HomePhysicsThe right way to Mannequin a Magnet Falling By a Solenoid

# The right way to Mannequin a Magnet Falling By a Solenoid

## Introduction

Modeling a magnet realistically is a job finest performed numerically.  Even the simplified mannequin of two separated disks with uniform floor magnetization ##pm~sigma_M## entails elliptic integrals simplifying assumptions. As a mannequin, the purpose dipole could also be unrealistic to some however the math is tractable and accessible.  The usefulness of the purpose dipole mannequin in electrodynamics is analogous to that of the purpose mass in kinematics and dynamics.  It brings forth the salient options and serves as a place to begin for extra concerned therapies.  Certainly, analyses of the falling magnet discovered within the literature depend on modifications of the purpose dipole equations with a watch towards bringing their predictions into higher settlement with experiments.

The aim of this text is to assemble in a single place what could be mentioned a few magnet dipole falling by a solenoid.  A second article, now in preparation, has the identical aim however for a degree magnetic dipole falling by a strong conducting pipe.

### The motional emf

The inverse Lorentz transformation equations for the parts of the electrical and magnetic fields are \$\$start{align}& mathbf{E}_{parallel}=mathbf{E’}_{parallel}~;~~mathbf{B}_{parallel}=mathbf{B’}_{parallel}nonumber & mathbf{E}_{perp} =gammaleft(mathbf{E’}_{perp} -mathbf{v}timesmathbf{B’}proper)~;~~mathbf{B}_{perp}=gammaleft(mathbf{B’}_{perp}-frac{1}{c^2}mathbf{v}timesmathbf{E’}proper).nonumber finish{align}\$\$

Take into account a magnetic level dipole ##mathbf{m}=m~hat z## falling with velocity ##mathbf{v}=v~hat z##.  Taking “down” as constructive  and for bizarre magnet velocities (##gamma approx 1##), we write in cylindrical coordinates,\$\$mathbf{E}_{perp}=-mathbf{v}occasions mathbf{B’}=-mathbf{v}occasions mathbf{B}=-v~hat ztimesleft(B_{rho}~hat {rho}+B_z~hat {z}proper)=-vB_{rho}~hat {theta}.\$\$The road integral ##int_a^b mathbf{E}_{perp}cdot dmathbf{l}## is the motional emf between factors ##a## and ##b##.

The relativistic therapy may be a bit an excessive amount of, however it helps get the indicators proper.  Anyway, from right here on we are going to discontinue using primes to indicate coordinates within the shifting body. As an alternative, we are going to use primes for supply coordinates and no primes for coordinates of factors of curiosity. Integrals might be taken over primed coordinates.

### A single ring

At focal point ##mathbf{r}(rho ,z)## when the dipole is at ##mathbf{r’}(rho’,z’)## the dipolar area is \$\$mathbf{B}=frac{mu_0}{4pi} left[ frac{ left(3 mathbf{m}cdot (mathbf{r}-mathbf{r}’))(mathbf{r}-mathbf{r}’)-|mathbf{r}-mathbf{r}’|^2 mathbf{m}right)}{|mathbf{r}-mathbf{r}’|^5}  -frac{mathbf{m}}{ |mathbf{r}-mathbf{r}’|^3}  right].\$\$We place the dipole on the ##z##-axis and orient it in order that ##mathbf{m}=m~hat z##. At  ##rho=a##, the radial element of the magnetic area is \$\$B_{rho}=frac{3mu_0ma}{4pi}  frac{ z-z’} {left[a^2+(z-z’)^2right]^{5/2}}.\$\$

The motional emf in a coaxial single ring of radius ##a## at ##z’## is \$\$start{align}textual content{emf}_{textual content{ring}}=int mathbf{E}_{perp}cdot dmathbf{l}=-vB_{rho}(2pi a) = -frac{3mu_0ma^2}{2}  frac{ (z-z’)~v} {left[a^2+(z-z’)^2right]^{5/2}}.finish{align}\$\$Observe that the ring might be an intangible loop, not essentially conducting, and the emf round it could nonetheless exist.  If, nevertheless, the ring is a conducting loop, there might be an induced present in it.  A plot of the spatial dependence of the emf is proven in Determine 1 under. We observe that when the dipole is farther than about 4 radii from the ring’s heart, the emf is under 1% of its peak worth. Determine 1. Single ring emf as a operate of dipole place.

Because the dipole falls in free area,  a wave of motional emf with the waveform proven in Determine 1 is touring with velocity ##v(t)## down an intangible tube consisting of  loops of radius ##a.##  Within the subsequent part, we are going to discover what to anticipate when a conducting finite solenoid of radius ##a## occupies a part of that area.

### From ring to solenoid

We think about the solenoid as a stack of  ##N## turns. The start of the ##okay##th flip is the top of the ##(k-1)##th and the top of the ##okay##th flip is the start of the ##(okay+1)##th.  Thus, the turns are a group of rings related in sequence.  We’re looking for an expression for the emf throughout the free ends of a solenoid of ##N## turns, radius ##a##, and size ##L##. It will likely be an integral of  ##textual content{emf}(z)## over the size of the solenoid.

The variety of rings in a stack of top ##dz’## is ##dn=dfrac{N}{L}dz’##. Integrating over the size, \$\$start{align}textual content{emf}(z)=frac{N}{L}int_{frac{-L}{2}}^{frac{L}{2}}textual content{emf}_{textual content{ring}}dz’=frac{mu _0 N m a^2}{2 L} left{ frac{v}{left[a^2+left(z-frac{L}{2}right)^2right]^{3/2}}-frac{v}{left[a^2+left(z+frac{L}{2}right)^2right]^{3/2}}proper}.finish{align}\$\$

We assume that the dipole has mass and is in free fall.  If the solenoid ends are open, there might be an emf throughout them however no present.  Even when a voltmeter is related to measure the emf, the present by it’s anticipated to be negligible . In both case, eddy present braking is negligible and we are going to ignore it along with air resistance.  Underneath these circumstances, the usual kinematic equations beneath fixed acceleration apply.  Taking the origin on the midpoint of the solenoid, we write an expression for the place of the dipole. We assume that it begins from relaxation at top ##h## above the highest of the solenoid (##z_0=-h-frac{L}{2}##.) Substituting \$\$z=(-h-frac{L}{2}+frac{1}{2}g~t^2)~textual content{ and }~v=g~t\$\$ in equation (2), we receive the emf throughout the solenoid as a operate of time. \$\$start{align}textual content{emf}(t)=frac{mu _0 N m a^2}{2 L} left{ frac{g~t}{left[a^2+left(-h-L+frac{1}{2}g~t^2right)^2right]^{3/2}}-frac{g~t}{left[a^2+left(-h+frac{1}{2}g~t^2right)^2right]^{3/2}}proper}finish{align}\$\$

### The emf profile

Determine 2 under is a plot of emf(t) vs ##t## with out the general multiplicative fixed.  For the plot, we used  ##g=980## for the acceleration, ##a=1## for the radius, ##h=6a## for the discharge top above the highest and ##L=10 a## for the size. Thus,  if the items for the radius ##a## are centimeters, the numbers on the abscissa might be seconds.  The strong vertical grid traces mark the occasions when the dipole is on the prime, midpoint, and backside of the solenoid. The pairs of pink dashed traces mark the occasions when the dipole is inside one solenoid diameter (##pm~2a##) from the highest (left) and the underside (proper). Determine 2. Solenoid emf as a operate of time.

Alternatively, we are able to receive an expression for the emf as a operate of place ##z## by substituting ##v=sqrt{2g[z-(-h-frac{L}{2})]}## in equation (2) to get \$\$start{align}textual content{emf}(z)=frac{mu _0 N m a^2}{2 L} left{frac{sqrt{2 g left(z+h+frac{L}{2}proper)}}{left[a^2+left(z-frac{L}{2}right)^2right]^{3/2}}-frac{sqrt{2 g left(z+h+frac{L}{2}proper)}}{left[a^2+left(z+frac{L}{2}right)^2right]^{3/2}}proper}.finish{align}\$\$A plot of the emf(##z##) vs. ##z## is just not proven as a result of it’s similar to the one in determine 2 aside from the labeling of the abscissa. Each plots characteristic extrema, one unfavourable and one constructive, with a area between when/the place the emf is close to zero. That is qualitatively in keeping with Faraday’s legislation. The magnetic flux modifications most quickly when the dipole is close to the ends and hardly in any respect when it’s close to the center of the solenoid.

### Time period corrections

Examination of determine 2 reveals that the extrema happen when the dipole is close to the ends of the solenoid. At ##z=+L/2## the primary time period in equation (4) has a most whereas the second time period is small; the reverse is true at ##z=-L/2##. An actual willpower of the extrema would contain fixing ##frac{d}{dz}textual content{emf}(z)=0##, a frightening job. Contemplating, nevertheless, that the true extrema are very near ##z=pm frac{L}{2}##, it’s expedient to do sequence expansions of equation (4) about ##z=pm frac{L}{2}## to second-order and discover corrections to the zeroth-order phrases.  Right here, we summarize the outcomes and relegate a few of the particulars to the Appendix.

The fractional corrections to the place and worth of the minimal are \$\$frac{delta z}{(-L/2)}=-5.5times 10^{-3}~;~~frac{delta(textual content{emf})}{textual content{emf}(-L/2)}=1.1times 10^{-3}.\$\$The corresponding numbers for the utmost are \$\$frac{delta z}{(+L/2)}=2.1times 10^{-3}~;~~frac{delta( textual content{emf})}{textual content{emf}(L/2)}=1.6times 10^{-4}.\$\$The numbers present that the corrections have a small impact on the zeroth order phrases. Thus, we are going to ignore them and make the algebra simpler.

### A relation between amplitudes

We outline as amplitudes the magnitudes of the extrema of ##textual content{emf}(t)## (Determine 2). Within the approximation that the extrema happen at ##z=pm ~frac{1}{2}##,  equation (4) offers, the amplitudes \$\$start{align} & A_1=left|textual content{emf}(-L/2)proper|=sqrt{2 g h}left[frac{1}{a^3}-frac{1}{left(a^2+L^2right)^{3/2}}right]nonumber &  A_2=textual content{emf}(+L/2)|=sqrt{2 g (h+L)}left[frac{1}{a^3}-frac{1}{left(a^2+L^2right)^{3/2}}right].nonumber finish{align}\$\$

We write the distinction between the 2 equations as \$\$start{align} &  A_2-A_1=textual content{const.} left(sqrt{frac{2 (h+L)}{g}}-sqrt{frac{2 h}{g}}proper) nonumber & textual content{the place  }textual content{const.}=frac{mu _0 N m a^2}{2 g L}left[frac{1}{a^3}-frac{1}{left(a^2+L^2right)^{3/2}}right].nonumber finish{align}\$\$We acknowledge the expression in parentheses multiplying the fixed because the transit time ##T## of the dipole from one finish of the solenoid to the opposite. This simplifies the equation much more to \$\$start{align}A_2-A_1=(textual content{const.})occasions T.finish{align}\$\$

It’s simple to think about an experimental protocol to check equation (5) with tools that samples and information the emf throughout the solenoid as a operate of time. Nonetheless, one ought to make sure that beginning heights ##h## are larger than the efficient dipole vary of 4 solenoid radii. This may decrease experimental uncertainties that develop into important at small values of ##h##. A PF thread posted by @billy_t right here describes such an experiment. Nonetheless, the evaluation targeted on the amplitudes themselves and never on their relation to the time of transit.

## Abstract

Modeling a magnet as a degree dipole falling by a solenoid is adequate to supply a smart description of the form and normal options of the emf(##t##) curve that’s in settlement with what one would count on qualitatively.  The flux varies most quickly when the dipole is coming into and exiting the solenoid and least quickly when it’s close to the solenoid’s center. The amplitude of the fitting peak is larger than the left. This occurs as a result of its width is narrower because the dipole is shifting sooner, but the realm beneath the curve is similar. It’s unlikely {that a} extra practical mannequin of the magnet will lead to a distinct interpretation of those options. A classy mannequin wants subtle information evaluation to match it, maybe one involving a point-by-point simulation of the complete curve.

## Acknowledgments

I thank @billyT_ for posting the thread quoted above; it seeded this perception. Additionally, the participation of @haruspex and @Charles Hyperlink honed my considering and is tremendously appreciated.

## Appendix

The work proven right here was Mathematica-assisted.  In equation (4) we first drop ##g## and substitute diminished lengths ##lambda=L/a,~eta=h/a##. Then broaden the equation for small values of ##epsilon = zpm lambda/2## to second order.  Every enlargement yielded a quadratic expression ##Aepsilon^2+Bepsilon+C.##  Setting the spinoff equal to zero and fixing for ##epsilon## offers ##epsilon=frac{-B}{2A}##.  Omitting the gory particulars, the corrections for the minimal on the left and the utmost on the fitting are:

\$\$start{align} & epsilon_{textual content{min}}  =frac{2 eta left(lambda ^2+1right) left[left(lambda ^2+1right)^{5/2}-left(6 eta lambda +lambda ^2+1right)right]}{left(12 eta ^2+1right) left(lambda ^2+1right)^{7/2}+12 eta ^2 left(4 lambda ^2-1right)+12 eta lambda left(lambda ^2+1right)-left(lambda ^2+1right)^2}nonumber & epsilon_{textual content{max}}  =frac{ 2 left(lambda ^2+1right) (eta +lambda ) left[6 eta lambda +left(lambda ^2+1right)^{5/2}+5 lambda ^2-1right]}    {12 eta ^2 left(4 lambda ^2-1right)+12 eta lambdaleft(7 lambda ^2-3right)  +left(lambda ^2+1right) left[12 left(lambda ^2+1right)^{5/2} (eta +lambda )^2+1right]+left(35 lambda ^4-26 lambda ^2-1right)}nonumber finish{align}\$\$Determine 3 reveals a element of determine 2 within the neighborhood of the utmost.  The vertical line is on the corrected worth of ##z/a##.  The zeroth order time period of the enlargement is to its left at ##z/a=5.00##. Determine 3. Element of the correction to the zeroth order time period.

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