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

Magnetic strength , B `= 0.8 ` T
Angle inclined with a magnetic field ` theta = 30^(@)`
Torque , `tau = 0.2 ` Nm .
(i) Magnetic moment of the magnet M = ?
` tau = MB sin theta `
`mu = tau/(B sin theta ) = (0.2)/(0.8 sin 30^(@))`
` = ( 0.2)/(0.8 xx 1/2) = 2/4 = 0 . 5 Am^(2) `
`mu = 0.5 Am^(2) `
(ii) Work done by an applied force in stable
`U_(i) = mu B cos theta `
` theta = 0^(@)`
Work done by an applied force in unstable configuration , `U_(f) = - mu B cos theta `
` theta = 180^(@)`
Work done by the applied magnetic field
` W = U_(f^(-))U_(i)`
` = - mu B cos 180^(@) - ( - mu B cos 0^(@))`
` = mu B + mu B `
` W = 2 mu B `
` = 2 xx 0.5 xx 0. 8 `
` W = 0. 85` .