JEE Main 2020 · 7 January · morning · Single correct · +4 / −1

If the magnetic field in a plane electromagnetic wave is given by
B⃗ = 3 × 10⁻⁸ sin(1.6 × 10³x + 48 × 10¹⁰t)ĵ T, then what will be expression for electric field ?

  1. (a) E⃗ = (9sin(1.6 × 10³x + 48 × 10¹⁰t)k̂ V/m)
  2. (b) E⃗ = (60sin(1.6 × 10³x + 48 × 10¹⁰t)k̂ V/m)
  3. (c) E⃗ = (3 × 10⁻⁸ sin(1.6 × 10³x + 48 × 10¹⁰t)î V/m)
  4. (d) E⃗ = (3 × 10⁻⁸ sin(1.6 × 10³x + 48 × 10¹⁰t)ĵ V/m)
Answer and solution

Answer (a) E⃗ = (9sin(1.6 × 10³x + 48 × 10¹⁰t)k̂ V/m)

Solution

Physics sir · 3 steps

Get the size from E₀ = cB₀, then make E⃗ × B⃗ point where the wave travels.

  1. Read the wave. The phase is kx + ω t with k = 1.6 × 10³ m⁻¹ and ω = 48 × 10¹⁰ rad s⁻¹. The plus sign means the wave travels along -x. The speed is ω / k = 3 × 10⁸ m s⁻¹, which is c, as it should be.

    E₀ = cB₀ = (3 × 10⁸) × (3 × 10⁻⁸) = 9 V m⁻¹

  2. Fix the direction. E⃗ must be perpendicular to both B⃗ (along ĵ) and the travel direction (along x), so it lies along ±k̂. With E⃗ along k̂, E⃗ × B⃗ points along k̂ × ĵ = -î, which matches the wave moving along -x.
  3. So. E⃗ = 9 sin(1.6 × 10³x + 48 × 10¹⁰t) k̂ V m⁻¹, with the same phase as B⃗. That is option (a).

The idea to keep. In an EM wave, E₀ = cB₀ and E⃗, B⃗ oscillate in phase. Their directions are fixed by E⃗ × B⃗ pointing along the direction the wave travels.

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 2020.