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.
T= pln X H
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