Electric field amplitude is given as $E_{0}=120 \mathrm{~N} / C$

Frequency of source is given as $v=50 \mathrm{MHz}=50 \times 10^{6} \mathrm{~Hz}$

Speed of light is known as $c=3 \times 10^{8} \mathrm{~m} / \mathrm{s}$

(a) Magnitude of magnetic field strength can be calculated from the expression,

$B_{0}=\frac{E_{0}}{c}$

$=\frac{120}{3 \times 10^{8}}$

$=40 \times 10^{-8}=400 \times 10^{-9} T=400 n T$

Angular frequency of source can be calculated from the expression,

$\omega=2 \pi v=2 \pi \times 50 \times 10^{6}=3.14 \times 10^{8} \mathrm{rads}^{-1}$

$=3.14 \times 10^{8} \mathrm{rad} / \mathrm{s}$

Propagation constant can be calculated as,

$k=\frac{\omega}{c}$

$=\frac{3.14 \times 10^{8}}{3 \times 10^{8}}=1.05 \mathrm{rad} / \mathrm{m}$

Wavelength of wave is given by the relation,

$\lambda=\frac{c}{v}$

$=\frac{3 \times 10^{8}}{50 \times 10^{6}}=6.0 \mathrm{~m}$

(b) Assume that the wave is moving in a positive x-direction. The magnetic field vector will then be positive z-direction and the electric field vector will be positive y-direction. This is due to the fact that all three vectors are perpendicular to one another.

Equation of electric field vector is given as:

$\bar{E}=E_{0} \sin (k x-\omega t) \hat{j}$

$=120 \sin \left[1.05 x-3.14 \times 10^{8} t\right] \widehat{j}$

And, magnetic field vector is given as:

$\bar{B}=B_{0} \sin (k x-\omega t) \widehat{k}$

$$
\bar{B}=\left(400 \times 10^{-9}\right) \sin \left[1.05 x-3.14 \times 10^{8} t\right] \widehat{k}
$$