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Problem 1
Create and Solve Magnetic Equivalent Circuit
A magnetic equivalent circuit (MEC) needs to be constructed from the given plunger geometry. A magnetomotive force of $ \mathcal{F} = Ni $ acts as an analogue of a DC voltage source. In general, a reluctance may be expressed as $ \mathcal{R} = \frac{\ell}{\mu A} $ and function as a resistance analogue. If the steel is infinitely permeable, then the reluctance associated with a path through steel is zero (ideal conductor equivalent). In this problem, the permeability of air (linear, isotropic, and homogeneous) is approximated as free space: $ \mu_{\text{air}} = \mu_{r,\text{air}} \mu_0 \approx \mu_0 $. Without any fringing or leakage considered, magnetic flux $ \Phi $ will more or less follow the direction of the steel.
The reluctances need to be calculated using the length and cross sectional area presented to each path in air.
- $ \mathcal{R}_1(g) = \frac{g}{\mu_0 w d} $
- $ \mathcal{R}_2 = \frac{c}{\mu_0 t d} $
The total magnetic flux passing through $ \mathcal{F} $ needs to be found. It should be apparent that $ \mathcal{R}_1 $ is in series with the parallel combination of $ \mathcal{R}_2 $ and $ \mathcal{R}_2 $.
$ \begin{align} \Phi_1(i,g) &= \frac{\mathcal{F}}{\mathcal{R}_{eq}} \\ \Phi_1(i,g) &= \frac{Ni}{\mathcal{R}_1 + \frac{1}{2}\mathcal{R}_2} \\ \Phi_1(i,g) &= \frac{Ni}{\frac{g}{\mu_0 w d} + \frac{c}{2\mu_0 t d}} \\ \Phi_1(i,g) &= \frac{Ni}{\frac{2 t g}{2\mu_0 w t d} + \frac{w c}{2\mu_0 w t d}} \\ \Phi_1(i,g) &= \frac{2N \mu_0 w t d i}{2 t g + w c} \end{align} $
The magnetic flux linkage of the coil follows readily.
$ \begin{align} \lambda_1(i,g) &= N \Phi_1(i,g) \\ \lambda_1(i,g) &= \frac{2N^2 \mu_0 w t d i}{2 t g + w c} \end{align} $
