A bar magnet of magnetic moment `1.5JT^-1` lies aligned with the direction of a uniform magnetic field of `0.22T`.
(a) What is the amount of work done to turn the magnet so as to align its mangetic 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) Magnetic moment, `M=1.5J T^(-1)`
Magnetic field strength, `B=0.22T`
(i) Initial angle between the axis and the magnetic field `theta_(1)=0^(@)`
Final angle between the axis and the magnetic field `theta_(2)=90^(@)`
The work required to make the magnetic moment normal to the direction of magnetic
field is given as:
`W=-MB(cos theta_(2)-costheta_(1))`
`=-1.5xx0.22(cos 90^(@)-cos0^(@))`
`=-0.33(0-1)`
`=0.33 J`
(ii) Initial angle between the axis and the magnetic field, `theta_(1)=0^(@)`
Final angle between the axis and the magnetic field, `theta_(2)=180^(@)`
The work required to make the magnetic moment opposite to the direction of magnetic field is given as:
`W=-MB(costheta_(2)-costheta_(1))`
`=-1.5xx0.22(cos 180-cos0^(@))`
`= -0.33(-1-1)`
`=0.66J`
(b) For case i: `theta=tehta_(2)=90^(@)`
`therefore` Torque, `r=MB sin theta`
`=1.5xx0.22sin90^(@)`
=0.33J
For case ii: `theta=theta_(2)=180^(@)`
`therefore` Torque, `r=MB sin theta`
`=MB sin 180^(@)=0J`