JEE MAIN - Physics (2024 - 4th April Evening Shift - No. 6)
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:
$$4 \times 10^{-2} \mathrm{~J}$$
$$16 \times 10^{-2} \mathrm{~J}$$
$$8 \times 10^{-2} \mathrm{~J}$$
Zero
Explanation
$$\begin{aligned}
W & =\int_0^{180} d \tau \cdot d \theta \\
& =m B(\cos 0-\cos 180) \\
& =0.5 \times 8 \times 10^{-2}(2) \\
& =8 \times 10^{-2} \mathrm{~J}
\end{aligned}
$$
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