A coil of wire of 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

Question

Dr. Abubacker · Accepted Answer

(b) $\frac{L_{2}}{L_{1}} = \frac{N_{2}^{2}}{N_{1}^{2}}$
$	herefore\ \ L_{2} = L_{1}\frac{N_{2}^{2}}{N_{1}^{2}} = 1.5\ \left( \frac{500}{100} ight)^{2} = 375\ $mH

Q. A coil of wire of 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

Explanation

Reviewed by Dr. Abubacker

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