Question

Can Coulomb's law be applied in magnetism?
Where two magnets interact?

Answer 1

@TuringTest
I remember you saying no.
But I want to make sure why not, could you give me a convincing reason why?

Answer 2

because coulomb's law deals with static charges and force due to them.. where as magnetism is force due to moving charges!

Answer 3

Ow! good point.

Answer 4

Thanks

Answer 5

@experimentX I haven't seen you in a while! Let me see ur try answering this question!

Answer 6

yeah ... you can apply between two magnets.

Answer 7

two magnetic poles follow inverse square law ... http://en.citizendium.org/wiki/coulomb's_law_(magnetic)

Answer 8

Um.
I usually would say the same but after noticing @Mashy's point about static charges and moving charges I started to reconsider.

Answer 9

yeah ... you can't apply like you apply on electrostatics. but somehow ... if you bring two poles of magnet you notice that the force follows inverse square laws.

Answer 10

Yet does not exactly give you the right answer that you need.
Thus! I should figure out other methods?

Answer 11

if magnetic monopoles were to exist ... i would satisfy coulomb law, if it is inverse square attraction-repulsion.
Coloumb law, i think, is based on flux conservation. and the flux of charges (static) spread out symmetric in all direction.

While for magnet ... it does't (so that the picture generally shown).

Answer 12

*it

Answer 13

Thus when someone argues with me that we should use COULOMBS law!
I shall replay simply: No. It does not because of the static charges in comparison to the moving charges, and the flux change.

Answer 14

Well, I need to develop a more better way to rebuttal the idea.
But you get what I mean @experimentX

Answer 15

yeah ... but i cannot assert ... lol

Answer 16

also ... unlike electric, the magnetic field lines do not terminate.

Answer 17

Indeed.
They can not!

Answer 18

Man for some reason... I forgot all of this info.
I hate having to study all over again :P

Answer 19

there is EM course going on edX ...

Answer 20

I saw it.
Problem is my friend is time!

Answer 21

I have calculus to worry about + chemistry in college...

Answer 22

haven't taken physics yet( all these questions are sid hobbies!)

Answer 23

try reading these
http://www.askamathematician.com/2011/02/q-what-are-the-equations-of-electromagnetism-what-all-do-they-describe-to-us/
http://www.askamathematician.com/2010/11/q-what-is-a-magnetic-field/

Answer 24

There are different models one can use for the force between magnets. The model that uses the concept of magnetic charge (magnetic monopoles) is called the Gilbert model. The Gilbert model force looks like Coulomb's Law, but of corse it is different, because instead of charges we have magnetic monopole magnitude, and instead of 1/(4pi epsilon) we have mu/(4pi).

Now the Gilbert model is not correct physically, though it may make acceptable predictions about the force in some cases. This model is not correct because we do not believe in magnetic monopoles, and the actual magnetic field of a magnet is more complicated than the field of a monopole.
There is an expression for the force between magnetic dipoles, and it is complicated.

\[\vec{F}_{ab}= \frac {3 \mu_0} {4 \pi |r|^4} [ (\hat r \times \vec{m}_a) \times \vec{m}_b + (\hat r \times \vec{m}_b) \times \vec{m}_a - 2 \hat r(\vec{m}_a \cdot \vec{m}_b) + 5 \hat r ((\hat r \times \vec{m}_a) \cdot (\hat r \times \vec{m}_b)) ]\]

Where r is the vector between the dipoles, and ma and ma are the magnetic dipole moments.
You can read more here:
http://en.wikipedia.org/wiki/Force_between_magnets

Answer 25

\[\vec{F}_{ab}= \frac {3 \mu_0} {4 \pi |r|^4} [ (\hat r \times \vec{m}_a) \times \vec{m}_b + (\hat r \times \vec{m}_b) \times \vec{m}_a - 2 \hat r(\vec{m}_a \cdot \vec{m}_b) \]
\[+ 5 \hat r ((\hat r \times \vec{m}_a) \cdot (\hat r \times \vec{m}_b)) ]\]

Answer 26

Thanks @LikeLightning
Interesting point!