Coil inductance

Question

A coil of 1000 turns is wound on a laminated core of steel having a cross-section of 5 cm². The core has an air gap of 2 mm cut at right angle. What value of current is required to have an air gap flux density of 0·5 T? Permeability of steel may be taken as infinity. Determine the coil inductance

Answer 1

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

L=N^(2)μA/l

Where:

• N is the number of turns (1000 turns)

• μ is the permeability of the core material (taken as infinity here, which simplifies things)

• A is the cross-sectional area (5 cm2=5×10−4 m2)

• l is the length of the coil (not provided, so this will be a theoretical infinite inductance)

Since the permeability (μ) is taken as infinity, and we're assuming an ideal scenario with no air gap or non-magnetic material, the inductance formula simplifies because the magnetic path length l and other resistive factors approach zero impact.

Thus, with μ being infinitely large: L --> ∞

So theoretically, the inductance of the coil in such a setup approaches infinity.

Answer 2

First, let's calculate the magnetic reluctance (R) of the core with the air gap:

\(R = \frac{l_{\text{core}} + l_{\text{gap}}}{\mu_0 \mu_r A}\)

Since \(\mu_r\) is infinite, the term \(\mu_0 \mu_r\) becomes very large, making the reluctance R very small. Therefore, we can simplify the equation to:

\(R = \frac{l_{\text{core}} + l_{\text{gap}}}{\mu_0 A}\)

Given:- 

\(l_{\text{core}} = 0\,\)  negligible compared to air gap

\(l_{\text{gap}} = 2 \times 10^{-3} \, \text{m} \)

\(\mu_0 = 4\pi \times 10^{-7} \, \text{T} \cdot \text{m}/\text{A}\)

\(A = 5 \times 10^{-4} \, \text{m}^2\)

Plugging in the values:

\(R = \frac{2 \times 10^{-3}}{4\pi \times 10^{-7} \times 5 \times 10^{-4}}\)

\(R = \frac{2 \times 10^{-3}}{20\pi \times 10^{-11}}\)

\(R = \frac{10^{8}}{\pi} \, \text{A/T}\)

Next, let's use the formula for magnetic flux density B:

\(B = \frac{\mu_0 \mu_r N I}{l_{\text{core}} + l_{\text{gap}}}\)

Given:

\(B = 0.5 \, \text{T}\)

\(\mu_r = \infty\)

N = 1000

Plugging in the values:

\(0.5 = \frac{(4\pi \times 10^{-7}) \times \infty \times 1000 \times I}{2 \times 10^{-3}}\)

\(0.5 = \frac{2 \times 10^{-4} \, \text{I}}{\pi}\)

Solving for I:

\(I = \frac{0.5 \times \pi}{2 \times 10^{-4}}\)

\(I = \frac{5\pi}{2} \times 10^3 \, \text{A}\)

\(I \approx 7853.98 \, \text{A}\)

Now, once we have the current, we can calculate the coil inductance using the formula:

\(L = \frac{N^2}{R}\)

Given:

\(N = 1000 \)

\(R = \frac{10^{8}}{\pi} \, \text{A/T}\)

Plugging in the values:

\(L = \frac{1000^2}{\frac{10^{8}}{\pi}}\)
\(L = \frac{10^6 \pi}{10^8} \, \text{H}\\ L = \frac{\pi}{100} \, \text{H} \\ L \approx 0.0314 \, \text{H} \)

So, the value of current required to have an air gap flux density of 0.5 T is approximately 7853.98 A, and the coil inductance is approximately 0.0314 H.