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The magnetic field of a plane electromagnetic wave is given by $\bar{B} = 3 \times 10^{- 8} [\sin (200 π (y + c t))] \hat{i} T$ . Where $C = 3 \times 10^{- 8} m / s$ is the speed of light. The corresponding electric field is

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$\bar{E} = - 9 \sin [200 π (y + c t)] \hat{k} V / m$

$\bar{E} = - 10^{- 6} \sin [\sin (200 π) (y + c t)] \hat{k} V / m$

$\bar{E} = 9 \sin [200 π (y + c t)] \hat{k} V / m$

$\bar{E} = 3 \times 10^{- 8} [\sin (200 π (y + c t))] \hat{k} V / m$

answer is D.

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Detailed Solution

$E_{0} = {cB}_{0} = 9 V / m$

$\vec{E} \times \vec{B}$ gives us the direction of propagation of wave.

Hence the electric field will be along -z-axis.

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