erm, well the magnetic field generated by a (cylindrical) solenoid is parallel to the axis of the cylinder the coils are wound around( and to a good approx uniform over the cylinder's cross section). If you reverse the current then the magnetic field will be in the opp direction, i.e. still parallel to axis but in other direction.
Magnetic fields interact with charged particles via the Lorentz Force law: F=q(v x B) (assuming no net electric fields present). Thus there is only a force on the particle if it moves perpendicular to the cylinder's axis (radially in cylindrical polar coords).
If a +ve particle initially moves radially outwards from the centre, and the current is such that the mag field inside the solenoid is to the right, the particle will experience a force that is again radially outward but at 90 degrees to the plane containing v and B. If the current was DC this would lead to the particle moving in a circle in the plane of the solenoid's cross section (I think as long as v isn't to great relative to field strength, otherwise it might leave the solenoid and be out of the mag field, hence it would fly off).
Now assuming the particle is moving along some orbital trajectory, reversing the field would mean the centripetal force also reverses direction so the particle begins moving on a *different orbital path* that touches the first.
I have attached a jpeg of what I think it would look like. To get the actual radius of curvature that the particle moves along you would have to use:
mv^2/r=qvB
=>r=qB/mv
and the portion of the circles orbit completed before "jumping onto another circular orbit" would depend on the AC frequency.
If particle was moving parallel to axis as well as radially, you'd probably end up with a kind of screwdriver motion, helix upon anti-helix, upon helix etc, that's my best guess anyway...
