Important Questions Class 12 Physics Chapter 8: Electromagnetic Waves

Electromagnetic waves are coupled electric and magnetic fields that travel through space with finite speed. In vacuum, every electromagnetic wave travels with speed c = 3 × 10^8 m/s.

Changing electric and magnetic fields explain light, radio signals, microwave heating, X-rays, and gamma radiation through one physical framework. Important Questions Class 12 Physics Chapter 8 help students practise displacement current, Ampere-Maxwell law, Maxwell’s equations, EM wave speed, field amplitudes, wave equations, and the electromagnetic spectrum. The CBSE 2026 chapter also connects Maxwell’s correction with communication systems, radar, remote controls, water purifiers, medical X-rays, and cancer treatment.

Key Takeaways

• Displacement Current: Maxwell added id = ε0 dΦE/dt to complete Ampere’s circuital law.
• EM Wave Source: Accelerated charges produce electromagnetic waves.
• Field Geometry: Electric field, magnetic field, and propagation direction stay mutually perpendicular.
• Spectrum Order: Wavelength decreases from radio waves to gamma rays.

Important Questions Class 12 Physics Chapter 8 Structure 2026

Important Questions Class 12 Physics Chapter 8 with Answers

The chapter starts with a flaw in Ampere’s circuital law and ends with the full electromagnetic spectrum.
Students should connect every formula to a physical idea, especially capacitor charging and wave propagation.
These electromagnetic waves class 12 important questions follow the NCERT 2026 chapter flow.

1. What does Important Questions Class 12 Physics Chapter 8 mainly test?

Important Questions Class 12 Physics Chapter 8 mainly test displacement current, Maxwell’s equations, EM wave nature, wave formulas, and spectrum uses. The chapter mixes derivation-based, conceptual, and numerical questions.

1. Theory Area: Displacement current and Maxwell equations.
2. Wave Area: Field directions, speed, frequency, and wavelength.
3. Spectrum Area: Radio, microwave, infrared, visible, UV, X-rays, and gamma rays.
4. Final Result: The chapter tests electromagnetic wave formation and applications.

2. Why did Maxwell need to correct Ampere’s circuital law?

Maxwell corrected Ampere’s circuital law because it gave contradictory results during capacitor charging. The correction added displacement current.

1. Problem: Different surfaces gave different enclosed currents.
2. Between Plates: No conduction current flows.
3. Changing Quantity: Electric flux changes with time.
4. Correction: id = ε0 dΦE/dt.
5. Final Result: Displacement current removes the contradiction.

3. Why is light treated as an electromagnetic wave?

Light is treated as an electromagnetic wave because Maxwell’s calculated wave speed matched light’s measured speed. Both values equal about 3 × 10^8 m/s.

1. Maxwell’s Speed: c = 1/√(μ0ε0).
2. Optical Speed: Light travels at 3 × 10^8 m/s in vacuum.
3. Conclusion: Light belongs to the electromagnetic wave family.
4. Final Result: Light is an electromagnetic wave.

Displacement Current Class 12 Physics Questions

A charging capacitor creates the exact situation where ordinary Ampere’s law fails.
The wire carries conduction current, while the gap between plates has changing electric flux.
This section uses displacement current class 12 physics and Ampere Maxwell law class 12 physics concepts.

4. What is displacement current in Class 12 Physics?

Displacement current is the current equivalent of a changing electric flux. Its formula is id = ε0 dΦE/dt.

1. Electric Flux: ΦE changes with time.
2. Constant Used: ε0 is permittivity of free space.
3. Formula Used: id = ε0 dΦE/dt.
4. Final Result: Displacement current acts as a source of magnetic field.

5. How is displacement current related to charging current in a capacitor?

Displacement current equals conduction current during capacitor charging. The equality follows from ΦE = Q/ε0.

1. Given Relation: ΦE = Q/ε0.
2. Differentiate: dΦE/dt = (1/ε0) dQ/dt.
3. Multiply by ε0: ε0 dΦE/dt = dQ/dt.
4. Current Relation: id = i.
5. Final Result: Displacement current equals charging current.

6. Why is there no conduction current between capacitor plates?

There is no conduction current between capacitor plates because charges do not cross the gap. The changing electric field fills the gap.

1. Wire Region: Charges move and create conduction current.
2. Gap Region: Charges do not cross the dielectric or air gap.
3. Field Region: Electric flux changes between plates.
4. Final Result: Only displacement current exists between ideal capacitor plates.

7. State Ampere-Maxwell law.

Ampere-Maxwell law states that magnetic circulation equals μ0 times total current. Total current includes conduction current and displacement current.

1. Formula Used: ∮ B · dl = μ0ic + μ0ε0 dΦE/dt.
2. Conduction Current: ic flows through conductors.
3. Displacement Current: id = ε0 dΦE/dt.
4. Final Result: ∮ B · dl = μ0(ic + id).

8. Why does Ampere-Maxwell law give the same magnetic field for different surfaces?

Ampere-Maxwell law gives the same magnetic field because it counts total current. It includes conduction current and displacement current.

1. Surface Through Wire: Encloses conduction current.
2. Surface Between Plates: Encloses displacement current.
3. Equality: ic = id during charging.
4. Final Result: Both surfaces give the same magnetic field.

9. What is the magnetic field around a wire carrying current i(t)?

The magnetic field at distance r is B = μ0i(t)/(2πr). The field circles the wire.

1. Given Data: Circular path radius = r.
2. Formula Used: ∮ B · dl = μ0i(t).
3. Calculation:
   B(2πr) = μ0i(t)
   B = μ0i(t)/(2πr)
4. Final Result: B = μ0i(t)/(2πr).

Maxwell Equations Class 12 Physics Questions

Maxwell’s equations combine electricity, magnetism, changing fields, and currents.
Their biggest result is the prediction of self-sustaining electromagnetic waves in free space.
Students search Maxwell equations class 12 physics often because the four laws define the chapter’s logic.

10. What are Maxwell’s equations in vacuum?

Maxwell’s equations in vacuum are four laws that describe electric and magnetic fields. They unify electricity, magnetism, and electromagnetic waves.

1. Gauss Law for Electricity: ∮ E · dA = Q/ε0.
2. Gauss Law for Magnetism: ∮ B · dA = 0.
3. Faraday’s Law: ∮ E · dl = −dΦB/dt.
4. Ampere-Maxwell Law: ∮ B · dl = μ0ic + μ0ε0 dΦE/dt.
5. Final Result: Maxwell’s equations express the basic laws of electromagnetism.

11. Which Maxwell equation proves that magnetic monopoles do not exist?

Gauss’s law for magnetism shows that magnetic monopoles do not exist. It states that net magnetic flux is zero.

1. Equation: ∮ B · dA = 0.
2. Meaning: No isolated magnetic source exists.
3. Field Pattern: Magnetic field lines form closed loops.
4. Final Result: Magnetic monopoles do not exist in NCERT physics.

12. Which law says that a changing magnetic field produces electric field?

Faraday’s law says that a changing magnetic field produces electric field. The induced electric field forms closed loops.

1. Equation: ∮ E · dl = −dΦB/dt.
2. Changing Quantity: Magnetic flux ΦB.
3. Effect: Induced electric field appears.
4. Final Result: Changing magnetic flux produces electric field.

13. Which law says that a changing electric field produces magnetic field?

Ampere-Maxwell law says that a changing electric field produces magnetic field. The extra term is displacement current.

1. Equation: ∮ B · dl = μ0ic + μ0ε0 dΦE/dt.
2. Changing Quantity: Electric flux ΦE.
3. Effect: Magnetic field appears.
4. Final Result: Changing electric flux produces magnetic field.

14. How are electromagnetic waves produced?

Electromagnetic waves are produced by accelerated charges. An oscillating charge is a basic source of EM waves.

1. Stationary Charge: Produces electrostatic field.
2. Uniform Motion: Produces steady magnetic field.
3. Acceleration: Produces electromagnetic radiation.
4. Final Result: Accelerated charges radiate electromagnetic waves.

15. Why are electromagnetic waves called transverse waves?

Electromagnetic waves are transverse because E and B are perpendicular to propagation. The two fields are also perpendicular to each other.

1. Electric Field: Perpendicular to magnetic field.
2. Magnetic Field: Perpendicular to propagation direction.
3. Wave Direction: Along E × B.
4. Final Result: E, B, and propagation direction are mutually perpendicular.

Electromagnetic Waves Class 12 Important Questions

Formula questions usually need unit conversion before substitution.
The most used relations are c = νλ, E0 = cB0, k = 2π/λ, and ω = 2πν.
This section covers electromagnetic wave speed formula class 12 and electromagnetic waves formula class 12 physics queries.

16. What is the speed of electromagnetic waves in vacuum?

The speed of electromagnetic waves in vacuum is 3 × 10^8 m/s. Its formula is c = 1/√(μ0ε0).

1. Formula Used: c = 1/√(μ0ε0).
2. Also Used: c = νλ.
3. Value: c = 3 × 10^8 m/s.
4. Final Result: EM waves travel at 3 × 10^8 m/s in vacuum.

17. What is the relation between electric and magnetic field amplitudes?

The relation is E0/B0 = c. It can also be written as E0 = cB0.

1. Formula Used: E0/B0 = c.
2. Rearranged Form: E0 = cB0.
3. Other Form: B0 = E0/c.
4. Final Result: Electric and magnetic field amplitudes are linked by c.

18. What is the relation between frequency and wavelength of an EM wave?

The relation is c = νλ. Frequency and wavelength are inversely related in vacuum.

1. Formula Used: c = νλ.
2. For Frequency: ν = c/λ.
3. For Wavelength: λ = c/ν.
4. Final Result: νλ = c.

19. What are k and ω in an electromagnetic wave equation?

k is wave number, and ω is angular frequency. They describe space and time variation of the wave.

1. Wave Number: k = 2π/λ.
2. Angular Frequency: ω = 2πν.
3. Wave Relation: ω = ck.
4. Final Result: k = 2π/λ and ω = 2πν.

20. Write the electric and magnetic field equations for a plane EM wave along z-axis.

The field equations are Ex = E0 sin(kz − ωt) and By = B0 sin(kz − ωt). The fields remain perpendicular.

1. Propagation Direction: z-axis.
2. Electric Field Direction: x-axis.
3. Magnetic Field Direction: y-axis.
4. Equations:
   Ex = E0 sin(kz − ωt)
   By = B0 sin(kz − ωt)
5. Final Result: The wave travels along the z-axis.

Electromagnetic Wave Speed Formula Class 12 Questions

Most numerical questions in this chapter use one main equation.
The challenge lies in choosing the correct relation from c = νλ, E0 = cB0, or k = 2π/λ.
These electromagnetic waves class 12 questions and answers follow NCERT exercise patterns.

21. A plane EM wave has E = 6.3 ĵ V/m and travels along x-axis. Find B.

The magnetic field is 2.1 × 10^-8 k̂ T. Use B = E/c and the right-hand rule.

1. Given Data:
   E = 6.3 ĵ V/m
   c = 3 × 10^8 m/s
2. Formula Used: B = E/c.
3. Calculation:
   B = 6.3/(3 × 10^8)
   B = 2.1 × 10^-8 T
4. Direction: E × B points along the x-axis.
5. Final Result: B = 2.1 × 10^-8 k̂ T.

22. Find the wavelength of an EM wave with frequency 30 MHz.

The wavelength is 10 m. Use λ = c/ν.

1. Given Data:
   ν = 30 MHz = 30 × 10^6 Hz
   c = 3 × 10^8 m/s
2. Formula Used: λ = c/ν.
3. Calculation:
   λ = (3 × 10^8)/(30 × 10^6)
   λ = 10 m
4. Final Result: λ = 10 m.

23. What wavelength band corresponds to 7.5 MHz to 12 MHz?

The wavelength band is 40 m to 25 m. Higher frequency gives shorter wavelength.

1. Formula Used: λ = c/ν.
2. For 7.5 MHz:
   λ = 3 × 10^8/(7.5 × 10^6)
   λ = 40 m
3. For 12 MHz:
   λ = 3 × 10^8/(12 × 10^6)
   λ = 25 m
4. Final Result: The wavelength band is 40 m to 25 m.

24. A charge oscillates with frequency 10^9 Hz. Find the EM wave frequency.

The EM wave frequency is 10^9 Hz. The wave frequency equals the oscillating charge’s frequency.

1. Given Data: Source frequency = 10^9 Hz.
2. Rule: Wave frequency equals source frequency.
3. Final Result: EM wave frequency = 10^9 Hz.

25. If B0 = 510 nT, find E0 in vacuum.

The electric field amplitude is 153 V/m. Use E0 = cB0.

1. Given Data:
   B0 = 510 nT = 510 × 10^-9 T
   c = 3 × 10^8 m/s
2. Formula Used: E0 = cB0.
3. Calculation:
   E0 = 3 × 10^8 × 510 × 10^-9
   E0 = 153 V/m
4. Final Result: E0 = 153 V/m.

26. If E0 = 120 N/C, find B0.

The magnetic field amplitude is 4.0 × 10^-7 T. Use B0 = E0/c.

1. Given Data:
   E0 = 120 N/C
   c = 3 × 10^8 m/s
2. Formula Used: B0 = E0/c.
3. Calculation:
   B0 = 120/(3 × 10^8)
   B0 = 4.0 × 10^-7 T
4. Final Result: B0 = 4.0 × 10^-7 T.

27. If frequency is 2.0 × 10^10 Hz and E0 = 48 V/m, find λ and B0.

The wavelength is 1.5 × 10^-2 m, and B0 is 1.6 × 10^-7 T.

1. Given Data:
   ν = 2.0 × 10^10 Hz
   E0 = 48 V/m
2. Wavelength:
   λ = c/ν
   λ = 3 × 10^8/(2.0 × 10^10)
   λ = 1.5 × 10^-2 m
3. Magnetic Field:
   B0 = E0/c
   B0 = 48/(3 × 10^8)
   B0 = 1.6 × 10^-7 T
4. Final Result: λ = 1.5 × 10^-2 m and B0 = 1.6 × 10^-7 T.

28. If By = 2 × 10^-7 sin(0.5 × 10^3 x + 1.5 × 10^11 t), find wavelength.

The wavelength is 1.26 cm. Compare the coefficient of x with k.

1. Given Data: k = 0.5 × 10^3 m^-1.
2. Formula Used: k = 2π/λ.
3. Calculation:
   λ = 2π/k
   λ = 2π/(0.5 × 10^3)
   λ = 1.26 × 10^-2 m
4. Final Result: λ = 1.26 cm.

29. For the same wave, find electric field amplitude.

The electric field amplitude is 60 V/m. Use E0 = cB0.

1. Given Data:
   B0 = 2 × 10^-7 T
   c = 3 × 10^8 m/s
2. Formula Used: E0 = cB0.
3. Calculation:
   E0 = 3 × 10^8 × 2 × 10^-7
   E0 = 60 V/m
4. Final Result: E0 = 60 V/m.

Electromagnetic Spectrum Class 12 Physics Questions

The electromagnetic spectrum has no sharp boundary between neighbouring regions.
NCERT classifies the regions mainly by frequency, wavelength, production, and detection method.
Students often search electromagnetic spectrum class 12 physics for order, uses, and source-based questions.

30. What is the order of the electromagnetic spectrum by increasing frequency?

The order is radio waves, microwaves, infrared, visible light, ultraviolet, X-rays, gamma rays. Frequency increases from radio to gamma rays.

1. Lowest Frequency: Radio waves.
2. Middle Regions: Infrared, visible, ultraviolet.
3. Highest Frequency: Gamma rays.
4. Final Result: Frequency increases from radio waves to gamma rays.

31. Which electromagnetic waves have the longest wavelength?

Radio waves have the longest wavelength. Their wavelengths are generally greater than 0.1 m.

1. Region: Radio waves.
2. Wavelength: Greater than 0.1 m.
3. Use: Radio and television communication.
4. Final Result: Radio waves have the longest wavelength.

32. Which electromagnetic waves have the highest frequency?

Gamma rays have the highest frequency. They occupy the upper end of the electromagnetic spectrum.

1. Region: Gamma rays.
2. Wavelength: Less than 10^-3 nm.
3. Source: Radioactive nuclei and nuclear reactions.
4. Final Result: Gamma rays have the highest frequency.

33. Which physical quantity is same for X-rays, red light, and radio waves in vacuum?

The speed is the same for X-rays, red light, and radio waves in vacuum. They all travel at c.

1. Common Nature: All are electromagnetic waves.
2. Vacuum Speed: c = 3 × 10^8 m/s.
3. Different Quantity: Their wavelengths and frequencies differ.
4. Final Result: All have the same speed in vacuum.

Radio Waves Class 12 Physics Questions

These three regions support communication, radar, heating, remote controls, and satellite sensing.
Their wavelengths are longer than visible light, but their applications differ sharply.
This section places radio waves class 12 physics, uses of microwaves in electromagnetic spectrum, and infrared waves class 12 physics questions naturally.

34. How are radio waves produced?

Radio waves are produced by accelerated charges in conducting wires. Receiver aerials detect them.

1. Source: Accelerating electrons in aerials.
2. Frequency Use: Radio and television communication.
3. Detection: Receiver aerials.
4. Final Result: Radio waves come from accelerated charges in conductors.

35. What are the AM and FM radio frequency ranges?

The AM band is 530 kHz to 1710 kHz, and the FM band is 88 MHz to 108 MHz.

1. AM Band: 530 kHz to 1710 kHz.
2. FM Band: 88 MHz to 108 MHz.
3. Use: Radio broadcasting.
4. Final Result: AM and FM use different radio frequency bands.

36. Why are microwaves used in radar?

Microwaves are used in radar because their short wavelengths help detect objects accurately. Aircraft navigation uses radar systems.

1. Wave Region: Gigahertz frequency range.
2. Production Devices: Klystrons, magnetrons, and Gunn diodes.
3. Use: Radar and speed guns.
4. Final Result: Microwaves suit radar because of short wavelengths.

37. Why do microwave ovens heat food?

Microwave ovens heat food because microwaves transfer energy to water molecules. Water molecules absorb the selected microwave frequency efficiently.

1. Main Absorber: Water molecules.
2. Energy Transfer: Waves increase molecular kinetic energy.
3. Effect: Food temperature rises.
4. Final Result: Microwave ovens heat food through water absorption.

38. Why are infrared waves called heat waves?

Infrared waves are called heat waves because molecules readily absorb them. Absorption increases thermal motion.

1. Absorbing Molecules: Water, CO2, and NH3.
2. Thermal Effect: Molecular motion increases.
3. Use: Infrared lamps in physical therapy.
4. Final Result: Infrared waves heat many materials.

39. How do infrared waves help in the greenhouse effect?

Infrared waves help in the greenhouse effect because Earth reradiates absorbed sunlight as infrared radiation. Greenhouse gases trap this radiation.

1. Incoming Radiation: Visible light reaches Earth’s surface.
2. Outgoing Radiation: Earth emits infrared radiation.
3. Trapping Gases: CO2 and water vapour absorb infrared radiation.
4. Final Result: Infrared radiation helps maintain Earth’s warmth.

Infrared Waves and Ultraviolet Rays Class 12 Physics Questions

Visible light forms only a small part of the electromagnetic spectrum.
Shorter wavelength regions interact more strongly with atoms, molecules, and living tissues.
This section includes ultraviolet rays class 12 physics and high-energy EM radiation questions.

40. What is the frequency range of visible light?

Visible light has frequency from about 4 × 10^14 Hz to 7 × 10^14 Hz. Human eyes detect this range.

1. Lower Frequency: About 4 × 10^14 Hz.
2. Higher Frequency: About 7 × 10^14 Hz.
3. Wavelength Range: About 700 nm to 400 nm.
4. Final Result: Visible light is the eye-detectable EM region.

41. What are ultraviolet rays?

Ultraviolet rays are electromagnetic waves with wavelengths from about 400 nm to 0.6 nm. They lie beyond violet light.

1. Wavelength Range: 400 nm to 0.6 nm.
2. Source: Sun, hot bodies, and special lamps.
3. Use: Water purifiers and precision applications.
4. Final Result: UV rays have shorter wavelength than visible violet light.

42. Why does ordinary glass prevent tanning through windows?

Ordinary glass prevents tanning because it absorbs ultraviolet radiation. UV exposure causes tanning through melanin production.

1. Cause of Tanning: UV radiation.
2. Glass Property: Ordinary glass absorbs UV.
3. Effect: UV does not reach skin through glass.
4. Final Result: Sunburn through ordinary glass does not occur.

43. Why do welders wear special glass goggles?

Welders wear special glass goggles because welding arcs produce intense UV radiation. The glass protects their eyes.

1. Source: Welding arc.
2. Hazard: Strong UV can damage eyes.
3. Protection: Special glass filters UV.
4. Final Result: Welders need goggles for UV protection.

44. Why is the ozone layer important for ultraviolet protection?

The ozone layer absorbs most ultraviolet radiation from the Sun. It lies about 40 to 50 km above Earth.

1. Layer: Ozone layer.
2. Height: About 40 to 50 km.
3. Function: Absorbs harmful UV radiation.
4. Final Result: The ozone layer protects life from excessive UV exposure.

45. How are X-rays produced and used?

X-rays are produced by bombarding a metal target with high-energy electrons. They are used in diagnosis and cancer treatment.

1. Production: High-energy electrons hit a metal target.
2. Wavelength Range: About 10 nm to 10^-4 nm.
3. Medical Use: Diagnosis and treatment.
4. Safety Point: Overexposure damages living tissues.
5. Final Result: X-rays need controlled medical use.

46. How are gamma rays produced and used?

Gamma rays are produced in nuclear reactions and radioactive decay. They are used in medicine to destroy cancer cells.

1. Source 1: Nuclear reactions.
2. Source 2: Radioactive nuclei.
3. Wavelength: Less than 10^-3 nm.
4. Use: Cancer cell destruction.
5. Final Result: Gamma rays are high-frequency nuclear radiation.

NCERT Class 12 Physics Electromagnetic Waves Questions

NCERT questions often combine direct theory with numerical substitution.
Students should write SI units before calculating values from Chapter 8 formulas.
These NCERT class 12 physics electromagnetic waves questions match the 2026 exercise style.

47. A capacitor has circular plates of radius 12 cm and separation 5 cm. What formula gives its capacitance?

The capacitance is found using C = ε0A/d. The plate area is A = πr².

1. Given Data:
   r = 12 cm = 0.12 m
   d = 5 cm = 0.05 m
2. Area: A = πr².
3. Formula Used: C = ε0A/d.
4. Final Result: C = ε0πr²/d.

48. If a capacitor charging current is 0.15 A, what is the displacement current?

The displacement current is 0.15 A. It equals the conduction current during capacitor charging.

1. Given Data: Charging current = 0.15 A.
2. Rule: id = ic during capacitor charging.
3. Final Result: id = 0.15 A.

49. Is Kirchhoff’s junction rule valid at each plate of a charging capacitor?

Kirchhoff’s junction rule is valid if displacement current is included. Total current remains continuous across the circuit.

1. At Wire: Conduction current reaches the plate.
2. Between Plates: Displacement current continues the current effect.
3. Equality: ic = id.
4. Final Result: Kirchhoff’s rule holds with displacement current.

50. What changes when an electromagnetic wave travels in a material medium?

The speed changes when an electromagnetic wave travels in a material medium. The speed becomes v = 1/√(με).

1. Vacuum Speed: c = 1/√(μ0ε0).
2. Medium Speed: v = 1/√(με).
3. Dependence: Speed depends on permeability and permittivity.
4. Final Result: EM wave speed depends on medium properties.

Class 12 Physics Chapter 8 Questions and Answers