$A$ plane electromagnetic wave of frequency $50 \ MHz$ travels in free space. If the average energy densities in the electric field and magnetic field are $K_{E}$ and $K_{B}$ respectively,then the correct option in the following is

  • A
    $K_{E} = K_{B}$
  • B
    $K_{E} = K_{B} = 0$
  • C
    $K_{E} > K_{B}$
  • D
    $K_{E} < K_{B}$

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Statement-$I$ :- During propagation of electromagnetic wave,$\vec{E}$,$\vec{B}$ and direction of propagation are perpendicular to each other.
Statement-$II$ :- During propagation of electromagnetic wave,the energy density due to electric and magnetic fields are equal.

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