JEE Main 2013PhysicsElectromagnetic InductionHardMCQ

JEE Main 2013 — Electromagnetic Induction Question with Solution

JEE Main 2013 (07 Apr)

Question

A circular loop of radius 0.3 cm lies parallel to a much bigger circular loop of radius 20 cm. The centre of the small loop is on the axis of the bigger loop. The distance between their centres is15 cm. If a current of 2.0 A flows through the smaller loop, then the flux linked with a bigger loop is:

Choose an option

Show full solutionCorrect option: C
Correct answer
C9.1×10-11weber

Step-by-step explanation

ϕ2=B1×A2=μ0IR122(R12+x2)32×πR22

= μ0(2)(20×10−2)22[(0.2)2+(0.15)2]32×π(0.3×10−2)

On solving

= 9.216 × 1 0 - 1 1

≈ 9.2 × 1 0 - 1 1  weber

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About this question

This is a previous-year question from JEE Main 2013, 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.