Electro Magnetic Induction And Alternating Currents Question 552

Question: A circular coil of radius 5 cm has 500 turns of a wire. The approximate value of the coefficient of self induction of the coil will be [MP PET 1996; Pb PET 2000]

Options:

A) 25 millihenry

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

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

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

Show Answer

Answer:

Correct Answer: A

Solution:

$ \varphi =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}=25\ mH $



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