Question: Q. 5. Two magnets of magnetic moment $M$ and $M \sqrt{ } 3$ are joined to form a cross. The combination is suspended in a uniform magnetic field $B$. The magnetic moment now makes an angle $\theta$ with the field direction. Find the value of angle $\theta$.
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Solution:
Ans.
[CBSE SQP, 2011]
In equilibrium condition, torque on both magnets are equal and opposite. Hence,
$$ \begin{align*} M B \sin \theta & =\sqrt{3} M \sin \left(90^{\circ}-\theta\right) \ M B \sin \theta & =\sqrt{3} M \cos \theta \ \frac{\sin \theta}{\cos \theta} & =\frac{\sqrt{3} M}{M} \ \tan \theta & =\sqrt{3} \text { or } \theta=60^{\circ} \tag{1} \end{align*} $$
