Question

The electric field in an electromagnetic wave in vacuum is given by:
\(\vec{E} = 6.3 \, N/C} \left[\cos \left(1.5 \, rad/m} \cdot y + 4.5 \times 10^8 \, rad/s} \cdot t\right)\right] \hat{i}\) % Part (a) Wavelength and Frequency Calculation (a) Find the wavelength and frequency of the wave:

Hint:

Remember, in electromagnetic waves, the electric and magnetic fields oscillate perpendicular to each other and the direction of wave propagation, following a right-handed coordinate system. The magnitudes of \( \vec{E} \) and \( \vec{B} \) are related through the speed of light in vacuum.

Answer 1

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