Radio station WCCO in Minneapolis broadcasts at a frequency of . At a point some distance from the transmitter, the magnetic-field amplitude of the electromagnetic wave from WCCO is $\mathbf{4}\mathbf{.}\mathbf{82}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{11}}\mathbf{T}$. Calculate (a) the wavelength; (b) the wave number; (c) the angular frequency; (d) the electric-field amplitude.

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

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

Where, T = 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 $3\times {10}^8\mathrm{m}/\mathrm{s}$, $\lambda$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 m/s.

The relation between wave number and wavelength is:

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

Where, k is wavenumber in rad/m.

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

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

Given that,$\mathrm{f}=830\times {10}^3\mathrm{Hz}$

The wavelength of wave is:

$$\begin{gathered} \mathrm{\lambda }=\frac{\mathrm{c}}{\mathrm{f}} \\ =\frac{3\times {10}^8}{830\times {10}^3} \\ =361\mathrm{m} \end{gathered}$$

The wave number of wave is:

role="math" $$\begin{gathered} \mathrm{k}=\frac{2\mathrm{\pi }}{\mathrm{\lambda }} \\ =\frac{2\mathrm{\pi }}{361} \\ =174\times {10}^{-4}\mathrm{rad}/\mathrm{s} \end{gathered}$$

Hence, wavelength of the electromagnetic wave is 361 m and the wave number of the electromagnetic wave is $174\times {10}^{-4}\mathrm{rad}/\mathrm{m}$.

The angular frequency of wave is:

$$\begin{gathered} \mathrm{\omega }=2\mathrm{\pi f} \\ =2\mathrm{\pi }\left(830\times {10}^3\right) \\ =5.22\times {10}^8\mathrm{rad}/\mathrm{s} \end{gathered}$$

Given that,${\mathrm{B}}_{\max }=4.82\times {10}^{-11}\mathrm{T}$

The formula used to determine the amplitude of electric field of the wave is:

$$\begin{gathered} {\mathrm{E}}_{\max }={\mathrm{cB}}_{\max } \\ =\left(3\times {10}^8\right)\left(4.82\times {10}^{-11}\right) \\ =144\times {10}^{-4}\mathrm{V}/\mathrm{m} \end{gathered}$$

Hence, the angular velocity of the electromagnetic wave is $5.22\times {10}^6\mathrm{rad}/\mathrm{s}$and the amplitude of electric field is$144\times {10}^{-4}\mathrm{V}/\mathrm{m}$ .