Electro Magnetic Induction And Alternating Currents Question 490

Question: A current of 1.5 A flows through a solenoid of length 20.0 cm, cross-section $ 20.0cm^{2} $ and 400 turns. The current is suddenly switched off in a short time of 1.0 millisecond. Ignoring the variation in the magnetic field the ends, the average back emf induced in the solenoid is:

Options:

A) 0.3 V

B) 9.6 V

C) 30.0 V

D) 3.0 V

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Answer:

Correct Answer: A

Solution:

  • $ e_{back}=L\frac{di}{dt} $

    where $ L=\frac{{\mu_{0}}N^{2}A}{l} $
    $ \therefore e_{back}=\frac{{\mu_{0}}N^{2}A}{l}[ \frac{1.5-0}{1\times 10^{3}} ] $

    $ =\frac{4\pi \times {{10}^{-7}}{{(400)}^{2}}\times 20\times {{10}^{-4}}}{20\times {{10}^{-2}}}\times (1.5\times 10^{3}) $ = 0.3 V



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