Suppose that the electric field amplitude of an electromagnetic wave propagating along x-direction is ${\mathrm{E}}_0=120 \mathrm{N} {\mathrm{C}}^{-1}$ and its frequency is $\nu =50 \mathrm{MHz}$. choose the correct statement(s).

Since the wave is propagating along x-direction, therefore, electric field $\overrightarrow{E}$is along y - direction and magnetic field $\overrightarrow{B}$is along z-direction.

$\because \mathrm{k}=\frac{2\mathrm{\pi }}{\mathrm{\lambda }}=\frac{2\mathrm{\pi }\nu }{\mathrm{c}}=\frac{2\mathrm{\pi }\times \left(\mathbf{50}\mathbf{\times }{\mathbf{10}}^{\mathbf{6}}\right)}{3\times {10}^8}=\frac{\mathrm{\pi }}{3}$

$\overrightarrow{\mathrm{E}}={\mathrm{E}}_0\sin (\mathrm{kx}-\mathrm{\omega t})\hat{\mathrm{j}}=120\sin \left(\frac{\mathrm{\pi }}{3}\mathrm{x}-100\mathrm{\pi }\times {10}^6\mathrm{t}\right)\hat{\mathrm{j}}$

As ${\mathrm{B}}_0=\frac{{\mathrm{E}}_0}{\mathrm{c}}=\frac{120}{3\times {10}^8}=40\times {10}^{-8} \mathrm{T}$

$\therefore \overrightarrow{\mathrm{B}}={\mathrm{B}}_0\sin \left(\mathrm{kx}-\mathrm{\omega t}\right)\hat{\mathrm{k}}=40\times {10}^{-8} \sin \left(\frac{\mathrm{\pi }}{3}\mathrm{x}-\mathrm{\pi }\times {10}^8\mathrm{t}\right)\hat{\mathrm{k}}$