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A coil of wire of a certain radius has 100 turns and a self inductance of $15 \mathrm{mH}$. The self inductance of a second similar coil of 500 turns will be
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Verified Answer
The correct answer is:
$375 \mathrm{mH}$
Inductance of a coil is given by
$\begin{aligned} & L=\frac{1}{2} \mu_0 \pi N^2 R \Rightarrow \frac{L_2}{L_1}=\frac{N_2^2}{N_1^2} \\ & \therefore L_2=L_1 \frac{N_2^2}{N_1^2}=\left(\frac{500}{100}\right)^2 15 \mathrm{mH}=375 \mathrm{mH}\end{aligned}$
$\begin{aligned} & L=\frac{1}{2} \mu_0 \pi N^2 R \Rightarrow \frac{L_2}{L_1}=\frac{N_2^2}{N_1^2} \\ & \therefore L_2=L_1 \frac{N_2^2}{N_1^2}=\left(\frac{500}{100}\right)^2 15 \mathrm{mH}=375 \mathrm{mH}\end{aligned}$
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