Answer Week 9 - 2 I |—l b l i Answers Week 9(i The...

Info icon This preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
Image of page 1

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

View Full Document Right Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 2; I |—l b.) l i Answers Week 9 (i) The elementary generators of [3 must be magnetic dipoles whose “size” is close to ZERO. (ii) 5 field lines form closed loops, there is no “beginning” or end. Non-Coulomb g - field lines also form closed loops surrounding the region where , the flux of g is changing with time so total flux of g through any closed Non~Coquinb surface will be zero. a = —0.5 V , bottom end of rod is positive because we need a clockwise “current” I. to oppose the increase in flux of Q out of page. We do not need to apply a force to move the rod. Suppose flip time is At , 5 m f in coil due to flip will be ZWFZB , . ‘ ZflrzB Ag 8 = ~ , causmg current I = — w = _ At _ AIR At 27rrZB V So A = . Q R FLUX “UP” IS REDUCING SO FLOW OF CHARGE will be counter clockwise T 0 make a generator, rotate coil at angular velocity 0) about y-axis. If so, flux offi through coil will vary with time and generate an a m f given by g = '— MB = cuABSi (cot) At for a single turn coil of area A. L=— g -—>.V01tixTime a Amp At ' R : Volt ‘ Amp L So —— ——> Time R When there is a current in an Inductor, there is a g - field inside it. The energy U B : é—L I 2 gets stored in this 13 - field. 9—15 r=1o*4s, Time for current to reach 90 percent of% will be t = 2.3 xlO'4 s, ...
View Full 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