Chapter 10, Problem 47P

(a)

To determine

Find the wave frequency and magnetic field intensity for the given EM wave.

Answer to Problem 47P

The wave frequency and magnetic field intensity for the given EM wave are $\underset{\_}{2.828\times {10}^8\text{rad}/\text{s}}$ and $\underset{\_}{\frac{0.225}{\rho }\sin \left(\omega t-2z\right)a_{\varphi }\text{A}/\text{m}}$, respectively.

Explanation of Solution

Calculation:

Write the expression for electric field component of given EM wave.

$E=\frac{40}{\rho }\sin \left(\omega t-2z\right)a_{\rho }\text{V}/\text{m}$

Rewrite the expression.

$E=\frac{40}{\rho }{e}^{-0z}\sin \left(\omega t-2z\right)a_{\rho }\text{V}/\text{m}$        (1)

Write the general form of the electric field component of EM wave.

$E=E_{\text{o}}{e}^{-0z}\sin \left(\omega t-\beta z\right)a_{\rho }\text{V}/\text{m}$        (2)

Compare Equation (1) with Equation (2) and write the values of $E_{\text{o}}$, $\alpha$, and $\beta$.

$\begin{array}{l}{E}_{\text{o}}=\frac{40}{\rho }\text{V}/\text{m}\\ \alpha =0\text{Np}/\text{m}\\ \beta =2\text{rad}/\text{m}\end{array}$

Consider the expression for phase constant in lossless dielectric medium.

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

Rewrite the expression for the given data.

$\beta =\omega \sqrt{{\mu }_{\text{o}}\left(4.5{\epsilon }_{\text{o}}\right)}\left\{\begin{array}{l}\because \mu ={\mu }_{\text{o}}\\ \because \epsilon =4.5{\epsilon }_{\text{o}}\end{array}\right\}$

Rearrange the expression for wave frequency.

$\omega =\frac{\beta }{\sqrt{{\mu }_{\text{o}}\left(4.5{\epsilon }_{\text{o}}\right)}}$

Substitute $4\pi \times {10}^{-7}\text{H}/\text{m}$ for ${\mu }_{\text{o}}$, $\frac{{10}^{-9}}{36\pi }\text{F}/\text{m}$ for ${\epsilon }_{\text{o}}$, and $2\text{rad}/\text{m}$ for $\beta$.

$\begin{array}{c}\omega =\frac{2\text{rad}/\text{m}}{\sqrt{\left(4\pi \times {10}^{-7}\text{H}/\text{m}\right)\left(4.5\right)\left(\frac{{10}^{-9}}{36\pi }\text{F}/\text{m}\right)}}\\ \cong 2.828\times {10}^8\text{rad}/\text{s}\end{array}$

Find the direction of wave propagation of magnetic field for the given EM wave.

$\begin{array}{c}{a}_H=a_k\times a_E\\ =a_z\times a_{\rho }\left\{\begin{array}{l}\because a_E=a_{\rho }\\ \because a_k=a_z\end{array}\right\}\\ =a_{\varphi }\left\{\because a_z\times a_{\rho }=a_{\varphi }\right\}\end{array}$

Write the expression to find the magnetic field $H$ for the given EM wave.

$H=\frac{E_{\text{o}}}{\left|\eta \right|}{e}^{-\alpha z}\sin \left(\omega t-\beta z+{\theta }_{\eta }\right)a_{\varphi }$        (3)

As $\alpha =0\text{Np}/\text{m}$, the value of phase angle of the intrinsic impedance is $0°$.

${\theta }_{\eta }=0°$

Consider the expression to find the magnitude of the intrinsic impedance in the lossless dielectric medium.

$\left|\eta \right|=\sqrt{\frac{\mu }{\epsilon }}$

Rewrite the expression for the given data.

$\left|\eta \right|=\sqrt{\frac{{\mu }_{\text{o}}}{4.5{\epsilon }_{\text{o}}}}$

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

$\begin{array}{c}\left|\eta \right|=\sqrt{\frac{4\pi \times {10}^{-7}\text{H}/\text{m}}{\left(4.5\right)\left(\frac{{10}^{-9}}{36\pi }\text{F}/\text{m}\right)}}\\ \cong 177.7\Omega \end{array}$

Substitute $\frac{40}{\rho }\text{V}/\text{m}$ for $E_{\text{o}}$, $177.7\Omega$ for $\left|\eta \right|$, $2.828\times {10}^8\text{rad}/\text{s}$ for $\omega$, $0\text{Np}/\text{m}$ for $\alpha$, $2\text{rad}/\text{m}$ for $\beta$, and $0°$ for ${\theta }_{\eta }$.

$\begin{array}{c}H=\frac{\frac{40}{\rho }\text{V}/\text{m}}{177.7\Omega }{e}^{-\left(0\text{Np}/\text{m}\right)z}\sin \left[\left(2.828\times {10}^8\text{rad}/\text{s}\right)t-\left(2\text{rad}/\text{m}\right)z+0°\right]a_{\varphi }\\ \cong \frac{0.225}{\rho }\sin \left(\omega t-2z\right)a_{\varphi }\text{A}/\text{m}\end{array}$

Conclusion:

Thus, the wave frequency and magnetic field intensity for the given EM wave are $\underset{\_}{2.828\times {10}^8\text{rad}/\text{s}}$ and $\underset{\_}{\frac{0.225}{\rho }\sin \left(\omega t-2z\right)a_{\varphi }\text{A}/\text{m}}$, respectively.

(b)

To determine

Find the Poynting vector for the given EM wave.

Answer to Problem 47P

The Poynting vector for the given EM wave is $\underset{\_}{\frac{9}{{\rho }^2}{\sin }^2\left(\omega t-2z\right)a_z\text{W}/{\text{m}}^2}$.

Explanation of Solution

Calculation:

Consider the expression for pointing vector.

$P=E\times H$

Use electric field intensity $\left(E\right)$ and magnetic field intensity $\left(H\right)$ and perform the cross product.

$\begin{array}{c}P=\left|\begin{array}{ccc}{a}_z& a_{\rho }& a_{\varphi }\\ 0& \frac{40}{\rho }\sin \left(\omega t-2z\right)& 0\\ 0& 0& \frac{0.225}{\rho }\sin \left(\omega t-2z\right)\end{array}\right|\\ =\left(\begin{array}{l}\left\{\left[\frac{40}{\rho }\sin \left(\omega t-2z\right)\right]\left[\frac{0.225}{\rho }\sin \left(\omega t-2z\right)\right]-0\right\}{a}_z\\ -\left(0-0\right)a_{\rho }+\left(0-0\right)a_{\varphi }\end{array}\right)\text{W}/{\text{m}}^2\\ =\frac{9}{{\rho }^2}{\sin }^2\left(\omega t-2z\right)a_z\text{W}/{\text{m}}^2\end{array}$

Conclusion:

Thus, the Poynting vector for the given EM wave is $\underset{\_}{\frac{9}{{\rho }^2}{\sin }^2\left(\omega t-2z\right)a_z\text{W}/{\text{m}}^2}$.

(c)

To determine

Find the total time-average power crossing the given surface.

Answer to Problem 47P

The total time-average power crossing the given surface is $\underset{\_}{11.46\text{W}}$.

Explanation of Solution

Calculation:

Write the expression for total time-average Poynting vector.

$P_{\text{ave}}=\frac{1}{2}\mathrm{Re}\left(E\times H^{\ast }\right)$

From Part (a) and Part (b), substitute $\left[\frac{40}{\rho }{e}^{-0z}\sin \left(\omega t-2z\right)a_{\rho }\text{V}/\text{m}\right]$ for $E$ and $\left[\frac{0.225}{\rho }{e}^{0z}\sin \left(\omega t-2z\right)a_{\varphi }\text{A}/\text{m}\right]$ for $H^{\ast }$.

$P_{\text{ave}}=\frac{1}{2}\mathrm{Re}\left\{\left[\frac{40}{\rho }{e}^{-0z}\sin \left(\omega t-2z\right)a_{\rho }\text{V}/\text{m}\right]\times \left[\frac{0.225}{\rho }{e}^{0z}\sin \left(\omega t-2z\right)a_{\varphi }\text{A}/\text{m}\right]\right\}$

Simplify the expression.

$\begin{array}{c}{P}_{\text{ave}}=\frac{1}{2}\mathrm{Re}\left|\begin{array}{ccc}{a}_z& a_{\rho }& a_{\varphi }\\ 0& \frac{40}{\rho }{e}^{-0z}\sin \left(\omega t-2z\right)& 0\\ 0& 0& \frac{0.225}{\rho }{e}^{0z}\sin \left(\omega t-2z\right)\end{array}\right|\\ =\frac{1}{2}\mathrm{Re}\left\{\frac{9}{{\rho }^2}{\sin }^2\left(\omega t-2z\right)a_z\right\}\text{W}/{\text{m}}^2\\ =\frac{4.5}{{\rho }^2}{a}_z\text{W}/{\text{m}}^2\end{array}$

Consider the expression to find the average power crossing the surface $z=1\text{m},2\text{mm}<\rho <3\text{mm},0<\varphi <2\pi$.

$P_{\text{ave}}={\displaystyle \underset{S}{\int }{P}_{\text{ave}}\cdot dS}$        (4)

Here,

$dS=\rho d\varphi d\rho a_z$.

Substitute $\frac{4.5}{{\rho }^2}{a}_z\text{W}/{\text{m}}^2$ for $P_{\text{ave}}$ and $\rho d\varphi d\rho a_z$ for $dS$ in Equation (4).

$P_{\text{ave}}={\displaystyle \underset{S}{\int }\left[\frac{4.5}{{\rho }^2}{a}_z\text{W}/{\text{m}}^2\right]\cdot \left[\rho d\varphi d\rho a_z\right]}$

Apply the limits and rewrite the expression.

$\begin{array}{c}{P}_{\text{ave}}=\left(4.5\right){\displaystyle \underset{\rho =2\text{mm}}{\overset{3\text{mm}}{\int }}{\displaystyle \underset{\varphi =0}{\overset{2\pi }{\int }}\frac{1}{\rho }d\rho d\varphi }}\text{W}\\ =\left(4.5\right){\displaystyle \underset{\varphi =0}{\overset{2\pi }{\int }}{\left[\ln \left(\rho \right)\right]}_{2\text{mm}}^{3\text{mm}}d\varphi }\text{W}\\ =\left(4.5\right)\left[\ln \left(\frac{3}{2}\right)\right]\left(2\pi \right)\text{W}\\ \cong 11.46\text{W}\end{array}$

Conclusion:

Thus, the total time-average power crossing the given surface is $\underset{\_}{11.46\text{W}}$.