Suppose that the electric field amplitude of an electromagnetic wave is E₀ = 120 N/C and that its frequency is v = 50.0 MHz. (a) Determine, B₀, ω, k, and λ. (b)    Find expressions for E and B.

Given,
Amplitude of EM waves, E_o = 120 N/C
Frequency of wave, v = 50.0 MHz 

(a) Using the relation, we have 

               $\boxed{\frac{{\mathrm{E}}_0}{{\mathrm{B}}_0} = \mathrm{C}}$

$\Rightarrow$              ${\mathrm{B}}_0 = \frac{{\mathrm{E}}_0}{\mathrm{C}} = \frac{120}{3 \times {10}^8}\mathrm{T}$

                               $$\begin{gathered} = 40 \times {10}^{-8}\mathrm{T} \\ = 400 \times {10}^{-9}\mathrm{T} \\ = 400 \mathrm{nT} \end{gathered}$$ 

which is the required amplitude of magnetic field.

Angular frequency is given by, 

                  $$\begin{gathered} \boxed{\mathrm{\omega } = 2\mathrm{\pi } \mathrm{v}} \\ = 2\mathrm{\pi } \times 50 \times {10}^6 \\ = 3.14 \times {10}^8 \mathrm{rad} {\mathrm{s}}^{-1} \end{gathered}$$

Wave vector k is given by, 

           $$\begin{gathered} \boxed{\mathrm{k} = \frac{2\mathrm{\pi }}{\mathrm{\lambda }}} \\ = \frac{2\mathrm{\pi } \mathrm{v}}{\mathrm{v\lambda }} = \frac{2\mathrm{\pi } \mathrm{v}}{\mathrm{C}} \\ =\frac{\mathrm{\omega }}{\mathrm{C}} = \frac{\mathrm{\pi } \times {10}^8}{3 \times {10}^8} \mathrm{rad} {\mathrm{m}}^{-1} \end{gathered}$$

              $= \frac{\mathrm{\pi }}{3}\mathrm{rad} {\mathrm{m}}^{-1} = 1.05 \mathrm{rad} {\mathrm{m}}^{-1}$ 

Wavelength of the EM wave is,
                 
                  $$\begin{gathered} \boxed{\mathrm{C}= \mathrm{v\lambda };} \\ \mathrm{\lambda } = \frac{\mathrm{C}}{\mathrm{v}} = \frac{3 \times {10}^8}{50 \times {10}^6}\mathrm{m} \\ = \frac{300}{50} = 6\mathrm{m}. \end{gathered}$$ 

(b) Let the electromagnetic wave travel along +x-axis, and $\overrightarrow{\mathrm{E}} \mathrm{and} \overrightarrow{\mathrm{B}}$ are along y-axis and z-axis respectivelty.
Then, 

$\boxed{\overrightarrow{{\mathrm{E}}_{\mathrm{y}}} = {\mathrm{E}}_0 \sin (\mathrm{kx}-\mathrm{\omega t}) \stackrel{⏜}{\mathrm{j}}}$
    $= 120 \sin (1.05 \mathrm{x} - 3.14 \times {10}^8\mathrm{t}) \hat{\mathrm{j}} {\mathrm{NC}}^{-1}$ and, 

$$\begin{gathered} \overrightarrow{{\mathrm{B}}_{\mathrm{z}}} = {\mathrm{B}}_0 \sin (\mathrm{kx} - \mathrm{\omega t}) \hat{\mathrm{k}} \\ = 400 \sin (1.05\mathrm{x} - 3.14 \times {10}^8 \mathrm{t}) \stackrel{⏜}{\mathrm{k}} \mathrm{nT} \end{gathered}$$ 

The above expressions represent electric and magnetic field.