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(A) The direction of propagation of an electromagnetic wave is given by the direction of the vector product $\overrightarrow{E} 	imes \overrightarrow

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$2 	imes 10^{-8} \; T$ along the $z$-direction

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(A) The direction of propagation of an electromagnetic wave is given by the direction of the vector product $\overrightarrow{E} 	imes \overrightarrow{B}$.Given that the wave propagates along the $x$-direction $(\hat{i})$ and the electric field is along the $y$-direction $(\hat{j})$, we have:$\hat{i} = \hat{j} 	imes \hat{B}$This implies that $\hat{B} = \hat{k}$, so the magnetic field is along the $z$-direction.The relationship between the magnitudes of the electric field $(E)$ and the magnetic field $(B)$ in free space is given by $C = \frac{E}{B}$, where $C = 3 	imes 10^8 \; ms^{-1}$ is the speed of light.Therefore, $B = \frac{E}{C} = \frac{6}{3 	imes 10^8} = 2 	imes 10^{-8} \; T$.Thus, the magnetic field component is $2 	imes 10^{-8} \; T$ along the $z$-direction.

$A$ plane electromagnetic wave travels in free space along the $x$-direction. The electric field component of the wave at a particular point of space and time is $E = 6 \; Vm^{-1}$ along the $y$-direction. Its corresponding magnetic field component, $B$, would be:

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