Magnetic field strength and SHM

  • Thread starter debwaldy
  • Start date
  • #1
38
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Homework Statement


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.


Homework Equations


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


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
Q.E.D???:tongue:
 

Answers and Replies

  • #2
38
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:biggrin: will i take it as being Q.E.D'd then ya?
 

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