The electric field in an electromagnetic wave is given by E = i 40 ω(t-z / c) NC^ -1 . The magnetic field induction of this wave is (in SI unit) :

Correct answer: D. B = j 40 C ω(t-z / c) E = i 40 ω(t- z c ) E is along +x direction v is along +z direction So direction of B will be along +y and magnitude of B will be (E)/(c) So answer is (40)/(c) ω(t-(z)/(c)) j

JEE Main 20244 AprilMorning Shift

Exam: JEE Main
Year: 2024
Date: 4 April
Shift: Morning Shift
Subject: Physics
Chapter: Electromagnetic Waves

JEE Main 2024 · 4 April · Morning Shift · Physics · Electromagnetic Waves

The electric field in an electromagnetic wave is given by E = i 40...

The electric field in an electromagnetic wave is given by

\overrightarrow{\mathrm{E}}=\hat{i} 40 \cos \omega(\mathrm{t}-z / \mathrm{c}) \mathrm{NC}^{-1}

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

Options

A $\overrightarrow{\mathrm{B}}=\hat{k} \frac{40}{\mathrm{C}} \cos \omega(\mathrm{t}-z / \mathrm{c})$
B $\vec{B}=\hat{j} 40 \cos \omega(t-z / c)$
C $\overrightarrow{\mathrm{B}}=\hat{i} \frac{40}{\mathrm{C}} \cos \omega(\mathrm{t}-z / \mathrm{c})$
D $\overrightarrow{\mathrm{B}}=\hat{j} \frac{40}{\mathrm{C}} \cos \omega(\mathrm{t}-z / \mathrm{c})$

Correct answer

D. $\overrightarrow{\mathrm{B}}=\hat{j} \frac{40}{\mathrm{C}} \cos \omega(\mathrm{t}-z / \mathrm{c})$

Step-by-step solution

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

\overrightarrow{\mathrm{E}}

is along

+\mathrm{x}

direction

\overrightarrow{\mathrm{v}}

is along

+\mathrm{z}

direction So direction of

\vec{B}

will be along +y and magnitude of

B

will be

\frac{E}{c}

So answer is

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

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