Properties of EM Waves: Practice Problem & Solution
The electric field part of an electromagnetic wave in a medium is represented by $E_x = 0$ $E_y = 2.5\cos\left[\left(2\pi \times 10^6 \frac{rad}{m}\right)t - \left(\pi \times 10^{-2} \frac{rad}{s}\right)x\right]$ $E_z = 0$ . The wave is: (2009)
Solution Explained:
To solve this problem, we apply the core principles of Properties of EM Waves. Understanding the underlying formula is key to arriving at the correct answer below:
Comparing with $E_y = E_0\cos(\omega t - kx)$ , we find $\omega = 2\pi \times 10^6 rad/s$ and $k = \pi \times 10^{-2} rad/m$ . Frequency $f = \frac{\omega}{2\pi} = 10^6 Hz$ . Wavelength $\lambda = \frac{2\pi}{k} = \frac{2\pi}{\pi \times 10^{-2}} = 200 m$ . The negative sign between the temporal and spatial terms indicates it moves in the +x direction.
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