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 $\left(\frac{1}{10} ight)^{ ext {th }}$. The value of self inductance is now

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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 $\left(\frac{1}{10}ight)^{	ext {th }}$. The value of self inductance is now

Aicharya · Accepted Answer

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

An air cored coil has self inductance of 0.1 H . A soft iron core of relative permeability 1000 is…

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