An air cored coil has a self inductance 0.1 H . A soft iron core of relative permeability 1000 is introduced…
An air cored coil has a self inductance 0.1 H . A soft iron core of relative permeability 1000 is introduced and the number of turns is reduced $\left(\frac{1}{10}\right)^{\text {th }}$. The value of self inductance is
0.1 H
1 mH
1 H
10 mH
Solution
For air core coil, $L=0.1=\frac{\mu_0 \mathrm{~N}^2 \mathrm{~A}}{l}$
For soft iron core coil, $\mathrm{L}^{\prime}=\frac{\mu_0 \mu_{\mathrm{r}}\left(\mathrm{N}^{\prime}\right)^2 \mathrm{~A}}{l}$
$\begin{aligned}
& \frac{L^{\prime}}{L}=\frac{1000 \times \mu_0 \times\left(\frac{N}{10}\right)^2 \mathrm{~A} \times l}{\mu_0 \mathrm{~N}_2 \mathrm{~A} \cdot l} \\
& \frac{\mathrm{~L}^{\prime}}{\mathrm{L}^{\prime}}=10 \\
& \mathrm{~L}^{\prime}=0.1 \times 10=1 \mathrm{H}
\end{aligned}$