A plane electromagnetic wave of frequency 1.50 x 1010 Hz is travelling along x-direction in free space. At a certain point and time electric field is \( E = (36\,\hat{\jmath}) \) V/m. The corresponding magnetic field \( \vec{B} \) (in T) at that point is:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → +12 x 10–8 \( \hat{k} \)
In a plane electromagnetic wave in free space, \( \vec{E} \) and \( \vec{B} \) are perpendicular to each other and to the direction of travel, their magnitudes are related by \( B = E/c \), and \( \vec{E}\times\vec{B} \) points along the direction of propagation.
\( \displaystyle B = \frac{E}{c} = \frac{36\,\mathrm{V\,m^{-1}}}{3\times10^{8}\,\mathrm{m\,s^{-1}}} \)
\( B = 12\times10^{-8}\,\mathrm{T} \)
The wave travels along the positive x-direction and \( \vec{E} \) is along \( +\hat{\jmath} \), so \( \vec{B} \) must point along the axis that makes \( \hat{\jmath}\times\vec{B} \) point along \( +\hat{\imath} \).
\( \hat{\jmath}\times\hat{k} = \hat{\imath} \)
\( \vec{B} = +12\times10^{-8}\,\hat{k}\ \mathrm{T} \)
The frequency is not needed.
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 – …
