JEE Main 2014PhysicsElectromagnetic InductionHardMCQ

JEE Main 2014 — Electromagnetic Induction Question with Solution

JEE Main 2014 (19 Apr Online)

Question

A square frame of side 10 cm and a long straight wire carrying current 1 A are in the plane of the paper. Starting from close to the wire, the frame moves towards the right with a constant speed of 10 m s-1 (see figure). The e.m.f induced at the time the left arm of the frame is at x=10 cm from the wire is

Choose an option

Show full solutionCorrect option: B
Correct answer
B1 μV

Step-by-step explanation

dϕ=μ0i2πyady

ϕ=μ0ia2π∫xx+adyy=μ0ia2πlnx+a-lnx

E.m.f=-dϕdt=-μ0ia2π1x+adxdt-1xdxdt

=μ0ia2πa.vxx+a=μ02π.ia2vxx+a=2×10-7×1×0.12×100.1×0.2=1μV

Alternative method:

Consider two sides which are perpendicular to the velocity as case of motional e.m.f.

E=E1-E2=B1–B2V l

=μ0I2π1x-1x+aV l

=2×10-7×110.1-10.2×10×0.1

=10-6 V=1 μV

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About this question

This is a previous-year question from JEE Main 2014, 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.