5.7 A bar magnet of magnetic moment 1.5 J T–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)?

Magnetic moment,
Magnetic field strength, B=0.22 T
${\theta }_{1}=0°$
(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 to normal to the direction of magnetic field is given as:$W=-MB\left(\mathrm{cos}{\theta }_{2}-\mathrm{cos}{\theta }_{1}\right)$
$=1.5×0.22\left(\mathrm{cos}90°-\mathrm{cos}0°\right)$
$=0.33\left(0-1\right)$

(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:
$W=-MB\left(\mathrm{cos}{\theta }_{2}-\mathrm{cos}{\theta }_{1}\right)$
$=-1.5×0.22\left(\mathrm{cos}180-\mathrm{cos}0°\right)$
$=-0.33\left(-1-1\right)$
$=0.66J$

(b) For case (i)

$\theta ={\theta }_{2}=90°$

For case (ii)
$\theta ={\theta }_{2}=180°$