How to Model a Magnet Falling Through a Conducting Pipe

[kuruman]

Introduction
In an earlier article, we examined a magnet falling through a solenoid. We argued that the point dipole model can account for the basic features of the induced emf across the solenoid ends. Here, we extend the model to a magnet falling through a conducting pipe along its axis.
With the falling dipole moment oriented along the vertical $z$-axis, the electric field $E(\rho,z)$ is tangent to circles centered on the axis.  The induced emf around a closed loop of radius $\rho$ is $\text{emf}(z)=2\pi \rho~E(\rho,z)$. There is a similarity and a difference between a solenoid and a conducting pipe placed in that space.  The similarity is in their modeling as a stack of conducting rings. The difference is in the conceptual connection of the constituent rings.  In the solenoid, the rings are in series and the overall emf across them is of interest.  In the pipe...

Continue reading...

 

[tech99]

As the magnet is continuously moving down the tube, the current would seem to follow a helical, rather than circular, path.

 

[kuruman]

As the magnet is continuously moving down the tube, the current would seem to follow a helical, rather than circular, path.

Can you elaborate? The assumption is that the point dipole is perfectly aligned along the z-axis and is moving along it. Therefore the emf is perfectly azimuthal. I can see a helical current in a solenoid if its ends are connected to something but not a pipe.

 

[vanhees71]

I guess it's a very complicated problem, if you consider the case that the magnet in addition to falling down in the specified orientation is in addition rotating. Then of course the current in the pipe becomes much more complicated too.

 

[Herman Trivilino]

If you cut a slit in the pipe in the axial direction the effect is still present.

 

[kuruman]

If you cut a slit in the pipe in the axial direction the effect is still present.

Sure, and if you flatten out the cut pipe into a rectangle, there will still be an induced current opposing the motion of the dipole. Free charges will always be pushed around if present and Lenz's law is here to stay.

 

[tech99]

I guess it's a very complicated problem, if you consider the case that the magnet in addition to falling down in the specified orientation is in addition rotating. Then of course the current in the pipe becomes much more complicated too.

The magnet makes a lot of noise when descending. When it drops out of the bottom it does not seem to be spinning (due to a helical current), but I can't be sure. Maybe the vertical axis of the magnet rotates (like precession) as it goes down?

 

[vanhees71]

Sure, I guess in practice if you release a bar magnet with its polarization pointing parallel to $\vec{g}$ it will stay in this orientation with sufficient accuracy. Also the experimental results in the papers quoted in this thread and in the Insights article are pretty well in agreement with the predictions (at least when the model of a homogeneously magnetized cylinder is used).

I'm pretty sure that a treatment including a possible spinning of the magnet is very complicated and, if satisfactorily treatable at all, only numerically possible.

 

[AndyG]

Simple experiment - put a paint mark on the top of the magnet, drop it down the tube and watch down the tube as the magnet falls. Your observation will indicate whether the magnet rotates or not.

 

[kuruman]

Simple experiment - put a paint mark on the top of the magnet, drop it down the tube and watch down the tube as the magnet falls. Your observation will indicate whether the magnet rotates or not.

That's a simple enough observation. It also seems to me that there is no reason for the magnet to rotate as it falls. We know that Lenz's law dictates that eddy currents will always form in conductors to generate forces and torques that oppose the motion of magnets relative to such conductors. So if the magnet is dropped without initial spin, why should it start spinning?

I used to do this convincing demonstration of Lenz's law. Roll a wheel-shaped neodymium magnet down an aluminum incline. We all expect it to lose the race against a geometrically identical non-magnetized cylinder. But not always. If the non-magnetized wheel is angled towards one of the edges of the incline, it will roll off. When the magnetized wheel reaches an edge, it will twist away, cross over to the opposite edge as it rolls down, twist again and thus zig-zag down the length of the incline. How does it "know" to do that? Simple: you can't beat Lenz's law.

 

[Herman Trivilino]

So if the magnet is dropped without initial spin, why should it start spinning?

Wouldn't it be very difficult to drop it bare-handed with negligible spin?