91Ó°ÊÓ

Intensity of sunlight,$I=1.40kW/m^2$

We use intensity relation of wave, square both the sides, and solve for electric field component. Finally, substitutingtheelectric field component value inthemagnetic field component relation, we can find its amplitude.

Formulae:

$c=\frac{E_m}{B_m}$

$I=\frac{E_m^2}{2{\mathrm{μ}}_{0}c}$

The time averaged rate per unit area at which energy is transported,$S_{avg},$is called the intensityof l the wave.

$I=\frac{E_{rms}^2}{c{\mathrm{μ}}_0}$

$E_{rms}=\frac{E_m}{\sqrt{2}}$

$I=\frac{E_m^2}{2{\mathrm{μ}}_{0}c}$

$\begin{array}{c}{E}_m=\sqrt{2{\mathrm{μ}}_{0}Ic}\\ =\sqrt{2(4\mathrm{π}×{10}^{-7}\frac{T.m}{A})(1.40×{10}^3\frac{W}{m^2})\left(3×{10}^8\frac{m}{s}\right)}\\ =1.03×{10}^{3}V/m\end{array}$

The electric field component $E_m$is$1.03×{10}^3\frac{V}{m}.$

The wave speed c and amplitudes of electric and magnetic fields are related as

$c=\frac{E}{B}$

$c=\frac{E_m}{B_m}$

The amplitude of magnetic field comopnent inthewave is given by

$\begin{array}{c}{B}_m=\frac{E_m}{c}\\ =\frac{1.03×{10}^3\frac{V}{m}}{3×{10}^8\text{ }\frac{m}{s}}\\ =3.43×{10}^{-6}T\end{array}$

The amplitude of the magnetic field component $B_m$is $3.43×{10}^{-6}T.$