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Electromagnetic induction

  1. May 25, 2014 #1
    1. The problem statement, all variables and given/known data
    Thin conducting disc, radius a, thickness b and resistivity p (assumed to be large enough induced currents produce no magnetic field). There is a uniform B field B0sin(wt) parallel to its axis.

    I first had to find the electric field a distance r from the disc axis in the plane of the disc. I did this by equating the induced emf to the integral of E.dl around a closed circular loop which gave E=-a2B0wcoswt/2r. Then using J=E/p, I had to find the induced current density, giving J=-a2B0wcoswt/2pr. This may be wrong.

    I now must find an expression for the time averaged power dissipated as heat over the disc.

    2. Relevant equations
    P=I2R

    3. The attempt at a solution
    I would like to find the resistance of the disc first, however I'm uneasy about this. The current flows in circles, so a thickness dr of the disc has resistance dR=2πrp/(bdr). I'm obviously going wrong there. I know how to find the resistance of a disc when the current flows radially outwards, but it should be different here right?

    Once I have that, I guess I can just integrate out the current density to get the total current at an instance, find the instantaneous power from P=I2R and then time average it over a cycle.
     
    Last edited: May 25, 2014
  2. jcsd
  3. May 25, 2014 #2


    since current flow and direction depend on drift velocity, so we cant expect the current to flow in circles. If such was the case the current flowing in the wire should also be circular because a wire is made up of a collection of disc. hence this cannot be the case here. I guess length should be equal to disc thickness 'b' and the circular part as the area of cross section.

    I Hope this helps
     
  4. May 25, 2014 #3
    In which way do you suppose current is flowing then? I think the question expects such an assumption, because the current density is calculated as to depend only on r.
     
  5. May 25, 2014 #4

    rude man

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    Why did you use the area of the entire disc in computing E around a path with radius r < a?
     
  6. May 25, 2014 #5
    Integral of E.dl around a circle of radius r is 2πrE. The induced emf is πa2B0wcoswt. Equating gives E as above. Where did I use the entire area?
     
  7. May 25, 2014 #6

    rude man

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    The induced emf is -d(flux)/dt where flux is B times the area circumscribed by your integration path.
     
  8. May 25, 2014 #7

    rude man

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    πa2 is the entire area of the disc.
     
  9. May 25, 2014 #8
    Ah why is that? I'm struggling to understand things so well in this scenario because it's not a clear cut circuit.

    Anyway, letting a->r in my previous results gives
    E=-rB0wcoswt/2
    J=-rB0wcoswt/2p.
     
  10. May 25, 2014 #9

    rude man

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    It is a clear-cut circuit. Your integration path is 2πr and your area is πr2 and your dB/dt is what you have.

    Which is much better! At least that way the E field does not → ∞ for small r! :-)

    I think J is correct also.

    So, for power, consider an annulus of radius r, width dr, thickness b, and you know J, so what is the current in this thin annulus? What is the dissipated power dP(r)?

    Then, integrate P(r) from r=0 to r=a.
     
  11. May 25, 2014 #10
    If I had a square wire loop, I would think of that as a clear cut circuit, because all of the flux through the circuit contributes to a single current around the loop.

    Here there's effectively many rings of current going around. Am I effectively treating each ring as a circular wire loop or radius r, say, thus the flux that contributes to it is that the links the loop of radius r only?

    And yes, I should have checked that :redface:

    Current is Jbdr. Power would then be Jbdr * πr2B0wcoswt i.e dP=VdI. Then I need to integrate it which I shall do now...

    Which gives
    Instantaneously: P=πbB02w2a4cos2(wt)/8p
    Averaged over a cycle: P=πbB02w2a4/16p
     
    Last edited: May 25, 2014
  12. May 25, 2014 #11

    rude man

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    OK, I went dP = (dI)2 R
    with dI = Jb dr so
    dP = J2b2(dr)2(2πrρ/b dr) so
    dP = 2πρrJ2b dr
    P = 2πρb∫0a J2(r) r dr
    but I did it quickly & may have slipped up. I did not do the integral. Ran out of time ...

    You could repeat your calculations using P = I2R instead of P=VR, although they should yield the same. In any case, you are very much on the right track.

    P.S. ever resolved the sliding rod problem?
     
  13. May 25, 2014 #12
    Thanks, I'll check that tomorrow.

    Ah the sliding rod - I'm convinced my answer is fine. Seems a bit peculiar though, because it was actually a 'show that the voltage between A and B is this' exam question, although I wouldn't put it past the people concerned to do that on an exam.
     
  14. May 26, 2014 #13
    Hi again.

    I was wondering if you would be able to quickly check my integral? I've stumbled upon this problem in a textbook and their mean power is 2* my answer, and I can't figure out why. They also ask for the current, and integrating my current density out over the disk gives the correct answer in that respect.
     
  15. May 26, 2014 #14

    rude man

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    I evaluated my integral starting from dP = d(I^2 R) and got exactly the same P you did.
     
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