If Electric field intensity of a uniform plane electro magnetic wave is given as E = - 301.6 sin(kz - ωt) \(\widehat{a}\)x + 452.4 sin (kz - ωt) \(\widehat{a}\)y \(\frac{V}{M}\). 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;

          \(\vec H = \frac{\vec B}{μ_o}\)

          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;

           \(\vec B = \frac{\vec E}{c} (\hat n \times \vec k)\)

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

CALCULATIONS:

Given :  E = - 301.6 sin(kz - ωt) \(\widehat{a}\)x + 452.4 sin (kz - ωt) \(\widehat{a}\)y \(\frac{V}{M}\)

Now, the magnetic field is written as;

\(\vec B = \frac{ - 301.6 sin(kz - ωt) (\vec x \times \vec k)+ 452.4 sin (kz - ωt) (\vec y \times \vec k) }{c} \)

⇒ \(\vec B = \frac{1}{c} (- 301.6 sin(kz - ωt) (\vec x \times \vec k)+ 452.4 sin (kz - ωt) (\vec y \times \vec k)) \)

⇒ \(\vec B = \frac{1}{c} (- 301.6 sin(kz - ωt) (-\hat a_y)+ 452.4 sin (kz - ωt)( - \hat a _x)) \)      -----(1)

Now magnetic intensity H is written as;

\(\vec H = \frac{\vec B}{μ_o}\)      -----(2)

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

\(\vec H = \frac{\frac{1}{c} (- 301.6 sin(kz - ωt) (-\hat a_y)+ 452.4 sin (kz - ωt)( - \hat a _x))}{μ_o}\)

Now, c = 3 \(\times\) 108 and μo = \(4\pi \times 10^{-7}\) N/A2

Now, \(\vec H = \frac{ (- 301.6 sin(kz - ωt) (-\hat a_y)+ 452.4 sin (kz - ωt)( - \hat a _x))}{3 \times 10 ^8 \times 4 \pi \times 10^{-7}}\)

⇒ \(\vec H \) = -0.8 sin (kz - ωt) \(\widehat{a}\)y - 1.2 sin (kz - ωt) \(\widehat{a}\)x

Hence, the correct answer is option 3).