A short bar magnet placed with its axis at `30^(@)` with an external magnetic field of 800 G experiences a torque of 0.2 N.m. Find the magnetic moment of the magnet.

Explanation

We know that,
Torque on a bar magnet, `tau = M xx B`
`|tau | = | M xxB| = MB sin theta implies tau = MB sin theta `
Magnetic moment,
`M = (tau )/( B sin theta ) = (0.2)/(( 800xx 10 ^(-4)) xx sin 30 ^(@))`
`[because 1` gauss `= 10 ^(-4)` tesla],
`= (0.4)/( 800 xx 10 ^(-4)) = ( 4 xx 10 ^(3)) /(800) = (1)/(2) xx 10 = 5 J//T`