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A plane electromagnetic wave travelling along the x-direction has a wavelength of 3 \mathrm{~mm}. The variation in the electric field occurs in the y - direction with an amplitude 66 \mathrm{~V} \mathrm{~m}^{-1}. The equations for the electric and magnetic fields as a function of \mathrm{x \: and\: t} are respectively:
 

Option: 1

\mathrm{E}_{\mathrm{y}}=33 \cos \pi \times 10^{11}\left(\mathrm{t}-\frac{\mathrm{x}}{\mathrm{c}}\right), \mathrm{B}_{\mathrm{z}}=1.1 \times 10^{-7} \cos \pi \times 10^{11}\left(\mathrm{t}-\frac{\mathrm{x}}{\mathrm{c}}\right)
 


Option: 2

\mathrm{E}_{\mathrm{y}}=11 \cos 2 \pi \times 10^{11}\left(\mathrm{t}-\frac{\mathrm{x}}{\mathrm{c}}\right), \mathrm{B}_{\mathrm{y}}=11 \times 10^{-7} \cos 2 \pi \times 10^{11}\left(\mathrm{t}-\frac{\mathrm{x}}{\mathrm{c}}\right)
 


Option: 3

\mathrm{E}_{\mathrm{x}}=33 \cos \pi \times 10^{11}\left(\mathrm{t}-\frac{\mathrm{x}}{\mathrm{c}}\right), \mathrm{B}_{\mathrm{x}}=11 \times 10^{-7} \cos \pi \times 10^{11}\left(\mathrm{t}-\frac{\mathrm{x}}{\mathrm{c}}\right)
 


Option: 4

\mathrm{E}_{\mathrm{y}}=66 \cos 2 \pi \times 10^{11}\left(\mathrm{t}-\frac{\mathrm{x}}{\mathrm{c}}\right), \mathrm{B}_{\mathrm{z}}=2.2 \times 10^{-7} \cos 2 \pi \times 10^{11}\left(\mathrm{t}-\frac{\mathrm{x}}{\mathrm{c}}\right)


Answers (1)

best_answer

Here, \mathrm{E}_0=66 \mathrm{~V} \mathrm{~m}^{-1}, \lambda=3 \mathrm{~mm}=3 \times 10^{-3} \mathrm{~m}

\mathrm{\therefore \quad \mathrm{B}_0 =\frac{\mathrm{E}_0}{\mathrm{c}}=\frac{66}{3 \times 10^8}=2.2 \times 10^{-7} \mathrm{~T} }

\mathrm{\omega =2 \pi v=\frac{2 \pi \mathrm{c}}{\lambda}=\frac{2 \pi \times 3 \times 10^8}{3 \times 10^{-3}}=2 \pi \times 10^{11} }

\mathrm{\overrightarrow{\mathrm{E}} \text{is along }\mathrm{y} } - direction and the wave propagates along \mathrm{\mathrm{x}} - direction. Therefore, \mathrm{\overrightarrow{\mathrm{B}}} should be in a direction perpendicular to both \mathrm{\mathrm{x} \: and\: \mathrm{y}} - axes. Using vector algebra \mathrm{\overrightarrow{\mathrm{E}} \times \overrightarrow{\mathrm{B}}} should be along \mathrm{\mathrm{x}}- direction.

Since \mathrm{(\hat{i})=(+\hat{j}) \times(+\hat{k}), \vec{B}} is along the \mathrm{z} direction.

\mathrm{\therefore} The equation for the electric field along \mathrm{y} - direction in the electromagnetic wave is given by

\mathrm{ \mathrm{E}_{\mathrm{y}}=\mathrm{E}_0 \cos \omega\left(\mathrm{t}-\frac{\mathrm{x}}{\mathrm{c}}\right)=66 \cos 2 \pi \times 10^{11}\left(\mathrm{t}-\frac{\mathrm{x}}{\mathrm{c}}\right) }
and the equation for the magnetic field along \mathrm{ z } - direction in the electromagnetic wave is given by
\mathrm{ \mathrm{B}_{\mathrm{z}}=\mathrm{B}_0 \cos \omega\left(\mathrm{t}-\frac{\mathrm{x}}{\mathrm{c}}\right)=2.2 \times 10^{-7} \cos 2 \pi \times 10^{11}\left(\mathrm{t}-\frac{\mathrm{x}}{\mathrm{c}}\right) }

Hence option 4 is correct.

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Sayak

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