Question

A bar magnet of magnetic moment 1.5 J/T 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)?

Answer 1

Hint:Magnetic moment is by magnetic dipole. The magnet is placed in an external field, so the two ends of the dipole will experience the force. Since, the magnet is placed such that it is parallel to the field, and had it been placed such that it made some angle with the field, then it will experience a torque.

Complete step by step answer:
Given values of magnetic moment, M= 1.5 J/T, Magnetic field strength, B= 0.22 T
(a) Initially the angle between the two is \[{{0}^{0}}\] and we want to align it normal to the field, so the final angle is \[{{90}^{0}}\]. We know work required to turn the magnet is given by:
\[W=-MB(\cos {{\theta }_{2}}-\cos {{\theta }_{1}})\]
$\Rightarrow W=-1.5\times 0.22(\cos 90-\cos 0) \\
\Rightarrow W=-0.33(0-1) \\
\therefore W=0.33J \\ $
Now the alignment is opposite to the field direction, so the angle is \[{{180}^{0}}\]
$W=-1.5\times 0.22(\cos 180-\cos 0) \\
\Rightarrow W=-0.33(-1-1) \\
\therefore W=-0.66J \\ $
(b) now we know torque is given by \[\tau =MB\sin \theta \]
For case (1)
$\Rightarrow MB\sin 90 \\
\Rightarrow 1.5\times 0.22\times 1 \\
\therefore 0.33Nm \\ $
For case (2)
$\tau =MB\sin \theta \\
\Rightarrow\tau = MB\sin 180 \\
\therefore\tau = 0 $

Note: Potential energy or the work both are the same because both are measured in units of Joules, for a magnetic dipole it comes out to be negative because the North pole of the bar magnet is placed at a location with a higher potential than the South pole. Also, negative value tells us that the configuration is stable. If we draw the potential energy vs distance graph, we see the curve is bounded.