A plane electromagnetic wave is travelling along +y-axis. The value of electric field at a point is \( \vec E=12.3\,\hat i\,\frac{\mathrm{N}}{\mathrm{C}} \). The magnitude and direction of magnetic field B at this point is:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( 4.1\times10^{-8}\,\mathrm{T} \), along -z axis
In an electromagnetic wave \( \vec{E} \) and \( \vec{B} \) are perpendicular to each other and to the direction of propagation, the magnitudes are related by \( E = cB \), and \( \vec{E}\times\vec{B} \) points along the direction of travel.
Magnitude
\( \displaystyle B = \frac{E}{c} = \frac{12.3\,\mathrm{N\,C^{-1}}}{3\times10^{8}\,\mathrm{m\,s^{-1}}} = 4.1\times10^{-8}\,\mathrm{T} \)
Direction (the wave travels along \( +\hat{\jmath} \), \( \vec{E} \) is along \( +\hat{\imath} \))
\( \hat{\imath}\times\hat{k} = -\hat{\jmath} \quad\text{and}\quad \hat{\imath}\times(-\hat{k}) = +\hat{\jmath} \)
So \( \vec{B} \) must be along \( -\hat{k} \), the negative z axis, for \( \vec{E}\times\vec{B} \) to point along \( +y \).
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 – …
