Chapter 34, Problem 34.46P

A large, flat sheet carries a uniformly distributed electric current with current per unit width J_s. This current creates a magnetic field on both sides of the sheet, parallel to the sheet and perpendicular to the current, with magnitude $B=\frac{1}{2}{\mu }_0{J}_s$. If the current is in the y direction and oscillates in time according to

$J_{\max }(\cos \omega t)\hat{j}=J_{\max }[\cos (-\omega t)]\hat{j}$

the sheet radiates an electromagnetic wave. Figure P33.28 shows such a wave emitted from one point on the sheet chosen to be the origin. Such electromagnetic waves arc emitted from all points on the sheet. The magnetic field of the wave to the right of the sheet is described by the wave function

$\stackrel{\to }{B}=\frac{1}{2}{\mu }_0{J}_{\max }[\cos (kx-\omega t)]\hat{k}$

(a) Find the wave function for the electric field of the wave to the right of the sheet. (b) Find the Poynting vector as a function of x and t. (c) Find the intensity of the wave. (d) What If? If the sheet is to emit radiation in each direction (normal to the plane of the sheet) with intensity 570 W/m², what maximum value of sinusoidal current density is required?

Figure P33.28

To determine

The wave function for the electric field of the wave to the right of the sheet.

Answer to Problem 34.46P

The wave function for the electric field of the wave to the right of the sheet is $\frac{1}{2}c{\mu }_0{J}_{\text{max}}\left[\cos \left(kx-\omega t\right)\right]\widehat{\text{j}}$.

Explanation of Solution

Given info: The wave function for the magnetic field of the wave to the right of the sheet is $\frac{1}{2}{\mu }_0{J}_{\text{max}}\left[\cos \left(kx-\omega t\right)\right]\widehat{\text{k}}$.

Explanation:

Write the Maxwell’s third equation,

$\frac{\partial E}{\partial x}=-\frac{\partial B}{\partial t}$

Here,

$\frac{\partial E}{\partial x}$ is the wave function of electric field.

$\frac{\partial B}{\partial t}$ is the wave function of the magnetic field.

Substitute $\frac{1}{2}{\mu }_0{J}_{\text{max}}\left[\cos \left(kx-\omega t\right)\right]\widehat{\text{k}}$ for $\frac{\partial B}{\partial t}$ in the above equation to find the value of $\frac{\partial E}{\partial x}$.

$\begin{array}{c}\frac{\partial E}{\partial x}=-\frac{\partial }{\partial t}\left(\frac{1}{2}{\mu }_0{J}_{\text{max}}\left[\cos \left(kx-\omega t\right)\right]\widehat{\text{k}}\right)\\ =\frac{\omega }{2}{\mu }_0{J}_{\text{max}}\sin \left(kx-\omega t\right)\end{array}$

Integrating the above equation with respect to $x$.

$\begin{array}{c}E={\displaystyle \int \frac{\omega }{2}{\mu }_0{J}_{\text{max}}\sin \left(kx-\omega t\right)dx}\\ =\frac{1}{2}\frac{\omega }{k}{\mu }_0{J}_{\text{max}}\cos \left(kx-\omega t\right)\end{array}$

Substitute $c$ for $\frac{\omega }{k}$ in the above equation.

$E=\frac{1}{2}c{\mu }_0{J}_{\text{max}}\cos \left(kx-\omega t\right)$

The direction of electric field must be perpendicular to the direction of propagation $\left(\widehat{\text{i}}\right)$ and the direction of magnetic field $\left(\widehat{\text{k}}\right)$. Thus the direction of the electric filed is $\left(\widehat{\text{j}}\right)$.

Conclusion:

Therefore, the wave function for the electric field of the wave to the right of the sheet is $\frac{1}{2}c{\mu }_0{J}_{\text{max}}\left[\cos \left(kx-\omega t\right)\right]\widehat{\text{j}}$.

To determine

The Poynting vector as a function of $x$ and $t$.

Answer to Problem 34.46P

The Poynting vector as a function of $x$ and $t$ is $\frac{1}{4}c{\mu }_0{J}_{\text{max}}^2{\cos }^2\left(kx-\omega t\right)\widehat{\text{i}}$.

Explanation of Solution

Given info: The wave function for the magnetic field of the wave to the right of the sheet is $\frac{1}{2}{\mu }_0{J}_{\text{max}}\left[\cos \left(kx-\omega t\right)\right]\widehat{\text{k}}$.

Explanation:

Write the formula to calculate the Poynting vector.

$\overrightarrow{\text{S}}=\frac{1}{{\mu }_0}\left(\overrightarrow{\text{E}}\times \overrightarrow{\text{B}}\right)$

Here,

$\overrightarrow{\text{S}}$ is the Poynting vector.

$\overrightarrow{\text{E}}$ is the wave function for the electric field of the wave.

$\overrightarrow{\text{B}}$ is the wave function for the magnetic field of the wave.

Substitute $\frac{1}{2}c{\mu }_0{J}_{\text{max}}\left[\cos \left(kx-\omega t\right)\right]\widehat{\text{j}}$ for $\overrightarrow{\text{E}}$ and $\frac{1}{2}{\mu }_0{J}_{\text{max}}\left[\cos \left(kx-\omega t\right)\right]\widehat{\text{k}}$ for $\overrightarrow{\text{B}}$ in the above equation.

$\begin{array}{c}\overrightarrow{\text{S}}=\frac{1}{{\mu }_0}\left(\frac{1}{2}c{\mu }_0{J}_{\text{max}}\left[\cos \left(kx-\omega t\right)\right]\widehat{\text{j}}\times \text{B}\frac{1}{2}{\mu }_0{J}_{\text{max}}\left[\cos \left(kx-\omega t\right)\right]\widehat{\text{k}}\right)\\ =\frac{1}{4}c{\mu }_0{J}_{\text{max}}^2{\cos }^2\left(kx-\omega t\right)\widehat{\text{i}}\end{array}$

Conclusion:

Therefore, the Poynting vector as a function of $x$ and $t$ is $\frac{1}{4}c{\mu }_0{J}_{\text{max}}^2{\cos }^2\left(kx-\omega t\right)\widehat{\text{i}}$.

To determine

The intensity of the wave.

Answer to Problem 34.46P

The intensity of the wave is $\frac{1}{8}c{\mu }_0{J}_{\text{max}}^2$.

Explanation of Solution

Given info: The wave function for the magnetic field of the wave to the right of the sheet is $\frac{1}{2}{\mu }_0{J}_{\text{max}}\left[\cos \left(kx-\omega t\right)\right]\widehat{\text{k}}$.

Explanation:

The wave function for the magnetic field of the wave is.

$B=\frac{1}{2}{\mu }_0{J}_{\text{max}}\left[\cos \left(kx-\omega t\right)\right]\widehat{\text{k}}$

The maximum value of $\cos \left(kx-\omega t\right)$ is $1$. Thus the maximum value of wave function for the magnetic field of the wave is

$B_{\text{max}}=\frac{1}{2}{\mu }_0{J}_{\text{max}}$

The wave function for the electric field of the wave is.

$E=\frac{1}{2}c{\mu }_0{J}_{\text{max}}\left[\cos \left(kx-\omega t\right)\right]\widehat{\text{j}}$

The maximum value of $\cos \left(kx-\omega t\right)$ is $1$. Thus the maximum value of wave function for the electric field of the wave is

$E_{\text{max}}=\frac{1}{2}c{\mu }_0{J}_{\text{max}}$

Write the formula to calculate the intensity of the wave is,

$I=\frac{E_{\text{max}}{B}_{\text{max}}}{2{\mu }_0}$

Here,

$I$ is the intensity of the wave.

$E_{\text{max}}$ is the maximum value of wave function for the electric field of the wave.

$B_{\text{max}}$ is the maximum value of wave function for the magnetic field of the wave.

Substitute $\frac{1}{2}c{\mu }_0{J}_{\text{max}}$ for $E_{\text{max}}$ and $\frac{1}{2}{\mu }_0{J}_{\text{max}}$ for $B_{\text{max}}$ in the above equation to find the value of $I$.

$\begin{array}{c}I=\frac{\left(\frac{1}{2}c{\mu }_0{J}_{\text{max}}\right)\left(\frac{1}{2}{\mu }_0{J}_{\text{max}}\right)}{2{\mu }_0}\\ =\frac{1}{8}c{\mu }_0{J}_{\text{max}}^2\end{array}$

Conclusion:

Therefore, the intensity of the wave is $\frac{1}{8}c{\mu }_0{J}_{\text{max}}^2$.

To determine

The maximum value of sinusoidal current density.

Answer to Problem 34.46P

The maximum value of sinusoidal current density is $3.47\text{A}/\text{m}$.

Explanation of Solution

Given info: The intensity of the wave is $570\text{W}/{\text{m}}^{\text{2}}$.

Explanation:

The intensity of the wave from part (c) is,

$I=\frac{1}{8}c{\mu }_0{}^2{J}_{\text{max}}^2$

Here,

$I$ is the intensity of the wave.

$c$ is the speed of the light.

$J_{\text{max}}$ is the maximum value of current density.

${\mu }_0$ is the permeability of the free space.

Rearrange the above expression for $J_{\text{max}}$.

$J_{\text{max}}=\sqrt{\frac{8I}{{\mu }_{0}c}}$

Substitute $570\text{W}/{\text{m}}^{\text{2}}$ for $I$, $3\times {10}^8\text{m}/\text{s}$ for $c$ and $4\pi \times {10}^{-7}\text{N}\cdot \text{kg}\cdot \text{m}\cdot {\text{s}}^{\text{-2}}\cdot {\text{A}}^{\text{-2}}$ for ${\mu }_0$ in the above equation to find the value of $J_{\text{max}}$.

$\begin{array}{c}{J}_{\text{max}}=\sqrt{\frac{8\left(570\text{W}/{\text{m}}^{\text{2}}\times \frac{1\text{V}\cdot \text{A}/{\text{m}}^{\text{2}}}{1\text{W}/{\text{m}}^{\text{2}}}\right)}{\left(4\pi \times {10}^{-7}\text{V}\cdot \text{s}/\text{A}\cdot \text{m}\right)\left(3\times {10}^8\text{m}/\text{s}\right)}}\\ =3.47\text{A}/\text{m}\end{array}$

Conclusion:

Therefore, the maximum value of sinusoidal current density is $3.47\text{A}/\text{m}$.