Question 5. A magnetic circuit of a plunger is shown in Figure 3. Calculate the force acting on the plunger when the distance r is 2cm and the current in the 100 - turn coil is 5A. The relative permeability of the magnetic material is 2000. Assume that the bushing is non magnetic Bushing Magnetic material 4.-2000 4,1 cm 100 20 cm Airgap cm 10:45 T cm Cross-sectional area of all parts - 5 cm -20 cm

A_{c}=5 \mathrm{~cm}^{2}=5 * 10^{-4} \mathrm{~m}^{2}   Simplify the parallel reluctances:   Calculate the inductance of the circuit: \mathcal{R}_{g}=\frac{l}{\mu_{0} A_{c}}=\frac{x}{4 \pi * 10^{-7} * 5 * 10^{-4}}=1.59 * 10^{9} x \mathcal{R}_{1}=\frac{l}{2000 \mu_{0} A_{c}}=\frac{0.05}{2000 * 4 \pi * 10^{-7} * 5 * 10^{-4}}=39788.74 \mathcal{R}_{4}=\frac{l}{2000 \mu_{0} A_{c}}=\frac{0.2-0.05-x}{2000 * 4 \pi * 10^{-7} * 5 * 10^{-4}}=119366.20-795774.72 x \mathcal{R}_{2}=\frac{l}{2000 \mu_{0} A_{c}}=\frac{0.4}{2000 * 4 \pi * 10^{-7} * 5 * 10^{-4}}=318309.89 \mathcal{R}_{b}=\frac{l}{\mu_{0} A_{c}}=\frac{0.005}{4 \pi * 10^{-7} * 5 * 10^{-4}}=7957747.15   Equivalent reluctance: \mathcal{R}_{e q}=\mathcal{R}_{1}+\mathcal{R}_{4}+\mathcal{R}_{g}+\frac{\mathcal{R}_{2}}{2}+\frac{\mathcal{R}_{b}}{2} \mathcal{R}_{e q}=39788.74+119366.20-795774.72 x+1.59 * 10^{9} x+\frac{318309.89}{2}+\frac{7957747.15}{2} \mathcal{R}_{e q}=4297183.46+1.589 * 10^{9} x   L=\frac{N}{\mathcal{R}_{e q}}=\frac{100}{4297183.46+1.589 * 10^{9} x} U=\frac{1}{2} L i^{2}=\frac{1}{2} \frac{100}{4297183.46+1.589 * 10^{9} x}(5)^{2} U=\frac{1250}{4297183.46+1.589 * 10^{9} x}   Force is: f(x)=-\frac{d}{d x} U(x)=-\frac{d}{d x} \frac{1250}{4297183.46+1.589 * 10^{9} x} f(x)=\frac{1250}{\left(4297183.46+1.589 * 10^{9} x\right)^{2}}\left(1.589 * 10^{9}\right)=\frac{1.98 * 10^{12}}{\left(4297183.46+1.589 * 10^{9} x\right)^{2}} At x=2cm f(0.02)=\frac{1.98 * 10^{12}}{\left(4297183.46+1.589 * 10^{9} * 0.02\right)^{2}}=0.0015 N ...