A bar magnet is placed in a uniform magnetic field whose strength is `0.8` T. Suppose the bar magnet orient an angle `30^(@)` with the external field experiences a torque of `0.2` Nm . Calculate :
(i) the magnetic moment of the magnet
(ii) the work done by an applied force in moving it form most stable configuration to the most unstable configuration and also compute the work done by the applied magnetic field in this case .

Explanation

Uniform magnetic field strength B = 0.8 T
Bar magnet orient an angle with magnetic field `theta = 30^(@)`
Torque `tau ` = 0.2 Nm
(i) Magnetic moment of the magnet
torque `tau = P_(m) B sin theta`
`therefore` Magnetic moment `, P_(m) = (tau)/("Bsin" theta ) = (0.2 )/(0.8 xx Sin 30^(@) ) = (0.2)/(0.4) `
`P = 0.5 Am^(2)`
(ii) Work done by external torque is stored in the magnet as potential energy .
W = U = `-P_(m) " B Cos " theta`
here, applied force acting on magnet its moving from most stable `theta`. to most unstable `theta`.
`theta. = 0^(@) and theta = 180^(@)`
So, workdone W = U = -`P_(m) " B "( Cos theta - Cos theta^(@) )`
= - `p_(m) B (Cos 180^(@) - Cos theta^(@)) = - 0.5 xx 0.8 ((-1)-1) = - 0.4 (-2) `
W = U = 0.8
W = 0.8 J