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 \(B=\frac{200}{2122.06}Wb/m^2=0.094Wb/m^2\)