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Two magnets of equal masses and moments M and $\sqrt{3} M$ are joined to form a cross (X). If this combination is pivoted so that it is free to rotate in the horizontal plane. The angle made by the magnet of smaller moment with the magnetic meridian in equilibrium is

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45º

60º

30º

0º

answer is B.

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Detailed Solution

Let be the angle made by the magnetic meridian with the magent of smaller moment. In the equilibrium position torque experienced by forst magnet is equal to the torque experienced by second magnet.

$∴ M B \sin θ = \sqrt{3} M B \sin (90 - θ)$

$\Rightarrow \sin θ = \sqrt{3} \cos θ \Rightarrow \tan θ = \sqrt{3}$

$∴ θ = 60 º$

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