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# How one can Mannequin a Magnet Falling By way of a Conducting Pipe

## Introduction

In an earlier article, we examined a magnet falling by 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 a conducting pipe alongside its axis.

With the falling dipole second oriented alongside the vertical ##z##-axis, the electrical area ##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 area.  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 form of in parallel and the whole present by them is of curiosity. “Form of” signifies that the emf, being position-dependent, isn’t 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 phase beneath the dipole, they are going to be in repulsion whereas within the phase 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 pressure 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 now have 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 optimistic ##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 area

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 area 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 optimistic ##z## course.  By Lenz’s regulation, the sphere and the dipole second should be antiparallel.

### Discovering the pressure

We are going to receive the pressure 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 pressure 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 pressure 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 pressure exerted by the ring on the dipole is adverse (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 pressure exerted on the ring by the dipole is per Newton’s third regulation and equation (3).

Determine 1 reveals the dependence of the magnitude of the pressure on the dipole as a perform of the decreased dipole-loop separation ##zeta=(z-z’)/a.## The pressure is zero when the dipole passes by the middle of the ring.  That’s as a result of the present goes by zero earlier than it reverses course. Nonetheless, the pressure additionally goes by zero on the middle, but it doesn’t reverse course and shows a attribute two hump profile.  Additionally, the pressure is critical solely inside a area of about ##pm## 2 ring radii from the middle. Determine 1. The spatial dependence of the pressure on the dipole exerted by a present loop as a perform 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 online pressure on the dipole is the vector sum of all ring contributions over the size of the pipe.  First, we convert equation (3) to a pressure 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 pressure issue ##mathcal{F}(z)## because the double integral with out the constants, the speed and the adverse 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}\$\$ In-built it’s the geometric info of the pipe: size ##L##, internal 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’## so as to match them on the display screen.  We now have,

\$\$ 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 pressure issue ##mathcal{F}(z).##

## The pressure issue

Determine 2 reveals two plots for the pressure 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 pressure issue rises quick, is flat, and rises larger for the thicker pipe.  We are going to cope with these options and what all of it means in FAQ format. Determine 2. The pressure issue as a perform of place for 2 values of the outer diameter, 1.25 a (blue) and a pair 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 internal radii, we’d count on ##mathcal{F}(z)## to alter appreciably inside that distance from the ends.  Determine 3 beneath 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 pressure issue is sort 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 pressure proven in Determine 1.  Very close to the purpose of entry the pressure because of the first infinitesimal ring and its close to neighbors is small and lowering, goes by zero, after which will increase.  The overall pressure 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 pressure 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 perform of the decreased outer radius ##b/a##. Determine 4 beneath reveals the ensuing plot. Evidently past ##bapprox 2a##, making the wall thicker produces diminishing returns. This is smart. On one hand, rising the wall thickness introduces extra cost carriers within the area. However, the magnetic area that induces these currents turns into weaker with distance at a quicker charge. Determine 4. The pressure issue at z=0 as a perform 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 pressure is proportional to the speed contained in the pipe.  Then we are able to write the magnitude of the pressure 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 pressure 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 when it comes to 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}\$\$

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 selections 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 is predicated on kinematics and is impartial of the magnet mannequin.

Furthermore, equation (5) reveals that  the pressure 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 by which the identical magnet is dropped by pipes of assorted wall thicknesses, lengths, and conductivities. When ##mathcal{F}(0)## and the conductivity are factored out of ##kappa##, a relentless shared by all experimental runs must 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 any different appropriate place within the article, the writer can’t resist mentioning the next statement. 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 specific 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 a pipe is ample to indicate mathematically that the magnetic pressure opposing the movement of the falling dipole contained in the solenoid is fast-rising and could be written as a relentless ##kappa## multiplied by the instantaneous velocity.  The mannequin additionally affords the calculation of a pressure issue that enables the separation of the pipe geometry and conductivity from the fixed ##kappa.## As soon as that separation is finished, one could extract a worth for an efficient dipole second that would function enter to a extra life like mannequin than the purpose dipole.

Moreover, if a practical mannequin is a objective, it appears that evidently a superb place to focus one’s experimental efforts are the entry and exit factors of the magnet. As Determine 3 reveals, at these factors, the pressure 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 a superb place to start out the mathematical evaluation of a magnet falling by a conducting pipe.

## Acknowledgment

The writer needs to thank consumer @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 pressure 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 record isn’t meant to be full.
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