JEE Main 2022 — Physics Electromagnetism

2022integereasy

The intensity of the light from a bulb incident on a surface is $0.22W{m}^{-2}$ . The amplitude of the magnetic field in this light-wave is _____ $\times {10}^{-9}T$

(Given : Permittivity of vacuum ${\epsilon }_{0}=8.85\times {10}^{-12}{C}^{2}{N}^{-1}{m}^{-2}$ , speed of light in vacuum $c=3\times {10}^{8}{ms}^{-1}$ )

Official previous-year question

Held on 29 Jun 2022 · Verified 6 Jul 2026.

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Solution

The intensity is defined as energy per unit time per unit area. Therefore, $I=\frac{dE}{Adt}$ .

Energy density is energy per unit volume. Therefore, ${U}_{d}=\frac{dE}{Adx}$ .

Dividing $\frac{I}{{U}_{d}}=\frac{dx}{dt}=c$

$\Rightarrow {U}_{d}=\frac{I}{c}=\frac{0.22}{3\times {10}^{8}}=\frac{22}{3}\times {10}^{-10}J{m}^{-3}$

Magnetic energy density will be half of total energy density, Therefore, ${U}_{b}=\frac{11}{3}\times {10}^{-10}J{m}^{-3}$ .

Now energy density in magnetic field is given by, ${U}_{b}=\frac{{B}^{2}}{2{\mu }_{0}}$ . Therefore,

$\frac{{B}_{\mathrm{rms}}^{2}}{2{\mu }_{0}}=\frac{11}{3}\times {10}^{-10}J{m}^{-3}$ where ${B}_{\mathrm{rms}}^{2}=(\frac{{B}_{0}}{\sqrt{2}})=\frac{{B}_{0}^{2}}{2}$

$\Rightarrow \frac{{B}_{0}^{2}}{4{\mu }_{0}}=\frac{11}{3}\times {10}^{-10} \Rightarrow {B}_{0}=\sqrt{4\times (4\pi \times {10}^{-7})\times \frac{11}{3}\times {10}^{-10}} \Rightarrow {B}_{0}=4.29\times {10}^{-8}T\simeq 43\times {10}^{-9}T$

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JEE Main 2022 — Physics Electromagnetism

2022integereasy

The intensity of the light from a bulb incident on a surface is $0.22W{m}^{-2}$ . The amplitude of the magnetic field in this light-wave is _____ $\times {10}^{-9}T$

(Given : Permittivity of vacuum ${\epsilon }_{0}=8.85\times {10}^{-12}{C}^{2}{N}^{-1}{m}^{-2}$ , speed of light in vacuum $c=3\times {10}^{8}{ms}^{-1}$ )

Official previous-year question

Held on 29 Jun 2022 · Verified 6 Jul 2026.

Did you get this right?

Solution

The intensity is defined as energy per unit time per unit area. Therefore, $I=\frac{dE}{Adt}$ .

Energy density is energy per unit volume. Therefore, ${U}_{d}=\frac{dE}{Adx}$ .

Dividing $\frac{I}{{U}_{d}}=\frac{dx}{dt}=c$

$\Rightarrow {U}_{d}=\frac{I}{c}=\frac{0.22}{3\times {10}^{8}}=\frac{22}{3}\times {10}^{-10}J{m}^{-3}$

Magnetic energy density will be half of total energy density, Therefore, ${U}_{b}=\frac{11}{3}\times {10}^{-10}J{m}^{-3}$ .

Now energy density in magnetic field is given by, ${U}_{b}=\frac{{B}^{2}}{2{\mu }_{0}}$ . Therefore,

$\frac{{B}_{\mathrm{rms}}^{2}}{2{\mu }_{0}}=\frac{11}{3}\times {10}^{-10}J{m}^{-3}$ where ${B}_{\mathrm{rms}}^{2}=(\frac{{B}_{0}}{\sqrt{2}})=\frac{{B}_{0}^{2}}{2}$

JEE Main 2022 — Physics Electromagnetism

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Solution

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JEE Main Electromagnetism in other years

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JEE Main 2022 — Physics Electromagnetism

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