If Electric field intensity of a uniform plane electro magnetic wave is given as E = - 301.6 sin(kz - ωt) x + 452.4 sin (kz - ωt) y . Then, magnetic intensity ‘H’ of this wave in Am^-1 will be :

[Given : Speed of light in vacuum c = 3 × 10⁸ ms^-1, Permeability of vacuum μ₀ = 4π × 10^-7 NA^-2]

CONCEPT:

• Magnetic intensity is written as;

          Here we have H as the magnetic intensity, B as the magnetic field and μ_o is the permeability of the free space.

• The magnetic field is written as;

           Here E is the electric field, and c is the velocity of light.

CALCULATIONS:

Given :  E = - 301.6 sin(kz - ωt) x + 452.4 sin (kz - ωt) y 

Now, the magnetic field is written as;

⇒ 

⇒       -----(1)

Now magnetic intensity H is written as;

      -----(2)

Now, on putting the value of equation (1) in (2) we have;

Now, c = 3  108 and μo =  N/A2

Now, 

⇒  = -0.8 sin (kz - ωt) y - 1.2 sin (kz - ωt) x

Hence, the correct answer is option 3).