Question

The intensity (I) of an em wave (having E₀ and B₀ as the ‘peak values’ for its electric and magnetic fields), can be expressed in the form

(1) \(\frac{ε_0E^2_0}{2c}\) or \(\frac{μ_0cB^2_0}{2}\)

(2) \(\frac{cε_0E^2_0}{2c}\) or \(\frac{μ_0cB^2_0}{2}\)

(3) \(\frac{c}{2}ε _0E^2_0\) or \(\frac{cB^2_0}{2μ_0}\)

(4) \(\frac{ε_0E^2_0}{2c}\) or \(\frac{cB^2_0}{2μ_0}\)

Answer 1

 (3) \(\frac{c}{2}ε _0E^2_0\) or \(\frac{cB^2_0}{2μ_0}\)

The intensity (at a given point), of an em wave, equals the energy received there per unit area per unit time. Consider an em wave, propagating along the x–axis; consider a cylinder, of unit area of cross section, having a length c (c = velocity of em waves). The axis of this cylinder is assumed to be parallel to the direction of propagation of the em wave. 

The volume of this cycilnder is (1 × c) = c 

The em wave (covering a distance c in one second) would transfer a total energy.

Through the cross section (= unity) of this cylinder in one second. 

Hence energy transferred per unit area per unit time.