Home exercise 4.3.1: Toroidal core

One flux tube of length $l$, area $S$ and permeability ${\mu }_0{\mu }_r$:

The mmf is $NI=500\times 0.2=100\mathrm{At}$, so

The gap is in series with the iron, and its reluctance is

nearly seven times that of the whole ring. The iron path is now $l-l_g=0.299\mathrm{m}$, $Z_c^{\prime }=2.97\times 10^5{\mathrm{H}}^{-1}$, and $Z=Z_c^{\prime }+Z_g=2.29\times 10^6{\mathrm{H}}^{-1}$:

To restore $\Phi =3.35\times 10^{-4}\mathrm{Wb}$ the mmf must be $N{I}^{\prime }=\Phi Z=3.35\times 10^{-4}\times 2.29\times 10^6=766\mathrm{At}$:

Directly, $B={\mu }_0{\mu }_{r}NI/l=4\pi \times 10^{-7}\times 2000\times 100/0.30=0.84\mathrm{T}$, as in (a). One millimetre of air ($1/300$ of the path) takes $Z_g/(Z_c^{\prime }+Z_g)=87\%$ of the mmf: the flux drops by a factor $7.7$, and the current must be raised by the same factor to bring it back. This is Section 4.3.5 with numbers: $Z_g=Z_c$ already at $l_g=l/{\mu }_r=0.15\mathrm{mm}$.