# Magnetic force and terminal velocity of falling loop and falling disc?

According to my textbook, if a square loop with mass $M$ is allowed to free fall (with a magnetic field at the bottom side and as soon as the square loop is dropped then the area inside the loop begins to be filled with more and more magnetic field) through a magnetic field, it will reach a terminal velocity. This is due to the increase in magnetic flux through the loop which creates a current in the loop opposing the change in magnetic flux and then a force is then exerted on the bottom side which balances out the gravitational force. The force on the loop sides balance each other out and by the time the top side enters the magnetic field then the flux is not changing. The terminal velocity is given as:

$$V=\frac{MgR}{B^2w^2}$$ (first picture below)

where $M$ is loop mass, $g$ is gravity, $R$ is resistance of loop, $B$ is magnetic field and $w$ is width of the loop

However, when a metal disc is dropped from just below an infinite wire carrying a current I, there is no terminal velocity even though the magnetic flux through the disc (falls so that magnetic field is perpendicular to the circular face where $A=\pi r^2$ is changing due to the change in magnetic field strength as the disc falls further from the wire. (Second picture below)

Why is this?  When you have a current-carrying wire, the rate of change of flux for a given velocity is a function of position. In fact, as you move further away it gets smaller (the disc is a finite size, and the field of the straight wire drops with $1/r$ ) - so the velocity at which the force is balanced keeps getting bigger. And that means there is no terminal velocity - because as you keep falling, that "terminal velocity" keeps getting bigger. Which is exactly what terminal velocity doesn't do...