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) → \( 90\,\mathrm{V/m} \) along the Y-direction
In a plane electromagnetic wave the field magnitudes are related by \( E = cB \), and \( \vec{E} \), \( \vec{B} \) and the direction of propagation are mutually perpendicular, with \( \vec{E}\times\vec{B} \) along the direction of travel.
\( \displaystyle B = 0.003\,\mathrm{G} = 0.003\times10^{-4}\,\mathrm{T} = 3\times10^{-7}\,\mathrm{T} \)
\( \displaystyle E = cB = 3\times10^{8} \times 3\times10^{-7} = 90\,\mathrm{V\,m^{-1}} \)
Direction: the wave travels along \( +X \) and \( \vec{B} \) is along \( +Z \). We need \( \vec{E}\times\vec{B} \) along \( +X \):
\( \displaystyle \hat{\jmath}\times\hat{k} = \hat{\imath} \)
So \( \vec{E} \) is along \( +Y \). The frequency of 50 MHz is not needed.