A uniform circular wire of radius a has an electric current i...

When the current $i$ enters the uniform circular wire at one point and exits at the diametrically opposite point, it divides equally into two semicircular paths. The current in each semicircular part is $\dfrac{i}{2}$.

The magnetic field at the centre due to the first semicircular part is $B_1 = \dfrac{\mu_0 (i/2)}{4a}$, directed perpendicular to the plane of the loop (say, inwards).

The magnetic field at the centre due to the second semicircular part is $B_2 = \dfrac{\mu_0 (i/2)}{4a}$, directed perpendicular to the plane of the loop in the opposite direction (outwards).

The net magnetic field at the centre of the circular wire is the vector sum of these two fields:

$B_{net} = B_1 - B_2 = 0$

The magnetic force on a charged particle moving with velocity $\vec{v}$ is given by $\vec{F} = q(\vec{v} \times \vec{B})$.

Since the net magnetic field at the centre is zero, the magnetic force acting on the particle at the instant it crosses the centre is also zero.