Question 1. A magnetic circuit of a plunger is shown below. Calculate the force acting on the plunger when the distance r = 2 cm and the current in the 100 - turn coil is 5 A. The relative permeability of the magnetic material is 2000. Assume that the bushing is nonmagnetic. 13 13 15 A 200 turns Thickness - 2 cm 0.5 21 Moving part 10.5 Nonmagnetic bushing 10 -21, 21-10 (All dimensions in centimeters)

The magnetic equivalent circunt can be drawn as.R - reluctance of magnetic path.R=\frac{l \cdot}{M_{0} M_{r} \cdot A}, H_{r_{\text {Dose }}}=2000, H_{1 \text { air }}: 1 \text {. }This implies that R_{\text {airgap }}>R_{\text {wore1, }}, R_{\text {core 2 }}, Rare 3\therefore R \sim R_{\text {airgap. (for easiness in calculation). }}^{\therefore \text {. }}.Ampere turns (NI) =100 \times 5=500 A T.\mathrm{Hg}=A T / \lg \quad \mathrm{lg}. air gap length.\lg _{g}=0.5 \mathrm{~cm}\therefore \mathrm{Hg}=500 /\left(0.5 \times 10^{-2}\right)=10^{5} \mathrm{AT} / \mathrm{m}\text { Now } B g=M_{0} H g=4 \pi \times 10^{-7} \times 10^{5}=0.1256 \mathrm{~T}Magnetic energy per anit volume stored in air gap is,E_{v}=\frac{1}{2} \frac{B_{q}}{H_{0}}Force exerted on plunger, F_{p}=E_{v} \cdot A_{g}\text { and } \begin{aligned}A_{g} & =\pi \cdot \frac{d^{2}}{4} \quad(d=0.02 \mathrm{~m}) \\A_{g} & =3.14 \times 10^{-4} \mathrm{~m}^{2} \\F_{p} & =E_{v} \cdot A_{g}=\frac{1}{2} \frac{B_{g}}{\mu_{D}} \times A_{g}=\frac{1}{2} \times \frac{0.1256}{4 \pi \times 10^{-7}} \times 3.14 \times 10^{-4} \\F_{p} & =15.7 \mathrm{~N}\end{aligned}Ans. Force exerted on plunger is 15.7 \mathrm{Nm}. ...