Suppose that the electric field amplitude of an electromagnetic wave is $E_{0}=120 \mathrm{~N} / \mathrm{C}$ and that its frequency is $v=50.0$ $\mathrm{MHz}$. (a) Determine, $B_{0}, \omega, k$, and $\lambda$. (b) Find expressions for $\mathbf{E}$ and $\mathbf{B}$.
Electric field amplitude, E0 = 120 N/C
Frequency of source, ν = 50.0 MHz = 50 × 106 Hz
Speed of light, c = 3 × 108 m/s
(a) Magnitude of magnetic field strength is given as:
$B_{0}=\frac{E_{0}}{c}$
$=\frac{120}{3 \times 10^{8}}$
$=4 \times 10^{-7} \mathrm{~T}=400 \mathrm{nT}$
Angular frequency of source is given as:
ω = 2πν
= 2π × 50 × 106
= 3.14 × 108 rad/s
Propagation constant is given 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 as:
$\lambda=\frac{c}{v}$
$=\frac{3 \times 10^{8}}{50 \times 10^{6}}=6.0 \mathrm{~m}$
(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:
$\vec{E}=E_{0} \sin (k x-\omega t) \hat{j}$
$=120 \sin \left[1.05 x-3.14 \times 10^{8} t\right] \hat{j}$
And, magnetic field vector is given as:
$\vec{B}=B_{0} \sin (k x-\omega t) \hat{k}$
$\vec{B}=\left(4 \times 10^{-7}\right) \sin \left[1.05 x-3.14 \times 10^{8} t\right] \hat{k}$