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

Step-by-step solution

Given, number of turns in the first ood, N₁=100 Its self-indioctanee, L_ t =15 mII =15 10⁻² 11 Number of turns in the socond eodl. N₂=500 Leg ²= ? We know that, self inductance of the coil is given by aligned L &= _ 0 N N² R 2 L & N² L₂ L₁ &= ( N₂ N₁ )²= ( 500 100 )²=25 L₂ &=25 L₄ &=25 15 10⁻² &=375 10⁻² H &=375 10⁻³ H =375 mH aligned