A bar magnet with a magnetic moment $5.0 \mathrm{Am}^{2}$ is placed in parallel position relative to a magnetic field of $0.4 \mathrm{~T}$. The amount of required work done in turning the magnet from parallel to antiparallel position relative to the field direction is _____________.
Solution
$W=-M B\left(\cos \theta_{2}-\cos \theta_{1}\right)$
<br/><br/>$$
\begin{aligned}
= & -0.4 \times 5\left[\cos 180^{\circ}-\cos 0\right] \\\\
= & 4 \mathrm{~J}
\end{aligned}
$$
About this question
Subject: Physics · Chapter: Magnetism · Topic: Bar Magnet and Magnetic Field
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