Question

A coil wound on an iron coil of permeability 400 has 150 turns and a cross sectional area of 5cm squared. Calculate the inductance of the coil. Given that a steady current of 3mA produces a magnetic field of 10 lines /cm2 when air is present as the medium

Answer 1

To calculate the inductance (\(L\)) of the coil, we can use the formula:

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

where:
- \(L\) = inductance in henries (H)
- \(N\) = number of turns (150 turns)
- \(\mu\) = permeability of the core material (H/m)
- \(A\) = cross-sectional area (m²)
- \(l\) = length of the coil (m)

However, we don't know the length of the coil directly from the information provided. We can determine the permeability \(\mu\) using the permeability of free space (\(\mu_0\)) and the relative permeability \(\mu_r\):

1. \(\mu_0\) (permeability of free space) is approximately \(4 \pi \times 10^{-7} \, \text{H/m}\).
2. \(\mu_r\) is given as 400.

Thus, the permeability \(\mu\) would be:

\[
\mu = \mu_0 \mu_r = (4 \pi \times 10^{-7} \, \text{H/m}) \times 400
\]

Calculating \(\mu\):

\[
\mu \approx 4 \pi \times 10^{-5} \, \text{H/m}
\]

Approximating \(\pi \approx 3.14\):

\[
\mu \approx 4 \times 3.14 \times 10^{-5} \approx 1.256 \times 10^{-4} \, \text{H/m}
\]

Next, we will convert the cross-sectional area \(A\) from cm² to m²:

\[
A = 5 \, \text{cm}^2 = 5 \times 10^{-4} \, \text{m}^2
\]

We need to determine the length of the coil \(l\). From the magnetic field (\(B\)) and current, we can find the magnetic field strength (\(H\)). Since the magnetic field strength \(B\) in air is given as \(10 \, \text{lines/cm}^2\), we first need to convert this to Teslas (1 line/cm² = 0.0001 T):

\[
B = 10 \, \text{lines/cm}^2 = 1 \times 10^{-3} \, \text{T}
\]

The relationship between \(B\), \(\mu\), and \(H\) is given by:

\[
B = \mu H
\]

Thus:

\[
H = \frac{B}{\mu}
\]

Substituting for \(H\):

\[
H = \frac{1 \times 10^{-3} \, \text{T}}{1.256 \times 10^{-4} \, \text{H/m}} \approx 7.96 \, \text{A/m}
\]

Given that the length \(l\) can be derived from \(H\) and and the number of turns (\(N\)), we have:

\[
l = \frac{N}{H}
\]

Substituting \(N = 150\) turns:

\[
l = \frac{150}{7.96} \approx 18.84 \, \text{m}
\]

Now substituting all the values back into the inductance formula, we can compute the inductance \(L\):

\[
L = \frac{N^2 \mu A}{l} = \frac{150^2 \times 1.256 \times 10^{-4} \times 5 \times 10^{-4}}{18.84}
\]

Calculating:

\[
150^2 = 22500
\]
\[
22500 \times 1.256 \times 10^{-4} \approx 2.832 \, \text{H} \cdot \text{m}^2
\]
Multiply by \(5 \times 10^{-4}\):

\[
2.832 \, \text{H} \cdot \text{m}^2 \times 5 \times 10^{-4} = 1.416 \times 10^{-3} \, \text{H} \cdot \text{m}^2
\]
Finally divide by \(18.84\):

\[
L \approx \frac{1.416 \times 10^{-3}}{18.84} \approx 7.50 \times 10^{-5} \, \text{H} \approx 75 \, \mu H
\]

Thus, the inductance of the coil is approximately \(75 \, \mu H\).