If Electric field intensity of… - JEE Main 2022 Physics PYQ

JEE Main 2022 — Physics Electromagnetism

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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 ${\mu }_{0}=4\pi \times {10}^{-7}N{A}^{-2}$ ]

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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.

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Options

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Solution

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JEE Main Electromagnetism in other years

More JEE Main Electromagnetism questions

Other Physics topics for JEE Main

Mechanics

Modern Physics

Optics

Thermodynamics

Waves & Oscillations

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