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

किसी एकवर्णी प्रकाश पुंज की आवृत्ति \(v=\frac{3}{2 \pi} \times 10^{12}\, Hz\) है और वह \(\frac{\hat{i}+\hat{j}}{\sqrt{2}}\) दिशा में गतिमान है। उसकी ध्रुवण की दिशा \(\hat{k}\) है। उसके चुम्बकीय क्षेत्र का स्वीकार्य स्वरूप होगा

  1. A \(\frac{{{E_0}}}{C}\left( {\frac{{\hat i - \hat j}}{{\sqrt 2 }}} \right)\cos \left[ {{{10}^4}\left( {\frac{{\hat i - \hat j}}{{\sqrt 2 }}} \right)\cdot \vec r - \left( {3 \times {{10}^{12}}} \right)t} \right]\)
  2. B \(\frac{{{E_0}}}{C}\left( {\frac{{\hat i - \hat j}}{{\sqrt 2 }}} \right)\cos \left[ {{{10}^4}\left( {\frac{{\hat i + \hat j}}{{\sqrt 2 }}} \right)\cdot \vec r - \left( {3 \times {{10}^{12}}} \right)t} \right]\)
  3. C \(\frac{{{E_0}}}{C}\hat k\cos \left[ {{{10}^4}\left( {\frac{{\hat i + \hat j}}{{\sqrt 2 }}} \right)\cdot \vec r + \left( {3 \times {{10}^{12}}} \right)t} \right]\)
  4. D \(\frac{{{E_0}}}{C}\frac{{\left( {\hat i + \hat j + \hat k} \right)}}{{\sqrt 3 }}\cos \left[ {{{10}^4}\left( {\frac{{\hat i + \hat j}}{{\sqrt 2 }}} \right)\cdot \vec r + \left( {3 \times {{10}^{12}}} \right)t} \right]\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(\frac{{{E_0}}}{C}\left( {\frac{{\hat i - \hat j}}{{\sqrt 2 }}} \right)\cos \left[ {{{10}^4}\left( {\frac{{\hat i - \hat j}}{{\sqrt 2 }}} \right)\cdot \vec r - \left( {3 \times {{10}^{12}}} \right)t} \right]\)

Step-by-step Solution

Detailed explanation

Poynting Vector - \(\vec{s}=\frac{\vec{E} \times \vec{B}}{\mu_{o}}\) -wherein It is total energy flowing perpendicularly per second per unit area into the surface in free space. \(\vec{E} \times \vec{B}\) should give a direction of wave propagation…
Same subject
Explore more questions on app
From JEE Mains
Explore more questions on app