Enthalpy Posted July 22, 2018 Share Posted July 22, 2018 Hello everybody! Because it is small, like 10-5, measuring the para- or diamagnetic susceptibility (chi = µr-1) of materials is uneasy. Overview of Methods for Magnetic Susceptibility Measurement P. Marcon and K. Ostanina describes several setups. The Gouy balance and the Evans balance measure a force, often for liquidshttps://en.wikipedia.org/wiki/Gouy_balancehttps://en.wikipedia.org/wiki/Evans_balancehttp://www.iiserkol.ac.in/~ph324/ExptManuals/quincke%27s%20manual.pdf An other method uses a Squid, which is sensitive and differential by nature, but demands cold. The less expected setup measures the tiny change of coil (or transformer) inductance due to the material, with an identical but unloaded coil for differential measurement. I propose instead to use a constant excitation magnetic field immobile versus the measurement coil and to move the sample in and out the sensitive volume. Then the induced voltage results only from susceptibility of the sample and nacelle. The electronics can easily integrate the voltage over time to obtain the flux variation in the measurement coils. Adequate arrangement of magnets and measurement coils, for instance as by Helmholtz, define a measurement volume where both the excitation field and the measurement sensitivity are uniform. In the sketch, permanent magnets make the excitation field, and a rotation passes the sample in and out the sensitive volume. Other possibilities exist. Figures should follow. Marc Schaefer, aka Enthalpy Link to comment Share on other sites More sharing options...
Enthalpy Posted July 22, 2018 Author Share Posted July 22, 2018 On the previous sketch, permanent magnets create the excitation field without iron. Rare-earth and ferrite magnets have a small permeability, like 1.05, which I neglect. They are equivalent to a current sheath I = e*H flowing at their rim. This lets arrange them as Helmholtz coilshttps://en.wikipedia.org/wiki/Helmholtz_coil 10mm thick Nd-Fe-B magnets with H=1MA/m are as strong as 10kA*turn, or 100A in 100 turns, wow. With D=60mm, the maximally flat induction distribution would result from 30mm spacing between the magnets if they were thin. This creates B=0.30T at the centre, dropping by 1% at +-5mm axial distance and a similar radial one. A tiny induction dip at the centre would slightly widen the good volume, but the maximally flat condition is algebraically simple: dB/dx = d2B/dx2 = d3B/dx3 where the odd derivatives vanish by symmetry and the second by adjustment. The magnets' thickness e spreads the equivalent current by the amount e. It's equivalent to a convolution of the current distribution by a square function of length e, and this square function is the integral of two opposed Dirac spaced by e. The convolution at the created induction incorporates some d2B/dx2 from x around the centre. For reasonable e, this can be compensated by increasing a bit the spacing. I expect an algebraic solution by this Diracs convolution method, writing d2B/dx2 with the degree change due to the integral, and solving for the maximally flat condition. I didn't check if this method is new, and don't plan to detail it. It applies to coils too. Permanent magnets are strong but dangerous. Electromagnets could replace them, with cores to achieve an interesting induction. The cores must allow easy flux variations. A uniform induction would result from FEM optimization and CNC machining of the poles. The measurement coils can surround the exciting coils; I'd keep them distinct to reduce noises. ========== The measurement coils too are arranged according to ol' Hermann. A sine current flowing in them would create a uniform induction in the measurement volume, hence induce a voltage in a small loop independent of the position there. Because the mutual induction is the same from object A to B and B to A, the same sine current in the small loop would induce the same voltage in the measurement coils from all positions in the measurement volume. This way, all points of the sample within this volume contribute equally to the measurement signal. To compute the signal induced by the sample, of which all points are equivalent, one can integrate over the measurement coils the vector potential A created by a small magnetic dipole, which has an algebraic expression. Instead, I compute here the mutual induction from the measurement coils in the sample. 1A*t in each R=50mm coil create 18µT so a 1cm2*1cm sample receives 1.8nWb and the mutual induction is 1.8nH, from the sample to the measurement coils too. If the 1cm2*1cm sample has a susceptibility chi=10-5 in 0.30T, it's equivalent to a dipole of 24mA and 2.4µA*2 that induces 43pV*s per turn in the coil pair, for instance 14nV over 3ms transitions of the sample in and out the excitation field. ========== If each coil has 2000 turns, the signal is 86nV*s or 29µV over 3ms. D=0.29mm enameled copper make 340 ohm resistance and about as much reactance at 40Hz. Integrated over 40Hz and with 2dB noise by the differential amplifier, the thermal background is 19nV, so the measure can be accurate even for smaller samples. Good amplifiers can cope with a smaller source resistance, that is, fewer turns. The measurement coils (and their feed cables) demand shielding against slow electric fields, especially at 50Hz and 60Hz. A nonmagnetic metal sleeve can surround each coil if the torus it makes is not closed, but its ends can overlap if insulated. It's better symmetric starting from the feed cable and connected to the feeder's shield. The situation is much easier than for electrocardiograms. 100mA*2mm*0.5m at 1m distance and 50Hz induce 0.2µV in the coil pair, so a decently calm environment needs no magnetic shielding. Averaging successive measures would further filter out interferences by the mains. The sample's cycle can be desynchronized from the mains or have a smart frequency ratio with it. ========== An electric motor can rotate the sample if it's shielded by construction or is far enough. An adjustable counterweight is useful. Or let the sample oscillate as a pendulum. The nacelle, or bottle for liquids and powders, should contribute little signal. Thin construction of polymer fibres is one logical choice. Its contribution can be measured separately and subtracted by computation. The parts holding the magnets and coils must be strong and also stiff. A 2000 turns coil moving by 1nm versus the magnet gets roughly 100nV*s, as big as the signal, so this shall not happen at the measurement timescale. Not very difficult, but needs computational attention. The nacelle's movements must be isolated from the magnets and coils. ========== If a sample keeps a permanent magnetization, it should be measured without the excitation magnets. Metals conduct too much, but most electrolytes fit. Marc Schaefer, aka Enthalpy Link to comment Share on other sites More sharing options...
Enthalpy Posted July 22, 2018 Author Share Posted July 22, 2018 Some crystals are anisotropic and their susceptibility is a tensor. By orientation of the measurement coils, my apparatus measures the nondiagonal terms of the susceptibility. Here a continuous rotation can't insert the sample in the measurement volume and extract it. A mechanical oscillation seems better, with enough amplitude to go over the centre or even exit the measurement volume in both directions, and can serve with the previous orientation too. The oscillatory pumps over oil wells may give inspiration. A spring can let exceed 1g if desireable. The sample's orientation is paramount, so parallel wires as sketched don't suffice: it needs at least a truss. Marc Schaefer, aka Enthalpy Link to comment Share on other sites More sharing options...
Enthalpy Posted July 25, 2018 Author Share Posted July 25, 2018 It is well known, but not by everybody, so here's a sketch of an electrostatically shielded coil. A closed loop would allow current induced in the shield to reduce the voltage induced in the windings, so the shield is interrupted. At such low frequencies, the ends of the shield can overlap if something prevents an electric contact, and the shield needs not be symmetric around the feed point. Nor is a symmetric cable vital here. ========== I have forgotten "=0" in the conditions of a maximally flat induction at a Helmholtz coil. Link to comment Share on other sites More sharing options...
Enthalpy Posted July 28, 2018 Author Share Posted July 28, 2018 "One can integrate over the measurement coils the vector potential A created by a small magnetic dipole" to compute the signal induced by the sample, but on 07/22/18 03:11 PM I took an other route. Here is the standard one, with most computations in the drawing. The A potential by a magnetic dipole has a know algebraic expression computed from Biot and Savarthttps://de.wikipedia.org/wiki/Biot-Savart-Gesetzhttps://en.wikipedia.org/wiki/Magnetic_dipole If I inject the former m=2.4µA*m2 and R=50mm I get 22pV*s per turn in the coil pair, precisely half as much as previously. I trust today's computation. The former had possibly a logic flaw because of the two coils. The setup has much noise margin anyway. Marc Schaefer, aka Enthalpy Link to comment Share on other sites More sharing options...
ALine Posted July 28, 2018 Share Posted July 28, 2018 Brah what is this mad scientist making? Link to comment Share on other sites More sharing options...
Enthalpy Posted July 29, 2018 Author Share Posted July 29, 2018 Signal, noise, electronics. If the sample rotates at 5Hz on R=0.1m, the enter and exit voltage bumps can be 20ms wide and spaced by 40ms, while 30ms or more are needed to integrate them completely. Helmholtz' region of uniform induction provides a time of zero voltage between the bumps where integration can stop and start. Software can integrate each bump over 30ms and compute the difference. Integration times and their distance are usefully multiples of 20ms to reject interferences at 50Hz, or 16.67ms at 60Hz, and the repetition period an odd multiple. As the sample's rotation or oscillation frequency can vary, a separate sensor can tell the instantaneous position and software determine the best start and duration of the integration windows. With the example times, the integration picks noise from 10Hz to 40Hz, with complicated limits. The window can be smoothened a bit, but we make metrology here. A few TL1028 make a good differential amplifier (instrumentation amplifiers are about as good: AD8229, AD8428, AD8429 and competitors)http://www.analog.com/media/en/technical-documentation/data-sheets/1028fd.pdf Noise per input is 1nV/sqrt(Hz) and 3pA/sqrt(Hz) at these frequencies. Each half-coil is essentially resistive: 340ohm and at 40Hz j200ohm. After integration and enter-exit difference, the differential noise is 24nV. Two 2000 turn coils pick 44nV*s from a 1cm3 chi=10-5 sample, or mean 1.5µV over 30ms. The enter-exit difference is 2.9µV or 100* the noise voltage. A resolution of chi=0.01*10-5 results from averaging few measures. The sample can be smaller. The analog circuit shall provide limited and fixed low-pass filtering. This leaves the mean value of a bump untouched, provided that the time of zero voltage between the bumps is kept. At identical transition and selectivity, inverse Chebychev and elliptic filters have a faster and quieter time response than Chebychev and Butterworth, don't believe books. The considered amplifiers have an offset smaller than a strong signal. The analog circuit can pass the DC to avoid measurement errors and to settle quickly despite the slow signals. Enter-exit bumps difference by software removes the DC and low frequency components. A PC distributes irregular >10A in unshielded cables and its processor draws >100A. It's a bad source of magnetic and conducted interferences. If box and distance don't suffice, consider a microcontroller instead. Marc Schaefer, aka Enthalpy Link to comment Share on other sites More sharing options...
Enthalpy Posted August 2, 2018 Author Share Posted August 2, 2018 On 7/28/2018 at 7:14 PM, Enthalpy said: I trust today's computation. I shouldn't have, because if now I compute the vector product properly using the sine and not the cosine, I get a signal twice as big that equals the other computation on 07/22/18 03:11 PM. On 7/29/2018 at 2:14 PM, Enthalpy said: the repetition period an odd multiple (of the mains' period) But this doesn't reject interferences. If P*Q measures are averaged, they should start at P different angles from the start of a mains' period, the angles being equally spread over a turn. Examples: 2Q measures can start at 10° and 190° from the start of a mains' period. 3Q at 20°, 140°, 260°. 6Q at 1°, 31°, 61°, 91°, 121°, 151°. This squashes the interferences at the mains' frequency and its harmonics not multiple of P (it's a sum of the roots of 1 in the complex plane) so a big P has some usefulness. Exact angles improve the rejection, so the measures should start when the sample's speed is stable, and even at big P*Q, precise timing improves over sampling asynchronous with the mains. Marc Schaefer, aka Enthalpy On 7/28/2018 at 7:33 PM, ALine said: Brah what is this mad scientist making? Sound science. As the title says, and using the most obvious method. Apparently it hasn't been developed up to now, I don't know why. Link to comment Share on other sites More sharing options...
Enthalpy Posted August 4, 2018 Author Share Posted August 4, 2018 Most sources show that the Helmholtz coil provides an induction uniform along the z axis because this is simple algebra. But how uniform is the induction radially? A less simple algebraic solution must exist off-axis, possibly with Bessel functions. Here I prefer to show more generally that, due to its properties in vacuum, if the induction is uniform along the symmetry axis, it's uniform radially too, and the deviation can be estimated. With approximation signs on the sketch, but a mathematician would do it cleanly with Taylor series and Cauchy remainder.. Here the on-axis induction Bz(0,z) is maximally flat, varying as beta*z4 approximately. div(B)=0 links the axial variation of Bz with the radial component Br near the axis, which is small, and if beta=0 it's zero. curl(B) aka rot(B)=0 links the axial variation of Br with the axial component Bz near the axis, which varies slowly with r, and if beta=0 it's uniform. So how wide can a sample be? Let's take a sphere centred on r=z=0 with radius Z. The radial variation is like r2z2 but on the sphere r2+z2=Z2, so the radial variation is maximum for r2=z2=Z2/2, or +0.75*beta*Z4, while the axial variation is -0.25*beta*Z4 there and -1.00*beta*Z4 at z=Z. That is, the sphere is a reasonable boundary. With the previous 2Z=1cm it provides 0.5cm3 to the sample. I've deduced function values on a surface or volume from the values on a line. That's common with a differential equation as we have here, and may even be accurate far from the line. We could have written Bz(0,z) with more powers of z, or as a Fourier series... I suppose a link with holomorph functions. In a vacuum domain not surrounding a current, div(B)=rot(B)=0 lets define a scalar potential grad(psi)=B with delta(psi)=0. Within a phi=const plane, delta(psi)=0 makes psi the real part of a holomorph function, so knowing its values on a line fully defines it on the domain. Marc Schaefer, aka Enthalpy Link to comment Share on other sites More sharing options...
Enthalpy Posted August 6, 2018 Author Share Posted August 6, 2018 On 07/29/18 02:14 PM, I meant "the sample rotates at 2.5Hz", not 5Hz. 5Hz is possible too. The duration and distance of the integration windows can still be multiples of 20ms or 16.67ms, and the signal-to-noise improves by 3dB for each pulse pair and by 6dB if averaging over an identical duration. Link to comment Share on other sites More sharing options...
Enthalpy Posted August 9, 2018 Author Share Posted August 9, 2018 Some materials, mainly man-made ones like polymers and technical ceramics, have homogeneous susceptibility and can be machined to accurate shape. Then, as the deviations of the induction have opposite signs along the axis and the radius, special proportions of the sample can let the deviations compensate an other. The measure is more accurate, and the sample can be bigger. Using the radial gradient of the induction obtained on 08/04/18, the image computes the compensation condition as Z/R=D/L=sqrt(2/5)~0.633 for a cylinder, within a domain where the on-axis deviation is written as -beta*z4, and the residue for a sphere of radius Z. The sphere isn't bad, provided it's accurately shaped and homogeneous: it compensates by 0.086, or to 0.086% if beta*Z4=1%. For every shape, an accurate position is necessary. A cylindrical or spherical container filled with a liquid or even a powder could keep the good properties if the filling is complete and uniform. Marc Schaefer, aka Enthalpy Link to comment Share on other sites More sharing options...
Enthalpy Posted August 11, 2018 Author Share Posted August 11, 2018 What is the consequence of a conductive sample on a measure? The drawing computes it for a cylindrical sample, but shape details matter little. The excitation induction is supposed essentially unaffected by the material since the main measured effect is about 10-5. The magnetic moment due to conductivity results from the excitation induction and must be smaller than the susceptibility effect. To achieve 0.01* the previous 2.4µA*m2 from 0.3T/3ms in 1cm2*1cm sample, the material must conduct less than 60S/m = 16mohm*m or 600mS/cm in the unit of the table: Electrical Conductivity of Aqueous Solutions, in recent CRC Handbook of Chemistry and Physics this allows all electrolytes listed there. Metals conduct much more, up to 106* the previous limit, but can still be measured if the sample consists of insulated bands, wires or coarse powder grains up to 103* smaller as needed: D=10µm. Maybe native or grown oxide suffices to insulate the grains, but I'd trust instead a suspension in a liquid or some solid wax. Apply a law of mixtures. Marc Schaefer, aka Enthalpy Link to comment Share on other sites More sharing options...
Recommended Posts
Create an account or sign in to comment
You need to be a member in order to leave a comment
Create an account
Sign up for a new account in our community. It's easy!
Register a new accountSign in
Already have an account? Sign in here.
Sign In Now