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Forces and fields: magnetic dipole vs electric monopole

  1. Aug 30, 2014 #1
    Two electrons, they have their electric fields and magnetic dipole moments. Their electric field is proportional to 1/r^2, and electric force is also proportional to 1/r^2.

    http://en.wikipedia.org/wiki/Magnetic_dipole–dipole_interaction

    However, their magnetic dipole field is proportional to 1/r^3, and magnetic force to 1/r^4. How does 1/r^3 come about and why it turns into 1/r^4 for the force equation?


    If electric force is proportional to 1/r^2 and magnetic dipole force to 1/r^4, does that mean if two electrons come close enough together this magnetic dipole force could overcome their electric repulsive force?
     
  2. jcsd
  3. Aug 30, 2014 #2

    Simon Bridge

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    The 1/r^2 rule applies to spherical symmetry.
    Other symmetries get other rules.

    i.e. compare the field for an electric dipole:
    http://hyperphysics.phy-astr.gsu.edu/hbase/electric/dipole.html

    But you also have to be careful when you read vector equations.
    i.e. The equation for electrostatic field of a point electron charge actually looks like this: $$\vec E = -\frac{ke}{r^3}\vec r$$ ... see why it's a 1/r^3 law this time?
     
  4. Aug 30, 2014 #3
    I really don't see. I thought it's supposed to look like this:

    6ef87591382929117d7f8e3bc1edc75e.png
    http://en.wikipedia.org/wiki/Coulomb's_law

    On the same page the force equation retains 1/r^2 in both scalar and vector form.
     
  5. Aug 30, 2014 #4

    Nugatory

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    What you've written is the formula for the magnitude of the electric field vector, not the formula for the vector itself. Write the vector ##\vec{r}## as the product of a scalar and a unit vector ##\hat{r}## and you'll see why the factor of ##r^3## is needed.
     
  6. Aug 30, 2014 #5
    Magnitude is all it matters. Unit vector does not cancel with the distance, it's just direction. Regardless of vector or scalar form, it should be 1/r^2 for both electric field and electric force equation.
     
  7. Aug 30, 2014 #6

    Simon Bridge

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    Yeah - when you see a bar on each side of the vector symbol like this |v| it means "the magnitude of the vector". I also put ##k=1/4\pi\epsilon_0## and ##q=-e## along with ##\vec r = r\hat r## so that ##r=|\vec r|## ;)

    In the (wikipedia) link you provided, down a bit from the equation you quote, there is a vector form for the force law - notice the unit vector in the numerator and the square-magnitude in the denominator? Scroll down some more to the "continuous charge distribution" section - there is an integral with complete vectors in the numerator and a cube magnitude in the denominator.

    However: Wikipedia is generally not a good source for learning physics. They don't explain enough and are often inconsistent... don't rely on it for insights.

    Did you follow my link though - so you can compare like with like: i.e. compare the electric dipole field with the magnetic dipole field?
     
  8. Aug 30, 2014 #7

    Dale

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    Magnitude is not all that matters, the direction is very important, particularly for fields. If magnitude were all that matters then you would never get destructive interference. It is a very clear physical fact that occurs in nature and cannot be replicated with simply looking at magnitude.

    Nugatory is rightly encouraging you to look at and understand the differences between two formulas.

    $$|\mathbf{\vec{E}}|=k\frac{|q|}{r^2}$$
    is a different formula from
    $$\mathbf{\vec{E}}=k\frac{q}{|\mathbf{\vec{r}}|^2}\mathbf{\hat{r}}=k \frac{q}{|\mathbf{\vec{r}}|^3} \mathbf{\vec{r}}$$

    Notice, in particular, that the vector equation with 1/r^2 is the same as the vector equation with 1/r^3. Both describe an inverse square force law, even though one has an inverse cube term written down.
     
    Last edited: Aug 30, 2014
  9. Aug 30, 2014 #8

    Simon Bridge

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    Take another look at what I wrote - I did not put a unit vector in there anywhere...
     
  10. Aug 30, 2014 #9
    I thought it was supposed to be. I see what you mean now, but I can't say the magnitude is inversely proportional to distance cubed, it's really proportional to distance squared. So writing that equation with unit vector would be preferred as it makes the relation more clear, right?
     
  11. Aug 30, 2014 #10

    jtbell

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    I usually prefer the version with the unit vector, myself. However, the other version might be more convenient sometimes for mathematical manipulations.
     
  12. Aug 30, 2014 #11

    vanhees71

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    Of course, the Coulomb law is that the magnitude goes with [itex]1/r^2[/itex], but
    [tex]|\vec{r}/r^3|=r/r^3=1/r^2.[/tex]
    So, of course, Simon's formula is correct. Don't mess with the vectors!
     
  13. Aug 30, 2014 #12
    electric e= 100, magnetic b= 50
    Fe= 100/r^2, Fb= 50/r^4

    We have electric force of say 100 N that changes with 1/r^2, and magnetic force of 50 N that changes with 1/r^4:

    10m: Fe= 1, Fb= 0.005
    1m: Fe= 100, Fb= 50
    0.1m: Fe= 10000, Fb= 500000

    So when two electrons move towards each other, the magnitude of magnetic force increases at higher rate than of electric force, and thus there is a point in distance where magnetic dipole force between the two electrons becomes stronger than their electric repulsive force?
     
  14. Aug 30, 2014 #13

    Simon Bridge

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    I think you need to set it out more carefully than that - write out the vector equations.
    If you really want to characterize the electromagnetic interaction between electrons you are not using a good model.
    Try setting it up for two classical point charges q heading towards each other, along the z axis, at speed v - and these charges also carry a magnetic dipole moment magnitude μ pointing along the direction of their motion (so like poles are pointing at each other).

    Now write out the equation for the net force of one charge on the other - remember to pay attention to the vectors.
     
  15. Aug 31, 2014 #14
    If their magnetic moment orientation is free to rotate, wouldn't they want to turn their opposite magnetic poles towards each other, so to attract rather than repel?


    2fa743e4f8027215904de3811e2d2a48.png

    I get 0 result from this equation. I was calculating force acting on electron "a" and I assumed all the unit vectors are in a->b direction. I'll try again later.
     
  16. Aug 31, 2014 #15

    Simon Bridge

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    ... if they were dead on, would they experience a torque? It would be an unstable configuration though.

    Rotation would kinda spoil your question though wouldn't it?
    In the e-e interaction, the electron spins would interact too wouldn't they? But you want to investigate the classical relationship.

    You only showed the first part of the calculation - did you decompose the vectors into their cartesian coordinates.
    You did not include terms for the electric field. Does it matter that each electron is a charged particle moving in the magnetic field of the other electron?

    Anyway - now you know how to work out the answer to your own question :)
     
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