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 $
