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 ε 0 = 8.85 × 10 - 12 C2 N ⁻¹ -m ⁻², speed of light in vacuum c = 3 × 108 ms ⁻¹) Numerical answer type. Work it out, then open the answer.

Answer: 43. I = (c B₀²)/(2μ₀) and μ₀ = (1)/(ε₀ c²), so B₀ = √((2I)/(ε₀ c³)) B₀ = √2 × 0.228.85 × 10⁻¹² × 27 × 10²⁴ = √1.84 × 10⁻¹⁵ B₀ ≈ 4.3 × 10⁻⁸ T = 43 × 10⁻⁹ T So the answer is 43.

The intensity of the light from a bulb incident on a surface is 0.22… | JEE Main 2022 (29 Jun, Shift 1)

Q16JEE Main 2022 · 29 JUN · SHIFT 1Medium

Energy Density & Intensity

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 ______________ $\times$ 10 $-$ 9 T.

(Given: Permittivity of vacuum $\varepsilon$ 0 = 8.85 $\times$ 10 $-$ 12 C2 N ${}^{-1}$ -m ${}^{-2}$ , speed of light in vacuum c = 3 $\times$ 108 ms ${}^{-1}$ )

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 43

$I = \dfrac{c B_0^2}{2\mu_0}$ and $\mu_0 = \dfrac{1}{\varepsilon_0 c^2}$ , so $B_0 = \sqrt{\dfrac{2I}{\varepsilon_0 c^3}}$
$B_0 = \sqrt{\dfrac{2 \times 0.22}{8.85 \times 10^{-12} \times 27 \times 10^{24}}} = \sqrt{1.84 \times 10^{-15}}$
$B_0 \approx 4.3 \times 10^{-8}\ \text{T} = 43 \times 10^{-9}\ \text{T}$
So the answer is $43$ .

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