Magnetic field strength and SHM

1. Feb 12, 2007

debwaldy

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
Q.E.D???:tongue:

2. Feb 13, 2007

debwaldy

will i take it as being Q.E.D'd then ya?