Intensity of sunlight is observed as 0.092 , | vedclass

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(D) The intensity $I$ of an electromagnetic wave is related to the peak magnetic field $B_0$ by the formula: $I = \frac{B_0^2 c}{2 \mu_0}$.Since $c =

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$2.77 	imes 10^{-8} \, T$

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(D) The intensity $I$ of an electromagnetic wave is related to the peak magnetic field $B_0$ by the formula: $I = \frac{B_0^2 c}{2 \mu_0}$.Since $c = \frac{1}{\sqrt{\mu_0 \epsilon_0}}$,we have $\frac{1}{\mu_0} = \epsilon_0 c^2$.Substituting this into the intensity formula: $I = \frac{B_0^2 c}{2} (\epsilon_0 c^2) = \frac{1}{2} \epsilon_0 c^3 B_0^2$.Rearranging for $B_0$: $B_0 = \sqrt{\frac{2I}{\epsilon_0 c^3}}$.Given $I = 0.092 \, W/m^2$,$\epsilon_0 = 8.85 	imes 10^{-12} \, F/m$,and $c = 3 	imes 10^8 \, m/s$:$B_0 = \sqrt{\frac{2 	imes 0.092}{8.85 	imes 10^{-12} 	imes (3 	imes 10^8)^3}}$.$B_0 = \sqrt{\frac{0.184}{8.85 	imes 10^{-12} 	imes 27 	imes 10^{24}}} = \sqrt{\frac{0.184}{238.95 	imes 10^{12}}} = \sqrt{0.00077 	imes 10^{-12}} = \sqrt{7.7 	imes 10^{-16}} \approx 2.77 	imes 10^{-8} \, T$.

Intensity of sunlight is observed as $0.092 \, W/m^2$ at a point in free space. What will be the peak value of the magnetic field at that point? (Given: $\epsilon_0 = 8.85 \times 10^{-12} \, C^2 N^{-1} m^{-2}$,$c = 3 \times 10^8 \, m/s$)

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