A plane electromagnetic wave of frequency 20 MHz propagates in free space along x-direction. At a particular point and time, the electric field is given by \( \vec{E} = 6.6\,\hat{\jmath} \) V/m. The magnetic field at this point and time is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( 2.2\times10^{-8}\,\hat{k} \) T
In a plane electromagnetic wave in free space, the magnitudes of the fields are related by B = E/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. The frequency is not needed.
\( \displaystyle B = \frac{E}{c} = \frac{6.6\,\mathrm{V\,m^{-1}}}{3\times10^{8}\,\mathrm{m\,s^{-1}}} \)
\( B = 2.2\times10^{-8}\,\mathrm{T} \)
The wave travels along the x-direction, taken as the positive x-direction, so \( \hat{\jmath} \times \hat{B} \) must be \( +\hat{\imath} \):
\( \hat{\jmath} \times \hat{k} = \hat{\imath} \)
So \( \vec{B} \) is along \( +\hat{k} \):
\( \vec{B} = 2.2\times10^{-8}\,\hat{k}\,\mathrm{T} \)
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 – …
