91Ó°ÊÓ

$E=E_{m}f\left(kxÂ\pm \mathrm{Ó\neg }t\right)$

$B=B_{m}f\left(kxÂ\pm \mathrm{Ó\neg }t\right)$

An electromagnetic wave traveling along an $\mathit{x}$axis has an electric field and a magnetic field with magnitudes that depend on $\mathit{x}$and$\mathit{t}$is given by equation 33-1 and equation 33-2. From using the result of 108, we find for this equation.

Formula:

$$\begin{gathered} \frac{{\mathbf{∂}}^{\mathbf{2}}\mathbf{E}}{\mathbf{∂}{\mathbf{t}}^{\mathbf{2}}}\mathbf{=}{\mathit{c}}^{\mathbf{2}}\frac{{\mathbf{∂}}^{\mathbf{2}}\mathbf{E}}{\mathbf{∂}{\mathbf{x}}^{\mathbf{2}}} \\ \frac{{\mathbf{∂}}^{\mathbf{2}}\mathbf{B}}{\mathbf{∂}{\mathbf{t}}^{\mathbf{2}}}\mathbf{=}{\mathit{c}}^{\mathbf{2}}\frac{{\mathbf{∂}}^{\mathbf{2}}\mathbf{B}}{\mathbf{∂}{\mathbf{x}}^{\mathbf{2}}} \end{gathered}$$

From equation 33-1 , we have

$E=E_m\sin \left(kx-\mathrm{Ó\neg }t\right)$

By finding its derivative twice with respect to $t$, we get

$\frac{{∂}^{2}E}{∂t^2}=-{\mathrm{Ó\neg }}^2{E}_m\sin \left(kx-\mathrm{Ó\neg }t\right)$ …(1)

Similarly, by finding its derivative twice with respect to$x$, we get

$\frac{{∂}^{2}E}{∂x^2}=-k^2{E}_m\sin \left(kx-\mathrm{Ó\neg }t\right)$

Multiplying$c^2$on both sides,

$c^2\frac{{∂}^{2}E}{∂x^2}=-c^2{k}^2{E}_m\sin \left(kx-\mathrm{Ó\neg }t\right)$

But$c=\frac{\mathrm{Ó\neg }}{k}$

Therefore,

$c^2\frac{{∂}^{2}E}{∂x^2}=-{\mathrm{Ó\neg }}^2{E}_m\sin \left(kx-\mathrm{Ó\neg }t\right)$ …(2)

From (1) and (2),

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

Thus, equation 33-1 satisfies the wave equation.

Now, from equation 33-2 , we have

$B=B_m\sin \left(kxÂ\pm \mathrm{Ó\neg }t\right)$

By finding its derivative twice with respect to $t$, we get

$\frac{{∂}^{2}B}{∂t^2}=-{\mathrm{Ó\neg }}^2{B}_m\sin \left(kx-\mathrm{Ó\neg }t\right)$ …(3)

Similarly, by finding its derivative twice with respect to$x$, we get

$\frac{{∂}^{2}B}{∂x^2}=-k^2{B}_m\sin \left(kx-\mathrm{Ó\neg }t\right)$

Multiplying$c^2$on both sides,

$c^2\frac{{∂}^{2}B}{∂x^2}=-c^2{k}^2{B}_m\sin \left(kx-\mathrm{Ó\neg }t\right)$

But$c=\frac{\mathrm{Ó\neg }}{k}$

Therefore,

$c^2\frac{{∂}^{2}B}{∂x^2}=-{\mathrm{Ó\neg }}^2{B}_m\sin \left(kx-\mathrm{Ó\neg }t\right)$ …(4)

From (3) and (4),

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

Thus, equation 33-2 satisfies the wave equation.

We have, $E=E_{m}f\left(kxÂ\pm \mathrm{Ó\neg }t\right)$

By finding its derivative twice with respect to $t$, we get

$\frac{{∂}^{2}E}{∂t^2}=-{\mathrm{Ó\neg }}^2{E}_{m}f\left(kxÂ\pm \mathrm{Ó\neg }t\right)$ …(5)

Similarly, by finding its derivative twice with respect to$x$, we get

$\frac{{∂}^{2}E}{∂x^2}=-k^2{E}_{m}f\left(kxÂ\pm \mathrm{Ó\neg }t\right)$

Multiplying$c^2$on both sides,

$c^2\frac{{∂}^{2}E}{∂x^2}=-c^2{k}^2{E}_{m}f\left(kxÂ\pm \mathrm{Ó\neg }t\right)$

But$c=\frac{\mathrm{Ó\neg }}{k}$

Therefore,

$c^2\frac{{∂}^{2}E}{∂x^2}=-{\mathrm{Ó\neg }}^2{E}_{m}f\left(kxÂ\pm \mathrm{Ó\neg }t\right)$ …(6)

From (5) and (6),

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

Thus,$E=E_{m}f\left(kxÂ\pm \mathrm{Ó\neg }t\right)$satisfies the wave equation.

Similarly, we have,$B=B_{m}f\left(kxÂ\pm \mathrm{Ó\neg }t\right)$

By finding its derivative twice with respect to $t$, we get

$\frac{{∂}^{2}B}{∂t^2}=-{\mathrm{Ó\neg }}^2{B}_{m}f\left(kxÂ\pm \mathrm{Ó\neg }t\right)$ …(7)

Similarly, by finding its derivative twice with respect to$x$, we get

$\frac{{∂}^{2}B}{∂x^2}=-k^2{B}_{m}f\left(kxÂ\pm \mathrm{Ó\neg }t\right)$

Multiplying$c^2$on both sides,

But$c=\frac{\mathrm{Ó\neg }}{k}$

Therefore,

$c^2\frac{{∂}^{2}B}{∂x^2}=-{\mathrm{Ó\neg }}^2{B}_{m}f\left(kxÂ\pm \mathrm{Ó\neg }t\right)$ …(8)

From (7) and (8),

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

Thus, $B=B_{m}f\left(kxÂ\pm \mathrm{Ó\neg }t\right)$ satisfies the wave equation.