Question:

If air core is replaced by an iron core in an inductor, its self-inductance is increased from 0.02 mH to 40 mH. The relative permeabiliry of iron is

Updated On: Apr 7, 2025

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Approach Solution - 1

Relative Permeability and Inductance

When an air core in an inductor is replaced by an iron core, the self-inductance increases from 0.02 mH to 40 mH. We want to find the relative permeability of the iron.

Step 1: Inductance and Permeability

The inductance ($L$) of an inductor is directly proportional to the permeability ($\mu$) of the core material. The formula for the inductance of a solenoid is:

$L = \frac{\mu N^2 A}{l}$

Where:

$L$ is the inductance.
$\mu$ is the permeability of the core material.
$N$ is the number of turns in the coil.
$A$ is the cross-sectional area of the coil.
$l$ is the length of the coil.

Step 2: Relative Permeability

The relative permeability ($\mu_r$) is the ratio of the permeability of a material ($\mu$) to the permeability of free space ($\mu_0$):

$\mu_r = \frac{\mu}{\mu_0}$

Therefore, $\mu = \mu_r \mu_0$.

Step 3: Ratio of Inductances

Let $L_{air}$ be the inductance with the air core, and $L_{iron}$ be the inductance with the iron core. Then:

$L_{air} = \frac{\mu_0 N^2 A}{l}$

$L_{iron} = \frac{\mu N^2 A}{l} = \frac{\mu_r \mu_0 N^2 A}{l}$

Taking the ratio of the two inductances:

$\frac{L_{iron}}{L_{air}} = \frac{\frac{\mu_r \mu_0 N^2 A}{l}}{\frac{\mu_0 N^2 A}{l}} = \mu_r$

Step 4: Calculate $\mu_r$

We are given $L_{air} = 0.02 \text{ mH}$ and $L_{iron} = 40 \text{ mH}$. Therefore:

$\mu_r = \frac{L_{iron}}{L_{air}} = \frac{40 \text{ mH}}{0.02 \text{ mH}} = 2000$

Conclusion

The relative permeability of the iron is 2000.

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Approach Solution -2

The self-inductance of a coil is given by the formula: \[ L = \mu \frac{N^2 A}{l} \] Where:
$ L $ is the inductance
$ \mu $ is the permeability of the core material
$ N $ is the number of turns
$ A $ is the cross-sectional area of the coil
$ l $ is the length of the coil

When air is replaced by iron, the inductance changes in proportion to the relative permeability of the core material. The inductance of a coil with an air core is given by: \[ L_{\text{air}} = \mu_0 \frac{N^2 A}{l} \] And with an iron core, the inductance becomes: \[ L_{\text{iron}} = \mu_{\text{iron}} \frac{N^2 A}{l} \] The relative permeability $ \mu_{\text{iron}} $ is related to the permeability of free space $ \mu_0 $ by: \[ \mu_{\text{iron}} = \mu_0 \times \mu_r \] Therefore, the ratio of the inductances gives the relative permeability $ \mu_r $: \[ \frac{L_{\text{iron}}}{L_{\text{air}}} = \mu_r \] Substitute the values for inductances: \[ \frac{40 \, \text{mH}}{0.02 \, \text{mH}} = \mu_r \] \[ \mu_r = 2000 \] Thus, the relative permeability of iron is 2000.