Magnetic force on a current element

For a small current element $d\vec{\ell}$ carrying current $I$ in an external magnetic field $\vec{B}$, the magnetic force is $d\vec{F} = I\, d\vec{\ell} \times \vec{B}.$

Effect on a flexible current-carrying loop

Since the wire is flexible, the magnetic forces can change its shape.

Each small element experiences a force perpendicular to both $d\vec{\ell}$ and $\vec{B}$. These forces tend to pull the loop outward in its own plane when the loop is oriented suitably with respect to the field.

Torque on the loop

A current loop in a magnetic field experiences torque $\vec{\tau} = \vec{m} \times \vec{B},$ where $\vec{m} = I\vec{A}$ is the magnetic moment, normal to the plane of the loop.

The stable equilibrium is when $\vec{m} \parallel \vec{B},$ i.e. the normal to the plane of the loop is along the field. Therefore, the plane of the loop becomes perpendicular to the magnetic field.

Why circular shape?

In this orientation, the magnetic force on each element lies radially outward in the plane of the loop. For a given wire length, the shape enclosing maximum area is a circle. The magnetic field tends to increase the area of the loop, so the flexible wire adjusts to a circular shape.

Checking options

• A: Loop assumes circular shape with its plane normal to the field. ✅ Correct.
• B: Plane parallel to the field would mean the magnetic moment is perpendicular to $\vec{B}$, not stable. ❌
• C: Wire becoming straight is not the equilibrium shape of a closed loop. ❌
• D: Shape does not remain unchanged because magnetic forces act on the flexible wire. ❌

Final answer

The loop becomes circular and its plane is normal to the magnetic field.