The electric field of a plane electromagnetic wave is given by $\vec{E}=E_{0}(\hat{x}+\hat{y})\sin(kz-\omega t)$. Its magnetic field will be given by:

(C) $\frac{E_{0}}{c}(-\hat{x}+\hat{y})\sin(kz-\omega t)$

(A) $\frac{E_{0}}{c}(\hat{x}-\hat{y})\sin(kz-\omega t)$

(B) $\frac{E_{0}}{c}(\hat{x}-\hat{y})\cos(kz-\omega t)$

(D) $\frac{E_{0}}{c}(\hat{x}+\hat{y})\sin(kz-\omega t)$

JEE Main · 2020 · 04 Sep 2020 · Shift 2 · EM Waves

The electric field of a plane electromagnetic wave is given by

E = E_{0} (\overset{x}{^} + \overset{y}{^}) sin (k z - ω t)

. Its magnetic field will be given by:

(A) $\frac{E _{0}}{c} (\overset{x}{^} - \overset{y}{^}) sin (k z - ω t)$
(B) $\frac{E _{0}}{c} (\overset{x}{^} - \overset{y}{^}) cos (k z - ω t)$
(C) $\frac{E _{0}}{c} (- \overset{x}{^} + \overset{y}{^}) sin (k z - ω t)$ ✓ Correct
(D) $\frac{E _{0}}{c} (\overset{x}{^} + \overset{y}{^}) sin (k z - ω t)$

Answer: (C)

For a plane electromagnetic wave the magnetic field follows from

B = \frac{1}{c} \hat{k} \times E

, where

\hat{k}

is the direction of propagation. Here the wave travels along

+ z

, so

\hat{k} = \overset{z}{^}

and

B = \frac{E _{0}}{c} (\overset{z}{^} \times (\overset{x}{^} + \overset{y}{^})) sin (k z - ω t)

. Since

\overset{z}{^} \times \overset{x}{^} = \overset{y}{^}

and

\overset{z}{^} \times \overset{y}{^} = - \overset{x}{^}

, this becomes

B = \frac{E _{0}}{c} (- \overset{x}{^} + \overset{y}{^}) sin (k z - ω t)

JEE Main · 2020 · 04 Sep 2020 · Shift 2 · Single correct option