$(a)$ $A$ current-carrying circular loop lies on a smooth horizontal plane. Can a uniform magnetic field be set up in such a manner that the loop turns around itself (i.e.,turns about the vertical axis)?
$(b)$ $A$ current-carrying circular loop is located in a uniform external magnetic field. If the loop is free to turn,what is its orientation of stable equilibrium? Show that in this orientation,the flux of the total field (external field $+$ field produced by the loop) is maximum.
$(c)$ $A$ loop of irregular shape carrying current is located in an external magnetic field. If the wire is flexible,why does it change to a circular shape?

(N/A) No,because that would require the torque $\tau$ to be in the vertical direction. Since the area vector $\vec{A}$ of the horizontal loop is in the vertical direction,the torque $\vec{\tau} = I(\vec{A} \times \vec{B})$ would always lie in the horizontal plane of the loop for any uniform magnetic field $\vec{B}$. Thus,it cannot rotate about the vertical axis.
$(b)$ The orientation of stable equilibrium is one where the area vector $\vec{A}$ of the loop is parallel to the external magnetic field $\vec{B}$. In this orientation,the magnetic field produced by the loop is in the same direction as the external field,both being normal to the plane of the loop. This alignment maximizes the total magnetic flux through the loop.
$(c)$ $A$ flexible current-carrying loop in a magnetic field tends to maximize the magnetic flux through it to reach a state of minimum potential energy. For a fixed perimeter,a circular shape encloses the maximum area. Therefore,the loop adopts a circular shape with its plane normal to the magnetic field to maximize the total flux.