Question

A sinusoidal electromagnetic wave having a magnetic field of amplitude$\mathbf{1}\mathbf{.}\mathbf{25}\mathbf{}\mathit{\mu }\mathit{T}$ and a wavelength of 432 nm is traveling in the +x- direction through empty space. (a) What is the frequency of this wave? (b) What is the amplitude of the associated electric field? (c) Write the equations for the electric and magnetic fields as functions of x and t in the form of Eqs. (32.17).

Short answer

a) The frequency of a sinusoidal electromagnetic wave is $6.94\times {10}^{14}\mathrm{Hz}$.

b) The amplitude of the associated electric field is 375 V/m.

c) The equations for the electric and magnetic fields as function of x and t are$\overrightarrow{\mathrm{E}}=(375\mathrm{V}/\mathrm{m})\cos \left(\right[1.45\times {10}^7\mathrm{rad}/\mathrm{m}]\mathrm{x}-[4.36\times {10}^{15}\mathrm{rad}/\mathrm{s}\left]\mathrm{t}\right)$ and$\overrightarrow{\mathrm{B}}=(1.25\times {10}^{-6}\mathrm{T})\cos \left(\right[1.45\times {10}^7\mathrm{rad}/\mathrm{m}]\mathrm{x}-[4.36\times {10}^{15}\mathrm{rad}/\mathrm{s}\left]\mathrm{t}\right)$ respectively.

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:

role="math" localid="1664345299622" $$\begin{gathered} \overrightarrow{E}=E_{max}cos(kx-\omega t) \\ \overrightarrow{B}=B_{max}cos(kx-\omega t) \end{gathered}$$

Determine the frequency of wave.

Given that,$\mathrm{\lambda }=432\times {10}^{-9}\mathrm{m}$

The wavelength-frequency relationship is

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

Substitute the values of variables

$$\begin{gathered} f=\frac{3\times {10}^8}{432\times {10}^{-9}} \\ =6.94\times {10}^{14}Hz \end{gathered}$$

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

Determine the amplitude of electric field.

Given that,$B_{max}=1.25\times {10}^{-6}\mathrm{T}$

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

$E_{max}=B_{max}.c$

Substitute the values of variables

$$\begin{gathered} E_{max}=\left(1.25\times {10}^6\right).\left(3\times {10}^8\right) \\ =375\mathrm{V}/\mathrm{m} \end{gathered}$$

Hence, the amplitude of the associated electric field is 375 V/m.

Determine the sinusoidal electromagnetic wave.

The angular frequency of wave is

$$\begin{gathered} \omega =2\mathrm{\pi f} \\ =2\mathrm{\pi }\left(6.94\times {10}^{14}\right) \\ =4.36\times {10}^{15}\mathrm{rad}/\mathrm{s} \end{gathered}$$

And the wavenumber of wave is

$$\begin{gathered} \mathrm{k}=\frac{2\mathrm{\pi }}{\mathrm{\lambda }} \\ =\frac{2\mathrm{\pi }}{432\times {10}^{-9}} \\ =1.45\times {10}^7\mathrm{rad}/\mathrm{m} \end{gathered}$$

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

role="math" localid="1664346679296" $$\begin{gathered} \overrightarrow{E}=E_{max}\cos \left(kx-\omega t\right) \\ \overrightarrow{E}=\left(375V/m\right)\cos \left(\left[1.45\times {10}^{7}rad/m\right]x-\left[4.36\times {10}^{15}rad/s\right]t\right) \\ \overrightarrow{B}=B_{max}\cos \left(ks-\omega t\right) \\ \overrightarrow{B}=\left(1.25\times {10}^{-6}T\right)\cos \left(\left[1.45\times {10}^{7}rad/m\right]x-\left[4.36\times {10}^{15}rad/s\right]t\right) \end{gathered}$$

Hence, the equations for the electric and magnetic fields as function of x and t are and respectively.

$$\begin{gathered} \overrightarrow{\mathrm{E}}=\left(375\mathrm{V}/\mathrm{m}\right)\cos \left(\left[1.45\times {10}^7\mathrm{rad}/\mathrm{m}\right]\mathrm{x}-\left[4.36\times {10}^{15}\mathrm{rad}/\mathrm{s}\right]\mathrm{t}\right)\mathrm{and} \\ \overrightarrow{\mathrm{B}}=\left(1.25\times {10}^{-6}\mathrm{T}\right)\cos \left(\left[1.45\times {10}^7\mathrm{rad}/\mathrm{m}\right]\mathrm{x}-\left[4.36\times {10}^{15}\mathrm{rad}/\mathrm{s}\right]\mathrm{t}\right)\mathrm{respectively} \end{gathered}$$