Chapter 34, Problem 34.77CP

A linearly polarized microwave of wavelength 1.50 cm is directed along the positive x axis. The electric field vector has a maximum value of 175 V/m and vibrates in the xy plane. Assuming the magnetic field component of the wave can be written in the form B = B_max sin (kx – ωt), give values for (a) B_max, (b) k, and (c) ω.(d) Determine in which plane the magnetic field vector vibrates. (e) Calculate the average value of the Poynting vector for this wave. (f) If this wave were directed at normal incidence onto a perfectly reflecting sheet, what radiation pressure would it exert? (g) What acceleration would be imparted to a 500-g sheet (perfectly reflecting and at normal incidence) with dimensions of 1.00 m × 0.750 m?

To determine

The maximum value of magnetic field.

Answer to Problem 34.77CP

The maximum value of magnetic field is $5.83\text{nT}$.

Explanation of Solution

Given Info: The wavelength of polarized microwave directed along positive $x$-axis is $1.50\text{cm}$. The maximum value of electric field vibrating in $xy$ plane is $175\text{V}/\text{m}$. The magnetic field component of wave is,

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

Here,

$k$ is the propogation vector.

$\omega$ is the angular frequency.

Formula to calculate the maximum value of magnetic field is,

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

Here,

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

$c$ is the speed of waves.

Substitute $175\text{V}/\text{m}$ for $E_{\text{max}}$ and $3\times {10}^8\text{m}/\text{s}$ for $c$ in above equation.

$\begin{array}{c}{B}_{\text{max}}=\frac{175\text{V}/\text{m}}{3\times {10}^8\text{m}/\text{s}}\\ =5.833\times {10}^{-7}\text{T}\\ \approx 583\text{nT}\end{array}$

Conclusion:

Therefore, the maximum value of magnetic field is $5.83\text{nT}$.

To determine

The value of propogation vector $k$.

Answer to Problem 34.77CP

The value of propogation vector $k$ is $\text{419}{\text{m}}^{-1}$.

Explanation of Solution

Given Info: The wavelength of polarized microwave directed along positive $x$-axis is $1.50\text{cm}$. The maximum value of electric field vibrating in $xy$ plane is $175\text{V}/\text{m}$. The magnetic field component of wave is,

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

Here,

$k$ is the propogation vector.

$\omega$ is the angular frequency.

Formula to calculate the propagation vector is,

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

Here,

$\lambda$ is the wavelength.

Substitute $1.50\text{cm}$ for $\lambda$ and 3.14 for $\pi$ in the above equation.

$\begin{array}{c}k=\frac{2\times 3.14}{1.50\text{cm}}\\ =\frac{2\times 3.14}{1.50\times {10}^{-2}\text{m}}\\ =418.666{\text{m}}^{-1}\\ \approx \text{419}{\text{m}}^{-1}\end{array}$

Conclusion:

Therefore, the value of propogation vector $k$ is $\text{419}{\text{m}}^{-1}$.

To determine

The value of angular frequency $\omega$.

Answer to Problem 34.77CP

The value of angular frequency $\omega$ is $1.26\times {10}^{11}{\text{s}}^{-1}$.

Explanation of Solution

Given Info: The wavelength of polarized microwave directed along positive $x$-axis is $1.50\text{cm}$. The maximum value of electric field vibrating in $xy$ plane is $175\text{V}/\text{m}$. The magnetic field component of wave is,

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

Here,

$k$ is the propogation vector.

$\omega$ is the angular frequency.

Formula to calculate the angular frequency is,

$\begin{array}{c}\omega =2\pi f\\ =\frac{2\pi c}{\lambda }\end{array}$

Here,

$f$ is the frequency.

$c$ is the speed of wave.

$\lambda$ is the wavelength.

Substitute $1.50\text{cm}$ for $\lambda$, $3\times {10}^8\text{m}/\text{s}$ for $c$ and 3.14 for $\pi$ in the above equation.

$\begin{array}{c}\omega =\frac{2\times 3.14\times 3\times {10}^8\text{m}/\text{s}}{1.50\text{cm}}\\ =\frac{2\times 3.14\times 3\times {10}^8\text{m}/\text{s}}{1.50\times {10}^{-2}\text{m}}\\ =1.26\times {10}^{11}{\text{s}}^{-1}\end{array}$

Conclusion:

Therefore, the value of angular frequency $\omega$ is $1.26\times {10}^{11}{\text{s}}^{-1}$.

To determine

The plane in which the magnetic field vector vibrates.

Answer to Problem 34.77CP

The plane in which the magnetic field vector vibrates is $xz$ plane.

Explanation of Solution

Given Info: The wavelength of polarized microwave directed along positive $x$-axis is $1.50\text{cm}$. The maximum value of electric field vibrating in $xy$ plane is $175\text{V}/\text{m}$. The magnetic field component of wave is,

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

Here,

$k$ is the propogation vector.

$\omega$ is the angular frequency.

The electromagnetic waves comprises of sinusoidally varying magnetic and electric fields which travel with a speed of light in vacuum. Both electric and magnetic field vectors oscillate perpendicular to each other as well as perpendicular to the direction of propagation of waves.

The wave is propagating along the positive $x$ diction and electric field vector is oscillating in $xy$ plane, so the magnetic field vector will oscillate perpendicular to both in $xz$ plane.

Conclusion:

Therefore, the plane in which the magnetic field vector vibrates is $xz$ plane.

To determine

The average value of Poynting vector for the microwave.

Answer to Problem 34.77CP

The average value of Poynting vector for the microwave is $40.6\widehat{\text{i}}\text{W}/{\text{m}}^2$.

Explanation of Solution

Given Info: The wavelength of polarized microwave directed along positive $x$-axis is $1.50\text{cm}$. The maximum value of electric field vibrating in $xy$ plane is $175\text{V}/\text{m}$. The magnetic field component of wave is,

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

Here,

$k$ is the propogation vector.

$\omega$ is the angular frequency.

Formula to calculate the magnitude of average value of poynting vector for the microwave is,

$S=\frac{E_{\text{max}}{B}_{\text{max}}}{2{\mu }_{\text{o}}}$

Here,

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

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

Substitute $5.833\times {10}^{-7}\text{T}$ for $B_{\text{max}}$, $4\pi \times {10}^{-7}\text{N}/{\text{A}}^2$ for ${\mu }_{\text{o}}$ and $175\text{V}/\text{m}$ for $E_{\text{max}}$ in the above equation.

$\begin{array}{c}S=\frac{\left(175\text{V}/\text{m}\right)\times \left(5.833\times {10}^{-7}\text{T}\right)}{2\times 4\pi \times {10}^{-7}\text{N}/{\text{A}}^2}\\ =40.6\text{W}/{\text{m}}^2\end{array}$

The vector notation of Poynting vector is,

$\overrightarrow{S}=40.6\widehat{\text{i}}\text{W}/{\text{m}}^2$

Conclusion:

Therefore, the average value of Poynting vector for the microwave is $40.6\widehat{\text{i}}\text{W}/{\text{m}}^2$.

To determine

The radiation pressure exerted on the perfectly reflecting sheet.

Answer to Problem 34.77CP

The radiation pressure exerted on the perfectly reflecting sheet is $271\text{nPa}$.

Explanation of Solution

Given Info: The wavelength of polarized microwave directed along positive $x$-axis is $1.50\text{cm}$. The maximum value of electric field vibrating in $xy$ plane is $175\text{V}/\text{m}$. The magnetic field component of wave is,

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

Here,

$k$ is the propogation vector.

$\omega$ is the angular frequency.

Formula to calculate the radiation pressure on a perfectly reflecting sheet is,

$P=\frac{2I}{c}$

Here,

$I$ is the radiation intensity.

$c$ is the speed of wave.

Substitute $40.6\text{W}/{\text{m}}^2$ for $I$ and $3\times {10}^8\text{m}/\text{s}$ for $c$ in the above equation.

$\begin{array}{c}P=\frac{2\times 40.6\text{W}/{\text{m}}^2}{3\times {10}^8\text{m}/\text{s}}\\ =2.71\times {10}^{-7}\text{Pa}\\ =\text{271}\text{nPa}\end{array}$

Conclusion:

Therefore, the radiation pressure exerted on the perfectly reflecting sheet is $\text{271}\text{nPa}$.

To determine

The acceleration imparted to a $500\text{g}$ sheet.

Answer to Problem 34.77CP

The acceleration imparted to a $500\text{g}$ sheet is $4.07\times {10}^{-6}\widehat{\text{i}}\text{m}/{\text{s}}^2$.

Explanation of Solution

Given Info: The wavelength of polarized microwave directed along positive $x$-axis is $1.50\text{cm}$. The maximum value of electric field vibrating in $xy$ plane is $175\text{V}/\text{m}$. Mass of sheet is $50\text{g}$ and dimensions of sheet are $1.00\text{m}\times \text{0}\text{.750}\text{m}$.

The magnetic field component of wave is,

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

Here,

$k$ is the propogation vector.

$\omega$ is the angular frequency.

Formula to calculate the area of sheet is,

$A=L\times B$

Here,

$L$ is the length of sheet.

$B$ is the width of sheet.

Substitute $1.00\text{m}$ for $L$ and $\text{0}\text{.750}\text{m}$ for $B$ in the above equation.

$\begin{array}{c}A=1.00\text{m}\times \text{0}\text{.750}\text{m}\\ =0.750{\text{m}}^2\end{array}$

Formula to calculate the acceleration imparted to sheet is,

$\begin{array}{c}a=\frac{F}{m}\\ =\frac{PA}{m}\end{array}$

Here,

$F$ is the force exerted on sheet.

$P$ is the radiation pressure.

$A$ is the area of sheet.

$m$ is the mass of sheet.

Substitute $2.7\times {10}^{-7}\text{Pa}$ for $P$, $0.750{\text{m}}^2$ for $A$ and $50\text{g}$ for $m$ in the above equation.

$\begin{array}{c}a=\frac{\left(2.7\times {10}^{-7}\text{Pa}\right)\times \left(0.750{\text{m}}^2\right)}{\left(50\text{g}\right)}\\ =\frac{\left(2.7\times {10}^{-7}\text{Pa}\right)\times \left(0.750{\text{m}}^2\right)}{\left(50\times {10}^{-3}\text{kg}\right)}\\ =4.07\times {10}^{-6}\text{m}/{\text{s}}^2\end{array}$

The vector notation of acceleration is,

$\overrightarrow{a}=4.07\times {10}^{-6}\widehat{\text{i}}\text{m}/{\text{s}}^2$

Conclusion:

Therefore, the acceleration imparted to a $500\text{g}$ sheet is $4.07\times {10}^{-6}\widehat{\text{i}}\text{m}/{\text{s}}^2$.