Question

a) Derive Eqs. 9.179, and from these obtain Eqs. 9.180.

(b) Put Eq. 9.180 into Maxwell's equations (i) and (ii) to obtain Eq. 9.181. Check that you get the same results using (i) and (iv) of Eq. 9.179.

Short answer

(a)

The value of electric field in x-axis direction is ${\mathrm{E}}_{\mathrm{x}}=\frac{\mathrm{i}}{{(\mathrm{\omega }/\mathrm{c})}^2-{\mathrm{k}}^2}\left(\mathrm{\omega }\frac{\partial {\mathrm{B}}_{\mathrm{z}}}{\partial \mathrm{y}}+\mathrm{k}\frac{\partial {\mathrm{E}}_{\mathrm{z}}}{\partial \mathrm{x}}\right)$.

The value of electric field in y-axis direction is ${\mathrm{E}}_{\mathrm{y}}=\frac{\mathrm{i}}{{(\mathrm{\omega }/\mathrm{c})}^2-{\mathrm{k}}^2}\left(\mathrm{k}\frac{\partial {\mathrm{E}}_{\mathrm{z}}}{\partial \mathrm{y}}-\mathrm{\omega }\frac{\partial {\mathrm{B}}_{\mathrm{z}}}{\partial \mathrm{x}}\right)$ .

The value of magnetic field in x-axis direction is ${\mathrm{B}}_{\mathrm{x}}=\frac{\mathrm{i}}{{(\mathrm{\omega }/\mathrm{c})}^2-{\mathrm{k}}^2}\left(\mathrm{k}\frac{\partial {\mathrm{B}}_2}{\partial \mathrm{x}}-\frac{\mathrm{\omega }}{{\mathrm{c}}^2}\frac{\partial {\mathrm{E}}_2}{\partial \mathrm{y}}\right)$.

The value of magnetic field in y-axis direction is ${\mathrm{B}}_{\mathrm{y}}=\frac{\mathrm{i}}{{(\mathrm{\omega }/\mathrm{c})}^2-{\mathrm{k}}^2}\left(\mathrm{k}\frac{\partial {\mathrm{B}}_{\mathrm{z}}}{\partial \mathrm{y}}+\frac{\mathrm{\omega }}{{\mathrm{c}}^2}\frac{\partial {\mathrm{E}}_{\mathrm{z}}}{\partial \mathrm{x}}\right)$ .

(b)

The value of use the results of 9.180) with (i) to obtain the same equations is $\left[\frac{{\partial }^2}{\partial {\mathrm{x}}^2}+\frac{{\partial }^2}{\partial {\mathrm{y}}^2}+{(\mathrm{\omega }/\mathrm{c})}^2-{\mathrm{k}}^2\right]{\mathrm{B}}_2=0$.

The value of use the results of 9.180) with (iv) to obtain the same equations is $\left[\frac{{\partial }^2}{\partial {\mathrm{x}}^2}+\frac{{\partial }^2}{\partial {\mathrm{y}}^2}+{(\mathrm{\omega }/\mathrm{c})}^2-{\mathrm{k}}^2\right]{\mathrm{E}}_2=0$.

Step-by-step solution

Write the given data from the question.

Consider the electric fields are$\overrightarrow{\mathrm{E}}={\overrightarrow{\mathrm{E}}}_0(\mathrm{x},\mathrm{y}){\mathrm{e}}^{\mathrm{i}(\mathrm{kz}-\mathrm{\omega t})}$.

Consider the electric field vector${\overrightarrow{\mathrm{E}}}_0={\mathrm{E}}_{\mathrm{x}}\widehat{\mathrm{x}}+{\mathrm{E}}_{\mathrm{y}}\widehat{\mathrm{y}}+{\mathrm{E}}_{\mathrm{z}}\widehat{\mathrm{z}}$.

Consider the magnetic fields are$\overrightarrow{\mathrm{B}}={\overrightarrow{\mathrm{B}}}_0(\mathrm{x},\mathrm{y}){\mathrm{e}}^{\mathrm{i}(\mathrm{kz}-\mathrm{\omega t})}$ .

Consider the magnetic fields are ${\overrightarrow{\mathrm{B}}}_0={\mathrm{B}}_{\mathrm{x}}\widehat{\mathrm{x}}+{\mathrm{B}}_{\mathrm{y}}\widehat{\mathrm{y}}+{\mathrm{B}}_{\mathrm{z}}\widehat{\mathrm{z}}$.

Determine the formula of electric and magnetic field in x-axis and y-axis direction.

Write the formula electric field in x-axis direction.

${\mathbf{B}}_y=\frac{1}{\mathrm{ik}}\mathbf{(}\frac{\mathbf{\partial }{\mathbf{B}}_{\mathbf{z}}}{\mathbf{\partial }\mathbf{y}}\mathbf{+}\frac{\mathbf{i\omega }}{{\mathbf{c}}^{\mathbf{2}}}{\mathbf{E}}_{\mathbf{z}}\mathbf{)}\to \left(\mathbf{2}\right)$ …… (1)

Here, $\mathbf{k}$is real, ${\mathbf{B}}_{\mathbf{z}}$is transverse electric wave, $\mathbf{\omega }$ is frequency and ${\mathbf{E}}_{\mathbf{z}}$ is transverse electric wave.

Write the formula of electric field in y-axis direction.

role="math" localid="1658734044435" ${\mathbf{B}}_x=-\frac{\omega }{{\mathbf{c}}^2}\frac{1}{k}{\mathbf{E}}_y-\frac{i}{k}\frac{\partial {\mathbf{B}}_z}{\partial \mathbf{x}}\to \left(\mathbf{1}\right)$ …… (2)

Here, $\mathrm{\omega }$is frequency, $\mathbf{k}$is real, ${\mathbf{E}}_y$is electric field in y-axis and ${\mathbf{B}}_z$is transverse electric wave.

Write the formula of magnetic field in x-axis direction.

${\mathbf{ikB}}_x-\frac{\partial {\mathbf{B}}_z}{\partial \mathbf{x}}$ …… (3)

Here, $\mathbf{k}$ is real, ${\mathbf{B}}_x$ is magnetic field in x-axis and ${\mathbf{B}}_z$ is transverse electric wave.

Write the formula of magnetic field in y-axis direction.

$\frac{\partial {\mathbf{B}}_z}{\partial \mathbf{y}}-{\mathbf{ikB}}_y$ …… (4)

Here, $\mathbf{k}$ is real, ${\mathbf{B}}_y$ is magnetic field in y-axis and ${\mathbf{B}}_z$ is transverse electric wave.

(a) Determine the value of electric and magnetic field in x-axis and y-axis direction.

From the 3^rd Maxwell equation by components:

$\begin{array}{c}\nabla \times =-\frac{\partial \overrightarrow{\mathrm{B}}}{\partial \mathrm{t}}\\ \frac{\partial {\mathrm{E}}_{\mathrm{z}}}{\partial \mathrm{y}}-\frac{\partial {\mathrm{E}}_{\mathrm{y}}}{\partial \mathrm{z}}=-\frac{\partial {\mathrm{B}}_{\mathrm{z}}}{\partial \mathrm{t}}\\ \frac{\partial {\mathrm{E}}_{\mathrm{z}}}{\partial \mathrm{y}}-{\mathrm{E}}_{\mathrm{y}}\mathrm{ik}=\mathrm{i\omega Bz}\end{array}$

Determine the 3^rd Maxwell equation by components in z-axis direction.

$\begin{array}{c}\frac{\partial {\mathrm{E}}_{\mathrm{z}}}{\partial \mathrm{z}}-\frac{\partial {\mathrm{E}}_{\mathrm{z}}}{\partial \mathrm{x}}=-\frac{\partial {\mathrm{B}}_{\mathrm{y}}}{\partial \mathrm{t}}\\ {\mathrm{E}}_{\mathrm{z}}\mathrm{ik}-\frac{\partial {\mathrm{E}}_{\mathrm{z}}}{\partial \mathrm{x}}={\mathrm{i\omega B}}_{\mathrm{y}}\end{array}$

Determine the 3^rd Maxwell equation by components in y-axis direction.

$\begin{array}{c}\frac{\partial {\mathrm{E}}_{\mathrm{y}}}{\partial \mathrm{x}}-\frac{\partial {\mathrm{E}}_{\mathrm{z}}}{\partial \mathrm{y}}=-\frac{\partial {\mathrm{B}}_{\mathrm{z}}}{\partial \mathrm{t}}\\ \frac{\partial {\mathrm{E}}_{\mathrm{y}}}{\partial \mathrm{x}}-\frac{\partial {\mathrm{E}}_{\mathrm{z}}}{\partial \mathrm{y}}=-{\mathrm{i\omega B}}_{\mathrm{z}}\end{array}$

From the 4^th Maxwell equation by components:

$\begin{array}{c}\nabla \times \overrightarrow{\mathrm{B}}=\frac{1}{{\mathrm{c}}^2}\frac{\partial \overrightarrow{\mathrm{E}}}{\partial \mathrm{t}}\\ \frac{\partial {\mathrm{B}}_{\mathrm{z}}}{\partial \mathrm{y}}-\frac{\partial {\mathrm{B}}_{\mathrm{y}}}{\partial \mathrm{z}}=-\frac{\partial {\mathrm{E}}_{\mathrm{z}}}{\partial \mathrm{t}}\\ \frac{\partial {\mathrm{B}}_{\mathrm{z}}}{\partial \mathrm{y}}-{\mathrm{B}}_{\mathrm{y}}\mathrm{ik}=-\frac{\mathrm{i\omega }}{{\mathrm{c}}^2}{\mathrm{E}}_{\mathrm{x}}\end{array}$

Determine the 4^th Maxwell equation by components in z-axis direction.

$\begin{array}{c}\frac{\partial {\mathrm{B}}_{\mathrm{z}}}{\partial \mathrm{z}}-\frac{\partial {\mathrm{B}}_{\mathrm{z}}}{\partial \mathrm{x}}=\frac{1}{{\mathrm{c}}^2}\frac{\partial {\mathrm{E}}_{\mathrm{y}}}{\partial \mathrm{t}}\\ {\mathrm{B}}_{\mathrm{z}}\mathrm{ik}-\frac{\partial {\mathrm{B}}_{\mathrm{z}}}{\partial \mathrm{x}}=-\frac{\mathrm{i\omega }}{{\mathrm{c}}^2}{\mathrm{E}}_{\mathrm{y}}\end{array}$

Determine the 4^th Maxwell equation by components in y-axis direction.

$\begin{array}{c}\frac{\partial {\mathrm{B}}_{\mathrm{y}}}{\partial \mathrm{x}}-\frac{\partial {\mathrm{B}}_{\mathrm{z}}}{\partial \mathrm{y}}=\frac{1}{{\mathrm{c}}^2}\frac{\partial {\mathrm{E}}_{\mathrm{z}}}{\partial \mathrm{t}}\\ \frac{\partial {\mathrm{B}}_{\mathrm{y}}}{\partial \mathrm{x}}-\frac{\partial {\mathrm{B}}_{\mathrm{z}}}{\partial \mathrm{y}}=-\frac{\mathrm{i\omega }}{{\mathrm{c}}^2}{\mathrm{E}}_{\mathrm{z}}\end{array}$

Now from (4) express ${\mathrm{B}}_{\mathrm{y}}$ and put this into (2), obtaining ${\mathrm{E}}_{\mathrm{x}}$:

Substitute ${\mathrm{E}}_{\mathrm{x}}\mathrm{ik}-\frac{\partial {\mathrm{E}}_{\mathrm{z}}}{\partial \mathrm{x}}$for ${\mathrm{B}}_{\mathrm{y}}$ into equation (1).

$\begin{array}{c}{\mathrm{E}}_{\mathrm{x}}\mathrm{ik}-\frac{\partial {\mathrm{E}}_{\mathrm{z}}}{\partial \mathrm{x}}=\mathrm{i\omega }\frac{1}{\mathrm{ik}}\left(\frac{\partial {\mathrm{B}}_{\mathrm{z}}}{\partial \mathrm{y}}+\frac{\mathrm{i\omega }}{{\mathrm{c}}^2}{\mathrm{E}}_{\mathrm{x}}\right)\\ {\mathrm{E}}_{\mathrm{x}}\left(\mathrm{ik}-\frac{\mathrm{i}}{\mathrm{k}}\frac{{\mathrm{\omega }}^2}{{\mathrm{c}}^2}\right)=\frac{\mathrm{\omega }}{\mathrm{k}}\frac{\partial {\mathrm{B}}_{\mathrm{z}}}{\partial \mathrm{y}}+\frac{\partial {\mathrm{E}}_{\mathrm{z}}}{\partial \mathrm{x}}\\ {\mathrm{E}}_{\mathrm{x}}\frac{\mathrm{i}}{\mathrm{k}}\left({\mathrm{k}}^2-\frac{{\mathrm{\omega }}^2}{{\mathrm{c}}^2}\right)=\frac{\mathrm{\omega }}{\mathrm{k}}\frac{\partial {\mathrm{B}}_{\mathrm{z}}}{\partial \mathrm{y}}+\frac{\partial {\mathrm{E}}_{\mathrm{z}}}{\partial \mathrm{x}}\\ {\mathrm{E}}_{\mathrm{x}}=\frac{\mathrm{i}}{{(\mathrm{\omega }/\mathrm{c})}^2-{\mathrm{k}}^2}\left(\mathrm{\omega }\frac{\partial {\mathrm{B}}_{\mathrm{z}}}{\partial \mathrm{y}}+\mathrm{k}\frac{\partial {\mathrm{E}}_{\mathrm{z}}}{\partial \mathrm{x}}\right)\end{array}$

Now express from (5) the ${\mathrm{B}}_x$ and put it into the (1) to get the ${\mathrm{E}}_{\mathrm{y}}$

Substitute $\frac{\partial {\mathrm{E}}_{\mathrm{z}}}{\partial \mathrm{y}}-{\mathrm{E}}_{\mathrm{y}}\mathrm{ik}$ for ${\mathrm{B}}_{\mathrm{x}}$ into equation (2).

$\begin{array}{c}\frac{\partial {\mathrm{E}}_{\mathrm{z}}}{\partial \mathrm{y}}-{\mathrm{E}}_{\mathrm{y}}\mathrm{ik}=-\frac{{\mathrm{\omega }}^2}{{\mathrm{c}}^2}\frac{1}{\mathrm{k}}{\mathrm{E}}_{\mathrm{y}}+\frac{\mathrm{\omega }}{\mathrm{k}}\frac{\partial {\mathrm{B}}_{\mathrm{z}}}{\partial \mathrm{x}}\\ {\mathrm{E}}_{\mathrm{y}}\frac{\mathrm{i}}{\mathrm{k}}\left(\frac{{\mathrm{\omega }}^2}{{\mathrm{c}}^2}-{\mathrm{k}}^2\right)=\frac{\mathrm{\omega }}{\mathrm{k}}\frac{\partial {\mathrm{B}}_{\mathrm{z}}}{\partial \mathrm{x}}-\frac{\partial {\mathrm{E}}_{\mathrm{z}}}{\partial \mathrm{y}}\\ {\mathrm{E}}_{\mathrm{y}}=\frac{\mathrm{i}}{(\mathrm{\omega }/{\mathrm{c}}^2)-{\mathrm{k}}^2}\left(\mathrm{k}\frac{\partial {\mathrm{E}}_{\mathrm{z}}}{\partial \mathrm{y}}-\mathrm{\omega }\frac{\partial {\mathrm{B}}_{\mathrm{z}}}{\partial \mathrm{x}}\right)\end{array}$

Now to get the other two equations put the result (7) into (4):

Substitute ${\mathrm{B}}_{\mathrm{y}}$ for $\frac{\partial {\mathrm{B}}_{\mathrm{z}}}{\partial \mathrm{y}}-{\mathrm{ikB}}_{\mathrm{y}}$ into equation (3).

$\begin{array}{c}{\mathrm{B}}_{\mathrm{y}}=\frac{1}{\mathrm{ik}}\frac{\partial {\mathrm{B}}_{\mathrm{z}}}{\partial \mathrm{y}}-\frac{1}{\mathrm{ik}}\frac{\mathrm{\omega }}{{\mathrm{c}}^2}\frac{1}{{(\mathrm{\omega }/\mathrm{c})}^2-{\mathrm{k}}^2}\left(\mathrm{\omega }\frac{\partial {\mathrm{B}}_{\mathrm{z}}}{\partial \mathrm{y}}+\mathrm{k}\frac{\partial {\mathrm{E}}_{\mathrm{z}}}{\partial \mathrm{x}}\right)\\ =\frac{1}{\mathrm{ik}}\left[\frac{\partial {\mathrm{B}}_{\mathrm{z}}}{\partial \mathrm{y}}\left(1-\frac{{(\mathrm{\omega }/\mathrm{c})}^2}{{(\mathrm{\omega }/\mathrm{c})}^2-{\mathrm{k}}^2}\right)-\frac{\mathrm{k\omega }}{{\mathrm{c}}^2}\frac{1}{{(\mathrm{\omega }/\mathrm{c})}^2-{\mathrm{k}}^2}\frac{\partial {\mathrm{E}}_{\mathrm{z}}}{\partial \mathrm{x}}\right]\\ {\mathrm{B}}_{\mathrm{y}}=\frac{\mathrm{i}}{{(\mathrm{\omega }/\mathrm{c})}^2-{\mathrm{k}}^2}\left(\mathrm{k}\frac{\partial {\mathrm{B}}_{\mathrm{z}}}{\partial \mathrm{y}}+\frac{\mathrm{\omega }}{{\mathrm{c}}^2}\frac{\partial {\mathrm{E}}_{\mathrm{z}}}{\partial \mathrm{x}}\right)\end{array}$

Now put the result (8) into (5) to finally reproduce all the eq. (9.180).

Substitute ${\mathrm{B}}_{\mathrm{x}}$ for ${\mathrm{ikB}}_{\mathrm{x}}-\frac{\partial {\mathrm{B}}_{\mathrm{z}}}{\partial \mathrm{x}}$ into equation (4).

$\begin{array}{c}{\mathrm{B}}_{\mathrm{x}}=-\frac{\mathrm{i}}{\mathrm{k}}\left[\frac{\partial {\mathrm{B}}_{\mathrm{z}}}{\partial \mathrm{x}}(1-\frac{{(\mathrm{\omega }/\mathrm{c})}^2}{{(\mathrm{\omega }/\mathrm{c})}^2-{\mathrm{k}}^2})+\frac{\mathrm{\omega k}}{{\mathrm{c}}^2}\frac{1}{{(\mathrm{\omega }/\mathrm{c})}^2-{\mathrm{k}}^2}\frac{\partial {\mathrm{E}}_{\mathrm{z}}}{\partial \mathrm{y}}\right]\\ {\mathrm{B}}_{\mathrm{x}}=\frac{\mathrm{i}}{{(\mathrm{\omega }/\mathrm{c})}^2-{\mathrm{k}}^2}\left(\mathrm{k}\frac{\partial {\mathrm{B}}_{\mathrm{z}}}{\partial \mathrm{x}}-\frac{\mathrm{\omega }}{{\mathrm{c}}^2}\frac{\partial {\mathrm{E}}_{\mathrm{z}}}{\partial \mathrm{y}}\right)\end{array}$

With this all of the Eq. 9.180 have been reproduced. Nothing fancy, just a lot of algebra.

(b) Determine the value of result of 9.180 with (i) and (iv) of 9.179.

The 1^st Maxwell equation is:

$\begin{array}{c}\nabla \cdot \overrightarrow{\mathrm{E}}=\frac{\partial {\mathrm{E}}_{\mathrm{x}}}{\partial \mathrm{x}}+\frac{\partial {\mathrm{E}}_{\mathrm{y}}}{\partial \mathrm{y}}+\frac{\partial {\mathrm{E}}_{\mathrm{z}}}{\partial \mathrm{z}}\\ \frac{\partial {\mathrm{E}}_{\mathrm{x}}}{\partial \mathrm{x}}+\frac{\partial {\mathrm{E}}_{\mathrm{y}}}{\partial \mathrm{y}}+\frac{\partial {\mathrm{E}}_{\mathrm{y}}}{\partial \mathrm{y}}+{\mathrm{ikE}}_{\mathrm{z}}=0\leftarrow \left(7\right),\left(8\right)\\ \frac{\mathrm{i}}{{(\mathrm{\omega }/\mathrm{c})}^2-{\mathrm{k}}^2}\mathrm{k}\left(\frac{{\partial }^2{\mathrm{E}}_{\mathrm{x}}}{\partial {\mathrm{x}}^2}+\frac{{\partial }^2{\mathrm{E}}_{\mathrm{z}}}{\partial {\mathrm{y}}^2}\right)+{\mathrm{ikE}}_{\mathrm{z}}=0/{(\mathrm{\omega }/\mathrm{c})}^2-{\mathrm{k}}^2\\ \left[\frac{{\partial }^2}{\partial {\mathrm{x}}^2}+\frac{{\partial }^2}{\partial {\mathrm{y}}^2}+{(\mathrm{\omega }/\mathrm{c})}^2-{\mathrm{k}}^2\right]{\mathrm{E}}_{\mathrm{z}}=0\end{array}$

Next do the similar thing with the 2^nd Maxwell equation:

$\begin{array}{c}\nabla \cdot \overrightarrow{\mathrm{B}}=\frac{\partial {\mathrm{B}}_{\mathrm{x}}}{\partial \mathrm{x}}+\frac{\partial {\mathrm{B}}_{\mathrm{y}}}{\partial \mathrm{y}}+\frac{\partial {\mathrm{B}}_{\mathrm{z}}}{\partial \mathrm{z}}=\frac{\partial {\mathrm{B}}_{\mathrm{x}}}{\partial \mathrm{x}}+\frac{\partial {\mathrm{B}}_{\mathrm{y}}}{\partial \mathrm{y}}+{\mathrm{ikB}}_{\mathrm{z}}=0\leftarrow \left(9\right),\left(10\right)\\ \frac{1}{{(\mathrm{\omega }/\mathrm{c})}^2-{\mathrm{k}}^2}\mathrm{k}\left(\frac{{\partial }^2{\mathrm{B}}_{\mathrm{z}}}{\partial {\mathrm{x}}^2}+\frac{{\partial }^2{\mathrm{B}}_{\mathrm{z}}}{\partial {\mathrm{y}}^2}\right)+{\mathrm{ikB}}_{\mathrm{z}}=0/\cdot {(\mathrm{\omega }/\mathrm{c})}^2-{\mathrm{k}}^2\\ \left[\frac{{\partial }^2}{\partial {\mathrm{x}}^2}+\frac{{\partial }^2}{\partial {\mathrm{y}}^2}+{(\mathrm{\omega }/\mathrm{c})}^2-{\mathrm{k}}^2\right]{\mathrm{B}}_{\mathrm{z}}=0\end{array}$

In both cases the crucial thing was the cancellation of the mixed second derivatives.

Finally use the results of 9.180 with (i) and (iv) of 9.179 to obtain the same equations:

(i)

$\begin{array}{c}\frac{\partial {\mathrm{E}}_{\mathrm{y}}}{\partial \mathrm{x}}-\frac{\partial {\mathrm{E}}_{\mathrm{x}}}{\partial \mathrm{y}}={\mathrm{i\omega B}}_{\mathrm{z}}\\ \frac{\partial {\mathrm{E}}_{\mathrm{y}}}{\partial \mathrm{x}}=\frac{\mathrm{i}}{{(\mathrm{\omega }/\mathrm{c})}^2-{\mathrm{k}}^2}\left(\mathrm{k}\frac{{\partial }^2{\mathrm{E}}_{\mathrm{z}}}{\partial \mathrm{x}\partial \mathrm{y}}-\mathrm{\omega }\frac{{\partial }^2{\mathrm{B}}_{\mathrm{z}}}{\partial {\mathrm{x}}^2}\right)\\ \frac{\partial {\mathrm{E}}_{\mathrm{y}}}{\partial \mathrm{x}}=\frac{\mathrm{i}}{{(\mathrm{\omega }/\mathrm{c})}^2-{\mathrm{k}}^2}\left(\mathrm{k}\frac{{\partial }^2{\mathrm{E}}_{\mathrm{z}}}{\partial \mathrm{x}\partial \mathrm{y}}+\mathrm{\omega }\frac{{\partial }^2{\mathrm{B}}_{\mathrm{z}}}{\partial {\mathrm{x}}^2}\right)\\ \frac{\partial {\mathrm{E}}_{\mathrm{y}}}{\partial \mathrm{x}}=\frac{\mathrm{i}}{{(\mathrm{\omega }/\mathrm{c})}^2-{\mathrm{k}}^2}\mathrm{\omega }\left(\frac{{\partial }^2{\mathrm{B}}_{\mathrm{z}}}{\partial {\mathrm{x}}^2}+\frac{{\partial }^2{\mathrm{B}}_{\mathrm{z}}}{\partial {\mathrm{y}}^2}\right)-{\mathrm{i\omega B}}_{\mathrm{z}}=0/\cdot {(\mathrm{\omega }/\mathrm{c})}^2-{\mathrm{k}}^2\end{array}$

Therefore, the value of use the results of 9.180) with (i) to obtain the same equations is $\left[\frac{{\partial }^2}{\partial {\mathrm{x}}^2}+\frac{{\partial }^2}{\partial {\mathrm{y}}^2}+{(\mathrm{\omega }/\mathrm{c})}^2-{\mathrm{k}}^2\right]{\mathrm{B}}_2=0$.

(iv)

$\begin{array}{c}\frac{\partial {\mathrm{B}}_{\mathrm{y}}}{\partial \mathrm{x}}-\frac{\partial {\mathrm{B}}_{\mathrm{x}}}{\partial \mathrm{y}}=-\frac{\mathrm{i\omega }}{{\mathrm{c}}^2}{\mathrm{E}}_{\mathrm{z}}\\ \frac{\partial {\mathrm{B}}_{\mathrm{y}}}{\partial \mathrm{x}}=\frac{\mathrm{i}}{{(\mathrm{\omega }/\mathrm{c})}^2-{\mathrm{k}}^2}\left(\mathrm{k}\frac{{\partial }^2{\mathrm{B}}_{\mathrm{z}}}{\partial \mathrm{x}\partial \mathrm{y}}+\frac{\mathrm{\omega }}{{\mathrm{c}}^2}\frac{{\partial }^2{\mathrm{E}}_{\mathrm{z}}}{\partial {\mathrm{x}}^2}\right)\\ \frac{\partial {\mathrm{B}}_{\mathrm{x}}}{\partial \mathrm{y}}=\frac{\mathrm{i}}{{(\mathrm{\omega }/\mathrm{c})}^2-{\mathrm{k}}^2}\left(\mathrm{k}\frac{{\partial }^2{\mathrm{B}}_{\mathrm{z}}}{\partial \mathrm{x}\partial \mathrm{y}}-\frac{\mathrm{\omega }}{{\mathrm{c}}^2}\frac{{\partial }^2{\mathrm{E}}_{\mathrm{z}}}{\partial {\mathrm{x}}^2}\right)\\ \frac{\mathrm{i}}{{(\mathrm{\omega }/\mathrm{c})}^2-{\mathrm{k}}^2}\frac{\mathrm{\omega }}{{\mathrm{c}}^2}\left(\frac{{\partial }^2{\mathrm{E}}_{\mathrm{z}}}{\partial {\mathrm{x}}^2}+\frac{{\partial }^2{\mathrm{E}}_{\mathrm{z}}}{\partial {\mathrm{y}}^2}\right)+\mathrm{i}\frac{\mathrm{\omega }}{{\mathrm{c}}^2}{\mathrm{E}}_{\mathrm{z}}/\cdot {(\mathrm{\omega }/\mathrm{c})}^2-{\mathrm{k}}^2\end{array}$

Therefore, the value of use the results of 9.180) with (iv) to obtain the same equations is $\left[\frac{{\partial }^2}{\partial {\mathrm{x}}^2}+\frac{{\partial }^2}{\partial {\mathrm{y}}^2}+{(\mathrm{\omega }/\mathrm{c})}^2-{\mathrm{k}}^2\right]{\mathrm{E}}_2=0$.