Given the wave number \( k = 1.5 \, rad/m} \), the wavelength \( \lambda \) is calculated using the relationship \( \lambda = \frac{2\pi}{k} \).
Calculating \( \lambda \):
\[
\lambda = \frac{2\pi}{1.5} \approx 4.19 \, m}
\]
The provided angular frequency is \( \omega = 4.5 \times 10^8 \, rad/s} \). The frequency \( f \) is determined by:
\[
f = \frac{\omega}{2\pi} = \frac{4.5 \times 10^8}{2\pi} \approx 7.16 \times 10^{-1} \, Hz}
\]
\textbf{(b) Determine the amplitude of the magnetic field of the wave.}
Solution:
The magnetic field amplitude \( B_0 \) is related to the electric field amplitude \( E_0 \) by:
\[
B_0 = \frac{E_0}{c}
\]
Given \( E_0 = 6.3 \, N/C} \) and \( c = 3 \times 10^8 \, m/s} \) (the speed of light):
\[
B_0 = \frac{6.3}{3 \times 10^8} = 2.1 \times 10^{-8} \, T}
\]
\textbf{(c) Formulate an expression for the magnetic field of this wave.}
Solution:
Since the magnetic field \( \mathbf{B} \) is perpendicular to \( \mathbf{E} \) and the wave propagates in the z-direction, the magnetic field vector can be represented as:
\[
\mathbf{B} = 2.1 \times 10^{-8} \cos \left( 1.5 \, rad/m} \cdot y + 4.5 \times 10^8 \, rad/s} \cdot t \right) \hat{j}
\]
The unit vector \( \hat{j} \) indicates that \( \mathbf{B} \) is perpendicular to \( \mathbf{E} \), consistent with the right-hand rule for electromagnetic waves.