Electron magnetic replusion compared to coulomb attraction

[Drakkith]

According to? Did anyone ever said that beside you just now? -- This is about free particles, not atoms and electron orbitals. Classical equations apply just fine. Google it.

Sure buddy, whatever you want to believe.

 

[zincshow]

Thanks for the formula, it is some years (more then I wish to say) since my university years, so please do not place a lot of confidence in my first attempt, I am open to corrections.

Assuming 2 magnets are pointed against each other we can take m1 = 1 and m2 = -1 wrt the direction of the vectors, we see the big term in brackets becomes -3r+5r=2r (3r-5r=-2r if they are attracting one another) so we get the force on one magnet:

F(r) = (3*µ0)/(2*ᴨ*r^4) times the strength (magnetic moment) of the 2 magnets

So indeed, we are getting the variance relates to r^4 where r is the distance between the magnets.

with the magnetic constant = µ0 = 1.3*10^−6 N/A^2 (Newtons over Amps squared)

First let's check units, magnetic moments are given in Joules/Tesla = Amps*meter^2 so
(N/A^2)*(A*m^2)*(A*m^2)/(m^4) = N so we indeed are in Newtons.

Magnetic moment of electron = 9.3*10^-24 J/T

So let's go: at 1.0 Angstoms = 1.0*10^-10 meters, the force in Newtons =
= 3*(1.3*10^-6)*(9.3*10^-24)*(9.3*10^-24) / (2*3.14*(1*10^-10)^4)
= (3.9)*(86.5)*(10^-56) / (6.28)*(10^-40)
= 53.7*10^-16 Newtons

Force at 2 angstroms = (53.7)*(10^-16)/(2^4) = 3.36*10^-16 Newtons
Force at 3 angstroms = (53.7)*(10^-16)/(3^4) = 0.66*10^-16 Newtons

I will think about this and attempt the coulomb force tomorrow.

PS: Attractive force if you flipped one electron would be the same with opposite direction.

 

[Vanadium 50]

Tris_d, as said before, context. The Lorentz force is the equation for the force of a charged particle in an external E and B field. You want the force on a magnetic dipole from another magnetic dipole.

You can't just change one equation for another willy-nilly.

 

[tris_d]

So indeed, we are getting the variance relates to r^4 where r is the distance between the magnets.

I'm afraid then something is wrong with that equation. Dipoles are supposed to fall off with the cube of the distance. I'm sure there was a different equation in place of that one few years ago, I'll try to find it somewhere else.

Force at 2 angstroms = (53.7)*(10^-16)/(2^4) = 3.36*10^-16 Newtons
Force at 3 angstroms = (53.7)*(10^-16)/(3^4) = 0.66*10^-16 Newtons

I will think about this and attempt the coulomb force tomorrow.

I'm looking forward to it.

 

[zincshow]

I'm afraid then something is wrong with that equation. Dipoles are supposed to fall off with the cube of the distance...

You may be correct, but you have to be very careful of the words that are used (field or force). Consider gravity from http://en.wikipedia.org/wiki/Newton's_law_of_universal_gravitation

"exists a gravitational potential field V(r) such that ... If m1 is a point mass or the mass of a sphere with homogeneous mass distribution, the force field g(r) outside the sphere is isotropic, i.e., depends only on the distance r from the center of the sphere. In that case"
V(r) = -G*m1/r ... ie. the "field" drops of relative to the distance (no square).

To calculate the gravitational force between two objects you would use
F(r) = -G*m1*m2/r^2 ... ie. the "force" from a second object drops off relative to the square of the distance.

Consider now the page you quoted http://en.wikipedia.org/wiki/Magnetic_dipole_moment, near the top:

"The vector potential of magnetic field produced by magnetic moment m is"
A(r) = (µ0)*(mxř)/(4*ᴨ*r^3)

I think the ř on top is only being used as a unit vector to give some direction to m, hence the "field" drops off with the cube of the distance.

Later on in the same article, just after your quoted equation "where F is the force acting on m2. The force acting on m1 is in opposite direction"
from which I think properly resolved to:
F(r) = (3*µ0)*m1*m2/(2*ᴨ*r^4)

hence the "force" on a close magnet drops off with the forth power of the distance.

Also, if I used r^3, the units would not work out properly to Newtons. For the moment, I am going to stick with the equation and my calculations.

 

[tris_d]

I think this is the equation I remember:
http://downloads.hindawi.com/archive/1998/079537.pdf

F= \frac {3 \mu_0} {4 \pi r^4} ( (\hat r \times m_a) \times m_b + (\hat r \times m_b) \times m_a - 2 \hat r(m_a \cdot m_b) + 5 \hat r ((\hat r \times m_a) \cdot (\hat r \times m_b)) )

In the paper they have derived it twice, with very subtle differences. The first one uses "hat" and "arrows", while the second equation uses only "arrows" above mass symbol. Well, mass is not a vector, so none of that funky notation makes sense to me and I removed it from the equation I wrote above. The question here is how this equation relates to the one from Wikipedia, but I don't see how could they produce the same result when this one uses both 'cross product' and 'dot product' while the one from Wikipedia has only 'dot product'.

http://img842.imageshack.us/img842/1382/magdipiole.jpg

 

[zincshow]

Thats the same equation. This guy is using the r-hat as a unit vector to show the direction of the magnetic moment, not an actual distance as r that needs to be factored out. It still shows force related to r^4, the confusion is all in the notation and different people do it differently.

 

[tris_d]

You may be correct, but you have to be very careful of the words that are used (field or force). Consider gravity from http://en.wikipedia.org/wiki/Newton's_law_of_universal_gravitation

"exists a gravitational potential field V(r) such that ... If m1 is a point mass or the mass of a sphere with homogeneous mass distribution, the force field g(r) outside the sphere is isotropic, i.e., depends only on the distance r from the center of the sphere. In that case"
V(r) = -G*m1/r ... ie. the "field" drops of relative to the distance (no square).

To calculate the gravitational force between two objects you would use
F(r) = -G*m1*m2/r^2 ... ie. the "force" from a second object drops off relative to the square of the distance.

I know what you mean. I don't consider my math skills are good enough for me to make any comment about dipoles though. You tell me how this second equation turns out and whether it boils down to 1/r^4 as well.

Consider now the page you quoted http://en.wikipedia.org/wiki/Magnetic_dipole_moment, near the top:

"The vector potential of magnetic field produced by magnetic moment m is"
A(r) = (µ0)*(mxř)/(4*ᴨ*r^3)

I think the ř on top is only being used as a unit vector to give some direction to m, hence the "field" drops off with the cube of the distance.

Yeah, r with hat is unit vector. In your example you have simple geometrical relation, so it will be either 1 or -1 depending on whether you're calculating force acting on dipole_1 or dipole_2. Meaning the force will be of the same magnitude but opposite direction.

Attraction: (dipole_1:r_hat= 1) ----> <---- (dipole_2:r_hat= -1)

Repulsion: (dipole_1:r_hat= -1) <---- ----> (dipole_2:r_hat= 1)

edit:
I should have said (1,0,0) and (-1,0,0) instead of 1 and -1.

Also, if I used r^3, the units would not work out properly to Newtons. For the moment, I am going to stick with the equation and my calculations.

Sure, units must work out.

 

[tris_d]

Thats the same equation. This guy is using the r-hat as a unit vector to show the direction of the magnetic moment, not an actual distance as r that needs to be factored out. It still shows force related to r^4, the confusion is all in the notation and different people do it differently.

Great, then r^4 it is, which is rather of obvious when I now look at this new equation from that paper. -- I don't think you realized my point though, they use vector notation for MASS, and that notation for MASS is different in those two equations. They must have had some reason for it, otherwise they would not bother to derive it twice. I will read the paper later on and hopefully that will answer that question.

Yes, I agree those two equations from that paper are the same, which is why I removed all the vector notation for mass, but are they the same as equation from Wikipedia?

PAPER:
F_{(r, m_a, m_b)}= \frac {3 \mu_0} {4 \pi r^4} [ (\hat r \times m_a) \times m_b + (\hat r \times m_b) \times m_a - 2 \hat r(m_a \cdot m_b) + 5 \hat r ((\hat r \times m_a) \cdot (\hat r \times m_b)) ]

WIKI:

 

[tris_d]

F(r) = (3*µ0)*m1*m2/(2*ᴨ*r^4)

F_{(r)}= \frac {3 \mu_0 (m_1 * m_2)} {2 \pi r^4} = 18.85 *10^{-7} \frac {m_1 * m_2} {r^4} =&gt; N/A^2 \frac {kg} {m^4}

This is my try to boil down bit more what you got there, I don't think units work out, and I hope that I'm wrong, but by having magnetic constant with units of N/A^2 I don't see how could we ever get rid of those amperes (coulomb per second) without having a 'charge' somewhere in the equation, two of them actually. How did you get units to work out?

 

[Vanadium 50]

The first one uses "hat" and "arrows", while the second equation uses only "arrows" above mass symbol. Well, mass is not a vector, so none of that funky notation makes sense to me and I removed it from the equation I wrote above. [/PLAIN]

And had you bothered to read that paper, you would have seen right at the top of the third page that they define m to be the magnetic dipole moment. You would also have seen in the abstract that they "assumed that the dipole sizes are small compared to their separation", which means it's a large distance approximation, and you want to use this as a small distance approximation.

This learning strategy of yours - posting a stream of demonstrably false statements and letting people correct them - is something we have found ineffective and extremely annoying to those who are trying to help you. You should abandon it and find another one.

 

[zincshow]

F_{(r)}= \frac {3 \mu_0 (m_1 * m_2)} {2 \pi r^4} = 18.85 *10^{-7} \frac {m_1 * m_2} {r^4} =&gt; N/A^2 \frac {kg} {m^4}

This is my try to boil down bit more what you got there, I don't think units work out, and I hope that I'm wrong, but by having magnetic constant with units of N/A^2 I don't see how could we ever get rid of those amperes (coulomb per second) without having a 'charge' somewhere in the equation, two of them actually. How did you get units to work out?

Good heavens, how did you get that? Please have a look at post #32,

First let's check units, magnetic moments are given in Joules/Tesla = Amps*meter^2 so
(N/A^2)*(A*m^2)*(A*m^2)/(m^4) = N so we indeed are in Newtons.

 

[tris_d]

And had you bothered to read that paper, you would have seen right at the top of the third page that they define m to be the magnetic dipole moment. You would also have seen in the abstract that they "assumed that the dipole sizes are small compared to their separation", which means it's a large distance approximation, and you want to use this as a small distance approximation.

This learning strategy of yours - posting a stream of demonstrably false statements and letting people correct them - is something we have found ineffective and extremely annoying to those who are trying to help you. You should abandon it and find another one.

Awww. Go ahead and ban me for thinking 'm' stands for mass, but I blame them for not using some other symbol. It's true, I haven't read Wikipedia article nor that paper yet and I apologize for my haste and silly assumptions.

But why is everyone so tense around here, people make mistakes, cut me some slack. I'm not superman, you know? I'd prefer if you just laughed at me. Anyway, shall I edit my posts now or leave it as it is so people can make fun of me and tell me how stupid I am?

 

[Vanadium 50]

Not stupid. Lazy, perhaps, but not stupid. You didn't bother reading the paper, and yet expected us to explain it to you. As far as "go ahead and ban" me, it won't help to be overdramatic.

People would cut you more slack if you put more effort in.

 

[tris_d]

Ok, but I did put effort, just in the wrong place. It took me 45 minutes to write that equation with latex. And to find it 4 hours at least.

Good heavens, how did you get that? Please have a look at post #32,

LOL. I thought 'm' stands for mass. Sorry about that. Now everything makes sense. Anyway, the question stands: do these two equations produce the same result:

http://img26.imageshack.us/img26/9388/magdipole2.jpg

...and if not, which one is the correct one?

 

[zincshow]

A calculation error in post #32 causes a revision to the numbers:

Corrected magnetic force:
at 1 Angstrom = 53.7*10^-14N
at 2 Angstrom = 3.36*10^-14N
at 3 Angstrom = 0.66*10^-14N

Coulomb force at 1 Angstrom:
= Ke*q1*q2/r^2
= (9*10^9)*(1.6*10^-19)*(1.6*10^-19)/(1*10^-10)^2
= 23*10^-9N

Therefore in response to the first post: at 1 angstrom the coulomb force is some 50000 times stronger then the magnetic force.

Further to the comment by Vanadium 50 that at some distance the magnetic force would match and become greater then the coulomb force, my calculations put that at 0.0048 angstroms or less then 0.5 picometers. I would suspect that QM comes into play in a big way at that distance scale.

Just to put the distance scale in perspective, I think in water the distance between the hydrogen proton and the oxygen nucleon is about 1 angstrom, many metals like copper or gold would have 2 or 3 angstroms between the nucleons. I suspect the separation of electrons based purely on density in some metals would not go below the 0.1 angstrom range.

 

[tris_d]

A calculation error in post #32 causes a revision to the numbers:

Corrected magnetic force:
at 1 Angstrom = 53.7*10^-14N
at 2 Angstrom = 3.36*10^-14N
at 3 Angstrom = 0.66*10^-14N

Coulomb force at 1 Angstrom:
= Ke*q1*q2/r^2
= (9*10^9)*(1.6*10^-19)*(1.6*10^-19)/(1*10^-10)^2
= 23*10^-9N

Therefore in response to the first post: at 1 angstrom the coulomb force is some 50000 times stronger then the magnetic force.

Further to the comment by Vanadium 50 that at some distance the magnetic force would match and become greater then the coulomb force, my calculations put that at 0.0048 angstroms or less then 0.5 picometers. I would suspect that QM comes into play in a big way at that distance scale.

Just to put the distance scale in perspective, I think in water the distance between the hydrogen proton and the oxygen nucleon is about 1 angstrom, many metals like copper or gold would have 2 or 3 angstroms between the nucleons. I suspect the separation of electrons based purely on density in some metals would not go below the 0.1 angstrom range.

Sweet. Just what I wanted to know for quite some time now. Since you are now expert in this calculation, would you mind to work out the result with that other equation as well? -- And to put some more perspective over distances, the most probable distance between proton and electron in a hydrogen (Bohr radius) is 0.529 Angstroms. In macromolecules (DNA and proteins) covalent bond length ranges between 1.0 and 1.8 Angstroms. Typical carbon-carbon (C-C) covalent bond has a bond length of 1.54 Angstroms.

 

[tris_d]

Here are my results using the second equation. I used this Wolfram input:

(3*(1.3*10^-6))/(4*pi*(1.0*10^-10)^4) * ( 5*((9.3*10^-24)*(9.3*10^-24)) )

http://img26.imageshack.us/img26/9388/magdipole2.jpg

Can someone with some math skills please verify this.

http://www.wolframalpha.comAt 1 Å= 1.34 x10^-12 N
At 2 Å= 8.39 x10^-14 N
At 3 Å= 1.66 x10^-14 N
At 4 Å= 5.24 x10^-15 N

Overcomes Coulomb:
At 0.0072 Å= 0.000499 N

Coulomb (0.0072 Å)= 0.000493 N
(9.99*10^9) * (1.6*10^-19)*(1.6*10^-19)/((0.0072*10^-10)^2)

 

[tris_d]

A calculation error in post #32 causes a revision to the numbers:

Can you explain how did you cancel distances and come up with 1/r^4?


--//--

This is my equation in Wolfram (overcomes Coulomb @ ~0.0072 Å):

(3*(1.3*10^-6))/(4*pi*(1.0*10^-10)^4) * ( 5*((9.3*10^-24)*(9.3*10^-24)) )

http://img26.imageshack.us/img26/9388/magdipole2.jpg


This is your equation in Wolfram (overcomes Coulomb @ ~0.0045 Å):

(3*(1.3*10^-6))/(4*pi*(1.0*10^-10)^4) * ( 2*((9.3*10^-24)*(9.3*10^-24)) )

http://www.wolframalpha.com

 

[tris_d]

http://downloads.hindawi.com/archive/1998/079537.pdf

30 April 1998:
- "...the force of one magnetic dipole on another has not yet been derived in electromagnetism textbooks or the periodical literature."

Very interesting.

 

[cragar]

What about starting with an infinite line charge and using gauss's law to find the E field. E=\frac{\lambda}{2\pi r \epsilon_0} And then figure out what speed the line charges need to move to overcome the force from the E field. I=\lambda v lambda is your charge per lenght.
And then use amperes law to find the B field of the moving line charge.
B=\frac{\mu_0\lambda v}{2\pi r} B field is in the phi direction.

 

[tris_d]

What about starting with an infinite line charge and using gauss's law to find the E field. And then figure out what speed the line charges need to move to overcome the force from the E field. I=\lambda v lambda is your charge per lenght.
And then use amperes law to find the B field of the moving line charge.
B=\frac{\mu_0\lambda v}{2\pi r}

Apparently this equation did not exist until 1998, so I guess the derivation is far from trivial. You would need circle instead of line. And since you need vectors instead of Gauss law you would have to use Biot-Savart law and Lorentz force equations, which in other words is Ampere's force law integrated over circle. That's how they did it anyway:

http://downloads.hindawi.com/archive/1998/079537.pdf

 

[cragar]

I was thinking of the B field created from the motion of the electrons and not their intrinisic dipole moment. Gauss's law would give you a vector.

 

[Vanadium 50]

This thread is rapidly filling with misinformation. Closed.

"...the force of one magnetic dipole on another has not yet been derived in electromagnetism textbooks or the periodical literature."[/I]

Very interesting.

And that's grossly out of context. The authors are discussing a closed-form analytic expression, and by their own admission, what they come up with is an approximation.