Question

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.

Short answer

1. The electric field component $E_m$is$1.03\times {10}^3\frac{V}{m}.$
2. The magnetic field component $B_m$is$3.43\times {10}^{-6}T.$

Step-by-step solution

Listing the given quantities

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

Understanding the concepts of amplitude

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:

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

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

(a) Calculations of themaximum value of electric field component

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$1.03\times {10}^3\frac{V}{m}.$

(b) Calculations ofamplitude of the magnetic field component Bm

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 $3.43\times {10}^{-6}T.$