The velocity of an EM wave in a given medium is \( V = 3 \times 10^{7}\,\hat{\jmath} \). When electric field oscillates along z-axis, then direction of oscillation of magnetic field at that time should be:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → +x-axis
In an electromagnetic wave the electric and magnetic fields are perpendicular to each other and to the direction of travel, and the wave travels along \( \vec{E} \times \vec{B} \).
The wave travels along \( +\hat{\jmath} \) and \( \vec{E} \) is along \( \hat{k} \). So \( \vec{B} \) must point along the unit vector that gives
\( \hat{k} \times \hat{B} = \hat{\jmath} \)
Among the axes, \( \hat{k} \times \hat{\imath} = \hat{\jmath} \), so the magnetic field oscillates along the x-axis (the \( +x \) direction when \( \vec{E} \) points along \( +z \)).
Why the other options are incorrect
Option (1) -y-axis — Incorrect. \( \vec{B} \) cannot lie along the y-axis, because the y-axis is the direction of travel and the wave is transverse.
Option (2) +z-axis — Incorrect. \( \vec{B} \) cannot be along \( z \), because it must be perpendicular to \( \vec{E} \).
Option (3) +y-axis — Incorrect. Same as Option (1): a field along the direction of travel is not possible in an electromagnetic wave.
NCERT (2026-27):
Nature of Electromagnetic Waves – “It can be shown from Maxwell’s equations that electric and magnetic fields in an electromagnetic wave are perpendicular to each other, and to the direction of propagation.”