Chapter 12, Problem 12.2P

To determine

(a)

The angular frequency, $\omega .$

Answer to Problem 12.2P

The angular frequency is $\omega =1.5\times {10}^9{\text{s}}^{\text{-1}}.$

Explanation of Solution

Given:

   $E_{x1}^{+}=10\cos (\omega t-15z)\text{V/m}$

And $\beta =15$

The plane $z=0$ defines the boundary which is there in between two dielectrics.

For $z<0$ ,

   ${\epsilon }_{r1}\text{=9,}{\epsilon }_{r1}^{\text{'}\text{'}}=0\text{and}{\mu }_1={\mu }_0$

For $z<0,$

   ${\epsilon }_{r2}^{\text{'}\text{'}}=3\text{,}{\epsilon }_{r2}^{\text{'}\text{'}}=0\text{and}{\mu }_2={\mu }_0$

Concept Used:

Use below formula to calculate the result:

   $\beta =\omega \sqrt{{\mu }_0{\epsilon }_1\text{'}}=\omega \sqrt{{\epsilon }_{r1}\text{'}}/c$

Calculation:

We have,

   $\begin{array}{l}\beta =\omega \sqrt{{\mu }_0{\epsilon }_1{}^{\text{'}}}=\omega \sqrt{{\epsilon }_{r1}{}^{\text{'}}}/c\\ \Rightarrow \omega =15c/\sqrt{{\epsilon }_{r1}{}^{\text{'}}}=15\times \left(3\times {10}^8\right)/\sqrt{9}=1.5\times {10}^9{\text{s}}^{\text{-1}}\end{array}$

To determine

(b)

The value of $<S_1{}^{+}>$.

Answer to Problem 12.2P

The value is $<S_1{}^{+}>=0.3978a_z{\text{W/m}}^{\text{2}}$.

Explanation of Solution

Given:

   $E_{x1}^{+}=10\cos (\omega t-15z)\text{V/m}$

And, $\beta =15$

The plane $z=0$ defines the boundary which is there in between two dielectrics.

For, $z<0$ ,

   ${\epsilon }_{r1}\text{=9,}{\epsilon }_{r1}^{\text{'}\text{'}}=0\text{and}{\mu }_1={\mu }_0$

For, $z<0$

   ${\epsilon }_{r2}^{\text{'}\text{'}}=3\text{,}{\epsilon }_{r2}^{\text{'}\text{'}}=0\text{and}{\mu }_2={\mu }_0$

Concept Used:

Use below formula to calculate the result:

   ${\eta }_1=\sqrt{\frac{{\mu }_0}{{\epsilon }_1^{\text{'}}}}={\eta }_0/\sqrt{{\epsilon }_{r1}^{\text{'}}}$

   $<S_1{}^{+}>=\frac{1}{2}\frac{{\left|E_1\right|}^2}{{\eta }_1}{a}_z$

Calculation:

We have,

   $E_{x1}^{+}=10\cos (\omega t-15z)\text{V/m}$

First, we need

   ${\eta }_1=\sqrt{\frac{{\mu }_0}{{\epsilon }_1^{\text{'}}}}={\eta }_0/\sqrt{{\epsilon }_{r1}^{\text{'}}}=377/\sqrt{9}=125.666\text{ohms}$

   $\left[{\eta }_0=377\right]$

Now, for the Poynting vector which has no loss value and no difference in phase between the electric and the magnetic field:

   $<S_1{}^{+}>=\frac{1}{2}\frac{{\left|E_1\right|}^2}{{\eta }_1}{a}_z=\frac{1}{2}\frac{{10}^2}{125.66}{a}_z=0.3978a_z{\text{W/m}}^{\text{2}}$

To determine

(c)

The value of $<S_1{}^{-}>$.

Answer to Problem 12.2P

The value is $<S_1{}^{-}>=-0.0285a_z{\text{W/m}}^{\text{2}}.$

Explanation of Solution

Given:

   $E_{x1}^{+}=10\cos (\omega t-15z)\text{V/m}$

And, $\beta =15$

The plane $z=0$ defines the boundary which is there in between two dielectrics.

For $z<0$ ,

   ${\epsilon }_{r1}\text{=9,}{\epsilon }_{r1}^{\text{'}\text{'}}=0\text{and}{\mu }_1={\mu }_0$

For $z<0,$

   ${\epsilon }_{r2}^{\text{'}\text{'}}=3\text{,}{\epsilon }_{r2}^{\text{'}\text{'}}=0\text{and}{\mu }_2={\mu }_0$

Concept Used:

Use the below formula to calculate the required value:

   $\Gamma =\frac{{\eta }_2-{\eta }_1}{{\eta }_2+{\eta }_1}=\frac{{\eta }_0/\sqrt{{\epsilon }_{r2}^{\text{'}}}-{\eta }_0/\sqrt{{\epsilon }_{r1}^{\text{'}}}}{{\eta }_0/\sqrt{{\epsilon }_{r2}^{\text{'}}}+{\eta }_0/\sqrt{{\epsilon }_{r1}^{\text{'}}}}$

Calculation:

First, we need to calculate the reflection coefficient:

   $\Gamma =\frac{{\eta }_2-{\eta }_1}{{\eta }_2+{\eta }_1}=\frac{{\eta }_0/\sqrt{{\epsilon }_{r2}^{\text{'}}}-{\eta }_0/\sqrt{{\epsilon }_{r1}^{\text{'}}}}{{\eta }_0/\sqrt{{\epsilon }_{r2}^{\text{'}}}+{\eta }_0/\sqrt{{\epsilon }_{r1}^{\text{'}}}}=\frac{\sqrt{{\epsilon }_{r1}^{\text{'}}}-\sqrt{{\epsilon }_{r2}^{\text{'}}}}{\sqrt{{\epsilon }_{r1}^{\text{'}}}+\sqrt{{\epsilon }_{r2}^{\text{'}}}}=\frac{\sqrt{9}-\sqrt{3}}{\sqrt{9}+\sqrt{3}}=0.2679$

Then,

   $<S_1{}^{-}>=-{\left|\Gamma \right|}^2<S_1{}^{+}>=-{\left(0.2679\right)}^2\left(0.3978\right)a_z=-0.0285a_z{\text{W/m}}^{\text{2}}$

To determine

(d)

The value of $<S_2{}^{+}>$.

Answer to Problem 12.2P

The value is $<S_2{}^{+}>=0.3693a_z{\text{W/m}}^{\text{2}}.$

Explanation of Solution

Given:

   $E_{x1}^{+}=10\cos (\omega t-15z)\text{V/m}$

And, $\beta =15$

The plane $z=0$ defines the boundary which is there in between two dielectrics.

For $z<0$ ,

   ${\epsilon }_{r1}\text{=9,}{\epsilon }_{r1}^{\text{'}\text{'}}=0\text{and}{\mu }_1={\mu }_0$

For $z<0,$

   ${\epsilon }_{r2}^{\text{'}\text{'}}=3\text{,}{\epsilon }_{r2}^{\text{'}\text{'}}=0\text{and}{\mu }_2={\mu }_0$

Concept Used:

First calculate the reflection coefficient by using the below formula:

   $\Gamma =\frac{{\eta }_2-{\eta }_1}{{\eta }_2+{\eta }_1}=\frac{{\eta }_0/\sqrt{{\epsilon }_{r2}^{\text{'}}}-{\eta }_0/\sqrt{{\epsilon }_{r1}^{\text{'}}}}{{\eta }_0/\sqrt{{\epsilon }_{r2}^{\text{'}}}+{\eta }_0/\sqrt{{\epsilon }_{r1}^{\text{'}}}}$

Then calculate $<S_1{}^{-}>=-{\left|\Gamma \right|}^2<S_1{}^{+}>$ and then the remaining power propagating in the forward z-direction by using the below formula:

   $<S_2{}^{+}>=<S_1{}^{+}>+<S_1{}^{-}>$

Calculation:

First, we need to calculate the reflection coefficient:

   $\Gamma =\frac{{\eta }_2-{\eta }_1}{{\eta }_2+{\eta }_1}=\frac{{\eta }_0/\sqrt{{\epsilon }_{r2}^{\text{'}}}-{\eta }_0/\sqrt{{\epsilon }_{r1}^{\text{'}}}}{{\eta }_0/\sqrt{{\epsilon }_{r2}^{\text{'}}}+{\eta }_0/\sqrt{{\epsilon }_{r1}^{\text{'}}}}=\frac{\sqrt{{\epsilon }_{r1}^{\text{'}}}-\sqrt{{\epsilon }_{r2}^{\text{'}}}}{\sqrt{{\epsilon }_{r1}^{\text{'}}}+\sqrt{{\epsilon }_{r2}^{\text{'}}}}=\frac{\sqrt{9}-\sqrt{3}}{\sqrt{9}+\sqrt{3}}=0.2679$

Then,

   $<S_1{}^{-}>=-{\left|\Gamma \right|}^2<S_1{}^{+}>=-{\left(0.2679\right)}^2\left(0.3978\right)a_z=-0.0285a_z{\text{W/m}}^{\text{2}}$

   $<S_2{}^{+}>$ is the remaining power propagating in the forward z -direction.

   $<S_2{}^{+}>=<S_1{}^{+}>+<S_1{}^{-}>=\left(0.3978\right)a_z-0.0285a_z=0.3693a_z{\text{W/m}}^{\text{2}}.$