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Electromagnetic Induction question

2018 · 16 Apr · Shift 1 · Q54
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Electromagnetic Induction question

2018 · 16 Apr · Shift 1 · Q54

JEE MainPhysicsElectromagnetic InductionMCQ+4 / −1
A coil of cross-sectional area A having n turns is placed in a uniform magnetic field B. When it is rotated with an angular velocity ω,\omega ,ω, the maxium e.m.f. induced in the coil will be:
  1. A
    3 nBA ω\omegaω
  2. B
    32{3 \over 2}23​ nBA ω\omegaω
  3. C
    nBA ω\omegaω
  4. D
    12{1 \over 2}21​ nBA ω\omegaω
View written solutionFree

Correct answer: C

  1. Magnetic flux through the rotating coil

For a coil of:

  • area AAA
  • number of turns nnn
  • placed in uniform magnetic field BBB

if it rotates with angular velocity ω\omegaω, the angle between the normal to the coil and the magnetic field changes as

θ=ωt\theta = \omega tθ=ωt

So the magnetic flux through the coil at time ttt is

Φ=nBAcos⁡θ=nBAcos⁡(ωt)\Phi = nBA\cos\theta = nBA\cos(\omega t)Φ=nBAcosθ=nBAcos(ωt)

  1. Induced e.m.f. using Faraday's law

Faraday's law gives

e=−dΦdte = -\frac{d\Phi}{dt}e=−dtdΦ​

Therefore,

e=−ddt(nBAcos⁡(ωt))e = -\frac{d}{dt}\big(nBA\cos(\omega t)\big)e=−dtd​(nBAcos(ωt))

e=−nBA(−ωsin⁡(ωt))e = -nBA\big(-\omega \sin(\omega t)\big)e=−nBA(−ωsin(ωt))

e=nBAωsin⁡(ωt)e = nBA\omega \sin(\omega t)e=nBAωsin(ωt)

  1. Maximum induced e.m.f.

The maximum value of sin⁡(ωt)\sin(\omega t)sin(ωt) is 111.

Hence,

emax⁡=nBAωe_{\max} = nBA\omegaemax​=nBAω

  1. Match with the options
  • A: 3nBAω3nBA\omega3nBAω ❌
  • B: 32nBAω\dfrac{3}{2}nBA\omega23​nBAω ❌
  • C: nBAωnBA\omeganBAω ✅
  • D: 12nBAω\dfrac{1}{2}nBA\omega21​nBAω ❌

So the correct option is C.

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