8.Electromagnetic waves
medium

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

A

$\mathrm{B}_{0} \frac{\hat{\mathrm{i}}-\hat{\mathrm{j}}}{\sqrt{2}} \cos \left(\omega \mathrm{t}-\mathrm{k} \frac{\hat{\mathrm{i}}+\hat{\mathrm{j}}}{\sqrt{2}}\right)$

B

$\mathrm{B}_{0} \frac{\hat{\mathrm{i}}+\hat{\mathrm{j}}}{\sqrt{2}} \cos \left(\omega \mathrm{t}-\mathrm{k} \frac{\hat{\mathrm{i}}+\hat{\mathrm{j}}}{\sqrt{2}}\right)$

C

$\mathrm{B}_{0} \hat{\mathrm{k}} \cos \left(\omega \mathrm{t}-\mathrm{k} \frac{\hat{\mathrm{i}}+\hat{\mathrm{j}}}{\sqrt{2}}\right)$

D

$\mathrm{B}_{0} \frac{\hat{\mathrm{j}}-\hat{\mathrm{i}}}{\sqrt{2}} \cos \left(\omega \mathrm{t}+\mathrm{k} \frac{\hat{\mathrm{i}}+\hat{\mathrm{j}}}{\sqrt{2}}\right)$

(JEE MAIN-2020)

Solution

Direction of polarisation $=\hat{\mathrm{E}}=\hat{\mathrm{k}}$

Direction of propagation $=\hat{\mathrm{E}} \times \hat{\mathrm{B}}=\frac{\hat{\mathrm{i}}+\hat{\mathrm{j}}}{\sqrt{2}}$

$\therefore \hat{\mathrm{E}} \times \hat{\mathrm{B}}=\frac{\hat{\mathrm{i}}+\hat{\mathrm{j}}}{\sqrt{2}}$

$\hat{\mathrm{B}}=\frac{\hat{\mathrm{i}}-\hat{\mathrm{j}}}{\sqrt{2}}$

Standard 12
Physics

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