Question: Q. 1. A bar magnet of magnetic moment $6 \mathrm{~J} / \mathrm{T}$ is aligned at $60^{\circ}$ 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).
R&U
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Solution:
Ans. (a) Formula and
Calculation of work done in the two cases
(b) Calculation of torque in case (ii)
(a) Work done $=m_{B}\left(\cos \theta_{1}-\cos \theta_{2}\right)$
(i) $\quad \theta_{1}=60^{\circ}, \theta_{2}=90^{\circ}$
$\therefore$ work done $=m B\left(\cos 60^{\circ}-\cos 90^{\circ}\right)$
$$ \begin{aligned} & =m B\left(\frac{1}{2}-0\right)=\frac{1}{2} m B \ & =\frac{1}{2} \times 6 \times 0.44 \mathrm{~J}=1.32 \mathrm{~J} \end{aligned} $$
(ii) $\theta_{1}=60^{\circ}, \theta_{2}=180^{\circ}$
$\therefore$ work done $=m B\left(\cos 60^{\circ}-\cos 180^{\circ}\right)$
$$ =m B\left(\frac{1}{2}-(-1)\right)=\frac{3}{2} m B $$
$$ =\frac{3}{2} \times 6 \times 0.44 \mathrm{~J}=3.96 \mathrm{~J} $$
[Also accept calculations done through changes in potential energy.]
(b)
For
$$ \text { Torque }=|\vec{m} \times \vec{B}|=m B \sin \theta $$
$\theta=180^{\circ}$, we have
$$ \text { Torque }=6 \times 0.44 \sin 180^{\circ}=0 $$
[If the student straight away writes that the torque is zero since magnetic moment and magnetic field are anti parallel in this orientation, award full marks] $1 / 2$
[CBSE Marking Scheme 2018]
