Chapter 22, Problem 22P

To determine

The average values of $\text{E}$ and $\text{B}$ in the beam.

Answer to Problem 22P

The average values of $\text{E}$ and $\text{B}$ given as $\text{1420}\text{V/m}$ , $\text{4}{\text{.72×10}}^{\text{-6}}\text{T}$ respectively.

Explanation of Solution

Given info:

Power, $\text{P=12}\text{.8}\text{mW}$

Diameter, $\text{d=1}\text{.75}\text{mm }$

Formula used:

  $\text{S=}\frac{\text{P}}{\text{A}}{\text{=cε}}_{\text{0}}{\text{E}}^{\text{2}}$

Where, $\text{P}$ is the power of the laser beam, $\text{c}$ is the speed of light and ${\text{ε}}_{\text{0}}$ is the constant, $\text{A}$ is the area, $\text{E}$ is the energy.

Also,

  ${\text{ε}}_{\text{0}}$ = $\text{8}{\text{.85×10}}^{\text{-12}}{\text{C}}^{\text{2}}\text{/N}\cdot {\text{m}}^{\text{2}}$

  $\text{c}$ = ${\text{3×10}}^{\text{8}}\text{m/s}$

Calculation:

Substituting the given values,

  $\text{S=}\frac{\text{P}}{\text{A}}{\text{=cε}}_{\text{0}}{\text{E}}^{\text{2}}$

  $\frac{\text{12}{\text{.8×10}}^{\text{-3}}\text{W}}{\left[\frac{\text{π}}{\text{4}}{\left(\text{1}{\text{.75×10}}^{\text{-3}}\text{m}\right)}^{\text{2}}\right]}$ $\text{=}\left({\text{3×10}}^{\text{8}}\text{m/s}\right)\left(\text{8}{\text{.85×10}}^{\text{-12}}{\text{C}}^{\text{2}}\text{/}\text{N}\cdot {\text{m}}^{\text{2}}\right){\text{E}}_{\text{rms}}{}^2$

  ${\text{E}}_{\text{rms}}\text{=1420}\text{V/m}$

The average value for B is calculating by the

  ${\text{B}}_{\text{rms}}\text{=}\frac{{\text{E}}_{\text{rms}}}{\text{c}}$

  $\text{=}\frac{\text{1420}\text{V/m}}{{\text{3×10}}^{\text{8}}\text{m/s}}\text{=}\text{4}{\text{.72×10}}^{\text{-6}}\text{T}$

  $\text{=14}{\text{.94×10}}^{\text{-6}}\text{T}$

Conclusion:

Theaverage value of E $\text{=1420}\text{V/m}$

The average value of B $\text{=}\text{4}{\text{.72×10}}^{\text{-6}}\text{T}$