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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: