NCERT Solutions for Class 12 Physics Chapter 8: Electromagnetic Waves

These Class 12 Physics Chapter 8 solutions cover Electromagnetic Waves from the NCERT textbook (2026–27). Every NCERT “Exercises” question is reproduced word-for-word and solved step by step — each numerical is worked out fully with units and cross-checked against the NCERT answer key. You also get the key formulas, extra practice, 10 MCQs and 5 Assertion–Reason questions for fast revision.

Class: 12 Subject: Physics Chapter: 8 Name: Electromagnetic Waves Exercises: 8.1 – 8.10 Session: 2026–27
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Class 12 Physics Chapter 8 – Overview

Chapter 8, Electromagnetic Waves, completes the story of electricity and magnetism. James Clerk Maxwell noticed that Ampere’s circuital law was incomplete: a changing electric field must also produce a magnetic field. He fixed this by adding the displacement current, id = ε0(dΦE/dt), giving the Ampere–Maxwell law. The four Maxwell’s equations together predict that accelerated charges radiate electromagnetic (EM) waves — coupled, mutually perpendicular oscillating electric and magnetic fields travelling through vacuum at speed c = 1/√(μ0ε0) ≈ 3×108 m/s. Since this equals the measured speed of light, light itself is an EM wave. The chapter ends with the electromagnetic spectrum — radio waves, microwaves, infrared, visible, ultraviolet, X-rays and gamma rays — arranged by wavelength and frequency, each with its own sources, detectors and uses.

Key Concepts & Definitions

Displacement current: a current associated with a time-varying electric field, id = ε0(dΦE/dt). It is not due to flowing charges but produces a magnetic field exactly like a conduction current. Inside a charging capacitor, the conduction current is zero but id = i.

Ampere–Maxwell law: the source of a magnetic field is the total current = conduction current + displacement current, so the magnetic field comes out the same for any surface bounded by the loop.

Nature of EM waves: the electric field E, magnetic field B and the direction of propagation are mutually perpendicular (transverse wave). In free space E, B and the propagation direction form a right-handed set, with E×B along the direction of travel.

Speed of EM waves: in vacuum, all EM waves travel at the same speed c, independent of wavelength. In a medium of permittivity ε and permeability μ, speed v = 1/√(με) < c.

Electromagnetic spectrum: the ordered range of EM waves by wavelength/frequency — from gamma rays (λ ~10−12 m) to long radio waves (λ ~106 m). All carry energy and momentum.

Important Formulas

Displacement current: id = ε0 (dΦE/dt),   ΦE = EA

Parallel-plate capacitance: C = ε0A/d

Capacitor relations: i = C(dV/dt);   id = ic (current is continuous)

Speed of EM wave: c = 1/√(μ0ε0) = 3×108 m/s;   in medium v = 1/√(με)

Wave relations: c = νλ;   ω = ck;   k = 2π/λ;   ω = 2πν

Field amplitudes: E0 = cB0 (so B0 = E0/c)

Energy densities: uE = ½ε0E2;   uB = B2/(2μ0);   average uE = uB

Photon energy: E = hν = hc/λ  (h = 6.63×10−34 J s)

NCERT Exercises (8.1–8.10) — Solutions

Questions are reproduced verbatim from the NCERT textbook; answers are original, step-by-step and verified with units against the NCERT answer key.

8.1 Figure 8.5 shows a capacitor made of two circular plates each of radius 12 cm, and separated by 5.0 cm. The capacitor is being charged by an external source (not shown in the figure). The charging current is constant and equal to 0.15 A. (a) Calculate the capacitance and the rate of change of potential difference between the plates. (b) Obtain the displacement current across the plates. (c) Is Kirchhoff’s first rule (junction rule) valid at each plate of the capacitor? Explain.

ANSWER Given: radius R = 12 cm = 0.12 m, separation d = 5.0 cm = 0.05 m, charging current i = 0.15 A, ε0 = 8.854×10−12 F m−1. (a) Capacitance: Plate area A = πR2 = π(0.12)2 = 0.04524 m2. C = ε0A/d = (8.854×10−12 × 0.04524) / 0.05 = 8.00×10−12 F = 8.00 pF. Rate of change of p.d.: Since i = C(dV/dt), dV/dt = i/C = 0.15 / (8.00×10−12) = 1.87×1010 V s−1. (b) Displacement current: id = ε0(dΦE/dt) = ε0A(dE/dt). Since E = Q/(ε0A), we get id = dQ/dt = i. So id = 0.15 A — equal to the conduction (charging) current. (c) Kirchhoff’s junction rule: Yes, it is valid at each plate, provided “current” means the sum of conduction and displacement currents. The conduction current arriving at a plate equals the displacement current leaving it through the gap, so the total current is continuous and the junction rule holds.

8.2 A parallel plate capacitor (Fig. 8.6) made of circular plates each of radius R = 6.0 cm has a capacitance C = 100 pF. The capacitor is connected to a 230 V ac supply with a (angular) frequency of 300 rad s−1. (a) What is the rms value of the conduction current? (b) Is the conduction current equal to the displacement current? (c) Determine the amplitude of B at a point 3.0 cm from the axis between the plates.

ANSWER Given: R = 6.0 cm = 0.06 m, C = 100 pF = 100×10−12 F, Vrms = 230 V, ω = 300 rad s−1, distance from axis r = 3.0 cm = 0.03 m. (a) rms conduction current: For a capacitor, XC = 1/(ωC), so Irms = Vrms·ωC = 230 × 300 × (100×10−12) = 6.9×10−6 A = 6.9 µA. (b) Yes. The displacement current equals the conduction current (as shown in 8.1b); this holds even when the current oscillates with time. (c) Amplitude of B at r = 3.0 cm: The Ampere–Maxwell law between the plates gives B = (μ0r/2πR2)·id. In amplitude terms, B0 = (μ0r/2πR2)·i0, where i0 = √2·Irms. Peak current: i0 = √2 × 6.9×10−6 = 9.76×10−6 A. B0 = (4π×10−7 × 0.03 × 9.76×10−6) / (2π × (0.06)2) = (2×10−7 × 0.03 × 9.76×10−6) / (3.6×10−3) = (5.856×10−14) / (3.6×10−3) = 1.63×10−11 T.

8.3 What physical quantity is the same for X-rays of wavelength 10−10 m, red light of wavelength 6800 Å and radiowaves of wavelength 500 m?

ANSWER All three are electromagnetic waves, so the physical quantity that is the same for all of them is their speed in vacuum, c = 3×108 m s−1. (Their wavelengths and frequencies differ, but the speed is common to every EM wave in free space.)

8.4 A plane electromagnetic wave travels in vacuum along z-direction. What can you say about the directions of its electric and magnetic field vectors? If the frequency of the wave is 30 MHz, what is its wavelength?

ANSWER Directions: EM waves are transverse, so both E and B are perpendicular to the direction of propagation (z-axis). Hence E and B lie in the x–y plane, and they are mutually perpendicular to each other. Wavelength: λ = c/ν = (3×108) / (30×106) = 10 m.

8.5 A radio can tune in to any station in the 7.5 MHz to 12 MHz band. What is the corresponding wavelength band?

ANSWER Using λ = c/ν: For ν = 7.5 MHz: λ = (3×108) / (7.5×106) = 40 m. For ν = 12 MHz: λ = (3×108) / (12×106) = 25 m. So the wavelength band is 40 m to 25 m (higher frequency → shorter wavelength).

8.6 A charged particle oscillates about its mean equilibrium position with a frequency of 109 Hz. What is the frequency of the electromagnetic waves produced by the oscillator?

ANSWER An oscillating (accelerated) charge radiates EM waves of the same frequency as its oscillation. Therefore the frequency of the EM waves produced is 109 Hz.

8.7 The amplitude of the magnetic field part of a harmonic electromagnetic wave in vacuum is B0 = 510 nT. What is the amplitude of the electric field part of the wave?

ANSWER Given: B0 = 510 nT = 510×10−9 T. For an EM wave, E0 = cB0 = (3×108) × (510×10−9) = 153 N C−1 (or 153 V m−1).

8.8 Suppose that the electric field amplitude of an electromagnetic wave is E0 = 120 N/C and that its frequency is ν = 50.0 MHz. (a) Determine, B0, ω, k, and λ. (b) Find expressions for E and B.

ANSWER Given: E0 = 120 N C−1, ν = 50.0 MHz = 5.0×107 Hz. (a) Magnetic field amplitude: B0 = E0/c = 120 / (3×108) = 4.0×10−7 T = 400 nT. Angular frequency: ω = 2πν = 2π × (5.0×107) = 3.14×108 rad s−1. Wave number: k = ω/c = (3.14×108) / (3×108) = 1.05 rad m−1. Wavelength: λ = c/ν = (3×108) / (5.0×107) = 6.00 m. (b) Taking the wave travelling along the x-direction with E along ĵ and B along k̂: E = {(120 N/C) sin[(1.05 rad m−1)x − (3.14×108 rad s−1)t]} ĵ B = {(400 nT) sin[(1.05 rad m−1)x − (3.14×108 rad s−1)t]} k̂

8.9 The terminology of different parts of the electromagnetic spectrum is given in the text. Use the formula E = hν (for energy of a quantum of radiation: photon) and obtain the photon energy in units of eV for different parts of the electromagnetic spectrum. In what way are the different scales of photon energies that you obtain related to the sources of electromagnetic radiation?

ANSWER Method: Photon energy E = hν = hc/λ. With h = 6.63×10−34 J s, c = 3×108 m s−1, and 1 eV = 1.6×10−19 J. Reference value (λ = 1 m): E = (6.63×10−34 × 3×108) / (1.6×10−19) eV ≈ 1.24×10−6 eV. Photon energy for any other wavelength is found by scaling: E(eV) ≈ 1.24×10−6/λ(m). Sample energies: For λ = 10−12 m (gamma rays): E ≈ 1.24×106 eV = 1.24 MeV. For visible light λ = 5×10−7 m: E ≈ 2.5 eV. Relation to sources: The photon energy indicates the spacing of energy levels of the source. Gamma-ray photons (~MeV) come from transitions between nuclear energy levels (spaced by ~1 MeV). Visible photons (~few eV) come from transitions between atomic/molecular energy levels spaced by a few eV. Thus higher-frequency radiation arises from sources with larger energy-level spacings.

8.10 In a plane electromagnetic wave, the electric field oscillates sinusoidally at a frequency of 2.0 × 1010 Hz and amplitude 48 V m−1. (a) What is the wavelength of the wave? (b) What is the amplitude of the oscillating magnetic field? (c) Show that the average energy density of the E field equals the average energy density of the B field. [c = 3 × 108 m s−1.]

ANSWER 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) Magnetic field amplitude: B0 = E0/c = 48 / (3×108) = 1.6×10−7 T. (c) Energy densities: Average energy density of the E field: uE = ½ε0E2. Average energy density of the B field: uB = B2/(2μ0). Using E = cB and c2 = 1/(μ0ε0): uE = ½ε0(cB)2 = ½ε0c2B2 = ½ε0·[1/(μ0ε0)]·B2 = B2/(2μ0) = uB. Hence the average energy density of the electric field equals that of the magnetic field, uE = uB. (The total energy of an EM wave is shared equally between its electric and magnetic fields.)

Extra Practice Questions

Short Answer Type Questions

Q1. What is displacement current and how does it differ from conduction current?

ANSWERDisplacement current, id = ε0(dΦE/dt), arises from a time-varying electric field, whereas conduction current is due to the actual flow of charges. They differ in origin but produce identical magnetic effects; inside a charging capacitor only displacement current exists.

Q2. Why are electromagnetic waves called transverse waves?

ANSWERBecause the oscillating electric field E and magnetic field B are both perpendicular to the direction of propagation (and to each other). The displacement of the field vectors is at right angles to the wave’s travel, which is the defining feature of a transverse wave.

Q3. State two uses each of microwaves and infrared waves.

ANSWERMicrowaves: radar systems for aircraft navigation and speed guns; microwave ovens for cooking. Infrared waves: physical therapy (infrared lamps) and remote-control switches for TVs and other electronic devices; they also keep the Earth warm via the greenhouse effect.

Q4. Calculate the speed of an EM wave in a medium of relative permittivity 4 and relative permeability 1.

ANSWERv = 1/√(με) = c/√(μrεr) = (3×108)/√(1×4) = (3×108)/2 = 1.5×108 m s−1.

Q5. Which EM radiation has (a) the highest frequency and (b) the longest wavelength in the spectrum?

ANSWER(a) Gamma rays have the highest frequency (shortest wavelength, ~10−12 m and below). (b) Radio waves have the longest wavelength (up to ~106 m) and the lowest frequency.

Long Answer Type Questions

Q1. Explain how Maxwell introduced the concept of displacement current, and write the Ampere–Maxwell law.

ANSWERApplying Ampere’s law to a charging capacitor, Maxwell found a contradiction: choosing a flat surface that cuts the connecting wire gives B = μ0i, but a pot-shaped surface passing between the plates carries no conduction current, giving B = 0 — even though both surfaces share the same boundary loop. To remove this, Maxwell noted that between the plates there is a changing electric flux. He added a term ε0(dΦE/dt), the displacement current, which equals the conduction current and restores a single, consistent value of B for every surface. The corrected (generalised) law is the Ampere–Maxwell law: ∮B·dl = μ0ic + μ0ε0(dΦE/dt). This established that a changing electric field is also a source of magnetic field, leading directly to the prediction of EM waves.

Q2. Describe the main characteristics of electromagnetic waves.

ANSWER(i) EM waves are produced by accelerated charges. (ii) They consist of mutually perpendicular oscillating electric and magnetic fields, both perpendicular to the direction of propagation (transverse). (iii) E, B and the propagation direction form a right-handed set, with E×B along the travel direction. (iv) They need no material medium and travel through vacuum at c = 1/√(μ0ε0) = 3×108 m/s, the same for all wavelengths. (v) Field amplitudes are related by E0 = cB0. (vi) They carry both energy and momentum, and their average electric and magnetic energy densities are equal.

Q3. Briefly describe the electromagnetic spectrum in order of increasing frequency, giving one use of each band.

ANSWERIn order of increasing frequency (decreasing wavelength): Radio waves — radio and TV communication; Microwaves — radar and microwave ovens; Infrared — heating, physical therapy and remote controls; Visible light — vision and illumination; Ultraviolet — sterilising water and LASIK eye surgery; X-rays — medical imaging and cancer treatment; Gamma rays — destroying cancer cells and studying nuclear reactions. All these are EM waves differing only in wavelength/frequency, and all travel at speed c in vacuum.

MCQs & Assertion–Reason

1. Displacement current is due to:

(a) flow of electrons    (b) a time-varying electric field    (c) a steady magnetic field    (d) a constant electric field

2. The speed of electromagnetic waves in vacuum equals:

(a) μ0ε0    (b) √(μ0ε0)    (c) 1/√(μ0ε0)    (d) 1/(μ0ε0)

3. In an EM wave, the electric and magnetic fields are:

(a) parallel to each other    (b) at 45°    (c) perpendicular to each other and to propagation    (d) along the direction of propagation

4. The ratio E0/B0 for an EM wave in vacuum is equal to:

(a) 1    (b) c    (c) 1/c    (d) c2

5. Which of the following has the shortest wavelength?

(a) radio waves    (b) microwaves    (c) visible light    (d) gamma rays

6. Electromagnetic waves are produced by:

(a) stationary charges    (b) charges moving with uniform velocity    (c) accelerated charges    (d) charges at rest in a field

7. Microwave ovens heat food because the microwave frequency matches the resonant frequency of:

(a) fat molecules    (b) water molecules    (c) salt molecules    (d) sugar molecules

8. The wavelength of an EM wave of frequency 30 MHz in vacuum is:

(a) 1 m    (b) 10 m    (c) 100 m    (d) 0.1 m

9. In an EM wave, the average energy density of the electric field is:

(a) greater than that of the magnetic field    (b) less than that of the magnetic field    (c) equal to that of the magnetic field    (d) zero

10. Which radiation is used to sterilise water in purifiers?

(a) infrared    (b) microwaves    (c) ultraviolet    (d) radio waves

Answer key: 1-(b), 2-(c), 3-(c), 4-(b), 5-(d), 6-(c), 7-(b), 8-(b), 9-(c), 10-(c).

For each Assertion–Reason question, choose: (A) Both true and the Reason correctly explains the Assertion; (B) Both true but the Reason is not the correct explanation; (C) Assertion true, Reason false; (D) Assertion false, Reason true.

A-R 1. Assertion: A changing electric field produces a magnetic field.

Reason: A time-varying electric flux constitutes a displacement current, which acts as a source of magnetic field.

A-R 2. Assertion: Electromagnetic waves do not require a material medium to travel.

Reason: EM waves are self-sustaining oscillations of electric and magnetic fields that can propagate through vacuum.

A-R 3. Assertion: All electromagnetic waves travel at the same speed in vacuum.

Reason: The speed of an EM wave in vacuum, c = 1/√(μ0ε0), depends only on the free-space constants and not on wavelength.

A-R 4. Assertion: Stationary charges can radiate electromagnetic waves.

Reason: Only accelerated charges produce electromagnetic waves.

A-R 5. Assertion: Gamma rays have higher photon energy than visible light.

Reason: Photon energy E = hν increases with frequency, and gamma rays have a higher frequency than visible light.

Answer key: 1-(A), 2-(A), 3-(A), 4-(D), 5-(A).

Common Mistakes to Avoid

Watch out for these

  • Writing the speed of light as c = √(μ0ε0) — it is the reciprocal, c = 1/√(μ0ε0).
  • Confusing E0 = cB0 with B0 = cE0 — the electric amplitude is the larger one (E0 = cB0).
  • Forgetting to convert peak and rms values: i0 = √2·Irms when finding the amplitude of B in capacitor problems.
  • Mixing up the order of the spectrum — remember wavelength increases from gamma → X-ray → UV → visible → IR → microwave → radio.
  • Saying displacement current carries charge — it does not; it is due to a changing electric field, yet it produces the same magnetic effect.
  • Dropping units or powers of ten in numericals — always carry SI units through every step.

How to score full marks in this chapter

Memorise the core relations — c = 1/√(μ0ε0) = νλ, E0 = cB0, and uE = uB — and quote them before substituting numbers. In numericals, write the formula, substitute with units, then simplify to the final value with correct powers of ten. For theory marks, be ready to derive the displacement current from a charging capacitor and to list the EM spectrum with one source, detector and use per band. Always state the mutual-perpendicularity of E, B and propagation when describing EM waves.

Frequently Asked Questions

What is Class 12 Physics Chapter 8 about?

Chapter 8, Electromagnetic Waves, introduces Maxwell’s displacement current and the Ampere–Maxwell law, shows how accelerated charges radiate electromagnetic waves that travel at the speed of light, explains the transverse nature and energy of EM waves, and describes the full electromagnetic spectrum from radio waves to gamma rays.

How many exercises are there in Class 12 Physics Chapter 8?

The NCERT textbook has 10 exercise questions (8.1 to 8.10). All are reproduced verbatim and solved step by step on this page, with numericals verified with units against the NCERT answer key.

What is the formula for the speed of electromagnetic waves?

In vacuum, the speed is c = 1/√(μ0ε0) ≈ 3×108 m/s, the same for all wavelengths. In a medium it becomes v = 1/√(με), which is less than c. The amplitudes are related by E0 = cB0.

Are these Class 12 Physics Chapter 8 solutions free?

Yes. All ClearStudy NCERT Solutions for Class 12 Physics are free and follow the official NCERT textbook for session 2026–27.

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