Suppose that the electric field amplitude of an electromagnetic wave is E₀ = 120 N/c and its frequency is v = 50 MHZ. The expression for $\vec{E}$ is:

$120 \sin \left(1.05 x-3.14 \times 10^8 t\right)$

$B_0=\frac{E_0}{c}= \frac{120}{3 \times 10^8}=4 \times 10^{-7} \mathrm{~T}$

$\omega=2 \pi v=3.14 \times 10^8$ rad/s

$\lambda =\frac{c}{v}=\frac{3 \times 10^8}{5 \times 10^7}$ = 6 m

$k =\frac{2 \pi}{\lambda}$ = 1.05 rad/m

E = $E_0 \sin (kx-\omega t) $

$=120 \sin \left(1.05 x-3.14 \times 10^8 t\right)$