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.

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