Chapter 34, Problem 72PQ

(a)

To determine

The frequency of the wave.

Answer to Problem 72PQ

The frequency of the wave is $8.82\times {10}^9\text{Hz}$.

Explanation of Solution

Write the expression for the frequency of an electromagnetic wave.

    $f=\frac{c}{\lambda }$                                                             (I)

Here $c$ is the speed of light and $f$ is the frequency of light and $\lambda$ is the wavelength.

Substitute $3\times {10}^8\text{m}/\text{s}$ for $c$ and $34\text{mm}$ for $\lambda$ in equation (I) to find $f$.

    $f=\frac{3\times {10}^8\text{m}/\text{s}}{\left(34\text{nm}\times \frac{{10}^{-3}\text{m}}{1\text{m}}\right)}$

  $=8.82\times {10}^9\text{Hz}$                                      (I)

Conclusion:

Thus, the frequency of the wave is $8.82\times {10}^9\text{Hz}$.

(b)

To determine

The magnitude and direction of electric field.

Answer to Problem 72PQ

The direction of the electric field is along the positive x-direction and the magnitude of the electric field is $28.8\text{V}/\text{m}$

Explanation of Solution

The direction of the electric field is along the positive x-direction, since the magnetic field is oscillating in the yz plane.

Write the expression for the maximum electric field of the electromagnetic wave.

    $E_{\max }=c{B}_{\max }$

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

Substitute $3\times {10}^8\text{m}/\text{s}$ for $c$ and $96.0\text{μT}$ for $B_{\max }$ in the above equation to find $E_{\max }$.

    $\begin{array}{c}{E}_{\max }=\left(3.0\times {10}^8\text{m}/\text{s}\right)\left(96.0\text{μT}\times \frac{{10}^{-9}\text{T}}{1\text{μT}}\right)\\ =28.8\text{V}/\text{m}\end{array}$

Conclusion:

Thus, the direction of the electric field is along the positive x-direction and the magnitude of the electric field is $28.8\text{V}/\text{m}$

(c)

To determine

The equation for the electric field of the electromagnetic wave.

Answer to Problem 72PQ

The equation for the electric field is $E=\left(28.8\text{V}/\text{m}\right)\sin \left[\left(185{\text{m}}^{-1}\right)z-\left(5.54\times {10}^{10}\text{rad}/\text{s}\right)t\right]$.

Explanation of Solution

Write the expression for the angular frequency of the wave.

    $\omega =\frac{2\pi c}{\lambda }$                                                                                                          (I)

Here $c$ is the speed of light and $\lambda$ is the wavelength.

Write the expression for the wave number of the wave.

    $k=\frac{2\pi }{\lambda }$                                                                                                              (II)

Here, $k$ is the wave number and $\lambda$ is the wavelength of the wave.

Write the expression for the electric field of an electromagnetic wave.

    $E_{\text{Z}}=E_{\max }\sin \left(kx-\omega t\right)\widehat{i}\text{V}/\text{m}$                                                                           (III)

Conclusion:

Substitute $3\times {10}^8\text{m}/\text{s}$ for $c$ and $34\text{mm}$ for $\lambda$ in (I) to find $\omega$.

    $\omega =\frac{2\pi \left(3.0\times {10}^8\text{m}/\text{s}\right)}{\left(34\text{mm}\times \frac{{10}^{-3}\text{m}}{1\text{mm}}\right)}$

  $=5.5\times {10}^{10}\text{rad}/\text{s}$

Substitute $34\text{mm}$ for $\lambda$ in the equation (II) to find $k$.

    $\begin{array}{c}k=\frac{2\pi }{\left(34\text{mm}\times \frac{{10}^{-3}\text{m}}{1\text{m}}\right)}\\ =185{\text{m}}^{-1}\end{array}$

Substitute $28.8\text{V}/\text{m}$ for $E_{\max }$, $185{\text{m}}^{-1}$ for $k$, and $5.54\times {10}^{10}\text{rad}/\text{s}$ for $\omega$ in the above equation to find $E_{\text{Z}}$

    $E=\left(28.8\text{V}/\text{m}\right)\sin \left[\left(185{\text{m}}^{-1}\right)x-\left(5.54\times {10}^{10}\text{rad}/\text{s}\right)t\right]\widehat{i}$

Therefore, the equation for the electric field is $E=\left(28.8\text{V}/\text{m}\right)\sin \left[\left(185{\text{m}}^{-1}\right)z-\left(5.54\times {10}^{10}\text{rad}/\text{s}\right)t\right]$