Question

(a) Show that Eqs. 33-1 land 33-2 satisfy the wave equations displayed in Problem 108. (b) Show that any expressions of the form$\mathit{E}\mathbf{=}\mathbf{}{\mathit{E}}_{\mathbf{m}}\mathbf{}\mathit{f}\mathbf{(}\mathbf{kx}\mathbf{}\mathbf{\pm }\mathbf{\omega t}\mathbf{)}$ and $\mathit{B}\mathbf{}\mathbf{=}{\mathit{B}}_{\mathbf{m}}\mathbf{}\mathit{f}\mathbf{(}\mathbf{kx}\mathbf{}\mathbf{\pm }\mathbf{\omega t}\mathbf{)}$ , where $\mathit{f}\mathbf{(}\mathbf{kx}\mathbf{}\mathbf{\pm }\mathbf{\omega t}\mathbf{)}$denotes an arbitrary function, also satisfy these wave equations.

Short answer

(a) It is proved that the equations 33-1 and 33-2 satisfy the wave equations.

(b) It is proved that any expressions of the form $E=E_{m}f\left(kx\pm \omega t\right)$ and

$B=B_{m}f\left(kx\pm \omega t\right)$, where $f\left(kx\pm \omega t\right)$denotes an arbitrary function, also satisfy the wave equations.

Step-by-step solution

Listing the given quantities

$E=E_{m}f\left(kx\pm \omega t\right)$

$B=B_{m}f\left(kx\pm \omega t\right)$

Understanding the concepts of electromagnetic wave 

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:

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

(a) Explanation

From equation 33-1 , we have

$E=E_m\sin \left(kx-\omega t\right)$

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

$\frac{{\partial }^{2}E}{\partial t^2}=-{\omega }^2{E}_m\sin \left(kx-\omega t\right)$ …(1)

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

$\frac{{\partial }^{2}E}{\partial x^2}=-k^2{E}_m\sin \left(kx-\omega t\right)$

Multiplying$c^2$on both sides,

$c^2\frac{{\partial }^{2}E}{\partial x^2}=-c^2{k}^2{E}_m\sin \left(kx-\omega t\right)$

But$c=\frac{\omega }{k}$

Therefore,

$c^2\frac{{\partial }^{2}E}{\partial x^2}=-{\omega }^2{E}_m\sin \left(kx-\omega t\right)$ …(2)

From (1) and (2),

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

Thus, equation 33-1 satisfies the wave equation.

Now, from equation 33-2 , we have

$B=B_m\sin \left(kx\pm \omega t\right)$

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

$\frac{{\partial }^{2}B}{\partial t^2}=-{\omega }^2{B}_m\sin \left(kx-\omega t\right)$ …(3)

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

$\frac{{\partial }^{2}B}{\partial x^2}=-k^2{B}_m\sin \left(kx-\omega t\right)$

Multiplying$c^2$on both sides,

$c^2\frac{{\partial }^{2}B}{\partial x^2}=-c^2{k}^2{B}_m\sin \left(kx-\omega t\right)$

But$c=\frac{\omega }{k}$

Therefore,

$c^2\frac{{\partial }^{2}B}{\partial x^2}=-{\omega }^2{B}_m\sin \left(kx-\omega t\right)$ …(4)

From (3) and (4),

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

Thus, equation 33-2 satisfies the wave equation.

(b) Explanation

We have, $E=E_{m}f\left(kx\pm \omega t\right)$

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

$\frac{{\partial }^{2}E}{\partial t^2}=-{\omega }^2{E}_{m}f\left(kx\pm \omega t\right)$ …(5)

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

$\frac{{\partial }^{2}E}{\partial x^2}=-k^2{E}_{m}f\left(kx\pm \omega t\right)$

Multiplying$c^2$on both sides,

$c^2\frac{{\partial }^{2}E}{\partial x^2}=-c^2{k}^2{E}_{m}f\left(kx\pm \omega t\right)$

But$c=\frac{\omega }{k}$

Therefore,

$c^2\frac{{\partial }^{2}E}{\partial x^2}=-{\omega }^2{E}_{m}f\left(kx\pm \omega t\right)$ …(6)

From (5) and (6),

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

Thus,$E=E_{m}f\left(kx\pm \omega t\right)$satisfies the wave equation.

Similarly, we have,$B=B_{m}f\left(kx\pm \omega t\right)$

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

$\frac{{\partial }^{2}B}{\partial t^2}=-{\omega }^2{B}_{m}f\left(kx\pm \omega t\right)$ …(7)

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

$\frac{{\partial }^{2}B}{\partial x^2}=-k^2{B}_{m}f\left(kx\pm \omega t\right)$

Multiplying$c^2$on both sides,

But$c=\frac{\omega }{k}$

Therefore,

$c^2\frac{{\partial }^{2}B}{\partial x^2}=-{\omega }^2{B}_{m}f\left(kx\pm \omega t\right)$ …(8)

From (7) and (8),

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

Thus, $B=B_{m}f\left(kx\pm \omega t\right)$ satisfies the wave equation.