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A bar magnet of magnetic moment M and moment of inertial I is freely suspended such that the magnetic axis lies in the magnetic meridian. If the magnet is displaced by a very small angle ( $θ$ ), the angular acceleration is (Magnetic induction of earth’s horizontal field = B_H)

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$\frac{{MB}_{H} θ}{I}$

$\frac{{IB}_{H} θ}{M}$

$\frac{Mθ}{{IB}_{H}}$

$\frac{Iθ}{{MB}_{H}}$

answer is A.

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Detailed Solution

$Iα = {MB}_{H} θ$

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