The intensity of the light from a bulb incident on a surface is 0.22 W/m². The amplitude of the magnetic field in this light-wave is ______________ × 10⁻⁹ T.
(Given : Permittivity of vacuum ∈₀ = 8.85 × 10⁻¹² C² N⁻¹-m⁻², speed of light in vacuum c = 3 × 10⁸ ms⁻¹)

Solution

Physics sir · 4 steps

Intensity gives you the peak electric field, and dividing that by c gives the peak magnetic field.

1. Intensity and amplitude. For an EM wave, the intensity is the average energy density times the speed of light. Written with the peak electric field E₀, it is:

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

2. Find E0. ε₀ c = 8.85 × 10⁻¹² × 3 × 10⁸ = 2.655 × 10⁻³. So E₀ = √(0.44/(2.655 × 10⁻³)) = √165.7 ≈ 12.87 V m⁻¹.
3. Find B0. The two amplitudes are linked by B₀ = E₀/c = 12.87/(3 × 10⁸) ≈ 4.29 × 10⁻⁸ T.
4. So. B₀ ≈ 42.9 × 10⁻⁹ T, which rounds to 43.

The common slip: 30. Dropping the 1/2 and using I = ε₀ c E₀² gives the rms field instead of the amplitude, so you get B ≈ 30 × 10⁻⁹ T, not 43.

The idea to keep. Intensity is 1/2ε₀ c E₀², and the 1/2 comes from averaging over time. Find E₀ first, then use B₀ = E₀/c.

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