# How to count magnetic repulsion

I have two equal flat round magnets. I know amount of force $F$ which attracts iron objects to one of them and geometric characteristics of magnets. I want to fix first of magnet and some additional mass in the air by second magnet. To do so, I am going to orient magnets so that second magnet repulse first and additional mass. But I need to establish dependence between distance between magnets and value of additional mass this construction can held. How to gain this dependence?

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Possible duplicate: physics.stackexchange.com/q/17309/2451 and links therein. –  Qmechanic Jan 10 '13 at 21:48

## 1 Answer

To a good approximation, normal magnets can be treated as dipole magnets, in which case the force between them can be found in this wikipedia article. To avoid link-only answers, here it is: $$\mathbf{F} = \dfrac{3 \mu_0}{4 \pi r^5}\left[(\mathbf{m}_1\cdot\mathbf{r})\mathbf{m}_2 + (\mathbf{m}_2\cdot\mathbf{r})\mathbf{m}_1 + (\mathbf{m}_1\cdot\mathbf{m}_2)\mathbf{r} - \dfrac{5(\mathbf{m}_1\cdot\mathbf{r})(\mathbf{m}_2\cdot\mathbf{r})}{r^2}\mathbf{r}\right].$$

However, be warned that while equilibrium levitation is possible using magnets, stable equilibria are impossible due to Earnshaw's theorem. Magnetic levitation is indeed possible but you need fancier schemes than just two magnets. The wikipedia article on it is a good starting point.

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How can i find $m_1$, $m_2$ and $\mu_0$? I'm planning to use some strings to establish equilibrium. –  freopen Dec 26 '12 at 12:48
$\mu_0$ is the magnetic permeability of free space. $\mathbf{m}_i$ are the magnetic dipole moment vectors of the two magnets; once you know their direction (e.g. using an iron-filing experiment) you only need the product $m_1m_2$ of their magnitudes and that you can find with a single force reading. –  Emilio Pisanty Dec 28 '12 at 1:13
However, please consider the stability of the equilibrium you are trying to build. If strings are involved, why not just hang the magnets? The equilibrium obtained using simple magnets is as stable as an up-ended pendulum. Do read the WP link! –  Emilio Pisanty Dec 28 '12 at 1:15