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Magnetic field strength and SHM

  1. Feb 12, 2007 #1
    1. The problem statement, all variables and given/known data
    Hi there.this is one of those proof dealies that i think i nearly have right but i was wondering if anyone could point out if i have made too many assumptions or anything thanks.
    Show that if a bar magnet is suspended in a uniform magnetic field, of strength H, and is displaced slightly from equilibrium, it undergoes SHM with period:
    T = 2*pi* (I/plH)^1/2, where I is the moment of inertia, p is the magnetic pole strength, and l is the separation between the poles of the bar magnet.

    2. Relevant equations
    T= pln X H
    P= 2*pi*(I/plH)^1/2

    3. The attempt at a solution
    I said: Torque on magnet T=pln X h
    T(alpha) = -plhsin(alpha) =-plH * alpha which is proportional to - alpha => SHM

    T/alpha = plH

    P= 2*pi* (moment of inertia about axis of rotation/restoring torque per unit angular displacement)^1/2
    => P = 2*pi*(I/plH)^1/2
  2. jcsd
  3. Feb 13, 2007 #2
    :biggrin: will i take it as being Q.E.D'd then ya?
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