91Ó°ÊÓ

Write the expression for electromagnetic momentum per unit volume.

$\mathbf{p}\mathbf{=}{\mathbf{Î\mu }}_{\mathbf{0}}\mathbf{∫}\left(E×B\right)\mathbf{»åÏ„}$ …… (1)

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

Substitute the known values in equation (1).

$p={\mathrm{Î\mu }}_0∫\left(-∇V\right)×Bd\mathrm{Ï„}$

Use integration parts in the above equation.

$$\begin{gathered} p=-{\mathrm{Î\mu }}_0∫\left[∇×\left(VB\right)-V∇×B\right]d\mathrm{Ï„} \\ p={\mathrm{Î\mu }}_0âˆ\circledR VB×da+{\mathrm{Î\mu }}_0{\mathrm{μ}}_0∫VJd\mathrm{Ï„} \\ p=0+{\mathrm{Î\mu }}_0{\mathrm{μ}}_0∫VJd\mathrm{Ï„} \\ p={\mathrm{Î\mu }}_0{\mathrm{μ}}_0∫VJd\mathrm{Ï„} \end{gathered}$$

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

localid="1657364211619" $$\begin{gathered} p={\mathrm{Î\mu }}_0{\mathrm{μ}}_{0}V\left(0\right)∫Jd\mathrm{Ï„}-{\mathrm{Î\mu }}_0{\mathrm{μ}}_0∫\left[E\left(0\right).r\right]Jd\mathrm{Ï„} \\ p=0-{\mathrm{Î\mu }}_0\mathrm{μ}∫\left[E\left(0\right).r\right]Jd\mathrm{Ï„} \end{gathered}$$ ……. (2)

For a current loop,

$∫Jd\mathrm{τ}→∫\left|\mathrm{d}\right|=I∫dl=0$

Solve the integral as,

$$\begin{gathered} ∫\left[E\left(0\right).r\right]Jd\mathrm{τ}=∫\left[E\left(0\right).r\right]Jd\mathrm{τ} \\ ∫\left[E\left(0\right).r\right]Jd\mathrm{τ}=∫\left[E\left(0\right).r\right]ldl \\ ∫\left[E\left(0\right).r\right]Jd\mathrm{τ}=∫\left[E\left(0\right).r\right]dl \\ ∫\left[E\left(0\right).r\right]Jd\mathrm{τ}=la×E\left(0\right) \end{gathered}$$

……. (3)

Write the equation for the magnetic dipole.

m = la

Substitute the known value in equation (3).

$∫\left[E\left(0\right).r\right]Jd\mathrm{τ}=m×E$

From equation (2),

$p=-{\mathrm{Î\mu }}_0{\mathrm{μ}}_0\left(m×E\right)$

Therefore, the field carry momentum is$p=-{\mathrm{Î\mu }}_0{\mathrm{μ}}_0\left(m×E\right)$.