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 $
BETA
