Question -
Answer -
Frequency of the electromagneticwave,┬а╬╜┬а= 2.0 ├Ч 1010┬аHz
Electricfield amplitude,┬аE0┬а= 48 V mтИТ1
Speed oflight,┬аc┬а= 3 ├Ч 108┬аm/s
(a)┬аWavelength of a wave is given as:
(b)┬аMagneticfield strength is given as:
(c)┬аEnergydensity of the electric field is given as:
And, energy density of themagnetic field is given as:
Where,
тИИ0┬а= Permittivity of free space
╬╝0┬а= Permeability offree space
We havethe relation connecting┬аE┬аand┬аB┬аas:
E┬а=┬аcB┬атАж (1)
Where,
┬атАж (2)
Puttingequation (2) in equation (1), we get
Squaring both sides, we get
