Consider an ideal stationary magnetic dipole m in a static electric field E. Show that the fields carry momentum

$\mathit{p}\mathbf{=}\mathbf{-}{\mathit{\epsilon }}_{\mathbf{0}}{\mathit{\mu }}_{\mathbf{0}}\left(m\times E\right)$ (8.45)

[Hint: There are several ways to do this. The simplest method is to start with $\mathit{p}\mathbf{=}{\mathit{\epsilon }}_{\mathbf{0}}\mathbf{\int }\left(E\times B\right)\mathit{d}\mathit{\tau }$, write $\mathit{E}\mathbf{=}\mathbf{-}\mathbf{\nabla }\mathit{V}$, and use integration by parts to show that

$\mathit{p}\mathbf{=}{\mathit{\epsilon }}_{\mathbf{0}}{\mathit{\mu }}_{\mathbf{0}}\mathbf{\int }\mathit{V}\mathit{J}\mathit{d}\mathit{\tau }$.

So far, this is valid for any localized static configuration. For a current confined to an infinitesimal neighbourhood of the origin, we can approximate $\mathit{V}\left(r\right)\mathbf{\approx }\mathit{V}\left(0\right)\mathbf{-}\mathit{E}\left(0\right)\mathbf{·}\mathit{r}$. Treat the dipole as a current loop, and use Eqs. 5.82 and 1.108.]²¹

Write the expression for electromagnetic momentum per unit volume.

$\mathbf{p}\mathbf{=}{\mathbf{\epsilon }}_{\mathbf{0}}\mathbf{\int }\left(E\times B\right)\mathbf{d\tau }$ …… (1)

Here, E is the electric field, and B is the magnetic field.

Substitute the known values in equation (1).

$p={\epsilon }_0\int \left(-\nabla V\right)\times Bd\tau$

Use integration parts in the above equation.

$$\begin{gathered} p=-{\epsilon }_0\int \left[\nabla \times \left(VB\right)-V\nabla \times B\right]d\tau \\ p={\epsilon }_0\oint VB\times da+{\epsilon }_0{\mu }_0\int VJd\tau \\ p=0+{\epsilon }_0{\mu }_0\int VJd\tau \\ p={\epsilon }_0{\mu }_0\int VJd\tau \end{gathered}$$

Use the given equation $V\left(r\right)\approx V\left(0\right)+\left(\nabla V\right).r=V\left(0\right)-E\left(0\right).r$.

localid="1657364211619" $$\begin{gathered} p={\epsilon }_0{\mu }_{0}V\left(0\right)\int Jd\tau -{\epsilon }_0{\mu }_0\int \left[E\left(0\right).r\right]Jd\tau \\ p=0-{\epsilon }_0\mu \int \left[E\left(0\right).r\right]Jd\tau \end{gathered}$$ ……. (2)

For a current loop,

$\int Jd\tau \to \int \left|\mathrm{d}\right|=I\int dl=0$

Solve the integral as,

$$\begin{gathered} \int \left[E\left(0\right).r\right]Jd\tau =\int \left[E\left(0\right).r\right]Jd\tau \\ \int \left[E\left(0\right).r\right]Jd\tau =\int \left[E\left(0\right).r\right]ldl \\ \int \left[E\left(0\right).r\right]Jd\tau =\int \left[E\left(0\right).r\right]dl \\ \int \left[E\left(0\right).r\right]Jd\tau =la\times E\left(0\right) \end{gathered}$$

……. (3)

Write the equation for the magnetic dipole.

m = la

Substitute the known value in equation (3).

$\int \left[E\left(0\right).r\right]Jd\tau =m\times E$

From equation (2),

$p=-{\epsilon }_0{\mu }_0\left(m\times E\right)$

Therefore, the field carry momentum is$p=-{\epsilon }_0{\mu }_0\left(m\times E\right)$.