Question: A coil of wire of a certain radius has 600 turns and a self-inductance of 108 mH. The self-inductanc...

Question

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

Answer 1

Solution

$L \propto N^{2}$ ⇒ $\frac{L_{1}}{L_{2}} = \left( \frac{N_{1}}{N_{2}} \right)^{2} \Rightarrow \frac{108}{L_{2}} = \left( \frac{600}{500} \right)^{2}$ ;

$L_{2} = 75mH$