Chapter 10, Problem 59P

To determine

Find the power densities in both media.

Answer to Problem 59P

The power densities in medium-1 and medium-2 are $\underset{\_}{38.197{\cos }^2\left(\omega t-1.2z\right)a_z-13.75{\cos }^2\left(\omega t+1.2z\right)a_z\text{W}/{\text{m}}^2}$ and $\underset{\_}{24.46{\cos }^2\left(\omega t-0.3z\right)a_z\text{W}/{\text{m}}^2}$, respectively.

Explanation of Solution

Calculation:

Write the expression for incident electric field in medium-1.

$E_i=60\cos \left(\omega t-{\beta }_{1}z\right)a_x\text{V}/\text{m}$        (1)

Consider the expression for phase constant ${\beta }_1$ in medium-1.

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

Rewrite the expression for the given data.

${\beta }_1=\omega \sqrt{{\mu }_{\text{o}}\left(16{\epsilon }_{\text{o}}\right)}\left\{\begin{array}{l}\because \mu ={\mu }_{\text{o}}\\ \because \epsilon =16{\epsilon }_{\text{o}}\end{array}\right\}$

Substitute $90\text{Mrad}/\text{s}$ 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 ${\epsilon }_{\text{o}}$.

$\begin{array}{c}{\beta }_1=\left(90\text{Mrad}/\text{s}\right)\sqrt{\left(4\pi \times {10}^{-7}\text{H}/\text{m}\right)\left(16\right)\left(\frac{{10}^{-9}}{36\pi }\text{F}/\text{m}\right)}\\ =\left(90\times {10}^6\right)\left(\frac{4}{3\times {10}^8}\right)\text{rad}/\text{m}\\ =1.2\text{rad}/\text{m}\end{array}$

Substitute $1.2\text{rad}/\text{m}$ for ${\beta }_1$ in Equation (1).

$E_i=60\cos \left(\omega t-1.2z\right)a_x\text{V}/\text{m}$

Consider the expression to find the total electric field in medium-1.

$E_1=E_i+E_r$        (2)

Here,

$E_r$ is the reflected electric field.

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

$E_r=E_{ro}\cos \left(\omega t+1.2z\right)a_x\text{V}/\text{m}$        (3)

Here,

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

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

$E_{ro}=\Gamma E_{io}$        (4)

Here,

$E_{io}$ is the magnitude of incident electric field at $z=0$, which is $60\text{V}/\text{m}$, and

$\Gamma$ is the reflection coefficient.

Write the expression to find the reflection coefficient.

$\Gamma =\frac{{\eta }_2-{\eta }_1}{{\eta }_2+{\eta }_1}$        (5)

Here,

${\eta }_1$ is the intrinsic impedance for medium-1 and

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

As medium-2 is free space, the intrinsic impedance is $120\pi \Omega$.

${\eta }_2=120\pi \Omega$

Find the intrinsic impedance for medium-1.

$\begin{array}{c}{\eta }_1=\sqrt{\frac{\mu }{\epsilon }}\\ =\sqrt{\frac{{\mu }_{\text{o}}}{16{\epsilon }_{\text{o}}}}\left\{\begin{array}{l}\because \mu ={\mu }_{\text{o}}\\ \because \epsilon =16{\epsilon }_{\text{o}}\end{array}\right\}\\ =\frac{1}{4}\sqrt{\frac{{\mu }_{\text{o}}}{{\epsilon }_{\text{o}}}}\\ =\frac{1}{4}\left(120\pi \Omega \right)\left\{\because \sqrt{\frac{{\mu }_{\text{o}}}{{\epsilon }_{\text{o}}}}\cong 120\pi \Omega \right\}\end{array}$

${\eta }_1=30\pi \Omega$

Substitute $30\pi \Omega$ for ${\eta }_1$ and $120\pi \Omega$ for ${\eta }_2$ in Equation (5).

$\begin{array}{c}\Gamma =\frac{120\pi \Omega -30\pi \Omega }{120\pi \Omega +30\pi \Omega }\\ =\frac{3}{5}\end{array}$

Substitute $\frac{3}{5}$ for $\Gamma$ and $60\text{V}/\text{m}$ for $E_{io}$ in Equation (4).

$\begin{array}{c}{E}_{ro}=\left(\frac{3}{5}\right)\left(60\text{V}/\text{m}\right)\\ =36\text{V}/\text{m}\end{array}$

Substitute $36\text{V}/\text{m}$ for $E_{ro}$ in Equation (3).

$E_r=36\cos \left(\omega t+1.2z\right)a_x\text{V}/\text{m}$

Consider the expression to find the incident magnetic field in medium-1.

$H_i=\frac{E_{io}}{{\eta }_1}\cos \left(\omega t-1.2z\right)a_{H_i}$        (6)

Find the vector $a_{H_i}$.

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

Substitute $60\text{V}/\text{m}$ for $E_{io}$, $30\pi \Omega$ for ${\eta }_1$, and $a_y$ for $a_{H_i}$ in Equation (6).

$\begin{array}{c}{H}_i=\frac{60\text{V}/\text{m}}{30\pi \Omega }\cos \left(\omega t-1.2z\right)a_y\\ =0.63661\cos \left(\omega t-1.2z\right)a_y\text{A}/\text{m}\end{array}$

Consider the expression to find the reflected magnetic field in medium-1.

$H_r=\frac{E_{ro}}{{\eta }_1}\cos \left(\omega t+1.2z\right)a_{H_r}$        (7)

Find the vector $a_{H_r}$.

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

Substitute $36\text{V}/\text{m}$ for $E_{io}$, $30\pi \Omega$ for ${\eta }_1$, and $-a_y$ for $a_{H_r}$ in Equation (7).

$\begin{array}{c}{H}_r=\frac{36\text{V}/\text{m}}{30\pi \Omega }\cos \left(\omega t+1.2z\right)\left(-a_y\right)\\ =-0.38197\cos \left(\omega t+1.2z\right)a_y\text{A}/\text{m}\end{array}$

From Equation (1), the total electric field in medium-1 is the sum of incident and reflected electric fields. Similarly, the total magnetic field in medium-1 is the sum of incident and reflected magnetic fields.

Write the expression to find the power density in medium-1.

$P_{\text{1}}=E_i\times H_i+E_r\times H_r$        (8)

Find the cross product $E_i\times H_i$.

$\begin{array}{c}{E}_i\times H_i=\left|\begin{array}{ccc}{a}_x& a_y& a_z\\ 60\cos \left(\omega t-1.2z\right)& 0& 0\\ 0& 0.63661\cos \left(\omega t-1.2z\right)& 0\end{array}\right|\\ =\left(\begin{array}{l}\left(0-0\right)a_x-\left(0-0\right)a_y\\ +\left\{\left[60\cos \left(\omega t-1.2z\right)\right]\left[0.63661\cos \left(\omega t-1.2z\right)\right]-0\right\}{a}_z\end{array}\right)\text{W}/{\text{m}}^2\\ =38.197{\cos }^2\left(\omega t-1.2z\right)a_z\text{W}/{\text{m}}^2\end{array}$

Find the cross product $E_r\times H_r$.

$\begin{array}{c}{E}_r\times H_r=\left|\begin{array}{ccc}{a}_x& a_y& a_z\\ 36\cos \left(\omega t+1.2z\right)& 0& 0\\ 0& -0.38197\cos \left(\omega t+1.2z\right)& 0\end{array}\right|\\ =\left(\begin{array}{l}\left(0-0\right)a_x-\left(0-0\right)a_y\\ +\left\{\left[36\cos \left(\omega t+1.2z\right)\right]\left[-0.38197\cos \left(\omega t+1.2z\right)\right]-0\right\}{a}_z\end{array}\right)\text{W}/{\text{m}}^2\\ =-13.75{\cos }^2\left(\omega t+1.2z\right)a_z\text{W}/{\text{m}}^2\end{array}$

Substitute $38.197{\cos }^2\left(\omega t-1.2z\right)a_z\text{W}/{\text{m}}^2$ for $E_i\times H_i$ and $-13.75{\cos }^2\left(\omega t+1.2z\right)a_z\text{W}/{\text{m}}^2$ for $E_r\times H_r$ in Equation (8).

$\begin{array}{c}{P}_{\text{1}}=\left(38.197{\cos }^2\left(\omega t-1.2z\right)a_z\text{W}/{\text{m}}^2\right)+\left(-13.75{\cos }^2\left(\omega t+1.2z\right)a_z\text{W}/{\text{m}}^2\right)\\ =38.197{\cos }^2\left(\omega t-1.2z\right)a_z-13.75{\cos }^2\left(\omega t+1.2z\right)a_z\text{W}/{\text{m}}^2\end{array}$

Consider the expression for transmitted electric field in medium-2.

$E_t=E_{to}\cos \left(\omega t-{\beta }_{2}z\right)a_x$        (9)

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

${\beta }_2=\omega \sqrt{{\mu }_{\text{o}}{\epsilon }_{\text{o}}}$

Substitute $90\text{Mrad}/\text{s}$ 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 ${\epsilon }_{\text{o}}$.

$\begin{array}{c}{\beta }_2=\left(90\text{Mrad}/\text{s}\right)\sqrt{\left(4\pi \times {10}^{-7}\text{H}/\text{m}\right)\left(\frac{{10}^{-9}}{36\pi }\text{F}/\text{m}\right)}\\ =\left(90\times {10}^6\right)\left(\frac{1}{3\times {10}^8}\right)\text{rad}/\text{m}\\ =0.3\text{rad}/\text{m}\end{array}$

Substitute $0.3\text{rad}/\text{m}$ for ${\beta }_2$ in Equation (9).

$E_t=E_{to}\cos \left(\omega t-0.3z\right)a_x$        (10)

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

$E_{to}=\tau E_{io}$        (11)

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 $30\pi \Omega$ for ${\eta }_2$.

$\begin{array}{c}\tau =\frac{\left(2\right)\left(120\pi \Omega \right)}{120\pi \Omega +30\pi \Omega }\\ =\frac{8}{5}\end{array}$

Substitute $\frac{8}{5}$ for $\tau$ and $60\text{V}/\text{m}$ for $E_{io}$ in Equation (11).

$\begin{array}{c}{E}_{to}=\left(\frac{8}{5}\right)\left(60\text{V}/\text{m}\right)\\ =96\text{V}/\text{m}\end{array}$

Substitute $96\text{V}/\text{m}$ for $E_{to}$ in Equation (10).

$E_t=96\cos \left(\omega t-0.3z\right)a_x\text{V}/\text{m}$

Consider the expression to find the transmitted magnetic field in medium-2.

$H_t=\frac{E_{to}}{{\eta }_2}\cos \left(\omega t-0.3z\right)a_{H_t}$        (12)

Find the vector $a_{H_t}$.

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

Substitute $96\text{V}/\text{m}$ for $E_{to}$, $120\pi \Omega$ for ${\eta }_2$, and $a_y$ for $a_{H_t}$ in Equation (12).

$\begin{array}{c}{H}_t=\frac{96\text{V}/\text{m}}{120\pi \Omega }\cos \left(\omega t-0.3z\right)\left(a_y\right)\\ =0.25464\cos \left(\omega t-0.3z\right)a_y\text{A}/\text{m}\end{array}$

Write the expression to find the power density in medium-2.

$\begin{array}{c}{P}_2=E_t\times H_t\\ =\left|\begin{array}{ccc}{a}_x& a_y& a_z\\ 96\cos \left(\omega t-0.3z\right)& 0& 0\\ 0& 0.25464\cos \left(\omega t-0.3z\right)& 0\end{array}\right|\\ =\left(\begin{array}{l}\left(0-0\right)a_x-\left(0-0\right)a_y\\ +\left\{\left[96\cos \left(\omega t-0.3z\right)\right]\left[0.25464\cos \left(\omega t-0.3z\right)\right]-0\right\}{a}_z\end{array}\right)\text{W}/{\text{m}}^2\\ \cong 24.46{\cos }^2\left(\omega t-0.3z\right)a_z\text{W}/{\text{m}}^2\end{array}$

Conclusion:

Thus, the power densities in medium-1 and medium-2 are $\underset{\_}{38.197{\cos }^2\left(\omega t-1.2z\right)a_z-13.75{\cos }^2\left(\omega t+1.2z\right)a_z\text{W}/{\text{m}}^2}$ and $\underset{\_}{24.46{\cos }^2\left(\omega t-0.3z\right)a_z\text{W}/{\text{m}}^2}$, respectively.