A coil of wire of a certain radius has 100 turns and a self inductance of 15 mH. The self inductance of a second similar coil of 500 turns will be

Options

A. 75 mH
B. 375 mH
C. 15 mH
D. none of these

Answer

B

Explanation

In ductance of a coil is given by
`L=(I)/(2)mu_(0)piN^(2)RrArr(L_(2))/(L_(1))=(N_(2)^(2))/(N_(1)^(2))`
`therefore L_(2)=L_(1)(N_(2)^(2))/(N_(1)^(2))=((500)/(100))^(2)" 15 mH = 375 mH"`