Chapter 33, Problem 51CP

(a)

To determine

The wavelength of the wave.

Answer to Problem 51CP

The wavelength of the wave is $3.33\text{m}$.

Explanation of Solution

Given info: The frequency of the wave is $90.0\text{MHz}$ and the peak value of the electric field is $2.00\text{mV}/\text{m}$ in positive y direction.

The formula to calculate the wavelength is,

$\lambda =\frac{c}{f}$

Here,

$c$ is the speed of the light.

$f$ is the frequency of the wave.

Substitute $3\times {10}^8\text{m}/\text{s}$ for $c$ and $90.0\text{MHz}$ for $f$ in the above equation to find the value of $\lambda$.

$\begin{array}{c}\lambda =\frac{3\times {10}^8\text{m}/\text{s}}{90.0\text{MHz}\times \frac{{10}^6\text{Hz}}{1\text{MHz}}}\\ =3.33\text{m}\end{array}$

Conclusion:

Therefore, the wavelength of the wave is $3.33\text{m}$.

(b)

To determine

The time period of the wave.

Answer to Problem 51CP

The time period of the wave is $11.1\text{ns}$.

Explanation of Solution

Given info: The frequency of the wave is $90.0\text{MHz}$ and the peak value of the electric field is $2.00\text{mV}/\text{m}$ in positive y direction.

The formula to calculate the time period is,

$T=\frac{1}{f}$

Substitute $90.0\text{MHz}$ for $f$ in the above equation to find the value of $T$ .

$\begin{array}{c}T=\frac{1}{\left(90.0\text{MHz}\times \frac{{10}^6\text{Hz}}{1\text{MHz}}\right)}\\ =11.1\times {10}^{-9}\text{s}\times \frac{{10}^9\text{ns}}{1\text{s}}\\ =11.1\text{ns}\end{array}$

Conclusion:

Therefore, the time period of the wave is $11.1\text{ns}$.

(c)

To determine

The maximum value of the magnetic field.

Answer to Problem 51CP

The maximum value of the magnetic field is $6.67\text{pT}$.

Explanation of Solution

Given info: The frequency of the wave is $90.0\text{MHz}$ and the peak value of the electric field is $2.00\text{mV}/\text{m}$ in positive y direction.

The formula to calculate the magnitude of the magnetic field is,

$B_{\max }=\frac{E_{\max }}{c}$

Here,

$E_{\max }$ is the peak value of the electric field.

Substitute $2.00\text{mV}/\text{m}$ for $E_{\max }$ and $3\times {10}^8\text{m}/\text{s}$ for $c$ in the above equation to find the value of $B_{\max }$ .

$\begin{array}{c}{B}_{\max }=\frac{\left(2.00\text{mV}/\text{m}\times \frac{{10}^{-3}\text{V}/\text{m}}{1\text{mV}/\text{m}}\right)}{\left(3\times {10}^8\text{m}/\text{s}\right)}\\ =6.67\times {10}^{-12}\text{T}\\ =6.67\text{pT}\end{array}$

Conclusion:

Therefore, the maximum value of the magnetic field is $6.67\text{pT}$.

(d)

To determine

The expression for electric field and the magnetic field.

Answer to Problem 51CP

The expression for electric field is $\overrightarrow{E}=\left(2.00\times {10}^{-3}\right)\cos 2\pi \left(\frac{x}{3.33}-90.0\times {10}^{6}t\right)\widehat{\text{j}}$ and the expression for magnetic field is $\overrightarrow{B}=\left(6.67\times {10}^{-12}\right)\cos 2\pi \left(\frac{x}{3.33}-90.0\times {10}^{6}t\right)\widehat{\text{k}}$.

Explanation of Solution

Given info: The frequency of the wave is $90.0\text{MHz}$ and the peak value of the electric field is $2.00\text{mV}/\text{m}$ in positive y direction.

The formula to calculate the angular frequency is,

$\omega =2\pi f$

Here,

$f$ is the frequency of the wave.

Substitute the $90.0\times {10}^6\text{Hz}$ for $f$ in the above equation to find the value of $\omega$,

$\omega =2\pi \left(90.0\times {10}^6\text{Hz}\right)$ (1)

The formula to calculate the angular constant is,

$k=\frac{2\pi }{\lambda }$

Here,

$\lambda$ is the wavelength of wave.

Substitute the $3.33\text{m}$ for $\lambda$ in the above equation to find the value of $k$,

$k=\frac{2\pi }{3.33\text{m}}$ (2)

The formula to calculate the electric field is,

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

Substitute $2\pi \left(90.0\times {10}^6\text{Hz}\right)$ for $\omega$, $\frac{2\pi }{3.33\text{m}}$ for $k$ and $2.00\text{mV}/\text{m}$ for $E_{\max }$ in the above equation to find the value of $E$ .

$\begin{array}{l}\overrightarrow{E}=\left(2.00\times {10}^{-3}\text{V}/\text{m}\right)\cos \left(\frac{2\pi }{3.33\text{m}}x-2\pi \left(90.0\times {10}^6\text{Hz}\right)t\right)\\ \overrightarrow{E}=\left(2.00\times {10}^{-3}\right)\cos 2\pi \left(\frac{x}{3.33}-90.0\times {10}^{6}t\right)\widehat{\text{j}}\end{array}$

The electric field is in the same direction of wave propagation.

The formula to calculate the magnetic field is,

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

Substitute $2\pi \left(90.0\times {10}^6\text{Hz}\right)$ for $\omega$, $\frac{2\pi }{3.33\text{m}}$ for $k$ and $\left(6.67\times {10}^{-12}\right)$ for $B_{\max }$ in the above equation to find the value of $B$ .

$\begin{array}{l}\overrightarrow{B}=\left(6.67\times {10}^{-12}\text{T}\right)\cos \left(\frac{2\pi }{3.33\text{m}}x-2\pi \left(90.0\times {10}^6\text{Hz}\right)t\right)\\ \overrightarrow{B}=\left(6.67\times {10}^{-12}\right)\cos 2\pi \left(\frac{x}{3.33}-90.0\times {10}^{6}t\right)\widehat{\text{k}}\end{array}$

The direction of propagation of the magnetic field is perpendicular to that of the electric field.

Conclusion:

Therefore, the expression for electric field is $\overrightarrow{E}=\left(2.00\times {10}^{-3}\right)\cos 2\pi \left(\frac{x}{3.33}-90.0\times {10}^{6}t\right)\widehat{\text{j}}$ and the expression for magnetic field is $\overrightarrow{B}=\left(6.67\times {10}^{-12}\right)\cos 2\pi \left(\frac{x}{3.33}-90.0\times {10}^{6}t\right)\widehat{\text{k}}$.

(e)

To determine

The average power per unit area the wave carries.

Answer to Problem 51CP

The average power per unit area the wave carries is $5.31\times {10}^{-9}\text{W}/{\text{m}}^2$ .

Explanation of Solution

Given info: The frequency of the wave is $90.0\text{MHz}$ and the peak value of the electric field is $2.00\text{mV}/\text{m}$ in positive y direction.

The formula to calculate the average power per unit area is,

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

Here,

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

$c$ is the speed of the light.

$E_{\max }$ is the maximum electric field.

Substitute $2.00\text{mV}/\text{m}$ for $E_{\max }$, $8.85\times {10}^{-12}{\text{C}}^{\text{2}}/\text{N}\cdot {\text{m}}^{\text{2}}$ for ${\epsilon }_{\circ }$ and $3\times {10}^8\text{m}/\text{s}$ for $c$ in the above equation to find the value of $I$ .

$\begin{array}{c}I=\frac{1}{2}\left(8.85\times {10}^{-12}{\text{C}}^{\text{2}}/\text{N}\cdot {\text{m}}^{\text{2}}\right)\left(3\times {10}^8\text{m}/\text{s}\right){\left(2.00\times {10}^{-3}\text{V}/\text{m}\right)}^2\\ =5.31\times {10}^{-9}\text{W}/{\text{m}}^2\end{array}$

Conclusion:

Therefore, the average power per unit area the wave carries is $5.31\times {10}^{-9}\text{W}/{\text{m}}^2$.

(f)

To determine

The average energy density in the radiation.

Answer to Problem 51CP

The average energy density in the radiation is $1.77\times {10}^{-17}\text{J}/{\text{m}}^{\text{2}}$.

Explanation of Solution

Given info: The frequency of the wave is $90.0\text{MHz}$ and the peak value of the electric field is $2.00\text{mV}/\text{m}$ in positive y direction.

The formula to calculate the average energy density is,

$e=\frac{I}{c}$

Substitute $5.31\times {10}^{-9}\text{W}/{\text{m}}^2$ for $I$ and $3\times {10}^8\text{m}/\text{s}$ for $c$ in the above equation to find the value of $B$ .

$\begin{array}{c}e=\frac{5.31\times {10}^{-9}\text{W}/{\text{m}}^2}{3\times {10}^8\text{m}/\text{s}}\\ =1.77\times {10}^{-17}\text{J}/{\text{m}}^{\text{2}}\end{array}$

Conclusion:

Therefore, the average energy density in the radiation is $1.77\times {10}^{-17}\text{J}/{\text{m}}^{\text{2}}$ .

(g)

To determine

The radiation pressure exerted by the wave.

Answer to Problem 51CP

The radiation pressure exerted by the wave is $3.54\times {10}^{-17}\text{J}/{\text{m}}^{\text{2}}$.

Explanation of Solution

Given info: The frequency of the wave is $90.0\text{MHz}$ and the peak value of the electric field is $2.00\text{mV}/\text{m}$ in positive y direction.

The formula to calculate the radiation pressure is,

$\begin{array}{c}P=2\frac{I}{c}\\ =2e\end{array}$

Substitute $1.77\times {10}^{-17}\text{J}/{\text{m}}^{\text{2}}$ for $e$ in above equation to find the value of $P$ .

$\begin{array}{c}P=2\left(1.77\times {10}^{-17}\text{J}/{\text{m}}^{\text{2}}\right)\\ =3.54\times {10}^{-17}\text{J}/{\text{m}}^{\text{2}}\end{array}$

Conclusion:

Therefore, the radiation pressure exerted by the wave is $3.54\times {10}^{-17}\text{J}/{\text{m}}^{\text{2}}$.