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Question

(A) The potential energy $U$ of a magnetic dipole (current-carrying coil) in a magnetic field $\vec{B}$ is given by $U = -\vec{M} \cdot \vec{B} = -MB

Vedclass · Accepted Answer

$I > III > II > IV$

Vedclass · Answer

(A) The potential energy $U$ of a magnetic dipole (current-carrying coil) in a magnetic field $\vec{B}$ is given by $U = -\vec{M} \cdot \vec{B} = -MB \cos 	heta$,where $	heta$ is the angle between the magnetic moment $\vec{M}$ (which is in the direction of the area vector $\hat{n}$) and the magnetic field $\vec{B}$.Given that the magnetic field $\vec{B}$ is directed towards the right:In position $I$,$\hat{n}$ is to the left,so $	heta = 180^\circ$,$U = -MB \cos(180^\circ) = +MB$ (Maximum).In position $II$,$\hat{n}$ is downwards,so $	heta = 90^\circ$,$U = -MB \cos(90^\circ) = 0$.In position $III$,$\hat{n}$ is at an obtuse angle,so $\cos 	heta$ is negative,making $U$ positive.In position $IV$,$\hat{n}$ is at an acute angle,so $\cos 	heta$ is positive,making $U$ negative (Minimum).Comparing the angles: $	heta_I = 180^\circ$,$	heta_{III} > 90^\circ$,$	heta_{II} = 90^\circ$,$	heta_{IV} Thus,the potential energy order is $U_I > U_{III} > U_{II} > U_{IV}$.

The figure represents four positions of a current-carrying coil in a magnetic field directed towards the right. $\hat{n}$ represents the direction of the area vector of the coil. The correct order of potential energy is:

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