# A bar magnet falling inside a vertical metal tube reaches a terminal velocity even if the tube is evacuated so that there is no air resistance. Explain?

May 9, 2017

As the magnet moves through a (non-magnetic) conducting metal tube, it will induce an EMF in the tube according to Faraday's Law. That creates an opposite-acting magnetic force that is recognised in Lenz's law .

$m a t h c a l E = - \frac{d \Phi}{\mathrm{dt}}$ (the negative sign recognises that the EMF acts to oppose the motion creating it, ie Lenz's law)

...then, we might suggest that the change in flux $\Phi$ is entirely due to the change in position $v$ of the magnet with time, as the strength of the magnet, and the geometry of the magnet and metal tube, are fixed.

On that basis:

$m a t h c a l E = - \alpha \frac{\mathrm{dx}}{\mathrm{dt}} = - \alpha v$

It also seems reasonable to argue that the back magnetic force is therefore a function of $v$. From there Newton's Law tells us that:

$m a = m g - \alpha v$.

This is a separable DE in form:

$\frac{\mathrm{dv}}{\mathrm{dt}} = g - \frac{\alpha}{m} v$

With $v \left(0\right) = 0$, this has solution:

$v \left(t\right) = \frac{m g}{\alpha} \left(1 - {e}^{- \frac{\alpha}{m} t}\right)$

And ${\lim}_{t \to \infty} v \left(t\right) = \frac{m g}{\alpha}$

That's very much a first stab but it suggest that there will be a trade off and that a terminal velocity exists.

Here's a really cool YouTube Vid on it

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