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Q. 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

AIIMSAIIMS 2009Electromagnetic Induction

Solution:

Inductance of a coil is given by
$L = \frac{1}{2}\mu_{0}\pi N^{2}R$
$ \Rightarrow \frac{L_{2}}{L_{1}} = \frac{N^{2}_{2}}{N^{2}_{1}}$
$\therefore \, L_{2} = L_{1} \frac{N^{2}_{2}}{N^{2}_{1}} = \left(\frac{500}{100}\right)^{2} \, 15\, mH = 375\,mH$