A bar magnet of magnetic moment \(1.8\ \mathrm{Am^2}\) is free to rotate about a vertical axis passing through its centre at a place where the vertical component of earth's magnetic field is
\[
0.3\times10^{-4}\text{ T}
\]
and the dip angle is \(45^\circ\). If the magnet at rest in east-west direction is released, then the kinetic energy (in \(\mu\)J) of the magnet when it reaches north-south direction is
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For a freely suspended magnet,
\[
\boxed{
U=-MB\cos\theta.
}
\]
If the magnet rotates about a vertical axis, only the horizontal component of Earth's magnetic field contributes:
\[
\boxed{
B_H=\frac{B_V}{\tan\delta}.
}
\]