Question

The energy flow per unit time per unit area (S) of an electromagnetic wave has an average value of 440 mW/m2. The maximum value of the magnetic field in the wave is closest:

Answer 1

To solve this problem we will proceed to convert the Intensity Units given to a normal system, that is, to standardize their value. Later we will use the value of the electric field according to the RMS electric field, to find it.

Finally the magnetic field would be given based on the previously found electric field, depending on the speed of light:

The value of the intensity is:

$I = 440mW/m^2 = 0.440W/m^2$

The maximum value of the electric field is,

$E_{max} = \sqrt{2}E_{rms}$

$E_{max} = \sqrt{2}\sqrt{Ic\mu_0}$

Where,

I = Intensity,

$\mu$= Permeability constant

c = Speed of light

Replacing,

$E_{max} = \sqrt{2}\sqrt{(0.440)(3*10^8)(4\pi*10^{-7})}$

$E_{max} = 18.2V/m$

The maximum value of the magnetic field is,

$B_{max} = \frac{E_{max}}{c}$

$B_{max} = \frac{18.2}{3*10^8}$

$B_{max} = 6.06*10^{-8}T$

Therefore the maximum magnetic field is $7.10*10^{-8}T$