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

vec B =hat j 40/c cos ω (t-z/c )

vec B =hat i 40/c cos ω (t-z/c )

overrightarrow B =hat j 40 cos ω (t-frac z c )

overrightarrow B =hat k frac 40 c cos ω ( t -frac z c )

Physics · Electromagnetic Waves

JEE Main 2024 — 4 April, Shift 1 — Question 32

The electric field in an electromagnetic wave is given by $\vec{E}=\hat{\mathrm{i}} 40 \cos \omega\left(t-\frac{z}{c}\right) N C^{-1} . \quad$

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

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

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

$\overrightarrow{\mathrm{E}}$ is along +x direction

$\overrightarrow{\mathrm{v}}$ is along +Z direction

So direction of $\overrightarrow{\mathrm{B}}$ will be along +y and magnitude of $B$ will be $\frac{E}{c}$

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

Answer key and solution verified before publishing.

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