Chapter 10, Problem 60P

(a)

To determine

Find the wavelength and wave frequency of the wave in air and the transmitted wave in the dielectric medium.

Answer to Problem 60P

The wavelength and wave frequency of the wave in air are $\underset{\_}{6.283\text{m}}$ and $\underset{\_}{3\times {10}^8\text{rad}/\text{s}}$, respectively.

The wavelength and wave frequency of the transmitted wave in the lossless dielectric medium are $\underset{\_}{3.6276\text{m}}$ and $\underset{\_}{3\times {10}^8\text{rad}/\text{s}}$, respectively.

Explanation of Solution

Calculation:

Write the expression for given incident electric field in medium-1 (air).

$E_i=10\cos \left(\omega t-1z\right)a_y\text{V}/\text{m}$        (1)

From the expression, the values of magnitude of incident electric field at $z=0$ and phase constant of medium-1 are $10\text{V}/\text{m}$ and $1\text{rad}/\text{m}$, respectively.

$\begin{array}{l}{E}_{io}=10\text{V}/\text{m}\\ {\beta }_1=1\text{rad}/\text{m}\end{array}$

Consider the expression to find the wavelength of the wave in air (medium-1).

${\lambda }_1=\frac{2\pi }{{\beta }_1}$

Substitute $1\text{rad}/\text{m}$ for ${\beta }_1$.

$\begin{array}{c}{\lambda }_1=\frac{2\pi }{1\text{rad}/\text{m}}\\ \cong 6.283\text{m}\end{array}$

Consider the expression for phase constant of medium-1.

${\beta }_1=\omega \sqrt{\mu \epsilon }$

Rewrite the expression for the given data.

${\beta }_1=\omega \sqrt{{\mu }_{\text{o}}{\epsilon }_{\text{o}}}\left\{\begin{array}{l}\because \mu ={\mu }_{\text{o}}\\ \because \epsilon ={\epsilon }_{\text{o}}\end{array}\right\}$

Rearrange the expression for wave frequency.

$\omega =\frac{{\beta }_1}{\sqrt{{\mu }_{\text{o}}{\epsilon }_{\text{o}}}}$

Substitute $1\text{rad}/\text{m}$ for ${\beta }_1$, $4\pi \times {10}^{-7}\text{H}/\text{m}$ for ${\mu }_{\text{o}}$, and $\frac{{10}^{-9}}{36\pi }\text{F}/\text{m}$ for ${\epsilon }_{\text{o}}$.

$\begin{array}{c}\omega =\frac{1\text{rad}/\text{m}}{\sqrt{\left(4\pi \times {10}^{-7}\text{H}/\text{m}\right)\left(\frac{{10}^{-9}}{36\pi }\text{F}/\text{m}\right)}}\\ =3\times {10}^8\text{rad}/\text{s}\end{array}$

The wave frequency is same for both the waves in air and the lossless dielectric medium.

Consider the expression for phase constant of medium-2 (lossless dielectric medium).

${\beta }_2=\omega \sqrt{\mu \epsilon }$

Rewrite the expression for the given data.

${\beta }_2=\omega \sqrt{{\mu }_{\text{o}}\left(3{\epsilon }_{\text{o}}\right)}\left\{\begin{array}{l}\because \mu ={\mu }_{\text{o}}\\ \because \epsilon =3{\epsilon }_{\text{o}}\end{array}\right\}$

Substitute $3\times {10}^8\text{rad}/\text{s}$ for $\omega$, $4\pi \times {10}^{-7}\text{H}/\text{m}$ for ${\mu }_{\text{o}}$, and $\frac{{10}^{-9}}{36\pi }\text{F}/\text{m}$ for ${\epsilon }_{\text{o}}$.

$\begin{array}{c}{\beta }_2=\left(3\times {10}^8\text{rad}/\text{s}\right)\sqrt{\left(4\pi \times {10}^{-7}\text{H}/\text{m}\right)\left(3\right)\left(\frac{{10}^{-9}}{36\pi }\text{F}/\text{m}\right)}\\ =\sqrt{3}\text{rad}/\text{m}\end{array}$

Consider the expression to find the wavelength of the transmitted wave in the lossless dielectric medium (medium-2).

${\lambda }_2=\frac{2\pi }{{\beta }_2}$

Substitute $\sqrt{3}\text{rad}/\text{m}$ for ${\beta }_2$.

$\begin{array}{c}{\lambda }_2=\frac{2\pi }{\sqrt{3}\text{rad}/\text{m}}\\ \cong 3.6276\text{m}\end{array}$

Conclusion:

Thus, the wavelength and wave frequency of the wave in air are $\underset{\_}{6.283\text{m}}$ and $\underset{\_}{3\times {10}^8\text{rad}/\text{s}}$, respectively and the wavelength and wave frequency of the transmitted wave in the lossless dielectric medium are $\underset{\_}{3.6276\text{m}}$ and $\underset{\_}{3\times {10}^8\text{rad}/\text{s}}$, respectively.

(b)

To determine

Find the incident magnetic field.

Answer to Problem 60P

The incident magnetic field is $\underset{\_}{-26.5\cos \left(\omega t-1z\right)a_x\text{mA}/\text{m}}$.

Explanation of Solution

Calculation:

Consider the expression to find the incident magnetic field.

$H_i=\frac{E_{io}}{{\eta }_1}\cos \left(\omega t-1z\right)a_{H_i}$        (2)

Here,

${\eta }_1$ is the intrinsic impedance of the medium-1.

As the medium-1 is free space, the value of ${\eta }_1$ is $120\pi \Omega$.

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

Find the vector $a_{H_i}$.

$\begin{array}{c}{a}_{H_i}=a_k\times a_{E_i}\\ =a_z\times a_y\left\{\begin{array}{l}\because a_{E_i}=a_y\\ \because a_k=a_z\end{array}\right\}\\ =-a_x\left\{\because a_z\times a_y=-a_x\right\}\end{array}$

Substitute $10\text{V}/\text{m}$ for $E_{io}$, $120\pi \Omega$ for ${\eta }_1$, and $\left(-a_x\right)$ for $a_{H_i}$ in Equation (2).

$\begin{array}{c}{H}_i=\frac{10\text{V}/\text{m}}{120\pi \Omega }\cos \left(\omega t-1z\right)\left(-a_x\right)\\ =-26.5\cos \left(\omega t-1z\right)a_x\text{mA}/\text{m}\end{array}$

Conclusion:

Thus, the incident magnetic field is $\underset{\_}{-26.5\cos \left(\omega t-1z\right)a_x\text{mA}/\text{m}}$.

(c)

To determine

Find the reflection and transmission coefficients.

Answer to Problem 60P

The reflection and transmission coefficients are $\underset{\_}{-0.268}$ and $\underset{\_}{0.732}$, respectively.

Explanation of Solution

Calculation:

Consider the expression to find the reflection coefficient.

$\Gamma =\frac{{\eta }_2-{\eta }_1}{{\eta }_2+{\eta }_1}$        (3)

Here,

${\eta }_2$ is the intrinsic impedance for medium-2.

Find the intrinsic impedance for medium-2.

$\begin{array}{c}{\eta }_2=\sqrt{\frac{\mu }{\epsilon }}\\ =\sqrt{\frac{{\mu }_{\text{o}}}{3{\epsilon }_{\text{o}}}}\left\{\begin{array}{l}\because \mu ={\mu }_{\text{o}}\\ \because \epsilon =3{\epsilon }_{\text{o}}\end{array}\right\}\\ =\frac{1}{\sqrt{3}}\sqrt{\frac{{\mu }_{\text{o}}}{{\epsilon }_{\text{o}}}}\\ =\frac{1}{\sqrt{3}}\left(120\pi \Omega \right)\end{array}$

${\eta }_2=69.2820\pi \Omega$

Substitute $120\pi \Omega$ for ${\eta }_1$ and $69.2820\pi \Omega$ for ${\eta }_2$ in Equation (3).

$\begin{array}{c}\Gamma =\frac{69.2820\pi \Omega -120\pi \Omega }{69.2820\pi \Omega +120\pi \Omega }\\ \cong -0.268\end{array}$

Consider for the expression for transmission coefficient.

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

Substitute $120\pi \Omega$ for ${\eta }_1$ and $69.2820\pi \Omega$ for ${\eta }_2$.

$\begin{array}{c}\tau =\frac{\left(2\right)\left(69.2820\pi \Omega \right)}{69.2820\pi \Omega +120\pi \Omega }\\ \cong 0.732\end{array}$

Conclusion:

Thus, the reflection and transmission coefficients are $\underset{\_}{-0.268}$ and $\underset{\_}{0.732}$, respectively.

(d)

To determine

Find the total electric field and time-average power in both the regions.

Answer to Problem 60P

The total electric field in the air and lossless dielectric medium are $\underset{\_}{10\cos \left(\omega t-1z\right)a_y-2.68\cos \left(\omega t+1z\right)a_y\text{V}/\text{m}}$ and $\underset{\_}{7.32\cos \left(\omega t-1z\right)a_y\text{V}/\text{m}}$, respectively.

The time-average power density in the air and lossless dielectric medium are $\underset{\_}{0.1231a_z\text{W}/{\text{m}}^2}$ and $\underset{\_}{0.1231a_z\text{W}/{\text{m}}^2}$, respectively.

Explanation of Solution

Calculation:

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

$E_1=E_i+E_r$        (4)

Here,

$E_r$ is the reflected electric field.

Consider the expression for reflected electric field in medium-1.

$E_r=E_{ro}\cos \left(\omega t+1z\right)a_y$        (5)

Here,

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

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

$E_{ro}=\Gamma E_{io}$

Substitute $-0.268$ for $\Gamma$ and $10\text{V}/\text{m}$ for $E_{io}$.

$\begin{array}{c}{E}_{ro}=\left(-0.268\right)\left(10\text{V}/\text{m}\right)\\ =-2.68\text{V}/\text{m}\end{array}$

Substitute $-2.68\text{V}/\text{m}$ for $E_{ro}$ in Equation (5).

$\begin{array}{c}{E}_r=\left(-2.68\text{V}/\text{m}\right)\cos \left(\omega t+1z\right)a_y\\ =-2.68\cos \left(\omega t+1z\right)a_y\text{V}/\text{m}\end{array}$

Substitute $\left[10\cos \left(\omega t-1z\right)a_y\text{V}/\text{m}\right]$ for $E_i$ and $\left[-2.68\cos \left(\omega t+1z\right)a_y\text{V}/\text{m}\right]$ for $E_r$ in Equation (4).

$\begin{array}{c}{E}_1=\left[10\cos \left(\omega t-1z\right)a_y\text{V}/\text{m}\right]+\left[-2.68\cos \left(\omega t+1z\right)a_y\text{V}/\text{m}\right]\\ =10\cos \left(\omega t-1z\right)a_y-2.68\cos \left(\omega t+1z\right)a_y\text{V}/\text{m}\end{array}$

The total electric field in the medium-2 is the transmitted electric field component.

Consider the expression for transmitted electric field.

$E_t=E_{to}\cos \left(\omega t-{\beta }_{2}z\right)a_y$        (6)

Here,

$E_{to}$ is the magnitude of transmitted electric field at $z=0$ and

${\beta }_2$ is the phase constant in medium-2.

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

$E_{to}=\tau E_{io}$

Substitute 0.732 for $\tau$ and $10\text{V}/\text{m}$ for $E_{io}$.

$\begin{array}{c}{E}_{to}=\left(0.732\right)\left(10\text{V}/\text{m}\right)\\ =7.32\text{V}/\text{m}\end{array}$

Substitute $7.32\text{V}/\text{m}$ for $E_{to}$ and $1.732\text{rad}/\text{m}$ for ${\beta }_2$ in Equation (6).

$\begin{array}{c}{E}_t=\left(7.32\text{V}/\text{m}\right)\cos \left(\omega t-1.732z\right)a_y\\ =7.32\cos \left(\omega t-1.732z\right)a_y\text{V}/\text{m}\end{array}$

Consider the expression to find the time-average power in region 1 (medium-1).

$P_{\text{ave1}}=\frac{E_{io}^2}{2{\eta }_1}\left(a_k\right)+\frac{E_{ro}^2}{2{\eta }_1}\left(-a_k\right)$        (7)

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

$a_k=a_z$

Substitute $10\text{V}/\text{m}$ for $E_{io}$, $-2.68\text{V}/\text{m}$ for $E_{ro}$, $120\pi \Omega$ for ${\eta }_1$, and $a_z$ for $a_k$ in Equation (7).

$\begin{array}{c}{P}_{\text{ave1}}=\frac{{\left(10\text{V}/\text{m}\right)}^2}{\left(2\right)\left(120\pi \Omega \right)}\left(a_z\right)+\frac{{\left(-2.68\text{V}/\text{m}\right)}^2}{\left(2\right)\left(120\pi \Omega \right)}\left(-a_z\right)\\ \cong 0.1231a_z\text{W}/{\text{m}}^2\end{array}$

Consider the expression to find the time-average power in region 2 (medium-2).

$P_{\text{ave2}}=\frac{E_{to}^2}{2{\eta }_2}{a}_k$

Substitute $7.32\text{V}/\text{m}$ for $E_{to}$, $69.2820\pi \Omega$ for ${\eta }_2$, and $a_z$ for $a_k$.

$\begin{array}{c}{P}_{\text{ave2}}=\frac{{\left(7.32\text{V}/\text{m}\right)}^2}{\left(2\right)\left(69.2820\pi \Omega \right)}{a}_z\\ \cong 0.1231a_z\text{W}/{\text{m}}^2\end{array}$

Conclusion:

Thus, the total electric field in the air and lossless dielectric medium are $\underset{\_}{10\cos \left(\omega t-1z\right)a_y-2.68\cos \left(\omega t+1z\right)a_y\text{V}/\text{m}}$ and $\underset{\_}{7.32\cos \left(\omega t-1z\right)a_y\text{V}/\text{m}}$, respectively and the time-average power density in the air and lossless dielectric medium are $\underset{\_}{0.1231a_z\text{W}/{\text{m}}^2}$ and $\underset{\_}{0.1231a_z\text{W}/{\text{m}}^2}$, respectively.