The electric field and magnetic field components of an electromagnetic wave going through vacuum is described by $\mathrm{E}_{\mathrm{x}}=\mathrm{E}_{0} \sin (\mathrm{kz}-\omega \mathrm{t})$ $\mathrm{B}_{\mathrm{y}}=\mathrm{B}_{0} \sin (\mathrm{kz}-\omega \mathrm{t})$ Then the correct relation between $\mathrm{E}_{0}$ and $\mathrm{B}_{0}$ is given by

(A) $\mathrm{kE}_{0}=\omega \mathrm{B}_{0}$

(B) $\mathrm{E}_{0} \mathrm{~B}_{0}=\omega \mathrm{k}$

(C) $\omega \mathrm{E}_{0}=\mathrm{kB}_{0}$

(D) $\mathrm{E}_{0}=\mathrm{kB}_{0}$

JEE Main · 2023 · 24 Jan 2023 · Shift 2 · EM Waves

The electric field and magnetic field components of an electromagnetic wave going through vacuum is described by

E_{x} = E_{0} sin (kz - ω t)

B_{y} = B_{0} sin (kz - ω t)

Then the correct relation between

E_{0}

and

B_{0}

is given by

(A) $kE_{0} = ω B_{0}$ ✓ Correct
(B) $E_{0} B_{0} = ω k$
(C) $ω E_{0} = kB_{0}$
(D) $E_{0} = kB_{0}$

Answer: (A)

Solution

In a plane electromagnetic wave travelling through vacuum the field amplitudes are linked to the speed of the wave by

\frac{E _{0}}{B _{0}} = c

, and for the given wave

c = \frac{ω}{k}

. Equating the two,

\frac{E _{0}}{B _{0}} = \frac{ω}{k}

, which rearranges to

k E_{0} = ω B_{0}

JEE Main · 2023 · 24 Jan 2023 · Shift 2 · Single correct option

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