JEE Main 2024 · 31 January · evening · Single correct · +4 / −1

Given below are two statements:
Statement I: Electromagnetic waves carry energy as they travel through space and this energy is equally shared by the electric and magnetic fields.
Statement II: When electromagnetic waves strike a surface, a pressure is exerted on the surface.
In the light of the above statements, choose the most appropriate answer from the options given below:

  1. (a) Statement I is incorrect but Statement II is correct.
  2. (b) Both Statement I and Statement II are correct.
  3. (c) Statement I is correct but Statement II is incorrect.
  4. (d) Both Statement I and Statement II are incorrect.
Answer and solution

Answer (b) Both Statement I and Statement II are correct.

Solution

Physics sir · 3 steps

Check each statement against one fact: the energy split for I, and the momentum carried for II.

  1. Energy is shared. The average energy density of the electric field is 1/4ε₀ E₀², and that of the magnetic field is 1/4B₀²/μ₀. Since B₀ = E₀/c and c² = 1/(μ₀ ε₀), these two are exactly equal.

    u_E = 1/4ε₀ E₀² = 1/4B₀²/μ₀ = u_B

  2. Radiation pressure. An EM wave carrying energy U also carries momentum p = U/c. When the wave hits a surface, this momentum is transferred, which means a force acts on the surface. Force per unit area is pressure, so Statement II is correct as well.
  3. So. Statement I is correct because the energy is split half and half between the electric and magnetic fields. Statement II is correct because EM waves exert radiation pressure. The answer is (b).

The idea to keep. In an EM wave the electric and magnetic fields always carry equal shares of the energy. The wave also carries momentum U/c, so it pushes on whatever it strikes.

Answer from the NTA final answer key. Solution by PrepWiser.

Study Electromagnetic Waves

Asked about 130 times in JEE Main since 2020. This question is from 2024.