A coil of wire of a certain radius has `600` turns and a self-inductance of `108 mH`. The self-inductance of a `2^(nd)` similar coil of `500` turns will be

Options

A. 80 mH
B. 75 mH
C. 108 mH
D. 90 mH

Answer

B

Explanation

Given, `L_(1)=108mH,N_(1)=600` turns, `N_(2)=500` turns and `L_(2)=?`
by slef inductance of a plane coil,
`L_(1)=(mu_(0)piN_(1)^(2)a_(1))/(2)` . . . (i)
and `L_(2)=(mu_(0)piN_(2)^(2)a_(2))/(2)` . . . (ii)
From the eqs. (i) and (ii), we get
`(L_(1))/(L_(2))=((N_(1))/(N_(2)))^(2)" "(because a_(1)=a_(2))`
`(108)/(L_(2))=((600)/(500))^(2)implies L_(2)=(108xx25)/(36)=3xx25=75mH`