The work done in rotating a bar magnet,which | vedclass

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(B) The work done in rotating a magnetic dipole in a uniform magnetic field is given by $W = MB(\cos 	heta_1 - \cos 	heta_2)$.Initially,the magnet i

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$W(\frac{\sqrt{3}-1}{\sqrt{2}-1})$

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(B) The work done in rotating a magnetic dipole in a uniform magnetic field is given by $W = MB(\cos 	heta_1 - \cos 	heta_2)$.Initially,the magnet is in the direction of the field,so $	heta_1 = 0^{\circ}$ and $	heta_2 = 45^{\circ}$.$W = MB(\cos 0^{\circ} - \cos 45^{\circ}) = MB(1 - \frac{1}{\sqrt{2}}) = MB(\frac{\sqrt{2}-1}{\sqrt{2}})$.Thus,$MB = \frac{W\sqrt{2}}{\sqrt{2}-1}$.Now,the additional work $W'$ to rotate it further by $15^{\circ}$ means rotating from $	heta_1 = 45^{\circ}$ to $	heta_2 = 45^{\circ} + 15^{\circ} = 60^{\circ}$.$W' = MB(\cos 45^{\circ} - \cos 60^{\circ}) = MB(\frac{1}{\sqrt{2}} - \frac{1}{2}) = MB(\frac{\sqrt{2}-1}{2\sqrt{2}})$.Substituting the value of $MB$:$W' = (\frac{W\sqrt{2}}{\sqrt{2}-1}) 	imes (\frac{\sqrt{2}-1}{2\sqrt{2}}) = \frac{W}{2}$.

The work done in rotating a bar magnet,which is initially in the direction of a uniform magnetic field,through $45^{\circ}$ is $W$. The additional work to be done to rotate the magnet further through $15^{\circ}$ is

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