The magnetic field of a plane electromagnetic wave is    \overrightarrow{B}=3\times 10^{-8}sin \left ( 200\pi \left ( y+ct \right ) \right ) \hat{i} \ T where c=3\times 10^{8}m/s is the speed of the light. The corresponding Electric field is
Option: 1 \overrightarrow{E}=9sin\left ( 200\pi \left ( y+ct \right ) \right) \hat{k} \ V/m
 
Option: 2 \overrightarrow{E}=-10^{-6}sin \left ( 200\pi \left ( y+ct \right ) \right )\hat{K} \ V/m
Option: 3 \overrightarrow{E}=-9sin \left ( 200\pi \left ( y+ct \right ) \right )\hat{K} \ V/m  
Option: 4 \overrightarrow{E}=3\times 10^{-8}sin \left ( 200\pi \left ( y+ct \right ) \right )\hat{K} \ V/m

Answers (1)

\\ \text{We know} |\mathrm{E}|=|\mathrm{B}| \times|\mathrm{C}| \\ where \ \mathrm{E} \ and \ \mathrm{B} \ \text{are the amplitude of electric and magnetic field intensities in an electromagnetic wave}. \\ Given \ \ |\mathrm{B}|=3 \times 10^{-8} \\ \therefore|\mathrm{E}|=|\mathrm{B}| \times|\mathrm{C}|=3 \times 10^{-8} \times 3 \times 10^{8}=9 \ V/m

As,

\overrightarrow{\mathbf{B}}=3 \times 10^{-8} \sin (200 \pi(y+ct)) \hat{\mathrm{i}} \ \mathrm{T}

Now,

\begin{aligned} &\text { and direction of wave propagation is given as }\\ &(\overrightarrow{\mathrm{E}} \times \overrightarrow{\mathrm{B}}) \| \overrightarrow{\mathrm{C}}\\ &\hat{\mathrm{B}}=\hat{\mathrm{i}} \quad \& \quad \hat{\mathrm{C}}=-\hat{\mathrm{j}}\\ \end{aligned}

-\hat{\mathbf{j}}=\hat{\mathbf{E}} \times \hat{\mathbf{i}}, \\ \therefore \hat{\mathbf{E}}=- \hat{\mathbf{K}}

So,

\therefore \mathrm{\vec{E}}= - 9 \sin (200 \pi(y+ct)) \hat{\mathrm{k}}

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