A plane E-M waves of frequency 48 MHz travels in free space along the positive x-direction. At a particular point in space and time \( \vec{E} = 10.2\,\hat{\jmath}\ \mathrm{V/m} \). The corresponding magnetic field \( \vec{B} \) at that point will be:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( 3.4\times10^{-8}\,\hat{k}\ \mathrm{T} \)
In an electromagnetic wave in free space the magnitudes of the fields are related by \( E/B = c \), and \( \vec{E} \), \( \vec{B} \) and the direction of propagation are mutually perpendicular, with \( \vec{E}\times\vec{B} \) along the direction of propagation.
\( \displaystyle B = \frac{E}{c} = \frac{10.2}{3\times10^{8}}\,\mathrm{T} = 3.4\times10^{-8}\,\mathrm{T} \)
The wave travels along \( +\hat{\imath} \) and \( \vec{E} \) is along \( +\hat{\jmath} \). Since \( \hat{\jmath}\times\hat{k} = \hat{\imath} \), the field \( \vec{B} \) must point along \( +\hat{k} \).
\( \vec{B} = 3.4\times10^{-8}\,\hat{k}\ \mathrm{T} \)
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 – …
