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(B) Given,$\vec{E} = 90 \sin (0.5 	imes 10^{3} x + 1.5 	imes 10^{11} t) \hat{k} 	ext{ V/m}$.Comparing this with the standard wave equation $\vec{E}

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$\vec{B} = 3 	imes 10^{-7} \sin (0.5 	imes 10^{3} x + 1.5 	imes 10^{11} t) \hat{j} 	ext{ T}$

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(B) Given,$\vec{E} = 90 \sin (0.5 	imes 10^{3} x + 1.5 	imes 10^{11} t) \hat{k} 	ext{ V/m}$.Comparing this with the standard wave equation $\vec{E} = E_{0} \sin (kx + \omega t) \hat{n}$,we get $E_{0} = 90 	ext{ V/m}$.The direction of propagation is along $-\hat{i}$ (since the argument is $kx + \omega t$).The amplitude of the magnetic field is $B_{0} = \frac{E_{0}}{c} = \frac{90}{3 	imes 10^{8}} = 3 	imes 10^{-7} 	ext{ T}$.The direction of propagation is given by the direction of $\vec{E} 	imes \vec{B}$.Here,$\vec{E}$ is along $\hat{k}$ and the direction of propagation is $-\hat{i}$.So,$\hat{k} 	imes \hat{B} = -\hat{i}$.Since $\hat{k} 	imes \hat{j} = \hat{i}$,it follows that $\hat{k} 	imes (-\hat{j}) = -\hat{i}$.Therefore,the magnetic field $\vec{B}$ must be along $\hat{j}$ (Wait,checking cross product: $\hat{k} 	imes \hat{j} = -\hat{i}$,so $\vec{B}$ is along $\hat{j}$).Thus,$\vec{B} = 3 	imes 10^{-7} \sin (0.5 	imes 10^{3} x + 1.5 	imes 10^{11} t) \hat{j} 	ext{ T}$.

For a plane electromagnetic wave,the electric field is given by $\overrightarrow{E} = 90 \sin (0.5 \times 10^{3} x + 1.5 \times 10^{11} t) \hat{k} \text{ V/m}$. The corresponding magnetic field $\vec{B}$ will be

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