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

Options

A. `25 millihenry`
B. `25xx10^(-3)` millihenry
C. `50xx10^(-3)` millihenry
D. `50xx10^(-3)` henry

Answer

A

Explanation

`varphi=LiimpliesNBA=LI`
since magnetic field at the centre of circular coil carrying current is given by `B=(mu_(0))/(4pi).(2piNi)/(R )`
`:. N.(mu_(0))/(4pi).(2piNi)/(R).pir^(2)=LiimpliesL(mu_(0)N^(2)pir)/(2)`
Hence self-inductance of a coil
`=(4pixx10^(-7)xx500xx500xxpixx0.05)/(2)=25 mH`