JEE Main 2021 · 22 July · evening · Single correct · +4 / −1

Intensity of sunlight is observed as 0.092 Wm⁻² at a point in free space. What will be the peak value of magnetic field at the point?
(ε ₀ = 8.85 × 10⁻¹²C²N⁻¹m⁻²)

  1. (a) 2.77 × 10⁻⁸ T
  2. (b) 1.96 × 10⁻⁸ T
  3. (c) 8.31 T
  4. (d) 5.88 T
Answer and solution

Answer (a) 2.77 × 10⁻⁸ T

Solution

Physics sir · 4 steps

Get the peak electric field from the intensity first, then divide by c to reach the peak magnetic field.

  1. Intensity and peak field. For an EM wave in free space, the average intensity depends on the peak electric field E₀. The factor 1/2 comes from averaging sin² over one cycle.

    I = 1/2 ε₀ c E₀² ⇒ E₀ = √(2I/(ε₀ c))

  2. Put in numbers. ε₀ c = 8.85 × 10⁻¹² × 3 × 10⁸ = 2.655 × 10⁻³. So E₀ = √(0.184/(2.655 × 10⁻³)) = √69.3 ≈ 8.32 V m⁻¹.
  3. Electric to magnetic. The peak fields are linked by B₀ = E₀/c = 8.32/(3 × 10⁸) ≈ 2.77 × 10⁻⁸ T.
  4. So. The peak magnetic field is 2.77 × 10⁻⁸ T, which is option (a).

The idea to keep. Intensity gives you E₀ through I = 1/2ε₀ c E₀². Then B₀ = E₀/c. Read carefully whether the question asks for the peak value or the rms value.

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 2021.