The electric field part of an electromagnetic wave in a medium is represented by aligned & E_x=0 & E_y=2.5 NC^ -1 [(2 π × 10^6 rad s^ -1 ) t-(π × 10^ -2 rad m^ -1 ) x] & E_z=0 aligned ) The wave is

Correct answer: A. moving along y direction with frequency 2 π × 10^6 ~Hz and wavelength 200 ~m . Comparing the given equation aligned & E_ y =2.5 N C [(2 π ×10^6 rad m ) t. &.-(π ×10^ -2 rad sec ) x] aligned With the standard equation E_ y =E_ 0 (ω t-kx)we get array ll & ω=2 π f=2 π ×10^6 & f=10^6 Hz array Moreover, we know that aligned (2 π)/(λ)& = k=π ×10^ -2 m^ -1 λ & =200 m aligned Hence, the wave is moving along positive x -direction with frequency 10^ 6 Hz and wavelength 200 m.

NEET 2009

Exam: NEET
Year: 2009
Subject: Physics
Chapter: Electromagnetic Waves

NEET 2009 · Physics · Electromagnetic Waves

The electric field part of an electromagnetic wave in a medium is...

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

Options

Amoving 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 $200 \mathrm{~m}$

Correct answer

A. moving along $y$ direction with frequency $2 \pi \times 10^6 \mathrm{~Hz}$ and wavelength $200 \mathrm{~m}$ .

Step-by-step solution

Comparing the given equation

\begin{aligned}& \mathrm{E}_{\mathrm{y}}=2.5 \frac{\mathrm{N}}{\mathrm{C}} \cos \left[\left(2 \pi \times10^6 \frac{\mathrm{rad}}{\mathrm{m}}\right) \mathrm{t}\right.\\ &\left.-\left(\pi \times10^{-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 \times10^6 \\ \therefore & \mathrm{f}=10^6 \mathrm{ Hz} \end{array}

Moreover, we know that \Rightarrow \quad

\begin{aligned} \frac{2 \pi}{\lambda}& = \mathrm{k}=\pi \times10^{-2} \mathrm{ m}^{-1}\\ \Rightarrow \quad \lambda & =200 \mathrm{ m} \end{aligned}

Hence,

the wave is moving along positive x

-direction with frequency 10^

\mathrm{ Hz}

and wavelength 200

\mathrm{ m}.

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