Question

The electric field, which is a part of an electromagnetic wave propagating in a medium, is represented by,

$E_x=0$

$E_y=2.5\cos [(2\pi \times {10}^6)t-(\pi \times {10}^{-2})x]$

$E_z=0$

The wave is -

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

Answer 1

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

Given:

$E_x=0$

$E_y=2.5\cos [(2\pi \times {10}^6)t-(\pi \times {10}^{-2})x]$

$E_z=0$

On comparing the above equation with, $E_y=E_0\cos (\omega t-kx)$, we get,

$\omega =2\pi \times {10}^6\text{rad/s}$

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

Also, the wave is moving along $x-$direction.

Further,

$\omega =2\pi f\Rightarrow f=\frac{\omega }{2\pi }=\frac{2\pi \times {10}^6}{2\pi }={10}^6\text{Hz}$

Also,

$k=\frac{2\pi }{\lambda }\Rightarrow \lambda =\frac{2\pi }{k}=\frac{2\pi }{\pi \times {10}^{-2}}=200\text{m}$

Hence, option $\left(B\right)$ is the correct answer.