Question

A coil of wire of a certain radius has 600 turns and a self-inductance is 108 mH. The self-inductance of a second similar coil of 500 turns will be:

पर्याय

• 74 mH

• 75 mH

• 76 H

• 77 mH

Answer 1

75 mH

Explanation:

As Ln²

∴ L = `108 xx (500/600)^2` mH = 75 mH.