It will experience some force but it won't participate in the group of forces which are making to coil to perform the circular motion.
Let's understand with this equation:-
$F = i(\vec L × \vec B)$
Here $i$ is the current, L is the vector in the direction of flow of current with the magnitude equal to the length of that side of the loop which we are focusing on. (But in the case of circular coil the case will be different) and $B$ here is the magnetic field vector with the magnitude equal to the field strenght at that point (assuming to be constant everywhere in this case).
So as we can see that in the case of $BC$ and $AD$ as you said when they will make the angle other than zero degree then in what direction the force will be exerted by the magnetic field? Let's jump on the above equation again and find out (assuming that you are aware of the concept of vector cross product):-
$F = i(\vec L × \vec B)$
$F = i(LB sinθ n̂)$
Where n̂ is unit vector perpendicular to $\vec L$ and $\vec B$.
So, when the side $BC$ and $AD$ will make some random angle other than zero with the magnetic field vector (According to your diagram at that instant let us assume that the direction from B to C is $î$, direction from A to B is $\hat{j}$ and direction vertically outwards is $k̂$) So current flowing in side $BC$ and $AD$ are in the directions $î$ and $- î$ respectively, and the magnetic field is in the direction that's north pole to south pole which is $î$ so if we find the force (assuming that $\vec L$ and $\vec B$ are perpendicular after some instant so our calculations will be a bit easier, non zero angle as you are asking for and very easy to digest the concept as well and assuming that when the $\vec L$ and $\vec B$ are perpendicular then the current flowing in $BC$ is in the direction $k̂$) we will get:-
$F = LB (k̂ × î)$
$F = LB \hat{j}$
So as we can see that the force acting on the side $BC$ is in the direction of $\hat{j}$ and similarly as the current flowing in the side $AD$ will be in the opposite direction as that of $BC$ i.e $- k̂$ So the force on side $AD$ will be in the direction $- \hat{j}$ so as these forces will be equal in magnitude (length of AD and BC are the same and magnetic field is already constant) and opposite in the directions as well so the net force will come out to be zero and because of having the same line of action the torque will come out to be zero too, because torque is nothing but the moment of force (i.e $|τ| = Fr sin θ$ in this case angle between the force vector and the radial vector is going to be be zero, hence the torque will also be zero) and therefore it won't do any circular motion or tend to do circular motion. 780