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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?

Square loop

Washer

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It's because with the square loop, it's the top or the bottom horizontal wire that is giving the force (depending on if the loop is entering into the field region from above or exiting on the bottom). The Lorentz force law shows F = q(v x B) where v is the velocity of the moving charges, not the velocity of the wire/disk.

In the case of the wire entering the field region from the top, the flux is increasing so a current will be induced to decrease the field within it, which means the current would flow clockwise. If the current is flowing clockwise then charge is moving to the left on the bottom part of the square loop. If you look at q*(v x B), that force vector points upward and opposes gravity.

In the case of the disk falling away from the wire the current would flow clockwise to increase the flux of the field that's pointed into the screen (which is decreasing as 1/r from the fall). In this case, the charge velocity is in the angular direction and the resulting q*(v x B) force will be in the radial direction and hence cancel out.

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If I understand correctly, in the first case the rate of change of flux is constant (the area of the loop intersecting the homogeneous field is increasing linearly with position). If the rate of change of flux is sufficient, then the induced current causes a force that opposes this change, and you reach terminal velocity.

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...

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  • $\begingroup$ I added pictures for clarification. In the first picture, I would think that once the loop fully enters the magnetic field, the change in magnetic flux is zero so it would start accelerating again. I understand that the flux is constantly changing for the disc falling from the infinite wire, but force is not taken into account at all in the velocity. Wouldn't it therefore fall at some rate given by F=Fg-Fb $\endgroup$ – Prevost Jul 4 '14 at 12:07
  • $\begingroup$ And with regards to the first paragraph, would the change in magnetic flux be constant while the loop is falling? I don't think it would be because the loop is accelerating due to gravity, which increases it's velocity so the rate of change in magnetic flux is....decreasing as it falls due to the magnetic force opposing the gravitational force. I think. $\endgroup$ – Prevost Jul 4 '14 at 13:35

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