A monochromatic light source with power output 60.0 W radiates light of wavelength 700 nm uniformly in all directions. Calculate ${\mathbf{E}}_{\mathbf{max}}$ and${\mathit{B}}_{\mathbf{max}}$ for the 700 - nm light at a distance of 5.00 m from the source.

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:

$E{\sqrt{\frac{2I}{{\epsilon }_{0}c}}}_{max}{\frac{B{E}_{max}}{c}}_{max}$

Where, ${\epsilon }_0=8.85\times {10}^{-12}{C}^2/N\cdot m^2$and is the speed of light that is equal to $3.0\times {10}^{8}m/s$.

Given that,

P = 60 W

r = 5 m

The formula used to calculate the intensity is:

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

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

The amplitude of electric field is:

$E{\sqrt{\frac{2I}{{\epsilon }_{0}c}}}_{max}$

Substitute the values

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

The amplitude of magnetic field is:

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

Substitute the values

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

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