Does circular motion of electron perpendicular to the magnetic field experiences a force?
Yes; instantaneously, the particle will always have a velocity perpendicular to the magnetic field, so there will be a Lorentz force \$q\vec{v}\times \vec{B}\$, regardless of what extra external forces you apply.

Any extra forces you apply externally to maintain a circular orbit perpendicular to the magnetic field will have the effect of either maintaining a larger or smaller orbit than the gyroradius.

If you think about it, there are only 2 non-zero options for this force which will maintain a perfectly circular orbit perpendicular to a uniform magnetic field:


It is perpendicular to the particle pointed towards the center of the orbit. This case is trying to maintain a smaller orbit than the natural gyroradius.
It is perpendicular and pointed away from the center of the orbit. This case is trying to maintain a larger orbit than the natural gyroradius.


Another simpler way to resolve the question is to notice that \$\frac{d\vec{v}}{dt}\$ is not zero for any circular orbit (it rotates); this is the definition of acceleration, and the only way to produce an acceleration is by applying a force.