Q138 / 200 · NEET 2009 (Screening) · Physics
The electric field part of an electromagnetic wave in a medium is represented by
\(\begin{aligned} & E_x=0 \\ & E_y=2.5 \mathrm{NC}^{-1} \cos \left[\left(2 \pi \times 10^6 \mathrm{rad} \mathrm{s}^{-1}\right) \mathrm{t}-\left(\pi \times 10^{-2} \mathrm{rad} \mathrm{m}^{-1}\right) \mathrm{x}\right] \\ & E_z=0\end{aligned}\)
The wave is
- Correct answerAmoving along $y$ direction with frequency $2 \pi \times 10^6 \mathrm{~Hz}$ and wavelength $200 \mathrm{~m}$.
- Bmoving along $x$ direction with frequency $10^6 \mathrm{~Hz}$ and wavelength $100 \mathrm{~m}$
- Cmoving along $x$ direction with frequency $10^6 \mathrm{~Hz}$ and wavelength $200 \mathrm{~m}$
- Dmoving along $-x$ direction with frequency $10^6 \mathrm{~Hz}$ and wavelength $200 \mathrm{~m}$
Solution
Comparing the given equation
$$
\begin{aligned}
& \mathrm{E}_{\mathrm{y}}=2.5 \frac{\mathrm{N}}{\mathrm{C}} \cos \left[\left(2 \pi \times 10^6 \frac{\mathrm{rad}}{\mathrm{m}}\right) \mathrm{t}\right. \\
&\left.-\left(\pi \times 10^{-2} \frac{\mathrm{rad}}{\mathrm{sec}}\right) \mathrm{x}\right]
\end{aligned}
$$
With the standard equation
$$
\mathrm{E}_{\mathrm{y}}=\mathrm{E}_0 \cos (\omega \mathrm{t}-\mathrm{kx})
$$
we get
$$
\begin{array}{ll}
& \omega=2 \pi \mathrm{f}=2 \pi \times 10^6 \\
\therefore & \mathrm{f}=10^6 \mathrm{~Hz}
\end{array}
$$
Moreover, we know that
$$
\Rightarrow \quad \begin{aligned}
\frac{2 \pi}{\lambda} & =\mathrm{k}=\pi \times 10^{-2} \mathrm{~m}^{-1} \\
\Rightarrow \quad \lambda & =200 \mathrm{~m}
\end{aligned}
$$
Hence, the wave is moving along positive $x$-direction with frequency $10^6 \mathrm{~Hz}$ and wavelength $200 \mathrm{~m}$.
Hence, the correct answer is option: 'A'.