Chapter 11, Problem 11.12P

Repeat Problem 11.11, except the wave now propagates in fresh water, having conductivity a = 10^-3 S/m, dielectric constant e_r' = 80.0, and permeability u₀.

To determine

(a)

The loss tangent and whether the material is good dielectric or good conductor.

Answer to Problem 11.12P

The value of loss tangent is $2.24\times {10}^{-3}$ and the material is a good dielectric.

Explanation of Solution

Calculation:

The line loss tangent is given by

   $\tan \theta =\frac{\sigma }{\omega \epsilon }$ ..... (1)

Here,

   $\sigma$ is the conductivity of the material

   $\epsilon$ is the permeability of the material

   $\omega$ is the radial frequency

The permeability $\epsilon$ is given by,

   $\epsilon ={\epsilon }_0{\epsilon }_r$ ..... (2)

Here,

   ${\epsilon }_0$ is the absolute permittivity with standard value of $8.854\times {10}^{-12}\text{F}/\text{m}$.

   ${\epsilon }_r$ is the relative permeability of the medium

The conversion from $\text{MHz}$ to $\text{Hz}$ is given by

   $1\text{MHz}={10}^6\text{Hz}$

The conversion of $100\text{MHz}$ in $\text{Hz}$ is given by

   $\begin{array}{c}100\text{MHz}=100\times {10}^6\text{Hz}\\ ={\text{10}}^8\text{Hz}\end{array}$

Hence, the frequency $f$ is

   $f={10}^8\text{Hz}$

The angular frequency $\omega$ is given by

   $\omega =2\pi \times f$

Here,

   $f$ is the frequency.

Substitute ${10}^8$ for $f$ in the above equation.

   $\omega =2\pi \times {10}^8\text{rad}/\text{s}$

Substitute ${\epsilon }_0{\epsilon }_r$ for $\epsilon$ in equation (1).

   $\tan \theta =\frac{\sigma }{\omega {\epsilon }_0{\epsilon }_r}$

Substitute $2\pi \times {10}^8$ for $\omega$ , ${10}^{-3}$ for $\sigma$ and $80$ for ${\epsilon }_r$ in the above equation.

   $\begin{array}{c}\tan \theta =\frac{{10}^{-3}}{2\pi \times {10}^8\times 8.854\times {10}^{-12}\times 80}\\ =\frac{{10}^{-3}}{4450.505\times {10}^{-4}}\\ =2.24\times {10}^{-3}\end{array}$

Since the value of loss tangent is less than 1, the material is a good dielectric.

Conclusion:

The value of loss tangent is $2.24\times {10}^{-3}$ and the material is a good dielectric.

To determine

(b)

The attenuation coefficient $\alpha$ , phase constant $\beta$ is and wave impedance $\eta$.

Answer to Problem 11.12P

The attenuation coefficient $\alpha$ is $0\text{Np/m}$ , phase constant $\beta$ is $18.73\text{rad}/\text{m}$ and the wave impedance $\eta$ is $42.149\angle 0°\Omega$.

Explanation of Solution

Calculation:

For a good dielectric, the value of $\alpha$ is zero

The value of phase constant $\beta$ is given by

   $\beta =\omega \sqrt{\mu \epsilon }$ ..... (3)

Here,

   $\mu$ is the permeability of the medium

The value of $\mu$ is given by

   $\mu ={\mu }_o{\mu }_r$ ..... (4)

   ${\mu }_0$ is the permeability of free space and has a standard value of $1.256\times {10}^{-6}\text{F}/\text{m}$

   ${\mu }_r$ is the relative permittivity of the medium

Substitute the value of equations (2) and (4) in equation (3).

   $\beta =\omega \sqrt{{\mu }_o{\mu }_r{\epsilon }_0{\epsilon }_r}$

Substitute $2\pi \times {10}^8$ for $\omega$ , $80$ for ${\epsilon }_r$ , $8.854\times {10}^{-12}$ for ${\epsilon }_0$ , $1.256\times {10}^{-6}$ for ${\mu }_r$ and $1$ for ${\mu }_r$ in the above equation.

   $\begin{array}{c}\beta =2\pi \times {10}^6\sqrt{1\times 1.256\times {10}^{-6}\times 8.854\times {10}^{-12}\times 80}\\ =2\pi \times {10}^6\sqrt{889.65\times {10}^{-18}}\\ =59.65\pi \times {10}^{-3}\\ =18.73\text{rad}/\text{m}\end{array}$

The value of wave impedance for a medium is given by

   $\eta ={\eta }_0\sqrt{\frac{{\mu }_r}{{\epsilon }_r}}$

Here,

   ${\eta }_0$ is the wave impedance at free space and its value is $377\Omega$

Substitute $80$ as ${\eta }_0$ , $80$ for ${\epsilon }_r$ and $1$ for ${\mu }_r$ in the above equation.

   $\begin{array}{c}\eta =377\sqrt{\frac{1}{80}}\\ =42.149\Omega \end{array}$

The impedance angle for a good dielectric is $0°$.

The value of $\eta$ is $42.149\angle 0°\Omega$.

Conclusion:

Therefore, the attenuation coefficient $\alpha$ is $0\text{Np/m}$ , phase constant $\beta$ is $18.73\text{rad}/\text{m}$ and wave impedance $\eta$ is $42.149\angle 0°\Omega$.

To determine

(c)

The electric field in phasor form.

Answer to Problem 11.12P

The phasor expression of electric field is $E_i={\widehat{a}}_x{E}_0{e}^{j18.73z}\text{V}/\text{m}$.

Explanation of Solution

Calculation:

The electric field $E_i$ in phasor form is

   $E_i={\widehat{a}}_x{E}_0{e}^{-\alpha z}{e}^{j\beta z}$

Here,

   $E_0$ is the electric field amplitude

   ${\widehat{a}}_x$ is unit vector in direction of positive x axis

Substitute $100$ for $E_0$ , $0$ for $\alpha$ and $18.73$ for $\beta$ in the above equation.

   $E_i={\widehat{a}}_x{E}_0{e}^{j18.73z}$

Conclusion:

The phasor expression of electric field is $E_i={\widehat{a}}_x{E}_0{e}^{j18.73z}\text{V}/\text{m}$.

To determine

(d)

The magnetic field in phasor form

Answer to Problem 11.12P

The magnetic field strength in phasor form is $H_i={\widehat{a}}_{y}2.37e^{-j18.73z}\text{A}/\text{m}$.

Explanation of Solution

Calculation:

The magnetic field strength $H_i$ in phasor form is

   $H_i={\widehat{a}}_y\frac{E_0}{\eta }{e}^{-\alpha z}{e}^{-j\beta z}{e}^{-j{\theta }_n}$

Here,

   ${\widehat{a}}_y$ is unit vector in direction of positive y direction

Substitute $100$ for $E_0$ , $0$ for $\alpha$ , $18.73$ for $\beta$ , $42.15$ for $\eta$ and $0$ for ${\theta }_n$ in the above equation.

   $\begin{array}{c}{H}_i={\widehat{a}}_y\frac{100}{42.15}{e}^{-j18.73z}\\ ={\widehat{a}}_{y}2.37e^{-j18.73z}\end{array}$

Conclusion:

Therefore, the magnetic field in phasor form is $H_i={\widehat{a}}_{y}2.37e^{-j18.73z}\text{A}/\text{m}$.

To determine

(e)

The time averaging Poynting vector.

Answer to Problem 11.12P

The time averaging Poynting vector is $P={\widehat{a}}_{z}118.65\text{W}/{\text{m}}^{\text{2}}$.

Explanation of Solution

Calculation:

The time averaging Poynting vector is given by

   $P={\widehat{a}}_z\frac{E_0^2}{2\eta }{e}^{-2\alpha z}\cos {\theta }_n$

Here,

   ${\widehat{a}}_z$ is a unit vector in the direction of positive z axis

Substitute $100$ for $E_0$ , $0$ for $\alpha$ , $42.15$ for $\eta$ and $0$ for ${\theta }_n$ in the above equation.

   $\begin{array}{c}P={\widehat{a}}_z\frac{{\left(100\right)}^2}{2\times 42.15}\cos 0°\\ ={\widehat{a}}_{z}118.65\text{W}/{\text{m}}^{\text{2}}\end{array}$

Conclusion:

Therefore, the time averaging Poynting vector is $P={\widehat{a}}_{z}118.65\text{W}/{\text{m}}^{\text{2}}$.

To determine

(f)

The 6-dB material thickness at which the wave power drops to 25 % of its value.

Answer to Problem 11.12P

There is no 6-dB material thickness.

Explanation of Solution

The attenuation constant $\alpha$ in dielectric has zero value. Thus, there will be no attenuation and thus, no power loss.