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A magnetic dipole is under the influence of two orthogonal magnetic fields, $B_1=0.5 \times 10^{-3} \mathrm{~T}$ and $B_2=0.866 \times 10^{-3} \mathrm{~T}$. If the dipole comes to stable equilibrium at an angle $\theta$ with respect to $B_2$ field, then the value of $\theta$ is
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The correct answer is:
$30^{\circ}$
From tangent law, we have, $m B_2 \sin \theta=m B_2 \sin \left(90^{\circ}-\theta\right)$
$$
\begin{aligned}
\Rightarrow \quad \tan \theta=\frac{B_2}{B_2} & =\frac{0.5 \times 10^{-3}}{0.866 \times 10^{-3}} \\
=\frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}} & =\frac{1}{\sqrt{3}} \\
\tan \theta & =\tan 30^{\circ} \\
\theta \quad \theta & =30^{\circ}
\end{aligned}
$$
$$
\begin{aligned}
\Rightarrow \quad \tan \theta=\frac{B_2}{B_2} & =\frac{0.5 \times 10^{-3}}{0.866 \times 10^{-3}} \\
=\frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}} & =\frac{1}{\sqrt{3}} \\
\tan \theta & =\tan 30^{\circ} \\
\theta \quad \theta & =30^{\circ}
\end{aligned}
$$
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