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JEE Main 20178 AprilOnline
Exam
JEE Main
Year
2017
Date
8 April
Mode
Online
Subject
Physics
Chapter
Electromagnetic Induction

JEE Main 2017 · 8 April · Online · Physics · Electromagnetic Induction

A small circular loop of wire of radius a is located at the centre...

A small circular loop of wire of radius aa is located at the centre of a much larger circular wire loop of radius bb. The two loops are in the same plane. The outer loop of radius bb carries an alternating current I=I0cos⁡(\omegat).I = {I}_{0}\cos \left(\omegat\right) . The emf induced in the smaller inner loop is nearly:

Options

  1. Aπμ0I0a2bωsin(\omegat)\pi{\mu}_{0}{I}_{0}\dfrac{{a}^{2}}{b}\omega sin \left(\omegat\right)
  2. Bπμ0I02.a2bωcos(\omegat)\dfrac{\pi{\mu}_{0}{I}_{0}}{2} . \dfrac{{a}^{2}}{b}\omega cos \left(\omegat\right)
  3. Cπμ0I0b2aωcos(\omegat)\dfrac{\pi{\mu}_{0}{I}_{0}{b}^{2}}{a}\omega cos \left(\omegat\right)
  4. Dπμ0I02.a2bωsin(\omegat)\dfrac{\pi{\mu}_{0}{I}_{0}}{2} . \dfrac{{a}^{2}}{b}\omega sin \left(\omegat\right)

Correct answer

D. πμ0I02.a2bωsin(\omegat)\dfrac{\pi{\mu}_{0}{I}_{0}}{2} . \dfrac{{a}^{2}}{b}\omega sin \left(\omegat\right)

Step-by-step solution

Induced emf,

ε=MdIdt,ε = \dfrac{MdI}{dt},

Current is given as, I=I0cos\omegatI = {I}_{0}cos\omegat

M=μ0πa22b,ε=−Mddt(I0cos\omegat)M = \dfrac{{\mu}_{0}\pi{a}^{2}}{2b}, ε = - M\dfrac{d}{dt} ( {I}_{0}cos\omegat )

c=MωI0sin⁡(\omegat)c = M\omega{I}_{0}\sin \left(\omegat\right)

=μ0πa22bωI0sin\omegat= \dfrac{{\mu}_{0}\pi{a}^{2}}{2b}\omega{I}_{0}sin \omegat

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