Chapter 34, Problem 16P

To determine

The verification for $E=E_{\text{max}}\cos \left(kx-\omega t\right)$ and $B=B_{\text{max}}\cos \left(kx-\omega t\right)$ being the solution to equations $34.15$ and $34.16$ respectively.

Answer to Problem 16P

The verification for $E=E_{\text{max}}\cos \left(kx-\omega t\right)$ and $B=B_{\text{max}}\cos \left(kx-\omega t\right)$ being the solution to equations $34.15$ and $34.16$ respectively is as stated below.

Explanation of Solution

The given equations are.

    $E=E_{\text{max}}\cos \left(kx-\omega t\right)$                                                                                                 (I)

Here, $E$ is the magnitude of electric field , $E_{\max }$ is the magnitude of the maximum electric field, $k$ is the wave vector and $\omega$ is the angular frequency.

    $B=B_{\text{max}}\cos \left(kx-\omega t\right)$                                                                                               (II)

Here, $B$ is the magnitude of magnetic field , $B_{\max }$ is the magnitude of the maximum magnetic field.

Differentiate equation (I) partially twice with respect to $x$.

    $\begin{array}{c}\frac{\partial E}{\partial x}=-E_{\text{max}}\sin \left(kx-\omega t\right)\left(k\right)\\ \frac{{\partial }^{2}E}{\partial x^2}=-E_{\text{max}}\cos \left(kx-\omega t\right)\left(k^2\right)\end{array}$                                                                                (III)

Differentiate equation (I) partially twice with respect to $t$.

    $\begin{array}{c}\frac{\partial E}{\partial t}=-E_{\text{max}}\sin \left(kx-\omega t\right)\left(-\omega \right)\\ \frac{{\partial }^{2}E}{\partial t^2}=-E_{\text{max}}\cos \left(kx-\omega t\right){\left(-\omega \right)}^2\end{array}$                                                                               (IV)

Differentiate equation (II) partially twice with respect to $x$.

    $\begin{array}{c}\frac{\partial B}{\partial x}=-B_{\text{max}}\sin \left(kx-\omega t\right)\left(k\right)\\ \frac{{\partial }^{2}B}{\partial x^2}=-B_{\text{max}}\cos \left(kx-\omega t\right)\left(k^2\right)\end{array}$                                                                                 (V)

Differentiate equation (II) partially twice with respect to $t$.

    $\begin{array}{c}\frac{\partial B}{\partial t}=-B_{\text{max}}\sin \left(kx-\omega t\right)\left(\omega \right)\\ \frac{{\partial }^{2}B}{\partial t^2}=-B_{\text{max}}\cos \left(kx-\omega t\right){\left(-\omega \right)}^2\end{array}$                                                                               (VI)

Divide equation (III) by equation (IV).

    $\begin{array}{c}\frac{\frac{{\partial }^{2}E}{\partial x^2}}{\frac{{\partial }^{2}E}{\partial t^2}}=\frac{-E_{\text{max}}\cos \left(kx-\omega t\right)\left(k^2\right)}{-E_{\text{max}}\cos \left(kx-\omega t\right){\left(-\omega \right)}^2}\\ \frac{{\partial }^{2}E}{\partial x^2}=\frac{k^2}{{\omega }^2}\left(\frac{{\partial }^{2}E}{\partial t^2}\right)\end{array}$                                                                          (VII)

Divide equation (V) by equation (VI).

    $\begin{array}{c}\frac{\frac{{\partial }^{2}B}{\partial x^2}}{\frac{{\partial }^{2}B}{\partial t^2}}=\frac{-B_{\text{max}}\cos \left(kx-\omega t\right)\left(k^2\right)}{-B_{\text{max}}\cos \left(kx-\omega t\right){\left(-\omega \right)}^2}\\ \frac{{\partial }^{2}E}{\partial x^2}=\frac{k^2}{{\omega }^2}\left(\frac{{\partial }^{2}B}{\partial t^2}\right)\end{array}$                                                                         (VIII)

Also,

    $\frac{k}{\omega }=\sqrt{{\mu }_0{\epsilon }_0}$                                                                                                              (IX)

Here, ${\mu }_0$ is the permeability of the medium and ${\epsilon }_0$ is the permittivity if the medium.

Substitute $\sqrt{{\mu }_0{\epsilon }_0}$ for $\frac{k}{\omega }$ in equation (VII).

    $\begin{array}{c}\frac{{\partial }^{2}E}{\partial x^2}={\left(\sqrt{{\mu }_0{\epsilon }_0}\right)}^2\left(\frac{{\partial }^{2}E}{\partial t^2}\right)\\ \frac{{\partial }^{2}E}{\partial x^2}={\mu }_0{\epsilon }_0\left(\frac{{\partial }^{2}E}{\partial t^2}\right)\end{array}$                                                                                           (X)

The equation $34.15$ is ,

    $\frac{{\partial }^{2}E}{\partial x^2}={\mu }_0{\epsilon }_0\left(\frac{{\partial }^{2}E}{\partial t^2}\right)$                                                                                                 (XI)

Substitute $\sqrt{{\mu }_0{\epsilon }_0}$ for $\frac{k}{\omega }$ in equation (VIII).

    $\begin{array}{c}\frac{{\partial }^{2}B}{\partial x^2}={\left(\sqrt{{\mu }_0{\epsilon }_0}\right)}^2\left(\frac{{\partial }^{2}B}{\partial t^2}\right)\\ \frac{{\partial }^{2}B}{\partial x^2}={\mu }_0{\epsilon }_0\left(\frac{{\partial }^{2}B}{\partial t^2}\right)\end{array}$                                                                                          (XII)

The equation $34.16$ is,

    $\frac{{\partial }^{2}B}{\partial x^2}={\mu }_0{\epsilon }_0\left(\frac{{\partial }^{2}B}{\partial t^2}\right)$                                                                                              (XIII)

Conclusion:

Equation (X) is equal to equation (XI) and equation (XII) equals to equal (XIII).

Hence, verified.