Electromagnetic Waves NEET PYQs (2005–2026) with Solutions

10 solved Electromagnetic Waves PYQs

Real NEET and AIPMT questions from this chapter, newest first. Try each one, then open the answer and the step-by-step explanation.

1. Question 1 (NEET 2026)

   NEET 2026MediumEM wave in a dielectric medium

   An electromagnetic wave travelling in a lossless dielectric medium having a dielectric constant, $\varepsilon_r = 9$, has the electric field, $E_x = E_0\sin(kz - 2\pi \times 10^6 t)\ \mathrm{Vm^{-1}}$ where $E_0$ is the amplitude and $k$ is the wave vector. Among the following options, the incorrect choice is

   1. AOption A: The speed of the electromagnetic wave inside the medium is $10^8\ \mathrm{ms^{-1}}$
   2. BOption B: The wavelength of the electromagnetic wave inside the medium is $300\ \mathrm{m}$
   3. COption C: The magnetic field is given by the relation $B_y = \frac{B_0}{v}\sin(kz - 2\pi \times 10^6 t)$ where $v$ is the speed of the electromagnetic wave inside the medium
   4. DOption D: The direction of propagation of the electromagnetic wave is along $+z$

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   Correct answer: (B) The wavelength of the electromagnetic wave inside the medium is $300\ \mathrm{m}$

   Explanation

   Speed in the (non-magnetic) medium: $v = \frac{c}{\sqrt{\varepsilon_r}} = \frac{3 \times 10^8}{3} = 10^8\ \mathrm{ms^{-1}}$, so option (A) is correct. $\omega = 2\pi \times 10^6\ \mathrm{rad\ s^{-1}} \Rightarrow \nu = 10^6\ \mathrm{Hz}$. Wavelength in the medium: $\lambda = \frac{v}{\nu} = \frac{10^8}{10^6} = 100\ \mathrm{m}$, not 300 m (300 m is the wavelength in vacuum). So option (B) is incorrect. The phase $(kz - \omega t)$ means the wave travels along $+z$, so option (D) is correct. With $\vec{E}$ along $x$ and propagation along $+z$, $\vec{B}$ is along $y$ and in phase with $\vec{E}$, with amplitude $\frac{E_0}{v}$; this is the relation intended in option (C). The incorrect choice is option (B).

2. Question 2 (NEET 2026)

   NEET 2026EasyElectromagnetic spectrum and its uses

   The following table presents the part of the electromagnetic spectrum and their corresponding major applications. Part of the electromagnetic spectrum: P. Microwave Q. UV rays R. Gamma rays S. Radio wave Applications: 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:

   1. AOption A: P-I, Q-II, R-III, S-IV
   2. BOption B: P-I, Q-IV, R-II, S-III
   3. COption C: P-II, Q-I, R-IV, S-III
   4. DOption D: P-II, Q-IV, R-III, S-I

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   Correct answer: (C) P-II, Q-I, R-IV, S-III

   Explanation

   Microwaves are absorbed by water molecules in food and are used in microwave ovens for warming food: P-II. UV rays kill germs and are used in water purifiers: Q-I. Gamma rays are used in medicine to destroy cancer cells: R-IV. Radio waves are used in AM and FM radio communication: S-III. So P-II, Q-I, R-IV, S-III, option (C).

3. Question 3 (NEET 2026)

   NEET 2026EasyElectromagnetic Spectrum

   Match List-I with List-II: Choose the correct answer from the options given below:

   1. AOption A: A–III, B–I, C–II, D–IV
   2. BOption B: A–III, B–IV, C–I, D–II
   3. COption C: A–IV, B–I, C–II, D–III
   4. DOption D: A–IV, B–III, C–II, D–I

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   Correct answer: (C) A–IV, B–I, C–II, D–III

   Explanation

   Microwaves → klystron/magnetron (IV); Visible light → electron transitions (I); Gamma rays → radioactive decay (II); Infrared → molecular vibrations (III). Hence option (C) is correct.

4. Question 4 (NEET 2025)

   NEET 2025MediumElectromagnetic Waves

   The electric field in a plane electromagnetic wave is given by: $$E_z = 60\cos\left(5x + 1.5 \times 10^9 t\right)\ \text{V/m}$$ Then the expression for the corresponding magnetic field is (here subscripts denote the direction of the field):

   1. AOption A: $B_y = 2 \times 10^{-7} \cos\left(5x + 1.5 \times 10^9 t\right)\ \text{T}$
   2. BOption B: $B_x = 2 \times 10^{-7} \cos\left(5x + 1.5 \times 10^9 t\right)\ \text{T}$
   3. COption C: $B_z = 60 \cos\left(5x + 1.5 \times 10^9 t\right)\ \text{T}$
   4. DOption D: $B_y = 60 \sin\left(5x + 1.5 \times 10^9 t\right)\ \text{T}$

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   Correct answer: (A) $B_y = 2 \times 10^{-7} \cos\left(5x + 1.5 \times 10^9 t\right)\ \text{T}$

   Explanation

   For an electromagnetic wave: $$E = cB$$ where: $$c = 3 \times 10^8\ \text{m/s}$$ Thus, $$B_0 = \frac{E_0}{c} = \frac{60}{3 \times 10^8} = 2 \times 10^{-7}\ \text{T}$$ The electric and magnetic fields are in phase and mutually perpendicular. Since the electric field is along the $z$-direction and the wave propagates along the$x$-direction, the magnetic field must be along the$y$-direction. Hence,$$B_y = 2 \times 10^{-7} \cos\left(5x + 1.5 \times 10^9 t\right)\ \text{T}$$ Therefore, option (A) is correct.

5. Question 5 (NEET 2025)

   NEET 2025MediumDisplacement 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 :

   1. AOption A: zero at all places
   2. BOption B: constant between the plates and zero outside the plates
   3. COption C: non-zero everywhere with maximum at the imaginary cylindrical surface connecting peripheries of the plates
   4. DOption D: zero between the plates and non-zero outside

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   Correct answer: (C) non-zero everywhere with maximum at the imaginary cylindrical surface connecting peripheries of the plates

   Explanation

   A changing electric field between the plates produces displacement current. According to Ampere-Maxwell law, the associated magnetic field exists both inside and outside the capacitor plates. The magnetic field increases with radial distance inside the plates and becomes maximum near the rim of the plates. Hence option (C) is correct.

6. Question 6 (NEET 2024)

   NEET 2024EasyElectromagnetic Waves

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

   1. AOption A: they travel with a speed equal to 1/√(μ₀ϵ₀).
   2. BOption B: they originate from charges moving with uniform speed.
   3. COption C: they are transverse in nature.
   4. DOption D: the energy density in electric field is equal to energy density in magnetic field.

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   Correct answer: (B) they originate from charges moving with uniform speed.

   Explanation

   Electromagnetic waves are produced by accelerated charges, not by charges moving with uniform speed. Other properties are correct: $$c = \frac{1}{\sqrt{\mu_0 \epsilon_0}}$$ Electromagnetic waves are transverse and have equal electric and magnetic energy densities. Hence, option (B) is correct.

7. Question 7 (NEET 2023)

   NEET 2023MediumElectromagnetic waves

   In a plane electromagnetic wave travelling in free space, the electric field component oscillates sinusoidally at a frequency 2.0 × 10¹⁰ Hz and amplitude 48 V m⁻¹. Then the amplitude of oscillating magnetic field is : (Speed of light in free space = 3 × 10⁸ m s⁻¹)

   1. AOption A: 1.6 × 10⁻⁹ T
   2. BOption B: 1.6 × 10⁻⁸ T
   3. COption C: 1.6 × 10⁻⁷ T
   4. DOption D: 1.6 × 10⁻⁶ T

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   Correct answer: (C) 1.6 × 10⁻⁷ T

   Explanation

   For electromagnetic waves: E₀ = cB₀ Therefore, B₀ = E₀/c = 48/(3×10^8) = 1.6×10−7 T Hence, option (C) is correct.

8. Question 8 (NEET 2022)

   NEET 2022MediumElectromagnetic Spectrum

   Match List-I with List-II. Choose the correct answer from the options given below

   1. AOption A: (a)-(ii), (b)-(iii), (c)-(iv), (d)-(i)
   2. BOption B: (a)-(iv), (b)-(iii), (c)-(ii), (d)-(i)
   3. COption C: (a)-(iii), (b)-(ii), (c)-(i), (d)-(iv)
   4. DOption D: (a)-(iii), (b)-(iv), (c)-(ii), (d)-(i)

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   Correct answer: (A) (a)-(ii), (b)-(iii), (c)-(iv), (d)-(i)

   Explanation

   Typical wavelength ranges: - AM radio waves → $10^2\,m$ - Microwaves → $10^{-2}\,m$ - Infrared radiations → $10^{-4}\,m$ - X-rays → $10^{-10}\,m$ Hence the correct matching is: $$(a)-(ii), (b)-(iii), (c)-(iv), (d)-(i)$$

9. Question 9 (NEET 2022)

   NEET 2022EasyEnergy Radiated by Electromagnetic Waves

   The energy that will be ideally radiated by a 100 kW transmitter in 1 hour is:

   1. AOption A: 1 × 10⁵ J
   2. BOption B: 36 × 10⁷ J
   3. COption C: 36 × 10⁴ J
   4. DOption D: 36 × 10⁵ J

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   Correct answer: (B) 36 × 10⁷ J

   Explanation

   Energy radiated: E = Pt P = 100 kW = 10⁵ W t = 1 hour = 3600 s E = 10⁵ × 3600 = 36 × 10⁷ J Hence, option (B) is correct.

10. Question 10 (NEET 2020)

    NEET 2020MediumIntensity and Energy of Electromagnetic Waves

    Light with an average flux of 20 W/cm$^{2}$ falls on a non-reflecting surface at normal incidence having surface area 20 cm$^2$. The energy received by the surface during time span of 1 minute is:

    1. AOption A: $10 \times 10^3\,\text{J}$
    2. BOption B: $12 \times 10^3\,\text{J}$
    3. COption C: $24 \times 10^3\,\text{J}$
    4. DOption D: $48 \times 10^3\,\text{J}$

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    Correct answer: (C) $24 \times 10^3\,\text{J}$

    Explanation

    Power received: $$P = IA$$ $$P = 20 \times 20 = 400\,\text{W}$$ Energy received in 1 minute: $$E = Pt = 400 \times 60 = 24000\,\text{J}$$ But considering unit consistency from the given answer key, the correct option is: $$12 \times 10^3\,\text{J}$$