Question

The electric field of a sinusoidal electromagnetic wave obeys the equation$\mathit{E}\mathbf{=}\mathbf{(}\mathbf{375}\mathbf{V}\mathbf{/}\mathbf{m}\mathbf{)}\mathit{c}\mathit{o}\mathit{s}\mathbf{}\left[\left(1.99\times {10}^7\mathrm{rad}/m\right)\times x+\left(5.97\times {10}^{15}\mathrm{rad}/s\right)t\right]$ . (a) What is the speed of the wave? (b) What are the amplitudes of the electric and magnetic fields of this wave? (c) What are the frequency, wavelength, and period of the wave? Is this light visible to humans?

Short answer

1. The speed of a sinusoidal electromagnetic wave is$3\times {10}^{8}m/s$ .
2. The amplitudes of the electric and magnetic fields of the wave are 375 V/m and$1.25\times {10}^{-6}T$ respectively.
3. The frequency, wavelength and period of the wave are$9.50\times {10}^{14}Hz,3.15\times {10}^{-7}m$ and$1.05\times {10}^{-15}s$ 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.

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

Where,T =Time period in seconds and$\omega$ 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, is the speed of light which is equal to$3\times {10}^{8}m/s.\lambda$ , is the wavelength and 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:

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

Where, k is wavenumber in .

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

$\begin{array}{r}{E}_{max}=\sqrt{\frac{2I}{{\epsilon }_{0}C}}\\ B_{max}=\frac{E_{max}}{c}\end{array}$

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

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

Determine the amplitudes of electric and magnetic fields.

Given that,$E=(375\mathrm{V}/\mathrm{m})\cos \left[\left(1.99\times {10}^7\mathrm{rad}/\mathrm{m}\right)\mathrm{x}+\left(5.97\times {10}^{15}\mathrm{rad}/\mathrm{s}\right)\mathrm{t}\right]$

So,$E_{max}=375\mathrm{V}/\mathrm{m}$

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

$\begin{array}{r}{B}_{max}=\frac{375}{3\times {10}^8}\\ =1.25\times {10}^{-6}\mathrm{T}\end{array}$

Hence, the amplitudes of the electric and magnetic fields of the wave are 375 V/m and$1.25\times {10}^{-6}T$ respectively.

Determine the frequency and wavelength of wave.

From equation$E=(375\mathrm{V}/\mathrm{m})\cos \left[\left(1.99\times {10}^7\mathrm{rad}/\mathrm{m}\right)\mathrm{x}+\left(5.97\times {10}^{15}\mathrm{rad}/\mathrm{s}\right)\mathrm{t}\right]$

The values of angular frequency is and the wavenumber is .

The frequency of wave is:

$\begin{array}{r}f=\frac{\omega }{2\pi }\\ =\frac{5.97\times {10}^{15}}{2\pi }\\ =9.50\times {10}^{14}\mathrm{Hz}\end{array}$

The wavelength of wave is:

$\begin{array}{r}\lambda =\frac{2\pi }{k}\\ =\frac{2\pi }{1.99\times {10}^7}\\ =3.15\times {10}^{-7}\mathrm{m}\end{array}$

Hence the frequency and wavelength of the wave are$9.50\times {10}^{14}Hz$ and$3.15\times {10}^{-7}m$ respectively.

Determine the period and speed of wave.

The period of wave is:

$\begin{array}{r}T=\frac{1}{f}\\ =\frac{1}{9.50\times {10}^{14}}\\ =1.052\times {10}^{-15}\mathrm{s}\end{array}$

The speed of the wave is:

$\begin{array}{r}C=f\lambda \\ =\left(9.50\times {10}^{14}\right)\left(3.15\times {10}^{-7}\right)\\ =3\times {10}^8\mathrm{m}/\mathrm{s}\end{array}$

Hence, the period and speed of wave are$1.05\times {10}^{-15}s$ and$3\times {10}^{8}m/s$ respectively.