Chapter 34, Problem 70AP

(a).

To determine

The sketch of the graph of electric field for the given wave at time $t=0$.

Answer to Problem 70AP

The sketch of the graph of electric field for the given wave at time $t=0$ is shown in figure below.

Explanation of Solution

The sketch the graph of electric field for the given wave at time $t=0$ travelling in the $x$ direction is shown in figure below.

Figure-(1)

(b).

To determine

The energy density in the electric field as a function of $x$ at the instant $t=0$.

Answer to Problem 70AP

The energy density in the electric field as a function of $x$ at the instant $t=0$ is $\frac{E_{\max }{}^2{\epsilon }_0{\left(\cos \left(kx\right)\right)}^2}{2}$.

Explanation of Solution

Write the equation for the electric field varying in $y$ direction.

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

Here, $E$ is the electric field varying in $y$ direction, $E_{\max }$ is the amplitude of maximum electric field, $k$ is the wave vector and $\omega$ is the angular frequency.

Write the expression for the energy density in terms of electric field.

    ${\mu }_{\text{E}}=\frac{1}{2}{\epsilon }_0{E}^2$                                                                                                              (II)

Here, ${\mu }_{\text{E}}$ is the energy density and ${\epsilon }_0$ is the permittivity.

Substitute $E_{\max }\cos \left(kx-\omega t\right)$ for $E$ in equation (II).

    ${\mu }_{\text{E}}=\frac{1}{2}{\epsilon }_0{\left(E_{\max }\cos \left(kx-\omega t\right)\right)}^2$                                                                                (III)

Conclusion:

Substitute $0$ for $t$ in equation (III) to get the expression for ${\mu }_{\text{E}}$.

    $\begin{array}{c}{\mu }_{\text{E}}\left(x\right)=\frac{1}{2}{\epsilon }_0{\left(E_{\max }\cos \left(kx-\omega \left(0\right)\right)\right)}^2\\ =\frac{1}{2}{\epsilon }_0{\left(E_{\max }\cos \left(kx\right)\right)}^2\\ =\frac{E_{\max }{}^2{\epsilon }_0{\left(\cos \left(kx\right)\right)}^2}{2}\end{array}$

Therefore, the energy density in the electric field as a function of $x$ at the instant $t=0$ is $\frac{E_{\max }{}^2{\epsilon }_0{\left(\cos \left(kx\right)\right)}^2}{2}$.

(c).

To determine

The energy density in the magnetic field as a function of $x$ at the instant $t=0$.

Answer to Problem 70AP

The energy density in the magnetic field as a function of $x$ at the instant $t=0$ is $\frac{B_{\max }{}^2{\left(\cos \left(kx\right)\right)}^2}{2{\mu }_0}$.

Explanation of Solution

Write the equation for the magnetic field varying in $z$ direction.

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

Here, $B$ is the magnetic field varying in $z$ direction, $B_{\max }$ is the amplitude of maximum magnetic field, $k$ is the wave vector and $\omega$ is the angular frequency.

Write the expression for the energy density in terms of magnetic field.

    ${\mu }_{\text{B}}=\frac{B^2}{2{\mu }_0}$                                                                                                                  (V)

Here, ${\mu }_{\text{B}}$ is the energy density and ${\mu }_0$ is the permeability.

Substitute $B_{\max }\cos \left(kx-\omega t\right)$ for $B$ in equation (V).

    ${\mu }_{\text{B}}=\frac{{\left(B_{\max }\cos \left(kx-\omega t\right)\right)}^2}{2{\mu }_0}$                                                                                      (VI)

Conclusion:

Substitute $0$ for $t$ in equation (VI) to get the expression for ${\mu }_{\text{B}}$.

    $\begin{array}{c}{\mu }_{\text{B}}\left(x\right)={\frac{\left(B_{\max }\cos \left(kx-\omega \left(0\right)\right)\right)}{2{\mu }_0}}^2\\ ={\frac{\left(B_{\max }\cos \left(kx\right)\right)}{2{\mu }_0}}^2\\ =\frac{B_{\max }{}^2{\left(\cos \left(kx\right)\right)}^2}{2{\mu }_0}\end{array}$

Therefore, the energy density in the magnetic field as a function of $x$ at the instant $t=0$ is $\frac{B_{\max }{}^2{\left(\cos \left(kx\right)\right)}^2}{2{\mu }_0}$.

(d).

To determine

The total energy density as a function of $x$ in terms of electric field amplitude.

Answer to Problem 70AP

The total energy density as a function of $x$ in terms of electric field amplitude is ${\epsilon }_0{E}_{\max }{}^2{\left(\cos \left(kx\right)\right)}^2$.

Explanation of Solution

Write the expression for the total energy density.

    $u\left(x\right)=u_{\text{E}}\left(x\right)+u_{\text{B}}\left(x\right)$                                                                                            (VII)

Substitute $\frac{B_{\max }{}^2{\left(\cos \left(kx\right)\right)}^2}{2{\mu }_0}$ for $u_{\text{B}}\left(x\right)$ and $\frac{E_{\max }{}^2{\epsilon }_0{\left(\cos \left(kx\right)\right)}^2}{2}$ for $u_{\text{E}}\left(x\right)$ in equation (VII).

    $\begin{array}{c}u\left(x\right)=\frac{E_{\max }{}^2{\epsilon }_0{\left(\cos \left(kx\right)\right)}^2}{2}+\frac{B_{\max }{}^2{\left(\cos \left(kx\right)\right)}^2}{2{\mu }_0}\\ u\left(x\right)=\frac{1}{2}\left({\epsilon }_0{E}_{\max }{}^2+\frac{B_{\max }{}^2}{2{\mu }_0}\right){\left(\cos \left(kx\right)\right)}^2\end{array}$                                                     (VIII)

Write the relation between the electric and the magnetic field.

    $\frac{E}{B}=\frac{1}{\sqrt{{\mu }_0{\epsilon }_0}}$                                                                                                             (IX)

Substitute $B_{\max }\cos \left(kx\right)$ for $B$ and $E_{\max }\cos \left(kx\right)$ for $E$ in equation (IX).

    $\begin{array}{c}\frac{E_{\max }\cos \left(kx\right)}{B_{\max }\cos \left(kx\right)}=\frac{1}{\sqrt{{\mu }_0{\epsilon }_0}}\\ B_{\max }=E_{\max }\left(\sqrt{{\mu }_0{\epsilon }_0}\right)\end{array}$

Substitute $E_{\max }\left(\sqrt{{\mu }_0{\epsilon }_0}\right)$ for $B_{\max }$ in equation (VIII) to get the expression for $u\left(x\right)$.

    $\begin{array}{c}u\left(x\right)=\frac{1}{2}\left({\epsilon }_0{E}_{\max }{}^2+\frac{{\left(E_{\max }\left(\sqrt{{\mu }_0{\epsilon }_0}\right)\right)}^2}{{\mu }_0}\right){\left(\cos \left(kx\right)\right)}^2\\ =\frac{1}{2}\left({\epsilon }_0{E}_{\max }{}^2+\frac{\left(E_{\max }{}^2{\mu }_0{\epsilon }_0\right)}{{\mu }_0}\right){\left(\cos \left(kx\right)\right)}^2\\ ={\epsilon }_0{E}_{\max }{}^2{\left(\cos \left(kx\right)\right)}^2\end{array}$

Therefore, the total energy density as a function of $x$ in terms of electric field amplitude is ${\epsilon }_0{E}_{\max }{}^2{\left(\cos \left(kx\right)\right)}^2$.

(e).

To determine

The energy in the "shoebox" in terms of $A$, $\lambda$, $E_{\max }$ and universal constant.

Answer to Problem 70AP

The energy in the "shoebox" in terms of $A$, $\lambda$, $E_{\max }$ and universal constant is $\frac{1}{2}{\epsilon }_0{E}_{\max }{}^{2}A\lambda$.

Explanation of Solution

Write the given equation for energy.

    $E_{\text{λ}}={\displaystyle {\int }_0^{\lambda }uAdx}$                                                                                                            (X)

Here, $E_{\text{λ}}$ is the energy in the shoebox of length $\lambda$ and $A$ is the frontal area.

Substitute ${\epsilon }_0{E}_{\max }{}^2{\left(\cos \left(kx\right)\right)}^2$ for $u$ in equation (X).

    $\begin{array}{c}{E}_{\text{λ}}={\displaystyle {\int }_0^{\lambda }\left({\epsilon }_0{E}_{\max }{}^2{\left(\cos \left(kx\right)\right)}^2\right)Adx}\\ ={\epsilon }_0{E}_{\max }{}^{2}A{\displaystyle {\int }_0^{\lambda }\left({\left(\cos \left(kx\right)\right)}^2\right)dx}\\ ={\epsilon }_0{E}_{\max }{}^{2}A{\left[\frac{\sin \left(2kx\right)+2kx}{4k}\right]}_0^{\lambda }\\ ={\epsilon }_0{E}_{\max }{}^{2}A\left(\frac{\sin \left(2k\lambda \right)+2k\lambda }{4k}\right)\end{array}$                                                                           (XI)

Write the expression for the wave vector.

    $k=\frac{2\pi }{\lambda }$                                                                                                                   (XII)

Substitute $\frac{2\pi }{\lambda }$ for $k$ in equation (XI).

`    $\begin{array}{c}{E}_{\text{λ}}={\epsilon }_0{E}_{\max }{}^{2}A\left(\frac{\sin \left(2\left(\frac{2\pi }{\lambda }\right)\lambda \right)+2\left(\left(\frac{2\pi }{\lambda }\right)\lambda \right)}{4\left(\frac{2\pi }{\lambda }\right)}\right)\\ ={\epsilon }_0{E}_{\max }{}^{2}A\left(\frac{\sin \left(\pi \right)+4\pi }{4\left(\frac{2\pi }{\lambda }\right)}\right)\\ ={\epsilon }_0{E}_{\max }{}^{2}A\left(\frac{4\lambda }{8}\right)\\ =\frac{1}{2}{\epsilon }_0{E}_{\max }{}^{2}A\lambda \end{array}$                                                    (XIII)

Therefore, the energy in the "shoebox" in terms of $A$, $\lambda$, $E_{\max }$ and universal constant is $\frac{1}{2}{\epsilon }_0{E}_{\max }{}^{2}A\lambda$.

(f).

To determine

The power the wave carries through the area $A$.

Answer to Problem 70AP

The power the wave carries through the area $A$ is $\frac{1}{2}{\epsilon }_0{E}_{\max }{}^{2}Ac$.

Explanation of Solution

Write the expression for power through an area $A$.

    $P=\frac{E_{\text{λ}}}{T}$                                                                                                                 (XIV)

Here, $P$ is the power through an area $A$ and $T$ is the period of a wave.

Substitute $\frac{1}{2}{\epsilon }_0{E}_{\max }{}^{2}A\lambda$ for $E_{\text{λ}}$ and $\frac{1}{f}$ for $T$ in equation (XIV).

    $\begin{array}{c}P=\frac{\frac{1}{2}{\epsilon }_0{E}_{\max }{}^{2}A\lambda }{\frac{1}{f}}\\ P=\frac{1}{2}{\epsilon }_0{E}_{\max }{}^{2}A\lambda f\end{array}$                                                                                                 (XV)

Also,

    $c=\lambda f$

Here, $c$ is the speed of light and $f$ is the frequency.

Substitute $c$ for $\lambda f$ in equation (XV).

    $P=\frac{1}{2}{\epsilon }_0{E}_{\max }{}^{2}Ac$

Therefore, the power the wave carries through the area $A$ is $\frac{1}{2}{\epsilon }_0{E}_{\max }{}^{2}Ac$.

(g).

To determine

The intensity in terms of $E_{\max }$ and universal constant.

Answer to Problem 70AP

The intensity in terms of $E_{\max }$ and universal constant is $\frac{E_{\max }{}^2}{2{\mu }_{0}c}$.

Explanation of Solution

Write the expression for intensity.

    $I=\frac{P}{A}$                                                                                                                   (XVI)

Here, $I$ is the intensity.

Substitute $\frac{1}{2}{\epsilon }_0{E}_{\max }{}^{2}Ac$ for $P$ in equation (XVI) to solve for $I$.

    $\begin{array}{c}I=\frac{\frac{1}{2}{\epsilon }_0{E}_{\max }{}^{2}Ac}{A}\\ =\frac{{\epsilon }_0{E}_{\max }{}^{2}c}{2}\end{array}$                                                                                                 (XVII)

Also,

    $\begin{array}{c}c=\frac{1}{\sqrt{{\mu }_0{\epsilon }_0}}\\ c^2=\frac{1}{{\mu }_0{\epsilon }_0}\\ c{\epsilon }_0=\frac{1}{c{\mu }_0}\end{array}$

Substitute $\frac{1}{c{\mu }_0}$ for $c{\epsilon }_0$ in equation (XVII).

    $I=\frac{E_{\max }{}^2}{2{\mu }_{0}c}$

Therefore, the intensity in terms of $E_{\max }$ and universal constant is $\frac{E_{\max }{}^2}{2{\mu }_{0}c}$.

(h).

To determine

The comparison of the result in part $g$ with equation $\left(34.24\right)$ $\left(I=\frac{E_{\max }{}^2}{2{\mu }_{0}c}\right)$.

Answer to Problem 70AP

On comparison of equation (XVIII) and (XIX), both the equation yields the same result for the intensity. Therefore, the intensity is same for both the conditions.

Explanation of Solution

The equation (34.24) is,

    $I=\frac{{\epsilon }_0{E}_{\max }{}^{2}c}{2}$                                                                                                     (XVIII)

The result obtained in part $\left(g\right)$ is,

    $I=\frac{{\epsilon }_0{E}_{\max }{}^{2}c}{2}$                                                                                                         (XIX)

On comparison of equation (XVIII) and (XIX), both the equation yields the same result for the intensity. Therefore, the intensity is same for both the conditions.