Here, we are given with:
[tex]{:\implies \quad \longrightarrow \begin{cases}\sf E_{0}= 50\: V/m\\ \sf \nu = 2\times 10^{7}Hz\end{cases}}[/tex]
(a) So now, we can thus obtain
[tex]{:\implies \quad \sf \omega =2\pi \nu}[/tex]
[tex]{:\implies \quad \sf \omega =2\pi (2\times 10^{7})}[/tex]
[tex]{:\implies \quad \boxed{\bf{\omega = 4\pi \times 10^{7}\: s^{-1}}}}[/tex]
Now, finding [tex]\lambda[/tex]
[tex]{:\implies \quad \sf \lambda =\dfrac{c}{\omega}=\dfrac{3\times 10^{8}}{4\pi \times 10^{7}}}[/tex]
[tex]{:\implies \quad \boxed{\bf{\lambda \approx 2.39\:m}}}[/tex]
Now, finding k
[tex]{:\implies \quad \sf k=\dfrac{2\pi}{\lambda}=\dfrac{200\pi}{239}}[/tex]
[tex]{:\implies \quad \boxed{\bf{k\approx 2.63\:m^{-1}}}}[/tex]
Thus, expression for the electric field is:
[tex]{:\implies \quad \boxed{\bf{E=50\sin \bigg(4\pi \times 10^{7}t-\dfrac{263x}{100}\bigg)}}}[/tex]
(b) Now, here
[tex]{:\implies \quad \sf B_{0}=\dfrac{E_{0}}{c}=\dfrac{50}{3\times 10^{8}}}[/tex]
[tex]{:\implies \quad \boxed{\bf{B_{0}\approx 16.67×10^{-7}\:T}}}[/tex]
Thus, expression for the magnetic field:
[tex]{:\implies \quad \boxed{\bf{B_{y}=16.67\times 10^{-7}\sin \bigg(4\pi \times 10^{7}t-\dfrac{263x}{100}\bigg)\:T}}}[/tex]
(c) The electromagnetic wave propagates along Z-axis
