Question

The number of turns in a coil of wire of fixed radius is 600 and its self inductance is 108 mH. The self inductance of a coil of 500 turns will be :

• A. 74 mH
• B. 75 mH
• C. 76 mH
• D. 77 mH

Answer 1

The correct option is B 75 mH

$L=\frac{\mu n^{2}A}{L}$.

$L\propto n^2$

Thus, for 500 turns, self inductance would be $108\times \left(500{)}^2/\right(600{)}^2=108\times 25/36=75mH$