Chapter 34, Problem 77PQ

To determine

The equations for the electric field and the magnetic field of the wave.

Answer to Problem 77PQ

The equation for the electric field is $E=\left(1.44\times {10}^4\text{V}/\text{m}\right)\sin \left[\left(1.01\times {10}^7{\text{m}}^{-1}\right)z-\left(3.02\times {10}^{15}\text{rad}/\text{s}\right)t\right]$ and the equation for the magnetic field is $B=\left(48.0\times {10}^{-6}\text{T}\right)\sin \left[\left(1.01\times {10}^7{\text{m}}^{-1}\right)z-\left(3.02\times {10}^{15}\text{rad}/\text{s}\right)t\right]$.

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.

Write the expression for the angular frequency of the wave.

    $\omega =2\pi f$                                                         (II)

Here, $\omega$ is the angular frequency and $f$ is the frequency of the wave.

Write the expression for the wave number of the wave.

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

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

Write the expression for the maximum electric field.

    $E_{\max }=c{B}_{\max }$                                                         (IV)

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

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

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

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

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

Conclusion:

Substitute $3\times {10}^8\text{m}/\text{s}$ for $c$ and $625\text{nm}$ for $\lambda$ in equation (I) to find $f$.

    $f=\frac{3\times {10}^8\text{m}/\text{s}}{\left(625\text{nm}\times \frac{{10}^{-9}\text{m}}{1\text{nm}}\right)}$

  $=4.8\times {10}^{14}\text{Hz}$

Substitute equation (I) in the equation (II) to find $\omega$.

    $\begin{array}{c}\omega =2\pi \left(4.8\times {10}^{14}\right)\\ =3.01\times {10}^{15}\text{rad}/\text{s}\end{array}$

Substitute $625\text{nm}$ for $\lambda$ in the (III)equation to find $k$.

    $\begin{array}{c}k=\frac{2\pi }{\left(625\text{nm}\times \frac{{10}^{-9}\text{m}}{1\text{nm}}\right)}\\ =1.01\times {10}^7{\text{m}}^{-1}\end{array}$

Substitute $48.0\text{μT}$ for $B_{\max }$ and $3\times {10}^8\text{m}/\text{s}$ for $c$ in (IV) to find $E_{\max }$.

    $\begin{array}{c}{E}_{\max }=\left(3.0\times {10}^8\text{m}/\text{s}\right)\left(48.0\text{μT}\times \frac{{10}^{-9}\text{T}}{1\text{μT}}\right)\\ =1.44\times {10}^4\text{V}/\text{m}\end{array}$

Substitute $1.44\times {10}^4\text{V}/\text{m}$ for $E_{\max }$, $1.01\times {10}^7{\text{m}}^{-1}$ for $k$, and $3.01\times {10}^{15}\text{rad}/\text{s}$ for $\omega$ in the (V) equation to find $E_{\text{Z}}$

    $E_{\text{Z}}=\left(1.44\times {10}^4\text{V}/\text{m}\right)\sin \left[\left(1.01\times {10}^7{\text{m}}^{-1}\right)z-\left(3.02\times {10}^{15}\text{rad}/\text{s}\right)t\right]$

Substitute $48.0\text{μT}$ for $B_{\max }$, $1.01\times {10}^7{\text{m}}^{-1}$ for $k$, and $3.01\times {10}^{15}\text{rad}/\text{s}$ for $\omega$ in(VI) equation to find $B_{\text{Z}}$

    $E_{\text{Z}}=\left(1.01\times {10}^{-6}\text{T}\right)\sin \left[\left(1.01\times {10}^7{\text{m}}^{-1}\right)z-\left(3.02\times {10}^{15}\text{rad}/\text{s}\right)t\right]$

Therefore, the equation for the electric field is $E=\left(1.44\times {10}^4\text{V}/\text{m}\right)\sin \left[\left(1.01\times {10}^7{\text{m}}^{-1}\right)z-\left(3.02\times {10}^{15}\text{rad}/\text{s}\right)t\right]$ and the equation for the magnetic field is $B=\left(48.0\times {10}^{-6}\text{T}\right)\sin \left[\left(1.01\times {10}^7{\text{m}}^{-1}\right)z-\left(3.02\times {10}^{15}\text{rad}/\text{s}\right)t\right]$.