Question

Suppose that the electric field amplitude of an electromagnetic wave is $E_0 = 120 \ N/C$ and that its frequency is $ν = 50.0 \ MHz$.

1. Determine, B0, ω, k, and λ.
2. Find expressions for E and B.

Answer 1

Electric field amplitude, $E_0 = 120\ N/C$
Frequency of source, $ν = 50.0 \ MHz = 50 × 10^6 Hz$
Speed of light, $c = 3 × 10^8 m/s$

---

(a) Magnitude of magnetic field strength is given as:

$B_o =\frac { E_o}{c}$

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

$Bo = 4 \times 10^{-7} T$

$B_o = 400 \ nT$

Angular frequency of source is given as:
$ω = 2\piν = 2\pi × 50 × 10^6$
$ω = 3.14 × 10^8 rad/s$
Propagation constant is given as:

$k = \frac {ω}{c}$

$k = \frac {3.14 \times 10^8}{3\times10^8}$

$k = 1.05 \ rad/m$

Wavelength of wave is given as:

$λ = \frac cv$

$λ =\frac { 3 \times 10^8}{50\times 10^6}$

$λ = 6.0 \ m$

---

(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:
$\vec E = E_o sin \ (kx-ωt)\hat j$
$\vec E= 120 sin\ (1.05x - 3.14 x 10^8t)\hat j$
And, magnetic field vector is given as:
$\vec B = B_o sin \ (kx-ωt)\hat k$
$\vec B = (4\times 10^{-7}) sin \ (1.05x - 3.14 \times 10^8t)\hat k$