JEE Mains · Physics · STD 12 - 8. Electromagnetic waves
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 \(\mathrm{B}_{0} \frac{\hat{\mathrm{j}}-\hat{\mathrm{i}}}{\sqrt{2}} \cos \left(\omega \mathrm{t}+\mathrm{k} \frac{\hat{\mathrm{i}}+\hat{\mathrm{j}}}{\sqrt{2}}\right)\)
Answer & Solution
Correct Answer
(A) \(\mathrm{B}_{0} \frac{\hat{\mathrm{i}}-\hat{\mathrm{j}}}{\sqrt{2}} \cos \left(\omega \mathrm{t}-\mathrm{k} \frac{\hat{\mathrm{i}}+\hat{\mathrm{j}}}{\sqrt{2}}\right)\)
Step-by-step Solution
Detailed explanation
Direction of polarisation \(=\hat{\mathrm{E}}=\hat{\mathrm{k}}\) Direction of propagation \(=\hat{\mathrm{E}} \times \hat{\mathrm{B}}=\frac{\hat{\mathrm{i}}+\hat{\mathrm{j}}}{\sqrt{2}}\)…
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