91Ó°ÊÓ

Write the expression for the electromagnetic momentum density.

${\mathbf{g}}_{\mathbf{em}}\mathbf{=}{\mathbf{Î\mu }}_{\mathbf{0}}\mathbf{(}\mathbf{E}\mathbf{×}\mathbf{B}\mathbf{)}$

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

Write the electromagnetic momentum density in terms of direction.

${\mathit{g}}_{\mathbf{e}\mathbf{m}}\mathbf{=}{\mathit{Î\mu }}_{\mathbf{0}}\mathit{E}\mathit{B}\widehat{\mathbf{y}}$ …… (1)

(a)

Write the expression for the electromagnetic momentum in the space between the plates.

$p_{\text{em}}=g_{\text{em}}V$

Here, V is the volume.

Substitute the known value of equation (1) in the above expression.

$\begin{array}{c}{p}_{\text{em}}={\mathrm{Î\mu }}_{0}EB\left(Ad\right)\widehat{y}\\ p_{\text{em}}={\mathrm{Î\mu }}_{0}EBAd\widehat{y}\end{array}$

Therefore, the electromagnetic momentum in the space between the plates is $p_{\text{em}}={\mathrm{Î\mu }}_{0}EBAd\widehat{y}$.

(b)

Write the expression for the total impulse delivered to the system during the discharge.

$I={∫}_0^{\mathrm{∞}}Fâ‹\dots dt$ …… (2)

Here, F is the magnetic force which is given as:

$F=I(l×B)$

Substitute the known value in equation (2).

$\begin{array}{c}I={∫}_0^{\mathrm{∞}}I(l×B)dt\\ I={∫}_0^{\mathrm{∞}}IBd(\widehat{z}×\widehat{x})dt\\ I=\left(Bd\widehat{y}\right){∫}_0^{\mathrm{∞}}\left(−\frac{dQ}{dt}\right)dt\end{array}$

On further solving,

$\begin{array}{c}I=−\left(Bd\widehat{y}\right){∫}_0^{\mathrm{∞}}dQ\\ I=−\left(Bd\widehat{y}\right){\left(Q\right)}_0^{\mathrm{∞}}\\ I=−\left(Bd\widehat{y}\right)[Q\left(\mathrm{∞}\right)−Q\left(0\right)]\\ I=BQd\widehat{y}\end{array}$

Now, as the original field was,

$\begin{array}{c}E=\frac{\mathrm{σ}}{{\mathrm{Î\mu }}_0}=\frac{Q}{{\mathrm{Î\mu }}_{0}A}\\ Q={\mathrm{Î\mu }}_{0}EA\end{array}$

So, the total impulse delivered to the system during the discharge will be,

$I={\mathrm{Î\mu }}_{0}EBAd\widehat{y}$

Therefore, the total impulse delivered to the system during the discharge is $I={\mathrm{Î\mu }}_{0}EBAd\widehat{y}$.