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  • Help me please, A plane electromagnetic wave of wavelength λ has an intensity I. It ispropagating along the positive Y-direction. The allowed expressions for the electricand magnetic fields are given by :

A plane electromagnetic wave of wavelength λ has an intensity I. It is
propagating along the positive Y-direction. The allowed expressions for the electric
and magnetic fields are given by :

  • Option 1)

    \vec{E}= \sqrt{2I/\epsilon _{0}c} \cos [\frac{2\pi }{\lambda }(y-ct)]\hat{k}; \vec{B} =+\frac{1}{c}E\hat{i}

  • Option 2)

    \vec{E}= \sqrt{2I/\epsilon _{0}c} \cos [\frac{2\pi }{\lambda }(y+ct)]\hat{k}; \vec{B} =+\frac{1}{c}E\hat{i}

  • Option 3)

    \vec{E}= \sqrt{I/\epsilon _{0}c} \cos [\frac{2\pi }{\lambda }(y-ct)]\hat{k}; \vec{B} =\frac{1}{c}E\hat{i}

  • Option 4)

    \vec{E}= \sqrt{I/\epsilon _{0}c} \cos [\frac{2\pi }{\lambda }(y-ct)]\hat{i}; \vec{B} =+\frac{1}{c}E\hat{i}

 

Answers (2)

best_answer

AS we have learned

Intensity of EM wave -

I = \frac{1}{2} \epsilon _{o} E_{o}^{2}c

- wherein

\epsilon _{o} = Permittivity of free space

E _{o} = Electric field amplitude

c = Speed of light in vacuum

 

 

 E_{0}=\sqrt{\left ( \frac{2I}{c\varepsilon _{0}} \right )}

B_{0}=\frac{E_{0}}{c}

direction of \vec{E}\times \vec{B}  is along \hat{j}

\vec{B} is along \hat{i}  and \vec{E} is along \hat{k}

 

 

 


Option 1)

\vec{E}= \sqrt{2I/\epsilon _{0}c} \cos [\frac{2\pi }{\lambda }(y-ct)]\hat{k}; \vec{B} =+\frac{1}{c}E\hat{i}

Option 2)

\vec{E}= \sqrt{2I/\epsilon _{0}c} \cos [\frac{2\pi }{\lambda }(y+ct)]\hat{k}; \vec{B} =+\frac{1}{c}E\hat{i}

Option 3)

\vec{E}= \sqrt{I/\epsilon _{0}c} \cos [\frac{2\pi }{\lambda }(y-ct)]\hat{k}; \vec{B} =\frac{1}{c}E\hat{i}

Option 4)

\vec{E}= \sqrt{I/\epsilon _{0}c} \cos [\frac{2\pi }{\lambda }(y-ct)]\hat{i}; \vec{B} =+\frac{1}{c}E\hat{i}

Posted by

Avinash

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