Electromagnetic WavesClass 12 Physics NCERT Solutions

Given: Radius of circular plates, r = 12 cm = 0.12 m Separation between plates, d = 5.0 cm = 0.05 m Charging current, I = 0.15 A

1. Capacitance (C): The capacitance of a parallel plate capacitor is given by: C = ε₀A / d where A is the area of the plates, A = πr².

C = (8.854 × 10⁻¹² F/m × π × (0.12 m)²) / 0.05 m C = (8.854 × 10⁻¹² × 3.1416 × 0.0144) / 0.05 C ≈ 8.01 × 10⁻¹² F = 8.01 pF

2. Rate of change of potential difference (dV/dt): The charge on the capacitor at any time t is Q = CV. The charging current is I = dQ/dt. I = d(CV)/dt Since C is constant, I = C (dV/dt). Therefore, dV/dt = I / C

dV/dt = 0.15 A / (8.01 × 10⁻¹² F) dV/dt ≈ 1.87 × 10¹⁰ V/s

The displacement current (i_d) is given by i_d = ε₀ (dΦ_E/dt). As derived in the chapter, for a charging capacitor, the displacement current between the plates is exactly equal to the conduction current (i_c) in the connecting wires. i_d = i_c = I = 0.15 A

Kirchhoff's first rule (junction rule) states that the algebraic sum of currents entering a junction is zero.

If we consider only the conduction current, the rule is not valid at the capacitor plate. Conduction current (0.15 A) flows towards one plate, but no conduction current flows out of it through the gap.

However, Maxwell's modification shows that the rule is valid if we include the displacement current. At the positive plate, the conduction current (i_c) flows in, and the displacement current (i_d) flows out from the plate into the space between the plates.

Since i_c = i_d = 0.15 A, the total current entering the junction (the plate) is i_c, and the total current leaving is i_d. Thus, the net current at the junction is i_c - i_d = 0. This makes Kirchhoff's junction rule valid at each plate of the capacitor.

Given: Radius of plates, R = 6.0 cm = 0.06 m Capacitance, C = 100 pF = 100 × 10⁻¹² F RMS voltage of AC supply, V_rms = 230 V Angular frequency, ω = 300 rad s⁻¹ Distance from the axis, r = 3.0 cm = 0.03 m

First, we calculate the capacitive reactance (X_C): X_C = 1 / (ωC) X_C = 1 / (300 rad/s × 100 × 10⁻¹² F) = 1 / (3 × 10⁻⁸) Ω ≈ 3.33 × 10⁷ Ω

The rms value of the conduction current (I_rms) is given by Ohm's law for AC circuits: I_rms = V_rms / X_C I_rms = 230 V / (3.33 × 10⁷ Ω) ≈ 6.9 × 10⁻⁶ A = 6.9 µA

Yes, the conduction current is equal to the displacement current. For a capacitor connected to an AC source, the conduction current flowing in the wires is always equal to the displacement current between the plates at any instant.

Between the plates, the magnetic field is produced only by the displacement current. Using the Ampere-Maxwell law for a circular loop of radius r (r < R) between the plates: ∮ B⋅dl = μ₀ × (current enclosed)

The displacement current enclosed by the loop of radius r is not the total displacement current, but a fraction of it, proportional to the area. I_d (enclosed) = I_d (total) × (πr² / πR²) = I_d × (r/R)²

The amplitude of the total displacement current is I₀ = I_rms × √2. I₀ = 6.9 × 10⁻⁶ A × √2 ≈ 9.76 × 10⁻⁶ A

From Ampere-Maxwell law, B × (2πr) = μ₀ × I₀ × (r²/R²) B = (μ₀ × I₀ × r) / (2πR²)

B = (4π × 10⁻⁷ T·m/A × 9.76 × 10⁻⁶ A × 0.03 m) / (2π × (0.06 m)²) B = (2 × 10⁻⁷ × 9.76 × 10⁻⁶ × 0.03) / (0.0036) B ≈ 1.63 × 10⁻¹¹ T

Thus, the amplitude of the magnetic field at that point is approximately 1.63 × 10⁻¹¹ T.

The physical quantity that is the same for all these electromagnetic waves (X-rays, red light, and radio waves) is their speed in a vacuum. All electromagnetic waves travel through a vacuum at the same speed, which is the speed of light, c.

c = 3 × 10⁸ m/s.

As stated in the chapter, for an electromagnetic wave, the electric field vector (E), the magnetic field vector (B), and the direction of propagation are mutually perpendicular to each other. Since the wave travels along the z-direction, the electric and magnetic field vectors must lie in the plane perpendicular to the z-direction, which is the x-y plane. They oscillate perpendicularly to each other and to the direction of propagation.

Given: Frequency, ν = 30 MHz = 30 × 10⁶ Hz Speed of wave in vacuum, c = 3 × 10⁸ m/s

The relationship between speed (c), frequency (ν), and wavelength (λ) is: c = νλ

Therefore, the wavelength λ is: λ = c / ν λ = (3 × 10⁸ m/s) / (30 × 10⁶ Hz) λ = (3 × 10⁸) / (3 × 10⁷) m λ = 10 m

The relationship between frequency (ν) and wavelength (λ) for electromagnetic waves is λ = c/ν, where c = 3 × 10⁸ m/s.

We need to find the wavelengths corresponding to the lower and upper frequency limits of the band.

For the lower frequency limit (ν₁ = 7.5 MHz): ν₁ = 7.5 MHz = 7.5 × 10⁶ Hz λ₁ = c / ν₁ = (3 × 10⁸ m/s) / (7.5 × 10⁶ Hz) λ₁ = 40 m

For the upper frequency limit (ν₂ = 12 MHz): ν₂ = 12 MHz = 12 × 10⁶ Hz λ₂ = c / ν₂ = (3 × 10⁸ m/s) / (12 × 10⁶ Hz) λ₂ = 25 m

Therefore, the corresponding wavelength band is from 25 m to 40 m.

According to Maxwell's theory, an accelerating charged particle radiates electromagnetic waves. An oscillating charged particle is an example of an accelerating charge.

The frequency of the electromagnetic waves produced by an oscillating charge is equal to the frequency of oscillation of the charge itself.

Given: Frequency of oscillation of the charged particle = 10⁹ Hz

Therefore, the frequency of the electromagnetic waves produced by the oscillator is also 10⁹ Hz.

Given: Amplitude of the magnetic field, B₀ = 510 nT = 510 × 10⁻⁹ T

The amplitudes of the electric field (E₀) and magnetic field (B₀) in an electromagnetic wave in a vacuum are related by the speed of light (c): E₀ = c × B₀

where c = 3 × 10⁸ m/s.

E₀ = (3 × 10⁸ m/s) × (510 × 10⁻⁹ T) E₀ = 1530 × 10⁻¹ V/m E₀ = 153 V/m

The amplitude of the electric field part of the wave is 153 V/m.

Given: Electric field amplitude, E₀ = 120 N/C Frequency, ν = 50.0 MHz = 50 × 10⁶ Hz

Let us assume the wave propagates along the positive z-direction, the electric field oscillates along the x-direction, and the magnetic field oscillates along the y-direction. The expressions for E and B are:

To calculate the photon energy in eV, we use the formula E = hν, where h (Planck's constant) = 6.63 × 10⁻³⁴ J·s. To convert from Joules to eV, we divide by the charge of an electron, e = 1.602 × 10⁻¹⁹ J/eV.

Here are the approximate photon energies for different parts of the spectrum:

The scale of photon energies is directly related to the energy transitions occurring in the source of the radiation. The mechanism that produces the radiation must involve energy changes of the same order of magnitude as the photon energy.

Given: Frequency, ν = 2.0 × 10¹⁰ Hz Electric field amplitude, E₀ = 48 V/m Speed of light, c = 3 × 10⁸ m/s

The relationship between wavelength, frequency, and speed is λ = c/ν. λ = (3 × 10⁸ m/s) / (2.0 × 10¹⁰ Hz) λ = 1.5 × 10⁻² m = 1.5 cm

The amplitudes of the electric and magnetic fields are related by B₀ = E₀ / c. B₀ = (48 V/m) / (3 × 10⁸ m/s) B₀ = 16 × 10⁻⁸ T = 1.6 × 10⁻⁷ T

The energy density of the electric field (u_E) is given by u_E = (1/2)ε₀E². The average energy density over one cycle for a sinusoidal wave is: ⟨u_E⟩ = (1/2)ε₀⟨E²⟩ = (1/2)ε₀(E₀²/2) = (1/4)ε₀E₀²

The energy density of the magnetic field (u_B) is given by u_B = (1/2μ₀)B². The average energy density over one cycle is: ⟨u_B⟩ = (1/2μ₀)⟨B²⟩ = (1/2μ₀)(B₀²/2) = (1/4μ₀)B₀²

To show that ⟨u_E⟩ = ⟨u_B⟩, we can substitute the relations E₀ = cB₀ and c² = 1/(μ₀ε₀) into the equation for ⟨u_E⟩.

⟨u_E⟩ = (1/4)ε₀E₀² ⟨u_E⟩ = (1/4)ε₀(cB₀)² ⟨u_E⟩ = (1/4)ε₀c²B₀²

Now substitute c² = 1/(μ₀ε₀): ⟨u_E⟩ = (1/4)ε₀ [1/(μ₀ε₀)] B₀² ⟨u_E⟩ = (1/4μ₀)B₀²

This is the expression for the average energy density of the magnetic field, ⟨u_B⟩. Thus, ⟨u_E⟩ = ⟨u_B⟩. The average energy density of the electric field is equal to the average energy density of the magnetic field.