Question

An electromagnetic wave of wavelength 435 nm is traveling in vacuum in the -z- direction. The electric field has amplitude$\mathbf{2}\mathbf{.}\mathbf{70}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{V}\mathbf{/}\mathbf{m}$and is parallel to the x-axis. What are (a) the frequency and (b) the magnetic-field amplitude? (c) Write the vector equations for$\overrightarrow{\mathbf{E}}\left(z,t\right)$ and $\overrightarrow{\mathbf{B}}\mathbf{(}\mathbf{z}\mathbf{,}\mathbf{t}\mathbf{)}$.

Short answer

a) The frequency of electromagnetic wave is $6.89\times {10}^{14}\mathrm{Hz}$.

b) The amplitude of magnetic field of electromagnetic wave is $9\times {10}^{-12}\mathrm{T}$.

c) The vector equation for$\overrightarrow{\mathrm{E}}(\mathrm{z},\mathrm{t})=\stackrel{⏜}{\mathrm{i}}[2.70\times {10}^{-3}\mathrm{V}/\mathrm{mcos}\left\{(1.44\times {10}^7\mathrm{m})\mathrm{z}-(4.3\times {10}^{15}\mathrm{rad}/\mathrm{s})\mathrm{t}\right\}]$ and$\overrightarrow{\mathrm{B}}(\mathrm{z},\mathrm{t})=-\stackrel{⏜}{\mathrm{j}}[9\times {10}^{-12}\mathrm{Tcos}\left\{(1.44\times {10}^7\mathrm{m})\mathrm{z}-(4.3\times {10}^{15}\mathrm{rad}/\mathrm{s})\mathrm{t}\right\}]$ .

Step-by-step solution

Step 1:Define frequency and write formulas.

The number of waves that moves a fixed point in unit time is known as frequency.

$f=\frac{1}{T}$

Where,T =Time period in s.

The frequency, wavelength, and speed of light of any wave are related by the wavelength-frequency relationship.

The wavelength-frequency relationship is:

$$\begin{gathered} c=\lambda f \\ \Rightarrow f=\frac{c}{\lambda } \end{gathered}$$

Where,c is the speed of light which is equal to $3\times {10}^8\mathrm{m}/\mathrm{s}$,$\lambda$ is the wavelength andf is the frequency.

The formula used to determine the amplitude of electric and magnetic fields of the wave are:

$$\begin{gathered} E_{max}=\sqrt{\frac{2l}{{\epsilon }_{0}c}} \\ {B}_{max}=\frac{E_{max}}{c} \end{gathered}$$

Where,${\mathrm{\epsilon }}_0=8.85\times {10}^{-12}{\mathrm{C}}^2/\mathrm{N}\cdot {\mathrm{m}}^2$ and c is the speed of light that is equal to $3.0\times {10}^8\mathrm{m}/\mathrm{s}$.

The equations of sinusoidal electromagnetic wave for electric and magnetic field are:

$$\begin{gathered} \overrightarrow{E}=E_{max}\stackrel{⏜}{j}\cos \left(kx-\omega t\right) \\ \overrightarrow{B}=B_{max}\stackrel{⏜}{k}\cos \left(kx-\omega t\right) \end{gathered}$$

Determine the frequency of wave.

Given that,$\lambda =435\times {10}^{-9}m$

The wavelength-frequency relationship is

$$\begin{gathered} c=\lambda f \\ \Rightarrow f=\frac{c}{\lambda } \end{gathered}$$

Substitute the values of variables

$$\begin{gathered} \mathrm{f}=\frac{3\times {10}^8}{435\times {10}^{-9}} \\ =6.89\times {10}^{14}\mathrm{Hz} \end{gathered}$$

Hence, the frequency of a sinusoidal electromagnetic wave is $6.94\times {10}^{14}\mathrm{Hz}$.

Determine the amplitude of magnetic field.

Given that,$E_{max}=2.70\times {10}^{-3}V/m$

The formula used to determine the amplitude of electric and magnetic fields of the wave are:

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

Substitute the values of variables

$$\begin{gathered} B_{max}=\frac{2.70\times {10}^{3}V/m}{3\times {10}^8} \\ =9\times {10}^{-12}T \end{gathered}$$

Hence, the amplitude of magnetic field of electromagnetic wave is $9\times {10}^{-12}\mathrm{T}$.

Determine the sinusoidal electromagnetic wave.

The angular frequency of wave is

$$\begin{gathered} \mathrm{\omega }=2\mathrm{\pi f} \\ =2\mathrm{\pi }\left(6.89\times {10}^{14}\right) \\ =4.3\times {10}^{15}\mathrm{rad}/\mathrm{s} \end{gathered}$$

And the wavenumber of wave is

$$\begin{gathered} \mathrm{k}=\frac{2\mathrm{\pi }}{\mathrm{\lambda }} \\ =\frac{2\mathrm{\pi }}{435\times {10}^{-9}} \\ =1.44\times {10}^7\mathrm{rad}/\mathrm{m} \end{gathered}$$

Substitute the variable in equations of sinusoidal electromagnetic wave for electric and magnetic field are:

$$\begin{gathered} \overrightarrow{E}=E_{max}=\cos \left(kx-\omega t\right) \\ \overrightarrow{E}=\left(2.70\times {10}^{-3}V/m\right)\cos \left(\left[1.44\times {10}^{7}m\right]z-\left[4.3\times {10}^{15}rad/s\right]t\right) \\ \overrightarrow{B}=B_{max}\cos \left(kx-\omega t\right) \\ \overrightarrow{B}=\left(9\times {10}^{-12}T\right)\cos \left(\left[1.44\times {10}^{7}rad/m\right]z-\left[4.3\times {10}^{15}rad/s\right]t\right) \end{gathered}$$

Hence, the vector equation for$\overrightarrow{E}\left(z,t\right)=\stackrel{⏜}{i}\left[2.70\times {10}^{-3}V/m\cos \left\{\left(1.44\times {10}^{7}m\right)z-\left[4.3\times {10}^{15}rad/s\right]t\right\}\right]$ and $\overrightarrow{\mathrm{B}}\left(\mathrm{z},\mathrm{t}\right)=-\stackrel{⏜}{\mathrm{j}}\left[9\times {10}^{-12}\mathrm{Tcos}\left\{\left(1.44\times {10}^7\mathrm{m}\right)\mathrm{z}-\left(4.3\times {10}^{15}\mathrm{rad}/\mathrm{s}\right)\mathrm{t}\right\}\right]$.