Electron magnetic replusion compared to coulomb attraction

[zincshow]

I know that the electron has an electric charge of 1.6x10^-19C and a magnetic moment of 9.3x10-24J/T. If you placed two electrons 1 angstrom apart in such a way as their north poles are pointing at each other, the coulomb force would attract them and the magnetic force would repel them. Which would win out?

I know how to calculate the force on the electrons based on the coulomb force, but I do not know how to calculate the force on the electrons based on the magnetic force. Does anyone know how to calculate the magnetic force of repulsion on two stationary magnets? or have a suggestion of a website that tells you how to calculate the force of repulsion on two stationary magnets?

 

[tris_d]

I know that the electron has an electric charge of 1.6x10^-19C and a magnetic moment of 9.3x10-24J/T. If you placed two electrons 1 angstrom apart in such a way as their north poles are pointing at each other, the coulomb force would attract them and the magnetic force would repel them. Which would win out?

I know how to calculate the force on the electrons based on the coulomb force, but I do not know how to calculate the force on the electrons based on the magnetic force. Does anyone know how to calculate the magnetic force of repulsion on two stationary magnets? or have a suggestion of a website that tells you how to calculate the force of repulsion on two stationary magnets?

I ditto that question. Why are you asking, what are you up to?

Here is my thread about something similar:
https://www.physicsforums.com/showthread.php?t=648966


http://en.wikipedia.org/wiki/Magnetic_dipole_moment

 

[Drakkith]

Why would two electrons be attracted towards each other?

 

[tris_d]

Why would two electrons be attracted towards each other?

When magnetic attractive force overcomes electric repulsive force.

http://en.wikipedia.org/wiki/Z-pinch

http://en.wikipedia.org/wiki/Lorentz_force

 

[Drakkith]

When magnetic attractive force overcomes electric repulsive force.

The OP specifically mentioned that the coloumb force was attracting them.

 

[tris_d]

The OP specifically mentioned that the coloumb force was attracting them.

Lapsus calami. It can't be the other way around, so he must have meant: "the coulomb force would repel them and the magnetic force would attract them".

 

[Drakkith]

Lapsus calami. It can't be the other way around, so he must have meant: "the coulomb force would repel them and the magnetic force would attract them".

I was going to assume so, but he also specifically placed both North poles towards each other.

 

[zincshow]

Why would two electrons be attracted towards each other?

Sorry, poorly worded and not checked. Better: 2 electrons 1 angstrom apart repelling each other through the coulomb force, but attracting each other through the magnetic force -or- 2 electrons 1 angstrom apart attracted by a proton inbetween them, hence the proton is attracting them and the electrons are repelling each other ... I was after the magnetic force calculation and clearly did not give enough thought to my layout. Darn ...

tris_d: Thanks, I had not seen that page, it will take a while to digest it.

The main question: at short distance which is the stronger force for an electron, the coulomb force or the magnetic force? Do they compare or is one "much" bigger then the other? Is the magnetic force an insignificant force at very small distances compared to the coulomb force?

Also: I have recently found this page http://www.kjmagnetics.com/calculator.asp, but I am not sure it helps me. I am looking for a formula I can punch in the distance apart and compare the coulomb force to the magnetic force for 2 electrons.

 

[tris_d]

I was going to assume so, but he also specifically placed both North poles towards each other.

Interesting. I haven't noticed any of that, and I see now there is the same mistake in the title as well.

 

[Drakkith]

I believe the coloumb force is far stronger than the magnetic force. But at this scale and distance Quantum Mechanical phenomena become dominant, so it may not be easy to explain.

 

[tris_d]

The main question: at short distance which is the stronger force for an electron, the coulomb force or the magnetic force? Do they compare or is one "much" bigger then the other? Is the magnetic force an insignificant force at very small distances compared to the coulomb force?

That I want to know too. Have no idea how to work it out.

Also: I have recently found this page http://www.kjmagnetics.com/calculator.asp, but I am not sure it helps me. I am looking for a formula I can punch in the distance apart and compare the coulomb force to the magnetic force for 2 electrons.

Formula I posted above takes mass and distance and gives you the force. The strange thing is I think that equation looked differently on that Wikipedia page few years ago. I think dipole magnetic moment should fall off with the cube of the distance, not square.

 

[tris_d]

I believe the coloumb force is far stronger than the magnetic force. But at this scale and distance Quantum Mechanical phenomena become dominant, so it may not be easy to explain.

Apart from QM I think chemistry and molecular dynamics have something to say about that as well.

 

[Drakkith]

Apart from QM I think chemistry and molecular dynamics have something to say about that as well.

I don't see how, but ok.

 

[tris_d]

I don't see how, but ok.

It just rings me a bell, probably due to molecular electric dipoles, but then there are molecular dynamics equations based on QM, so if that magnetic dipole moment of electron has some impact greater then insignificant it could be a part of some of those equations.

 

[Vanadium 50]

The force you wrote above goes as 1/r⁴ and the Coulomb force goes as 1/r². So classically, there is a region where the magnetic force is stronger. Quantum mechanically, there is not.

 

[tris_d]

I don't see how, but ok.

Ok, I found it now where I got that thought from.

http://en.wikipedia.org/wiki/Pauli_exclusion_principle
- The Pauli exclusion principle helps explain a wide variety of physical phenomena. One particularly important consequence of the principle is the elaborate electron shell structure of atoms and the way atoms share electrons, explaining the variety of chemical elements and their chemical combinations. An electrically neutral atom contains bound electrons equal in number to the protons in the nucleus. Electrons, being fermions, cannot occupy the same quantum state, so electrons have to "stack" within an atom, i.e. have different spins while at the same place.

- An example is the neutral helium atom, which has two bound electrons, both of which can occupy the lowest-energy (1s) states by acquiring opposite spin; as spin is part of the quantum state of the electron, the two electrons are in different quantum states and do not violate the Pauli principle.


It seems to me Pauli exclusion principle describes the same thing as in the question from opening post, where two electrons would pair up by having their opposite magnetic poles turned towards each other.

 

[tris_d]

The force you wrote above goes as 1/r⁴ and the Coulomb force goes as 1/r². So classically, there is a region where the magnetic force is stronger. Quantum mechanically, there is not.

Can you explain that a bit more, with some example and actual numbers?


By the way, don't you mean 1/r^3 instead of 1/r^4? It says here:

http://en.wikipedia.org/wiki/Magnetic_dipole_moment
The dipole component of an object's magnetic field is symmetric about the direction of its magnetic dipole moment, and decreases as the inverse cube of the distance from the object.

 

[DrZoidberg]

The force you wrote above goes as 1/r⁴ and the Coulomb force goes as 1/r². So classically, there is a region where the magnetic force is stronger. Quantum mechanically, there is not.

Classically, if you model an electron as a spinning electrically charged sphere with a radius > 0, i think it would have to spin faster than the speed of light for there to be a region where the attraction is stronger than the repulsion.

 

[tris_d]

Classically, if you model an electron as a spinning electrically charged sphere with a radius > 0, i think it would have to spin faster than the speed of light for there to be a region where the attraction is stronger than the repulsion.

They have no size, being point particles, so spin rate is not really defined. And it's not important, we just say that magnetic dipole moment is INTRINSIC property of an electron and we are only concerned by the force produced due to interaction of those magnetic fields compared to force between electric fields.

 

[Vanadium 50]

Can you explain that a bit more, with some example and actual numbers?

If the electric force is 8/r^2 and the magnetic force is 2/r^4, at r < 0.5, the magnetic force is stronger. If the electric force is a/r^2 and the magnetic force is b/r^4, at r < sqrt(b/a) , the magnetic force is stronger. So there is always a line where the Classical magnetic force is stronger.

By the way, don't you mean 1/r^3 instead of 1/r^4?

Read what I wrote. I said "the force you wrote above". I didn't check to see that it is correct or not.

 

[tris_d]

If the electric force is 8/r^2 and the magnetic force is 2/r^4, at r < 0.5, the magnetic force is stronger. If the electric force is a/r^2 and the magnetic force is b/r^4, at r < sqrt(b/a) , the magnetic force is stronger. So there is always a line where the Classical magnetic force is stronger.

I wish I was not stupid for math, but I get all we have to do now is figure out what is the actual value of 'a' and 'b' and we can calculate at what distance will magnetic attractive force overcome electric repulsive force. Can you help us work that out?

Mass: 9.109×10^−31 kilograms
Electric charge: −1.602×10^−19 coulombs
Magnetic moment: 1.001 bohr magnetons

What now? What equation do we use for magnetic moment and how do we fit "1.001 bohr magnetons" into it? Would this equation be the one to use here:

Read what I wrote. I said "the force you wrote above". I didn't check to see that it is correct or not.

My math is really terrible, and I see lots of 'r' in that equation, but I have no idea how they cancel out. Are you saying that after cancellation that equation comes up as 1/r^4? Would that be incorrect then according to their statement that dipole magnetic moment drops off with the cube of the distance?

 

[Drakkith]

I wish I was not stupid for math, but I get all we have to do now is figure out what is the actual value of 'a' and 'b' and we can calculate at what distance will magnetic attractive force overcome electric repulsive force. Can you help us work that out?

You realize that in reality the magnetic force will never overcome the electric force right?

 

[tris_d]

You realize that in reality the magnetic force will never overcome the electric force right?

How did you arrive to that conclusion? In z-pinch it happens when Lorentz force overcomes Coulomb force. And as for electron dipole magnetic moment, if you consider what Vanadium said it's just a matter of distance.

Did you see what I posted about Pauli exclusion principle? What else, what other force could it be that "stacks" electrons in the same orbit and aligns them to have opposite spin?

 

[Drakkith]

How did you arrive to that conclusion? In z-pinch it happens when Lorentz force overcomes Coulomb force.

We are talking about 2 particles, not a Z-Pinch.

And as for electron dipole magnetic moment, if you consider what Vanadium said it's just a matter of distance.

Incorrect. Quantum mechanics tells us that this will not happen.

Did you see what I posted about Pauli exclusion principle? What else, what other force could it be that "stacks" electrons in the same orbit and aligns them to have opposite spin?

I wouldn't say the magnetic force does this, but it is a consequence of not being able to fall into a lower energy state with another electron unless this happens. The wavefunction of fermions interfere destructively when two fermions try to occupy the same state at the same time, and switching the spin means the two fermions are no longer in the same state.

 

[tris_d]

We are talking about 2 particles, not a Z-Pinch.

You said magnetic force will never overcome electric force, and I gave you an example where it happens. Lorentz force applies to 2 particles just as well as to many.

How many electrons do you imagine there needs to be for it to start working?

Incorrect. Quantum mechanics tells us that this will not happen.

Where does it say so, what are you referring to? -- Are you trying to say Coulomb's law does not apply to two electrons or that this equation below does not apply to electron magnetic dipole moment?

I wouldn't say the magnetic force does this, but it is a consequence of not being able to fall into a lower energy state with another electron unless this happens. The wavefunction of fermions interfere destructively when two fermions try to occupy the same state at the same time, and switching the spin means the two fermions are no longer in the same state.

Never mind.

 

[Vanadium 50]

Tris_d, you can't just start posting equations out of the air and expect that to make sense. I suggest you start by carefully rereading what has already been posted on this thread.

 

[Drakkith]

You said magnetic force will never overcome electric force, and I gave you an example where it happens. Lorentz force applies to 2 particles just as well as to many.

Context man! Context!

 

[tris_d]

Tris_d, you can't just start posting equations out of the air and expect that to make sense. I suggest you start by carefully rereading what has already been posted on this thread.

What does that mean? What is wrong?

Do you mean to say this equation does not apply where there are only 2 electrons?
http://en.wikipedia.org/wiki/Lorentz_force

Do you mean to say this equation does not apply to electron dipole moment?
http://en.wikipedia.org/wiki/Magnetic_dipole_moment

 

[Drakkith]

Those do not apply in Quantum Mechanics, which is what you need to accurately describe two electrons at extremely close range.

 

[tris_d]

Those do not apply in Quantum Mechanics, which is what you need to accurately describe two electrons at extremely close range.

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.

 

[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.