JEE Mains · Physics · STD 12 - 8. Electromagnetic waves
The electric field component of a monochromatic radiation is given by \(\vec E = 2{E_0}\,\hat i\,\cos\, kz\,\cos\, \omega t\) Its magnetic field \(\vec B\) is then given by
- A \(\frac{{2{E_0}}}{c}\hat j\,\sin\, kz\,\cos\, \omega t\)
- B \(-\frac{{2{E_0}}}{c}\hat j\,\sin\, kz\,\sin\, \omega t\)
- C \(\frac{{2{E_0}}}{c}\hat j\,\sin\, kz\,\sin\, \omega t\)
- D \(\frac{{2{E_0}}}{c}\hat j\,\cos\, kz\,\cos\, \omega t\)
Answer & Solution
Correct Answer
(C) \(\frac{{2{E_0}}}{c}\hat j\,\sin\, kz\,\sin\, \omega t\)
Step-by-step Solution
Detailed explanation
Given, Electric field component of monochromatic radiation, \((\overrightarrow{\mathrm{E}})=2 \mathrm{E}_{0} \hat{\mathrm{i}} \cos \mathrm{kz} \cos \omega \mathrm{t}\) We know that, \(\frac{d E}{d z}=-\frac{d B}{d t}\)…
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