NEET Electromagnetic Waves PYQs & Practice Questions

10 free NEET Physics questions on Electromagnetic Waves, starting with previous year questions (2017–2026). Tap “Show answer” to check each one. No login needed.

Topics: EM Wave Relations · Displacement Current · Nature and Properties of EM Waves · Speed of EM Waves · Electromagnetic Spectrum · Energy and Intensity of EM Waves · Electromagnetic Spectrum - Applications · Displacement Current and Modified Ampere's Law · Photon Energy and the EM Spectrum · Reflection of an EM Wave from a Conducting Wall · Momentum Delivered by Light (Radiation Pressure) · Electric Field Intensity vs Bulb Power

1. Q1NEET 2026EM Wave Properties in a Dielectric Medium

   An electromagnetic wave travelling in a lossless dielectric medium having a dielectric constant, epsilon_r = 9, has the electric field, Ex = E0 sin(kz - 2*pi*10^6 t) V/m, where E0 is the amplitude and k is the wave vector. Among the following options, the INCORRECT choice(s) is/are: (1) The wavelength of the electromagnetic wave inside the medium is 300 m. (2) The magnetic field is given by the relation By = (B0/v) sin(kz - 2*pi*10^6 t), where v is the speed of the electromagnetic wave inside the medium. (3) The direction of propagation of the electromagnetic wave is along +z. (4) The speed of the electromagnetic wave inside the medium is 10^8 m/s.

   • A. Statement (1) only
   • B. Statement (2) only
   • C. Statements (1) and (2)
   • D. Statements (3) and (4)
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   Answer: B — Statement (2) only

   Refractive index n = sqrt(epsilon_r) = 3, so speed in medium v = c/n = 10^8 m/s (statement 4 is correct). Frequency f = 10^6 Hz, so wavelength = v/f = 10^8/10^6 = 100 m, NOT 300 m, making statement (1) incorrect. The wave travels in the +z direction (statement 3 is correct). The correct relation for the magnetic field amplitude is B0 = E0/v, so the field should be written as By = (E0/v) sin(...), not (B0/v) sin(...) which is dimensionally inconsistent as stated, making statement (2) also incorrect. So both (1) and (2) are the incorrect choices (officially marked as a multi-select answer '1, 2').

2. Q2NEET 2026Applications of the Electromagnetic Spectrum

   The following table presents the part of the electromagnetic spectrum and their corresponding major applications: P: Microwave, Q: UV rays, R: Gamma rays, S: Radio wave I: For purifying the water, II: For warming the food, III: For AM and FM communication systems, IV: For treating the Cancer cells The correct option is:

   • A. P-I, Q-IV, R-II, S-III
   • B. P-II, Q-I, R-IV, S-III
   • C. P-II, Q-IV, R-III, S-I
   • D. P-I, Q-II, R-III, S-IV
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   Answer: A — P-I, Q-IV, R-II, S-III

   Microwaves are used for warming food (P-II, as in microwave ovens). UV rays are used for purifying water (Q-I, UV water purification). Gamma rays are used for treating cancer cells (R-IV, radiotherapy). Radio waves are used for AM and FM communication systems (S-III). This matches P-II, Q-I, R-IV, S-III.

3. Q3NEET 2026Matching production of EM waves

   Match List I with List II: List I — A. Microwave, B. Visible light, C. Gamma rays, D. Infra-red rays. List II — I. Electrons in atoms emit light when they move from a higher to a lower energy level, II. Radioactive decay of nucleus, III. Vibration of atoms and molecules, IV. Klystron valve or magnetron valve.

   • A. A-IV, B-III, C-II, D-I
   • B. A-III, B-IV, C-I, D-II
   • C. A-IV, B-I, C-II, D-III
   • D. A-IV, B-I, C-II, D-III
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   Answer: C — A-IV, B-I, C-II, D-III

   Microwave produced by klystron/magnetron (IV); visible light from electron transitions (I); gamma rays from radioactive decay (II); infra-red from vibration of atoms/molecules (III). A-IV, B-I, C-II, D-III.

4. Q4NEET 2025EM Wave Relations

   The electric field in a plane electromagnetic wave is given by Ez = 60 cos(5x + 1.5×10⁹t) V/m. Then the expression for the corresponding magnetic field is (subscripts denote the direction of the field):

   • A. By = 2×10⁻⁷ cos(5x + 1.5×10⁹t) T
   • B. Bx = 2×10⁻⁷ cos(5x + 1.5×10⁹t) T
   • C. Bz = 60 cos(5x + 1.5×10⁹t) T
   • D. By = 60 sin(5x + 1.5×10⁹t) T
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   Answer: —

   B = E/c = 60/(3×10⁸) = 20×10⁻⁸ = 2×10⁻⁷ T. Since B must be perpendicular to both E (z-direction) and the direction of propagation (x-direction), B is along the y-direction.

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5. Q5NEET 2025Displacement Current

   A parallel plate capacitor made of circular plates is being charged such that the surface charge density on its plates is increasing at a constant rate with time. The magnetic field arising due to displacement current is :

   • A. zero at all places
   • B. constant between the plates and zero outside the plates
   • C. non-zero everywhere with maximum at the imaginary cylindrical surface connecting peripheries of the plates
   • D. zero between the plates and non-zero outside
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   Answer: B — constant between the plates and zero outside the plates

   Displacement current Id = ε0A(dσ/ε0dt)/... is constant (uniform over the plate area), so by Ampere-Maxwell law the magnetic field increases with distance from the axis inside the plate region and is maximum at the edge (the imaginary cylindrical surface joining the plate peripheries), then decreases outside like a long straight wire's field — non-zero everywhere with max at that surface.

6. Q6NEET 2024Displacement Current in Charging Capacitor

   A parallel plate capacitor is charged by connecting it to a battery through a resistor. If I is the current in the circuit, then in the gap between the plates:

   • A. displacement current of magnitude equal to I flows in a direction opposite to that of I.
   • B. displacement current of magnitude greater than I flows but can be in any direction.
   • C. there is no current.
   • D. displacement current of magnitude equal to I flows in the same direction as I.
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   Answer: C — there is no current.

   By Maxwell's concept of displacement current, the displacement current in the gap between the capacitor plates equals the conduction current I in the circuit and flows in the same direction, maintaining continuity of current.

7. Q7NEET 2024Properties of EM Waves in Free Space (NOT property)

   The property which is not of an electromagnetic wave travelling in free space is that:

   • A. they travel with a speed equal to 1/√(μ0ε0)
   • B. they originate from charges moving with uniform speed
   • C. they are transverse in nature
   • D. the energy density in electric field is equal to energy density in magnetic field.
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   Answer: A — they travel with a speed equal to 1/√(μ0ε0)

   Electromagnetic waves are produced by accelerating (not uniformly moving) charges. A charge moving with uniform speed does not radiate EM waves. So option (2) is NOT a correct property of an EM wave.

8. Q8NEET 2023Electric and Magnetic Field Amplitude Relation

   In a plane electromagnetic wave travelling in free space, the electric field component oscillates sinusoidally at a frequency of 2.0×10^10 Hz and amplitude 48Vm^-1. Then the amplitude of oscillating magnetic field is: (Speed of light in free space = 3 × 10^8 ms^-1)

   • A. 1.6 × 10^-8 T
   • B. 1.6 × 10^-7 T
   • C. 1.6 × 10^-6 T
   • D. 1.6 × 10^-9 T
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   Answer: A — 1.6 × 10^-8 T

   Using E0/B0 = c, B0 = E0/c = 48/(3×10^8) = 16×10^-8 = 1.6×10^-7 T.

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9. Q9NEET 2023Displacement Current in Capacitor

   An ac source is connected to a capacitor C. Due to decrease in its operating frequency:

   • A. displacement current increases
   • B. displacement current decreases
   • C. capacitive reactance remains constant
   • D. capacitive reactance decreases
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   Answer: A — displacement current increases

   Displacement current is proportional to frequency (Id = CdV/dt, and for AC, this scales with frequency). As frequency decreases, the displacement current decreases.

10. Q10NEET 2022Electromagnetic Spectrum

    Match List-I with List-II: List-I (Electromagnetic waves): (a) AM radio waves (b) Microwaves (c) Infrared radiations (d) X-rays -- List-II (Wavelength): (i) 10^-10 m (ii) 10^2 m (iii) 10^-2 m (iv) 10^-4 m. Choose the correct answer from the options given below:

    • A. a-iv, b-iii, c-ii, d-i
    • B. a-iii, b-ii, c-i, d-iv
    • C. a-iii, b-iv, c-ii, d-i
    • D. a-ii, b-iii, c-iv, d-i
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    Answer: C — a-iii, b-iv, c-ii, d-i

    AM radio waves have wavelengths of order 10^2 m, microwaves around 10^-2 m, infrared radiations around 10^-4 m, and X-rays around 10^-10 m. This matches a-ii, b-iii, c-iv, d-i.