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
Answer & explanation
Correct answer: option 3
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 $+y$