An air cored coil has self inductance of 0.1 H . A soft iron core of relative permeability 1000 is introduced and the number of turns is reduced to \(\left(\frac{1}{10}\right)^{\text {th }}\). The value of self inductance is now

\(\begin{aligned}
& \mathrm{L}=\frac{\mu_0 \mathrm{~N}^2 \mathrm{~A}}{l} \\
& \mathrm{~L}^{\prime}=\frac{\mu_0 \mu_{\mathrm{r}}\left(\mathrm{~N}^{\prime}\right)^2 \mathrm{~A}}{l}=\frac{\mu_0 \times 1000 \times\left(\frac{\mathrm{N}}{10}\right)^2 \mathrm{~A}}{l}
\end{aligned}\)
When, \(\mu_{\mathrm{r}}=1000\) and \(\mathrm{N}=\frac{1}{10}\),
\(\therefore \frac{\mathrm{L}}{\mathrm{~L}^{\prime}}=\frac{0.1}{\mathrm{~L}^{\prime}}=\frac{\mu_0 \mathrm{~N}^2 \mathrm{~A}}{l}\) \(\times \frac{l}{\mu_0 \times 1000 \times\left(\frac{\mathrm{N}}{10}\right)^2 \mathrm{~A}}\)
\(\therefore \frac{0.1}{\mathrm{~L}^{\prime}}=\frac{1}{10}\)
\(\therefore \mathrm{~L}^{\prime}=0.1 \times 10=1 \mathrm{H}\)