Question

An electromagnetic wave has a magnetic field given by $\overrightarrow{\mathbf{B}}\left(x,t\right)\mathbf{=}\mathbf{-}\left(8.25\times {10}^{-9}T\right)\hat{\mathbf{j}}\mathbf{cos}\left[\left(1.38\times {10}^4\mathrm{rad}/m\right)x+\mathrm{\omega t}\right]$. (a) In which direction is the wave traveling? (b) What is the frequency of the wave? (c) Write the vector equation for $\overrightarrow{\mathbf{E}}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{t}\mathbf{)}$.

Short answer

1. In - x - direction wave is traveling.
2. The frequency of the wave is $6.59\times {10}^{11}\mathrm{Hz}$.

3. The vector equation for $\overrightarrow{\mathrm{E}}(\mathrm{x},\mathrm{t})$is

$\overrightarrow{\mathrm{E}}=\left(-248\mathrm{V}/\mathrm{m}\right)\cos \left[\left(1.38\times {10}^4\mathrm{rad}/\mathrm{m}\right)\mathrm{x}+\left(4.14\times {10}^{12}\mathrm{rad}/\mathrm{s}\right)\mathrm{t}\right]$ .

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.

$f\mathit{=}\frac{\mathit{1}}{\mathit{T}}or\frac{\mathit{\omega }}{\mathit{2}\mathit{\pi }}$

Where, Time period in seconds and is angular frequency in .

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

The wavelength-frequency relationship is:

$c=\lambda f$

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

The wavelength is inversely proportional to the frequency.

$\lambda =\frac{c}{f}or\frac{v}{f}$

Where, v is velocity of wave in .

The relation between wave number and wavelength is:

$k=\frac{2\pi }{\lambda }$

Where, k is wavenumber in .

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

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

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

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

Determine the direction of wave and frequency of wave.

$\overrightarrow{\mathrm{B}}(\mathrm{x},\mathrm{t})=-(8.25\times {10}^{-9}\mathrm{T})\hat{\mathrm{j}}\cos [(1.38\times {10}^4\mathrm{rad}/\mathrm{m})\mathrm{x}+\mathrm{\omega t}]$Given that,

The values of wavenumber$\left(k\right)$ is$1.38\times {10}^4\mathrm{rad}/\mathrm{m}$ and the maximum magnetic field is .

${\mathrm{B}}_{\max }\mathrm{is}-8.25\times {10}^{-9}\mathrm{T}$

$\overrightarrow{\mathrm{B}}=-{\mathrm{B}}_{\max }\cos \left(\mathrm{kx}+\mathrm{\omega t}\right)$

Since, if the phase of wave is$kx-\omega t$ then wave travels in positive x- direction. Therefore, the above equation defines the wave is travelling in negative x-direction as the phase of wave is $kx-\omega t$.

The frequency of wave is:

$$\begin{gathered} \mathrm{f}=\frac{\mathrm{c}}{\mathrm{\lambda }} \\ =\frac{\mathrm{kc}}{2\mathrm{\pi }} \\ =\frac{\left(1.38\times {10}^4\right)\left(3.0\times {10}^8\right)}{2\mathrm{\pi }} \\ =6.59\times {10}^{11}\mathrm{Hz} \end{gathered}$$

Determine the vector equation.

The angular velocity of wave is:

$$\begin{gathered} \mathrm{\omega }=2\mathrm{\pi f} \\ =2\mathrm{\pi }\left(6.59\times {10}^{11}\right) \\ =4.14\times {10}^{12}\mathrm{rad}/\mathrm{s} \end{gathered}$$

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

$$\begin{gathered} {\mathrm{E}}_{\max }={\mathrm{cB}}_{\max } \\ =\left(3\times {10}^8\right)\left(-8.28\times {10}^9\right) \\ =-2.48\mathrm{V}/\mathrm{m} \end{gathered}$$

The equation of sinusoidal electromagnetic wave for magnetic field is:

$$\begin{gathered} \overrightarrow{\mathrm{E}}={\mathrm{E}}_{\max }\cos \left(\mathrm{kx}+\mathrm{\omega t}\right) \\ =\left(-2.48\mathrm{V}/\mathrm{m}\right)\cos \left[\left(1.38\times {10}^4\mathrm{rad}/\mathrm{m}\right)\mathrm{x}+\left(4.14\times {10}^{12}\mathrm{rad}/\mathrm{s}\right)\mathrm{t}\right] \end{gathered}$$

Hence, the vector equation for is $\overrightarrow{\mathrm{E}}=\left(\mathrm{x},\mathrm{t}\right)$.

$\overrightarrow{\mathrm{E}}=\left(-2.48\mathrm{V}/\mathrm{m}\right)\cos \left[\left(1.38\times {10}^4\mathrm{rad}/\mathrm{m}\right)\mathrm{x}+\left(4.14\times {10}^{12}\mathrm{rad}/\mathrm{s}\right)\mathrm{t}\right]$