Home exercise 4.3.3: Lifting electromagnet

The plate hangs from two pole faces, each pulling with $F_1=B^{2}S/(2{\mu }_0)$, so the total force is $F=B^{2}S/{\mu }_0$. It must equal the weight $mg=100\times 9.81=981\mathrm{N}$:

Loop rule round the flux path, one iron path and two gaps in series (equal area everywhere, so $B$ is the same in iron and gap):

With $l_g=1\mathrm{mm}$ the gap term doubles, $2\times 1.11\times 10^{-3}/{\mu }_0=1767\mathrm{At}$, while the iron term stays $236\mathrm{At}$:

Units: $B^{2}S/{\mu }_0$ is ${\mathrm{T}}^2{\mathrm{m}}^2/(H/m)=\mathrm{W}{\mathrm{b}}^2/(\mathrm{H}{\mathrm{m}}^2)\times {\mathrm{m}}^2=J/m=\mathrm{N}$. Even at $0.5\mathrm{mm}$ the two gaps take $79\%$ of the mmf, and the $40\mathrm{cm}$ of iron (${\mu }_r=1500$) are worth only $0.27\mathrm{mm}$ of air. Conversely, at a fixed current the induction, and with it the force $\propto B^2$, collapses as the gap opens: with $I=2.8\mathrm{A}$ at $l_g=1\mathrm{mm}$ one gets $B\approx 0.62\mathrm{T}$ and $F\approx 310\mathrm{N}$, less than a third of the weight. A lifting magnet therefore needs clean, flat contact, and a safety factor in the current.