# Force over a dipole

### In an arbitrary field

The easiest way to get the force over a dipole is to consider it as the limit of two oppositely charged monopoles that are closely spaced. If the dipole has moment $\mathbf{m}$ and is at position $\mathbf{r_0}$, it can be considered as the limit of two monopoles, one with charge $-\epsilon^{-1}$ at position $\mathbf{r_0} - (\epsilon / 2) \mathbf{m}$ and the other with charge $\epsilon^{-1}$ at position $\mathbf{r_0} + (\epsilon / 2) \mathbf{m}$, when $\epsilon$ goes to zero.

If we consider the finite size dipole immersed in a (let’s say magnetic) field $\mathbf{B}(\mathbf{r})$, the total force will be $\displaystyle \mathbf{F} = -\epsilon^{-1}\mathbf{B}(\mathbf{r_0} - (\epsilon / 2) \mathbf{m}) +\epsilon^{-1}\mathbf{B}(\mathbf{r_0} + (\epsilon / 2) \mathbf{m})$ $\displaystyle \mathbf{F} = \frac{\mathbf{B}(\mathbf{r_0} + (\epsilon / 2) \mathbf{m}) - \mathbf{B}(\mathbf{r_0} - (\epsilon / 2) \mathbf{m})}{\epsilon}$.

We can get the force for the infinitesimal dipole by taking the limit when $\epsilon$ goes to zero $\displaystyle \mathbf{F} = \lim_{\epsilon \to 0}\frac{\mathbf{B}(\mathbf{r_0} + (\epsilon / 2) \mathbf{m}) - \mathbf{B}(\mathbf{r_0} - (\epsilon / 2) \mathbf{m})}{\epsilon}$ $\displaystyle \mathbf{F} = \mathbf{m} \cdot (\mathbf{\nabla B})(\mathbf{r_0})$,

where $\mathbf{\nabla B}$ is the (tensor) derivative of the magnetic field.

### Between two antialigned dipoles

The general field of a magnetic dipole of moment $\mathbf{m}$ at position $\mathbf{r_0}$ is $\displaystyle \mathbf{B}_d(\mathbf{r}) = \frac{\mu_0}{4\pi}\left(\frac{3\mathbf{m}\cdot(\mathbf{r}-\mathbf{r_0})(\mathbf{r}-\mathbf{r_0})-\mathbf{m}|\mathbf{r}-\mathbf{r_0}|^2}{|\mathbf{r}-\mathbf{r_0}|^5}\right)$.

If we assume we have one dipole at $x = 0$ with its moment $+m\mathbf{\hat{x}}$ and the other at $x = +R$ with its moment $-m\mathbf{\hat{x}}$, we get a field at $x = +R$ of $\displaystyle \mathbf{B}(+R\mathbf{\hat{x}}) = \frac{\mu_0}{4\pi}\left(\frac{3m\mathbf{\hat{x}}\cdot(+R\mathbf{\hat{x}})(+R\mathbf{\hat{x}})-m\mathbf{\hat{x}}|+R\mathbf{\hat{x}}|^2}{|+R\mathbf{\hat{x}}|^5}\right)$ $\displaystyle \mathbf{B}(+R\mathbf{\hat{x}}) = \frac{\mu_0}{4\pi}\left(\frac{3mR^2-mR^2}{R^5}\right)\mathbf{\hat{x}}$ $\displaystyle \mathbf{B}(+R\mathbf{\hat{x}}) = \frac{\mu_0 m}{2\pi R^3}\mathbf{\hat{x}}$.

By symmetry, we are only interested in the x-component of the x-derivative of the field, $\displaystyle \left(\frac{\partial\mathbf{B}}{\partial x}\right)_x = -\frac{3\mu_0 m}{2\pi R^4}$.

And the force on the dipole at $x = +R$ will be $\displaystyle \mathbf{F} = -m\mathbf{\hat{x}} \cdot (\mathbf{\nabla B})(+R\mathbf{\hat{x}})$ $\displaystyle \mathbf{F} = -m \left(-\frac{3\mu_0 m}{2\pi R^4}\right)\mathbf{\hat{x}}$ $\displaystyle \mathbf{F} = \frac{3\mu_0 m^2}{2\pi R^4}\mathbf{\hat{x}}$,

a repulsive force as expected of antialigned dipoles.