Magnetic Repulsion - Untrained Idiot Question

[Guest nigel]

Hello

 

I've no grounding whatsoever in physics but am trying to workout a magnetic strengh problem for my own amusement. I've spent several hours reviewing physics web sites but am totally confused. How do I go about modeling the following situation?

 

I have to permanent bar magnets on end. Both are held vertically within a tighly fitting tube so that they cannot move sideways. The tube is sealed at one end and open at the other. The magnets have their like poles pointed at one another so that they repel each other and (assuming their strong enough) one floats above the other. If I begin adding weight (something non-ferrometalic) to the floating magnet the air gap will shrink until eventually the weight is to much and the two magnets touch one another.

 

How do I calculate the size of the gap for magnets of a given strength (Gauss?) and weight/shape? How do I calculate the change in gap as more weight is added to the floating magnet? If I've missed anything important from my description gravity, pressure, temprature, etc please assume normal conditions on earth at sea level.

 

I'm not really looking for an exact answer, more a list of the things I need to know and the formula invloved so I can go away have a go at calculating it myself. I'm pretty sure that the physics site I found http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html has all the information I need but I don't understand enough to apply it.

 

Thank you, for you guidance

 

Nigel

[Meir Achuz]

The force needed to push the two magnets together is easy.

F=2 pi M^2 A, where M is the strength of each magnet in gauss,

and A is the cross-sectional area of the tube.

Calculating how the force decreases as the magnets move apart is a bit messy,

until they get so far apart that you can make some approximations.

To calculate the force when they are still close together requires using a Legendre polynomial expansion of the field from the end of one magnet.

[Meir Achuz]

The force will be in dynes.

[labview1958]

Does the repulsive follows the inverse square law? Or does it follow the inverse cube law? or does it the inverse cube law?

 

If z is the separation distance & x is a constant, does it follow F=AEXP(-zx) or does it follow F=AZ^x?

[Meir Achuz]

The formula I gave is for two touching magnets.

It will be approximately true as long as the distance z between the magnet faces

satifies z<< r, where r is a radius of the face.

As the magents get further apart, the force starts to fall off like 1/z, then 1/z^2, and eventually (when z>>r and z>>L, the length of the magnet) like 1/z^3.

[Meir Achuz]

At large z, the forced falls off like 1/r^4 (not r^3).

It is the force between two magnetic dipoles

[labview1958]

"To calculate the force when they are still close together requires using a Legendre polynomial expansion of the field from the end of one magnet."

 

What about my formula. Is it reasonable?

"If z is the separation distance & x is a constant, does it follow F=AEXP(-zx)"

 

With your reasoning the force Vs distance is NOT a power series, then mine is more logical?

[Meir Achuz]

This force is never exponential in z.

It can be calculated explicitly for any distance.

It is always expandable in a power series in z. I just gave the leading terms.

It is not a question of logic, but of explicit calculaton.

[labview1958]

What I undestand is that if r=radius of the pole and z is the separation distance then:

 

If z << r, then F = ..1/z^1

If z = r , then F =..1/z^2

If z > r, then F = ....1/z^3

If z>> r, then F = ..1/z^4

 

Definetly, F = A EXP(-zx)

[Meir Achuz]

What I understand is that if r=radius of the pole and z is the separation distance then:

 

If z << r' date=' then F = ..1/z^1

If z = r , then F =..1/z^2

If z > r, then F = ....1/z^3

If z>> r, then F = ..1/z^4

 

Definitely, F = A EXP(-zx)[/quote']

 

The force never varies like exp(-zx). Also, not all of the formulae you give are correct. I will give the approximate force (in dynes) for several regions for two identical bar magnets of length L, with circular cross-sections of radius R (R<<L), each having a strength of M gauss:

 

1) z<<R: F=2pi M^2 A (A=pi R^2). This force does not vary appreciably with distance, as long as z<<R. It is just like the force between two uniformly charged disks close together.

 

2) R<<z<<L: F=(MA/z)^2, just like the Coulomb force between two charges.

 

3) z>>L: F=6(MAL)^2/z^4, like the force between two dipoles.

 

In between these regions, the force is more complicated, but can be treated using a Legendre polyomial expansion.

[labview1958]

Here is a site that is interesting!

 

http://www.magnetsales.com/Design/Calc_filles/PullAndPushBetween2RectMagnets.asp

There is some empirical data here, however I find that the FORCE rapidly becomes zero.

[Meir Achuz]

That is an interesting website. I have several comments to make:

 

* They consider magnets with L~R, while I had L>>R. This means that only my case (3), with F~1/x^4 for large x, is appropriate for their magnets. For one case, they get a pull of 7.2# at 10" and 0.6# at 20". This is approaching a 1/x^4 fall off.

 

* Their attraction does not equal their repulsion. This is because there is a demagnetizing effect with oppposing magnets that weakens the effective magnetization. I considered an ideal magnet for which the magnetization never changes. This gives equal push and pull. I don't understand why they still have pull>push at large distances where the demagnetization should not be effective.

It may be that they used the same demagnetizing factor at large distances as at small distances. (Their ratio is a constant 5/3.)

 

* At one point, they mention "The strength of a magnetic field drops off roughly exponentially over distance." But the standard formula they give is the same one I used. It gives a 1/x^3 dependence to the field at large x, leading to 1/x^4 for the force. I call 1/x^4 a "power law" dependence.

[labview1958]

Is there a limit of F = ..1/x^4 for larger distance? Should an even larger distance give F = ..1/x^5 where x is the separation. At what point does F = 0? Is it possible to calculate that?

[DQW]

Is there a limit of F = ..1/x^4 for larger distance? Should an even larger distance give F = ..1/x^5 where x is the separation.

No, the farther apart you go, the closer you get to an ideal dipole-dipole interaction (and demagnetizing fields vanish too). The limiting behavior will not be x^-5

 

At what point does F = 0? Is it possible to calculate that?

Sure, it happens where 1/x = 0

[labview1958]

If the limiting factor is not 1/x^5 then what it is?

[Meir Achuz]

1/x^4

[labview1958]

How do you know that is 1/x^4? Any proof?

[Meir Achuz]

"3) z>>L: F=6(MAL)^2/z^4, like the force between two dipoles.

In between these regions, the force is more complicated, but can be treated using a Legendre polyomial expansion."

 

A power series on the symmetry axis shows that the leadilng term at large x is 1/x^4.

The Legendre polynomial expansion can extend this to all angles.

This is done in many EM texts.

[labview1958]

Here is an interesting site.

 

http://www.exo.net/~pauld/activities/magnetism/forcebetweenmagnets.html

 

According to this site the forces between two magnets can be F = ...1/x^7

[Meir Achuz]

"The force between two pairs of poles falls off proportional to the inverse fourth power of the distance between them. (The like poles repel and the unlike poles attract. The nearer poles exert stronger forces on each other than the more distant ones.) So, at large distances, the force between these two magnets should fall off proportional to the inverse fourth power of distance."

If a web page is "proof", that is your proof.

 

THEN:

 

"Measure the force between a magnet and a steel washer on the balance.

Notice that the dependence of force on distance is different. In particular the force drops off more quickly for the steel washer than it does for the magnet. The washer becomes magnetized when it is near a magnet. The magnetization of the washer falls off in proportion to the strength of the magnetic field from magnet. The magnetic field of the first magnet falls off as the inverse third power of distance, combined with the inverse fourth power fall off of the force between two magnets the resulting force falls off as the inverse seventh power. (However if the steel washer is not a perfect soft magnet, that is if it retains magnetization when removed from the presence of a magnet then the force will fall off somewhat more slowly than the inverse seventh power.)"

 

That is NOT the force betrween two magnets, but between a magnet and a piece of soft iron. The point implied buy the site is the the induced magnetization of the iron washer depneds on the field of the magnet which falls off like 1/x^3, which is where the other power comes in.

[labview1958]

"The magnetic field of the first magnet falls off as the inverse third power of distance, "

 

If we have two similiar magnets with the magnetic field falling off as the inverse third power then the force beteen magnets would be ....1/x^3 * 1/x^3 = ....1/x^6 !

[labview1958]

"* They consider magnets with L~R, while I had L>>R. This means that only my case (3), with F~1/x^4 for large x, is appropriate for their magnets. For one case, they get a pull of 7.2# at 10" and 0.6# at 20". This is approaching a 1/x^4 fall off. "

 

Actuall y I was thinking of the cases where L<<<R and L<z<R and also the cases where L<<<R and z>>R>>L. Again z = separation distance between magnets.

[Meir Achuz]

For z>>R>>L, the force ~1/z^4. Anytime z>> than anything else, the force is

dipole-dipole and ~1/z^4.

For L<<<R, and z<R, the force is the same as the electric force between two uniformly charged discs. F will be almost constant when z<<R, and then fall off slowly with z

as it becomes comparable with R. The force can be calculated for z~R using a Legendre expansion.

[Meir Achuz]

"The magnetic field of the first magnet falls off as the inverse third power of distance' date=' "

If we have two similiar magnets with the magnetic field falling off as the inverse third power then the force between magnets would be ....1/x^3 * 1/x^3 = ....1/x^6 ![/quote']

 

The force does NOT depend on the product of the two fields.

The force at large distance is given by mu*dB/dz, where mu is the magnetic moment of one magnet, and B the field due to the other.

[labview1958]

"The force needed to push the two magnets together is easy.

F=2 pi M^2 A, where M is the strength of each magnet in gauss,

and A is the cross-sectional area of the tube."

 

The force DOES depend on the product of the two fields!