A bar magnet having a magnetic movement of $2 \times 10^4 \mathrm{JT}^{-1}$ is free to rotate in a…

A bar magnet having a magnetic movement of $2 \times 10^4 \mathrm{JT}^{-1}$ is free to rotate in a horizontal plane. A horizontal magnetic field $B=6 \times 10^{-4}$ $T$ exists in the space. The work done in taking the magnet slowly from a direction parallel to the field to a direction $60^{\circ}$ from the field is :
  1. $2 \mathrm{~J}$
  2. $0.6 \mathrm{~J}$
  3. $12 \mathrm{~J}$
  4. $6 \mathrm{~J}$

Solution

$\begin{aligned} \mathrm{W} & =\mathrm{MB}\left(\cos \theta_1-\cos \theta_2\right) \\ & =2 \times 10^4 \times 6 \times 10^{-4}\left(\cos \theta-\cos 60^{\circ}\right) \\ & =12 \times \frac{1}{2}=6 \mathrm{~J} \end{aligned}$

Asked in: NEET 2009 (Mains)

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