The electric field of a plane electromagnetic wave is given by $$\overrightarrow{\mathrm{E}}=\mathrm{E}_{0} \hat{\mathrm{i}} \cos (\mathrm{kz}) \cos (\omega \mathrm{t})$$ The corresponding magnetic field $\vec{B}$ is then given by:

(D) $\vec{B}=\frac{E_{0}}{C} \hat{j} \sin (k z) \sin (\omega t)$

(A) $\vec{B}=\frac{E_{0}}{C} \hat{j} \sin (k z) \cos (\omega t)$

(B) $\vec{B}=\frac{E_{0}}{C} \hat{k} \sin (k z) \cos (\omega t)$

(C) $\vec{B}=\frac{E_{0}}{C} \hat{j} \cos (k z) \sin (\omega t)$

JEE Main · 2019 · 10 Apr 2019 · Shift 1 · EM Waves

The electric field of a plane electromagnetic wave is given by

E = E_{0} \hat{i} cos (kz) cos (ω t)

The corresponding magnetic field

B

is then given by:

Answer: (D)

For this wave

E

lies along

\hat{i}

and the

k z

dependence means it travels along

+ z

, so

E \times B

points along

+ z

and

B

must lie along

\hat{j}

. Faraday's law in the form

\frac{\partial E _{x}}{\partial z} = - \frac{\partial B _{y}}{\partial t}

gives

\frac{\partial B _{y}}{\partial t} = k E_{0} sin (k z) cos (ω t)

. Integrating this over time brings in a factor

\frac{1}{ω}

and turns

cos ω t

into

sin ω t

, so

B_{y} = \frac{k E _{0}}{ω} sin (k z) sin (ω t) = \frac{E _{0}}{c} sin (k z) sin (ω t)

, that is

B = \frac{E _{0}}{c} sin (k z) sin (ω t) \hat{j}

JEE Main · 2019 · 10 Apr 2019 · Shift 1 · Single correct option