Chapter 24, Problem 73P

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 value for $B_{\max }$.

Answer to Problem 73P

The value for $B_{\max }$ is $5.83\times {10}^{-7}\text{T}$.

Explanation of Solution

Write the expression to calculate the peak value of magnetic field.

  $B_{\max }=\frac{E_{\max }}{c}$

Here, $E_{\max }$ is the peak value for electric field, $B_{\max }$ is the peak value of magnetic field and c is the speed of light.

Conclusion:

Substitute $175\text{V}/\text{m}$ for $E_{\max }$ and $3.00\times {10}^8\text{m}/\text{s}$ for c in the above equation to calculate $B_{\max }$.

  $\begin{array}{c}{B}_{\max }=\frac{175\text{V}/\text{m}}{3.00\times {10}^8\text{m}/\text{s}}\\ =5.83\times {10}^{-7}\text{T}\end{array}$

Thus, the value for $B_{\max }$ is $5.83\times {10}^{-7}\text{T}$.

To determine

The magnitude of k.

Answer to Problem 73P

The magnitude of k is $419{\text{m}}^{-1}$.

Explanation of Solution

Write the expression to calculate the wavenumber of k.

  $k=\frac{2\pi }{\lambda }$

Here, $\lambda$ is the wavelength.

Conclusion:

Substitute $1.50\text{cm}$ for $\lambda$ to calculate k.

  $\begin{array}{c}k=\frac{2\pi }{1.50\text{cm}\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)}\\ =419{\text{m}}^{-1}\end{array}$

Thus, the magnitude of k is $419{\text{m}}^{-1}$.

To determine

The magnitude of $\omega$.

Answer to Problem 73P

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

Explanation of Solution

Write the expression to calculate the angular frequency or $\omega$.

  $\omega =kc$

Conclusion:

Substitute $419{\text{m}}^{-1}$ for k and $3.00\times {10}^8\text{m}/\text{s}$ for c in the above equation to calculate $\omega$.

  $\begin{array}{c}\omega =\left(419{\text{m}}^{-1}\right)\left(3.00\times {10}^8\text{m}/\text{s}\right)\\ =1.26\times {10}^{11}{\text{s}}^{-1}\end{array}$

Thus, the magnitude of $\omega$ is $1.26\times {10}^{11}{\text{s}}^{-1}$.

To determine

The magnitude of $\omega$.

Answer to Problem 73P

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

Explanation of Solution

Write the expression to calculate the angular frequency or $\omega$.

  $\omega =kc$

Conclusion:

Substitute $419{\text{m}}^{-1}$ for k and $3.00\times {10}^8\text{m}/\text{s}$ for c in the above equation to calculate $\omega$.

  $\begin{array}{c}\omega =\left(419{\text{m}}^{-1}\right)\left(3.00\times {10}^8\text{m}/\text{s}\right)\\ =1.26\times {10}^{11}{\text{s}}^{-1}\end{array}$

Thus, the magnitude of $\omega$ is $1.26\times {10}^{11}{\text{s}}^{-1}$.

To determine

The plane at which magnetic field vector vibrates.

Answer to Problem 73P

The plane of vibration of magnetic field vector is z direction.

Explanation of Solution

Here, both the electric and magnetic field vectors are vibrates in xy-plane. The direction of pointing vector is same as that of the direction of the wave since the wave carries the energy along the direction of propagation.

From the expression of the magnetic field given, the electric field vibrates along y direction. Therefore, the magnetic field must vibrate in z direction since the wave travels along the x direction which is mutually normal to both the electric and magnetic vibrations.

Conclusion:

Substitute $419{\text{m}}^{-1}$ for k and $3.00\times {10}^8\text{m}/\text{s}$ for c in the above equation to calculate $\omega$.

  $\begin{array}{c}\omega =\left(419{\text{m}}^{-1}\right)\left(3.00\times {10}^8\text{m}/\text{s}\right)\\ =1.26\times {10}^{11}{\text{s}}^{-1}\end{array}$

Thus, the plane of vibration of magnetic field vector is z direction.

To determine

The average value of pointing vector in the wave.

Answer to Problem 73P

The average value of pointing vector is $40.6\text{W}/{\text{m}}^{\text{2}}$.

Explanation of Solution

Write the expression to calculate the average value of pointing vector.

  $S_{\text{avg}}=\frac{E_{\max }{B}_{\max }}{{\mu }_0}$

Here, $S_{\text{avg}}$ is the average pointing vector and ${\mu }_0$ is the permeability of free space.

Conclusion:

Substitute $175\text{V}/\text{m}$ for $E_{\max }$, $4\pi \times {10}^{-7}\text{T}\cdot \text{m}/\text{A}$ for ${\mu }_0$ and $5.83\times {10}^{-7}\text{T}$ for $B_{\max }$ in the above equation to calculate $S_{\text{avg}}$.

  $\begin{array}{c}{S}_{\text{avg}}=\frac{\left(175\text{V}/\text{m}\right)\left(5.83\times {10}^{-7}\text{T}\right)}{4\pi \times {10}^{-7}\text{T}\cdot \text{m}/\text{A}}\\ =40.6\text{W}/{\text{m}}^{\text{2}}\end{array}$

Thus, the average value of pointing vector is $40.6\text{W}/{\text{m}}^{\text{2}}$.

To determine

The radiation pressure exerted by the wave.

Answer to Problem 73P

The radiation pressure exerted by the wave is $2.71\times {10}^{-7}\text{N}/{\text{m}}^{\text{2}}$.

Explanation of Solution

Write the expression to calculate the radiation pressure.

  $P_r=\frac{2S_{avg}}{c}$

Here, $P_r$ is the radiation pressure.

Conclusion:

Substitute $40.6\text{W}/{\text{m}}^{\text{2}}$ for $S_{\text{avg}}$ and $3.00\times {10}^8\text{m}/\text{s}$ for c in the above equation to calculate $P_r$.

  $\begin{array}{c}{P}_r=\frac{2\left(40.6\text{W}/{\text{m}}^{\text{2}}\right)}{3.00\times {10}^8\text{m}/\text{s}}\\ =2.71\times {10}^{-7}\text{N}/{\text{m}}^{\text{2}}\end{array}$

Thus, the radiation pressure exerted by the wave is $2.71\times {10}^{-7}\text{N}/{\text{m}}^{\text{2}}$.

To determine

The acceleration imparted to the given sheet.

Answer to Problem 73P

The acceleration is $4.07\times {10}^{-7}\text{m}/{\text{s}}^{\text{2}}$.

Explanation of Solution

The area of the sheet is $0.750{\text{m}}^2$.

Write the expression to calculate the acceleration.

  $a=\frac{P_r}{m}A$

Here, a is the acceleration, A is the area and m is the mass.

Conclusion:

Substitute $2.71\times {10}^{-7}\text{N}/{\text{m}}^{\text{2}}$ for $P_r$, $0.750{\text{m}}^2$  for A and $500\text{g}$ for m in the above equation to calculate a.

  $\begin{array}{c}a=\frac{\left(2.71\times {10}^{-7}\text{N}/{\text{m}}^{\text{2}}\right)}{500\text{g}\left(\frac{{10}^{-3}\text{kg}}{1\text{g}}\right)}\left(0.750{\text{m}}^2\right)\\ =4.07\times {10}^{-7}\text{m}/{\text{s}}^{\text{2}}\end{array}$

Thus, the acceleration is $4.07\times {10}^{-7}\text{m}/{\text{s}}^{\text{2}}$.