Question

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

Short answer

(a)The electromagnetic momentum in the space between the plates is $p_{\text{em}}={\epsilon }_{0}EBAd\widehat{y}$.

(b) The total impulse delivered to the system during the discharge is $I={\epsilon }_{0}EBAd\widehat{y}$.

Step-by-step solution

Expression for the electromagnetic momentum density:

Write the expression for the electromagnetic momentum density.

${\mathbf{g}}_{\mathbf{em}}\mathbf{=}{\mathbf{\epsilon }}_{\mathbf{0}}\mathbf{(}\mathbf{E}\mathbf{\times }\mathbf{B}\mathbf{)}$

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

Write the electromagnetic momentum density in terms of direction.

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

Determine the electromagnetic momentum in the space between the plates:

(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 $p_{\text{em}}={\epsilon }_{0}EBAd\widehat{y}$.

Determine the total impulse delivered to the system during the discharge:

(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:

$F=I(l\times B)$

Substitute the known value in equation (2).

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

On further solving,

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

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