The work done in turning a magnet of magnetic | vedclass

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(B) The work done in rotating a magnetic dipole in a uniform magnetic field $B$ from an angle $	heta_1$ to $	heta_2$ is given by $W = MB(\cos 	heta

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$2$

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(B) The work done in rotating a magnetic dipole in a uniform magnetic field $B$ from an angle $	heta_1$ to $	heta_2$ is given by $W = MB(\cos 	heta_1 - \cos 	heta_2)$.For the first case,the magnet is turned from the meridian $(	heta_1 = 0^{\circ})$ to $	heta_2 = 90^{\circ}$:$W_1 = MB(\cos 0^{\circ} - \cos 90^{\circ}) = MB(1 - 0) = MB$.For the second case,the magnet is turned from the meridian $(	heta_1 = 0^{\circ})$ to $	heta_2 = 60^{\circ}$:$W_2 = MB(\cos 0^{\circ} - \cos 60^{\circ}) = MB(1 - 0.5) = 0.5MB = \frac{MB}{2}$.Given that $W_1 = n W_2$,we have:$MB = n 	imes \frac{MB}{2}$.Solving for $n$,we get $n = 2$.

The work done in turning a magnet of magnetic moment $M$ by an angle of $90^{\circ}$ from the magnetic meridian is $n$ times the corresponding work done to turn it through an angle of $60^{\circ}$. The value of $n$ is:

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