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A bar magnet is held perpendicular to a uniform magnetic field. If the couple acting on the magnet is to be halved by rotating it, then the angle by which it is to be rotated is:

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45^o

90^o

60^o

30^o

answer is C.

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Detailed Solution

$\begin{array}{l} τ = MBsinθ \Rightarrow τ \propto sinθ \\ \Rightarrow \frac{τ_{1}}{τ_{2}} = \frac{{sinθ}_{1}}{{sinθ}_{2}} \Rightarrow \frac{τ}{τ / 2} = \frac{\sin 90 °}{{sinθ}_{2}} \end{array}$

$\begin{array}{l} \Rightarrow {sinθ}_{2} = \frac{1}{2} \Rightarrow θ_{2} = 30 ° \\ \Rightarrow angle of rotation = 90 ° - 30 ° = 60 ° \end{array}$

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