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  • The electric field part of an electromagnetic wave in a medium is represented by <img alt="\mathrm{\mathrm{E}_{\mathrm{x}}=0; \mathrm{E}_{\mathrm{y}}=2.5 \mathrm{NC}^{-1} \times \cos \left[\left(2

The electric field part of an electromagnetic wave in a medium is represented by \mathrm{\mathrm{E}_{\mathrm{x}}=0; \mathrm{E}_{\mathrm{y}}=2.5 \mathrm{NC}^{-1} \times \cos \left[\left(2 \pi \times 10^6 \mathrm{rads}^{-1}\right) \mathrm{t}-\left(\pi \times 10^{-2} \mathrm{radm}^{-1}\right) \mathrm{x}\right] ; \mathrm{E}_{\mathrm{z}}=0}. The wave is:
 

Option: 1

moving along the \mathrm{x} - direction with frequency \mathrm{10^6 \mathrm{~Hz}} and wavelength \mathrm{100 \mathrm{~m}}


 


Option: 2

moving along \mathrm{x} - direction with frequency 10^6 \mathrm{~Hz} and wavelength 200 \mathrm{~m}
 


Option: 3

moving along \mathrm{x} - direction with frequency \mathrm{10^6 \mathrm{~Hz}} and wavelength 200 \mathrm{~m}
 


Option: 4

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


Answers (1)

Given, \mathrm{E}_{\mathrm{x}}=0,

\mathrm{E}_{\mathrm{y}}=2.5 \mathrm{NC}^{-1} \times \cos \left[\left(2 \pi \times 10^6 \mathrm{rads}^{-1}\right) \mathrm{t}-\left(\pi \times 10^{-2} \mathrm{radm}^{-1}\right) \mathrm{x}\right], \mathrm{E}_{\mathrm{z}}=0This shows that the wave is propagating along \mathrm{x}-axis. Comparing the given equation with \mathrm{E}=\mathrm{E}_0 \cos (\omega \mathrm{t}-\mathrm{kx}), we have

\omega=2 \pi \times 10^6

\mathrm{or \quad 2 \pi \mathrm{v}=2 \pi \times 10^6 \Rightarrow \mathrm{v}=10^6 \mathrm{~Hz}}

\mathrm{and \frac{2 \pi}{\lambda}=\mathrm{k}=\pi \times 10^{-2} \Rightarrow \lambda=\frac{2 \pi}{\pi \times 10^{-2}}=200 \mathrm{~m}}

Hence option 2 is correct.

Posted by

Kshitij

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