The magnetic field of a plane electromagnetic wave is $$\overrightarrow{\mathrm{B}}=3 \times 10^{-8} \sin [200 \pi(\mathrm{y}+\mathrm{ct})] \hat{\mathrm{i}} T$$ where $\mathrm{c}=3 \times 10^{8} \mathrm{~ms}^{-1}$ is the speed of light. The corresponding electric field is

(B) $\overrightarrow{\mathrm{E}}=-9 \sin [200 \pi(\mathrm{y}+\mathrm{ct})] \hat{\mathrm{k}} \mathrm{V} / \mathrm{m}$

(A) $\overrightarrow{\mathrm{E}}=9 \sin [200 \pi(\mathrm{y}+\mathrm{ct})] \hat{\mathrm{k}} \mathrm{V} / \mathrm{m}$

(C) $\overrightarrow{\mathrm{E}}=3 \times 10^{-8} \sin [200 \pi(\mathrm{y}+\mathrm{ct})] \hat{\mathrm{k}} \mathrm{V} / \mathrm{m}$

(D) $\overrightarrow{\mathrm{E}}=-10^{-6} \sin [200 \pi(\mathrm{y}+\mathrm{ct})] \hat{\mathrm{k}} \mathrm{V} / \mathrm{m}$

JEE Main · 2020 · 03 Sep 2020 · Shift 1 · EM Waves

The magnetic field of a plane electromagnetic wave is

B = 3 \times 1 0^{- 8} sin [200 π (y + ct)] \hat{i} T

where

c = 3 \times 1 0^{8} ms^{- 1}

is the speed of light.
The corresponding electric field is

Answer: (B)

For a plane electromagnetic wave the field amplitudes satisfy

E_{0} = c B_{0} = (3 \times 1 0^{8}) (3 \times 1 0^{- 8}) = 9

V m

^{- 1}

. The argument

(y + c t)

shows the wave travels along

- \hat{j}

, and since

E \times B

must point along the direction of propagation with

B

along

\hat{i}

, the electric field must lie along

- \hat{k}

. Hence

E = - 9 sin [200 π (y + c t)] \hat{k}

V m

^{- 1}

JEE Main · 2020 · 03 Sep 2020 · Shift 1 · Single correct option