JEE Main 2022PhysicsElectromagnetic WavesMediumMCQ

JEE Main 2022 — Electromagnetic Waves Question with Solution

JEE Main 2022 (24 Jun Shift 2)

Question

An electric bulb is rated as 200 W. What will be the peak magnetic field at 4 m distance produced by the radiations coming from this bulb? Consider this bulb as a point source with 3.5% efficiency.

Choose an option

Show full solutionCorrect option: B
Correct answer
B1.71×10-8T

Step-by-step explanation

Intensity of light is I=PowerArea=200×3.51004π42=0.034 W m-2

Using the relation between magnetic field and intensity, I=B02c2μ0

We have, magnetic field as B0=I×2μ0c=0.034×2×4π×10-73×108=1.69×10-8 T≈1.71×10-8 T

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About this question

This is a previous-year question from JEE Main 2022, covering the Electromagnetic Waves chapter of Physics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.