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M and $M \sqrt{3}$ are the magnetic dipole moments of the two magnets, which are joined to form a cross as shown in figure. The inclination of the system with the field, if their combination is suspended freely in a uniform external magnetic field B is
M and are the magnetic dipole moments of the two magnets, which are joined to form a cross figure. The inclination of the system with the field, if their combination is suspended freely in a uniform external magnetic field B is

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$θ = 30^{o}$

$θ = 45^{o}$

$θ = 60^{o}$

$θ = 15^{o}$

answer is C.

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Detailed Solution

$T o r q u e a c t i n g o n m a g n e t (1) i s τ_{1} = M B \sin θ$

$T o r q u e a c t i n g o n m a g n e t (2) i s τ_{2} = \sqrt{3} M B \sin (90 - θ)$

$F o r e q u i l i b r i u m, τ_{1} = τ_{2}$

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

$\Rightarrow \tan θ = \sqrt{3}$

$\Rightarrow θ = 60^{o}$

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