Question

A bar magnet of magnetic moment 6 J/T is aligned at ${60}^0$ with a uniform external magnetic field of 0.44 T. Calculate (a) the work done in turning the magnet to align its magnetic moment (i) normal to the magnetic field, (ii) opposite to the magnetic field, and
(b) the torque on the magnet in the final orientation
in case (ii).

Answer 1

Given, magnetic momentum, m = 6 $J{T}^{-1}$

External magnetic field, B = 0.44 T

${\theta }_1={60}^0\Rightarrow cos{\theta }_1=cos{60}^0=\frac{1}{2}.$

(a) Work done in turning the magnet normal to the field,

$W=-mB(cos{\theta }_2-cos{\theta }_1)$

(i) Here, ${\theta }_2={90}^0$

$\therefore W=+mBcos{\theta }_1$

= $6\times 0.44\times \frac{1}{2}=1.32J$

(ii) Here ${\theta }_2={180}^0$

$\therefore W=-mB(cos{\theta }_2-cos{\theta }_1)$

W=$-6\times 0.44(-1-\frac{1}{2})$

= 3.96 J

(b) Torque on magnet when moment is aligned opposite to the field,

$\tau =mBsin\theta$

putting the values,

= $6\times 0.44\times sin{180}^{\circ }$

= $0(\because sin{180}^{\circ }=0)$