The electric field in an electromagnetic wave is given by $\overrightarrow{\mathrm{E}}=\hat{\mathrm{i}} 40 \cos \omega\left(\mathrm{t}-\frac{\mathrm{z}}{\mathrm{c}}\right) \mathrm{NC}^{-1}$. The magnetic field induction of this wave is (in SI unit):

(D) $\vec{B}=\hat{j} \frac{40}{c} \cos \omega\left(t-\frac{z}{c}\right)$

(A) $\overrightarrow{\mathrm{B}}=\hat{\mathrm{i}} \frac{40}{\mathrm{c}} \cos \omega\left(\mathrm{t}-\frac{\mathrm{z}}{\mathrm{c}}\right)$

(B) $\vec{B}=\hat{j} 40 \cos \omega\left(t-\frac{z}{c}\right)$

(C) $\vec{B}=\hat{k} \frac{40}{c} \cos \omega\left(t-\frac{z}{c}\right)$

JEE Main · 2024 · 04 Apr 2024 · Shift 1 · EM Waves

The electric field in an electromagnetic wave is given by

E = \hat{i} 40 cos ω (t - \frac{z}{c}) NC^{- 1}

. The magnetic field induction of this wave is (in SI unit):

Answer: (D)

For a plane electromagnetic wave,

E

B

and the direction of propagation

\overset{n}{^}

form a right-handed set with

B = \frac{n ^ \times E}{c}

.

Here

E = \hat{i} 40 cos ω (t - \frac{z}{c})

, so

E

is along

+ x

and, from the phase

(t - \frac{z}{c})

, the wave travels along

+ z

. Then

\overset{n}{^} \times \hat{E} = \hat{k} \times \hat{i} = \hat{j} .

B

points along

+ y

with magnitude

E / c

B = \hat{j} \frac{40}{c} cos ω (t - \frac{z}{c}) .

Option D.

JEE Main · 2024 · 04 Apr 2024 · Shift 1 · Single correct option