A circular loop of radius a carries an electric current i . A very...

The north pole of a very long bar magnet acts approximately as a magnetic monopole. When placed at the centre of the circular loop, the magnetic field $\vec{B}$ produced by it at any point on the periphery of the wire is directed radially outward.

Let the current element of the circular loop be $d\vec{l}$. Since the wire is circular and the field is radial, the angle between the current element $d\vec{l}$ (which is tangential) and the magnetic field $\vec{B}$ is $90^{{}^{{}^{\circ}}}$.

The magnetic force on the small current element is given by:
$d\vec{F} = i(d\vec{l} \times \vec{B})$

The magnitude of this force is:
$dF = i \cdot dl \cdot B \sin 90^{{}^{{}^{\circ}}} = i B dl$

By the properties of the cross product, the direction of $d\vec{F}$ is perpendicular to both $d\vec{l}$ and $\vec{B}$, which means it is perpendicular to the plane of the circular wire.

Integrating this force over the entire circumference of the loop, the total force is:
$F = \int i B dl = i B \int dl = i B (2\pi a) = 2\pi a i B$

Since a real bar magnet is not an ideal point monopole (the pole is distributed and slightly inside the magnet), the magnetic field may have a very small component non-parallel to the plane. Hence, the net force is very nearly $2\pi a i B$ and is directed perpendicular to the plane of the wire.