 2 . The electric field between the two plates is E = Q ² 0 A . Then the rate of change of electric flux be- tween the plates is d Φ E dt = d dt ( E A ) = I ² 0 = 0 . 0568 A 8 . 85419 × 10 - 12 C 2 / N m 2 = 6 . 41504 × 10 9 V m / s . 003 (part 2 of 2) 4 points Find the displacement current between the plates. Correct answer: 0 . 0568 A. Explanation: The displacement current is I d = ² 0 d Φ E dt = I = 0 . 0568 A . keywords: 004 (part 1 of 2) 4 points A straight, horizontal rod slides along parallel conducting rails at an angle with the horizon- tal, as shown below. The rails are connected at the bottom by a horizontal rail so that the rod and rails forms a closed rectangular loop.
Image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern