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A short bar magnet placed with its axis at $30^{\circ}$ with a uniform external magnetic field of $0.25 \mathrm{~T}$ experiences a torque of magnitude equal to $4.5 \times 10^{-2} \mathrm{~J}$. What is the magnitude of magnetic moment of the magnet?
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Verified Answer
Given: magnetic field $\mathrm{B}=0.25 \mathrm{~T}$,
Torque $\tau=4.5 \times 10^{-2} \mathrm{~J}$,
Angle between axis of magnet and field $\theta=30^{\circ}$
To find: Magnetic moment $=$ ?
Formula used: $\tau=\mathrm{MB} \sin \theta$
$$
\begin{aligned}
\mathrm{M} &=\frac{\tau}{\mathrm{B} \sin \theta}=\frac{4.5 \times 10^{-2}}{0.25 \times \sin 30^{\circ}} \\
&=\frac{18 \times 10^{-2}}{1 / 2}=0.36 \mathrm{~A} \mathrm{~m}^2
\end{aligned}
$$
Torque $\tau=4.5 \times 10^{-2} \mathrm{~J}$,
Angle between axis of magnet and field $\theta=30^{\circ}$
To find: Magnetic moment $=$ ?
Formula used: $\tau=\mathrm{MB} \sin \theta$
$$
\begin{aligned}
\mathrm{M} &=\frac{\tau}{\mathrm{B} \sin \theta}=\frac{4.5 \times 10^{-2}}{0.25 \times \sin 30^{\circ}} \\
&=\frac{18 \times 10^{-2}}{1 / 2}=0.36 \mathrm{~A} \mathrm{~m}^2
\end{aligned}
$$
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