91Ó°ÊÓ

The intensity of sunlight $I=1.40{\text{kW/m}}^{\text{2}}=1.40×{10}^3{\text{W/m}}^{\text{2}}$.

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

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

The maximum value of the electric component can be written as,

$E_m=\sqrt{2{\mathrm{μ}}_{0}Ic}$

Here ${\mathrm{μ}}_0$ is the permeability of the free space, and c is the speed of the light.

Substitute the values in the above expression, and we get,

$\begin{array}{rcl}{E}_m& =& \sqrt{2\left(4\mathrm{π}×{10}^{-7}\frac{\text{T}·\text{m}}{\text{A}}\right)\left(1.40×{10}^3\frac{\text{W}}{{\text{m}}^{\text{2}}}\right)\left(3×{10}^8\frac{\text{m}}{\text{s}}\right)}\\ & =& \sqrt{2\left(4\mathrm{π}×{10}^{-7}\right)\left(1.40×{10}^3\right)\left(3×{10}^8\right)\left(1\frac{\text{T}·\text{m}}{\text{A}}×1\frac{\text{W}}{{\text{m}}^{\text{2}}}×1\frac{\text{m}}{\text{s}}×\frac{1\text{V}·\text{A}}{1\text{W}}×\frac{\text{1}\text{V}·{\text{s/m}}^{\text{2}}}{1\text{T}}\right)}\\ & =& 1.03×{10}^3\text{V/m}\end{array}$

The amplitude of the electric field component is $1.03×{10}^3\frac{\mathrm{V}}{\mathrm{m}}$.

The amplitude of the magnetic field component in the wave is given by,

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

Substitute the values in the above expression, and we get,

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

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