Chapter 34, Problem 32PQ

To determine

The equations for the electric field and the magnetic field for an electromagnetic wave.

Answer to Problem 32PQ

The equation for the electric field of the electromagnetic wave is given by:

    $E=\left(0.050\text{V}/\text{m}\right)\sin \left[\left(1.132\times {10}^7{\text{m}}^{-1}\right)x-\left(3.4\times {10}^{15}\text{rad}/\text{s}\right)t\right]\widehat{k}$.

The equation for the magnetic field of the electromagnetic wave is given by:

    $B=\left(1.67\times {10}^{-10}\text{T}\right)\sin \left[\left(1.132\times {10}^7{\text{m}}^{-1}\right)x-\left(3.4\times {10}^{15}\text{rad}/\text{s}\right)t\right]\left(-\widehat{j}\right)$.

Explanation of Solution

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

    $E=E_{\max }\sin \left(kx-\omega t\right)\widehat{k}$                                                                         (I)

Here, $E_{\max }$ is the amplitude of the electric field, $k$ is the propagation constant and $\omega$ is the angular frequency of the wave.

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

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

Here, $B_{\max }$ is the amplitude of the magnetic field, $k$ is the propagation constant and $\omega$ is the angular frequency of the wave.

Write the expression for the propagation constant.

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

Here, $k$ is the propagation constant and $\lambda$ is the wavelength.

Write the expression for the angular frequency of the wave.

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

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

Write the expression for the amplitude of the magnetic field in terms of magnitude of electric field.

    $B_{\max }=\frac{E_{\max }}{c}$                                                                                          (V)

Conclusion:

Substitute $555\text{nm}$ for $\lambda$ in (III) to find $k$.

    $\begin{array}{c}k=\frac{2\pi }{\left(555\text{nm}\times \frac{{10}^{-9}\text{m}}{1\text{nm}}\right)}\\ =1.132\times {10}^7{\text{m}}^{-1}\end{array}$

Substitute, $3\times {10}^8\text{m}/\text{s}$ for $c$ and $555\text{nm}$ for $\lambda$ in (IV)to find $\omega$.

    $\begin{array}{c}\omega =2\pi \frac{3\times {10}^8\text{m}/\text{s}}{\left(555\text{nm}\times \frac{{10}^{-9}{\text{m}}^{-1}}{1\text{nm}}\right)}\\ =3.4\times {10}^{15}\text{rad}/\text{s}\end{array}$

Substitute $1.132\times {10}^7{\text{m}}^{-1}$ for $k$ , $3.4\times {10}^{15}\text{rad}/\text{s}$ for $\omega$ and $0.050\text{V}/\text{m}$ for $E_{\max }$ in equation(I) to find $E$.

    $E=\left(0.050\text{V}/\text{m}\right)\sin \left(\left(1.132\times {10}^7{\text{m}}^{-1}\right)x-\left(3.4\times {10}^{15}\text{rad}/\text{s}\right)t\right)\widehat{k}$.

Substitute $3\times {10}^8\text{m}/\text{s}$ for $c$ and $0.050\text{V}/\text{m}$ for $E_{\max }$ in (V) to find $B_{\max }$.

    $\begin{array}{c}{B}_{\max }=\frac{0.050\text{V}/\text{m}}{3\times {10}^8\text{m}/\text{s}}\\ =1.67\times {10}^{-10}\text{T}\end{array}$

Substitute $1.132\times {10}^7{\text{m}}^{-1}$ for $k$ , $3.4\times {10}^{15}\text{rad}/\text{s}$ for $\omega$ and $1.67\times {10}^{-10}\text{T}$ for $B_{\max }$ in equation(II) to find $B$.

    $B=\left(1.67\times {10}^{-10}\text{T}\right)\sin \left[\left(1.132\times {10}^7{\text{m}}^{-1}\right)x-\left(3.4\times {10}^{15}\text{rad}/\text{s}\right)t\right]\left(-\widehat{j}\right)$.

Thus, the equation for the electric field of the electromagnetic wave is given by:

    $E=\left(0.050\text{V}/\text{m}\right)\sin \left[\left(1.132\times {10}^7{\text{m}}^{-1}\right)x-\left(3.4\times {10}^{15}\text{rad}/\text{s}\right)t\right]\widehat{k}$.

The equation for the magnetic field of the electromagnetic wave is given by:

    $B=\left(1.67\times {10}^{-10}\text{T}\right)\sin \left[\left(1.132\times {10}^7{\text{m}}^{-1}\right)x-\left(3.4\times {10}^{15}\text{rad}/\text{s}\right)t\right]\left(-\widehat{j}\right)$.