Two magnets of equal mass are joined at right angles to each other as shown. Magnet $1$ has a magnetic moment $3$ times that of magnet $2$ . This arrangement is pivoted so that it is free to rotate in the horizontal plane. In equilibrium what angle will the magnet $1$ subtend with the magnetic meridian

{tan}^{- 1} (\frac{1}{2})

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{tan}^{- 1} (\frac{1}{3})

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{tan}^{- 1} (1)

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0^{o}

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Solution

The correct option is B ${tan}^{- 1} (\frac{1}{3})$
$\begin{matrix} F o r e q u i l i b r i u m o f t h e s y s t e m t o r q u e s o n M_{1} a n d M_{2} d u e t o B_{H} m u s t c o u n t e r e a c h o t h e r, i . e ., M_{1} \times B_{H} = M_{2} \times B_{H} . I f q i s t h e a n g l e b e t w e e n M_{1} a n d B_{H} w i l l b e (90 - θ), S o, M_{1} B_{^{H}} sin θ = M_{2} B_{^{H}} sin (90 - θ) \Rightarrow tan θ = \frac{M_{2}}{M_{1}} = \frac{M}{3 M} = \frac{1}{3} \Rightarrow θ = {tan}^{- 1} (\frac{1}{3}) \end{matrix}$

Hence,

option $(B)$ is correct answer.

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Q. Two magnets of equal mass are joined at right angles to each other as shown. The magnet 1 has a magnetic moment 3 times that of magnet 2. This arrangement is pivoted so that it is free to rotate in the horizontal plane. In equilibrium, angle with which the magnet 1 subtend with the magnetic meridian is

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Two magnets of equal mass are joined at right angles to each other as shown. Magnet 1 has a magnetic moment 3 times that of magnet 2. This arrangement is pivoted so that it is free to rotate in the horizontal plane. In equilibrium what angle will the magnet 1 subtend with the magnetic meridian

The correct option is B tan−1(13)ForequilibriumofthesystemtorquesonM1andM2duetoBHmustcountereachother,i.e.,M1×BH=M2×BH.IfqistheanglebetweenM1andBHwillbe(90−θ),So,M1BHsinθ=M2BHsin(90−θ)⇒tanθ=M2M1=M3M=13⇒θ=tan−1(13) Hence, option (B) is correct answer.