(a) A current-carrying circular loop lies on a smooth horizontal … - Vidyadip

Question

(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?

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Answer

(a) No, because that would require $\tau$ to be in the vertical direction. But $\tau=I A \times B$, and since $A$ of the horizontal loop is in the vertical direction, $\tau$ would be in the plane of the loop for any $B$.
(b) Orientation of stable equilibrium is one where the area vector $A$ of the loop is in the direction of external magnetic field. In this orientation, the magnetic field produced by the loop is in the same direction as external field, both normal to the plane of the loop, thus giving rise to maximum flux of the total field.
(c) It assumes circular shape with its plane normal to the field to maximise flux, since for a given perimeter, a circle encloses greater area than any other shape.

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