NCERT Solutions for Class 12 Physics Chapter 8: Electromagnetic Waves

Concept and context. In a charging capacitor the conduction current i_c flows in the connecting wires but there is no flow of charge through the vacuum/dielectric between the plates; yet the electric field between the plates changes with time. Maxwell showed that a changing electric field is equivalent to a current, the displacement current i_d = ε₀ dΦ_E/dt (Φ_E = electric flux through the surface between the plates). For a parallel plate capacitor E = Q/(ε₀A), so Φ_E = Q/ε₀ and i_d = dQ/dt, which is exactly equal to the conduction current i_c in the wires. The Ampère–Maxwell law, ∮B·dl = μ₀(i_c + i_d), then makes the magnetic field computed from any surface bounded by a loop the same — removing the contradiction that arises when only conduction current is used (NCERT Fig. 8.1).

Given: plate radius R = 12 cm = 0.12 m; separation d = 5.0 cm = 0.050 m; charging current I = 0.15 A (constant); ε₀ = 8.854 × 10⁻¹² C² N⁻¹ m⁻².

(a) Capacitance and the rate of change of potential difference

1. Area of each plate: A = πR² = 3.14 × (0.12)² = 4.52 × 10⁻² m².
2. Capacitance: C = ε₀A/d = (8.854 × 10⁻¹² × 4.52 × 10⁻²)/0.050 = 8.0 × 10⁻¹² F = 80.1 pF.
3. Rate of change of p.d.: since Q = CV, I = dQ/dt = C dV/dt ⇒ dV/dt = I/C = 0.15/(80.1 × 10⁻¹²) = 1.87 × 10⁹ V s⁻¹.

(b) Displacement current across the plates

1. i_d = ε₀ dΦ_E/dt, with Φ_E = EA = (V/d)A: i_d = ε₀ (A/d) dV/dt = C dV/dt.
2. Hence i_d = C dV/dt = 80.1 × 10⁻¹² × 1.87 × 10⁹ = 0.15 A.
3. The displacement current equals the conduction current: i_d = i_c = 0.15 A.

(c) Is Kirchhoff’s junction rule valid at each plate?

• Yes — if the displacement current is included. At a plate, the conduction current i_c = 0.15 A flows in through the wire, and no conduction current leaves through the gap; but a displacement current i_d = 0.15 A “flows out” of the plate through the gap (via the changing electric field). The total current (conduction + displacement) entering equals the total leaving, so the junction rule (conservation of charge: the sum of currents into a junction is zero) is valid for the total current at each plate.
• Without the displacement current (considering only conduction current), the rule would appear to fail — the conduction current ends abruptly at the plate — which is exactly the inconsistency Maxwell removed by introducing i_d.

• Remark: a rate of rise of 10⁹ V per second is enormous — the charging current of 0.15 A into such a tiny capacitor makes the potential difference rise extremely fast (the capacitor would be charged to 100 V in about 50 ns).
• Magnetic field between the plates: the displacement current produces a magnetic field B = μ₀ i_d r/(2πR²) at distance r < R from the axis (see Exercise 8.2), identical in form to the field that conduction current would produce.

Conclusion: (a) C ≈ 80 pF and dV/dt ≈ 1.9 × 10⁹ V s⁻¹; (b) the displacement current is 0.15 A, equal to the conduction current; (c) yes, Kirchhoff’s junction rule holds at each plate when the displacement current is included along with the conduction current.

Concept and context. When an ac voltage drives a capacitor, the capacitor is continually charged and discharged: a conduction current flows in the leads, while between the plates a changing electric field E(t) constitutes a displacement current. Both are equal at every instant, so the circuit current is continuous. The displacement current sets up a magnetic field between the plates, found by applying the Ampère–Maxwell law ∮B·dl = μ₀ i_d(enclosed) = μ₀ε₀ dΦ_E/dt to a circle of radius r (r < R) coaxial with the plates — the same as for a conduction current of the same value (NCERT Fig. 8.2, E between the plates and B along circles).

Given: R = 6.0 cm = 0.060 m; C = 100 pF = 10⁻¹⁰ F; V_rms = 230 V; ω = 300 rad s⁻¹; r = 3.0 cm = 0.030 m (inside the plates, r < R).

(a) rms value of the conduction current

1. Capacitive reactance X_C = 1/(ωC) = 1/(300 × 10⁻¹⁰) = 3.33 × 10⁷ Ω.
2. I_rms = V_rms/X_C = 230/(3.33 × 10⁷) = 6.9 × 10⁻⁶ A = 6.9 μA (= V_rms ωC).

(b) Is the conduction current equal to the displacement current?

• Yes. The displacement current between the plates i_d = ε₀ dΦ_E/dt = dQ/dt (since Φ_E = Q/ε₀) is equal at each instant to the conduction current i_c = dQ/dt in the wires, so i_d = i_c (the rms value of i_d is also 6.9 μA). The total current is continuous around the circuit, as it must be by conservation of charge.

(c) Amplitude of B at 3.0 cm from the axis

1. Amplitude of the current: I₀ = √2 I_rms = 1.414 × 6.9 × 10⁻⁶ = 9.76 × 10⁻⁶ A.
2. Displacement current through a circle of radius r: the uniform field E acts over the whole plate area, so the displacement current through the smaller circle of area πr² is a fraction r²/R² of the total: i_d(r) = I₀ (πr²/πR²) = I₀ r²/R².
3. Apply the Ampère–Maxwell law: B (2πr) = μ₀ I₀ r²/R² ⇒ B₀ = μ₀ I₀ r/(2πR²).
4. Substitute: B₀ = (4π × 10⁻⁷ × 9.76 × 10⁻⁶ × 0.030)/(2π × (0.060)²) = (2 × 10⁻⁷ × 9.76 × 10⁻⁶ × 0.030)/(3.6 × 10⁻³) = 1.63 × 10⁻¹¹ T.

B₀ = μ₀ I₀ r / (2π R²) ≈ 1.6 × 10⁻¹¹ T

• Why so tiny? The current is only microamperes and the dimension is centimetres: B ~ 10⁻¹¹ T is far weaker than the Earth’s field (3 × 10⁻⁵ T). Displacement currents produce measurable magnetic effects only at very high frequencies or large currents.
• Check: field just outside (r = R): B = μ₀I₀/2πR = 3.25 × 10⁻¹¹ T; B grows linearly with r inside (B ∝ r), as for a uniform current density in a wire.

Conclusion: (a) I_rms = 6.9 μA; (b) yes, the conduction current equals the displacement current at every instant; (c) the amplitude of B at 3.0 cm from the axis is ≈ 1.6 × 10⁻¹¹ T.

Concept and context. X-rays, visible light and radio waves belong to the same family — the electromagnetic spectrum. They differ enormously in wavelength (10⁻¹⁰ m, 6.8 × 10⁻⁷ m and 500 m — a span of about 12 orders of magnitude), in frequency, in photon energy, in the way they are produced and detected, and in their interaction with matter. What they have in common follows from the nature of electromagnetic waves: they are self-sustaining oscillations of electric and magnetic fields that need no medium, and their speed in vacuum is given by Maxwell’s relation c = 1/√(μ₀ε₀), independent of wavelength.

• Answer: the speed in vacuum (c = 3 × 10⁸ m s⁻¹) is the same for all three. All electromagnetic waves — whatever their wavelength or frequency — travel with this same speed in free space (vacuum); the relation c = νλ simply tells us that waves of smaller wavelength have proportionally larger frequency.
• Other common properties: all are transverse waves with E ⟂ B ⟂ direction of propagation; E₀/B₀ = c; all are produced by accelerated charges; all carry energy and momentum; none needs a medium; all can be polarised, reflected, refracted, diffracted and interfere.
• What differs: wavelength/frequency, photon energy, production, detection, penetration and interaction with matter (X-rays penetrate soft tissue, visible light is absorbed/reflected, radio waves pass through walls).
• In a material medium the speed of each wave differs from c (v = 1/√(με)) and depends on the wavelength (dispersion) — the common value c holds only in vacuum.

Conclusion: the physical quantity that is the same for X-rays of wavelength 10⁻¹⁰ m, red light of wavelength 6800 Å and radio waves of wavelength 500 m is their speed in vacuum, c = 3 × 10⁸ m s⁻¹.

Concept and context. Maxwell’s equations show that an electromagnetic wave is transverse: the oscillating electric field E and magnetic field B are perpendicular to each other and both perpendicular to the direction of propagation; moreover E, B and the propagation direction form a right-handed set — the direction of propagation is that of E × B. The two fields oscillate in phase (they reach maxima and minima together) and their magnitudes are related by E/B = c. The speed of the wave in vacuum is c = νλ, so the wavelength is λ = c/ν.

Part 1 — Directions of E and B

• For a wave travelling along the z-direction, the electric and magnetic field vectors lie in the xy-plane (the plane perpendicular to z), and are perpendicular to each other. For example E may be along x (E_x = E₀ sin(kz − ωt)) and then B is along y (B_y = B₀ sin(kz − ωt)) so that E × B (x̂ × ŷ = ẑ) is along +z. If E is along y, B is along −x, and so on. (A general polarisation has E at any angle in the xy-plane with B at 90° to it.)
• E and B oscillate in phase and their magnitudes obey E₀/B₀ = c (so B is very small in SI units).
• Hence the wave is transverse and there is no component of the fields along the direction of propagation.

Part 2 — Wavelength of a 30 MHz wave

1. Given ν = 30 MHz = 30 × 10⁶ Hz = 3 × 10⁷ Hz; c = 3 × 10⁸ m s⁻¹.
2. λ = c/ν = (3 × 10⁸)/(3 × 10⁷) = 10 m.

λ = c/ν = 10 m (a short-wave radio wave)

Conclusion: the electric and magnetic field vectors are mutually perpendicular and both lie in the plane perpendicular to the z-direction of propagation (E × B along z); the wavelength of the 30 MHz wave is 10 m.

Concept and context. Radio waves are electromagnetic waves, so their frequency ν and wavelength λ are related by c = νλ, with c = 3 × 10⁸ m s⁻¹. A higher frequency corresponds to a shorter wavelength (inverse relation). Tuning a radio (varying the capacitance of the tuning circuit so that its resonant frequency equals the station frequency) therefore selects a particular wavelength; a band of frequencies maps to a band of wavelengths. In the NCERT classification, the short-wave radio bands lie up to 54 MHz.

Given: ν₁ = 7.5 MHz = 7.5 × 10⁶ Hz; ν₂ = 12 MHz = 12 × 10⁶ Hz; c = 3 × 10⁸ m s⁻¹.

1. At 7.5 MHz: λ = c/ν = (3 × 10⁸)/(7.5 × 10⁶) = 40 m.
2. At 12 MHz: λ = (3 × 10⁸)/(12 × 10⁶) = 25 m.
3. Because λ ∝ 1/ν, the lower frequency gives the longer wavelength: the wavelength band is 25 m to 40 m.

Wavelength band: 25 m – 40 m

• Note: the band 25–40 m corresponds to the shortwave broadcasting bands (the 25 m, 31 m and 41 m bands in common use), consistent with the “short wave” band of NCERT section 8.4.1.
• Tuning: a radio tunes by changing C of an LC circuit so that ν₀ = 1/(2π√LC) equals the carrier frequency; the capacitance range required is (12/7.5)² = 2.56 times (C ∝ 1/ν²).

Conclusion: the radio tuned to 7.5–12 MHz receives waves in the wavelength band 25 m to 40 m.

Concept and context. According to Maxwell’s theory, accelerated charges radiate electromagnetic waves. A charge oscillating harmonically is accelerating, so it produces an oscillating electric field in space, which produces an oscillating magnetic field, which in turn regenerates the electric field, and so on — the fields regenerate each other as the wave propagates. The wave frequency is determined entirely by the source: the frequency of the electromagnetic wave equals the frequency of oscillation of the charge (the energy carried comes at the expense of the energy of the oscillator).

• Answer: the electromagnetic waves produced have the same frequency, 10⁹ Hz (1 GHz).
• Wavelength: λ = c/ν = 3 × 10⁸/10⁹ = 0.30 m = 30 cm — a microwave (UHF) wave, as used in mobile phones and radar.
• Why equal? In steady state the fields at any point in space oscillate at the rate at which the source charge oscillates, since the fields are the “image” of the source; a wave cannot be of a frequency different from that of its source (in vacuum, with no non-linear processes).
• Energy: the radiated power comes from the oscillator — without continued driving the oscillations die down (radiation damping).

Conclusion: the electromagnetic waves produced by the oscillating charge have a frequency of 10⁹ Hz, the same as the frequency of oscillation of the charge.

Concept and context. In an electromagnetic wave the magnitudes of the electric and magnetic fields are related at every point and every instant by E = cB; correspondingly the amplitudes satisfy E₀ = c B₀ (NCERT Eq. 8.10), where c = 1/√(μ₀ε₀) = 3 × 10⁸ m s⁻¹ is the speed of the wave in vacuum. Because c is so large, the magnetic field is numerically tiny (in tesla) compared with the electric field (in V m⁻¹) — nanotesla magnetic fields accompany electric fields of hundreds of V m⁻¹.

Given: B₀ = 510 nT = 510 × 10⁻⁹ T = 5.1 × 10⁻⁷ T; c = 3 × 10⁸ m s⁻¹.

1. E₀ = c B₀ = (3 × 10⁸ m s⁻¹) × (5.1 × 10⁻⁷ T).
2. E₀ = 15.3 × 10¹ = 153 V m⁻¹ (N C⁻¹).

E₀ = cB₀ = 153 N C⁻¹ (V m⁻¹)

• Unit check: m s⁻¹ × T = m s⁻¹ × N A⁻¹ m⁻¹ = N C⁻¹ (since A = C s⁻¹) ✓.
• Intensity (extra): I = ½ε₀cE₀² = 0.5 × 8.85 × 10⁻¹² × 3 × 10⁸ × (153)² ≈ 31 W m⁻², or I = cB₀²/2μ₀ — comparable to the Sun’s radiation intensity at the Earth’s surface (~10³ W m⁻² for sunlight — here about 30 times lower).
• Direction: E, B and the propagation direction form a right-handed system: E × B points along the wave direction.

Conclusion: the amplitude of the electric field part of the wave is E₀ = cB₀ = 153 N C⁻¹.

Concept and context. A plane harmonic electromagnetic wave travelling along +x with its electric field along y has E_y = E₀ sin(kx − ωt) and B_z = B₀ sin(kx − ωt), where ω = 2πν is the angular frequency, k = 2π/λ = ω/c the wave number (magnitude of the propagation vector), λ = c/ν the wavelength and B₀ = E₀/c. The two fields are in phase and mutually perpendicular, with E × B along the direction of propagation (ĵ × k̂ = î) (NCERT Fig. 8.3 shows the wave along z with E along x and B along y).

Given: E₀ = 120 N C⁻¹; ν = 50.0 MHz = 5.0 × 10⁷ Hz; c = 3 × 10⁸ m s⁻¹.

(a) B₀, ω, k and λ

1. Magnetic field amplitude: B₀ = E₀/c = 120/(3 × 10⁸) = 4.0 × 10⁻⁷ T (400 nT).
2. Angular frequency: ω = 2πν = 2 × 3.14 × 5.0 × 10⁷ = 3.14 × 10⁸ rad s⁻¹.
3. Wave number: k = ω/c = (3.14 × 10⁸)/(3 × 10⁸) = 1.05 rad m⁻¹.
4. Wavelength: λ = c/ν = (3 × 10⁸)/(5.0 × 10⁷) = 6.0 m (check: 2π/k = 6.28/1.05 = 6.0 m ✓).

(b) Expressions for E and B (taking propagation along +x, E along y, and B along z so that E × B ∥ x̂)

E = {120 sin[1.05 x − 3.14 × 10⁸ t]} ĵ N/C

B = {4 × 10⁻⁷ sin[1.05 x − 3.14 × 10⁸ t]} k̂ T

• The directions chosen are an example: any other mutually perpendicular choice (E along z, B along −y; or propagation along −x with kx + ωt) is equally valid, provided E × B points along the propagation direction.
• Both fields are in phase and have the same argument (kx − ωt); the ratio E/B = c = 3 × 10⁸ m s⁻¹ at every point.
• Check of the speed: ω/k = 3.14 × 10⁸/1.05 = 3.0 × 10⁸ m s⁻¹ ✓.

Conclusion: (a) B₀ = 4.0 × 10⁻⁷ T, ω = 3.14 × 10⁸ rad s⁻¹, k = 1.05 rad m⁻¹, λ = 6.0 m; (b) E = 120 sin(1.05x − 3.14 × 10⁸t) ĵ N/C and B = 4 × 10⁻⁷ sin(1.05x − 3.14 × 10⁸t) k̂ T.

Concept and context. The electromagnetic spectrum (NCERT Fig. 8.4 and Table 8.1) is a continuous range of waves of different frequencies ν and wavelengths λ, all travelling at c in vacuum (c = νλ). In the quantum description, electromagnetic radiation consists of photons, each carrying energy E = hν = hc/λ, where Planck’s constant h = 6.63 × 10⁻³⁴ J s. In electron-volt units (1 eV = 1.6 × 10⁻¹⁹ J): E(eV) = 4.14 × 10⁻¹⁵ ν(Hz) = 1240/λ(nm). Because the frequency (hence the energy) varies over more than twenty orders of magnitude, the different regions of the spectrum correspond to very different photon-energy scales — and these are intimately connected with the energy scales of the physical processes that emit the radiation.

Method: E = hν = hc/λ with h = 6.63 × 10⁻³⁴ J s, c = 3 × 10⁸ m s⁻¹, 1 eV = 1.6 × 10⁻¹⁹ J ⇒ hc = 1.24 × 10⁻⁶ eV m = 1240 eV nm.

Relation of the photon-energy scales to the sources of radiation

• γ-rays (MeV–GeV): originate in the atomic nucleus — nuclear reactions and radioactive decay; the energy scale of nuclear energy levels is ~ MeV, so nuclear transitions emit γ-ray photons of 10⁵–10⁷ eV.
• X-rays (keV): emitted when high-energy electrons strike a metal target — transitions of the inner-shell electrons of heavy atoms (and the deceleration of electrons), where the binding energies are 10³–10⁵ eV.
• Ultraviolet and visible (≈ 1–100 eV): produced by transitions of the outer (valence) electrons between the energy levels of atoms and molecules (the energy scale of atomic energy levels is a few eV); the Sun and very hot bodies emit them.
• Infrared (10⁻³–1 eV): produced by hot bodies and by the vibrations and rotations of molecules and whole atoms — molecular energy scales are far smaller than electronic scales.
• Microwaves (10⁻⁵–10⁻³ eV): generated by electronic devices (klystrons, magnetrons, Gunn diodes) and by rotational transitions in molecules.
• Radio waves (< 10⁻⁵ eV): produced by the accelerated motion of charges in conducting wires/antenna circuits (macroscopic oscillating currents) — very low photon energy; each radio photon carries negligible energy, so radio waves behave essentially as classical waves.
• General principle: the characteristic energy of a source determines the photon energy (E = hν), i.e. the part of the spectrum: the smaller the size of the system that radiates (nucleus ≈ 10⁻¹⁵ m ≪ atom ≈ 10⁻¹⁰ m ≪ circuit ≈ m), the larger the energies involved and the shorter the wavelength emitted; the wavelength of the radiation is often correlated with the characteristic size of the radiating system.

Conclusion: photon energies range from about 10⁻⁹–10⁻⁵ eV for radio waves, through ~1–3 eV for visible light, to keV for X-rays and MeV for γ-rays; these scales correspond to the energy scales of the sources — circuits, molecular vibrations, outer-electron transitions, inner-shell transitions and nuclear transitions respectively.

Concept and context. In a plane electromagnetic wave in vacuum the fields satisfy E = cB, and the wavelength and frequency are related by c = νλ. The wave transports energy: the energy density (energy per unit volume) of the electric field is u_E = ½ε₀E² and that of the magnetic field is u_B = B²/(2μ₀). Averaging over a cycle (⟨sin²⟩ = ½) gives ⟨u_E⟩ = ¼ε₀E₀² and ⟨u_B⟩ = B₀²/(4μ₀). Using c² = 1/(μ₀ε₀) and B₀ = E₀/c, the two averages turn out to be equal — the energy in an electromagnetic wave is shared equally between the electric and magnetic fields. The wave of this exercise (2 × 10¹⁰ Hz = 20 GHz) lies in the microwave region.

Given: ν = 2.0 × 10¹⁰ Hz; E₀ = 48 V m⁻¹; c = 3 × 10⁸ m s⁻¹.

(a) Wavelength

λ = c/ν = (3 × 10⁸)/(2.0 × 10¹⁰) = 1.5 × 10⁻² m = 1.5 cm

(b) Amplitude of the magnetic field

B₀ = E₀/c = 48/(3 × 10⁸) = 1.6 × 10⁻⁷ T

(c) Equality of the average energy densities

1. Average energy density of the E field: ⟨u_E⟩ = ½ε₀⟨E²⟩ = ½ε₀(E₀²/2) = ¼ ε₀ E₀².
2. Average energy density of the B field: ⟨u_B⟩ = ⟨B²⟩/(2μ₀) = B₀²/(4μ₀).
3. Substitute B₀ = E₀/c: ⟨u_B⟩ = E₀²/(4μ₀c²). Using c² = 1/(μ₀ε₀): ⟨u_B⟩ = E₀² μ₀ε₀/(4μ₀) = ¼ ε₀ E₀² = ⟨u_E⟩. Hence ⟨u_E⟩ = ⟨u_B⟩.
4. Numerical check: ⟨u_E⟩ = ¼ × 8.85 × 10⁻¹² × (48)² = 5.1 × 10⁻⁹ J m⁻³; ⟨u_B⟩ = (1.6 × 10⁻⁷)²/(4 × 4π × 10⁻⁷) = 2.56 × 10⁻¹⁴/5.03 × 10⁻⁶ = 5.1 × 10⁻⁹ J m⁻³ ✓.

• Meaning: half of the wave’s energy is stored in the electric field and half in the magnetic field at every point; the total average energy density is ½ε₀E₀² = 1.0 × 10⁻⁸ J m⁻³ and the intensity (energy crossing unit area per second) is I = c × (total energy density) ≈ 3.1 W m⁻².
• This equality is an expression of the symmetry between the electric and magnetic fields in an electromagnetic wave, a consequence of Maxwell’s equations.

Conclusion: (a) λ = 1.5 cm; (b) B₀ = 1.6 × 10⁻⁷ T; (c) using B₀ = E₀/c and c² = 1/(μ₀ε₀), ⟨u_B⟩ = B₀²/4μ₀ = ¼ε₀E₀² = ⟨u_E⟩, so the average energy densities of the electric and magnetic fields are equal.