The work done in rotating a magnet of magneti | vedclass

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(B) The work done $W$ in rotating a magnetic dipole in a magnetic field is given by the formula $W = MB(\cos 	heta_1 - \cos 	heta_2)$.Here,the initi

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$2 	imes 10^{-2} \, J$

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(B) The work done $W$ in rotating a magnetic dipole in a magnetic field is given by the formula $W = MB(\cos 	heta_1 - \cos 	heta_2)$.Here,the initial angle $	heta_1 = 0^\circ$ (along the field) and the final angle $	heta_2 = 180^\circ$ (opposite to the field).Substituting the values: $W = MB(\cos 0^\circ - \cos 180^\circ)$.Since $\cos 0^\circ = 1$ and $\cos 180^\circ = -1$,we get $W = MB(1 - (-1)) = 2MB$.Given $M = 2 \, A-m^2$ and $B = 5 	imes 10^{-3} \, T$.$W = 2 	imes (2) 	imes (5 	imes 10^{-3}) = 20 	imes 10^{-3} = 2 	imes 10^{-2} \, J$.

The work done in rotating a magnet of magnetic moment $2 \, A-m^2$ in a magnetic field of $5 \times 10^{-3} \, T$ from the direction along the magnetic field to the opposite direction to the magnetic field is:

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