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  • Stuck here, help me understand: - Electromagnetic Induction and alternative Currents - NEET-2

A long solenoid of diameter 0.1m has 2\times 10^{4} turns per meter. At the centre of the solenoid, a coil of 100 turns and radius 0.01 m is placed with its axis coinciding with the solenoid axis. The current in the solenoid reduces at a constant rate to 0A from 4A in 0.05s.  If the resistance of the coil is 10\pi ^{2}\Omega. the total charge flowing through the coil during this time is:

  • Option 1)

    16\mu C

  • Option 2)

    32\mu C

  • Option 3)

    16 \pi \mu C

  • Option 4)

    32 \pi \mu C

 

Answers (1)

best_answer

 

Faraday Second Law of Induction emf -

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

-

 

 B=\mu ni\ \left ( For \ soleroid \right )

\left | \varepsilon =\frac{d\Phi }{dt} \right |=\frac{d}{dt}\left ( B.A \right )=\left ( M_{o}nA \right )\frac{di}{dt}

=4\pi\times 10^{-7}\times 2\times 10^{4}\times 100\times \pi\times \left ( 0.01 \right )^{2}\times \frac{4}{.05}

=8\pi^{2}\times 10^{-1}\times \frac{4\times 10^{-4}}{.05}=6.4\pi^{2}\times 10^{-3}V

I=\frac{\varepsilon }{R}=\frac{6.4\pi^{2}\times 10^{-3}}{10\pi^{2}}A=6.4\times 10^{-4}A

\therefore Total\ charge=It=6.4\times 10^{-4}\times 0.05=32\mu c


Option 1)

16\mu C

This is incorrect option

Option 2)

32\mu C

This is correct option

Option 3)

16 \pi \mu C

This is incorrect option

Option 4)

32 \pi \mu C

This is incorrect option

Posted by

Aadil

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