EXERCISES

This exercise involves calculating the capacitance of a capacitor and understanding the displacement current. A capacitor stores electrical energy in the electric field between its plates. The capacitance (C) can be determined using the formula C = ε₀(A/d), where A is the area of the plates and d is the distance between them. The displacement current can be computed using the changing electric field, which relates to the same current flowing through the capacitor. Kirchhoff’s junction rule states that the total current entering a junction must equal the total current leaving, which must be analyzed at the capacitor plates considering both conduction and displacement currents.

In this exercise, we analyze a parallel plate capacitor connected to an alternating current (AC) supply. The root mean square (RMS) value of the conduction current can be calculated using the formula I = C * V * ω, where ω is the angular frequency. We need to check if the conduction current equals the displacement current by understanding the relationship established by Maxwell. Finally, knowing the electric field (E) can help determine the magnetic field amplitude (B) at a given distance using the relationship B = E/c.

This exercise prompts students to think about a fundamental property of electromagnetic waves: all waves, regardless of their type or wavelength, travel at the speed of light in vacuum. Thus, no matter if they are X-rays, visible light, or radio waves, their behavior under vacuum circumstances will be consistent, with each type having its own unique applications and energy characteristics.

In a plane electromagnetic wave, the electric field (E) and the magnetic field (B) are always perpendicular to each other and to the direction of propagation of the wave. The wavelength (λ) can be computed using the relationship c = fλ, where c is the speed of light and f is the frequency. For a frequency of 30 MHz, the corresponding wavelength can be calculated as λ = c/f.

To find the wavelength band for radio frequencies, we can convert frequencies to wavelengths using the formula λ = c/f. By inputting the frequencies for the range provided, we can calculate the minimum and maximum wavelengths for this band of radio frequencies.

In this case, the frequency of the oscillating charge directly relates to the frequency of the electromagnetic waves it produces. Therefore, if the charged particle oscillates at 10^9 Hz, it emits electromagnetic waves of the same frequency.

To find the electric field amplitude (E) from the magnetic field amplitude (B), we can use the relationship E = cB, where c is the speed of light in vacuum. Substituting the values will yield the electric field amplitude.

For this exercise, we calculate several parameters of the electromagnetic wave, including the magnetic field amplitude (B₀) using B₀ = E₀/c, angular frequency (ω) as ω = 2πf, the wave number (k) as k = 2π/λ, and the wavelength (λ) using the relation c = fλ. We will find the expressions for E and B using their sinusoidal expressions relating to the given frequencies and amplitudes.

This exercise requires the use of the relationship E = hn to calculate the energy of photons for various electromagnetic radiation parts. Different energy levels indicate the type of radiation produced—from radio waves (low energy) to gamma rays (high energy)—and this energy relates directly to their production sources and interactions with matter.

To solve these questions, we first calculate the wavelength using λ = c/f. Next, we find the magnetic field amplitude using B = E/c. The average energy density for both fields can be derived using the formulas for energy density in an electromagnetic field to show they are equal, emphasizing the energy's distribution across electric and magnetic components.