Question 239: Electromagnetic Induction

Question

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

Accepted Answer

The self-inductance $ L $ of an air-cored coil is given by:
$$ L_1 = \frac{\mu_0 N^2 A}{l} $$
where:
$ \mu_0 $ is the permeability of free space,
$ N $ is the number of turns,
$ A $ is the cross-sectional area,
$ l $ is the length of the coil.
When a soft iron core with a relative permeability $ \mu_r $ is introduced, the inductance becomes:
$$ L_2 = \frac{\mu_0 \mu_r N'^2 A}{l} $$
where $ N' = \frac{N}{10} $. Given that $ \mu_r = 1000 $, the ratio of the old inductance $ L_1 $ to the new inductance $ L_2 $ is:
$$ \frac{L_2}{L_1} = \frac{\mu_r (N')^2}{N^2} = \frac{1000 \left(\frac{N}{10}ight)^2}{N^2} $$
Simplifying:
$$ \frac{L_2}{L_1} = \frac{1000 \cdot \frac{N^2}{100}}{N^2} = \frac{1000}{100} = 10 $$
Therefore:
$$ L_2 = 10 	imes L_1 = 10 	imes 0.1 \, 	ext{H} = 1 \, 	ext{H} $$
Thus, the value of the self-inductance after introducing the soft iron core and reducing the number of turns is
1 H
. The correct option is
Option C: 1 H
.

Answer

0.1 H

Answer

1 mH

Answer

1 H

Answer

10 mH

Electromagnetic Induction

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

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