An iron ring of mean length 50 cm, has an air – gap of 1 mm and winding of 200 turns. The relative permeability of iron is 300. When 1A current follows through the coil, determine flux density.
Given that
The mean length of the iron ring
l=50 cm=0.5 m
\(l_{ag}=1\ mm=1\times{10}^{-3}m\)
I=1 Amp.
Number of turns N=200
Relative permeability of iron \(\mu_r=300\)
For finding the flux density, the ampere turns for iron and air gape are to be determined. Leakage and fringing effects are neglected
At iron=\(H\times l\)
Where H is the magnetic field intensity
\(H=\frac{B}{\mu_0\mu_r}=\frac{B}{4\pi\times{10}^{-7}\times300}=2652.58BAT/m\)
Where B is the flux density.
Hence AT for iron is \(HlAT=2652.58B\times0.5AT=26.29BAT\)
At for air gap is \(H_{air\ gap}\times l_{ag}AT=\frac{B}{4\pi\times{10}^{-7}}\times{10}^{-3}=795.77B\ AT\)
Total ampere turns is \(B\left(1326.29+795.77\right)AT=2122.06B\ AT\)
Total ampere turns can be expressed as \(N\times I=200\times1AT200\times1=2122.06B\)
Hence flux density \(B=\frac{200}{2122.06}Wb/m^2=0.094Wb/m^2\)
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Fig. 1
