91Ó°ÊÓ

$E\left(x,t\right)$ and $B\left(x,t\right)$ are the electric and magnetic field components.

We have to take the derivative of equation 33-11 with respect to $\mathit{x}$ and the derivative of equation 33-17 with respect to $\mathit{t}$. By comparing these two equations, we can prove that the electric field satisfies the wave equation. We can use the same concept to prove that the magnetic field satisfies the wave equation.

Formula:

$\frac{∂E}{∂x}=-\frac{∂B}{∂t}$

$c=\frac{1}{\sqrt{{\mathrm{μ}}_0{\mathrm{Î\mu }}_0}}$

$-\frac{∂B}{∂x}={\mathrm{μ}}_0{\mathrm{Î\mu }}_0\frac{∂E}{∂t}$

From equation 33-11 , we have

$\frac{∂E}{∂x}=-\frac{∂B}{∂t}$

Differentiating this with respect to $x$ we get

$\frac{∂}{∂x}\left(\frac{∂E}{∂x}\right)=\frac{∂}{∂x}\left(-\frac{∂B}{∂t}\right)$

$\frac{{∂}^{2}E}{∂x^2}=-\frac{{∂}^{2}B}{∂x∂t}······\left(1\right)$

Now, from equation , we have

$-\frac{∂B}{∂x}={\mathrm{μ}}_0{\mathrm{Î\mu }}_0\frac{∂E}{∂t}$

Differentiating this equation with respect to $t$, we get

$$\begin{gathered} \frac{∂}{∂t}\left(-\frac{∂B}{∂x}\right)={\mathrm{μ}}_0{\mathrm{Î\mu }}_0\frac{∂}{∂t}\left(\frac{∂E}{∂t}\right) \\ \left(-\frac{{∂}^{2}B}{∂t∂x}\right)={\mathrm{μ}}_0{\mathrm{Î\mu }}_0\left(\frac{{∂}^{2}E}{∂t^2}\right)······\left(2\right) \end{gathered}$$

Comparing (1) and (2) , we get

$$\begin{gathered} {\mathrm{μ}}_0{\mathrm{Î\mu }}_0\frac{{∂}^{2}E}{∂t^2}=\frac{{∂}^{2}E}{∂x^2} \\ \frac{{∂}^{2}E}{∂t^2}=\frac{1}{{\mathrm{μ}}_0{\mathrm{Î\mu }}_0}\frac{{∂}^{2}E}{∂x^2} \end{gathered}$$

But, $c=\frac{1}{\sqrt{{\mathrm{μ}}_0{\mathrm{Î\mu }}_0}}$, so

$\frac{{∂}^{2}E}{∂t^2}=c^2\frac{{∂}^{2}E}{∂x^2}$

Thus, the electric field satisfies the wave equation.

Now, differentiating equation 33-11 with respect to $t$ we get

$$\begin{gathered} \frac{∂}{∂t}\left(\frac{∂E}{∂x}\right)=\frac{∂}{∂t}\left(-\frac{∂B}{∂t}\right) \\ \left(\frac{{∂}^{2}E}{∂t∂x}\right)=\left(-\frac{{∂}^{2}B}{∂t^2}\right)······\left(3\right) \end{gathered}$$

Now, differentiating equation 33-17 with respect to $x$,

$$\begin{gathered} -\frac{∂B}{∂x}={\mathrm{μ}}_0{\mathrm{Î\mu }}_0\frac{∂E}{∂t} \\ \frac{∂}{∂x}\left(-\frac{∂B}{∂x}\right)={\mathrm{μ}}_0{\mathrm{Î\mu }}_0\frac{∂}{∂x}\left(\frac{∂E}{∂t}\right) \\ \left(-\frac{{∂}^{2}B}{∂x^2}\right)={\mathrm{μ}}_0{\mathrm{Î\mu }}_0\left(\frac{{∂}^{2}E}{∂x∂t}\right)·····\left(4\right) \end{gathered}$$

By comparing equations (3) and (4) , we get

$$\begin{gathered} \left(-\frac{{∂}^{2}B}{∂x^2}\right)={\mathrm{μ}}_0{\mathrm{Î\mu }}_0\left(-\frac{{∂}^{2}B}{∂t^2}\right) \\ \frac{{∂}^{2}B}{∂x^2}={\mathrm{μ}}_0{\mathrm{Î\mu }}_0\frac{{∂}^{2}B}{∂t^2} \\ \frac{{∂}^{2}B}{∂t^2}=\frac{1}{{\mathrm{μ}}_0{\mathrm{Î\mu }}_0}\frac{{∂}^{2}B}{∂x^2} \end{gathered}$$

But, $c=\frac{1}{\sqrt{{\mathrm{μ}}_0{\mathrm{Î\mu }}_0}}$, so

$\frac{{∂}^{2}B}{∂t^2}=c^2\frac{{∂}^{2}B}{∂x^2}$

Thus, the magnetic field satisfies the wave equation.