Verified Solution • 2025–26 Session moderate

QJEEM17Ph4868: JEE Main 2017 Physics - Class 12 JEEM17Ph4868 Solution: A small circular loop of wire of radius a is located at the…

A small circular loop of wire of radius a is located at the centre of a much larger circular wire loop of radius b. The two loops are in the same plane. The outer loop of radius b carries an alternating current I = Io cos ( ω\omega t). The emf induced in the smaller inner loop is nearly :

Multiple Choice Options

  • A πμoIo2.a2b ωsin⁡(ωt){{\pi {\mu _o}{I_o}} \over 2}.{{{a^2}} \over b}\,\omega \sin \left( {\omega t} \right)
  • B πμoIo2.a2b ωcos⁡(ωt){{\pi {\mu _o}{I_o}} \over 2}.{{{a^2}} \over b}\,\omega \cos \left( {\omega t} \right)
  • C πμoIo a2bωsin⁡(ωt)\pi {\mu _o}{I_o}\,{{{a^2}} \over b}\omega \sin \left( {\omega t} \right)
  • D πμoIo b2aωcos⁡(ωt){{\pi {\mu _o}{I_o}\,{b^2}} \over a}\omega \cos \left( {\omega t} \right)

Step-by-Step Verified Solution

Mutual inductance,

M = μ0πN1N2 a22b{{{\mu _0}\pi {N_1}{N_2}\,{a^2}} \over {2b}}

here N1{{N_1}} = N2 = 1

∴   \therefore\,\,\, M = μ0πa22b{{{\mu _0}\pi {a^2}} \over {2b}}

Current I = I0 cos ( ω\omega t)

According to Faraday's law,

e = −- M dIdt{{dI} \over {dt}}

= −- μ0πa22b{{{\mu _0}\pi {a^2}} \over {2b}} ddt{d \over {dt}} (I0 cos ω\omega t)

= + μ0πa22b{{{\mu _0}\pi {a^2}} \over {2b}} I0 ω\omega sin ω\omega t

= πμ0I02{{\pi {\mu _0}{I_0}} \over 2} . a2b{{{a^2}} \over b} ω\omega sin ω\omega t

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