Magnetic Circuit

Problem Statement

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Problem 2.16 from Electric Machinery and Transformers, 3rd ed:

A magnetic circuit is given in Figure P2.16. What must be the current in the 1600-turn coil to set up a flux density of 0.1 T in the air-gap? All dimensions are in centimeters. Assume that magnetic flux density varies as

$B=[1.5H/(750+H)]$

.<ref>Guru and Huseyin, Electric Machinery and Transformers, 3rd ed. (New York: Oxford University Press, 2001), 129.</ref>

Solution

First, we note that the problem statement is incomplete. Assume that the core has a relative permeability of 500. Hence, for all magnetic sections excluding the air gap,

$\mu ={\mu }_r{\mu }_0=(500)(4\pi \times 10^{-7})$

Also, as recommended in the text, we will neglect fringing.

The lengths and areas of each of the sections to be evaluated are given in the following table.

Table 1: Lengths and Areas for the pertinent secions of the magnetic circuit.

Section	fg	def	ghc	dc	dabc
Length $l$ (m)	0.01	0.555	0.555	0.48	1.4
Area $A$ (m2)	0.0032	0.0032	0.0032	0.0096	8.0e-4

We must now work backward from the air-gap, since the value of the flux-density is given there. All units are understood to be standard units.
Air Gap:
${ℛ}=\frac{l_{fg}}{\mu A_{f}g}=4973.6$
${\mathrm{\Phi }}_{fg}=B_{fg}{A}_{fg}=3.2\times 10^{-4}$
${ℱ}={{ℛ}}_{fg}{\mathrm{\Phi }}_{fg}=1.5915$

Right Arms:
${\mathrm{\Phi }}_{def}={\mathrm{\Phi }}_{ghc}={\mathrm{\Phi }}_{fg}=3.2\times 10^{-4}$
${{ℛ}}_{def}={{ℛ}}_{ghc}=\frac{l_{def}}{\mu A_{def}}=276030$
${{ℱ}}_{def}={{ℱ}}_{ghc}={{ℛ}}_{def}{\mathrm{\Phi }}_{def}=88.331$

Center Column:
${{ℱ}}_{dc}={{ℱ}}_{def}+{{ℱ}}_{fg}+{{ℱ}}_{ghc}=178.25$
${{ℛ}}_{dc}=\frac{l_{dc}}{\mu A_{dc}}=79577$
${\mathrm{\Phi }}_{dc}=\frac{{{ℱ}}_{dc}}{{{ℛ}}_{dc}}=0.0022$

Left Arm:
${\mathrm{\Phi }}_{dabc}={\mathrm{\Phi }}_{dc}-{\mathrm{\Phi }}_{def}=0.0019$
${{ℛ}}_{dabc}=\frac{l_{dabc}}{\mu A_{dabc}}=2.785\times 10^6$
Failed to parse (syntax error): {\displaystyle \mathcal{F}_{dabc}=\mathcal_{dabc}\Phi_{dabc}=5347.6}

Conclusions:
Failed to parse (syntax error): {\displaystyle \mathcal{F}_{Total}=\mathcal{F}_{dabc}+\mathcal{_{dc}+\mathcal{F}_{def}+\mathcal{F}_{fg}+\mathcal{F}_{ghc}=57041}

$\mathbf{i}\mathbf{=}\frac{{{ℱ}}_{Total}}{N}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{5}\mathbf{7}\mathbf{A}$

Calculations were performed using the following Magnetic Circuit Matlab Script.

References

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