Chapter 7, Problem 1P

The magnetic field of a wave propagating through a certain nonmagnetic material is given by

$H\text{ = }\hat{z}$30 cos(10⁸ t − 0.5y) (m/Am).

Find the following:

1. (a) The direction of wave propagation.
2. (b) The phase velocity.
3. (c) The wavelength in the material.
4. (d) The relative permittivity of the material.
5. (e) The electric field phasor.

To determine

The direction of wave propagation for the given condition.

Answer to Problem 1P

The direction of wave propagation for the given condition is $+y$ direction.

Explanation of Solution

Given data:

The magnetic field intensity of the wave is $H=\widehat{z}30\cos \left({10}^{8}t-0.5y\right)\text{mA}/\text{m}$.

Calculation:

Write the standard expression for the magnetic field phasor of TEM wave propagating in the $\widehat{y}$ direction and polarized in the $\widehat{x}$ direction.

$\tilde{H}\left(y,t\right)=\widehat{z}{a}_{\text{x}}\cos \left(\omega t\pm ky+{\delta }_{\text{x}}\right)$ (1)

Here,

$\tilde{H}\left(y,t\right)$ is the magnetic field intensity.

$a_{\text{x}}$ is the magnitude of the $x-$ component of the electric field.

${\delta }_{\text{x}}$ is the phase angle of the electric field in the $x-$ direction.

$\omega$ is the angular frequency of the wave.

$k$ is the wave number.

In the above equation positive sign is used when the wave is travelling in the $-y$ direction and negative sign is used when the wave is travelling in the $+y$ direction.

In the given magnetic field negative sign is used so the direction of wave propagation is along the $+y$ direction.

Conclusion:

Therefore, the direction of wave propagation for the given condition is $+y$ direction.

To determine

The phase velocity for the given condition.

Answer to Problem 1P

The phase velocity for the given condition is $2\times {10}^8\text{m}/\text{s}$.

Explanation of Solution

Calculation:

Write the standard relation of phase velocity, wave number and angular frequency of uniform plane wave.

$u_{\text{p}}=\frac{\omega }{k}\text{m}/\text{s}$ (2)

Here,

$u_{\text{p}}$ is the phase velocity of the wave.

Compare the given magnetic field and equation (1) to calculate the value of $\omega$ and $k$ that is,

$\omega ={10}^8\text{rad}/\text{sec}$ and $k=0.5\text{rad}/\text{m}$.

Substitute $0.5\text{rad}/\text{m}$ for $k$ and ${10}^8\text{rad}/\text{sec}$ for $\omega$ in the equation (2).

$\begin{array}{c}{u}_{\text{p}}=\frac{{10}^8\text{rad}/\text{sec}}{0.5\text{rad}/\text{m}}\\ =2\times {10}^8\text{m}/\text{s}\end{array}$

Conclusion:

Therefore, the phase velocity for the given condition is $2\times {10}^8\text{m}/\text{s}$.

To determine

The wavelength in the material.

Answer to Problem 1P

The wavelength in the material is $12.6\text{m}$.

Explanation of Solution

Calculation:

Write the standard relation between wavelength and wave number of plane wave.

$\text{λ}=\frac{2\pi }{k}$

Here,

$\text{λ}$ is the wavelength of the wave.

Substitute $0.5\text{rad}/\text{m}$ for $k$ in the above equation.

$\begin{array}{c}\text{λ}=\frac{2\pi }{0.5\text{rad}/\text{m}}\\ =4\pi \\ =12.6\text{m}\end{array}$

Conclusion:

Therefore, the wavelength in the material is $12.6\text{m}$.

To determine

The relative permittivity of the medium.

Answer to Problem 1P

The relative permittivity $\left({\epsilon }_{\text{r}}\right)$ of the medium is $2.25$.

Explanation of Solution

Calculation:

Write the standard relation between relative permittivity and speed of light and phase velocity.

${\epsilon }_{\text{r}}={\left(\frac{c}{u_{\text{p}}}\right)}^2$

Here,

${\epsilon }_{\text{r}}$ is the relative permittivity of the medium.

$c$ is the speed of light, which has standard value as $3\times {10}^8\text{m}/\text{s}$.

Substitute $3\times {10}^8\text{m}/\text{s}$ for $c$ and $2\times {10}^8\text{m}/\text{s}$ for $u_{\text{p}}$ in the above equation.

$\begin{array}{c}{\epsilon }_{\text{r}}={\left(\frac{3\times {10}^8\text{m}/\text{s}}{2\times {10}^8\text{m}/\text{s}}\right)}^2\\ ={\left(1.5\right)}^2\\ =2.25\end{array}$

Conclusion:

Therefore, the relative permittivity $\left({\epsilon }_{\text{r}}\right)$ of the medium is $2.25$.

To determine

The electric field phasor.

Answer to Problem 1P

The electric field phasor is $\tilde{E}=-\widehat{x}7.54\cos \left({10}^{8}t-0.5y\right)\text{V}/\text{m}$.

Explanation of Solution

Calculation:

Write the standard relation between electric field and magnetic field intensity.

$\tilde{E}=-\eta \widehat{k}\times \tilde{H}$ (3)

Here,

$\tilde{E}$ is the electric field intensity phasor of the plane wave.

$\tilde{H}$ is the magnetic field phasor of the plane wave.

$\widehat{k}$ is the direction of the wave.

$\eta$ is the intrinsic impedance of the medium.

Write the standard expression for the intrinsic impedance of the medium.

$\eta =\sqrt{\frac{\mu }{\epsilon }}$

Here,

$\mu$ is permeability of the medium.

$\epsilon$ is the permittivity of the medium.

Write the standard expression for the electrical permittivity of any medium as,

$\epsilon ={\epsilon }_{\text{r}}{\epsilon }_0$

Here,

${\epsilon }_{\text{r}}$ is the relative permeability of the medium.

${\epsilon }_0$ is the permeability of the air, which has standard value as $8.85\times {10}^{-12}\text{F}/\text{m}$.

Substitute $2.25$ for ${\epsilon }_{\text{r}}$ and $8.85\times {10}^{-12}\text{F}/\text{m}$ for ${\epsilon }_0$ in the above equation.

$\epsilon =2.25\times \left(8.85\times {10}^{-12}\text{F}/\text{m}\right)$

The permeability of the given non magnetic medium is,

${\mu }_0=4\pi \times {10}^{-7}\text{H}/\text{m}$

Substitute $4\pi \times {10}^{-7}\text{H}/\text{m}$ for ${\mu }_0$ and $2.25\times \left(8.85\times {10}^{-12}\text{F}/\text{m}\right)$ for $\epsilon$ in the above equation.

$\begin{array}{c}\eta =\sqrt{\frac{4\pi \times {10}^{-7}\text{H}/\text{m}}{2.25\times \left(8.85\times {10}^{-12}\text{F}/\text{m}\right)}}\\ =251.21\Omega \end{array}$

Substitute $\widehat{y}$ for $\widehat{k}$, $251.21\Omega$ for $\eta$ and $\widehat{z}30\cos \left({10}^{8}t-0.5y\right)\text{mA}/\text{m}$ for $\widehat{H}$ in equation (3).

$\begin{array}{c}\tilde{E}=-\left(251.21\Omega \right)\left(\widehat{y}\times \left(\widehat{z}30\cos \left({10}^{8}t-0.5y\right)\text{mA}/\text{m}\right)\right)\\ =-\left(251.21\right)\left(\widehat{y}\times \widehat{z}30\cos \left({10}^{8}t-0.5y\right)\right)\text{mV}/\text{m}\end{array}$

Simplify the above expression.

$\tilde{E}=-\widehat{x}7536\cos \left({10}^{8}t-0.5y\right)\text{mV}/\text{m}$ (4)

The conversion from $1\text{mV}/\text{m}$ into $\text{V}/\text{m}$ is done as,

$1\text{mV}/\text{m}=1\times {10}^{-3}\text{V}/\text{m}$

So, the conversion from $7536\text{mV}/\text{m}$ into $\text{V}/\text{m}$ is given as,

$\begin{array}{c}7536\text{mV}/\text{m}=7536\times {10}^{-3}\text{V}/\text{m}\\ =7.54\text{V}/\text{m}\end{array}$

Substitute $7.54\text{V}/\text{m}$ for $7536\text{mV}/\text{m}$ in equation (4).

$\tilde{E}=-\widehat{x}7.54\cos \left({10}^{8}t-0.5y\right)\text{V}/\text{m}$

Conclusion:

Therefore, the electric field phasor is $\tilde{E}=-\widehat{x}7.54\cos \left({10}^{8}t-0.5y\right)\text{V}/\text{m}$.