Q171 / 180 · COMEDK 2020 · Physics
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
- A$75 \mathrm{mH}$
- Correct answerB$375 \mathrm{mH}$
- C$15 \mathrm{mH}$
- DNone of the above
Solution
Given, number of turns in the first ood, $N_{1}=100$
Its self-indioctanee, $L_{\mathrm{t}}=15 \mathrm{mII}=15 \times 10^{-2} \mathrm{11}$
Number of turns in the socond eodl.
$$
N_{2}=500
$$
$$
\therefore \quad \mathrm{Leg}^{2}=\text { ? }
$$
We know that, self inductance of the coil is given by
$$
\begin{aligned}
L &=\frac{\mu_{0 \pi N} N^{2} R}{2} \\
\Rightarrow \quad L & \propto N^{2} \\
\therefore \quad \frac{L_{2}}{L_{1}} &=\left(\frac{N_{2}}{N_{1}}\right)^{2}=\left(\frac{500}{100}\right)^{2}=25 \\
\Rightarrow \quad L_{2} &=25 L_{4} \\
&=25 \times 15 \times 10^{-2} \\
&=375 \times 10^{-2} \mathrm{H} \\
&=375 \times 10^{-3} \mathrm{H}=375 \mathrm{mH}
\end{aligned}
$$
Hence, the correct answer is option: 'B'.