Monday, October 3, 2022
HomePhysicsLearn how to Mannequin a Magnet Falling By means of a Conducting...

# Learn how to Mannequin a Magnet Falling By means of 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 essential options of the induced emf throughout the solenoid ends. Right here, we prolong 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 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 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 full present by way of them is of curiosity. “Form of” signifies that the emf, being position-dependent, isn’t the identical from one ring to the following.

Nonetheless, Lenz’s legislation 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 search out the web electromagnetic power on the magnet. We are going to then resolve Newton’s second legislation equation to acquire an expression appropriate for elementary information evaluation.

## A single ring

We now have seen that the motional emf in a coaxial single ring of radius ##a## situated 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## path.  By Lenz’s legislation, the sector and the dipole second have to be antiparallel.

### Discovering the power

We are going to acquire the power on the dipole utilizing power issues. 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 power 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 power 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 power 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} instances mathbf{B}_{dip}## confirms that the power exerted on the ring by the dipole is in line with Newton’s third legislation and equation (3).

Determine 1 exhibits the dependence of the magnitude of the power on the dipole as a operate of the lowered dipole-loop separation ##zeta=(z-z’)/a.## The power 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 path. Nonetheless, the power additionally goes by way of zero on the heart, but it doesn’t reverse path and shows a attribute two hump profile.  Additionally, the power is critical solely inside a area of about ##pm## 2 ring radii from the middle. Determine 1. The spatial dependence of the power on the dipole exerted by a present loop as a operate of the lowered distance between the 2.

## From ring to conducting pipe

We place the origin on the midpoint of a pipe of size ##L##. The online power on the dipole is the vector sum of all ring contributions over the size of the pipe.  First, we convert equation (3) to a power aspect ##dmathbf{F}##.  We contemplate a hoop of radius ##rho’## and rectangular cross-sectional space ##dA’ =dz’instances 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 power 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##, inside 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’## with the intention 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 power issue ##mathcal{F}(z).##

## The power issue

Determine 2 exhibits two plots for the power 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 power issue rises quick, is flat, and rises greater for the thicker pipe.  We are going to cope with these options and what all of it means in FAQ format. Determine 2. The power issue as a operate 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?

Because the efficient vary of the dipole is 4 inside radii, we’d anticipate ##mathcal{F}(z)## to vary appreciably inside that distance from the ends.  Determine 3 under exhibits 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 contemplate the dipole to be “simply absolutely inside” the pipe.   Clearly, the power 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 might be understood by referring to the double-humped power proven in Determine 1.  Very close to the purpose of entry the power as a result of first infinitesimal ring and its close to neighbors is small and reducing, goes by way of zero, after which will increase.  The entire power stays flat till the dipole is way sufficient inside for the second hump to kick in.  As soon as absolutely 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 p.c ratio ##mathcal{F}(-48)/mathcal{F}(0)##.  It’s the ratio of the worth of the power 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 lowered outer radius ##b/a##. Determine 4 under exhibits the ensuing plot. It appears that evidently past ##bapprox 2a##, making the wall thicker produces diminishing returns. This is sensible. On one hand, growing the wall thickness introduces extra cost carriers within the house. Alternatively, the magnetic area that induces these currents turns into weaker with distance at a sooner charge. Determine 4. The power 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 power is proportional to the speed contained in the pipe.  Then we will write the magnitude of the power 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 power 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 forged 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 might 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 urged experiment

An apparent experiment to carry out and analyze could be to drop a magnet right into a pipe and use photogates and information 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 search 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.##  Observe that this dedication relies on kinematics and is impartial of the magnet mannequin.

Furthermore, equation (5) exhibits that  the power issue (decided by the pipe geometry), and the conductivity (decided by the selection of pipe materials), might be mathematically separated out of ##kappa.##  One can check the validity of this separation by performing a sequence of experiments during which the identical magnet is dropped by way of pipes of assorted wall thicknesses, lengths, and conductivities. When ##mathcal{F}(0)## and the conductivity are factored out of ##kappa##, a continuing shared by all experimental runs must be left.  One can then resolve equation (5) to search out an efficient magnetic second acceptable to a given magnet.

As an apart and since there is no such thing as a different appropriate place within the article, the creator can’t resist mentioning the next remark. 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 way of a pipe is ample to point out mathematically that the magnetic power opposing the movement of the falling dipole contained in the solenoid is fast-rising and might be written as a continuing ##kappa## multiplied by the instantaneous velocity.  The mannequin additionally affords the calculation of a power issue that enables the separation of the pipe geometry and conductivity from the fixed ##kappa.## As soon as that separation is completed, one could extract a price for an efficient dipole second that might function enter to a extra real looking mannequin than the purpose dipole.

Moreover, if a practical mannequin is a purpose, plainly an excellent place to focus one’s experimental efforts are the entry and exit factors of the magnet. As Determine 3 exhibits, at these factors, the power profile is most delicate to the form of the magnet and its efficient vary. No matter one’s purpose, the purpose dipole mannequin has fairly a bit to say; its simplicity makes it an excellent 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 lowered lengths ##lambda= L/a## and ##beta = b/a##, the worth of the power 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 listing isn’t 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

RELATED ARTICLES