Sunlight just outside Earth’s atmosphere has an intensity of$\mathbf{1}\mathbf{.}\mathbf{40}\mathbf{}\mathit{k}\mathit{W}\mathbf{/}{\mathit{m}}^{\mathbf{2}}$.

(a) Calculate${\mathit{E}}_{\mathbf{m}}$for sunlight there, assuming it to be a plane wave and

(b) Calculate${\mathit{B}}_{\mathbf{m}}$for sunlight there, assuming it to be a plane wave.

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:

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

$I=\frac{E_m^2}{2{\mu }_{0}c}$

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

$I=\frac{E_{rms}^2}{c{\mu }_0}$

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

$I=\frac{E_m^2}{2{\mu }_{0}c}$

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

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

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

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

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