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A short bar magnet placed at 30^o with an external magnetic field 0.25T experiences a torque of 45 mJ. The workdone to turn it from this position to a position of maximum torque is

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45 $\sqrt{3}$ mJ

45 mJ

90 mJ

90 $\sqrt{3}$ mJ

answer is C.

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Detailed Solution

$\begin{array}{l} τ = MBsin θ \Rightarrow \frac{MB}{2} = 45 mJ \\ W = MB (\cos θ_{1} - \cos θ_{2}) \\ = MB (\cos 30 - \cos 90) = \frac{\sqrt{3} MB}{2} \\ \Rightarrow W = 45 \sqrt{3} mJ \end{array}$

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