Suppose that the electric field amplitude of an electromagnetic wave is $E_{0} = 120 \; N/C$ and that its frequency is $\nu = 50.0 \; MHz$.
$(a)$ Determine $B_{0}, \omega, k,$ and $\lambda$.
$(b)$ Find expressions for $E$ and $B$.

(N/A) Electric field amplitude,$E_{0} = 120 \; N/C$.
Frequency of source,$\nu = 50.0 \; MHz = 50 \times 10^{6} \; Hz$.
Speed of light,$c = 3 \times 10^{8} \; m/s$.
$(a)$ Magnitude of magnetic field strength is given as:
$B_{0} = \frac{E_{0}}{c} = \frac{120}{3 \times 10^{8}} = 4 \times 10^{-7} \; T = 400 \; nT$.
Angular frequency of source is given as:
$\omega = 2 \pi \nu = 2 \pi \times 50 \times 10^{6} = 3.14 \times 10^{8} \; rad/s$.
Propagation constant is given as:
$k = \frac{\omega}{c} = \frac{3.14 \times 10^{8}}{3 \times 10^{8}} = 1.05 \; rad/m$.
Wavelength of wave is given as:
$\lambda = \frac{c}{\nu} = \frac{3 \times 10^{8}}{50 \times 10^{6}} = 6.0 \; m$.
$(b)$ Suppose the wave is propagating in the positive $x$-direction. Then,the electric field vector will be in the positive $y$-direction and the magnetic field vector will be in the positive $z$-direction. This is because all three vectors are mutually perpendicular. Equation of electric field vector is given as:
$\vec{E} = E_{0} \sin(kx - \omega t) \hat{j} = 120 \sin(1.05x - 3.14 \times 10^{8}t) \hat{j} \; V/m$.
And,magnetic field vector is given as:
$\vec{B} = B_{0} \sin(kx - \omega t) \hat{k} = (4 \times 10^{-7}) \sin(1.05x - 3.14 \times 10^{8}t) \hat{k} \; T$.