Need a little help (transformer)

[NascentOxygen]

I wasn't counting the flux lines, it's their distribution that doesn't look right. Insertion of the iron core should bring a concentration of field lines with some emerging from the ends of the rod.

The pattern in your image looks appropriate for an air core.

 

[jim hardy]

I wasn't counting the flux lines, it's their distribution that doesn't look right. Insertion of the iron core should bring a concentration of field lines with some emerging from the ends of the rod.

The pattern in your image looks appropriate for an air core.

You're quite right

sorry , N.O. i thought you were teasing me about the primitive drawing

My thinking is iron core roughly doubles the flux density in its vicinity
because around any loop that traverses it, roughly half the path is low reluctance iron and the other half remains high reluctance air

so yes of course replacing my cardboard core with an iron one should add a few lines to my sketch, like this blue one,
and roughly double the number of purple ones.

I've done that experiment on a 12 foot long coil. We too were surprised flux only doubles.

Thanks for your sharp eye !

PS _{i removed the snickering kitty}

 

[sophiecentaur]

@Ruby: It's a shame we couldn't be there to help you during your 'confrontation'.

With bifilar on a rod, cardboard core gives roughly half as much flux per amp as iron. That's counterintuitive because iron's relative permeability is so high.

Could it be that's the point of the experiment ?

I don't think there would be any point in presenting that to students at the apparent level of the OP.
I think it's time to sit back and wait for some feedback - hopefully with information from the teacher, even.

 

[Merlin3189]

Re. #30 and #31 of Jim and NO : my understanding was that a solenoid had a uniform field inside and nearly all the flux passed through the ends, as shown on the hyperphysics solenoid page or

This would seem to suggest that if the secondary were wound on top of the centre portion of the primary (as shown in Ruby's photo), then all the flux (caused by primary current) would link both coils irrespective of the core or bifilar winding. (OTOH, perhaps not all of the flux due to any secondary current would link the primary, as the secondary is much shorter. But I (we?) haven't got that far yet.)

Re. #32 Jim, I agree with the nearly doubling of flux for any decent permeability core. The length of the air path is nearly halved and the air contributes nearly all the reluctance.

Meanwhile I've found an 8mm bolt and a couple of metres of copper wire and made my own transformer. It was about 1.7Ω primary, but I didn't have time to measure the reactance, because, my step-daughter decided to have her son a few days early and I've been whisked away to do something useful instead of playing with my toys. At least I've got my laptop, so I can keep up with the chat.

 

[jim hardy]

Re. #30 and #31 of Jim and NO : my understanding was that a solenoid had a uniform field inside and nearly all the flux passed through the ends, as shown on the hyperphysics solenoid page or

agreed. nearly all but there is leakage flux , integrate h dot dl around any single turn ?

https://sharepoint.umich.edu/lsa/ph...et Library/Two solenoids, B-field_5H15.40.JPG

This would seem to suggest that if the secondary were wound on top of the centre portion of the primary (as shown in Ruby's photo),

I didnt read his photo that way

i take yellow tape as underneath primary and green tape as underneath two halves of secondary.
But i didnt think about hooking secondaries in series with turns all going same direction...

Might that be another trouble..?

 

[NascentOxygen]

i take yellow tape as underneath primary and green tape as underneath two halves of secondary.
But i didnt think about hooking secondaries in series with turns all going same direction...

EDITED x2, incorporating corrections
The concealed winding is the primary, I believe; it looks like it has been covered with a cardboard tube.

In OP's photo, the southern-most end of the primary seems like it's the beginning of clockwise wrapping (let's call it clockwise), but, unless my eyes deceive me, the upper end seems to emerge through the green tape after a counter-clockwise wrapping. So somewhere out of view, perhaps midway, the primary winding has deliberately been reversed?

Looks like Jim might be close to unravelling the Mystery of the Missing Volts! [emoji859][emoji361]

P.S. Jim, the 25 turn winding is denoted the secondary in OP's setup, but in your copy of the image you have labelled it Primary?

 

[Merlin3189]

Oh! I saw it oppositely. The green bits as start and end of the primary winding, with an extra bit of grey insulation in the middle. Then the secondary wound on top between the two yellow bits, about 1/4 the length of the primary.

If your interpretation is right, then two windings (either primary or secondary) accidentally connected antiphase would certainly explain zero output. (I see NO has already picked up on this as well.)

Coming back to your picture(*), if the turns were very widely spaced, then it looks 'obvious' that some of the flux from each turn must do a 'short' circuit and reduce the summed flux down the solenoid. I don't know. Although I sort of remembered learning this idea of all the flux being contained inside the solenoid, I did the quick check .with Hyperphys before I posted, but didn't look past it's apparent confirmation. I must go back and read the full explanation.

Again I'd need to check to be sure, but if there is a lot of leakage, isn't that like having a fully coupled transformer with a separate inductance in series with each winding? And wouldn't that mean here that as well as part of the primary voltage being across the winding resistance, some would also be across the leakage inductance, leaving an even smaller proportion across the ideal transformer primary?.

(*) Nice source btw. I must remember this when I'm looking for useful illustrations.

 

[jim hardy]

P.S. Jim, the 25 turn winding is labelled the secondary in OP's photo, but in your copy of the image you have labelled it Primary?

Did i miss a photo ? Labels on OP's ? I don't see them in post # 18.

Are there four wires there or six? Am i mistaking shadows of wires for wires ?

I make those kind of mistakes often enough I'm not ashamed to ask.

 

[NascentOxygen]

The primary coil has 100 turns and the secondary has 25.

Did i miss a photo ? Labels on OP's ? I don't see them in post # 18.

Sorry, I should be saying that OP "denoted" the 25 turn winding as the secondary. My count of the turns on the outer-most coil is 25, so I conclude it to be the secondary. (Editing my post now to replace "labelled" with "denoted".)

Are there four wires there or six? Am i mistaking shadows of wires for wires ?

Only 4 wire ends.

 

[jim hardy]

Only 4 wire ends.

oops...

And i had to blow it up 200% to see what you meant about reversed winding direction - Good Eye !

Again I'd need to check to be sure, but if there is a lot of leakage, isn't that like having a fully coupled transformer with a separate inductance in series with each winding? And wouldn't that mean here that as well as part of the primary voltage being across the winding resistance, some would also be across the leakage inductance, leaving an even smaller proportion across the ideal transformer primary?.

I think that's so.

Air core means not much inductance
so
using Wiki's transformer model
https://en.wikipedia.org/wiki/Transformer

X_M is quite small and I_M large, lots of drop across R_P and X_P leaves not much for E_P
just as you said.

 

[Merlin3189]

Well, I'm back in my play room. First result for my iron bolt transformer:
Primary winding: 100t ± , R_P=1.7Ω ±0.1 Secondary 25t ± , R_S= 0.2Ω ±0.2 (different wire)
I set it up with 0.506A AC 50Hz and measured a primary voltage of 0.96V and secondary voltage of 0.053V
The voltage across the resistance should be 0.86V , which, assuming only an inductive component at 90^o, gives a voltage across the inductance of 0.43V.
The secondary voltage of 0.053V is a ratio of 1:8 rather than the ideal 1:4. I may well have miscounted the turns a bit, but I don't think I can have been that inaccurate. (I'm about to get winding again more carefully.) But there's enough output to measure - with a digital multimeter, but not my analogue meter. (Though whether even the digital multimeter is accurate at such low voltages, I wonder?)

I therefore doubt my claim that this setup would behave like a fully coupled transformer after allowing for the low inductance to resistance ratio.

Further, I notice that the flux in a solenoid does not increase with the number of turns in the primary (putting aside for the moment the idea that an ideal solenoid has an infinite number of turns!) The flux in a solenoid, either infinite (or sufficiently long to approximate that) is determined by the turns per unit length (and of course the current and μ.) So as you add turns, you would simply approach the ideal more closely.

So I guess that a transformer based on the form of a solenoid (surrounded by uniform permeability) might have a voltage ratio equal to the turns per unit length ratio. This would equal the turns ratio if they were wound the same length. But if the flux is uniform along the length of the solenoid, why would the secondary need to be spread out? Each secondary turn would link the same amount of primary flux wherever it was on the solenoid (well, best near the centre rather than near the ends.) So the secondary might best be pile wound at the centre? Then I'm back with no explanation for my 1:8 ratio.

Final worry! If I believe Hyperphys, adding a core to a solenoid multiplies the flux by k (or μ/μ₀), rather than simply halving it as I had supposed. The permeability of the surroundings appears immaterial. Consistent with the notion that the solenoid generates no flux outside, but a bit puzzling for people who like flux lines to be closed loops. I console myself with the idea that space is very large, so there are very many paths for the external flux lines - effectively many parallel magnetic resistances, who's sum (edit: cross out "must" and substitute "might") approach 0.

But all this speculation is tiring my poor old brain. I'll go back to some more coil winding and experiments.

 

[The Electrician]

I used a slightly different bolt; mine is 3.5 inches long and .5 inches in diameter (instead of .3 inches). I used larger wire (26 AWG), and here's what it looks like with only the primary (100 turns) on. There is a layer of Nomex over the primary; that's what the white material is. I wound 25 turns over the nomex for a secondary:

Now I measured the parameters without the bolt (air core, in other words) using an impedance analyzer at 50 Hz. The imaginary component is on top with the real component below:

Then with the bolt in place:

The inductance with an "air core" was about 43 uH, and with the bolt in place it was 608 uH.

I used much larger wire than the OP, and the resistive part of my primary impedance is 3.5 times as large as the reactive part. For the OP the ratio is much larger.

A very telling thing I measured is the change in primary impedance when the secondary is shorted. In a "real" transformer, this will cause a large decrease in the primary impedance. For this bolt transformer, there was only about a 1.7% change. This means that the coupling between primary and secondary is very weak, so we shouldn't expect very much output on the secondary.

Most of the voltage applied to the OP's transformer is lost across the resistance of the wire. I wonder if the OP has a millivolt range on his DVM?

 

[jim hardy]

In a non laminated core like a bolt or a rod, the core itself conducts eddy currents and that's rather like having a shorted secondary. So it's a long way from an ideal inductor.

If you guys have oscilloscopes handy , display primary current and secondary voltage together , in chop mode so you get a true picture of phase..
In an inductor, current and flux should be in phase, voltage and flux 90 degrees out of phase (dsin wt = wcoswt, that's 90 degree shift) .

So
IF:
you see other than 90 degrees between primary current and flux as measured by an unloaded coil linking same flux,
THEN
there's another current flowing that affects your flux field.

That's your eddy currents.

Try it both with and without a core, i wager you'll see 90 degrees air core and something less iron core. That's because the iron core let's eddy current flow and that changes the flux.

Next, if you have a function generator, try same experiment with a triangle wave current. I don;t know what to suggest for a half amp triangle wave source, maybe an audio amp ?

Since voltage is d(flux)/d(time), voltage wave should resemble derivative of current wave.
Triangle wave current should give square wave voltage.
Here's some 'scope photos i took in 1992
in all three,
Top trace is current approximately 20 ma p-p through a 12 foot long coil with an iron rod maybe 2½ inches diameter for core.
Bottom trace is voltage induced in another winding surrounding same core.

First photo is at low frequency of 3 hz
you see that triangle wave current gives approximate square wave voltage, as expected. Rounded corners of square wave are eddy currents reversing as the iron magnetizes from outside in. That's called "Retardation of magnetization" in my 1901 textbook.

Center photo is at ten hz, you can see we don't have time for square wave to flatten out.

Last photo is at 60 hz. That particular steel rod made a poor transformer core for 60 hz operation.
Above 400 hz the coil didn't even notice the core being inserted .

I don't know how the analyzer calculates L and R .

L is proportional to flux per amp, does it allow for the amps from eddy currents? That was one mistake i made in '92 doing the calculations longhand.

Electrician- try heating your bolt with a hair dryer and see if it changes the readings? Eddy currents go down as temperature goes up, so it should show more inductance. At least that's what we saw in '92.

old jim

PS i know I've posted that photo before, sorry for being repetitious. That experimenting taught me to think in terms of flux whenever messing with inductors.
Hope it helps somebody..

 

[The Electrician]

Jim, the impedance analyzer I'm using is this: http://www.waynekerrtest.com/global/html/products/LCR/6430%206440.htm

It makes a measurement by applying an AC voltage (or current) and measuring (4 terminals) the resulting current (or voltage) with a phase sensitive detector and thereby getting an in phase (real) and out of phase (reactive) value, then calculating things like impedance, admittance, Q, D, etc. I can also make a 4 terminal DC measurement.

Here's what I get for real part of the impedance (ohms) at several frequencies, with and without the bolt:

Frequency   Air core      Iron core
DC            .655         .655
50            .655         .706
1k            .655          1.522
10k           .658          4.865
100k          .852          14.61

We see that the AC resistance of the wire alone (air core) doesn't increase much (due to skin effect) with frequency until about 10 kHz, but with the iron in place, it increases rapidly.

I applied 1 VAC to the primary and with the bolt in place I measured .112 VAC on the secondary. Without the bolt (air core in other words), the output was .00646 VAC. I used the millivolt range on my DVM.

 

[The Electrician]

The OP said the resistance of his primary is about 6 ohms. I don't have a 6 ohm resistor handy, but I wired a 10 ohm resistor in series with my primary to approximate the OP's situation. I then applied 1 VAC to the resulting "high resistance primary". I measured 6.31 millivolts at the secondary with the bolt in place and .437 millivolts with no bolt (air core).

Earlier in the thread it was suggested that this was the OP's problem, and it seems clear that it is.

The OP needs to make a new transformer with larger wire so resistance doesn't dominate the primary impedance, and use a DVM with a millivolt range.

 

[jim hardy]

Interesting, electrician !

Here's what I get for real part of the impedance (ohms) at several frequencies, with and without the bolt:

by real, you mean the resistance ?
If it applies a constant voltage and Ohms of resistance go up with frequency, does that suggest less iron loss ?

Going back to the transformer model from Wikipedia

Winding resistance is Rp and easily measured
but Rc is core loss from eddy current and hysteresis, mostly
so is not easy to measure.
X_M is transformer's inductance, not so easy to separate from Xp i think ??

Es is produced by flux in the core, if you integrate it you get flux. I used a simple RC integrator at first but switched to triangle waves...
Es 's volts per turn ought to be same as Ep's
but we don't know Ep because of drop across Xp
If with a 'scope one could get magnitude phase of Es wrt Vp and Ip , and just assume same volts per turn

might one be able to estimate what is Z of X_M(inductance of transformer) and R_C(effective resistance representing core losses) ? Thereby separating X_M from Xp ?

An unlaminated bar is lossy so has low R_C, and a half air core is hard to magnetize so has low X_M . So there's unexpected drop across Rp and Xp.
I think that's why Merlin's voltage ratio ≠ turns ratio.

As you increase frequency skin effect in iron makes less and less of it appear to be "there" so R_C goes up. But with less iron effectively there due to skin effect, X won't be linear with frequency.
I don't know if that's why your machine reported more ohms at higher frequency. Does it give phase angles too ? I'd have to watch it with a 'scope. I saw one computerized Z meter that applied voltage ramps and steps, .never could figure out what it was up to.
Yours shows a nonlinear ohms to frequency relation, as X should be...

In a transformer with a nice core that's closed and laminated , R_C and X_M are both decently high and we get away with equating current and voltage ratios to turns ratio. But it's an approximation nonetheless. OP's gizmo is an imperfect inductor with another coil in proximity. We can use that coil to measure flux but not much else i think...

I console myself with the idea that space is very large, so there are very many paths for the external flux lines - effectively many parallel magnetic resistances, who's sum (edit: cross out "must" and substitute "might") approach 0.

Yes, some of those loops go clear out of the solar system i suppose. But does their lumped magnetic resistance approach zero, or does it approach that of solenoid?

 

[jim hardy]

The OP needs to make a new transformer with larger wire so resistance doesn't dominate the primary impedance, and use a DVM with a millivolt range.

and plot secondary (milli)volts versus primary amps...

 

[The Electrician]

If the impedance consists of a real part R and imaginary part X, then the impedance can be written Z = R + jX.

The analyzer measures R and X directly with a phase sensitive detector. It then does some calculations and can display the result as a real and imaginary part, or as impedance magnitude |Z| and angle, or as admittance magnitude |Y| and angle, or as inductance and resistive (ESR) part. It can display the result as a series equivalent or a parallel equivalent. The images in post #42 show the result as a series equivalent; you can see the little picture of an inductor in series with a resistor (European style). In the Wikipedia model, Rp should be measured as a series equivalent, and Rc as a parallel equivalent.

My analyzer measurement of the primary lumps all those various resistances and inductances together. My "air" core measurement only shows Rp and Xp because that's all there is with no iron core. Rp increases with frequency as shown in post #44. Knowing Rp from the "air" core measurement, and the combined value of the primary impedances with the bolt in place, we could calculate what Rc must be.

Both Rp and Rc result in losses, but for high losses, Rp is large, and Rc is small. The analyzer lumps them together, and since it's displaying a series equivalent, that series equivalent resistance increases for more losses, If Rp were zero, and as Rc increased, the measured series equivalent resistance would increase as Rc decreased, giving more loss.

 

[The Electrician]

I also did the hot bolt test. I measured the inductance with the bolt at room temperature and got 232 uH. I heated the bolt with a hot air gun and got it quite hot; the inductance then measured 291 uH.

The bolt was so hot that it heated up the windings and their resistance increased also.

 

[jim hardy]

also did the hot bolt test. I measured the inductance with the bolt at room temperature and got 232 uH. I heated the bolt with a hot air gun and got it quite hot; the inductance then measured 291 uH.

Hmm . thinking aloud

Inductance is flux linkages per ampere NΦ/I ,
and N didn't change...
So, heating the bolt increased flux per ampere by 291/232 = nearly 25% ?
That suggests a surprising fraction of the total current flowing is eddy currents in the iron bolt (cancelling flux by Lenz's law..) That's a long way from an ideal inductor...
(Edit-- In all honesty though, steel might have a temperature coefficient of permeability on order of 0.1%/°C , just found this while double-checking myself)

http://nvlpubs.nist.gov/nistpubs/bulletin/12/nbsbulletinv12n1p1_A2b.pdf

My analyzer measurement of the primary lumps all those various resistances and inductances together. My "air" core measurement only shows Rp and Xp because that's all there is with no iron core. Rp increases with frequency as shown in post #44. Knowing Rp from the "air" core measurement, and the combined value of the primary impedances with the bolt in place, we could calculate what Rc must be.

Both Rp and Rc result in losses, but for high losses, Rp is large, and Rc is small. The analyzer lumps them together, and since it's displaying a series equivalent, that series equivalent resistance increases for more losses, If Rp were zero, and as Rc increased, the measured series equivalent resistance would increase as Rc decreased, giving more loss.

Rp is virtually constant up to 10khz... so adding the bolt will increase X_M and Xp, and decrease R_C. I still don't see how to separate X_M from Xp.

But it's past my bedtime and I'm foggy...

TTFN

old jim

 

[jim hardy]

im, the impedance analyzer I'm using is this: http://www.waynekerrtest.com/global/html/products/LCR/6430%206440.htm

It makes a measurement by applying an AC voltage (or current) and measuring (4 terminals) the resulting current (or voltage) with a phase sensitive detector and thereby getting an in phase (real) and out of phase (reactive) value, then calculating things like impedance, admittance, Q, D, etc. I can also make a 4 terminal DC measurement.

That's quite a machine you have there.
From its datasheet

Have you looked at a graph of inductance (or impedance) versus frequency?
Over audio range
Air core inductance should be pretty constant but iron core should not, it'll start higher and approach air value as less and less of the iron participates. Remember magnetism proceeds from outside inward and at high frequency it doesn't have time to get beyond skin deep.
At least that's how i explained my results from those photos. My core was an exotic stainless steel used for control rod lifting mechanisms in the reactor.. 400 series stainless makes good solenoid cores because its high resistivity means less eddy currents.

I sure wish we'd had such an instrument when we did our tests back in '92. I neglected to get enough phase measurements to completely characterize our sensor..my bad.
We were investigating temperature sensitivity of the control rod position indicators, basically they measure 60 hz inductance of the 12 foot tall coil as the steel shaft moves up inside it carrying the rod below. It woulda worked a lot better at 3hz, i will always believe. Those photos were an in-situ test from control room . .

As the song goes - "Those were the days my friend! "

------------------------------------------------------------------------------------------------------------------------------------------------------------------------

But i have digressed a long way from the OP's subject. I apologize. To get back to it...
A transformer is an inductor with another coil sharing the region where it makes flux. To understand transformers it is first necessary to understand the humble inductor and that's why i digressed. Sorry if i derailed the thread...

Merlin's voltage readings gives us a number for how much flux couples the secondary coil, but we don't know how much more flux the primary coil made. And we don't know how much flux got canceled by eddy currents in the core.

I set it up with 0.506A AC 50Hz and measured a primary voltage of 0.96V and secondary voltage of 0.053V
The voltage across the resistance should be 0.86V , which, assuming only an inductive component at 90°, gives a voltage across the inductance of 0.43V.

I think if we knew the phase angles between primary amps and both windings' voltages we'd be better positioned to figure it out.

Anyhow, as electrician said it's going to take a lot of current to make significant flux in OP's device.
And if as N O observed OP's primary is wound half one way and half opposing, well he's built a low ohm non-inductive resistor.

old jim

 

[NascentOxygen]

The bolt was so hot that it heated up the windings and their resistance increased

If the copper coil expanded in diameter then its L would increase. If the wire's gauge expanded, the compression tending to fill in small spacings between turns, I think this would increase inter-turn flux linkage and slightly increase L.

 

[The Electrician]

If the wire diameter increases, that effect alone, ignoring compression of the entire coil, will decrease the inductance. There may be other effects not considered.
All that is easy to test. With the bolt at room temperature, I measured the following @50 Hz:
R = .647 ohms, L = 230.1 uH

with bolt removed:
R = .647 ohms, L = 43.017 uH
------------------------------------------------------------
With hot bolt inserted in form for 10 seconds:
R = .783 ohms, L = 294.3 uH

Remove bolt and quickly measure:
R = .748 ohms, L = 42.998 uH

The resistance of the wire slowly decreased back to the pre-heat value of .647 ohms.

The inductance of the coil alone decreased (very little) as a result of heating.

 

[Rive]

... I used an ac source of 16 volt...I found that voltage across the primary coil is about 2 volt

I don't know if it's still 'alive' or not, but I think the problem would be about that AC source, which is most likely some small transformer itself, probably capable only up to a few VA output. With such setup the limiting factor would be the resistance of the primary of the source transformer.

Another limiting factor might be the measurement device - cheap multimeters are not really good with measuring low voltage AC (below 1.2 or 0.6V, depending on how cheap the stuff is).

 

[jim hardy]

If the wire diameter increases, that effect alone, ignoring compression of the entire coil, will decrease the inductance. There may be other effects not considered.
All that is easy to test. With the bolt at room temperature, I measured the following @50 Hz:
R = .647 ohms, L = 230.1 uH

with bolt removed:
R = .647 ohms, L = 43.017 uH
------------------------------------------------------------
With hot bolt inserted in form for 10 seconds:
R = .783 ohms, L = 294.3 uH

Remove bolt and quickly measure:
R = .748 ohms, L = 42.998 uH

The resistance of the wire slowly decreased back to the pre-heat value of .647 ohms.

The inductance of the coil alone decreased (very little) as a result of heating.

to slide rule accuracy of 3 significant figures :
coil by itself without bolt: 43.0 uH regardless hot or cold
with bolt cold 230 uH
with bolt hot 294 uH

the about six fold increase in inductance by adding the core surprises me, i suspect that for a longer coil the increase would be less dramatic

but clearly temperature affects the iron core's effective permeability
U_r 294/43 = 6.84 hot
U_r 230/43 = 5.35 cold

There's a pretty good derivation here for a long thin solenoid
he starts with Biot Savart, simplifies to a current sheet instead of individual turns
and comes to the familiar equation we all use which as Merlin said gives flux density as function of just mmf amp-turns per unit length,
http://info.ee.surrey.ac.uk/Workshop/advice/coils/air_coils.html

If b >> a (the solenoid is long and thin) then the angles ψ1 and ψ2 will both tend to zero for a point in the middle
B = μ₀ Fl teslas Equation ACT (F1 he derives earlier as mmf amps/meter)

At the end of a long coil the flux will be exactly half this value. Only towards the ends does the flux 'leak' through the sides.

In a short coil considerable flux "leaks" out at both ends.

I haven't tried to calculate how much as a function of solenoid's length. I rely on my observations that inserting a core into my 12 foot coil just about doubled its flux. Its length do diameter ratio i'd estimate from memory around thirty-five.

length to diameter : 150/46 = 3.2 , if i understood .Ohhh the things they never mentioned in undergrad class ...
http://www.g3ynh.info/zdocs/magnetics/Solenoids.pdf page 32 of 97

 

[The Electrician]

I misspoke in post #53; the frequency of the measurement was 1000 Hz, not 50 Hz. Refer back to post #43 where the 50 Hz inductance with the bolt in place was 608 uH. At 50 Hz, the measured inductance is increased about 14 times by the insertion of the bolt.

You asked about the inductance as a function of frequency. Here's a sweep of the inductance and AC resistance measured at the primary (the 100 turn winding). The frequency is swept from 10 Hz to 1 MHz. The scale for the inductance curve (green) varies from 1 uH at the bottom of the image to 1000 uH at the top. The scale for AC resistance (real part of the impedance) curve (yellow) varies from .1 ohm at the bottom to 100 ohms at the top. There are two markers, A and B. The values of inductance and resistance at the marker frequencies (50 Hz and 10 kHz) can be read at the top right of the image. This sweep is for the coil without the bolt inserted ("air" core, in other words):

Notice how the inductance is constant with frequency and the resistance is constant until it begins to rise in the neighborhood of 50 kHz due to skin and proximity effect.

Now with the bolt inserted:

We see inductance decreasing drastically with increasing frequency, and resistance increasing.

Here are both sweep results superimposed:

Notice how the inductance with the bolt inserted actually becomes less than the "air" core value as the frequency increases above about 45 kHz. At such high frequencies, the bolt just looks a shorted turn with very little penetration of flux into the bolt.

 

[The Electrician]

Another limiting factor might be the measurement device - cheap multimeters are not really good with measuring low voltage AC (below 1.2 or 0.6V, depending on how cheap the stuff is).

No doubt. Have a look at the values I measured in post #45. The OP definitely needs to use a DVM with a millivolt scale.