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## Homework Statement

Consider an ideal stationary magnetic dipole ##\vec{m}## in a static electric field ##\vec{E}##. Show that the fields carry momentum ##\vec{p} = -\epsilon_0\mu_0\left(\vec{m}\times\vec{E}\right)##.

## Homework Equations

Momentum stored in electromagnetic fields: ##\vec{p} = \epsilon_0\int(\vec{E}\times\vec{B})d\tau## where ##d\tau## is an infinitesimal volume element.

For static electric fields: ##\vec{E} = -\nabla V## where ##V## is the electric potential.

A product rule I used in my attempt: ##\nabla\times(f\vec{A}) = f(\nabla\times\vec{A}) - \vec{A}\times(\nabla f)## where ##f## is a scalar function while ##\vec{A}## is a vector function.

For static magnetic fields: ##\nabla\times\vec{B} = \mu_0\vec{J}## where ##\vec{J}## is the volume current density.

Also, for steady currents: ##\nabla\cdot\vec{J} = 0##.

Another identity I used: ##\oint(\vec{c}\cdot\vec{r})d\vec{l} = \vec{a}\times\vec{c}## where ##\vec{c}## is a constant vector and ##\vec{a} = \int_S d\vec{a}## is the vector area.

Yet another identity: ##\int_{\nu}(\nabla\times\vec{v})d\tau = -\oint_S\vec{v}\times d\vec{a}## where ##\nu## is a volume and ##S## is its surface.

Finally, ##\int\vec{J}d\tau = \frac{d\vec{P}}{dt}## where ##\vec{P}## is the total

*electric*dipole moment (0 in our problem)

## The Attempt at a Solution

First, I substitute ##\vec{E} = -\nabla V## into ##\vec{p} = \epsilon_0\int(\vec{E}\times\vec{B})d\tau## to get ##\vec{p} = \epsilon_0\int(\vec{B}\times\nabla V)d\tau##. From here, I use the product rule quoted above to get that ##\int\vec{B}\times\nabla V d\tau = \int V(\nabla\times\vec{B})d\tau - \int\nabla\times(V\vec{B})d\tau##. Now, by another identity: ##- \int\nabla\times(V\vec{B})d\tau = -\oint_S\vec{v}\times d\vec{a}##. But what is the volume/surface we're integrating over? All of space! Thus, the surface is at infinity and ##\vec{B}=0## there (since it is far away from the dipole) and this integral vanishes. We're thus left with ##\vec{p} = \epsilon_0\int V(\nabla\times\vec{B})d\tau = \epsilon_0\mu_0\int V\vec{J}d\tau##. Since the only current is confined to the infinitesimal volume near the origin (i.e. the dipole), this integral is nonzero only there. Thus, ##V## can be expanded with a Taylor series for vectors about the origin: ##V ≈ V(0) - \vec{E}(0)\cdot\vec{r}##. With this, our expression becomes: ##\vec{p} = \epsilon_0\mu_0\int (V(0) - \vec{E}(0)\cdot\vec{r})\vec{J}d\tau##. Focusing our attention on the first integral, ##\int V(0)\vec{J}d\tau = V(0)\int\vec{J}d\tau = 0## as noted above. Thus, ##\vec{p} = -\epsilon_0\mu_0\int (\vec{E}(0)\cdot\vec{r})\vec{J}d\tau##.

Now I am stuck. I wanted to just say that ##\vec{J}d\tau → \vec{I}dl = Id\vec{l}## since the dipole can be treated as a current loop. Then, everything works out by using the other identities involving line integrals I quoted. But my math conscientiousness didn't allow me to. So can anyone give me any hint as to how to proceed with this proof or suggest a simpler method for doing it?

Any suggestions/comments will be greatly appreciated!