NEETAceToday's NEET Daily

Characteristics and transverse nature of electromagnetic waves: NEET MCQs with answers

81 questions on this topic of Electromagnetic Waves (NEET 2027 syllabus). Here are 12; tap one to see its answer and explanation.

Practise Electromagnetic Waves freeAll Electromagnetic Waves questions

Free with a Google sign-in; NEET Daily needs no sign-in at all.

easy

1. A radio transmitter emits electromagnetic waves of frequency 450 MHz in free space. The wavelength of these waves is:

  1. 0.667 m
  2. 1.5 m
  3. 0.334 m
  4. 667 m
Show answer and explanation

Answer: (A) 0.667 m

λ = c/f = (3 × 10⁸)/(450 × 10⁶) = 0.667 m.

Quick tip: λ = c/f. Convert MHz to Hz first.

NCERT Class 12, Chapter 8

medium

2. The peak electric field of a plane electromagnetic wave in free space is 48 V/m. Its peak magnetic field is:

  1. 1600 nT
  2. 16 nT
  3. 160 nT
  4. 480 nT
Show answer and explanation

Answer: (C) 160 nT

B₀ = E₀/c = 48/(3 × 10⁸) T = 160 × 10⁻⁹ T = 160 nT.

Quick tip: B₀ = E₀/c, so the magnetic field is tiny in tesla.

NCERT Class 12, Chapter 8

hard

3. A point source radiates electromagnetic waves isotropically with an average power of 1500 W. What is the peak electric field amplitude at a distance of 5.0 m from the source? (Take 1/(4πε₀) = 9 × 10⁹ N·m²/C² and c = 3 × 10⁸ m/s)

  1. 12 V/m
  2. 36 V/m
  3. 120 V/m
  4. 60 V/m
Show answer and explanation

Answer: (D) 60 V/m

Intensity at distance r is I = P / (4πr²) = (1/2) ε₀ c E₀². Thus, E₀² = 2P / (4πr² ε₀ c) = (2P / (r² c)) × (1 / (4πε₀)). Substituting the given values: E₀² = [2 × 1500 × 9 × 10⁹] / [25 × 3 × 10⁸] = [27000 × 10⁹] / [75 × 10⁸] = 3600. Therefore, E₀ = √3600 = 60 V/m.

medium

4. The electric field part of an electromagnetic wave in vacuum is given by E = 60 cos[(1.2 × 10⁷)z - (3.6 × 10¹⁵)t] î N/C, where z is in metres and t is in seconds. The expression for the corresponding magnetic field vector B is:

  1. B = 2.0 × 10⁻⁷ cos[(1.2 × 10⁷)z - (3.6 × 10¹⁵)t] (-ĵ) T
  2. B = 2.0 × 10⁻⁷ cos[(1.2 × 10⁷)z - (3.6 × 10¹⁵)t] ĵ T
  3. B = 1.8 × 10¹⁰ cos[(1.2 × 10⁷)z - (3.6 × 10¹⁵)t] ĵ T
  4. B = 2.0 × 10⁻⁷ cos[(1.2 × 10⁷)z - (3.6 × 10¹⁵)t] k̂ T
Show answer and explanation

Answer: (B) B = 2.0 × 10⁻⁷ cos[(1.2 × 10⁷)z - (3.6 × 10¹⁵)t] ĵ T

The wave propagates along the +z direction (k̂). Amplitude B₀ = E₀ / c = 60 / (3 × 10⁸) = 2.0 × 10⁻⁷ T. Since E is along î and wave propagation is along k̂, the magnetic field must be along ĵ because î × ĵ = k̂.

easy

5. In a plane electromagnetic wave in vacuum, the electric field amplitude is 3 V/m. The magnetic field amplitude is:

  1. 1 × 10⁻⁸ T
  2. 9 × 10⁸ T
  3. 2 × 10⁻⁸ T
  4. 1 × 10⁻⁶ T
Show answer and explanation

Answer: (A) 1 × 10⁻⁸ T

In an EM wave in vacuum E₀ = cB₀, so B₀ = E₀/c = 3/(3 × 10⁸) = 1 × 10⁻⁸ T.

hard

6. In a non-magnetic medium having relative permittivity ε_r = 4.0 and relative permeability μ_r = 1.0, the amplitude of the electric field of an electromagnetic wave is 60 V/m. The amplitude of its magnetic field in this medium is:

  1. 4.0 × 10⁻⁷ T
  2. 2.0 × 10⁻⁷ T
  3. 1.0 × 10⁻⁷ T
  4. 8.0 × 10⁻⁷ T
Show answer and explanation

Answer: (A) 4.0 × 10⁻⁷ T

The speed of the wave in the medium is v = c / √(ε_r μ_r) = (3 × 10⁸) / √4 = 1.5 × 10⁸ m/s. In any medium, the ratio of amplitudes is E₀ / B₀ = v. Therefore, B₀ = E₀ / v = 60 / (1.5 × 10⁸) = 4.0 × 10⁻⁷ T.

medium

7. The electric field component of a plane electromagnetic wave propagating in a non-magnetic medium is given by E_y = 120 sin(1.2 × 10⁷ t - 0.08 x) V/m, where x is in metres and t is in seconds. The refractive index of the medium is: (Take c = 3.0 × 10⁸ m/s)

  1. 1.5
  2. 2.0
  3. 1.33
  4. 2.5
Show answer and explanation

Answer: (B) 2.0

The speed of the wave in the medium is v = ω / k = (1.2 × 10⁷) / 0.08 = 1.5 × 10⁸ m/s. The refractive index is n = c / v = (3.0 × 10⁸) / (1.5 × 10⁸) = 2.0.

easy

8. A radio transmitter emits electromagnetic waves of frequency 600 MHz in free space. The wavelength of these waves is:

  1. 0.5 m
  2. 500 m
  3. 0.25 m
  4. 2 m
Show answer and explanation

Answer: (A) 0.5 m

λ = c/f = (3 × 10⁸)/(600 × 10⁶) = 0.5 m.

Quick tip: λ = c/f. Convert MHz to Hz first.

NCERT Class 12, Chapter 8

medium

9. The peak electric field of a plane electromagnetic wave in free space is 3 V/m. Its peak magnetic field is:

  1. 1 nT
  2. 100 nT
  3. 10 nT
  4. 30 nT
Show answer and explanation

Answer: (C) 10 nT

B₀ = E₀/c = 3/(3 × 10⁸) T = 10 × 10⁻⁹ T = 10 nT.

Quick tip: B₀ = E₀/c, so the magnetic field is tiny in tesla.

NCERT Class 12, Chapter 8

easy

10. Which of the following characteristics does NOT change when an electromagnetic wave enters from air into a dielectric medium of refractive index n?

  1. Wavelength
  2. Speed
  3. Frequency
  4. Amplitude
Show answer and explanation

Answer: (C) Frequency

The frequency of an electromagnetic wave is determined by the source generating the wave and remains strictly constant when the wave passes across different media. Wavelength, speed, and amplitude all vary.

medium

11. In a plane electromagnetic wave in vacuum, the amplitude of the electric field is 6 V/m. The amplitude of the magnetic field is:

  1. 60 nT
  2. 20 nT
  3. 2 nT
  4. 200 nT
Show answer and explanation

Answer: (B) 20 nT

B₀ = E₀/c = 6/(3 × 10⁸) T = 20 × 10⁻⁹ T = 20 nT.

Quick tip: B₀ = E₀/c, so the magnetic field is tiny in tesla.

NCERT Class 12, Chapter 8

easy

12. An electromagnetic wave of frequency 600 MHz travels in vacuum. Its wavelength is:

  1. 0.25 m
  2. 500 m
  3. 2 m
  4. 0.5 m
Show answer and explanation

Answer: (D) 0.5 m

λ = c/f = (3 × 10⁸)/(600 × 10⁶) = 0.5 m.

Quick tip: λ = c/f. Convert MHz to Hz first.

NCERT Class 12, Chapter 8

Keep going with Electromagnetic Waves. All 232 questions with instant answers, your mistakes saved for revision and your accuracy by topic.
Start practising

Other topics in Electromagnetic Waves