Question

The phasor of the electric field of a plane wave that propagates in air is given by the following expression:

E = (4aₓ - j2aᵧ)e - j20π
a). Determine the frequency and the wavelength of this plane wave

Answer 1

The amplitude of the electric field is √20 V/m, the frequency is 10 Hz, and the wavelength is 3 x 10^7 m. The wave propagates forward in time.

Step-by-step explanation:

The phasor of the electric field of a plane wave propagating in air is given by E = (4a₁ - j2aₒ)e^(-j20πt). From this expression, we can analyze the properties of the wave:

• The amplitude of the electric field is given by the magnitude of the vector (4a₁ - j2aₒ), which can be calculated using the Pythagorean theorem, resulting in a magnitude of √(4² + 2²) = √20 V/m.
• The frequency (f) can be determined from the angular frequency (ω) in the exponent of the phasor. Since ω = 20π, we can use the relation ω = 2πf to find that f = 10 Hz.
• The wavelength (λ) can be found using the relationship between speed (c), frequency (f), and wavelength (λ): c = fλ. In air, the speed of light c is approximately 3 x 108 m/s, so λ = c/f which is 3 x 108/10 m.
• The wave is propagating in the direction determined by the phase factor e^(-j20πt). This shows that the wave is traveling in the direction of increasing time, or forward in time.