Magnetic Circuit

Author: John Hawkins

Problem Statement

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,

$\mu =\mu _{r}\mu _{0}=(500)(4\pi \times 10^{-7})=6.2832\times 10^{-4}$

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	1.38
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. We need only employ the analagous equations to Ohm's Law, KVL, and KCL. All units are standard units.
Air Gap:

${\mathcal {R}}_{fg}={\frac {l_{fg}}{\mu A_{fg}}}=4973.6$
$\Phi _{fg}=B_{fg}A_{fg}=3.20\times 10^{-4}$
${\mathcal {F}}_{fg}={\mathcal {R}}_{fg}\Phi _{fg}=1.5915$

Right Arms:

$\Phi _{def}=\Phi _{ghc}=\Phi _{fg}=3.20\times 10^{-4}$
${\mathcal {R}}_{def}={\mathcal {R}}_{ghc}={\frac {l_{def}}{\mu A_{def}}}=2.7603\times 10^{5}$
${\mathcal {F}}_{def}={\mathcal {F}}_{ghc}={\mathcal {R}}_{def}\Phi _{def}=88.331$

Center Column:

${\mathcal {F}}_{dc}={\mathcal {F}}_{def}+{\mathcal {F}}_{fg}+{\mathcal {F}}_{ghc}=178.25$
${\mathcal {R}}_{dc}={\frac {l_{dc}}{\mu A_{dc}}}=79,577$
$\Phi _{dc}={\frac {{\mathcal {F}}_{dc}}{{\mathcal {R}}_{dc}}}=0.0022$

Left Arm:

$\ \Phi _{dabc}=\Phi _{dc}-\Phi _{def}=0.0019$
${\mathcal {R}}_{dabc}={\frac {l_{dabc}}{\mu A_{dabc}}}=2.745\times 10^{6}$
${\mathcal {F}}_{dabc}={\mathcal {F}}_{dabc}\Phi _{dabc}=5,271.2$

Conclusions:

${\mathcal {F}}_{Total}={\mathcal {F}}_{dabc}+{\mathcal {F}}_{dc}+{\mathcal {F}}_{def}+{\mathcal {F}}_{fg}+{\mathcal {F}}_{ghc}=5,627.7$
$\mathbf {i={\frac {{\mathcal {F}}_{Total}}{N}}=3.52A}$

Which is the quantity we were looking for.

Calculations were performed using the following Magnetic Circuit Matlab Script.

References

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Comments from reviewers

So I still don't understand how this works but I figured I would just put my comments here and then you can delete them. -Amy

• I believe your length for dabc is incorrect and should be 1.38 unless I am understanding it wrong. if so it will prob change a few numbers :(
• On your first equation with A sub fg, it looks like the g is too big... just a formatting thing
• The first equation dealing with the left arm is all small, can you fix that?

Other then that it looks good!

(John): Thanks! Fixed them.