OHW 25 - terry(ect328 oldhomework 25 Turner(59130 This...

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

View Full Document Right Arrow Icon
terry (ect328) – oldhomework 25 – Turner – (59130) 1 This print-out should have 13 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 10.0 points A rectangular loop coplanar to and located a distance 3 . 46 cm from a long wire carrying a current of 0 . 0905 A is shown in the figure. The wire is parallel to the longest side of the loop. 3 . 46 cm 7 . 38 cm 30 . 8 cm 0 . 0905 A Find the total magnetic flux through the loop. Correct answer: 6 . 36628 × 10 9 Wb. Explanation: Let : c = 3 . 46 cm , a = 7 . 38 cm , b = 30 . 8 cm , and I = 0 . 0905 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 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 (it depends on r ), so it cannot be removed from the integral. In order to integrate, the area element shaded in the figure as dA = b dr . Since r is the only variable that now appears in the integral, the magnetic flux is 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 . 0905 A)(0 . 308 m) 2 π ln parenleftbigg a + c c parenrightbigg = μ 0 (0 . 0905 A)(0 . 308 m) 2 π (1 . 14197) = 6 . 36628 × 10 9 Wb . 002 (part 2 of 2) 10.0 points What is the direction of the magnetic field through the rectangular loop? 1. into the plane of the paper correct 2. Parallel to the plane of the paper and parallel to the current direction in the wire. 3. out of the plane of the paper 4. Parallel to the plane of the paper and perpendicular to the current direction in the wire. Explanation:
Image of page 1

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

View Full Document Right Arrow Icon
terry (ect328) – oldhomework 25 – Turner – (59130) 2 Using the right-hand rule on the long straight wire, if the current flows upward (downward), the magnetic field would be into (out of) the plane of the paper. Consequently the magnetic field is directed into the plane of the paper. 003 (part 1 of 4) 10.0 points The circular loop of wire shown in the figure is placed in a spatially uniform magnetic field such that the plane of the circular loop is per- pendicular to the direction for the magnetic field as shown in the figure. The magnetic field vector B ( t ) varies with time, with the time de- pendence given by B ( t ) = a + b t , where a = 0 . 17 T and b = 0 . 028 T / s. The acceleration due to gravity is 9 . 8 m / s 2 . r = 7 . 9 cm radius B ( t ) What is the magnetic flux through the loop as a function of time? 1. Φ B ( t ) = 2 a π r 2. Φ B ( t ) = b π r 2 3. Φ B ( t ) = 2 ( a + b ) π r 4. Φ B ( t ) = 2 parenleftBig a t + b parenrightBig π r 5. Φ B ( t ) = parenleftBig a t + b parenrightBig π r 2 6. Φ B ( t ) = a π r 2 7. Φ B ( t ) = ( a + b ) π r 2 8. Φ B ( t ) = 2 ( a + b t ) π r 9. Φ B ( t ) = ( a + b t ) π r 2 correct 10. Φ B ( t ) = 2 b π r Explanation: Faraday’s Law of Induction E ind = - d Φ B dt = - ΔΦ B Δ t .
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