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

`(L_(2))/(L_(1))=(N_(2)^(2))/(N_(1)^(2))`
`therefore L_(2)=L_(1)(N_(2)^(2))/(N_(1)^(2))=15((500)/(100))^(2)=375mH`