HomePhysicsLearn how to Mannequin a Magnet Falling Via a Conducting Pipe

Learn how to Mannequin a Magnet Falling Via a Conducting Pipe

Introduction

In an earlier article, we examined a magnet falling by way of a solenoid. We argued that the purpose dipole mannequin can account for the fundamental options of the induced emf throughout the solenoid ends. Right here, we lengthen the mannequin to a magnet falling by way of a conducting pipe alongside its axis.

With the falling dipole second oriented alongside the vertical ##z##-axis, the electrical discipline ##E(rho,z)## is tangent to circles centered on the axis.  The induced emf round a closed loop of radius ##rho## is ##textual content{emf}(z)=2pi rho~E(rho,z)##. There’s a similarity and a distinction between a solenoid and a conducting pipe positioned in that house.  The similarity is of their modeling as a stack of conducting rings. The distinction is within the conceptual connection of the constituent rings.  Within the solenoid, the rings are in sequence and the general emf throughout them is of curiosity.  Within the pipe, the rings are type of in parallel and the whole present by way of them is of curiosity. “Kind of” signifies that the emf, being position-dependent, is just not the identical from one ring to the subsequent.

Nonetheless, Lenz’s regulation dictates that the induced currents will oppose the movement of the falling magnet.  Within the section beneath the dipole, they are going to be in repulsion whereas within the section above the dipole they are going to be in attraction with the magnet.  We are going to add these forces to seek out the online electromagnetic drive on the magnet. We are going to then clear up Newton’s second regulation equation to acquire an expression appropriate for elementary knowledge evaluation.

A single ring

We have now seen that the motional emf in a coaxial single ring of radius ##a## positioned at ##z’## is \$\$start{align}textual content{emf}_{textual content{ring}}= -frac{3mu_0ma^2}{2}  frac{ (z-z’)~v} {left[a^2+(z-z’)^2right]^{5/2}}.finish{align}\$\$Right here, the constructive ##z## axis is “down”.  The dipole with magnetic second ##mathbf{m}=m~hat{z}## is on the axis at ##z##.

Discovering the present and its discipline

The present within the ring is \$\$start{align}I=frac{textual content{emf}}{R}= -frac{3mu_0ma^2}{2R}  frac{ (z-z’)~v} {left[a^2+(z-z’)^2right]^{5/2}}finish{align}\$\$the place ##R## is the ring’s resistance. The on-axis magnetic discipline generated by this present on the location of the dipole is \$\$mathbf{B}_{textual content{ring}}=frac{mu_0 a^2 I~hat{z}}{left[(a^2+(z-z’)^2right]^{3/2}}.\$\$ We assume that the ring is beneath the dipole (##z-z'<0##) and that the dipole factors within the constructive ##z## route.  By Lenz’s regulation, the sphere and the dipole second have to be antiparallel.

Discovering the drive

We are going to receive the drive on the dipole utilizing power concerns. The potential power is \$\$U=-mathbf{m}cdotmathbf{B}_{textual content{ring}}=+m{B}_{textual content{ring}}=frac{mu_0 m a^2 I}{left[(a^2+(z-z’)^2right]^{3/2}}\$\$and the drive on the dipole is \$\$mathbf{F}=-mathbf{nabla}U=-mu_0 m a^2 Ifrac{partial}{partial z}left{frac{1}{left[(a^2+(z-z’)^2right]^{3/2}}proper}~hat z=frac{3 mu _0 a^2 I (z-z’)}{2 left[a^2+(z-z’)^2right]^{5/2}}~hat z.\$\$With the present from equation (2) the magnetic drive on the dipole is \$\$start{align}mathbf{F}=-frac{ 9 mu _0^2 m^2 a^4 (z-z’) ^2 ~v ~hat{z}}{4 R left[a^2+(z-z’) ^2right]^5}.finish{align}\$\$The drive exerted by the ring on the dipole is detrimental (up) for all values of the dipole-loop separation ##z-z’.## A separate calculation (not proven) of the integral ##oint I dmathbf{l} occasions mathbf{B}_{dip}## confirms that the drive exerted on the ring by the dipole is in step with Newton’s third regulation and equation (3).

Determine 1 reveals the dependence of the magnitude of the drive on the dipole as a operate of the decreased dipole-loop separation ##zeta=(z-z’)/a.## The drive is zero when the dipole passes by way of the middle of the ring.  That’s as a result of the present goes by way of zero earlier than it reverses route. Nevertheless, the drive additionally goes by way of zero on the heart, but it doesn’t reverse route and shows a attribute two hump profile.  Additionally, the drive is important solely inside a area of about ##pm## 2 ring radii from the middle.

Determine 1. The spatial dependence of the drive on the dipole exerted by a present loop as a operate of the decreased distance between the 2.

From ring to conducting pipe

We place the origin on the midpoint of a pipe of size ##L##. The web drive on the dipole is the vector sum of all ring contributions over the size of the pipe.  First, we convert equation (3) to a drive factor ##dmathbf{F}##.  We think about a hoop of radius ##rho’## and rectangular cross-sectional space ##dA’ =dz’occasions drho’##. The resistance of this ring is \$\$dR=frac{2pi rho’}{sigma~drho’~dz’}\$\$the place ##sigma## is the conductivity.  Then \$\$dmathbf{F}=-frac{ 9sigma mu _0^2 m^2 {rho’}^3 (z-z’) ^2 ~v }{8pi left[{rho’}^2+(z-z’) ^2right]^5}~drho’~ dz’~hat{z}\$\$

We outline a drive issue ##mathcal{F}(z)## because the double integral with out the constants, the rate and the detrimental signal:

\$\$start{align}mathcal{F}(z)equivint_{-L/2}^{L/2}(z-z’)^2~dz’int_{a}^{b}frac{{rho’}^3}{left[{rho’}^2+(z-z’) ^2right]^5}~drho’.finish{align}\$\$ Inbuilt it’s the geometric info of the pipe: size ##L##, interior diameter ##a## and outer diameter ##b##.

The integrals

Assisted by Mathematica, we discover the radial integral,

\$\$int_a^b frac{ {rho’}^3 ~drho’}{left[{rho’}^2+(z-z’) ^2right]^5}=frac{4 a^2+(z-z’) ^2}{24 left[a^2+(z-z’) ^2right]^4}-frac{4 b^2+(z-z’) ^2}{24 left[b^2+(z-z’) ^2right]^4}.\$\$

We are going to break up and compact the outcomes of the axial integration over ##z’## as a way to match them on the display.  We have now,

\$\$ start{align}& int_{-L/2}^{L/2}frac{left(z-z’)^2[4 a^2+(z-z’) ^2right]}{24 left[a^2+(z-z’) ^2right]^4}dz’=nonumber & left .frac{1}{384a^3}left{frac{a (z-z’) left[5 a^4-8 a^2 left(z-z’right)^2-5 (z-z’)^4right]}{left[a^2+left(z-z’right)^2right]^3}-5 tan ^{-1}left(frac{z-z’}{a}proper)proper}
proper |_{z’=-L/2}^{z’=L/2}nonumber finish{align}\$\$

and

\$\$ start{align}&- int_{-L/2}^{L/2}frac{left(z-z’)^2[4 b^2+(z-z’) ^2right]}{24 left[b^2+(z-z’) ^2right]^4}dz’=nonumber & left . -frac{1}{384b^3}left{frac{b (z-z’) left[5 b^4-8 b^2 left(z-z’right)^2-5 (z-z’)^4right]}{left[b^2+left(z-z’right)^2right]^3}-5 tan ^{-1}left(frac{z-z’}{b}proper)proper} proper |_{z’=-L/2}^{z’=L/2}nonumber finish{align}\$\$

The sum of the evaluated phrases above is the drive issue ##mathcal{F}(z).##

The drive issue

Determine 2 reveals two plots for the drive issue ##mathcal{F}(z)## for 2 outer radii, ##b=1.25a## (blue line) and ##b=2a## (orange line). The pipe size was set at ##L=100a##.  The plots point out that the drive issue rises quick, is flat, and rises increased for the thicker pipe.  We are going to take care of these options and what all of it means in FAQ format.

Determine 2. The drive issue as a operate of place for 2 values of the outer diameter, 1.25 a (blue) and a couple of.0 a (orange). The size of the pipe is 100 a.

How briskly does it rise?

For the reason that efficient vary of the dipole is 4 interior radii, we might anticipate ##mathcal{F}(z)## to alter appreciably inside that distance from the ends.  Determine 3 under reveals a element of Determine 2 and confirms this. The primary vertical line is on the entry level at ##z/a=-L/(2a)=-50.## The second vertical line is at at ##z/a=-L/(2a)+2=-48.##  It’s the place we think about the dipole to be “simply totally inside” the pipe.   Clearly, the drive issue is kind of flat there. That’s fairly quick.

Determine 3. Element of Determine 2 close to the purpose of entry at -50.

The plateau on the level of entry could be understood by referring to the double-humped drive proven in Determine 1.  Very close to the purpose of entry the drive because of the first infinitesimal ring and its close to neighbors is small and reducing, goes by way of zero, after which will increase.  The whole drive stays flat till the dipole is much sufficient inside for the second hump to kick in.  As soon as totally inside there’s a steady-state.  New rings forward of the dipole come into vary whereas previous ones behind the dipole fade out of vary.

Sure, however how flat is “flat”?

We take as a measure of flatness is the % ratio ##mathcal{F}(-48)/mathcal{F}(0)##.  It’s the ratio of the worth of the drive issue at 2 ring radii in to the worth within the center.  For the pipe with ##b=1.25a## (blue line),  the ratio is 99.3%; for the pipe with ##b=2~a## it’s 98.2%.  That’s fairly flat.

How does the wall thickness have an effect on it?

We calculated the worth on the midpoint ##mathcal{F}(0)## as a operate of the decreased outer radius ##b/a##. Determine 4 under reveals the ensuing plot. Evidently past ##bapprox 2a##, making the wall thicker produces diminishing returns. This is smart. On one hand, growing the wall thickness introduces extra cost carriers within the house. Then again, the magnetic discipline that induces these currents turns into weaker with distance at a quicker charge.

Determine 4. The drive issue at z=0 as a operate of the outer radius for ##lambda##=100. See equation (8) within the Appendix.

What’s the backside line?

To an excellent approximation, ##mathcal{F}(z) approx mathcal{F}(0)## for ##-L/2 leq z leq L/2##. This validates the assertion that the magnetic drive is proportional to the rate contained in the pipe.  Then we will write the magnitude of the drive on the dipole as \$\$start{align}F(z) =frac{ 9sigma mu _0^2 m^2 v }{8pi}mathcal{F}(0).finish{align}\$\$ The analytic expression for ##mathcal{F}(0)## seems within the Appendix as equation (8) and is plotted in Determine 4 for ##lambda=L/a=100.##

The kinematics

We write the magnetic drive on the dipole as ##F=-kappa~v## and begin with the differential equation \$\$Mfrac{dv}{dt}=-kappa~v+Mg~~~~left(kappa equiv frac{ 9sigma mu _0^2 m^2}{8pi}mathcal{F}(0)proper)\$\$ the place ##M## is the mass of the dipole.

We solid the well-known resolution by way of the terminal velocity, ##v_{textual content{ter}}=dfrac{Mg}{kappa}##: \$\$v=v_{textual content{ter}}+(v_0-v_{textual content{ter}})e^{-g~t/v_{textual content{ter}}}implies e^{-g~t/v_{textual content{ter}}}=frac{(v-v_{textual content{ter}})}{(v_0-v_{textual content{ter}})}.\$\$

Additionally, \$\$z=int_0^television~dt=v_{textual content{ter}}t+frac{v_{textual content{ter}}(v_0-v_{textual content{ter}})}{g}left[1-e^{-g~t/v_{text{ter}}}right].\$\$ By eliminating the exponential, the 2 equations could be mixed to yield \$\$start{align}z=v_{textual content{ter}}t-frac{v_{textual content{ter}}left(v-v_0right)}{g}.finish{align}\$\$

A prompt experiment

An apparent experiment to carry out and analyze can be to drop a magnet right into a pipe and use photogates and knowledge logging software program to measure the entry velocity ##v_0##, exit velocity ##v_{!f}##, and transit time ##T##. Equation (6) turns into \$\$start{align}L=v_{textual content{ter}}T-frac{v_{textual content{ter}}left(v_{!f}-v_0right)}{g}.finish{align}\$\$It may be solved to seek out the terminal velocity. Repeating for different decisions of entry velocity ought to confirm the fidelity of the terminal velocity. The worth of the terminal velocity would then decide the worth of ##kappa.##  Notice that this dedication relies on kinematics and is impartial of the magnet mannequin.

Furthermore, equation (5) reveals that  the drive issue (decided by the pipe geometry), and the conductivity (decided by the selection of pipe materials), could be mathematically separated out of ##kappa.##  One can check the validity of this separation by performing a sequence of experiments wherein the identical magnet is dropped by way of pipes of varied wall thicknesses, lengths, and conductivities. When ##mathcal{F}(0)## and the conductivity are factored out of ##kappa##, a continuing shared by all experimental runs ought to be left.  One can then clear up equation (5) to seek out an efficient magnetic second acceptable to a given magnet.

As an apart and since there isn’t a different appropriate place within the article, the creator can not resist mentioning the next commentary. With the usual definitions of common velocity ##bar v=L/T##, common acceleration ##bar a=(v_{!f}-v_0)/T## and exponential time fixed ##tau=v_{textual content{ter}}/g##, equation (7) turns into the curiously suggestive relation \$\$v_{textual content{ter}}=bar v+bar atau.\$\$It brings collectively two express constants, ##v_{textual content{ter}}## and ##kappa##, an implicit impartial variable, ##v_0## and two implicit dependent variables, ##v_{!f}## and ##T##. Thus, regardless of its SUVAToid look, it’s a constraint relation on the dependent variables.

Abstract and afterthoughts

The magnetic dipole mannequin of a magnet falling by way of a pipe is enough to point out mathematically that the magnetic drive opposing the movement of the falling dipole contained in the solenoid is fast-rising and could be written as a continuing ##kappa## multiplied by the instantaneous velocity.  The mannequin additionally affords the calculation of a drive issue that enables the separation of the pipe geometry and conductivity from the fixed ##kappa.## As soon as that separation is finished, one might extract a worth for an efficient dipole second that might function enter to a extra real looking mannequin than the purpose dipole.

Moreover, if a sensible mannequin is a objective, evidently place to focus one’s experimental efforts are the entry and exit factors of the magnet. As Determine 3 reveals, at these factors, the drive profile is most delicate to the form of the magnet and its efficient vary. No matter one’s objective, the purpose dipole mannequin has fairly a bit to say; its simplicity makes it place to start out the mathematical evaluation of a magnet falling by way of a conducting pipe.

Acknowledgment

The creator needs to thank person @Delta2 for useful feedback and remarks on an earlier model of this text.

Appendix

With decreased lengths ##lambda= L/a## and ##beta = b/a##, the worth of the drive issue is

\$\$start{align}mathcal{F}(0)= & frac{lambda}{96a^3}left[frac{80 beta ^4-32 beta ^2 lambda ^2-5 lambda ^4}{beta ^2 left(4 beta ^2+lambda ^2right)^3}-frac{80-32 lambda ^2-5 lambda ^4}{left(lambda ^2+4right)^3}right] nonumber & + frac{5}{192a^3}left[tan ^{-1}left(frac{lambda }{2}right)-frac{1}{beta ^3}tan ^{-1}left(frac{lambda }{2 beta }right)right]
.finish{align}\$\$

References

For readers who want to discover different approaches. The checklist is just not meant to be full.
C.S. MacLatchy, P. Backman and L. Bogan,  Am. J. Phys.  61, 1096 (1993); doi: 10.1119/1.17356
Ok.D. Hahn, E.M. Johnson, A. Brokken and Steven Baldwinc,  Am. J. Phys.  66,  1066 (1998); doi: 10.1119/1.19060
Y. Levin, F.L. da Silveira and F.B. Rizzato,  Am. J. Phys. 74, 815 (2006); doi: 10.1119/1.2203645
G. Donoso, C.L. Ladera and P. Martin,  Eur. J. Phys.  30 (2009),  855–869 doi:10.1088/0143-0807/30/4/018

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