Question

An industrial laser is used to burn a hole through a piece of metal. The average intensity of the light is = 2.93 × 109 W/m2. What is the rms value of (a) the electric field and (b) the magnetic field in the electromagnetic wave emitted by the laser?

Answer 1

To solve this problem, it is necessary to apply the concepts related to the electric field according to the intensity of the wave, the permittivity constant in free space and the speed of light.

As well as the expression of the rms of the magnetic field as a function of the electric field and the speed of light.

PART A) The expression for the rms of electric field is

$E_{rms} = \sqrt{\frac{S}{\epsilon_0 c}}$

Where,

S= Intensity of the wave

$\epsilon_0$= Permitivitty at free space

c = Light speed

Replacing we have that,

$E_{rms} = \sqrt{\frac{(2.93*10^9)}{(8.85*10^{-12})(3*10^8)}}$

$E_{rms} = 1.05*10^6N/C$

The RMS value of electric field is $1.05*10^6N/C$

PART B) The expression for the RMS of magnetic field is,

$B_{rms} = \frac{E_{rms}}{c}\\B_{rms} = \frac{1.05*10^6}{3*10^8}\\B_{rms} =3.5*10^{-3}T$

The RMS of the magnetic field is $3.5*10^{-3}T$