Magnetic induction of displacement current

Author Topic: Magnetic induction of displacement current (Read 3733 times)

Topic Options

F6FLT

Magnetic induction of displacement current « on: 2019-02-28, 16:28:22 »

Group: Experimentalist
Hero Member

Posts: 2161

Can we magnetically induce displacement currents in a dielectric and observe their effects?

The purpose of this thread is to make it work, or to explain the impossibility.

All my attempts have failed so far, but the reason is not yet clear.
I just tried an idea of Smudge that he provided in this document.

Principle: a magnetic field is created in a first toroid by a current in a winding.
A second toroid used as a dielectric is supposed to receive the induction of the first toroid and produce an induced displacement current. It is used as a current loop.
This displacement current should therefore induce a variable magnetic field in the last toroid, and causes a voltage to appear in the output winding.

My first setup (see the last attached file inductTorBadSetup.jpg) had experimental biases that were impossible to overcome (capacitive couplings, direct couplings outside the ferrite, resonance frequencies...), so I opted for a more serious setup.

To reduce capacitive effects, the first 10-turn winding is coaxial. It is connected to the generator and the other end of the cable is connected to ground. This coaxial cable is from a defective scope probe that I recovered. The central conductor is very thin, limiting the parasitic capacitance.
With a loop of a first probe around this first toroid, I measure the voltage induced by this winding, on channel 1 of the scope (therefore in volt/turn).
I do the same thing with the second probe by looping it around the last toroid and see the result on the second channel. The intermediate toroid is here the yoke (see photo inductTor.jpg).

I measure just over 12 mV/tour pp for the first toroid (see scope view).
I only measure about 500 µV/tour pp output when we should have about the same voltage because the intermediary toroid plays the role of a single turn current loop.

Since the induced voltage represents only 4.1% of the expected voltage (this % does not depend on the frequency), it can be said that the experiment either invalidates the idea of a magnetically induced displacement current or indicates that it is very weak, for unknown reasons.

Any ideas?

PS Although it is only used as a dielectric, nevertheless by curiosity I measured the voltage magnetically induced by the intermediary toroid, it is 300 µV/tour.

-----------------------------------------------
Update 01/03

I had inadvertently connected the end of the coaxial cable to the unused input of the generator, instead of the ground (see corrected photo).
The cable was therefore loaded by a very high impedance, which explains the weakness of the signal.

After correction today, the input signal is much stronger (0.5 V pp instead of 12 mV), but the ratio between input and output is still very low: about 2.2% whatever the frequency between 100 KHz and 2 MHz.

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

Magnetic induction of displacement current

inductDispCurr1.jpg (130.48 kB, 800x480 - viewed 773 times.)

Magnetic induction of displacement current

inductTor.jpg (335.81 kB, 1282x855 - viewed 700 times.)

Magnetic induction of displacement current

inductTorBadSetup.jpg (169.09 kB, 1078x541 - viewed 650 times.)

« Last Edit: 2019-03-01, 08:58:59 by F6FLT »

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

"Open your mind, but not like a trash bin"

gyula

Re: Magnetic induction of displacement current « Reply #1 on: 2019-02-28, 17:49:13 »

Group: Tech Wizard
Hero Member

Posts: 1259

Hi F6FLT,

Have you checked the DC resistance of the thin center conductor that is inside the 10 turn winding ?

I mention this because for most scope probes' coax cable, the thin center conductor is made of resistive wire (150-250 Ohm) to reduce ringing. This is not a problem when the terminating resistance is the 1 MOhm scope input but may cause loss when your generator's internal resistance terminates it.

Sorry if you already checked this and the center conductor in the cable used for the 10 turn input winding has low resistance as would be expected from a 'normal' coax cable.

Gyula

Smudge

Re: Magnetic induction of displacement current « Reply #2 on: 2019-02-28, 18:53:47 »

Group: Professor
Hero Member

Posts: 1979

Hi F6FLT,,

The dielectric toroid will not act like a wire conductor current loop so you won't get anywhere near full voltage at the output. The actual displacement current will depend upon the dielectric constant, and of course the magnetization of the output toroid hence output voltage will depend on that current. The fact that the output did not change with frequency goes against this but there could be an explanation for this. When a transformer sees a capacitive load, that reflects as a negative reluctance in series with the core reluctance hence it affects the input circuit. I will try to model this later.

Smudge

F6FLT	Re: Magnetic induction of displacement current « Reply #3 on: 2019-03-01, 08:48:06 »
Group: Experimentalist Hero Member Posts: 2161	Quote from: Smudge on 2019-02-28, 18:53:47 Hi F6FLT,, The dielectric toroid will not act like a wire conductor current loop so you won't get anywhere near full voltage at the output. The actual displacement current will depend upon the dielectric constant, and of course the magnetization of the output toroid hence output voltage will depend on that current. The fact that the output did not change with frequency goes against this but there could be an explanation for this. When a transformer sees a capacitive load, that reflects as a negative reluctance in series with the core reluctance hence it affects the input circuit. I will try to model this later. Smudge Hi Smudge, I didn't say that "the output did not change with frequency". It's only the ratio input/ouput that didn't change. The input signal is about proportional to the frequency, probably due to the resistance of the coaxial wire and the impedance of the circuit that loads the generator, so that it can't impose its voltage as a certain number of V/turn in the first toroid winding. The dielectric toroid will not act like a wire conductor current but it's the same voltage that is induced along it as if it would be the case in a conductor (E depends only on dA/dt integrated along the path). And we have no load on the output, we should expect for a stronger voltage. --------------------------- "Open your mind, but not like a trash bin"

F6FLT	Re: Magnetic induction of displacement current « Reply #4 on: 2019-03-01, 08:55:26 »
Group: Experimentalist Hero Member Posts: 2161	Quote from: gyula on 2019-02-28, 17:49:13 Hi F6FLT, Have you checked the DC resistance of the thin center conductor that is inside the 10 turn winding ? I mention this because for most scope probes' coax cable, the thin center conductor is made of resistive wire (150-250 Ohm) to reduce ringing. This is not a problem when the terminating resistance is the 1 MOhm scope input but may cause loss when your generator's internal resistance terminates it. Sorry if you already checked this and the center conductor in the cable used for the 10 turn input winding has low resistance as would be expected from a 'normal' coax cable. Gyula Hi Gyula, Thank you for this information. I didn't know that. I measured the resistance: about 400 Ω in my case (I had another concern, please reread my first post that I corrected). When it comes to transferring a certain amount of power from input to output, I will change the cable. I'm keeping it for now because reducing the ringing is what I need, and as we measure the input signal after the resistance, it's not really annoying. --------------------------- "Open your mind, but not like a trash bin"

Smudge	Re: Magnetic induction of displacement current « Reply #5 on: 2019-03-01, 10:35:33 »
Group: Professor Hero Member Posts: 1979	Quote from: F6FLT on 2019-03-01, 08:48:06 The dielectric toroid will not act like a wire conductor current but it's the same voltage that is induced along it as if it would be the case in a conductor (E depends only on dA/dt integrated along the path). And we have no load on the output, we should expect for a stronger voltage. But it is not the voltage that couples to the output core, it is the current. You can't look upon the dielectric ring as a voltage driven single turn. For all transformers it is the current that drives the core. I think you are looking at the output toroid as a single turn voltage driven transformer, but it is not driven from a low impedance voltage source. If it were the flux in that core would be determined by the voltage, obeying Phi=int(V) and that then sets the induced output voltage, but not in this case. The current in the dielectric (displacement current) is determined by the induced voltage, this current creates an opposing induced voltage from the inductance of the single turn around the output core and it is that voltage that you should see as your open circuit output. But of course to determine this theoretically your dielectric current calculation has to take account of the induced voltage from the input core and the inductive induced voltage from the output core, so its not as simple as first thought. Smudge

F6FLT	Re: Magnetic induction of displacement current « Reply #6 on: 2019-03-01, 11:42:51 »
Group: Experimentalist Hero Member Posts: 2161	Quote from: Smudge on 2019-03-01, 10:35:33 But it is not the voltage that couples to the output core, it is the current. You can't look upon the dielectric ring as a voltage driven single turn. For all transformers it is the current that drives the core. I think you are looking at the output toroid as a single turn voltage driven transformer, but it is not driven from a low impedance voltage source... Suppose an ideal transformer except the wire resistance. If there is a 1 turn primary and a 1 turn secondary and you put 1 v on the primary, you get 1 v at the secondary. If the wire has a 1 MΩ resistance, you will still get 1 v at the secondary because in an ideal transformer with resistive wires, there is no losses when there is no current in the wires. So if it's not loaded, the EMF remains the same around a magnetic circuit. The one that creates the field is also the one we can recover. If the nested toroids were such ideal components, whatever displacement current in the dielectric toroid would guarantee the same EMF induced in the output toroid as that having caused the displacement current, whatever the current. The question of current appears only when there are losses in the system or when there is a load at the output. A variable current flowing in a winding should induce eddy displacement currents in any ferrite, and losses in its dielectric. However, this is not the case, the dielectric losses of ferrites are low if they are used at suitable frequencies and without saturation. So the idea of creating induced displacement currents with observable effect remains questionable or the real explanation of their weakness is not yet given, a too weak current being not enough to explain, for the reason I gave, a so weak output voltage in a system with low losses and no load (or this has to be quantified). --------------------------- "Open your mind, but not like a trash bin"

F6FLT	Re: Magnetic induction of displacement current « Reply #7 on: 2019-03-01, 13:15:25 »
Group: Experimentalist Hero Member Posts: 2161	I realized that there was an electric coupling between input and output ferrites. They are a bit conductive and the current induced in the coaxial shield makes "hot" its not connected end so that we retrieve the potentials of the induced emf along the coaxial shield while it is capacitively coupled to the input toroid which is capacitively coupled to the output one (it's a very common problem, we can't screen induced EMF). So I have shielded the first toroid with aluminium connected to the ground. In this better setup, not only the output voltage is much weaker (1mv out for 500 mV in, ratio 0.2%) but removing half the output toroid (it's in two parts) doesn't change much the voltage. We can infer that the output voltage is still partly due to remaining unwanted couplings. So if a displacement current participates in something in the output voltage, it's probably for less than 0.1%. Another method is needed. --------------------------- "Open your mind, but not like a trash bin"

Smudge	Re: Magnetic induction of displacement current « Reply #8 on: 2019-03-01, 16:00:43 »
Group: Professor Hero Member Posts: 1979	Quote from: F6FLT on 2019-03-01, 11:42:51 Suppose an ideal transformer except the wire resistance. If there is a 1 turn primary and a 1 turn secondary and you put 1 v on the primary, you get 1 v at the secondary. If the wire has a 1 MΩ resistance, you will still get 1 v at the secondary because in an ideal transformer with resistive wires, there is no losses when there is no current in the wires. So if it's not loaded, the EMF remains the same around a magnetic circuit. Only for the ideal transformer that has no primary inductance and doesn't need a magnetizing current. Such a transformer does not exist. Have you tried your 1 MΩ primary resistance? You will not get 1 v at the secondary because the magnetizing current is significantly reduced. The problem with the standard teaching on transformers is that dratted equivalent circuit that has an ideal transformer at its centre. That ideal transformer magically works without any magnetic fields, so we have to then add another component to it to account for the field in the non-ideal transformer. Unfortunately this gives a false impression on how transformers actually work Quote The one that creates the field is also the one we can recover. If the nested toroids were such ideal components, whatever displacement current in the dielectric toroid would guarantee the same EMF induced in the output toroid as that having caused the displacement current, whatever the current. So you believe but you are wrong. You are fixated on the ideal transformer that magically transforms voltage by the turns ratio and magically transforms current by the inverse of the turns ratio. The magnetization(hence flux) in a transformer core is proportional to primary current, not voltage. Admittedly when driven from a voltage source the low impedance source forces the situation where it appears that the flux is proportional to voltage but that only applies to a primary coil which has negligible wire resistance. Pu 1 MΩ in series and the story is different. Quote The question of current appears only when there are losses in the system or when there is a load at the output. Only for the magical ideal transformer that has zero magnetizing current. For any load on the transformer the output voltage is related directly to that current. Have you seen my paper "Analyzing Transformers in the Magnetic Domain"? Smudge

verpies

Re: Magnetic induction of displacement current « Reply #9 on: 2019-03-01, 16:43:40 »

Group: Professor
Hero Member

Posts: 3695

As far as my experience goes - analyzing inductors and coupled inductors in terms of voltage has always led me down a garden path.

F6FLT	Re: Magnetic induction of displacement current « Reply #10 on: 2019-03-01, 20:38:51 »
Group: Experimentalist Hero Member Posts: 2161	Quote from: Smudge on 2019-03-01, 16:00:43 Only for the ideal transformer that has no primary inductance and doesn't need a magnetizing current. Such a transformer does not exist. Have you tried your 1 MΩ primary resistance? You will not get 1 v at the secondary because the magnetizing current is significantly reduced. The problem with the standard teaching on transformers is that dratted equivalent circuit that has an ideal transformer at its centre. That ideal transformer magically works without any magnetic fields, so we have to then add another component to it to account for the field in the non-ideal transformer. Unfortunately this gives a false impression on how transformers actually work So you believe but you are wrong. You are fixated on the ideal transformer that magically transforms voltage by the turns ratio and magically transforms current by the inverse of the turns ratio. The magnetization(hence flux) in a transformer core is proportional to primary current, not voltage. Admittedly when driven from a voltage source the low impedance source forces the situation where it appears that the flux is proportional to voltage but that only applies to a primary coil which has negligible wire resistance. Pu 1 MΩ in series and the story is different. Only for the magical ideal transformer that has zero magnetizing current. For any load on the transformer the output voltage is related directly to that current. Have you seen my paper "Analyzing Transformers in the Magnetic Domain"? Smudge "you believe... " "You are fixated....", not at all! I don't believe anything, that's why I'm experimenting. On the contrary, the experiments make me doubt, after having thought that these induced displacement currents would be possible, rather easily observable, and of interest. It is you who firmly believe that these currents exist when all the attempts I make fail, when I have not seen in the literature that this has been done, and when I have not seen from you any experimental demonstration either. Of course there is no such thing as a perfect transformer, I only assumed it to talked about the principle, as you do in your PDFs where reality is also quite toned down. With good precautions, we can see that the effect, if it exists, represents less than 0.1% of the signal measurement bias. How can you say that these currents are not detected because they would be negligible compared to the measurement biases related to imperfections, when it seems you have no idea of the order of magnitude of the expected currents? It doesn't make sense and it sounds like an excuse. Otherwise, what is the order of magnitude of the current we should expect in the dielectric of the intermediate toroid? 1 fA? 1 pA? 1 nA? 1 µA? Can you say that? For me, the question of these currents is open both theoretically and practically. For example, whereas in a conductor only free electrons are concerned, in a dielectric all electrons are concerned, and not only electrons but also all protons. All charges should therefore respond en masse to induction, but obviously this is not the case or there are counter-effects. I think there are things we may not understand here. --------------------------- "Open your mind, but not like a trash bin"

partzman

Re: Magnetic induction of displacement current « Reply #11 on: 2019-03-02, 00:03:37 »

Group: Experimentalist
Hero Member

Posts: 1862

Well, I don't really know but here is a test that may be of interest or, maybe not!

Your opinions may vary.

The schematic is below and basically the transformer shown is a wound on a 2" diameter ferrite toroid with an intitial perm of ~3500. The primary is 16 ga magnet wire and the single turn secondary is pvc insulated 22ga.

The first scope pix shows the operation with mosfet M1 driven by a pulse at 1.5MHz with a 25% duty cycle as seen on CH1.

CH2 is the output voltage across the secondary and CH3 is the voltage on the drain of M1.

The current probe on CH4 is placed as shown and remains there during the current measurements with jumpers X1 and X1 connected as follows: With X1 removed and X2 connected, the resulting current is shown in the R1 reference trace. With X1 connected and X2 removed, the current thru the primary is now seen in the CH4 trace. I have offset the R1 for clarity but it can be seen that for all practical purposes, they appear to be the same.

IMO, the current spike seen in both R1 and CH4 is primarily created by the discharge of the current probe capacitance as it is common in both measurements. It does not appear to be a current generated in the primary winding. However, we do see the generated secondary output voltage on CH2.

The second scope pix is with CH3 connected at the junction of R1, X2, and the primary and we do not see any voltage change that would represent the transient currents seen by the current probe. IMO, the voltage drop is created by R1 in series with the inductance reactance of the primary.

Regards,
Pm

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

Magnetic induction of displacement current

Induction Test Schematic.png (15.54 kB, 712x630 - viewed 541 times.)

Magnetic induction of displacement current

Induction Test.png (28.94 kB, 800x480 - viewed 547 times.)

Magnetic induction of displacement current

Induction Test2.png (22.32 kB, 800x480 - viewed 275 times.)

F6FLT	Re: Magnetic induction of displacement current « Reply #12 on: 2019-03-02, 09:48:42 »
Group: Experimentalist Hero Member Posts: 2161	Hi Pm Case with X1 open: when the transistor switches on, the end of the primary connected to the drain has its potential changing. This winding has a certain capacitance relative to the ground and another relative to the secondary winding. A small current will therefore flow through the primary until the entire winding is at the same potential as the drain. The current which depends on the inductance and both capacitances, can induce a current in the secondary by magnetic induction. In addition, and I think this effect is predominant, the rapid change of the primary potential will also influence the secondary by the capacitive coupling even if it is low, due to the short pulse rise time. The probe capacitance surely adds concerns, it's difficult to analyse and distinguish these effects. --------------------------- "Open your mind, but not like a trash bin"

F6FLT	Re: Magnetic induction of displacement current « Reply #13 on: 2019-03-02, 10:00:57 »
Group: Experimentalist Hero Member Posts: 2161	Induction in a dielectric tends to separate negative charges from positive charges, this is the classic polarization effect. As opposite charges move in opposite directions, they create a current in the same direction. That's what we wanted to detect. But the Coulomb force between positive and negative charges very quickly limits this displacement by exerting a restoring force, and the system balances itself. The equilibrium is probably reached in a very short time, very short compared to the signal period, so that the displacement current is not observable with the means used here. This is the explanation I propose for the inconclusive outcome of this experiment. However, I am not really satisfied: why would this current be so low, while a displacement current in the dielectric of a capacitor can be extremely high? --------------------------- "Open your mind, but not like a trash bin"

Smudge	Re: Magnetic induction of displacement current « Reply #14 on: 2019-03-02, 16:28:35 »
Group: Professor Hero Member Posts: 1979	Quote from: F6FLT on 2019-03-01, 20:38:51 Otherwise, what is the order of magnitude of the current we should expect in the dielectric of the intermediate toroid? 1 fA? 1 pA? 1 nA? 1 µA? Can you say that? Given all the data on cores and dielectric ring I can calculate it. If L is the inductance of a single turn on your output toroid and C is the capacitance of the dielectric ring then the current is given by VomegaC/(1-omega²LC) where V is your measured 1 turn voltage on your input toroid. The capacitance C is KenoughtA/l where A is the area and l is the length of the dielectric (in the same manner that the permeance of a ferro ring core is muRmunoughtA/l). Smudge

F6FLT	Re: Magnetic induction of displacement current « Reply #15 on: 2019-03-02, 18:12:50 »
Group: Experimentalist Hero Member Posts: 2161	Quote from: Smudge on 2019-03-02, 16:28:35 Given all the data on cores and dielectric ring I can calculate it. If L is the inductance of a single turn on your output toroid and C is the capacitance of the dielectric ring then the current is given by VomegaC/(1-omega²LC) where V is your measured 1 turn voltage on your input toroid. The capacitance C is KenoughtA/l where A is the area and l is the length of the dielectric (in the same manner that the permeance of a ferro ring core is muRmunoughtA/l). Smudge I'll take your hypothesis. From what I understand you see the toroid as a capacitor that would be stretched and bent circularly so that its 2 plates coincide and can be removed. In the circuit its impedance is Z and the current is I=V/Z where V is the induced voltage. Let's take an elementary section d=1mm thick that can be assimilated to a parallel plate capacitor c. So we have c = Ɛ.A/d. A=3 cm² for my toroid, let's take Ɛr=9, then c = 9 * 8.85 × 10^-12 * 310^-4 / 10^-3 ≈ 24 pF The average diameter of the toroid is 5.5 cm, the circumference is 170 mm. The toroid is therefore equivalent to 170 capacities in series as calculated, which gives us: C= 24/170 = 0.14 pF. At 1 Mhz, the inductance can be neglected. It's to be verified but I'm here just interested in the order of magnitude. With 1 v, this gives us a current of I = V.ω.C = 1 2pi10⁶* 0.14 * 10-¹² = 0.8 µA. It's really low but the effects should be within our reach: to work with higher frequencies and especially at a resonant frequency, provided that the starting hypothesis is correct. However, I am not sure that an induced EMF has the same effect as a potential difference through a dielectric. A few years ago I had done tests in water, permitivity 80, with a single transistor LC oscillator at radio frequency. I had no difference whether the coil was in the air or in the water, whereas the currents induced in the water should have affected the frequency or amplitude of the oscillator. --------------------------- "Open your mind, but not like a trash bin"

Smudge	Re: Magnetic induction of displacement current « Reply #16 on: 2019-03-04, 18:00:11 »
Group: Professor Hero Member Posts: 1979	Quote from: F6FLT on 2019-03-02, 18:12:50 However, I am not sure that an induced EMF has the same effect as a potential difference through a dielectric. One definition of displacement current density is J_D=dD/dt where D is the electric displacement given by KenoughtE, E being the electric field and K the relative permittivity. E can come from a potential difference or from a changing A field. Quote A few years ago I had done tests in water, permitivity 80, with a single transistor LC oscillator at radio frequency. I had no difference whether the coil was in the air or in the water, whereas the currents induced in the water should have affected the frequency or amplitude of the oscillator. The near field for a coil has a wave impedance that is (a) reactive and (b) increases in magnitude at increasing distance from the coil, being zero at zero distance and (c) increases with frequency. Having worked with these near fields I used to have a figure of merit of x Ohms per meter-Mhz but I can't remember what x is! The size of the coil compared to the wavelength determines what the local wave impedance is. I'll try to work it out again from first principles. What I am getting at is the characteristic impedance of water is about 42 ohms and if your coil's local wave impedance is very low compared to that then it will not have any discernible effect. My guess is you used coils on ferrite rods at frequencies up to medium wave, but had you used air coils at VHF then you would have noticed an effect. Smudge

F6FLT	Re: Magnetic induction of displacement current « Reply #17 on: 2019-03-13, 10:47:19 »
Group: Experimentalist Hero Member Posts: 2161	Quote from: Smudge on 2019-03-04, 18:00:11 One definition of displacement current density is J_D=dD/dt where D is the electric displacement given by KenoughtE, E being the electric field and K the relative permittivity. E can come from a potential difference or from a changing A field. The near field for a coil has a wave impedance that is (a) reactive and (b) increases in magnitude at increasing distance from the coil, being zero at zero distance and (c) increases with frequency. Having worked with these near fields I used to have a figure of merit of x Ohms per meter-Mhz but I can't remember what x is! The size of the coil compared to the wavelength determines what the local wave impedance is. I'll try to work it out again from first principles. What I am getting at is the characteristic impedance of water is about 42 ohms and if your coil's local wave impedance is very low compared to that then it will not have any discernible effect. My guess is you used coils on ferrite rods at frequencies up to medium wave, but had you used air coils at VHF then you would have noticed an effect. Smudge "increases in magnitude at increasing distance from the coil"? I agree that the magnetic field is less important near a long solenoid than further away, it is even assumed to be zero if the solenoid is infinite. This has a limit, because at a greater distance from a finite size coil the field decreases. However, the validity of your statement could be maintained by setting a limit to the near field. But we see that at zero frequency, the near field is infinite, and yet the field decreases with the distance for a finite size coil. So for me the maximum field strength is at a distance that will depend on the size of the coil in relation to the wavelength, the shorter the coil, the closer the max field, but never zero. For the middle impedance, I suppose you mean Z=√µ/Ɛ. For the "coil's local wave impedance", I don't see, except the classical impedance Z=r+jlω. If the coil is immersed in the medium, then L changes proportionally to µ, and if it is not immersed but close, its field extends into the medium of permeability µ, so the impedance must be somewhere in between. Possible as you say, that my experiment was done at too low a frequency to notice anything with the submerged coil, the displacement currents in the dielectric being too low for them to have an impact, by Lenz's law, on the oscillator. After further reflection on this subject, I finally wonder whether even strong displacement currents could have an influence. Indeed, if the losses in the dielectric are low, even large currents are not dissipated, we remain in a purely reactive system, and this could also be an explanation of the negative result. --------------------------- "Open your mind, but not like a trash bin"

verpies	Re: Magnetic induction of displacement current « Reply #18 on: 2019-03-13, 11:39:13 »
Group: Professor Hero Member Posts: 3695	Quote from: F6FLT on 2019-03-02, 18:12:50 A few years ago I had done tests in water, permitivity 80, with a single transistor LC oscillator at radio frequency. I had no difference whether the coil was in the air or in the water, whereas the currents induced in the water should have affected the frequency or amplitude of the oscillator. This is in conflict with the experiment described on Pg.60 (Sect. 8) of the attached document: "When cylindrical blocks of dielectric material are inserted into air solenoids suspended on the jig, a substantial reduction of the Self Resonance Frequency" occurs." ------------------------ Coil Self-Resonace.pdf (1315.01 kB - downloaded 514 times.)

Smudge	Re: Magnetic induction of displacement current « Reply #19 on: 2019-03-13, 16:12:35 »
Group: Professor Hero Member Posts: 1979	Quote from: F6FLT on 2019-03-13, 10:47:19 "increases in magnitude at increasing distance from the coil"? It is the imaginary impedance (reactance) that increases with distance. For a magnetic dipole of moment m, in the radiative direction (at right angles to the axis of the coil) the H field is given by m/(4pir³). The A field is at right angles to the H field and is given by munoughtm/(4pir²). Hence for an alternating dipole at a sinusoidal frequency the E field is at right angles to the H field, is at 90 degrees phase and is given by omegamunoughtm/(4pir²). The reactive wave impedance given by E/H is therefore omegamunought*r which may be stated as 8 ohms per meter-Mhz. This applies if the retarded effects can be ignored. Smudge

F6FLT	Re: Magnetic induction of displacement current « Reply #20 on: 2019-03-13, 20:49:22 »
Group: Experimentalist Hero Member Posts: 2161	Quote from: Smudge on 2019-03-13, 16:12:35 It is the imaginary impedance (reactance) that increases with distance. For a magnetic dipole of moment m, in the radiative direction (at right angles to the axis of the coil) the H field is given by m/(4pir³). The A field is at right angles to the H field and is given by munoughtm/(4pir²). Hence for an alternating dipole at a sinusoidal frequency the E field is at right angles to the H field, is at 90 degrees phase and is given by omegamunoughtm/(4pir²). The reactive wave impedance given by E/H is therefore omegamunought*r which may be stated as 8 ohms per meter-Mhz. This applies if the retarded effects can be ignored. Smudge Okay, I didn't read it carefully enough, sorry for my misunderstanding on the issue of what increases with distance and thank you for the clarifications. --------------------------- "Open your mind, but not like a trash bin"

F6FLT	Re: Magnetic induction of displacement current « Reply #21 on: 2019-03-13, 20:54:02 »
Group: Experimentalist Hero Member Posts: 2161	Quote from: verpies on 2019-03-13, 11:39:13 This is in conflict with the experiment described on Pg.60 (Sect. of the attached document: "When cylindrical blocks of dielectric material are inserted into air solenoids suspended on the jig, a substantial reduction of the Self Resonance Frequency" occurs." I am impressed by the quality of this study and its originality, particularly on page 16. I didn't expect such a short wavelength when the helical wave is seen along its propagation axis. This opens new doors for me, because it is obvious that we can concentrate standing waves in space and get almost any wavelength on an axis. I thought about my experiment with the submerged coil, made a few years ago. I remembered that the coil was wound on a ferrite rod, and that I saw water as a dielectric loop around the coil, constituting as the secondary of a transformer. At the time I didn't know that the permittivity of ferrites is very important. My negative experiment could therefore be explained by the fact that the displacement currents induced in water could be negligible compared to those induced in ferrite, at the working frequency of about 1 to 2 MHz. However, water is supposed to have a relative permittivity of about 80, even at radio frequencies, so I remain doubtful. « Last Edit: 2019-03-14, 08:44:41 by F6FLT » --------------------------- "Open your mind, but not like a trash bin"

F6FLT	Re: Magnetic induction of displacement current « Reply #22 on: 2019-03-14, 09:35:03 »
Group: Experimentalist Hero Member Posts: 2161	I did the experiment again this morning. It is an LC circuit wound on a ferrite cylinder, resonating at the frequency of 1670 Khz. When immersed in distilled water, the frequency drops to 1210 KHz. So the difference is huge (27,5%), I probably made a big mistake during my first experiment. With an air coil recovered from a loudspeaker: 1000 KHz in air, 910 KHz in water! Less difference (9%) than with the ferrite coil when we would expect more. With an air coil made of enamelled wire 1 mm in diameter with non-jointed turns: 25 MHz in air, 19 MHz in water, difference 24%, same order of magnitude as the first case. « Last Edit: 2019-03-14, 10:11:40 by F6FLT » --------------------------- "Open your mind, but not like a trash bin"

F6FLT	Re: Magnetic induction of displacement current « Reply #23 on: 2022-04-27, 09:36:25 »
Group: Experimentalist Hero Member Posts: 2161	Three years ago I was trying to demonstrate the possibility of creating an induced displacement current from a flux variation, or the opposite, creating an EMF in a magnetic circuit from a displacement current. This had given the following setup, where the 2 square magnetic circuits are classically used. If the central yoke was replaced by a conductive loop, we understand that we have a double transformer and that we should measure an EMF in the loop formed by the second probe of the oscilloscope, on the right. The one on the left is used to visualize the input current generated by the coil in coaxial cable. The yoke is used here as a dielectric, because ferrites have the advantage not only of having a high permeability, but also a high permittivity. The yoke is supposed to be traversed by a displacement current (as would be the conductive loop) to couple the 2 square magnetic circuits. I also tried other setups (see in the history of this thread), some of which are not described here, for example the immersion of a coil in water where despite the permittivity of 80 of the water surrounding it, there is no effect on the inductance. I have never been able to detect any signal whatever the experiment. I had put that at the time on the fact that the positive charges of the dielectric act in opposite direction of the negative charges, cancelling the effect. Yesterday I came across a paper on vixra that was extremely enlightening (beware, this site does not filter, there are many crazy and frankly false theses, but also some valuable works). The author assumes a different permittivity for dielectrics subjected to a field deriving from a potential (for example between the plates of a capacitor) or subjected to an induced field of the type E=-∂A/∂t as around a coil with a variable current. He proposes a method for measuring both. His two pictures on page 5 (fig.3 and 4) were the revelation. The attached image is taken from it. If the orientation of the electric dipoles along the electric field is the same in both cases, there is a force in the direction of the field exerted on the dipoles only in the case of the capacitor. We see in the case of the induced circular field, that the equilibrium of the dipoles along the circle implies the absence of force along the circle, contrary to the one on the free electrons in a looped conductor where a current is induced. The conclusion is clear: an induced electric field does not create displacement currents. On the other hand, it does rotate the dipoles around their center in the medium to align them. Is this effect detectable and could it be useful, this is my new question. Another one is: wouldn't an electrically charged dielectric allow to obtain a displacement current like in the case of the capacitor? I think it does. Any idea is welcome. ------------------------ Magnetic induction of displacement current Courant de déplacement induit, permitivité cas champ rotationnel ou pas.png (168.7 kB, 1195x1496 - viewed 159 times.) --------------------------- "Open your mind, but not like a trash bin"