A bar magnet of magnetic moment `1.5 JT^(-1)` lies aligned with the direction of a uniform magnetic field of 0.22 T.
(a) What is the amount of work required by an external torque to turn the magnet so as to align its magnetic moment: (i) normal to the field direction (ii) opposite to the field direction?
(b) What is the torque on the magnet in cases (i) and (ii)?

Explanation

(a) (i) Required amount of work done,
`W= mB ( cos theta_(1)- cos theta_(2) )`
`= (1.5 ) (0.22) ( cos 0^(@) -cos 90^(@) )`
`=(1.5)(0.22) (1-0)`
`=0.33` J
(ii) In this case,
`W= mB ( cos theta_(1)- cos theta_(2))`
`therefore W= (1.5) ( 0.22) ( cos0^(@) - cos180^(@) )`
`therefore = (1.5) (0.22) (1- ( -1) )`
`therefore W= (1.5) (0.22) (2)`
`therefore W= 0.66` J
(b) (i) Torque exerted,
`tau = m B sin theta`
`therefore tau = (1.5) (0.22) sin 90^@`
`therefore tau= (1.5) (0.22) (1)`
`therefore tau = 0.33` Nm
(ii) In this case,
`tau= m B sin theta`
`therefore tau= (1.5) (0.22) (0)`
`therefore tau=0`