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Significantly lowering the temperature of a neutrally charged rectangular semiconductor can cause the charges on the semiconductor to polarize. (Removing energy from a semiconductor will influence the molecular structure, apparently, quasi-analogous to spin polarization, which might be the mechanism for charge polarization in this problem.) Moving the polarized rectangular semiconductor into a "stationary" magnetic field is same as moving a "stationary" magnetic field toward, or away, from the stationary electrostatic field of the polarized rectangular semiconductor. The external energy supplied to the system by the motion of bringing the magnetic field and the polarized semiconductor closer together will cause the rectangular semiconductor to rotate about its center of axis via Lorentz's Force equation.

The electric field of the rectangular semiconductor can be quantified in coulombs as +q and -q. That is, +q at one end and -q at the other end of the rectangular semiconductor. A semiconductor glass rod that has static charges of +q and-q is similar to the charged rectangular semiconductor in this problem. Think of the charged rectangular semiconductor as being supported on a string at its center of gravity. Now apply Lorentz's Force equation as the rectangular semiconductor is being moved closer toward the "stationary" magnetic field. The resulting Lorentz's forces on the rectangular semiconductor will cause the rectangular semiconductor to rotate about its center of gravity virtually along the axis of the string. (Assume that the magnetic field intensity is perpendicular to the plane of the rectangular semiconductor, and that the rectangular semiconductor moves closer to the field at an angle with respect to the string axis.)

Now that the polarized rectangular semiconductor is rotating about its center of gravity because of Lorentz's forces, a counter magnetic field is induced in accordance with Faraday's law. That induced magnetic field opposes the driving magnetic field in accordance with Lenz's law. So, it will appear to float on top of the magnet. (Resistance to air will eventually cause the rectangular semiconductor to reach a rotational equilibrium, and eventually use-up the external energy supplied to the system in moving the magnetic field and electric fields closer together. I.e., a terminal rotational speed, and eventual slow down.)

As the rectangular semiconductor warms-up to ambient temperature, the polarized charges of the rectangular semiconductor should return to a neutral condition similar to that prior to emersion into liquid nitrogen. The neutrally charged rectangular semiconductor will also fall with respect to gravity as the electrical polarization decreases, because there is no longer an opposite magnetic field being induced. (There could be an electric "hysteresis" curve, with respect to time, with temperature normalization, but that would apparently be within the equations of spin polarization.)

As mentioned earlier, the polarized rectangular semiconductor will fall with respect to gravity as the external energy supplied to the system (when the electrically charged rectangular semiconductor was moved to closer proximity with the magnetic field intensity) is used-up by friction from air resistance. In other words, a significant portion of the mechanical energy supplied to the system will eventually be transferred to the surrounding air and space as the semiconductor rotor, slows down, quasi-analogous to a mechanical flywheel slowing down.