The magnetic moment of a bar magnet is $0.5 \mathrm{~Am}^2$. It is suspended in a uniform magnetic field of $8 \times 10^{-2} \mathrm{~T}$. The work done in rotating it from its most stable to most unstable position is:
Solution
<p>$$\begin{aligned}
W & =\int_0^{180} d \tau \cdot d \theta \\
& =m B(\cos 0-\cos 180) \\
& =0.5 \times 8 \times 10^{-2}(2) \\
& =8 \times 10^{-2} \mathrm{~J}
\end{aligned}
$$</p>
About this question
Subject: Physics · Chapter: Magnetism · Topic: Bar Magnet and Magnetic Field
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