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JEE Mains · Physics · STD 12 - 8. Electromagnetic waves

एक समतल विधुत चुम्बकीय तरंग, जो निर्वात में \(x\) दिशा में चल रही है, का विधुत क्षेत्र \(\overrightarrow{ E }= E _{0} \hat{ j } \cos (\omega t - kx )\). है। समय \(t =0\) पर इसका चुम्बकीय क्षेत्र होगा ।

  1. A \(\mathrm{B}_{0} \frac{\hat{\mathrm{i}}-\hat{\mathrm{j}}}{\sqrt{2}} \cos \left(\omega \mathrm{t}-\mathrm{k} \frac{\hat{\mathrm{i}}+\hat{\mathrm{j}}}{\sqrt{2}}\right)\)
  2. B \(\mathrm{B}_{0} \frac{\hat{\mathrm{i}}+\hat{\mathrm{j}}}{\sqrt{2}} \cos \left(\omega \mathrm{t}-\mathrm{k} \frac{\hat{\mathrm{i}}+\hat{\mathrm{j}}}{\sqrt{2}}\right)\)
  3. C \(\mathrm{B}_{0} \hat{\mathrm{k}} \cos \left(\omega \mathrm{t}-\mathrm{k} \frac{\hat{\mathrm{i}}+\hat{\mathrm{j}}}{\sqrt{2}}\right)\)
  4. D \(\mathrm{B}_{0} \frac{\hat{\mathrm{j}}-\hat{\mathrm{i}}}{\sqrt{2}} \cos \left(\omega \mathrm{t}+\mathrm{k} \frac{\hat{\mathrm{i}}+\hat{\mathrm{j}}}{\sqrt{2}}\right)\)
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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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