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(C) The potential energy $U$ of a magnetic dipole in a uniform magnetic field $B$ is given by $U = -m \cdot B = -mB \cos 	heta$, where $	heta$ is th

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$I, IV, II, III$

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(C) The potential energy $U$ of a magnetic dipole in a uniform magnetic field $B$ is given by $U = -m \cdot B = -mB \cos 	heta$, where $	heta$ is the angle between the magnetic moment vector $m$ (which is along $\hat{n}$) and the magnetic field $B$.$(i)$ For orientation $I$, the angle $	heta = 180^{\circ}$. Thus, $U_I = -mB \cos 180^{\circ} = mB$.(ii) For orientation $II$, the angle $	heta = 90^{\circ}$. Thus, $U_{II} = -mB \cos 90^{\circ} = 0$.(iii) For orientation $III$, the angle $	heta$ is between $0^{\circ}$ and $90^{\circ}$ (acute angle). Thus, $U_{III} = -mB \cos 	heta$, which is negative (between $-mB$ and $0$).(iv) For orientation $IV$, the angle $	heta$ is between $90^{\circ}$ and $180^{\circ}$ (obtuse angle). Thus, $U_{IV} = -mB \cos 	heta$, which is positive (between $0$ and $mB$).Comparing the values: $U_I = mB$, $U_{IV} > 0$, $U_{II} = 0$, and $U_{III} Therefore, the decreasing order of potential energy is $I > IV > II > III$.

$A$ current-carrying loop is placed in a uniform magnetic field $B$ in different orientations $I$, $II$, $III$, and $IV$ as shown in the figure. The correct order of decreasing potential energy is ($\hat{n}$ is the unit vector normal to the plane of the loop).

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