# 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,

$\mu=\mu_r\mu_0=(500)(4\pi\times10^{-7})=6.2832\times10^{-4}$

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:

$\mathcal{R}_{fg}=\frac{l_{fg}}{\mu A_{fg}} = 4973.6$
$\Phi_{fg}=B_{fg}A_{fg}=3.20\times10^{-4}$
$\mathcal{F}_{fg}=\mathcal{R}_{fg}\Phi_{fg}=1.5915$

Right Arms:

$\Phi_{def}=\Phi_{ghc} = \Phi_{fg} = 3.20\times10^{-4}$
$\mathcal{R}_{def}=\mathcal{R}_{ghc}=\frac{l_{def}}{\mu A_{def}}=2.7603\times10^5$
$\mathcal{F}_{def}=\mathcal{F}_{ghc}=\mathcal{R}_{def}\Phi_{def}=88.331$

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:

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

Which is the quantity we were looking for.

Calculations were performed using the following Magnetic Circuit Matlab Script.

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Amy Crosby