A plane electromagnetic wave of frequency 50 MHz travels in free space in the X- direction. At a particular point in space and time, the magnetic field is 0.003 G in the Z-direction. The magnitude and direction of the electric field is

90 V/m along the Y-direction

The correct answer is Option (3) → 90 V/m along the Y-direction

Given:

Frequency: $f = 50~\text{MHz} = 50 \times 10^6~\text{Hz}$

Magnetic field: $B = 0.003~\text{G} = 3 \times 10^{-7}~\text{T}$

Propagation direction: $x$-direction

Magnetic field direction: $z$-direction

For a plane EM wave in free space:

$E = c B$, where $c = 3 \times 10^8~\text{m/s}$

Magnitude of electric field:

$E = (3 \times 10^8) \cdot (3 \times 10^{-7}) = 90~\text{V/m}$

Direction of electric field: EM wave has **$\vec{E} \perp \vec{B}$ and $\vec{E} \times \vec{B}$ along propagation direction**.

Given $\vec{k}$ along $x$ and $\vec{B}$ along $z$, then $\vec{E}$ is along $y$-direction.

Answer: Magnitude $E = 90~\text{V/m}$, direction along $+y$