An Analytical Approach to Eddy Current in Electromagnetic Damping
AAn Analytical Approach toEddy Current in Electromagnetic Damping
Hao Chen ∗ Department of Physics, Yale University,P.O. Box 202736, New Haven, Connecticut 06520, USA † Yawen Xiao ‡ Department of Physics, University of IllinoisUrbana-Champaign, Urbana, Illinois 61801-3003, USA † Jinze Shi § and Weishi Wan ¶ School of Physical Science and Technology,ShanghaiTech University, Pudong, Shanghai 201210, China
Abstract
An analytical method of calculating eddy current in a metallic spinning gyroscope in externalmagnetic field is presented. With reasonable assumptions, the problem is simplified from thetime-dependent one governed by Maxwell equations to the boundary value problem of Poisson’sequation, which yields a closed-form expression of the eddy current. The rotation frequency as afunction of time is calculated, compared with experiment and the relative error is found to be 8.61% . ∗ First author; [email protected]; [email protected] † School of Physical Science and Technology, ShanghaiTech University, Pudong, Shanghai 201210, China ‡ First author; [email protected] § [email protected] ¶ Corresponding author; [email protected] a r X i v : . [ phy s i c s . c l a ss - ph ] M a y . INTRODUCTION The International Young Physicist Tournament (2019) contains a problem of a spin-ning gyroscope made from a conducting, yet non-ferromagnetic material that slows downwhen placed in a magnetic field due to the interaction between induced eddy current andthe magnetic field[1]. Such a phenomenon, which is usually referred to as ´electromagneticbreaking’ or ´electromagnetic damping’, is widely discussed qualitatively in college and highschool level electromagnetism textbooks as a direct result of Faraday’s law of electromag-netic induction[3]. Quantitatively, the eddy current in electromagnetic damping has alsobeen studied rather extensively by physicists and electrical engineers over the past decades.W. R. Smythe calculated the eddy current in a conducting disk rotating in an externalmagnetic field generated by one cylidrical permenant magnetic pole on each side[4]. Later,another approach to this problem was developed by D. Schieber, which gave the result thatwas in agreement with Smythe’s[5]. In both of the two papers, the Maxwell equations aresolved in the frame of reference that moves together with the rotating disk. Recently, M.A. Nurge et al. studied the distribution of eddy current in a rotating sphere in externalmagnetic field in the lab frame[6]. Although their derivation was rigorous, they expressedthe result as a sum of a series rather than a closed-form formula, and each term of the serieswas a multiple of an associated Legendre polynomial. Such a result is rather complicatedfor application to a practical case. In addition, one has to expand the external magneticfield with associated Legendre polynomials in advance, making their result even more com-plicated. It is advantageous to develop a method of solving the eddy current problem in labframe which gives a closed-form result that is easy to use.In this paper, we present a general method of calculating eddy current based on solvingthe Maxwell equations with some reasonable assumptions, which can be further simplifiedto a boundary value problem of a Poisson’s equation. In general cases, the solution of thePoisson’s equation can only be expressed as a sum of a series. However, in order to havea better understanding of the underlying physical mechanism and make the result easyto be applied to practical cases, deriving a closed-form formula is of great significance, ifachievable. Fortunately, in our gyroscope problem, the closed-form formula of eddy currentcan be obtained with a simplified expression of the magnetic field. Meanwhile, all theidealized conditions in our theory can be realized with a simple apparatus in experiments.2hile it is hard to directly measure eddy currents in conducting materials, which is therotator of our gyroscope, it is straightforward to compare the deceleration processes deducedin theory and observed in experiment to verify our closed-form result of the eddy current.In this article, to show the whole picture of our work, we firstly stated the basic logic ofour theory of calculating eddy currents in a block of conducting material. Then, we applythe theory to the gyroscope problem to get a theoretical result for the deceleration process.The comparison of the theoretical and experimental results shows that they fit each otherextremely well. Lastly, we discuss the justification of all the assumptions of our model andgive a qualitative explanation of the remaining 8.61% error.
II. BASIC LOGIC OF THE THEORY AND ITS APPLICATION
In this section, we will discuss how to apply electromagnetism laws to calculate eddycurrents, which starts with Maxwell equations[7]: (cid:126) ∇ · (cid:126)E = ρε , (1) (cid:126) ∇ × (cid:126)E = − ∂ (cid:126)B∂t , (2) (cid:126) ∇ · (cid:126)B = 0 , (3) c (cid:126) ∇ × (cid:126)B = (cid:126)jε + ∂ (cid:126)E∂t , (4)where (cid:126)E, (cid:126)B are electric and magnetic fields, while ρ,(cid:126)j are the densities of charge and current,respectively. (We may call them ’charge’ and ’current’ in the rest of the article to be concise.)Since we are dealing with current in a piece of conductive material, Ohm’s law must betaken into account to describe it: (cid:126)j = σ ( (cid:126)E + (cid:126)f ) , (5)where (cid:126)f is the ’non-electrostatic force’, which is usually inside batteries. Its loop integralalong the circuit is ’electromotive force’, usually denoted as ’emf’.[8]In the problem of the gyroscope, the magnetic field is produced by permanent magneticblocks and hence independent of time. (The magnetic field produced by eddy current isomitted as usual.) As a result, the electric field in the conducting material (rotator of the3yroscope) obeys the reduced form of Equation (1) & (2): (cid:126) ∇ · (cid:126)E = ρε , (6) (cid:126) ∇ × (cid:126)E = 0 . (7)To make the above equations easier to solve, it is convenient to set up a scalar field: ϕ ( (cid:126)r ),which is called scalar potential in electromagnetism, whose minus gradient is the electricfield ( − (cid:126) ∇ ϕ = (cid:126)E , and it satisfies (7) automatically). Plugging this into Equation (6), wehave: − ∇ ϕ = ρε . (8)Now, it appears that Equation (8), together with proper boundary conditions, is sufficientto determine the scalar potential ϕ (and then electric field). However, one cannot calculatescalar potential ϕ from Equation (8) yet, since we do not have a priori knowledge of thecharge distribution. Therefore, a different perspective must be brought in to constructanother relation between ϕ and ρ , in order to get a set of solvable equations of scalarpotential. It is worth noting that (cid:126)f together with (cid:126)E in Equation(5) drives the motion ofcharge, and with this, the conservation law of charge: (cid:126) ∇ · (cid:126)j = ∂ρ∂t , (9)can be used to establish the relationship between current and charge. By plugging Equation(5) into (9), and then replacing (cid:126)E with − (cid:126) ∇ ϕ , another equation linking electric field andcharge density is derived: σ ( −∇ ϕ + (cid:126) ∇ · (cid:126)f ) = ∂ρ∂t . (10)Then, by solving Equations (8) & (10), a result of ϕ and then (cid:126)E and finally current (cid:126)j can be obtained. Note that, in order to solve these equations, proper boundary and initialconditions are needed. Since initial conditions are unique and straightforward in differentproblems, we only list the boundary condition below, which is derived from the fact thatthe current cannot go through the surface of the material: (cid:126)j · ˆ n | boundary = j n | boundary = 0 , (11)where ˆ n is the unit normal vector of the boundary.In summary, we construct a general method to calculate current in conducting material.For static magnetic field, the electric field has to be “irrotational” ( (cid:126) ∇ × (cid:126)E = − ∂ (cid:126)B/∂t = 0).4ther parameters, such as the shape of the conductor, the conductivity or the distributionof non-electrostatic force can be arbitrary. While difficult, the second order time-dependingpartial differential equations below are solvable: −∇ ϕ = ρε ,σ ( −∇ ϕ + (cid:126) ∇ · (cid:126)f ) = ∂ρ∂t ,j n | boundary = 0 . (12)In order to solve these equations in our problem, we are going to introduce several assump-tions to simplify the calculation (justification for these assumptions will be discussed in thelast section of this article):1. The current varies so slowly with time that it can be treated as nearly time-independent,which means ∂ρ/∂t = 0.2. The non-electrostatic force here is due to the rotation of the gyroscope, which can beexpressed as: (cid:126)f = (cid:126)v rot × (cid:126)B = ( (cid:126)ω × (cid:126)r ) × (cid:126)B, (13)where (cid:126)v rot represents the velocity of a particular point on the rotator, (cid:126)ω is the angularvelocity of the rotator and (cid:126)r is the position vector from the center of the gyroscope.3. The magnetic field supplies the current, and this force acts on the gyroscope directlycausing its rotation slow down, but does not affect current distribution.With these assumptions, we can use a much more concise Poisson’s Equation to calculatethe current in the gyroscope: ∇ ϕ = (cid:126) ∇ · [( (cid:126)ω × (cid:126)r ) × (cid:126)B ] . (14)We then calculate the deceleration process by evaluating the magnetic torque on it.We will show in the next section that the current has a rotational form and can beconsidered as ’eddy current’. Additionally, it is worth pointing out that such ’eddy current’is a common phenomenon when non-electrostatic force is originally produced by magneticfield (both magnetic force and the force given by induced electric field). Such ’eddy current’is harmful in many cases since it cause extra loss of energy.5 II. MODELING
We turn to the simplest case to solve the problem analytically. Here we only take thecomponent of the field perpendicular to the plane of the rotator into consideration. Thein-plane magnetic field, whose role will be discussed in the last part of the article, has amuch smaller contribution to eddy current due to the fact that the radius of the rotator ismuch larger than its thickness.Qualitative analysis shows that there will be no stable current in the gyroscope placed inan even magnetic field because the non-electrostatic force distribution has a rotational sym-metry (the free electrons will immediately form an electric field to cancel non-electrostaticforce precisely). Therefore, a magnetic field distributed anti-symmetrically is the simplestform for us to setup the experiments and analyze the problem theoretically. (a)The schematic diagram of our model (b)Vertical component of the field
FIG. 1. Figure (a) gives a schematic view of our model. It shows the relative position between themagnets (red-blue blocks) and the rotator of our gyroscope (yellow cylinder). Red and blue partsrepresent the north and south poles of the magnets respectively. Figure (b) is a plot of the verticalcomponent (y component) of the field at the height of the rotator. We find that the magneticfield is anti-symmetric about the x-y plane. (We will discuss the horizontal component in the lastsection, since it contributes to our rest 8.61% error.)
To produce such an anti-symmetric field, we set two identical cuboid magnets beside eachother with the north polarity of one matched to the south polarity of the other. When we dothis the magnets attract and stick together. Such a configuration forms an anti-symmetricalmagnetic field above it, in the vertical direction. (The picture of our model and the verticalcomponent of field distribution is shown in FIG.1).The gyroscope is placed on the centerline where the two magnets meet, with the gyroscopes brass rotator parallel to the magnetic6urface. The center of the brass rotator is on the middle of the line between the two magnets.These equipments and experiment process are shown in FIG.2 and its illustration.
FIG. 2. Here shows how the experiment was conducted. The blue rectangular device consists ofa small rotating motor, and we use the device to initialize the rotation of the gyroscope, as themagnetic field is not strong enough to start the rotation. Attached to the rotator are four identicalpieces of tape, which form a cross that we use to measure the rotation of the gyroscope. Then,using a 20.00 Hz strobe light, we measure the angular velocity of the rotator during the processof deceleration. When the adhesive tape appears to rotate at a diminishing rate, then begins torotate in the other direction, at the instance of the directional change, the frequency of the rotatorhas synchronized with the strobe light. When the strobe light synchronizes, the revolution of therotator is an integral multiple of 5.000 r/sec (a quarter of 20.00 Hz).Notice: 20.00 Hz was chosen as an appropriate frequency because higher frequencies make videorecording difficult. Conversely, lower frequencies make us unable to catch the directional changewith the strobe light.Detailed information of our apparatus: 1. Magnets: 10.00cm * 10.00cm * 2.00cm; 2. Rotator:made of brass (type: H59), 5.30cm in diameter; 3. Distance between top surface of the magnetsand the rotator: h = 3 . cm . V. DISTRIBUTION OF MAGNETIC FIELD
Before dealing with the gyroscope problem, we first focus on the magnetic field. In ourmodel, the magnetic field is merely an external parameter that can be measured point bypoint via experiment. However, the measurement can be rather difficult. The difficulty isthat the magnetic field is a vector field, hence we have to ensure that the surface of theHall sensor on the Teslameter is precisely parallel to the surface of the magnets, in orderto get the correct measurement of the vertical component. Unfortunately, due to the smallsize of the Hall sensor chip, it is nearly impossible to ensure that it is horizontal within theaccuracy we need.To get around the above problem, we decide to use a theoretical formula of the magneticfield produced by magnet blocks, which can provide us a credible result of the distributionof magnetic field with only a few external parameters that are not hard to obtain. Schlueterand Marks[9] have derived the distribution of a magnetic field produced by a rectangularpermanent magnet block using the uniform magnetic charge sheet model. For our particularproblem, we only consider the case that magnetization is along the y axis. As a result, thecharge sheets exist only on the surfaces parallel to the x-z plane, i.e., the normal of thecharge sheets are of the same amplitude but opposite signs. For a single charge sheet, whosecorners are ( x , y , z ) , ( x , y , z ) , ( x , y , z ) and ( x , y , z ), the components of magneticfield are: B x = B r π ln [ z − z + r ( x , y , z )][ z − z + r ( x , y , z )][ z − z + r ( x , y , z )][ z − z + r ( x , y , z )] , (15) B y = B r π { tan − [ ( z − z )( x − x )( y − y ) r ( x , y , z ) ] + tan − [ ( z − z )( x − x )( y − y ) r ( x , y , z ) ] − tan − [ ( z − z )( x − x )( y − y ) r ( x , y , z ) ] − tan − [ ( z − z )( x − x )( y − y ) r ( x , y , z ) ] } , (16) B z = B r π ln [ x − x + r ( x , y , z )][ x − x + r ( x , y , z )][ x − x + r ( x , y , z )][ x − x + r ( x , y , z )] , (17)where, r ( x i , y j , z k ) = (cid:113) ( x i − x ) + ( y j − y ) + ( z k − z ) , (18)and quantity B r is the remanence of the magnet, which can be treated as a fitting parameterin practice. For our magnets, B r is about 600 mT, which leads a field whose verticalcomponent is around 110 mT right on the top surface. (We use the word ’around’ herebecause this parameter is checked every time before an experiment is taken.)8or the other single charge sheet of the block, whose corners are ( x , y , z ) , ( x , y , z ) , ( x , y , z )and ( x , y , z ), the field components are similar to the previous ones: B x = − B r π ln [ z − z + r ( x , y , z )][ z − z + r ( x , y , z )][ z − z + r ( x , y , z )][ z − z + r ( x , y , z )] , (19) B y = − B r π { tan − [ ( z − z )( x − x )( y − y ) r ( x , y , z ) ] + tan − [ ( z − z )( x − x )( y − y ) r ( x , y , z ) ] − tan − [ ( z − z )( x − x )( y − y ) r ( x , y , z ) ] − tan − [ ( z − z )( x − x )( y − y ) r ( x , y , z ) ] } , (20) B z = − B r π ln [ x − x + r ( x , y , z )][ x − x + r ( x , y , z )][ x − x + r ( x , y , z )][ x − x + r ( x , y , z )] . (21)The total field is: (cid:126)B = (cid:126)B + (cid:126)B . (22)The resulting magnetic field, is obtained assuming µ r = 1, where µ r is the relative per-meability. In reality, µ r (cid:107) = 1 . − .
08 and µ r ⊥ = 1 . − .
08 [10] [11]. This is the mainsource of error in the model. Yet our method of calibrating B r with measurements removesthe error to a large extent, replacing it with measurement error. Of course the anisotropyof the real magnet is still unaccounted for, but it is in general smaller.Now we have an analytical formula of the magnetic field distribution, but it is still hard toapply to further calculations of eddy currents, since such a complicated formula will appearas a nonhomogeneous term in Equation (14). Therefore, we seek for a simpler formula toreplace it. By entering the external parameters of our magnets into the precise formulaof the magnetic field, we find that in the region occupied by the gyroscope, the verticalcomponent of the field is solely dependent on coordinate z, as shown in FIG.3.Therefore, it is reasonable to describe the magnetic field (vertical component) by a one-parameter function. A conventional way to find a simpler local formula of a function is themethod of approximation, which is similar to polynomial curve fitting in experiments. Tobe specific, we use the following function as an approximation of the field distribution: B y ( x, y = h, z ) = B z = k ( z − z ) + k ( z − z ) . (23)where k & k are undetermined coefficients, and z (which is zero in our model) is theposition of the boundary of the two magnets.Then, with the help of Wolfram Mathematica, the values of k & k are obtained. Theresult shown in FIG.4 indicates that in the region of the rotator, the analytical and approx-imate formulas correspond to each other very well.9 IG. 3. The vertical component of the field is almost solely dependent on coordinate z in the regionof gyroscope.FIG. 4. The blue line is the result of the original formula, while the orange one is that of Equa-tion(23) with the correct values of k = − . mT · cm − & k = 2 . mT · cm − . It can be seenthat they fit each other very well. . CALCULATION OF THE CURRENT Based on this preliminary work, we can transform our ’gyroscope teslameter’ probleminto a boundary value problem of Poisson’s equation: ∇ ϕ = (cid:126) ∇ · [( (cid:126)ω × (cid:126)r ) × (cid:126)B ] ,j n | boundary = σ ( (cid:126)E + (cid:126)f ) · ˆ n | boundary = 0 . (24)Since the rotator of our gyroscope is not simply a cylinder, but a combination of tworings and a circular plate (see FIG.5), we decide to solve the Poisson’s equation in the ringsand the plate separately, and then combine them together. As a result, the problem ineach region reduces to 2-dimensions (since they are translational invariant along the y axis).Therefore, we solve the following equations: ∇ ϕ = (cid:126) ∇ · [ ωr ( k r sin θ + k r sin θ )ˆ r ]= ω [3 k r sin θ + k r (3 sin θ − sin 3 θ )] , ∂ϕ∂r | r = a − ωa ( k a sin θ + k a sin θ ) = 0 , ∂ϕ∂r | r = b − ωb ( k b sin θ + k b sin θ ) = 0 , (25)where a and b are inside and outside diameter of the ring. For the plane region, we just take a = 0. Here we make use of cylindrical coordinates (or polar coordinates since the problemis 2D) them. The relations between them and the original cartesian coordinates are: x = r cos θ,y = y,z = r sin θ. (26)To solve the Equation (25), we use the superposition principle to split this Poisson’s equa-tion with the complicated boundary conditions into two equations. One of these equations isthe same Poisson’s equation with all boundary conditions equal to zero, i.e. a inhomogeneouspartial differential equation with homogeneous boundary conditions: ∇ ϕ P = ω [3 k r sin θ + k r (3 sin θ − sin 3 θ )] , ∂ϕ P ∂r | r = a = 0 , ∂ϕ P ∂r | r = b = 0 . (27)11 IG. 5. This is a close-up view of the gyroscope. It can be seen that the rotator is a combinationof two rings and a plate (there is only one ring can be seen on the picture, and the other one is onthe bottom side).
This equation can be solved by assuming that the solution is a summation of infinite terms: ϕ P = (cid:80) ∞ n =0 ϕ n , where ϕ n = A n ( r ) cos( nθ ) + B n ( r ) sin( nθ ). By plugging it into the Equa-tion (27), the partial differential equation is reduced to second order ordinary differentialequations of variable r , which are named ’Euler’s equation’ in mathematical physics, eachassociated with one function: A n ( r ) or B n ( r ) with a particular n . There is a systematic wayto solve Euler’s equation, and Equation (27) is then solved. It is worth noticing that theform of solution we assumed above is not arbitrary, but is based on the fact that cos( nθ ) andsin( nθ ) with integer n are the ’angular eigenfunctions’ of Poisson’s equation (and Laplace’sequation) in such a region.The other equation (Equation 27), then becomes a Laplace’s equation with the sameboundary conditions of the initial equation (25), i.e. a homogeneous equation with inhomo-geneous boundary condition: ∇ ϕ L = 0 , ∂ϕ L ∂r | r = a = ωa ( k a sin θ + k a sin θ ) , ∂ϕ L ∂r | r = b = ωb ( k b sin θ + k b sin θ ) , (28)which can be solved the same way as above.The solutions ϕ P and ϕ L to these two equations can be added to get the solution ϕ to12he initial equation: ϕ = ϕ P + ϕ L . (29)Finally, we obtain an analytical result of the scalar potential distribution. Subsequently,we can get the electric field by evaluating (cid:126)E = − (cid:126) ∇ ϕ , and then plug (cid:126)E back into Equation(5) to get the current distribution. Since the final formula is rather long, we list the currentinduced by the first and second term of the magnetic field (see Equation 23) separately,naming them (cid:126)j and (cid:126)j respectively: (cid:126)j ( r, θ ) = σωk [ r sin θ + 18 ( a + b − a b r − r ) sin θ ]ˆ r + σωk [ 18 ( a + b − a b r − r ) cos θ ]ˆ θ, (30) (cid:126)j ( r, θ ) = σωk ( a − r )( r − b )64( a + a b + b ) r [2( a + a b + b )( a + b + r ) r sin θ − r ( a + a b + b ) + r a b ( a + b ) + a b ) sin 3 θ ]ˆ r + σωk r [2 r ( a b ( a + b ) + r ( a + a b + b ) − r ) cos θ − r ( a + a b + a b + b ) − r ( a + a b + b ) + 9 a b a + a b + b cos 3 θ ]ˆ θ. (31)The above two formulas represent the current distributions in the rotator analytically. Togive a more intuitive picture, we use a built-in function of Wolfram Mathematica to generatethe following stream plots of the currents (see FIG.6, the formulas of boundary conditions arealso listed there). These stream plots are slightly different from the conventional ones thatillustrate electric and magnetic fields, since the density of curves here does not represent theintensity of current. Only the directions of arrows have physical significance as the directionof current at each point. VI. COMPARISON WITH EXPERIMENTS
With the formulas of current distribution, Equation (30) & (31), we can easily calculatethe magnetic torque (cid:126)M on the rotator of gyroscope via the following integral: (cid:126)M = (cid:90) V (cid:126)j × (cid:126)Bdv = − λ(cid:126)ω, (32)where coefficient λ is independent of angular velocity. That the torque is proportional toangular velocity is obvious since the current in the integral is proportional to it as well. The13 a) (b)(c) (d) FIG. 6. These are stream plots of the currents generated by Wolfram Mathematica. Figure (a)represents eddy current in a circle plate (or a cylinder) induced by a ’linear magnet field’, thatis, the first term of Equation (23) (Equation(30)). Figure (b) represents eddy current in a ringinduced by the same field. And figures (c) and (d) show the currents induced by a ’cubic magneticfield’, that is, the second term of Equation (23) (Equation(31)). One interesting feature of theseplots is the degree of symmetry. There are left-right symmetry and top-down anti-symmetry, whichreflects the symmetry of the magnetic field. following law of mechanics is taken here:
I dωdt = M = − λω. (33)In the solution ω ( t ) = ω e − t/τ , (34) τ = I/λ is the ’characteristic decaying time’.14
IG. 7. Measurement of the deceleration process, which is displayed as orange points, is fit bya curve (brown) with the form of Equation (35). In our experiments, the initial angular velocityis ω − ω rest = 377 . rad/s , which is 60.00 Hz, and the constant term is ω rest = 32 . rad/s . Theresult of decaying time (given by the fitting function of OriginLab) is τ exp = 5 . s . Moreover,the correlation coefficient is r = 0 . Entering the dimensions and the conductivity of our gyroscope, we get the result that τ = 7 . s .On the other hand, the measurement of deceleration of the rotator above magnets isdone with the favor of a strobe light (see FIG.2 for details). However, we realize that theexperimental result is not an exponential decay, but with a minus constant added, as shownbelow: ω ( t ) = ω e − t/τ exp − ω rest , (35)where the subscript ’exp’ (refers to ’experiment’) is to distinguish from the theoretical one.Such a result is reasonable since an ideal exponential decay indicates that the gyroscopewill not stop spinning forever, which is unrealistic. After examining Equation (33) carefully,15e found that if a constant frictional force is added to the right hand side, a deceleratingcurve of the same form with the experimental one is obtained.Consequently, we turn to investigate the frictional force here. Through experiments, wemeasure the deceleration curve of the gyroscope without the magnets, which means the onlytorque on the rotator is caused by frictional force. Such a deceleration curve gives directlythe form of the torque generated by the frictional force, which is linear to the angularvelocity: M f = − M − αω. (36)However, this is not the end of the story. An intrinsic shortcoming of our gyroscope isthat its shaft is made of steel, which is magnetic. The attraction between the shaft andthe magnets exerts additional pressure between the rotator and the frame which enlargesfrictional force. Therefore, the frictional torque we measured away from the magnets isnot identical to that in the presence of the magnets. Fortunately, the empirical formula offrictional force: F f = µF N indicates that it is proportional to pressure, hence the torque weare looking for is likely to be a multiple of the one measured in absence of the magnets.In our model, we treat this multiplier as a fitting parameter. By choosing it properly, weare able to make the constant term in the theoretical deceleration curve the same as thatof the experimental curve. In the end, we achieve good agreement between the theory andexperiments (see FIG.8), the error in the characteristic decaying time is about 8.61% . VII. REASONABILITY OF THE ASSUMPTIONS
In this section, we give a brief review of the assumptions we introduced in Section II anddiscuss possible sources of error.The first assumption we introduced is that we treat the current (as well as electric field) asnearly time-independent. As all the fields, charge and current distribution varying slowly?astime goes on, quantity ∂ρ/∂t will be small enough to be omitted, hence the divergence ofthe current is zero. To be more specific, the condition of nearly time-independent that canbe expressed as the partial derivative of electric field (or current) is much less than that ofelectric field itself (of course with some constants to scale them to the same dimension):[12] ε ∂ (cid:126)E∂t (cid:28) σ (cid:126)E. (37)16
IG. 8. A comparison between the theory and the experiment. As stated below FIG.7, the valueof decaying time of our experiment is τ exp = 5 . s . Our calculation of the interaction betweenmagnetic field and eddy current, together with the calibration of the frictional torque, gives thatwith the same initial conditions, the decaying time is τ = 5 . s . The error in the characteristicdecaying time is about 8.61 % . As for our brass rotator, σ(cid:15) = 1 . × s − , which has a huge dimension so that the electricfield and the current can be treated as time-independent.Our second assumption is straight forward, and is what people usually take when calcu-lating dynamic electromotive force.[6]Regarding the third assumption, which is that the magnetic field gives a force to thecurrent, and this force acts on the gyroscope directly to make it slow down without affectcurrent distribution, is justified by the fact that under normal circumstance, where electronsmove at non-relativistic velocity electric force on a moving charge will be much more signifi-cant than magnetic force, hence the distribution of the current will be determined primarilyby the electric field. Being a higher order perturbation, magnetic force will drag the movingelectrons aside slightly, but such a transversal movement will be prevented by rapid collisionbetween electrons and lattice. As a result, the magnetic force is transfer to lattice of theconductor.Finally, we would like to discuss a possible source of the 8.61 % error in τ . Besides all17he approximations we made in process of calculation, the most significant origin of theerror is that the magnetic field is not actually vertical. In fact, there is a strong horizontalcomponent of the same dimension with the vertical one, which may also induce a verticalcomponent of current and then another torque. Due to the fact that the thickness of thegyroscope is small relative to its diameter, the energy loss is small compared to the maineffect. However, such a current is not easy to calculate, so only a limit of error is here givenby coarse estimation, which is about 10 % in torque and also in characteristic decaying time. VIII. CONCLUSION
The piece of work introduced here provides an analytical method to obtain the distribu-tion of the eddy current in a thin metal disk or ring and the damping time of the rotation,whose validity has been verified by experiments. Compared to numerical simulation, thismethod has the advantage of much easier access and shorter computation time. Neitherspecialized professional software package nor high performance computers are necessary.Despite the fact that the configuration of the magnetic field used in the paper is non-generic, the result presented in the paper can be readily applied to arbitrary distributionof magnetic field. One only has to decompose the magnetic field component perpendicularto the disk in its vicinity into symmetric and anti-symmetric parts and then follow steps inthe paper to obtain the damping time. In case that analytical expression of the magneticfield distribution is not available, measurement or numerical data can be used. After thedecomposition is done, a third-order polynomial fit can be done. Again, the damping timecan be calculated based on the method of this paper.We hope the work can assist students around the world to learn the eddy currents andgive them a deeper understanding of Maxwell equations and how to apply them in real cases.We intentionally omit many details of solving the Poisson’s equation since we do not wantthe main logic of physics to be obstructed by mathematical details.
ACKNOWLEDGMENTS
Our program is strongly supported by School of Physical Science and Technology atShanghaiTech University. The authors would like to thank Dr. Yintao You for giving a18easible plan to measure the angular velocity. Especially, the authors should appreciate Prof.Ramamurti Shankar from Yale University who helped to revise the whole paper carefully,as well as Ms. Bingrui Wu from College of Physics, Jilin Uiversity, who helped us a greatdeal in making use of computer programs to do calculation, designing the figures to betterillustrate the results, and editing this article with LATEX. In addition, we wish to thankDr. Zhongkai Liu, Dr. Jiamin Xue, Dr. Fuxiang Han, and Dr. Xiaoping Liu for manyuseful discussions and advice, and Mrs. Kangning Wang and Dean Xuguang Xu for helpingus contact with different departments in the university for support. [1] Problems for 32nd IYPT 2019: 12. Gyroscope Teslameter. https://iypt.org/index.php/problems/problems-for-the-32nd-iypt-2019/ [2] Youtube video: VIDEO 111 UNCOVERING SECRETS OF MAGNETISM. Magnet/ Gyroscope MYSTERY! Solve this unseen video. [3] Richard P. Feynman, Robert B. Leighton, and Matthew Sands,
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