A plane electromagnetic wave of frequency 500 MHz is traveling in vacuum along y-direction. At a particular point in space and time, $\vec{B}=8.0 \times 10^{-8} \hat{z} T$. The value of electric field at this point is: $\left(\right.$ speed of light $\left.=3 \times 10^{8} \mathrm{~ms}^{-1}\right) \hat{x}, \hat{y}, \hat{z}$ are unit vectors along $x, y$ and $z$ directions.

(C) $-24 \hat{x} \mathrm{~V} / \mathrm{m}$

(A) $24 \hat{x} \mathrm{~V} / \mathrm{m}$

(B) $2.6 \hat{x} \mathrm{~V} / \mathrm{m}$

(D) $-2.6 \hat{y} \mathrm{~V} / \mathrm{m}$

JEE Main · 2021 · 16 Mar 2021 · Shift 1 · EM Waves

A plane electromagnetic wave of frequency 500 MHz is traveling in vacuum along y-direction. At a particular point in space and time,

B = 8.0 \times 1 0^{- 8} \overset{z}{^} T

. The value of electric field at this point is:

(

speed of light

= 3 \times 1 0^{8} ms^{- 1}) \overset{x}{^}, \overset{y}{^}, \overset{z}{^}

are unit vectors along

x, y

and

z

directions.

Answer: (C)

In a plane electromagnetic wave the field magnitudes are tied by

E = c B

E = 3 \times 1 0^{8} \times 8.0 \times 1 0^{- 8} = 24 V/m

The direction follows from the requirement that

E \times B

point along the direction of travel. Here

B

lies along

\overset{z}{^}

and the wave travels along

\overset{y}{^}

, so the unit vector

\hat{E}

must satisfy

\hat{E} \times \overset{z}{^} = \overset{y}{^}

. Since

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

, the field points along

- \overset{x}{^}

.
The electric field at that point is

- 24 \overset{x}{^}

V/m.

JEE Main · 2021 · 16 Mar 2021 · Shift 1 · Single correct option