the electric field vector in an electromagnetic wave travelling in free space has amplitude 120N/C.determine electrostatic energy density and the amplitude of the magnett field vector.

Some one please give this answer

Given,

E₀ = 120N/C, Let frequency V = 50. 0 MHZ

B₀ = \(\frac{E_0}{c}\)

B₀ = \(\frac{120}{3\times10^8}\)

B₀ = 40 x 10^-8

B₀ = 400 x 10^-9 T

ω = \(2\pi v\)

ω = 2\(\pi\) x 50 x 10⁶

ω = 3.14 x 10⁸ rad/5

\(K=\frac{w}{c}\)

\(K = \frac{3.14\times10^8}{3\times10^8}\)

K = 1.04 rad/m

Suppose electromagnetic wave is moving x direction, electric field be along y - direction and magnetic field alomg the z direction then,

E = E₀ sin (kx - ωt) \(\hat{j}\)

E = 120 sin (1.04 x - 3.14 x 10⁸ t) \(\hat{j}\)

and,

B = B₀ sin (kx - ωt) \(\hat{k}\)

B = 400 x 10^-9 sin (1.05 x - 3.14 x 10⁸ + t) \(\hat{k}\)