Electromagnetic Waves Class 12 Physics NCERT Solutions

Step-by-step solution

1. 1(a) A = πr² = π(0.12)² = 4.52 × 10⁻² m²; d = 5.0 cm = 0.05 m.
2. 2C = ε₀A/d = 8.85 × 10⁻¹² × 4.52 × 10⁻²/0.05 = 8.0 × 10⁻¹² F ≈ 80 pF.
3. 3dV/dt = I/C = 0.15/8.0 × 10⁻¹² = 1.87 × 10¹⁰ V s⁻¹.
4. 4(b) Across the plates the conduction current cannot flow, but the changing field keeps the current continuous: I_d = ε₀ dΦE/dt = C dV/dt = I = 0.15 A.
5. 5(c) For conduction currents alone the junction rule fails at the plates — charge keeps accumulating there. The rule is restored only when the displacement current 0.15 A across the gap is included in the balance.

Step-by-step solution

1. 1(a) X_C = 1/(ωC), so I_rms = V_rms ωC = 230 × 300 × 100 × 10⁻¹².
2. 2I_rms = 230 × 3.0 × 10⁻⁸ = 6.9 × 10⁻⁶ A = 6.9 µA.
3. 3(b) Yes — the total current is continuous; the conduction current in the leads equals the displacement current between the plates at every instant.
4. 4(c) Use Ampere–Maxwell: at radius r (inside the plate, r < R), B·2πr = μ₀ I_d(r²/R²), so B = μ₀ I_d r/(2πR²).
5. 5Peak displacement current: I₀ = √2 I_rms = √2 × 6.9 × 10⁻⁶ = 9.76 × 10⁻⁶ A.
6. 6B_max = (4π × 10⁻⁷ × 9.76 × 10⁻⁶ × 0.03)/(2π × 0.06²).
7. 7B_max = (2 × 10⁻⁷ × 9.76 × 10⁻⁶ × 0.03)/3.6 × 10⁻³ = 1.63 × 10⁻¹¹ T.

Step-by-step solution

1. 1All three are electromagnetic waves travelling in vacuum.
2. 2The common physical quantity is their speed c = 3 × 10⁸ m s⁻¹.
3. 3(Their wavelengths, frequencies, photon energies and penetration power all differ enormously.)

Step-by-step solution

1. 1The wave is transverse: E and B are perpendicular to each other and both are perpendicular to the direction of propagation (the z-axis).
2. 2If E oscillates along one transverse axis (say x), then B oscillates along the other (y) so that E × B points along the propagation direction z.
3. 3λ = c/ν = 3 × 10⁸/30 × 10⁶ = 10 m.

Step-by-step solution

1. 1λ = c/ν.
2. 2At 12 MHz: λ = 3 × 10⁸/12 × 10⁶ = 25 m.
3. 3At 7.5 MHz: λ = 3 × 10⁸/7.5 × 10⁶ = 40 m.
4. 4So the tuner covers the wavelength band 25 m to 40 m.

Step-by-step solution

1. 1Accelerated charge radiates EM waves of the same frequency at which it oscillates.
2. 2Hence the EM waves have frequency 10⁹ Hz.

Step-by-step solution

1. 1B₀ = 510 nT = 5.10 × 10⁻⁷ T.
2. 2E₀ = c B₀ = 3 × 10⁸ × 5.10 × 10⁻⁷.
3. 3E₀ = 1.53 × 10² V m⁻¹ = 153 V m⁻¹.

Step-by-step solution

1. 1(a) B₀ = E₀/c = 120/3 × 10⁸ = 4.0 × 10⁻⁷ T.
2. 2ω = 2πν = 2π × 50 × 10⁶ = 3.14 × 10⁸ rad s⁻¹.
3. 3λ = c/ν = 3 × 10⁸/50 × 10⁶ = 6.0 m.
4. 4k = 2π/λ = 2π/6.0 = 1.05 rad m⁻¹ (equivalently k = ω/c).
5. 5(b) Taking the wave along the z-axis with E along x and B along y: E = 120 sin(1.05z – 3.14 × 10⁸t) x̂ V m⁻¹.
6. 6B = 4.0 × 10⁻⁷ sin(1.05z – 3.14 × 10⁸t) ŷ T.

Step-by-step solution

1. 1E = hν = hc/λ, and E(eV) = 1240 eV·nm/λ(nm).
2. 2Radio (λ ≈ 500 m): E = 6.63 × 10⁻³⁴ × 3 × 10⁸/500 = 3.98 × 10⁻²⁸ J ≈ 2.5 × 10⁻⁹ eV.
3. 3Microwave (λ ≈ 10⁻² m): E ≈ 1.2 × 10⁻⁴ eV.
4. 4Infrared (λ ≈ 10⁻⁵ m): E ≈ 0.12 eV.
5. 5Visible (λ ≈ 5 × 10⁻⁷ m): E ≈ 2.5 eV.
6. 6Ultraviolet (λ ≈ 10⁻⁸ m): E ≈ 1.2 × 10² eV.
7. 7X-rays (λ ≈ 10⁻¹⁰ m): E = 6.63 × 10⁻³⁴ × 3 × 10⁸/10⁻¹⁰ = 1.99 × 10⁻¹⁵ J ≈ 1.24 × 10⁴ eV.
8. 8Gamma rays (λ ≈ 10⁻¹² m): E ≈ 1.24 × 10⁶ eV (MeV range).
9. 9Interpretation: photon energy rises by many orders of magnitude from radio to gamma, exactly matching the scale of the emitting processes — oscillating circuits and molecular rotation (radio–IR), atomic and molecular transitions (visible–UV), inner-shell and nuclear processes (X-ray–gamma).

Step-by-step solution

1. 1(a) λ = c/ν = 3 × 10⁸/2.0 × 10¹⁰ = 1.5 × 10⁻² m.
2. 2(b) B₀ = E₀/c = 48/3 × 10⁸ = 1.6 × 10⁻⁷ T.
3. 3(c) Average energy density of E: ū_E = ½ε₀⟨E²⟩ = ¼ε₀E₀².
4. 4Average energy density of B: ū_B = ⟨B²⟩/(2μ₀) = B₀²/(4μ₀).
5. 5Since B₀ = E₀/c and c² = 1/(ε₀μ₀): B₀²/μ₀ = E₀²/(c²μ₀) = ε₀E₀².
6. 6Hence ū_B = ¼ε₀E₀² = ū_E. The two average energy densities are exactly equal.