A plane electromagnetic wave of frequency 30 MHz travels in free space along the x-direction. At a particular point in space and time \( \vec{E} = 6.3\,\hat{\jmath}\,\mathrm{V/m} \). The \( \vec{B} \) at this point would be:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( 2.1 \times 10^{-8}\,\hat{k}\,\mathrm{T} \)
In an electromagnetic wave in free space the magnitudes are related by \( E = cB \), and \( \vec{E} \), \( \vec{B} \) and the direction of travel are mutually perpendicular, with \( \vec{E}\times\vec{B} \) along the direction of propagation.
\( \displaystyle B = \frac{E}{c} = \frac{6.3\,\mathrm{V\,m^{-1}}}{3\times10^{8}\,\mathrm{m\,s^{-1}}} = 2.1\times10^{-8}\,\mathrm{T} \)
The wave travels along \( +x \) and \( \vec{E} \) is along \( \hat{\jmath} \). Since \( \hat{\jmath}\times\hat{k} = \hat{\imath} \), \( \vec{B} \) must be along \( \hat{k} \).
\( \displaystyle \vec{B} = 2.1\times10^{-8}\,\hat{k}\,\mathrm{T} \)
The frequency is not needed for this.
Note: NTA picked this question from an NCERT (2026-27) example and changed the figures. Do not ignore NCERT while practising numericals.
NCERT (2026-27):
Nature of Electromagnetic Waves – …
