(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)?
Answer –
(a)
We are given that the magnetic moment, M = 1.5 JT-1
And the magnetic field strength, B = 0.22 T
(i) Initial angle between the magnetic field and the axis, θ1 = 0°
Final angle between the magnetic field and the axis, θ2 = 90°
Hence, the work required to make the magnetic moment normal to the direction of the magnetic field will be given by the relation –
\[W=-MB(\cos {{\theta }_{2}}-\cos {{\theta }_{1}})\]
\[W=-1.5\times 0.22(\cos {{90}^{\circ }}-\cos {{0}^{\circ }})=-0.33(0-1)\]
\[W=0.33J\]
(ii) Initial angle between the magnetic field and the axis, θ1 = 0°
Final angle between the magnetic field and the axis, θ2 = 180°
Hence, the work required to make the magnetic moment normal to the direction of the magnetic field will be given by the relation –
\[W=-MB(\cos {{\theta }_{2}}-\cos {{\theta }_{1}})\]
\[W=-1.5\times 0.22(\cos {{180}^{\circ }}-\cos {{0}^{\circ }})=-0.33(-1-1)\]
\[W=0.66J\]
(b)
For case (i)
\[\theta ={{\theta }_{2}}={{90}^{\circ }}\]
\[\therefore T=MB\sin \theta \]
\[\therefore T=1.5\times 0.22\times \sin {{90}^{\circ }}=0.33J\]
The torque tends to align the magnitude moment vector along B.
For case (ii):
\[\theta ={{\theta }_{2}}={{180}^{\circ }}\]
\[\therefore T=MB\sin \theta \]
\[\therefore T=1.5\times 0.22\times \sin {{180}^{\circ }}=0J\]