A charged parallel-plate capacitor (with uniform electric field $E=E\widehat{z}$) is placed in a uniform magnetic field $B=B\widehat{x}$, as shown in Fig. 8.6.

Figure 8.6

(a) Find the electromagnetic momentum in the space between the plates.

(b) Now a resistive wire is connected between the plates, along the z-axis, so that the capacitor slowly discharges. The current through the wire will experience a magnetic force; what is the total impulse delivered to the system, during the discharge?

Write the expression for the electromagnetic momentum density.

$g_{em}={\epsilon }_0\left(E\times B\right)$

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

Write the electromagnetic momentum density in terms of direction.

$g_{em}={\epsilon }_{0}EB\widehat{y}$ …… (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.

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

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

(b)

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

$I={\int }_0^{\infty }F\cdot dt$ …… (2)

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

$F=I\left(l\times B\right)$

Substitute the known value in equation (2).

$\begin{array}{c}I={\int }_0^{\infty }I\left(l\times B\right)dt\\ I={\int }_0^{\infty }IBd\left(\widehat{z}\times \widehat{x}\right)dt\\ I=\left(Bd\widehat{y}\right){\int }_0^{\infty }\left(-\frac{dQ}{dt}\right)dt\end{array}$

On further solving,

$\begin{array}{c}I=-\left(Bd\widehat{y}\right){\int }_0^{\infty }dQ\\ I=-\left(Bd\widehat{y}\right){\left(Q\right)}_0^{\infty }\\ I=-\left(Bd\widehat{y}\right)\left[Q\left(\infty \right)-Q\left(0\right)\right]\\ I=BQd\widehat{y}\end{array}$

Now, as the original field was,

$\begin{array}{c}E=\frac{\sigma }{{\epsilon }_0}=\frac{Q}{{\epsilon }_{0}A}\\ Q={\epsilon }_{0}EA\end{array}$

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

$I={\epsilon }_{0}EBAd\widehat{y}$

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