A plane electromagnetic wave travels in vacuum along y-direction. At…

The correct answer is Option (3) → E along z-axis; B along x-axis

In a plane electromagnetic wave, \( \vec{E} \) and \( \vec{B} \) are perpendicular to each other and to the direction of travel, and the wave travels along \( \vec{E}\times\vec{B} \).

For travel along \( +y \), both fields must lie in the x-z plane, and their cross product must point along \( +\hat{\jmath} \):

    \( \hat{k}\times\hat{\imath} = +\hat{\jmath} \)

So \( \vec{E} \) along z and \( \vec{B} \) along x gives a wave moving along \( +y \).

Option (1) E along z-axis ; B along y-axis — Incorrect. \( \vec{B} \) along y would lie along the direction of travel; the fields of an electromagnetic wave are transverse.

Option (2) E along x-axis; B along z-axis — Incorrect. Both fields are transverse, but \( \hat{\imath}\times\hat{k} = -\hat{\jmath} \), so this wave would travel along \( -y \).

Option (3) E along z-axis; B along x-axis — Correct. \( \hat{k}\times\hat{\imath} = +\hat{\jmath} \): the direction of \( \vec{E}\times\vec{B} \) is the direction of travel.

Option (4) E along y-axis; B along z-axis — Incorrect. \( \vec{E} \) along y would lie along the direction of travel, which is not possible.

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.”

Nature of Electromagnetic Waves – “… To find the direction, we note that E is along y-direction and the wave propagates along x-axis. Therefore, B should be in a direction perpendicular to both x- and y-axes. Using vector algebra, E × B should be along x-direction.”