The magnetic field of a plane electromagnetic Wave is $\overrightarrow{ B }=3 \times 10^{-8} \sin [200 \pi( y + ct )] \hat{ i }\, T$ Where $c=3 \times 10^{8} \,ms ^{-1}$ is the speed of light. The corresponding electric field is

  • [JEE MAIN 2020]
  • A

    $\overrightarrow{ E }=-10^{-6} \sin [200 \pi( y + ct )] \hat{ k }\, \;V / m$

  • B

     $\overrightarrow{ E }=-9 \sin [200 \pi( y + ct )] \hat{ k }\, \;V / m$

  • C

    $\overrightarrow{ E }=9 \sin [200 \pi( y + ct )] \hat{ k }\, \;V / m$

  • D

    $\overrightarrow{ E }=3 \times 10^{-8} \sin [200 \pi( y + ct )] \hat{ k }\, \;V / m$

Similar Questions

A plane $EM$ wave travelling in vacuum along $z-$ direction is given by $\vec E = {E_0}\,\,\sin (kz - \omega t)\hat i$ and $\vec B = {B_0}\,\,\sin (kz - \omega t)\hat j$.

$(i)$ Evaluate $\int {\vec E.\overrightarrow {dl} } $ over the rectangular loop $1234$ shown in figure.

$(ii)$ Evaluate $\int {\vec B} .\overrightarrow {ds} $ over the surface bounded by loop $1234$.

$(iii)$ $\int {\vec E.\overrightarrow {dl}  =  - \frac{{d{\phi _E}}}{{dt}}} $ to prove $\frac{{{E_0}}}{{{B_0}}} = c$

$(iv)$ By using similar process and the equation $\int {\vec B} .\overrightarrow {dl}  = {\mu _0}I + { \in _0}\frac{{d{\phi _E}}}{{dt}}$ , prove that  $c = \frac{1}{{\sqrt {{\mu _0}{ \in _0}} }}$ 

The electric field of a plane electromagnetic wave propagating along the $x$ direction in vacuum is $\overrightarrow{ E }= E _{0} \hat{ j } \cos (\omega t - kx )$. The magnetic field $\overrightarrow{ B },$ at the moment $t =0$ is :

  • [JEE MAIN 2020]

Ratio of electric field and magnetic field gives which physical quantity ? 

An $LC$ resonant circuit contains a $400 pF$ capacitor and a $100\mu H$ inductor. It is set into oscillation coupled to an antenna. The wavelength of the radiated electromagnetic waves is

Which of the following statement is true for displacement current