NEET (UG) · Physics, Class 12
Electromagnetic Waves: NEET PYQ, 2018–2026
Electromagnetic Waves (Physics, Class 12) has been asked 13 times in the 7 NEET papers from 2018, 2020, 2021, 2023, 2024, 2025 and 2026: about 1.9 questions a paper, 4% of all Physics questions. It is the 15th most-asked of 20 Physics chapters. NEET 2026 had 2 from it. Last checked 28 October 2026.
Questions per paper
Year by year.
| Year | Questions |
|---|---|
| 2018 | 1 |
| 2020 | 2 |
| 2021 | 2 |
| 2023 | 2 |
| 2024 | 2 |
| 2025 | 2 |
| 2026 | 2 |
From 2021 to 2024 each paper had 200 questions (a Section B with a choice); the others had 180.
Inside the chapter
Topics asked most.
- Properties of electromagnetic waves8
- Displacement current4
- EM spectrum1
Try it
10 Electromagnetic Waves questions to practise.
Our own questions, written in the NEET pattern and checked, each with an explanation. Pick an option to see the answer.
Q1 · Physics · Electromagnetic Waves
A parallel-plate capacitor in vacuum is being charged by a steady conduction current of 3.0 A in the connecting wires. What is the displacement current between its plates?
Answer: (A)
Displacement current is I_d = ε0 dΦE/dt. For a capacitor, the flux between the plates changes as the plate charge changes, and dQ/dt equals the conduction current I in the leads. Since ΦE = Q/ε0 between the plates, ε0 dΦE/dt = dQ/dt = I. So the displacement current equals 3.0 A, the conduction current, in magnitude.
Q2 · Physics · Electromagnetic Waves
A parallel-plate capacitor has circular plates of radius 4.0 cm separated by a small air gap. At an instant when the charging current in the wires is 2.0 A, the magnitude of the magnetic field at a point between the plates, 2.0 cm from the axis of the plates, is (μ₀ = 4π × 10⁻⁷ T·m/A):
Answer: (A)
Apply the Ampere–Maxwell law to a circular loop of radius r = 2 cm coaxial with the plates, with a flat surface lying in the gap. There is no conduction current through it; the enclosed displacement current is i_enc = I (r²/R²) because the field is uniform over the plates. Then B(2πr) = μ₀ I r²/R², so B = μ₀ I r /(2πR²) = (2 × 10⁻⁷)(2.0)(0.02)/(0.0016) = 5.0 × 10⁻⁶ T. The field grows linearly with r inside and falls as 1/r outside the plate edge.
Q3 · Physics · Electromagnetic Waves
A beam of light of power 30 W falls for 10 s on a surface that absorbs it completely. How much momentum does the beam deliver to the surface? (c = 3 × 10⁸ m s⁻¹.)
Answer: (C)
The energy delivered is U = Pt = 30 × 10 = 300 J. An electromagnetic wave of energy U carries momentum p = U/c, and a surface that absorbs it completely takes up all of this: p = 300 / 3 × 10⁸ = 1 × 10⁻⁶ kg m s⁻¹. The smallness of this number is why the pressure of light is not noticed in daily life.
Q4 · Physics · Electromagnetic Waves
A parallel-plate capacitor in vacuum is being charged by a steady conduction current of 2.5 A in its connecting leads. Neglecting edge effects, the displacement current between the plates at that instant is:
Answer: (A)
Maxwell added the displacement current Id = ε₀ dΦ_E/dt so that the total current (conduction plus displacement) is continuous. In the leads the current is purely conduction, I = 2.5 A. In the gap between the plates it is purely displacement. Continuity requires Id = I = 2.5 A, independent of plate separation and independent of whether any charge crosses the gap.
Q5 · Physics · Electromagnetic Waves
Which of the following lists radiations in order of increasing wavelength?
Answer: (B)
In order of increasing wavelength (decreasing frequency) the spectrum runs: gamma rays, X-rays, ultraviolet, visible, infrared, microwaves, radio waves.
Q6 · Physics · Electromagnetic Waves
In vacuum, a radio photon from a 1.0 MHz transmitter and an X-ray photon of wavelength 0.10 nm are compared. What is the ratio (momentum of X-ray photon) : (momentum of radio photon)? (c = 3 × 10⁸ m/s.)
Answer: (A)
Every photon travels at c in vacuum, but its momentum is p = E/c = h/λ. The momentum ratio is therefore the inverse of the wavelength ratio. First find the radio wavelength: λ = c/f = (3 × 10⁸)/(1.0 × 10⁶) = 300 m. The X-ray wavelength is 0.10 nm = 1.0 × 10⁻¹⁰ m. Then p(X)/p(radio) = λ(radio)/λ(X) = 300/(1.0 × 10⁻¹⁰) = 3 × 10¹².
Q7 · Physics · Electromagnetic Waves
A light wave of frequency 6.0 × 10¹⁴ Hz travels from air into glass of refractive index 1.5. What is its frequency in the glass?
Answer: (B)
The frequency of a wave is determined by the source that produces it and is unchanged at a boundary. In a medium of refractive index n the speed becomes v = c/n and the wavelength becomes λ/n, so that v = fλ still holds with the same f. The frequency in glass is therefore 6.0 × 10¹⁴ Hz.
Q8 · Physics · Electromagnetic Waves
Which of the following charge motions will radiate electromagnetic waves?
Answer: (A)
Electromagnetic waves are produced by accelerated (for example oscillating) charges. A charge at rest, a charge in uniform straight-line motion and a steady current all produce fields that are constant in time and so do not radiate. A charge in uniform circular motion has centripetal acceleration, so it radiates.
Q9 · Physics · Electromagnetic Waves
A circular parallel-plate capacitor with plates of radius 4.0 cm and capacitance 2.0 nF is charged by a 10 V battery through a 1.0 kΩ resistor. The charging starts at t = 0. Neglecting edge effects, the magnitude of the magnetic field in the gap at a distance 2.0 cm from the axis at time t = RC (the time constant) is closest to (μ₀ = 4π × 10⁻⁷ T m A⁻¹, e ≈ 2.718):
Answer: (C)
While charging through a resistor, the current is I(t) = (V/R) e^(−t/RC). At t = RC it is (10/1000)/e ≈ 3.68 mA. This equals the displacement current in the gap. The displacement current is uniform over the plate area, so a coaxial loop of radius r < R encloses I_d r²/R². Ampère–Maxwell gives B(2πr) = μ₀ I r²/R², so B = μ₀ I r/(2πR²) = (2 × 10⁻⁷)(3.68 × 10⁻³)(0.02)/(0.04)² ≈ 9.2 × 10⁻⁹ T.
Q10 · Physics · Electromagnetic Waves
A parallel-plate capacitor with plates A and B is discharging through an external wire. Plate A carries positive charge and plate B carries negative charge, so the electric field in the gap points from A to B. Which statement correctly describes the displacement current in the gap at this instant?
Answer: (C)
The displacement current density is J_d = ε₀ ∂E/∂t, so its direction is that of the rate of change of E. During discharge E (from A to B) decreases, so ∂E/∂t points from B to A. Outside, positive charge flows through the wire from A to B. Inside the gap the displacement current runs from B to A, which completes a continuous loop of current.
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All 13 real NEET questions from Electromagnetic Waves.
The questions from the actual papers, with the official answer and an explanation for each — chapter by chapter, or as full timed papers.
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About Electromagnetic Waves in NEET.
How many questions come from Electromagnetic Waves in NEET?
Electromagnetic Waves (Physics, Class 12) has been asked 13 times in the 7 NEET papers from 2018, 2020, 2021, 2023, 2024, 2025 and 2026: about 1.9 questions a paper, 4% of all Physics questions. It is the 15th most-asked of 20 Physics chapters. NEET 2026 had 2 from it.
Which topics of Electromagnetic Waves are asked most in NEET?
Properties of electromagnetic waves (8), Displacement current (4), EM spectrum (1), counted over the papers from 2018–2026.
Is Electromagnetic Waves important for NEET 2027?
It is in the syllabus in force, so it can be asked. By past papers it ranks 15 of 20 in Physics; use that to decide how much time it gets, not whether to study it.
Where can I solve Electromagnetic Waves NEET previous year questions?
On Nikaalle, free: sign up and practise every past NEET question from Electromagnetic Waves with an explanation for each, or sit the full papers timed. The ten questions on this page are our own, written in the same pattern.
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