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Problem from Griffiths confused me

  1. Aug 24, 2009 #1
    I'll explain the problem.

    [tex]\cdots[/tex] [tex]\cdots[/tex] [tex]\cdots[/tex] [tex]\cdots[/tex]
    [tex]\cdots[/tex] [tex]\cdots[/tex] [tex]\ldots[/tex]=====|======
    [tex]\cdots[/tex] [tex]\cdots[/tex] || LOOP | LOOP || --> V
    [tex]\cdots[/tex] [tex]\cdots[/tex] ||=====|=====||
    [tex]\cdots[/tex] [tex]\cdots[/tex] <-x-> ..|
    [tex]\cdots[/tex] [tex]\cdots[/tex]

    [Dotted area indicates field]

    In the usual discussion of motional emf, all authors go through the usual loop moving through a magnetic field which is perpendicular to the plane in which the loop is moving. Usually, this loop have resistance, say R and starts out with a bit of the loop already outside the field. The emf can be shown to be Bhv wherein B is the strength of the magnetic field, h is the length of the rod and v is the velocity with which the loop is being pulled along. Therefore the current can be calculated as BLV/R.

    Now he introduces a new twist in the question. Assume the rod has no resistance, i.e. that it is a perfect conductor. I have to show that the rod will now execute simple harmonic motion.

    This being said, I got the answer.
    Qualitatively, we can say that the current will not shoot to infinity as the self-inductance will not allow that to happen. The back EMF associated with the loop will save the day.

    In this case the flux can be taken to be Bhx where x is the length of the loop still inside the field. Refer to my "diagram". The flux for this loop is also given by LI where I is the current flowing through the loop and L is its self-inductance.

    Equating the two, we have current I = Bhx/L
    The force on this loop is given by (IhB).
    So,
    mx'' = -Kx, where K is B2h2/mL.

    It is quite obvious that the rod will execute simple harmonic motion. Hence proved.

    Now what confuses me is that the above equation holds for any loop with or without resistance. He says something here in the problem that in general cases the self-inductance thing is negligible. I don't get that. Shouldn't all loops execute SHM?
     
    Last edited: Aug 24, 2009
  2. jcsd
  3. Aug 24, 2009 #2
    Hold on. I think I'm confusing two different things. One is the EMF generated by the change in flux of the magnetic field through the loop and the other is due to the induced current. I think I'm along the lines of the answer now, but still can't put my finger on it. Can someone finish it off?
     
  4. Aug 24, 2009 #3
    I think I'm done. Listen to this. By Ohm's law,

    [tex]\epsilon[/tex] - L (dI/dt) = IR = 0, where [tex]\epsilon[/tex] is the EMF.

    On the other hand, [tex]\epsilon[/tex] = d[tex]\phi[/tex]/dt is the line integral of the force per unit charge = Bhv.

    Therefore,
    [tex]\epsilon[/tex] = Bhv = L (dI/dt).

    Now that we've found dI/dt in terms of v, the force acting on the loop mdv/dt = - I(hxB)

    Differentiating on both sides, we have our answer.
    That now results in the above equation that I'd originally written down in my first post and thus leads to simple harmonic motion. There! I'm done. Something about this forum that makes me able to answer my own questions. Love this place.
     
    Last edited: Aug 24, 2009
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