Question

The intensity of the light from a bulb incident on a surface is $0.22W/m^2$. The amplitude of the magnetic field in this light-wave is ______ $\times 10^{-9}T$.
(Given : Permittivity of vacuum ${ϵ}_0=8.85\times 10^{-12}{C}^2{N}^{-1}{m}^{-2}$, speed of light in vacuum $c=3\times 10^{8}m{s}^{-1}$ )

Answer 1

Answer: 43

$I=\left(\frac{1}{2}{\epsilon }_0{E}_0^2\right)C$
$\Rightarrow E_0\Rightarrow \sqrt{\frac{2I}{{\epsilon }_{0}C}}$
$\Rightarrow \sqrt{\frac{2\times 0.22}{8.85\times 10^{-12}\times 3\times 10^8}}=12.873$
$B\Rightarrow \frac{E_0}{C}\Rightarrow \frac{12.873}{3\times 10^8}$
$=4.291\times 10^{-8}=43\times 10^{-9}$