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.