Question

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.

Short answer

The values of $E_{max}$and $B_{max}$are 12 V/m and$4\times {10}^{-8}T$ respectively.

Step-by-step solution

Define the intensity (l) and define the formulas.

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 $3.0\times {10}^{8}m/s$.

Determine the intensity of visible-light.

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 $0.19W/m^2$.

Determine the values of Emax and Bmax.

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.