Consider the following statements about propagation of an electromagnetic wave in a medium and choose the correct statements.
(A) Its electric field vector is perpendicular to its direction of propagation
(B) Its magnetic field vector is parallel to its direction of propagation
(C) Both its electric and magnetic field vectors are perpendicular to each other
(D) The speed of propagation of an electromagnetic wave in a medium is \( \displaystyle\frac{1}{\sqrt{\mu\epsilon}} \)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A), (C) and (D) only
An electromagnetic wave is a transverse wave: its electric and magnetic fields are perpendicular to each other and both are perpendicular to the direction of propagation, which is along \( \vec{E} \times \vec{B} \).
(A) Its electric field vector is perpendicular to its direction of propagation — Correct. \( \vec{E} \) oscillates at right angles to the direction of travel.
(B) Its magnetic field vector is parallel to its direction of propagation — Incorrect. \( \vec{B} \) is also perpendicular to the direction of propagation.
(C) Both its electric and magnetic field vectors are perpendicular to each other — Correct. \( \vec{E} \perp \vec{B} \) at every point of the wave.
(D) The speed of propagation of an electromagnetic wave in a medium is \( 1/\sqrt{\mu\epsilon} \) — Correct. In a medium of permeability \( \mu \) and permittivity \( \epsilon \) the speed is \( 1/\sqrt{\mu\epsilon} \); in vacuum it becomes \( c = 1/\sqrt{\mu_0\varepsilon_0} \).
NCERT (2026-27):
Summary – “… The oscillating electric and magnetic fields, E and B are perpendicular to each other, and to the direction of propagation of the electromagnetic wave.”
Summary – “… The speed of light, or of electromagnetic waves in a material medium is given by
\( \displaystyle v = \frac{1}{\sqrt{\mu\varepsilon}} \)
where \( \mu \) is the permeability of the medium and \( \varepsilon \) its permittivity.”