JEE Main 2020PhysicsElectromagnetic InductionMediumMCQ

JEE Main 2020 — Electromagnetic Induction Question with Solution

JEE Main 2020 (02 Sep Shift 2)

Question

An inductance coil has a reactance of 100 Ω . When an AC signal of frequency 1000 Hz is applied to the coil, the applied voltage leads the current by 45°. The self-inductance of the coil is

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Show full solutionCorrect option: A
Correct answer
A1.1×10-2 H

Step-by-step explanation

tanθ=xLR=tan45°
xL=R
=100=xL2+R2
100=R2+R2
2R=100
R=502
∴xL=502
Lω=502
L=5022π×1000=252πmH
=1.1×10-2H

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About this question

This is a previous-year question from JEE Main 2020, covering the Electromagnetic Induction 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.