8.Electromagnetic waves
medium

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

A

$\vec E\, = \,\sqrt {\frac{I}{{{\varepsilon _0}C}}} \cos \left[ {\frac{{2\pi }}{\lambda }(y - ct)} \right]\,\hat i\,;\,\vec B\, = \,\frac{1}{c}E\hat k$

B

$\vec E\, = \,\sqrt {\frac{I}{{{\varepsilon _0}C}}} \cos \left[ {\frac{{2\pi }}{\lambda }(y - ct)} \right]\,\hat k\,;\,\vec B\, =  - \,\frac{1}{c}E\hat i$

C

$\vec E\, = \,\sqrt {\frac{{2I}}{{{\varepsilon _0}C}}} \cos \left[ {\frac{{2\pi }}{\lambda }(y - ct)} \right]\,\hat k\,;\,\vec B\, =  + \frac{1}{c}E\hat i$

D

$\vec E\, = \,\sqrt {\frac{{2I}}{{{\varepsilon _0}C}}} \cos \left[ {\frac{{2\pi }}{\lambda }(y + ct)} \right]\,\hat k\,;\,\vec B\, = \frac{1}{c}E\hat i$

(JEE MAIN-2018)

Solution

If $E_0$  is magnitude of electric field then  $\frac{1}{2}\,{\varepsilon _0}{E_0}^2\, \times \,C\, = \,I\, \Rightarrow {\kern 1pt} {E_0}\, = \sqrt {\frac{{2I}}{{C{\varepsilon _0}}}} $

${B_0} = {\kern 1pt} \frac{{{E_0}}}{C}\,$  

Direction of $\vec E \times \vec B$ will be along  $ + \hat j.$

Standard 12
Physics

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