Someone having fun with magnets

[belwar]

Magnet drops super-slowly through an eddy tube - Snotr

More of a side thread. I cant even begin to comprehend the magnetic theory behind this!

Someone explain it in plain commoner english for me please!

b.

[David King]

http://www.physics.ubc.ca/~outreach/...eandmagnet.mpg

Well it's basically some form of dynamic breaking. You have a very heavy, single turn coil of highly conductive copper and it's shorted out at the ends so whatever current the magnet generates falling through it is going into pushing the magnet back up the tube.

Check out magnet dampers on triple beam balance scales.

[Possum]

...

Thats a cool video and classic demonstration of eddy currents. The magnet generates current in the copper through motion, these are eddy currents that oppose the magnetic field so the magnet slows way down. There are elevator brakes that work this way, and this why the more wire you put on a pickup the more treble gets lost...

[Joe Gwinn]

That's a cool video and classic demonstration of eddy currents. The magnet generates current in the copper through motion, these are eddy currents that oppose the magnetic field so the magnet slows way down. There are elevator brakes that work this way,

In bullet form:

Moving magnet drags its magnetic field with it, generating a moving magnetic field.

A stationary observer sees this as a changing field - first it increases, then it decreases. (It may do more complicated things, depending on details, but the simplest case suffices for this explanation.)

Changing magnetic fields induce currents in nearby conductors, like the pipe. One can do much the same demo by sliding the magnet down a teflon-coated aluminum cookie sheet, so it is not essential that a tube be used.

By Lenz's Law, the induced currents will generate magnetic fields that oppose the change that generated the currents. Lenz's law - Wikipedia, the free encyclopedia

Lenz's Law is a consequence of the Conservation of Energy. With the magnet falling through the pipe, the magnet arrives at the bottom with far less kinetic energy than it would have absent the pipe. Where did the missing energy go? Into the pipe. The eddy currents dissipate energy as heat in the metal.


If one spins an aluminum disk and arranges the disk edge to move vertically between the poles of a strong magnet, the apparent weight of the magnet assembly will vary with disk RPM and magnetic field strength at the disk, and if the apparent weight variation is large, the disk will become warm to the touch. (I have done this experiment using my research winder and a pair of rare earth magnets in an old drill press vice. The vertically part is needed only if one will measure drag torque using a lab scale.)

...and this why the more wire you put on a pickup the more treble gets lost...

Not exactly. While eddy currents in the metal parts of the pickup do affect highs more than lows, this is independent of the amount of wire wound on the pickup. The wire is too fine for eddy currents in the wire itself to be important at audio frequencies.

[David Schwab]

This is also how they break roller coaster cars.

[Mike Sulzer]

Not exactly. While eddy currents in the metal parts of the pickup do affect highs more than lows, this is independent of the amount of wire wound on the pickup. The wire is too fine for eddy currents in the wire itself to be important at audio frequencies.

Well, right, the inductance is the reason, but that is caused by the same law of magnetic induction.

But I do not think what you have said about the frequency dependence of eddy currents in pickups is correct:

1. The coil-core structure in a pickup leaks a lot of flux because the core is open. Therefore the correct model has a leakage inductance in series with the core resistance. At high frequencies the high inductive impedance means that reduced current flows. (This is why it is hard to make a very wide bandwidth transformer. You need high permeability and lots of turns to get a large magnetizing inductance for good low frequencies, but you have to keep the leakage flux small for it to work at high frequencies.)

2. The high permeability of the core contributes to the small skin depth. The skin depth decreases with increasing frequency (square root) raising the resistance and further reducing the current.

This has been verified experimentally; both those effects must be accounted for in order to get a model of the a humbucker impedance to match the measured results.

[Joe Gwinn]

Well, right, the inductance is the reason, but that is caused by the same law of magnetic induction.

But I do not think what you have said about the frequency dependence of eddy currents in pickups is correct:

1. The coil-core structure in a pickup leaks a lot of flux because the core is open. Therefore the correct model has a leakage inductance in series with the core resistance. At high frequencies the high inductive impedance means that reduced current flows. (This is why it is hard to make a very wide bandwidth transformer. You need high permeability and lots of turns to get a large magnetizing inductance for good low frequencies, but you have to keep the leakage flux small for it to work at high frequencies.)

2. The high permeability of the core contributes to the small skin depth. The skin depth decreases with increasing frequency (square root) raising the resistance and further reducing the current.

This has been verified experimentally; both those effects must be accounted for in order to get a model of the a humbucker impedance to match the measured results.

It's the eddy currents in the non-wire metal that does not depend on the winding.

[Mike Sulzer]

It's the eddy currents in the non-wire metal that does not depend on the winding.

The currents induced in the cores of humbucker are considered eddy currents. They do not affect the highs more than the lows, as I explained in my previous post.

[David Schwab]

What's everyone think of this quote?

Eddy currents alter sound and output of a pickup and play an important role in pickup design.

Eddy currents are induced in metals in the vicinity of an AC magnetic field, creating a secondary magnetic field which opposes the inducing magnetic field of the coil. The dimensions, conductivity and permeability of the metal, along with the frequency of the current in the coil, determine the magnitude and phase relation of the eddy currents. An internal short in a pickup coil forms a conductive loop which, also, becomes the source for internal eddy current interference.

Bill Lawrence Website

[Joe Gwinn]

The currents induced in the cores of humbucker are considered eddy currents. They do not affect the highs more than the lows, as I explained in my previous post.

Heh? Eddy current loading is proportional to the square root of frequency, so they most definitely affect highs more than lows.

[Mike Sulzer]

Heh? Eddy current loading is proportional to the square root of frequency, so they most definitely affect highs more than lows.

Under what conditions? I think I have shown that that is not the case for humbucker pickups. The pickup magnetic circuit with its open pole pieces is nothing like the closed core of a transformer or many other types of magnetic circuits. There is no reason why eddy currents shouold have the same behavior.

[Joe Gwinn]

Under what conditions? I think I have shown that that is not the case for humbucker pickups. The pickup magnetic circuit with its open pole pieces is nothing like the closed core of a transformer or many other types of magnetic circuits. There is no reason why eddy currents should have the same behavior.

It does not matter if the magnetic circuit is closed or not. Eddy current loading varies with the skin depth, which varies in inverse proportion to the square root of frequency.

Laminating the transformer iron sharply reduces eddy currents below some frequency that depends on the lamination thickness, but most pickups have solid magnets, poles, and covers.

[Mike Sulzer]

It does not matter if the magnetic circuit is closed or not. Eddy current loading varies with the skin depth, which varies in inverse proportion to the square root of frequency.

Laminating the transformer iron sharply reduces eddy currents below some frequency that depends on the lamination thickness, but most pickups have solid magnets, poles, and covers.

"skin depth, which varies in inverse proportion to the square root of frequency": Exactly, this means that as the frequency goes up, the skin depth goes down and the resistance goes up. If this resistance is driven by a voltage source through some impedance, then less energy is dissipated as the frequency goes up. In the case of a humbucker pickup core, the skin depth is small enough so that we can think of the voltage induced around the core by the current flow in the pickup coil as resulting in a current around the outer layer of the core, that is, its "skin". This current decreases at higher frequencies because the skin depth decreases, increasing the resistance of the path.

The complicating factor is that the flux leakage due to the open core means that there is a "leakage inductance" appearing in series with this resistance. The impedance of this inductance also increases with frequency, meaning that the energy dissipated in the core decreases with frequency faster than from the increasing resistance alone.

This was included in the last discussion on modeling a humbucker pickup. Both effects have to be included to in the circuit model in order to get good agreement with the measured results.

[David Schwab]

Laminating the transformer iron sharply reduces eddy currents below some frequency that depends on the lamination thickness, but most pickups have solid magnets, poles, and covers.

Right. Laminating the pickup's core also reduces high frequency loss, assuming you are using a blade.

It's clear that adding a metal cover, or even a brass baseplate reduces the high end.

[Mike Sulzer]

Right. Laminating the pickup's core also reduces high frequency loss, assuming you are using a blade.

It's clear that adding a metal cover, or even a brass baseplate reduces the high end.

Yes, but let's look at the reason why. The high end is set by the pickup resonance in its operating circuit, including the cable capacitance. The resonance frequency has the highest impedance; it is where the impedance of the L and C cancel out. This is the frequency where a resistor across the pickup has the most effect. (For example, the volume control affects the high frequencies for just this reason. If you use a pot with a smaller than normal value, you get less highs.) The losses due to eddy currents are like a resistor across the pickup, and so they affect the high frequencies most because the impedance is high at these high frequencies emphasized by the resonance.

But this does not tell us how the equivalent resistance of the eddy current effect varies with frequency. It turns out that this resistance is lower at lower frequencies, low enough to affect a humbucker's impedance at frequencies well below the resonance. This is something one might not expect, but in order to model the total impedance of the pickup, including the effects of eddy currents, one must use a resistor that increases in value with frequency. But a resistance is not sufficient. One must also use a series inductance.