91Ó°ÊÓ

The power transported per unit area is known as the intensity (l).

The formula used to calculate the intensity (l) is:

$I=\frac{P}{A}$

Where, A is area measured in the direction perpendicular to the energy and P is the power in watts.

The formula used to determine the amplitude of electric and magnetic fields of the wave are:

$E{\sqrt{\frac{2I}{{\mathrm{Î\mu }}_{0}c}}}_{max}{\frac{B{E}_{max}}{c}}_{max}$

Where, ${\mathrm{Î\mu }}_0=8.85×{10}^{-12}{C}^2/Nâ‹\dots m^2$and is the speed of light that is equal to $3.0×{10}^{8}m/s$.

Given that,

P = 60 W

r = 5 m

The formula used to calculate the intensity is:

$$\begin{gathered} l=\frac{P}{A} \\ =\frac{60}{4\mathrm{Ï€}{\left(5\right)}^2} \\ =0.19W/m^2 \end{gathered}$$

Hence, the intensity of monochromatic light source is $0.19W/m^2$.

The amplitude of electric field is:

$E{\sqrt{\frac{2I}{{\mathrm{Î\mu }}_{0}c}}}_{max}$

Substitute the values

$$\begin{gathered} E_{max}=\sqrt{\frac{2×\left(0.19\right)}{\left(8.85×{10}^{-12}\right)\left(3×{10}^8\right)}} \\ =12V/m \end{gathered}$$

The amplitude of magnetic field is:

${\frac{B{E}_{max}}{c}}_{max}$

Substitute the values

$$\begin{gathered} B_{max}=\frac{12}{3×{10}^8} \\ =4×{10}^{-8}T \end{gathered}$$

Hence, values of $E_{max}\mathrm{and}{B}_{max}$are 12V/m and$4×{10}^{-8}T$ respectively.