Verified Solution • 2025–26 Session moderate

QJEEM18Ph5fee: JEE Main 2018 Physics - Class 12 JEEM18Ph5fee Solution: At the center of a fixed large circular coil of radius R, a…

At the center of a fixed large circular coil of radius R, a much smaller circular coil of radius r is placed. The two coils are concentric and are in the same plane. The larger coil carries a current I. The smaller coil is set to rotate with a constant angular velocity ω\omega about an axis along their common diameter. Calculate the emf induced in their smaller coil after a time t of its start of rotation.

Multiple Choice Options

  • A μoI2 R{{{\mu _o}{\rm I}} \over {2\,R}} ω\omega π\pi r2 sin ω\omega t
  • B μoI4 R{{{\mu _o}{\rm I}} \over {4\,R}} ω\omega π\pi r2 sin ω\omega t
  • C μoI4 R{{{\mu _o}{\rm I}} \over {4\,R}} ω\omega r2 sin ω\omega t
  • D μoI2 R{{{\mu _o}{\rm I}} \over {2\,R}} ω\omega r2 sin ω\omega t

Step-by-Step Verified Solution

We know that electric flux ϕ=B→ . A→\phi = \overrightarrow B \,.\,\overrightarrow A

⇒ϕ=BAcos⁡ωt\Rightarrow \phi = BA\cos \omega t

Now, B=μ02IRB = {{{\mu _0}} \over 2}{I \over R} is magnetic field due to circular coil of radius R and A=πr2A = \pi {r^2} is area of circular coil of radius r. Therefore,

ϕ=μ02IRπr2cos⁡ωt\phi = {{{\mu _0}} \over 2}{I \over R}\pi {r^2}\cos \omega t

Now induced emf ε=−dϕdt=−ddt(μ02IRπr2cos⁡ωt)\varepsilon = {{ - d\phi } \over {dt}} = {{ - d} \over {dt}}\left( {{{{\mu _0}} \over 2}{I \over R}\pi {r^2}\cos \omega t} \right)

⇒ε=μ02IRπr2sin⁡ωt\Rightarrow \varepsilon = {{{\mu _0}} \over 2}{I \over R}\pi {r^2}\sin \omega t

Related Learning Resources & Exam Preparation

JEE Main PYQ Question Bank
Browse year-wise authentic previous years' questions with solutions
Authentic CBT Mock Tests
Simulate official exam environment with live timer and percentile scoring