A plane electromagnetic wave is propagating along the direction i^+j^2 , with its polarization along the direction k^ . The correct form of the magnetic field of the wave would be (here B0 is an appropriate constant)

Generally B→=B0.sinwt−kx in which direction of propagation is x axis.As per question direction of propagation is along i^+j^2 and Direction of polarization (direction of electric field) is k^ Using, relation E^×B^=C^ or B^=C^×E^ ⇒B^=i^+j^2×k^ ⇒B^=i^−j^2

A plane electromagnetic wave is propagating along the direction i^+j^2 , with its polarization along the direction k^ . The correct form of the magnetic field of the wave would be (here B0 is an appropriate constant)

Generally B→=B0.sinwt−kx in which direction of propagation is x axis.As per question direction of propagation is along i^+j^2 and Direction of polarization (direction of electric field) is k^ Using, relation E^×B^=C^ or B^=C^×E^ ⇒B^=i^+j^2×k^ ⇒B^=i^−j^2

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A plane electromagnetic wave is propagating along the direction $\frac{\hat{i} + \hat{j}}{\sqrt{2}}$ , with its polarization along the direction $\hat{k}$ . The correct form of the magnetic field of the wave would be (here $B_{0}$ is an appropriate constant)

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$B_{0} \hat{k} \cos (ω t - k \frac{\hat{i} + \hat{j}}{\sqrt{2}})$

$B_{0} \frac{\hat{i} + \hat{j}}{\sqrt{2}} \cos (ω t - k \frac{\hat{i} + \hat{j}}{\sqrt{2}})$

$B_{0} \frac{\hat{i} - \hat{j}}{\sqrt{2}} \cos (ω t - k \frac{\hat{i} + \hat{j}}{\sqrt{2}})$

$B_{0} \frac{\hat{j} - \hat{i}}{\sqrt{2}} \cos (ω t - k \frac{\hat{i} + \hat{j}}{\sqrt{2}})$

answer is C.

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Detailed Solution

Generally $\vec{B} = B_{0} . \sin (w t - k x)$ in which direction of propagation is x axis.
As per question direction of propagation is along $\frac{\hat{i} + \hat{j}}{\sqrt{2}}$
and Direction of polarization (direction of electric field) is $\hat{k}$
$U \sin g, r e l a t i o n \hat{E} \times \hat{B} = \hat{C} o r \hat{B} = \hat{C} \times \hat{E} \Rightarrow \hat{B} = (\frac{\hat{i} + \hat{j}}{\sqrt{2}}) \times \hat{k} \Rightarrow \hat{B} = \frac{\hat{i} - \hat{j}}{\sqrt{2}}$