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Home/Practice questions/Physics
PhysicsElectromagnetic Waves

In a plane electromagnetic wave, the phase difference between electri…

In a plane electromagnetic wave, the phase difference between electric and magnetic field vectors is

Answer & explanation+

Correct answer: option 1

The correct answer is Option (1) → zero

In a plane electromagnetic wave, E and B are perpendicular in direction but oscillate in phase: they reach their maxima and zeros at the same place and time.

    \( E_x = E_0\sin(kz - \omega t) \)
    \( B_y = B_0\sin(kz - \omega t) \)

Both have the same phase \( (kz - \omega t) \), with \( E_0 = cB_0 \), so the phase difference is zero.

Why the other options are incorrect

  • Option (2) \( \pi/4 \) — Incorrect. There is no phase lag of \( \pi/4 \) between the fields; both follow the same sine function.

  • Option (3) \( \pi/2 \) — Incorrect. \( \pi/2 \) (90°) is the angle between the directions of E and B, not their phase difference.

  • Option (4) \( \pi \) — Incorrect. A phase difference of \( \pi \) would make B zero or reversed when E is maximum, but E and B peak together.

NCERT (2026-27):
Nature of Electromagnetic Waves – “… The electric and magnetic fields \( E_x \) and \( B_y \) are perpendicular to each other, and to the direction \( z \) of propagation. We can write \( E_x \) and \( B_y \) as follows:

    \( \displaystyle E_x = E_0 \sin (kz - \omega t) \qquad [8.7(a)] \)

    \( \displaystyle B_y = B_0 \sin (kz - \omega t) \qquad [8.7(b)] \)”

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