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(D) The work done $W$ in rotating a magnetic dipole in a magnetic field is given by the formula: $W = MB(\cos 	heta_1 - \cos 	heta_2)$.Initially, th

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$ + 2MB$

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(D) The work done $W$ in rotating a magnetic dipole in a magnetic field is given by the formula: $W = MB(\cos 	heta_1 - \cos 	heta_2)$.Initially, the magnet is along the direction of the magnetic field, so the initial angle $	heta_1 = 0^{\circ}$.We need to rotate it by an angle of $180^{\circ}$, so the final angle $	heta_2 = 180^{\circ}$.Substituting these values into the formula:$W = MB(\cos 0^{\circ} - \cos 180^{\circ})$Since $\cos 0^{\circ} = 1$ and $\cos 180^{\circ} = -1$, we get:$W = MB(1 - (-1)) = MB(1 + 1) = 2MB$.Therefore, the work done is $2MB$.

$A$ magnet of magnetic moment $M$ is situated with its axis along the direction of a magnetic field of strength $B$. The work done in rotating it by an angle of $180^{\circ}$ will be

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