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JEE Main 2006 (AIEEE) · Magnetic Flux & Faraday's Law

Magnetic Flux & Faraday's Law question from JEE Main 2006 (AIEEE)

Q84JEE Main 2006 · AIEEEEasy

Magnetic Flux & Faraday's Law

The flux linked with a coil at any instant ’t’\text{'}t\text{'} is given by

ϕ=10t2−50t+250\phi = 10t^{2} - 50t + 250

The induced emfemf at t=3 st = 3\text{ s} is

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Answer: (b)

  1. e=−dϕdt=−(20t−50)e = -\frac{d\phi}{dt} = -(20t - 50)
  2. At t=3t = 3 s: e=−(60−50)=−10e = -(60 - 50) = -10 V
  3. So the answer is (b) −10-10 V.
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