JEE MAIN - Physics (2023 - 31st January Morning Shift - No. 2)
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 _____________.
zero
1 J
2 J
4 J
Explanation
$W=-M B\left(\cos \theta_{2}-\cos \theta_{1}\right)$
$$ \begin{aligned} = & -0.4 \times 5\left[\cos 180^{\circ}-\cos 0\right] \\\\ = & 4 \mathrm{~J} \end{aligned} $$
$$ \begin{aligned} = & -0.4 \times 5\left[\cos 180^{\circ}-\cos 0\right] \\\\ = & 4 \mathrm{~J} \end{aligned} $$
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