Question:

Published on: 22 June, 2022

**A flux of 0.0006Wb is required in the air – gap of an iron ring of cross – section 5.0 cm2and mean length 2.7 m with an air – gap of 4.5 mm. Determine the ampere turns required. Six H values and corresponding B values are noted from the magnetization curve of iron and given below.**

## H(At/m) |
## 200 |
## 400 |
## 500 |
## 600 |
## 800 |
## 1000 |

## B(Wb/m2) |
## 0.4 |
## 0.8 |
## 1.0 |
## 1.09 |
## 1.17 |
## 1.19 |

Answer:

Given that, Flux=0.0006 Wb

Cross section area \(A=5\ {cm}^2=5\times{10}^{-4}m^2\)

The flux density \(B=\frac{0.0006}{5\times{10}^{-4}}=1.2\ Wb/m^2\)

From the given data, we have

\(H(AT/m)=\frac{1000AT}{m}forB=1.2Wb/m^2\)

Hence ampere turns required for the iron part producing the given flux density.

Flux density =\(Hl=1000\times2.7\ At=2700\ AT\)

Extra ampere turns necessary to produce the same flux in the air gap is

\(B=\mu H\)

Here \(H=\frac{B}{\mu_0}=\frac{1.2}{4\pi\times{10}^{-7}}=954930\ AT\)

Extra ampere turns required is \(954930\times4.5\times{10}^{-3}AT=4297.185\ AT\)

The total number of ampere turns necessary is \(\left(2700+4297.185\right)=6997.185\ AT\)

Neglecting leakage and fringing field.

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