{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

HW 13 solution - song(shs546 Homework 13 weathers(22202...

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

View Full Document Right Arrow Icon
song (shs546) – Homework 13 – weathers – (22202) 1 This print-out should have 11 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001(part1of3)10.0points An infinite wire is bent as shown in the figure. The current is I . It consists of 3 segments AB , BDE , and EF . AB is vertical with A extending to infinity. EF is horizontal with F extending to infinity. BDE is one-quarter of a circular arc. We use the Roman numerals I, II, III and IV to denote the direction of a vector in the xy plane, which begins at the origin and is pointing into the quadrants I, II, III and IV respectively. B O D C E F A O y x I II III IV The direction of the magnetic vector at O due to the current segment BD , that due to DE and that due to EF are given by respectively (“in, out” mean “into, or out of”the xy plane) 1. I, II, B ED = in 2. B BD = in, B DE = out, B ED = out 3. B BD = in, B DE = in, B ED = in 4. IV, IV, B ED = out 5. B BD = out, B DE = in, B ED = out 6. B BD = out, B DE = out, B ED = in 7. B BD = in, B DE = in, B ED = out 8. B BD = in, B DE = out, B ED = in 9. B BD = out, B DE = out, B ED = out correct 10. B BD = out, B DE = in, B ED = in Explanation: The Biot-Savart Law is given by d vector B = μ 0 4 π I d vector l × ˆ r r 2 . Since both d vector l and ˆ r lie in the xy plane, their cross product must be perpendicular to this plane. Apply the Biot-Savart Law to each line element of the wire, d vector B is found always to direct out the xy plane at O . 002(part2of3)10.0points What is the magnitude of vector B at the center of the arc due to the arc BDE alone? 1. B = μ 0 I 2 r 2. B = μ 0 I π r 3. B = μ 0 I 8 π r 4. B = μ 0 I 4 π r 5. B = μ 0 I r 6. B = μ 0 2 π I r 7. B = μ 0 I 8 r correct 8. B = μ 0 I 4 r 9. B = μ 0 π I r 10. B = μ 0 I 2 π r Explanation: Note: The distance r from a current el- ement of BDE to O is a constant and the current element I d vector l is always perpendicular to ˆ r . Hence, the magnitude of the vector B at the center of the arc due to the arc BDC alone is
Image of page 1

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

View Full Document Right Arrow Icon
song (shs546) – Homework 13 – weathers – (22202) 2 B BDE = μ 0 4 π I r 2 integraldisplay dl = μ 0 4 π I r 2 π 2 r = μ 0 I 8 r . 003(part3of3)10.0points Let r = 0 . 14 m and I = 0 . 29 A. The permeability of free space is 1 . 25664 × 10 6 N / A 2 . Find the magnitude of vector B at O due to the entire wire ABDEF ? Correct answer: 7 . 39665 × 10 7 T. Explanation: Let : μ 0 = 1 . 25664 × 10 6 N / A 2 , I = 0 . 29 A , and r = 0 . 14 m . First calculate the magnetic field of a straight wire carrying a current(See figure be- low). I r s y y O θ B From the Biot-Savart Law, the contribution to the magnetic field at O due to a current element I y at y is given by B = μ 0 4 π I y s 2 sin θ = μ 0 I 4 π θ r sin θ . Upon integration, the magnetic field con- tributed by a current segment from θ 1 to θ 2 is given by B = μ 0 I 4 π r (cos θ 1 cos θ 2 ) For the wire segment AB , we set θ 1 = π 2 and θ 2 = π . Hence B AB = μ 0 I 4 π r parenleftBig cos π 2 cos π parenrightBig = μ 0 I 4 π r .
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