# Calculate how long it takes a neodymium magnet to fall down a copper tube? [closed]

Assuming we know the dimensions of the copper tube and the magnet, what other variables would one need to calculate the amount of time it takes a neodymium magnet to fall down the copper tube and how would one actually calculate it.

Edit: What I've tried:

I found this video https://www.youtube.com/watch?v=2-iEVFICIqM which explains how to calculate the velocity as a function of time.

From here, though, I'm not quite sure how to proceed. I know that taking the integral of the velocity function would give me the distance as a function of time, but I've never actually seen someone do this as the equation is the same as a terminal velocity equation. Also, when I tried to do it anyway, I'm not sure how to isolate the time variable as it is present within the exponent of e and outside of the exponent.

I'm not sure if I'm on the right track by taking the integral of the solution provided in the Youtube video or if I'm completely off track.

Edit 2: Also, if the exact equation is not possible, is there a way to mathematically show that an increase in the magnetic field (or just the strength of the magnet) would translate into an increase in the time it takes to fall down the copper tube?

## closed as off-topic by DanielSank, Kyle Kanos, ZeroTheHero, Cosmas Zachos, Aaron StevensJun 27 at 16:55

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• There is no general solution to this problem without knowing (in detail) the field of the magnet. I suspect that in most cases there will be no closed form solution at all, though if we assume, say, an ideal dipole and a sufficient level of symmetry to the problem (magnet dropped down the symmetry axis with its own axis aligned) we might get a problem that that at least has a closed form solution. I, however, am not masochistic enough to try it. – dmckee Jun 13 at 0:00
• @dmckee Is there way to just to correlate an increase in magnetic field with an increase in time required to fall down the copper tube? – AmazingGrace Jun 13 at 0:11
• The full mechanics of the problem depend on the detailed 3D shape and strength of the field (including orientation), which is not a trivial problem. Getting an effective value for a given magnet is an easy experimental problem, but if you want a first principles calculation rather than a semi-empirical fit you are into the realm where most people would reach for a numeric code. – dmckee Jun 13 at 0:23

## The commenters are correct, but we can still get somewhere.

• It's true that you need to know the complete 3D field of the magnet in the same way you can't calculate how a non-magnetic object will fall without knowing the complete 3D gravitational field of Earth. Start by modeling an ideal dipole, and add complexity as needed.
• Assuming that the magnet falls axially down the pipe is probably what you meant, and not too hard arrange in practice.
• Pneumatic forces and friction may complicate matters, but you can just keep getting heavier and stronger magnets until electromagnetic and gravitational forces dominate. (or just glue some lead weights to your magnet)
• Calculating the acceleration of the weight probably is relatively hard, but I bet we can skip it. If we can show that the weight reaches a terminal velocity quickly after entering the pipe, then maybe we can just ignore the time during which it's accelerating.

## Assuming a spherical cow in a vacuum...

Once the weight reaches terminal velocity, it will be falling down (moving from higher to lower gravitational potential), but not accelerating (gaining kinetic energy).
So where is that energy going?

The energy is going to resistive heating of the pipe.

What's the power?

$$E_{potential/distance} = mg$$ $$\therefore P = mgv_t$$ where $$g$$ is the force of gravity and $$v_t$$ is the terminal velocity.

We know (don't we?) that all the electrical fields and currents will be around the pipe (there will be no $$V$$ or $$I$$ component along the axis of the pipe), so we can treat the pipe as an infinitely long stack of infinitesimally thin conducting loops. Assuming pipe-radius $$r$$ substantially greater than the wall-thickness $$w$$, and a uniform material resistivity $$\rho$$, we have the conductance/pipe-length $$C_l$$ $$C_l = \frac{w}{2\pi\rho r}$$ Therefore the resistive power per pipe-length will be $$P_l = \mathcal{E}^2C_l$$ Faraday's law: $$\mathcal{E} = \frac{d\Phi}{dt}$$

Now we can start combining things: $$mgv_t = P$$ $$= \int_{-\infty}^{\infty}P_l \;dl$$ $$= \int_{-\infty}^{\infty}\mathcal{E}^2C_l \;dl$$ $$= \int_{-\infty}^{\infty}\left(\frac{d\Phi}{dt}\right)^2C_l \;dl$$

Under the conditions we're assuming, taking the derivative of the flux through a given loop with respect to time is sort of like taking the derivative of the flux through adjacent loops with respect to axial offset $$l$$. Specifically, $$\frac{dx}{dt} \equiv v_t\frac{dx}{dl}$$. $$mgv_t = \int_{-\infty}^{\infty}v_t^2\left(\frac{d\Phi}{dl}\right)^2C_l \;dl$$ $$\therefore v_t = \frac{mg}{\int_{-\infty}^{\infty}\left(\frac{d\Phi}{dl}\right)^2C_l \;dl}$$

So if we really want a value for terminal velocity, we'll need a tractable function for the flux $$\Phi$$ through these loops in terms of their axial offset from the weight. For our idealized dipole that should be doable, but it's more calculus than I'm in the mood for. But we can already start to make predictions about the system:

• If we double our mass but keep the same magnet strength, it'll fall twice as fast.
• If we double our pipe-wall thickness, the weight will fall half as fast.
• If we use two identical magnets while keeping the same total mass (by removing some of the lead weight), the weight will fall a quarter as fast (because fluxes add linearly).
• We can't yet say what will happen if we change the pipe radius. The affect on our conductance equation is easy enough, but you'll have to figure out what will happen to the flux...

## EDIT:

Something must be wrong above!

• If we double our mass but keep the same magnet strength, it'll fall twice as fast.
• If we use two identical magnets while keeping the same total mass, the weight will fall a quarter as fast.
• That means that if you just stick two identical magnets to each other, they'd fall half as fast as either one alone. That... might be right? I don't have good magnets or pipes on hand to test with :(