91Ó°ÊÓ

To determine the frequency of electromagnetic waves, we rely on the fundamental relationship between speed, frequency, and wavelength. The formula \( f = \frac{v}{\lambda} \) is our key to finding frequency \( f \). Here, \( v \) represents the speed of the wave, which for electromagnetic waves like light, is the speed of light, approximately \( 3 \times 10^8 \, \mathrm{m/s} \). The symbol \( \lambda \) stands for the wavelength of the wave.
For the problem at hand, the wavelength is provided as 35.4 cm, which we need to convert to meters—0.354 m. By substituting these values into our formula, the frequency calculation becomes clear:
\[f = \frac{3 \times 10^8 \, \mathrm{m/s}}{0.354 \, \mathrm{m}} \approx 8.475 \times 10^8 \, \mathrm{Hz}\]
This means our wave vibrates at a high frequency of approximately 847.5 million cycles per second, a common frequency range for cellular phones and other wireless technologies.

Once you know the electric field amplitude, calculating the magnetic field amplitude is straightforward using the equation \( c = \frac{E_0}{B_0} \). \( E_0 \) is the electric field amplitude, which is given in the problem, and \( c \) is the constant representing the speed of light.
Rearrange the equation to solve for the magnetic field amplitude \( B_0 \):
\[B_0 = \frac{E_0}{c}\]
Substituting the given electric field amplitude of \( 5.40 \times 10^{-2} \, \mathrm{V/m} \) and the speed of light \( 3 \times 10^8 \, \mathrm{m/s} \), we find:
\[B_0 = \frac{5.40 \times 10^{-2} \, \mathrm{V/m}}{3 \times 10^8 \, \mathrm{m/s}} \approx 1.8 \times 10^{-10} \, \mathrm{T}\]
This tiny magnetic field amplitude, measured in teslas, is typical for electromagnetic waves, indicating how strongly the magnetic field varies over time and space.

Wave intensity gives us an understanding of the power carried by the wave per unit area. In electromagnetics, intensity \( I \) is calculated using the formula:
\[I = \frac{1}{2} \cdot \epsilon_0 \cdot c \cdot E_0^2\]
Here, \( \epsilon_0 \) is the vacuum permittivity (\( 8.85 \times 10^{-12} \, \mathrm{F/m} \)), and \( E_0 \) is the electric field amplitude.
By plugging in the known values of electric field amplitude and speed of light, the calculation proceeds as follows:
\[I = \frac{1}{2} \times 8.85 \times 10^{-12} \, \mathrm{F/m} \times 3 \times 10^8 \, \mathrm{m/s} \times (5.40 \times 10^{-2} \, \mathrm{V/m})^2 \approx 3.87 \times 10^{-3} \, \mathrm{W/m^2}\]
This shows the wave's intensity, giving an estimate of how much power is transmitted over a given area. Such a level of intensity is typical for the type of electromagnetic waves emitted by a cellular phone, indicating a relatively low power needed for transmission.