Electro Magnetic Induction And Alternating Currents Question 499

Question: A circular coil is radius 5 cm has 500 turns of a wire. The approximate value of the coefficient of self-induction of the coil will be-

Options:

A) $ 25mH $

B) $ 25\times {{10}^{-3}},mH $

C) $ 50\times {{10}^{-3}},mH $

D) $ 50\times {{10}^{-3}},H $

Show Answer

Answer:

Correct Answer: A

Solution:

  • $ \phi =Li\Rightarrow NBA=Li $

    Since magnetic field at the centre of circular coil carrying current is given by $ B=\frac{{\mu_{0}}}{4\pi } $ ,

    $ \frac{2\pi Ni}{r} $
    $ \therefore N.\frac{{\mu_{0}}}{4\pi }.\frac{2\pi Ni}{r}.\pi r^{2}=Li\Rightarrow L=\frac{{\mu_{0}}N^{2}\pi r}{2} $

    Hence self-inductance of a coil

    $ \frac{4\pi \times {{10}^{-7}}\times 500\times 500\times \pi \times 0.05}{2}=25mH $



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