Question

The electric field part of an electromagnetic wave in a medium is represented by
$E_x=0;E_y=2.5{\text{NC}}^{-1}\cos [(2\pi \times {10}^6{\text{rad s}}^{-1})t-(\pi \times {10}^{-2}{\text{rad m}}^{-1})x];E_z=0.$ Then, the waves is

• A. Moving along $x-$direction with frequency ${10}^6\text{Hz}$ and wavelength $200\text{m}$
  
• B. Moving along $y-$direction with frequency $2\pi \times {10}^6\text{Hz}$ and wavelength $200\text{m}$
  
• C. Moving along negative $x-$ direction with frequency ${10}^6\text{Hz}$ and wavelength $200\text{m}$
• D. Moving along $x-$direction with frequency ${10}^6\text{Hz}$ and wavelength $100\text{m}$

Answer 1

The correct option is A Moving along $x-$direction with frequency ${10}^6\text{Hz}$ and wavelength $200\text{m}$


The standard equation of electromagnetic wave is,

$E_y=E_0\cos (\omega t-kx)[\text{along}x-direction]$

The given equation,

$E_y=2.5{\text{NC}}^{-1}\cos [(2\pi \times {10}^6{\text{rad s}}^{-1})t-(\pi \times {10}^{-2}{\text{rad m}}^{-1})x]$

Comparing with standard equation, we get,

$\omega =2\pi f=2\pi \times {10}^6\text{and}k=\pi \times {10}^{-2}$

$\Rightarrow f={10}^6\text{Hz}$

$\because k=\frac{2\pi }{\lambda }=\pi \times {10}^{-2}{\text{m}}^{-1}$

$\Rightarrow \lambda =200\text{m}$

Hence, the wave is moving along positive $x-$direction with frequency ${10}^6\text{Hz}$ and wavelength $200\text{m}.$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(A\right)$ is the correct answer.