Suppose that the electric field amplitude of an electromagnetic

Solution:

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:

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

$\vec{B}=\left(4 \times 10^{-7}\right) \sin \left[1.05 x-3.14 \times 10^{8} t\right] \hat{k}$