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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30o

45o

60o

90o

answer is C.

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Detailed Solution

τ=MBsinθ ⇒τ∝sinθ ⇒τ1τ2=sinθ1sinθ2 ⇒ττ2=sin90sinθ2 ⇒sinθ2=12 ⇒θ2=300 So, bar magnet should be rotated by 600.

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