JEE (Advanced) · Physics, Class 12
Electromagnetic Waves: JEE Advanced PYQ, 2007–2026
Electromagnetic Waves (Physics, Class 12) has been asked 5 times in the 40 JEE (Advanced) papers from 2007–2026: about 0.1 questions a paper, 1% of all Physics questions. It is the 19th most-asked of 19 Physics chapters. The two 2026 papers had 1 from it. Last checked 28 October 2026.
Questions per year, both papers
Year by year.
| Year | Questions |
|---|---|
| 2007 | 0in 2 papers |
| 2008 | 0in 2 papers |
| 2009 | 0in 2 papers |
| 2010 | 0in 2 papers |
| 2011 | 0in 2 papers |
| 2012 | 0in 2 papers |
| 2013 | 1in 2 papers |
| 2014 | 0in 2 papers |
| 2015 | 0in 2 papers |
| 2016 | 0in 2 papers |
| 2017 | 0in 2 papers |
| 2018 | 0in 2 papers |
| 2019 | 1in 2 papers |
| 2020 | 0in 2 papers |
| 2021 | 0in 2 papers |
| 2022 | 0in 2 papers |
| 2023 | 1in 2 papers |
| 2024 | 0in 2 papers |
| 2025 | 1in 2 papers |
| 2026 | 1in 2 papers |
Each year has two papers, and the number of questions has fallen over time, so compare per paper and expect some drift.
Inside the chapter
Topics asked most.
- Properties of electromagnetic waves5
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10 Electromagnetic Waves questions to practise.
Our own questions, written in the JEE Advanced pattern and checked, each with an explanation. Pick an option to see the answer.
Q1 · Physics · Electromagnetic Waves
A plane electromagnetic wave travels in vacuum along the positive z-axis. At a certain point and instant its magnetic field is B = (3î − 4ĵ) × 10⁻⁸ T. What is the y-component of its electric field at that point and instant? (c = 3 × 10⁸ m/s)
Answer: (B)
E is perpendicular to B and to the z-axis, with |E| = c|B| = 3 × 10⁸ × 5 × 10⁻⁸ = 15 V/m, and E × B must point along +z. This gives E = c(B × k̂) = 3 × (3î − 4ĵ) × k̂ = 3(−3ĵ − 4î) = (−12î − 9ĵ) V/m. As a test, the z-component of E × B is (−12)(−4) − (−9)(3) = 75, which is positive. The y-component is −9 V/m.
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
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.
Q4 · 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.
Q5 · 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.
Q6 · Physics · Electromagnetic Waves
Telescopes for observing X-rays and ultraviolet radiation from stars are usually placed on satellites above the Earth's atmosphere, whereas optical and radio telescopes work from the ground. What is the correct reason?
Answer: (A)
The atmosphere is not equally transparent to all regions of the spectrum. Ozone and atmospheric gases absorb nearly all ultraviolet radiation and X-rays from space, so detectors must be above the atmosphere. Visible light and radio waves penetrate the atmosphere, so ground-based optical and radio telescopes work. Electromagnetic waves travel through vacuum, so space is no obstacle.
Q7 · Physics · Electromagnetic Waves
A microwave source emitting radiation of wavelength 30 cm and an ultraviolet source emitting radiation of wavelength 300 nm each radiate the same power in vacuum. What is the ratio of the number of photons emitted per second by the microwave source to that emitted by the ultraviolet source?
Answer: (A)
Principle: a photon of wavelength λ carries energy E = hf = hc/λ. Power P = N E, where N is the number of photons per second, so for equal power N = Pλ/(hc), which means N ∝ λ. Then N_micro/N_UV = λ_micro/λ_UV = (30 × 10⁻² m)/(300 × 10⁻⁹ m) = 10⁻¹/10⁻⁷ = 10⁶. Check: the microwave has f = 10⁹ Hz and the UV has f = 10¹⁵ Hz, so each UV photon has 10⁶ times the energy and the microwave source must emit 10⁶ times as many photons.
Q8 · Physics · Electromagnetic Waves
Light of frequency 6.0 × 10¹⁴ Hz travels from vacuum into glass of refractive index 1.5. (c = 3 × 10⁸ m/s.) Which statement about the light inside the glass is correct?
Answer: (A)
Frequency is set by the source and does not change when the wave enters a medium. In vacuum, λ₀ = c/f = 3 × 10⁸ / 6 × 10¹⁴ = 500 nm, which is green-yellow visible light. In glass, v = c/n = 2 × 10⁸ m/s, so λ = v/f = 2 × 10⁸ / 6 × 10¹⁴ ≈ 333 nm. The numerical value of 333 nm lies below 400 nm, but the region of the spectrum is fixed by frequency, so the radiation is still visible light.
Q9 · Physics · Electromagnetic Waves
Two photons have energies 1.55 eV and 12.4 keV respectively. Using hc ≈ 1240 eV·nm and taking the visible region as 400 nm to 700 nm and X-rays as 10⁻³ nm to 1 nm, which option correctly names the regions of the electromagnetic spectrum to which the 1.55 eV photon and the 12.4 keV photon belong, in that order?
Answer: (A)
The wavelength of a photon follows from λ = hc/E = 1240 eV·nm / E. For E = 1.55 eV, λ = 1240/1.55 = 800 nm. This is longer than the red limit of 700 nm, so it is infrared, although it is close to the visible edge. For E = 12.4 keV = 12400 eV, λ = 1240/12400 = 0.10 nm = 10⁻¹⁰ m. This lies between 10⁻³ nm and 1 nm, so it is an X-ray. The pair is therefore infrared and X-ray.
Q10 · Physics · Electromagnetic Waves
A parallel-plate capacitor whose air-gap capacitance is 1.0 μF is completely filled with a dielectric of dielectric constant K = 4 and is then connected across an a.c. source v = 100 sin(1000 t) volt (t in seconds). The peak value of the displacement current between the plates is:
Answer: (A)
The displacement current across the gap always equals the conduction current in the leads, i_d = C dv/dt. With the dielectric, C = KC₀ = 4 × 1.0 μF = 4.0 μF. Then i_d = C ω V₀ cos(ωt), whose peak is ωCV₀ = (1000 rad/s)(4.0 × 10⁻⁶ F)(100 V) = 0.40 A. (Equivalently, i_d = ε dΦ_E/dt with ε = Kε₀ inside the dielectric.)
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All 5 real JEE Advanced 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 JEE Advanced.
How many questions come from Electromagnetic Waves in JEE Advanced?
Electromagnetic Waves (Physics, Class 12) has been asked 5 times in the 40 JEE (Advanced) papers from 2007–2026: about 0.1 questions a paper, 1% of all Physics questions. It is the 19th most-asked of 19 Physics chapters. The two 2026 papers had 1 from it.
Which topics of Electromagnetic Waves are asked most in JEE Advanced?
Properties of electromagnetic waves (5), counted over the papers from 2007–2026.
Is Electromagnetic Waves important for JEE Advanced 2027?
It is in the syllabus in force, so it can be asked. By past papers it ranks 19 of 19 in Physics; use that to decide how much time it gets, not whether to study it.
Where can I solve Electromagnetic Waves JEE Advanced previous year questions?
On Nikaalle, free: sign up and practise every past JEE Advanced 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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