Chapter 34, Problem 34.70AP

You may wish to review Sections 16.4 and 16.8 on the transport of energy by string waves and sound. Figure P33.46 is a graphical representation of an electromagnetic wave moving in the x direction. We wish to find an expression for the intensity of this wave by means of a different process from that by which Equation 33.27 was generated. (a) Sketch a graph of the electric field in this wave at the instant t = 0, letting your flat paper represent the xy plane. (b) Compute the energy density u_E in the electric field as a function of x at the instant t = 0. (c) Compute the energy density in the magnetic field u_B as a function of x at that instant. (d) Find the total energy density u as a function of x, expressed in terms of only the electric field amplitude. (e) The energy in a “shoebox” of length λ and frontal area A is $E_{\lambda }={\displaystyle {\int }_0^{\lambda }uAdx}$. (The symbol E_λ for energy in a wavelength imitates the notation of Section 16.4.) Perform the integration to compute the amount of this energy in terms of A, λ, E_max, and universal constants. (f) We may think of the energy transport by the whole wave as a series of these shoeboxes going past as if carried on a conveyor belt. Each shoebox passes by a point in a time interval defined as the period T = 1/f of the wave. Find the power the wave carries through area A. (g) The intensity of the wave is the power per unit area through which the wave passes. Compute this intensity in terms of E_max and universal constants. (h) Explain how your result compares with that given in Equation 33.27.

Figure P33.46

To determine

To draw: The sketch a graph of the electric field at the instant $t=0$, the flat paper representing the $xy$ plane.

Answer to Problem 34.70AP

The graph of the electric field is shown in the diagram below.

Figure (1)

Explanation of Solution

Given info: The expression for the average intensity of the wave is,

$S_{\text{avg}}=\frac{E_{\max }{B}_{\max }}{2{\mu }_{\circ }}=\frac{E^2{}_{\max }}{2{\mu }_{\circ }c}=\frac{c{B}^2{}_{\max }}{2{\mu }_{\circ }}$

Introduction: The electric filed is a region around a charged particle within which a force would be experienced by other particles that may be attractive force or repulsive force.

Explanation:

The electric field in the figure is perpendicular to the magnetic field and the sinusoidal wave is travelling with the speed of light. The waves move on the positive x axis and the electric field is in the $xy$-plane. The wavelength of the of the wave is $\lambda$.

The expression for the electric field is,

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

Here,

$E_{\text{max}}$ is the maximum value of the magnitude of electric field.

$k$ is the angular wave number.

$\omega$ is the angular frequency.

Substitute $0$ for $t$ in the above equation to find the value of $\text{E}$.

$E=E_{\text{max}}\cos \left(kx\right)$ (1)

The diagram of the electric field is shown below.

Figure (1)

To determine

The energy density $u_{\text{E}}$ in the electric field as a function of $x$ at an instant $t=0$ .

Answer to Problem 34.70AP

The energy density $u_{\text{E}}$ in the electric field as a function of $x$ at an instant $t=0$ is $\frac{1}{2}{\epsilon }_{\circ }{E}^2{}_{\text{max}}{\cos }^2\left(kx\right)$.

Explanation of Solution

Given info: The expression for the average intensity of the wave is,

$S_{\text{avg}}=\frac{E_{\max }{B}_{\max }}{2{\mu }_{\circ }}=\frac{E^2{}_{\max }}{2{\mu }_{\circ }c}=\frac{c{B}^2{}_{\max }}{2{\mu }_{\circ }}$ .

The formula to calculate the energy density in the electric field is,

$u_{\text{E}}=\frac{1}{2}{\epsilon }_{\circ }{E}^2$

Here,

${\epsilon }_{\circ }$ is the permittivity of the free space.

$E$ is the magnitude of electric field.

Substitute $E_{\text{max}}\cos \left(kx\right)$ for $E$ in the above equation to find the value of $u_{\text{E}}$.

$\begin{array}{c}{u}_{\text{E}}=\frac{1}{2}{\epsilon }_{\circ }{\left(E_{\text{max}}\cos \left(kx\right)\right)}^2\\ =\frac{1}{2}{\epsilon }_{\circ }{E}^2{}_{\text{max}}{\cos }^2\left(kx\right)\end{array}$

Thus, the energy density $u_{\text{E}}$ in the electric field as a function of $x$ at an instant $t=0$ is $\frac{1}{2}{\epsilon }_{\circ }{E}^2{}_{\text{max}}{\cos }^2\left(kx\right)$.

Conclusion:

Therefore, the energy density $u_{\text{E}}$ in the electric field as a function of $x$ at an instant $t=0$ is $\frac{1}{2}{\epsilon }_{\circ }{E}^2{}_{\text{max}}{\cos }^2\left(kx\right)$.

To determine

The energy density $u_{\text{B}}$ in the magnetic field as a function of $x$ at an instant $t=0$.

Answer to Problem 34.70AP

The energy density $u_{\text{B}}$ in the magnetic field as a function of $x$ at an instant $t=0$ is $\frac{1}{2{\mu }_{\circ }}{B}^2{}_{\text{max}}{\cos }^2\left(kx\right)$.

Explanation of Solution

Given info: The expression for the average intensity of the wave is,

$S_{\text{avg}}=\frac{E_{\max }{B}_{\max }}{2{\mu }_{\circ }}=\frac{E^2{}_{\max }}{2{\mu }_{\circ }c}=\frac{c{B}^2{}_{\max }}{2{\mu }_{\circ }}$ .

The expression for the magnetic field is,

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

Here,

$B_{\text{max}}$ is the maximum value of the magnitude of magnetic field.

$k$ is the angular wave number.

$\omega$ is the angular frequency

Substitute $0$ for $t$ in the above equation to find the value of $B$.

$B=B_{\text{max}}\cos \left(kx\right)$ (2)

The formula to calculate the energy density in the magnetic is,

$u_{\text{B}}=\frac{1}{2{\mu }_{\circ }}{B}^2$

Here,

${\mu }_{\circ }$ is the permissibility of the region.

$B$ is the magnetic field.

Substitute $B_{\text{max}}\cos \left(kx\right)$ for $B$ in the above equation to find the value of $u_{\text{B}}$.

$\begin{array}{c}{u}_{\text{B}}=\frac{1}{2{\mu }_{\circ }}{\left(B_{\text{max}}\cos \left(kx\right)\right)}^2\\ =\frac{1}{2{\mu }_{\circ }}{B}^2{}_{\text{max}}{\cos }^2\left(kx\right)\end{array}$

Thus, the energy density $u_{\text{B}}$ in the magnetic field as a function of $x$ at an instant $t=0$ is $\frac{1}{2{\mu }_{\circ }}{B}^2{}_{\text{max}}{\cos }^2\left(kx\right)$.

Conclusion:

Therefore, the energy density $u_{\text{B}}$ in the magnetic field as a function of $x$ at an instant $t=0$ is $\frac{1}{2{\mu }_{\circ }}{B}^2{}_{\text{max}}{\cos }^2\left(kx\right)$.

To determine

The total energy density in terms of electric field amplitude.

Answer to Problem 34.70AP

The total energy density in terms of electric field amplitude is ${\in }_{\circ }{E}^2{}_{\text{max}}{\cos }^2\left(kx\right)$.

Explanation of Solution

Given info: The expression for the average intensity of the wave is,

$S_{\text{avg}}=\frac{E_{\max }{B}_{\max }}{2{\mu }_{\circ }}=\frac{E^2{}_{\max }}{2{\mu }_{\circ }c}=\frac{c{B}^2{}_{\max }}{2{\mu }_{\circ }}$ .

The expression for the energy due to magnetic field at instant $t=0$ is,

$u_{\text{B}}=\frac{1}{2{\mu }_{\circ }}{B}^2{}_{\text{max}}{\cos }^2\left(kx\right)$ (3)

The value of $B_{\text{max}}$ is given by,

$\begin{array}{c}{B}_{\text{max}}=\frac{E_{\text{max}}}{c}\\ =\frac{E_{\text{max}}}{\frac{1}{\sqrt{{\epsilon }_{\circ }{\mu }_{\circ }}}}\end{array}$

Substitute $\frac{E_{\text{max}}}{\frac{1}{\sqrt{{\epsilon }_{\circ }{\mu }_{\circ }}}}$ for $B_{\text{max}}$ in equation (3) to find the value of $u_{\text{B}}$.

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

The formula to calculate the total energy density is,

$u=u_{\text{E}}+u_{\text{B}}$

Substitute $\frac{1}{2}{\epsilon }_{\circ }{E}^2{}_{\text{max}}{\cos }^2\left(kx\right)$ for $u_{\text{B}}$ and $\frac{1}{2}{\epsilon }_{\circ }{E}^2{}_{\text{max}}{\cos }^2\left(kx\right)$ for $u_{\text{E}}$ in the above equation to find the value of $u$.

$\begin{array}{c}u=\frac{1}{2}{\epsilon }_{\circ }{E}^2{}_{\text{max}}{\cos }^2\left(kx\right)+\frac{1}{2}{\epsilon }_{\circ }{E}^2{}_{\text{max}}{\cos }^2\left(kx\right)\\ ={\epsilon }_{\circ }{E}^2{}_{\text{max}}{\cos }^2\left(kx\right)\end{array}$ (4)

Thus, the total energy density in terms of electric field amplitude is ${\epsilon }_{\circ }{E}^2{}_{\text{max}}{\cos }^2\left(kx\right)$.

Conclusion:

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

To determine

The amount of energy in the shoebox.

Answer to Problem 34.70AP

The amount of energy in the shoebox is $\frac{1}{2}{\epsilon }_{\circ }{E}^2{}_{\text{max}}A\lambda$.

Explanation of Solution

Given info: The expression for the average intensity of the wave is,

$S_{\text{avg}}=\frac{E_{\max }{B}_{\max }}{2{\mu }_{\circ }}=\frac{E^2{}_{\max }}{2{\mu }_{\circ }c}=\frac{c{B}^2{}_{\max }}{2{\mu }_{\circ }}$ .

The expression for the energy in the shoebox is,

$E_{\text{λ}}={\displaystyle \underset{0}{\overset{\text{λ}}{\int }}uAdx}$

Here,

$A$ is the area of the shoebox.

Substitute ${\epsilon }_{\circ }{E}^2{}_{\text{max}}{\cos }^2\left(kx\right)$ for $u$ in the above equation to find the value of $E_{\text{λ}}$.

$E_{\text{λ}}={\displaystyle \underset{0}{\overset{\text{λ}}{\int }}{\epsilon }_{\circ }{E}^2{}_{\text{max}}{\cos }^2\left(kx\right)Adx}$

Integrate the above equation to find the value of $E_{\text{λ}}$.

$\begin{array}{c}{E}_{\text{λ}}={\epsilon }_{\circ }{E}^2{}_{\text{max}}A{\displaystyle \underset{0}{\overset{\text{λ}}{\int }}\left[\frac{1}{2}+\frac{1}{2}\cos \left(2kx\right)\right]dx}\\ =\frac{1}{2}{\epsilon }_{\circ }{E}^2{}_{\text{max}}A{\left[x\right]}_0^{\text{λ}}+\frac{1}{2\left(2k\right)}{\epsilon }_{\circ }{E}^2{}_{\text{max}}A{\left[\sin \left(2kx\right)\right]}_0^{\text{λ}}\\ =\frac{1}{2}{\epsilon }_{\circ }{E}^2{}_{\text{max}}A\lambda +\frac{1}{4k}{\epsilon }_{\circ }{E}^2{}_{\text{max}}A\left[\sin \left(4\pi \right)-\sin \left(0\right)\right]\\ =\frac{1}{2}{\epsilon }_{\circ }{E}^2{}_{\text{max}}A\lambda \end{array}$

Thus, the amount of energy in the shoebox is $\frac{1}{2}{\epsilon }_{\circ }{E}^2{}_{\text{max}}A\lambda$.

Conclusion:

Therefore, the amount of energy in the shoebox is $\frac{1}{2}{\epsilon }_{\circ }{E}^2{}_{\text{max}}A\lambda$.

To determine

The power the wave carries through area $A$ .

Answer to Problem 34.70AP

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

Explanation of Solution

Given info: The expression for the average intensity of the wave is,

$S_{\text{avg}}=\frac{E_{\max }{B}_{\max }}{2{\mu }_{\circ }}=\frac{E^2{}_{\max }}{2{\mu }_{\circ }c}=\frac{c{B}^2{}_{\max }}{2{\mu }_{\circ }}$ .

The expression for the energy in the shoebox is,

$E_{\text{λ}}=\frac{1}{2}{\epsilon }_{\circ }{E}^2{}_{\text{max}}A\lambda$

The formula to calculate the power is,

$P=\frac{E_{\lambda }}{T}$ (5)

Substitute $\frac{1}{2}{\epsilon }_{\circ }{E}^2{}_{\text{max}}A\lambda$ for $E_{\lambda }$ and $\frac{1}{f}$ for $T$ in the above equation to find value of $P$.

$\begin{array}{c}P=\frac{\frac{1}{2}{\epsilon }_{\circ }{E}^2{}_{\text{max}}A\lambda }{\frac{1}{f}}\\ =\frac{1}{2}{\epsilon }_{\circ }{E}^2{}_{\text{max}}A\lambda f\\ =\frac{1}{2}{\epsilon }_{\circ }c{E}^2{}_{\text{max}}A\end{array}$

Thus, the power the wave carries through area $A$  is $\frac{1}{2}{\epsilon }_{\circ }c{E}^2{}_{\text{max}}A$.

Conclusion:

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

To determine

The intensity of the wave .

Answer to Problem 34.70AP

The intensity of the wave is $\frac{1}{2}{\epsilon }_{\circ }c{E}^2{}_{\text{max}}$.

Explanation of Solution

Given info: The expression for the average intensity of the wave is,

$S_{\text{avg}}=\frac{E_{\max }{B}_{\max }}{2{\mu }_{\circ }}=\frac{E^2{}_{\max }}{2{\mu }_{\circ }c}=\frac{c{B}^2{}_{\max }}{2{\mu }_{\circ }}$ .

The formula to calculate the intensity of the wave is,

$I=\frac{P}{A}$

Substitute $\frac{1}{2}{\epsilon }_{\circ }c{E}^2{}_{\text{max}}A$ for $P$ in the above equation to find the value of $I$.

$\begin{array}{c}I=\frac{\frac{1}{2}{\epsilon }_{\circ }c{E}^2{}_{\text{max}}A}{A}\\ =\frac{1}{2}{\epsilon }_{\circ }c{E}^2{}_{\text{max}}\end{array}$

Thus, the intensity of the wave is $\frac{1}{2}{\epsilon }_{\circ }c{E}^2{}_{\text{max}}$.

Conclusion:

Therefore, the intensity of the wave is $\frac{1}{2}{\epsilon }_{\circ }c{E}^2{}_{\text{max}}$.

To determine

The comparison of the result with the equation $S_{\text{avg}}=\frac{E_{\max }{B}_{\max }}{2{\mu }_{\circ }}=\frac{E^2{}_{\max }}{2{\mu }_{\circ }c}=\frac{c{B}^2{}_{\max }}{2{\mu }_{\circ }}$ .

Answer to Problem 34.70AP

The expression of the result is same as that of the expression.

Explanation of Solution

Given info: The expression for the average intensity of the wave is,

$S_{\text{avg}}=\frac{E_{\max }{B}_{\max }}{2{\mu }_{\circ }}=\frac{E^2{}_{\max }}{2{\mu }_{\circ }c}=\frac{c{B}^2{}_{\max }}{2{\mu }_{\circ }}$ .

The formula to calculate the average intensity is,

$I=\frac{1}{2}{\epsilon }_{\circ }c{E}^2{}_{\text{max}}$

Substitute the average value as $\frac{1}{{\mu }_{\circ }{c}^2}$ for ${\epsilon }_{\circ }$ in the above equation to find the average intensity.

$\begin{array}{c}I=\frac{1}{2}\frac{1}{{\mu }_{\circ }{c}^2}c{E}^2{}_{\text{max}}\\ =\frac{E^2{}_{\max }}{2{\mu }_{\circ }c}\end{array}$

Thus, the average value of the intensity is same as that given in the equation.

Conclusion:

Therefore, the expression of the result is same as that of the expression.