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JEE Main 2022 · 29 Jun · Shift 1, Physics question 52

The intensity of the light from a bulb incident on a surface is 0.22 W/m2^2. The amplitude of the magnetic field in this light-wave is ______ ×10−9\times10^{-9} T.

(Given : Permittivity of vacuum ϵ0=8.85×10−12\epsilon_0=8.85\times10^{-12} C2^2 N−1^{-1}-m−2^{-2}, speed of light in vacuum c=3×108c=3\times10^8ms−1^{-1})

Answer and worked solution

Official answer: 43

Intensity. I=12ε0cE02I=\frac12\varepsilon_0cE_0^2: E0=0.448.85×10−12×3×108≈12.9E_0=\sqrt{\frac{0.44}{8.85\times10^{-12}\times3\times10^8}}\approx12.9 V m−1^{-1}.

Magnetic amplitude. E0c≈4.3×10−8=43×10−9\frac{E_0}c\approx4.3\times10^{-8}=43\times10^{-9} T.

Answer: 43

Topics:Energy and momentum of EM waves
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