Check: the dimension lines on Figure 6 give the outer-arc length ($60\text{ cm}$), a leg cross-section of $5\text{ cm}$, a $1\text{ mm}$ gap, and a $30\text{ cm}\times6\text{ cm}$ armature -- but the drawing does not label the core's out-of-page depth or the legs' full vertical run separately from the arc. Following the standard convention for this figure style, the solution below takes the core (and the flux-carrying portion of the gap/armature) as a square $5\text{ cm}\times5\text{ cm}$ cross-section (area $A=25\text{ cm}^2$), and the mean core path as the $60\text{ cm}$ arc plus one $5\text{ cm}$ leg-length allowance on each side ($l_{core}=60+2(5)=70\text{ cm}$); fringing at the gap is neglected, as is standard at this level.
Given data
Quantity
Value
$\mu_r$
$2000$
$N$
$1000\text{ turns}$
$i$
$1\text{ A}$
Mean core length $l_{core}$
$70\text{ cm}$
Cross-sectional area $A$
$25\text{ cm}^2$
Air gap $l_g$
$1\text{ mm}$
Find. Total mmf; core and gap reluctances; flux, flux density, and field intensity in the gap; force on the armature.
Figure 6: Horseshoe core + relay armature -- N=1000, μ_r=2000, i=1A, 1 mm gap
Approach. Model the horseshoe core and air gap as two reluctances in series driven by one mmf source; solve for flux, then $B$ and $H$ in the gap from the (uniform) cross-sectional area; finally use the magnetic force formula for the attractive force across an air gap.
a) Total mmf. $$\mathcal{F} = Ni = (1000)(1) = \boxed{1000\text{ A}\cdot\text{t}}$$
b) Reluctances. Core (relative permeability $\mu_r$, length $l_{core}$, area $A$): $$\mathcal{R}_{core} = \dfrac{l_{core}}{\mu_0\mu_rA} = \dfrac{0.70}{(4\pi\times10^{-7})(2000)(25\times10^{-4})} = \boxed{1.114\times10^{5}\text{ A}\cdot\text{t/Wb}}$$ Air gap (non-magnetic, $\mu_r=1$, same area $A$, length $l_g$): $$\mathcal{R}_{gap} = \dfrac{l_g}{\mu_0A} = \dfrac{0.001}{(4\pi\times10^{-7})(25\times10^{-4})} = \boxed{3.183\times10^{5}\text{ A}\cdot\text{t/Wb}}$$ Total (series): $$\mathcal{R}_{tot} = \mathcal{R}_{core}+\mathcal{R}_{gap} = \boxed{4.297\times10^{5}\text{ A}\cdot\text{t/Wb}}$$
c) Flux, flux density, field intensity in the gap. $$\phi = \dfrac{\mathcal{F}}{\mathcal{R}_{tot}} = \dfrac{1000}{4.297\times10^5} = \boxed{2.327\text{ mWb}}$$ $$B_g = \dfrac{\phi}{A} = \dfrac{2.327\times10^{-3}}{25\times10^{-4}} = \boxed{0.9308\text{ T}}$$ $$H_g = \dfrac{B_g}{\mu_0} = \dfrac{0.9308}{4\pi\times10^{-7}} = \boxed{7.407\times10^{5}\text{ A/m}}$$
d) Force on the armature. The attractive force across an air gap of area $A$ carrying flux density $B_g$ is $$F = \dfrac{B_g^2A}{2\mu_0} = \dfrac{(0.9308)^2(25\times10^{-4})}{2(4\pi\times10^{-7})} = \boxed{861.9\text{ N}}$$ directed to close the gap (pulling the armature up toward the fixed core).
