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JEE Main 2006
Exam
JEE Main
Year
2006
Subject
Physics
Chapter
Electromagnetic Induction

JEE Main 2006 · Physics · Electromagnetic Induction

The flux linked with a coil at any instant ' t ' is given by φ=10...

The flux linked with a coil at any instant ' tt ' is given by ϕ=10t2−50t+250\phi=10 t^2-50 t+250 The induced emf at t=3 st=3 \mathrm{~s} is

Options

  1. A190 V190 \mathrm{~V}
  2. B−190 V-190 \mathrm{~V}
  3. C−10 V-10 \mathrm{~V}
  4. D10 V10 \mathrm{~V}

Correct answer

C. −10 V-10 \mathrm{~V}

Step-by-step solution

e=−dϕdt=−(20t−50)=−10e=-\frac{d \phi}{d t}=-(20 t-50)=-10 volt
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