JEE Main 2022 — Physics Electromagnetism

2022mcqmedium

If Electric field intensity of a uniform plane electro magnetic wave is given as $E=-301.6\mathrm{sin}(kz-\omega t){\hat{a}}_{x}+452.4\mathrm{sin}(kz-\omega t){\hat{a}}_{y}V{m}^{-1}.$ Then, magnetic intensity $H$ of this wave in $A{m}^{-1}$ will be [Given : Speed of light in vacuum $c=3\times {10}^{8}{ms}^{-1}$ , Permeability of vacuum ${\mu }_{0}=4\pi \times {10}^{-7}N{A}^{-2}$ ]

Official previous-year question

Held on 26 Jun 2022 · Verified 6 Jul 2026.

Options

A
$+0.8\mathrm{sin}(kz-\omega t){\hat{a}}_{y}+0.8\mathrm{sin}(kz-\omega t){\hat{a}}_{x}$
B
$+1.0\times {10}^{-6}\mathrm{sin}(kz-\omega t){\hat{a}}_{y}+1.5\times {10}^{-6}(kz-\omega t){\hat{a}}_{x}$
C
$-0.8\mathrm{sin}(kz-\omega t){\hat{a}}_{y}-1.2\mathrm{sin}(kz-\omega t){\hat{a}}_{x}$
D
$-1.0\times {10}^{-6}\mathrm{sin}(kz-\omega t){\hat{a}}_{y}-1.5\times {10}^{-6}\mathrm{sin}(kz-\omega t){\hat{a}}_{x}$

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Solution

The relation between magnetic field and electric field is given by,

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

And magnetic intensity is given by,

${H}_{0}=\frac{{B}_{0}}{{\mu }_{0}} \Rightarrow {H}_{0}=\frac{{E}_{0}}{c{\mu }_{0}}$

Also, $\vec{E}$ and $\vec{B}$ are perpendicular to each other.

Therefore,

$|{H}_{y0}|=\frac{301.6}{3\times {10}^{8}\times 4\pi \times {10}^{-7}}=0.8A{m}^{-1} |{H}_{x0}|=\frac{452.4}{3\times {10}^{8}\times 4\pi \times {10}^{-7}}=1.2A{m}^{-1}$

Hence, C is the answer.

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Options

$+0.8\mathrm{sin}(kz-\omega t){\hat{a}}_{y}+0.8\mathrm{sin}(kz-\omega t){\hat{a}}_{x}$

$+1.0\times {10}^{-6}\mathrm{sin}(kz-\omega t){\hat{a}}_{y}+1.5\times {10}^{-6}(kz-\omega t){\hat{a}}_{x}$

$-0.8\mathrm{sin}(kz-\omega t){\hat{a}}_{y}-1.2\mathrm{sin}(kz-\omega t){\hat{a}}_{x}$

$-1.0\times {10}^{-6}\mathrm{sin}(kz-\omega t){\hat{a}}_{y}-1.5\times {10}^{-6}\mathrm{sin}(kz-\omega t){\hat{a}}_{x}$

Solution

The relation between magnetic field and electric field is given by,

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

And magnetic intensity is given by,

${H}_{0}=\frac{{B}_{0}}{{\mu }_{0}} \Rightarrow {H}_{0}=\frac{{E}_{0}}{c{\mu }_{0}}$

Also, $\vec{E}$ and $\vec{B}$ are perpendicular to each other.

Therefore,

$|{H}_{y0}|=\frac{301.6}{3\times {10}^{8}\times 4\pi \times {10}^{-7}}=0.8A{m}^{-1} |{H}_{x0}|=\frac{452.4}{3\times {10}^{8}\times 4\pi \times {10}^{-7}}=1.2A{m}^{-1}$

Hence, C is the answer.

JEE Main 2022 — Physics Electromagnetism

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JEE Main 2022 — Physics Electromagnetism

Options

Did you get this right?

Solution

Your note

JEE Main Electromagnetism in other years

More JEE Main Electromagnetism questions

Other Physics topics for JEE Main

Mechanics

Modern Physics

Optics

Thermodynamics

Waves & Oscillations

Keep practicing Electromagnetism