Chapter 10, Problem 55P

(a)

To determine

Find the total electric field in the air.

Answer to Problem 55P

The total electric field in the air is $\underset{\_}{-1.508\sin \left(\omega t-5x\right)a_z+0.503\sin \left(\omega t+5x\right)a_z\text{kV}/\text{m}}$.

Explanation of Solution

Calculation:

Write the expression for given magnetic field in the air.

$H=4\sin \left(\omega t-5x\right)a_y\text{A}/\text{m}$        (1)

Consider the expression to find the total electric field in the air.

$E_1=E_i+E_r$        (2)

Here,

$E_i$ is the incident electric field and

$E_r$ is the reflected electric field.

Consider the expression for incident electric field for the given data.

$E_i=E_{io}\sin \left(\omega t-5x\right)a_E\text{A}/\text{m}$        (3)

Here,

$E_{io}$ is the magnitude of incident electric field at $x=0$.

Find the vector $a_E$.

$\begin{array}{c}{a}_E=-\left(a_k\times a_H\right)\\ =-\left(a_x\times a_y\right)\left\{\begin{array}{l}\because a_H=a_y\\ \because a_k=a_x\end{array}\right\}\\ =-a_z\left\{\because a_x\times a_y=a_z\right\}\end{array}$

Consider the expression to find magnitude of incident electric field at $x=0$.

$E_{io}=H_{io}{\eta }_1$        (4)

Here,

$H_{io}$ is the magnitude of the given magnetic field, which is $4\text{A}/\text{m}$ and

${\eta }_1$ is the intrinsic impedance of medium-1 (free space).

As the medium-1 is free space (air), the intrinsic impedance is $120\pi \Omega$.

${\eta }_1=120\pi \Omega$

From Equation (4), substitute $\left(H_{io}{\eta }_1\right)$ for $E_{io}$ and $-a_z$ for $a_E$ in Equation (3)

$E_i=\left(H_{io}{\eta }_1\right)\sin \left(\omega t-5x\right)\left(-a_z\right)\text{A}/\text{m}$

Substitute $4\text{A}/\text{m}$ for $H_{io}$ and $120\pi \Omega$ for ${\eta }_1$.

$\begin{array}{c}{E}_i=\left(4\text{A}/\text{m}\right)\left(120\pi \Omega \right)\sin \left(\omega t-5x\right)\left(-a_z\right)\\ \cong -1.508\sin \left(\omega t-5x\right)a_z\text{kV}/\text{m}\end{array}$

Consider the expression for reflected electric field for the given data.

$E_r=E_{ro}\sin \left(\omega t+5x\right)a_E\text{A}/\text{m}$        (5)

Here,

$E_{ro}$ is the magnitude of reflected electric field at $x=0$.

Consider the expression for reflected electric field at $x=0$.

$E_{ro}=\Gamma E_{io}$        (6)

Here,

$\Gamma$ is the reflection coefficient.

Write the expression to find the reflection coefficient.

$\Gamma =\frac{{\eta }_2-{\eta }_1}{{\eta }_2+{\eta }_1}$        (7)

Here,

${\eta }_2$ is the intrinsic impedance for medium-2 (plastic region).

Find the intrinsic impedance for medium-2.

$\begin{array}{c}{\eta }_2=\sqrt{\frac{\mu }{\epsilon }}\\ =\sqrt{\frac{{\mu }_{\text{o}}}{4{\epsilon }_{\text{o}}}}\left\{\begin{array}{l}\because \mu ={\mu }_{\text{o}}\\ \because \epsilon =4{\epsilon }_{\text{o}}\end{array}\right\}\\ =\frac{1}{2}\sqrt{\frac{{\mu }_{\text{o}}}{{\epsilon }_{\text{o}}}}\\ =\frac{1}{2}\left(120\pi \Omega \right)\left\{\because \sqrt{\frac{{\mu }_{\text{o}}}{{\epsilon }_{\text{o}}}}\cong 120\pi \Omega \right\}\end{array}$

${\eta }_2=60\pi \Omega$

Substitute $120\pi \Omega$ for ${\eta }_1$ and $60\pi \Omega$ for ${\eta }_2$ in Equation (7).

$\begin{array}{c}\Gamma =\frac{60\pi \Omega -120\pi \Omega }{60\pi \Omega +120\pi \Omega }\\ =-\frac{1}{3}\end{array}$

From Equation (6), substitute $\left(\Gamma E_{io}\right)$ for $E_{ro}$ and $-a_z$ for $a_E$ in Equation (5)

$E_r=\left(\Gamma E_{io}\right)\sin \left(\omega t+5x\right)\left(-a_z\right)\text{A}/\text{m}$

From Equation (4), substitute $\left(H_{io}{\eta }_1\right)$ for $E_{io}$.

$E_r=\left(\Gamma \right)\left(H_{io}{\eta }_1\right)\sin \left(\omega t+5x\right)\left(-a_z\right)\text{A}/\text{m}$

Substitute $\left(-\frac{1}{3}\right)$ for $\Gamma$, $4\text{A}/\text{m}$ for $H_{io}$, and $120\pi \Omega$ for ${\eta }_1$.

$\begin{array}{c}{E}_r=\left(-\frac{1}{3}\right)\left(4\text{A}/\text{m}\right)\left(120\pi \Omega \right)\sin \left(\omega t+5x\right)\left(-a_z\right)\\ \cong 0.503\sin \left(\omega t+5x\right)a_z\text{kV}/\text{m}\end{array}$

Substitute $\left[-1.508\sin \left(\omega t-5x\right)a_z\text{kV}/\text{m}\right]$ for $E_i$ and $\left[0.503\sin \left(\omega t+5x\right)a_z\text{kV}/\text{m}\right]$ for $E_r$ in Equation (2).

$\begin{array}{c}{E}_1=\left[-1.508\sin \left(\omega t-5x\right)a_z\text{kV}/\text{m}\right]+\left[0.503\sin \left(\omega t+5x\right)a_z\text{kV}/\text{m}\right]\\ =-1.508\sin \left(\omega t-5x\right)a_z+0.503\sin \left(\omega t+5x\right)a_z\text{kV}/\text{m}\end{array}$

Conclusion:

Thus, the total electric field in the air is $\underset{\_}{-1.508\sin \left(\omega t-5x\right)a_z+0.503\sin \left(\omega t+5x\right)a_z\text{kV}/\text{m}}$.

(b)

To determine

Find the time-average power density in the plastic region.

Answer to Problem 55P

The time-average power density in the plastic region is $\underset{\_}{2.68a_x\text{kW}/{\text{m}}^2}$.

Explanation of Solution

Calculation:

Consider the expression to find the time-average power density in the plastic region.

$P_{\text{ave}}=\frac{E_{to}^2}{2{\eta }_2}{a}_k$        (8)

Here,

$E_{to}$ is the magnitude of transmitted electric field at $x=0$.

From Equation (1), the vector $a_k$ is $a_x$.

$a_k=a_x$

Consider the expression to find the magnitude of transmitted electric field at $x=0$.

$E_{to}=\tau E_{io}$        (9)

Here,

$\tau$ is the transmission coefficient.

Consider for the expression for transmission coefficient.

$\tau =\frac{2{\eta }_2}{{\eta }_2+{\eta }_1}$

Substitute $120\pi \Omega$ for ${\eta }_1$ and $60\pi \Omega$ for ${\eta }_2$.

$\begin{array}{c}\tau =\frac{\left(2\right)\left(60\pi \Omega \right)}{60\pi \Omega +120\pi \Omega }\\ =\frac{2}{3}\end{array}$

From Equation (4), substitute $\left(H_{io}{\eta }_1\right)$ for $E_{io}$ in Equation (9).

$E_{to}=\tau H_{io}{\eta }_1$

Substitute $\frac{2}{3}$ for $\tau$, $4\text{A}/\text{m}$ for $H_{io}$, and $120\pi \Omega$ for ${\eta }_1$.

$\begin{array}{c}{E}_{to}=\left(\frac{2}{3}\right)\left(4\text{A}/\text{m}\right)\left(120\pi \Omega \right)\\ =320\pi \text{V}/\text{m}\end{array}$

Substitute $320\pi \text{V}/\text{m}$ for $E_{to}$, $60\pi \Omega$ for ${\eta }_2$, and $a_x$ for $a_k$ in Equation (8).

$\begin{array}{c}{P}_{\text{ave}}=\frac{{\left(320\pi \text{V}/\text{m}\right)}^2}{\left(2\right)\left(60\pi \Omega \right)}{a}_x\\ \cong 2.68\text{kW}/{\text{m}}^2\end{array}$

Conclusion:

Thus, the time-average power density in the plastic region is $\underset{\_}{2.68a_x\text{kW}/{\text{m}}^2}$.

(c)

To determine

Find the standing wave ratio.

Answer to Problem 55P

The standing wave ratio is 2.

Explanation of Solution

Calculation:

Consider the expression to find the standing wave ratio.

$s=\frac{1+\left|\Gamma \right|}{1-\left|\Gamma \right|}$

From Part (a), substitute $\left(-\frac{1}{3}\right)$ for $\Gamma$.

$\begin{array}{c}s=\frac{1+\left|-\frac{1}{3}\right|}{1-\left|-\frac{1}{3}\right|}\\ =\frac{1+\frac{1}{3}}{1-\frac{1}{3}}\\ =2\end{array}$

Conclusion:

Thus, the standing wave ratio is 2.