Question 267: Electromagnetic Induction

Question

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

Accepted Answer

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

Answer

0.1 H

Answer

1 mH

Answer

1 H

Answer

10 mH

Electromagnetic Induction

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

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