A) \[2\sqrt{3}\,J\]
B) 3J
C) \[\sqrt{3}\,J\]
D) \[\frac{3}{2}\,J\]
Correct Answer: B
Solution :
According to work energy theorem \[W={{U}_{final}}-{{U}_{initial}}=MB\,(cos0-\cos 60{}^\circ )\] \[W=\frac{MB}{2}=\sqrt{3}\,J\] \[\tau =\vec{M}\times \vec{B}=\,\,MB\,\,sin\,\,60{}^\circ =\,\,\left( \frac{MB\sqrt{3}}{2} \right)\] ... (ii) From equation (i) and (ii) \[\tau \,\,=\,\,\frac{2\sqrt{3}\times \sqrt{3}}{2}=3\,J\]
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