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  • Can someone help me with this, - Electromagnetic Induction and alternative Currents - NEET

 A coil of resistance 400\Omega is placed in a magnetic field. If the magnetic flux \phi (wb) linked with the coil varies with time t (\sec) as \phi=50t^{2}+4 . The current in the coil at t=2 sec is:

 

  • Option 1)

    0.5A

  • Option 2)

    0.1A

  • Option 3)

    2A

  • Option 4)

    1A

 

Answers (1)

 

Faraday Second Law of Induction emf -

varepsilon = frac{-dphi }{dt}= -Lfrac{dI}{dt}

-

 

 \Phi =50t^{2}+4

\Rightarrow \left | \varepsilon \right |=\left | \frac{d\Phi }{dt} \right |=100t

=200v\left ( at\ t=2sec \right )

\Rightarrow Current I=\frac{\varepsilon }{R}=\frac{200}{400}A=0.5A


Option 1)

0.5A

This is correct option

Option 2)

0.1A

This is incorrect option

Option 3)

2A

This is incorrect option

Option 4)

1A

This is incorrect option

Posted by

Vakul

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