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A bar magnet has a magnetic moment of $200 \mathrm{Am}^{2}$. The magnet is suspended in a magnetic field of $0.30 \mathrm{NA}^{-1} \mathrm{m}^{-1}$. The torque required to rotate the magnet from its equilibrium position through an angle of $30^{\circ},$ will be
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The correct answer is:
$30 \mathrm{N}$ -m
Given, $\mathrm{M}=200 \mathrm{A} \mathrm{m}^{2}$
$\mathrm{B}=0.30 \mathrm{NA}^{-} \mathrm{M}^{-1}$
and $\theta=30^{\circ}$
We know that the Torque,
$$
\begin{aligned}
\tau &=M \times B \\
\Rightarrow \quad|t| &=M B \sin \theta=200 \times 0.3 \times \frac{1}{2} \\
&=100 \times 0.3=30 \mathrm{N} \mathrm{m}
\end{aligned}
$$
$\mathrm{B}=0.30 \mathrm{NA}^{-} \mathrm{M}^{-1}$
and $\theta=30^{\circ}$
We know that the Torque,
$$
\begin{aligned}
\tau &=M \times B \\
\Rightarrow \quad|t| &=M B \sin \theta=200 \times 0.3 \times \frac{1}{2} \\
&=100 \times 0.3=30 \mathrm{N} \mathrm{m}
\end{aligned}
$$
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