The electric field of a plane electromagnetic wave propagating along the $x$ direction in vacuum is $\vec{E}=E_{0} \hat{j} \cos (\omega t-k x)$. The magnetic field $\vec{B}$, at the moment $t=0$ is:

(B) $\vec{B}=E_{0} \sqrt{\mu_{0} \in_{0}} \cos (k x) \hat{k}$

(A) $\vec{B}=E_{0} \sqrt{\mu_{0} \in_{0}} \cos (k x) \hat{j}$

(C) $\vec{B}=\frac{E_{0}}{\sqrt{\mu_{0} \epsilon_{0}}} \cos (k x) \hat{k}$

(D) $\vec{B}=\frac{E_{0}}{\sqrt{\mu_{0} \epsilon_{0}}} \cos (k x) \hat{j}$

JEE Main · 2020 · 03 Sep 2020 · Shift 2 · EM Waves

The electric field of a plane electromagnetic wave propagating along the

x

direction in vacuum is

E = E_{0} \hat{j} cos (ω t - k x)

. The magnetic field

B

, at the moment

t = 0

is:

Answer: (B)

In a plane electromagnetic wave the magnetic field is perpendicular to both the electric field and the direction of travel, and its amplitude follows

B_{0} = E_{0} / c

with

c = 1/ μ_{0} ϵ_{0}

, so

B_{0} = E_{0} μ_{0} ϵ_{0}

. The wave advances along

\overset{x}{^}

and

E

points along

\hat{j}

, and since

E \times B

must point along the direction of propagation

\overset{x}{^}

, the field

B

has to lie along

\hat{k}

. Carrying the same phase as the electric field, at

t = 0

it is

B = E_{0} μ_{0} ϵ_{0} cos (k x) \hat{k}

JEE Main · 2020 · 03 Sep 2020 · Shift 2 · Single correct option