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  • Need clarity, kindly explain! - Electromagnetic Induction and Alternating currents - JEE Main-3

The flux linked with a coil at any instant t is given by \phi = 10t2 - 50t + 250. The induced emf at t = 3s is

  • Option 1)

    190 V

  • Option 2)

    -190 V

  • Option 3)

    -10 V

  • Option 4)

    10 V

 

Answers (1)

As we learnt in 

Rate of change of magnetic Flux -

\varepsilon = \frac{-d\phi }{dt}
 

- wherein

d\phi\rightarrow \phi _{2}-\phi _{1}

\phi _{2}-\phi _{1}= change in flux

\phi = 10t2 - 50t + 250

\therefore \; \; \frac{d\phi }{dt}=20t-50

Induced\; emf\; \; e=\frac{-d\phi }{dt}

or\; e=-(20t-50)=-\left [ (20\times 3)-50 \right ]=-10\; volt

or\; \; e=-10\; volt

 


Option 1)

190 V

This option is incorrect

Option 2)

-190 V

This option is incorrect

Option 3)

-10 V

This option is correct

Option 4)

10 V

This option is incorrect

Posted by

Sabhrant Ambastha

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